Applied Engineering Mathematics 293s29

), where A= lzl is the modulus and¢ is the phase shift. Therefore, we have z = Aei(wt+ )' (7.23) where (7.24) and ¢=tan

_1

-2-Xw

. (7.25) w02 -w 2 As the amplitude of the forcing is a, the gain G(w) of the oscillation is

A

G(w) = - = a

1

J(w5 - w2)2 + 4.X2w2

,

(7.26)

which is shown in Figure 7.5. Finally, the solution of the original equation (7.14) is the real part. That is

y(t)

=

Acos(wt + ¢).

115

(7.27)

7.3 Normal Modes

Vibration and Harmonic A1otion

Some special cases where w--+ 0 and w--+ oo are very interesting. For w << wo, the driving force is at very low frequency, we have --+ 0. (7.28) That is

y(t)

a

~,

wo

coswt.

(7.29)

The system is in the same phase with the forcing. If w >> wo, the forcing is at very high frequency. We have --+ -7r.

(7.30)

The oscillator is completely out of phase with the forcing. If w ~ wo, we have a A--+ 2Awo'

(7.31)

and

y(t)

=

: 2

w;

sinwot.

(7.32)

0

At resonance frequency = w~- 2A 2 , the amplitude of the oscillations increases dramatically.

7.3

Normal Modes

In the above harmonic oscillations, we know wo is the natural frequency of the concerned system, which is relatively simple. In general, there may be many natural frequencies or modes in a system, and the natural frequencies are in fact determined from the eigenvalue problem resulting from the system. Now let us study a more complicated system with three mass blocks attached in by two springs as shown in Figure 7.6. This system can be thought of as a car attached to two caravans on a flat road. Let u 1, u2, u3 be the displacement of the three mass blocks m 1 , m 2 , m 3 , respectively. Then, their accelerations will be ii.1 , 116

Vibration and Harmonic 1\lfotion

7.3 Normal !v!odes

u2, ih where ii. = d 2 uj dt 2 • From the balance of forces and Newton's law, we have mtii.t =kt(u2-ul),

(7.33)

m2u2 = k2(u3- u2)- kt(U2- Ut),

(7.34)

ffi3U3 = -k2(u3- u2).

(7.35)

These equations can be written in a matrix form as

(I +

-;1

( kt

-kt kt + k2 -k2

0 0 ) ( Ut ii.2 ) m2 0 0 m3 u3

~:2)(::)=(~}

(7.36)

or

Mii+Ku = 0, where

uT

(7.37)

= (ut, u2, u3). The mass matrix is

0 (7.38) and the stiffness matrix is

Equation (7.37) is a second-order ordinary differential equation in of matrices. This homogeneous equation can be solved by substituting Ui = Ui cos(wt) where Ui(i = 1, 2, 3) are constants and w2 can have several values which correspond to the natural frequencies. Now we have

-w[MUi cos(wt) + KUi cos(wt) 117

= 0,

(7.40)

7.3 Normal Modes

Vibration and Harmonic A1otion

~1

~3

Figure 7.6: Harmonic vibrations. where i = 1, 2, 3. Dividing both sides by cos(wt), we have (7.41)

This is essentially an eigenvalue problem because any nontrivial solutions for ui require (7.42)

Therefore, the eigenvalues of this equation give the natural frequencies. 0 Example 7.1: For the simplest case when m 1 = m 2 = m 3 = m and k 1 = k 2 = k, we have

=0, or

-w 2 (k- w2 m)(3km- w 2 m 2 ) = 0. This is a. cubic equation in of w2 , and it has three solutions. Therefore, the three natural frequencies are 2

w~ = 0,

3k

w3 = -:;;·

*(1,

For w~ = 0, we have (U1 , U2 : U3 ) = 1, 1), which is the rigid body motion. For w 2 = k/m, the eigenvector is determined by

-k k

-k 118

Vibration and Harmonic 1\lfotion 7.4 Small Amplitude Oscillations which leads to U2 = 0, and U1 = U3 . \-Vritten in normalized form, it becomes (U 1 , U2 , U3 ) = ~(1,0, -1). This means that block 1 moves in the opposite direction away from block 3, and block 2 remains stationary. For wi = 3k/m, we have (U1 , U2, U3 ) = "*(1, -2, 1). That is to say, block 2 moves in the different direction from block 3 which is at the same pace with block 1. 0

7.4

Small Amplitude Oscillations

For a mechanically conservative system, its total energy E = T + V is conserved, where Tis its total kinetic energy and V is its total potential energy. The configuration of the mechanical system can be described by its general coordinates q = (q1, q2, ... , qn). The general coordinates can be distance and angles. Thus, the velocities of the system will be q = 41, ti2, ... , tinIf we consider the system consists of many small particles or even imaginary parts, then the total kinetic energy T is a function of q and sometimes q, but the potential energy V is mainly a function of q only. As we are only concerned with small amplitude oscillations near equilibrium Vmin = V(O) = Vo, we can always take q = 0 at the equilibrium so that we can expand V in of q as a Taylor series

V(q) =

Vmin

+I: ~~ q; +I: I; K;iq;qi +..., l

(7.43)

J

l

where the stiffness matrix is

K l..J_- ~ 8 Vo 2

2 8qi8qj

I qi=O,qj=O

.

(7.44)

Since potential energy is always relative to an arbitrary reference point, we can thus take the potential energy at equilibrium V min to be zero. In addition, the equilibrium or the minimum value of V requires ~ = 0 at the equilibrium point qi = 0, and the force Fi = g~ shall be zero. This is correct because the resultant force must be zero at equilibrium, otherwise, the 119

7.4 Small Amplitude Oscillations Vibration and Harmonic A1otion

systen1 will be driven away by the resultant force. The component of the resultant force along the general coordinate qi should also be zero. Therefore, the total potential energy is now simplified as

v = L L qiKijqj = qTKq,

(7.45)

j

which is a quadratic form. For any small oscillation, the velocity is linear in of Qi, and thus the corresponding kinetic energy is ~mql. The total kinetic energy is the sum of all the components over all particles or parts, forming a quadratic form. That is to say,

. . = q. TM q, . T = """' L- """' L- ffiijqiqj i

(7.46)

j

where M = [mij] is the mass matrix. For a conservative system, the total mechanical energy E = T +Vis conserved, and thus time-independent. So we have d(T

+ V)

dt

=

!!:_ [ .T M q. + q TK q ] = dt q

0

.

(7.47)

Since M and K are symmetric matrices, this above equation becomes Mq+Kq=O. (7.48) This is a second order ordinary differential equation for matrices. Assuming the solution in the form qT = (q1 , q2 , ... , qn) = (Ut coswt, U2 coswt, ... , Un coswt) and substituting it into the above equation, we have (7.49) which is an eigenvalue problem. As an application, let us solve the same system of three mass blocks discussed earlier as shown in Figure 7.6. The total potential energy T is the sum of each mass block

120

Vibration and Harmonic 1\lfotion 7.4 Small Amplitude Oscillations 0

(7.50)

which can be written as a quadratic form

T

·TM·U, =U

(7.51)

0

(7.52)

We see that M is a symmetric matrix. For a spring, the force is f = kx, thus the potential energy stored in a spring is (7.53)

Therefore, the total potential energy of the two-spring system is (7.54) Since interchange of u 1 and u 2 does not change V, it is thus symmetric in of u1, u2 etc, which implies that Kij should be symmetric as well. The stiffness matrix K = [Kij] can be calculated using (7.55)

For example, (7.56)

and

121

7.4 Small Amplitude Oscillations Vibration and Harmonic A1otion

1 {)

kt

= 2{)ul [kt (u2- ut) + k2(u3- u2)] = - 2 .

(7.57)

Following the similar calculations, we have

(7.58)

which is exactly 1/2 multiplying the stiffness matrix we obtained earlier in equation (7.39). Thus, the equation for small amplitude oscillation is

Mii+Ku= 0. For the special case of mt = m2 = m3 its eigenvalues are determined by

(7.59)

= m and

=0,

kt

= k2 =

k,

(7.60)

which is exactly the problem we solved in the previous section (see example 7.1). For a simple system such as a pendulum, equation (7.39) is equivalent to the following simple formula for calculating the natural frequency V"(q) (7.61) w=

M(q)'

where V" > 0 because the potential energy at equilibrium is minimum. 0 Example 7.2:

A simple pendulum with a mass m is hanged vertically from a ceiling with a distance L from the fixed point. Let 8 be the small angle from its equilibrium, then the kinetic energy is 2 2 T= = (0) 2 • The potential energy is

!mv

!m£

V

= mgL(1- cosO). 122

Vibration and Harmonic 1\lfotion 7.4 Small Amplitude Oscillations Therefore, the stiffness is K = !V"(8) = !mgLcos8le=O = mgL/2. The equivalent mass is 1\tf ( 8) = !mL 2 • The governing equation becomes 1 2 .. 1 mL 8 + mgL8,

2

2

or

..

L

8+ -8 = 0. g

The natural frequency for small oscillations is

The period of this pendulum is T

=

2 1r = 21r w

!I.. y; 0

123

Chapter 8

Integral Transforms The mathematical transform is a method of changing one kind of functions and equations into another kind, often simpler or solvable one. In general, the transform is essentially a mathematical operator that produces a new function F( s) by integrating the product of an existing function f(t) and a kernel function K(t, s) between suitable limits

F(s) =

J

K(t, s)f(t)dt.

(8.1)

In the Laplace transform, the kernel is simply exp( -st) and integration limits are from 0 to oo. In the Fourier transform, the kernel is exp(±ist) with a normalized factor 1/../2-ff. The Fourier transform maps the time domain of a timedependent series such as a signal into a frequency domain, which is common practice in signal processing. The Laplace transform is a very powerful tool in solving differential equations. In this chapter, we will focus on the three major transforms: Fourier, Laplace and Wavelet transforms. They are commonly encountered in engineering and computational sciences. 125

8.1 Fourier Transform

8.1 8.1.1

Integral Transforms

Fourier Transform Fourier Series

For a function f (t) on an interval t E [-T, T], the Fourier series is defined as

where

11T 11T = 11T f(T) . T T ao = T

and

an

T

-T

-T

(8.3)

f(t)dt, n1rt

f(t) cos( T

)dt,

n1rt s1n(-)dt, (n = 1, 2, ... ).

bn = -

-T

(8.4) (8.5)

Here an and bn are the Fourier coefficients of f (t) on [- T, T]. The function f (t) can be continuous or piecewise continuous with a jump discontinuity. For a jump discontinuity at t =to, if !'(to-) and !'(to+) both exist, but !(to-) f f(to+ ), then the Fourier series converges to [!(to-)+ !(to+ )]/2. The Fourier series in general tends to converge slowly. From the coefficients an and bn, one can easily see that bn = 0 for an even function f( -t) = f(t). Similarly, ao = an = 0 for an odd function f (- t) = - f (t). In both cases, only oneside [0, T] of the integration can be used due to the symmetry. Thus, for an even function f (t), we have the Fourier cosine series on [0, T]

(8.6) For an odd function f (t), we have the Fourier sine series 00

f(t) =

L

n1rt

sin( T

n=l

126

).

(8.7)

Integral Transforms


f

t

-4

-2

0

2

4

Figure 8.1: Triangular wave with a period of 2.

0 Example 8.1: The triangular wave is defined by f(t) = ltl for t E [-1~ 1] with a period of2 or f(t+2) = f(t). Using the coefficients of the Fourier series, we have

ao =

11 2

110 (

1

2[

ltldt =

-1

Since both

-t)dt +

11

-1

It I and cos(n1rt) are even functions, 1

tdtJ =

0

1

2·

we have for any n

~

1,

1

ltl cos(n7rt)dt = 21 t cos(n7rt)dt }_1 0

an = {

1

= 2-t-sin(n1rt)l

n1r

-

0

~ n1r }

1

sin(n1rt)dt =

{

0

2 n 2 7r 2

[cos(n1r) -1].

Because ltl sin(n1rt) is an odd function, we have 1

ltl sin(n1rt)dt = 0.

bn = / -1

Hence, the Fourier series for the triangular wave can be written as f(t) =

1

~ cos(n1r)- 1

2 + 2 L-

n 2 1r2

cos(n1rt).

n=O

0

The n-term of the Fourier series, that is an cos(n1rtjT) + bn sin(n1rtjT), is called the n-th harmonic. The energy of the n harmonic is defined by A~ = a~ + b~. The sequence of An forms the energy or power spectrum of the Fourier series. The energy spectrum of the triangular wave is shown in Figure 8.2.

127


Integral Transforms

0.2

0.15

i;;

!

0.1

0.05

Figure 8.2: Energy spectrum of the triangular wave.

8.1.2

Fourier Integral

For the Fourier coefficients of a function defined on the whole real axis [-oo, oo], we can take the limits

a(wn) = /_: f(t) cos(wnt)dt, and n1f'

Wn=T'

as the limits T

--+

oo and

Wn --+

0. We have ao

--+

(8.8)

0 if

(8.9) In this case, the Fourier series becomes the Fourier integral

f(t) =

fooo [a(w) cos(wt) + b(w) sin(wt)dw,

where

a(w)

= ;1

1 b(w) = 1f'

!00-oo f(t) cos(wt)dt, !00-oo f(t) sin(wt)dt. 128

(8.10)

(8.11) (8.12)

Integral Transforms


Following the similar discussion for even and odd functions, we know that even functions lead to Fourier cosine integrals and odd functions lead to Fourier sine integrals.

8.1.3

Fourier Transform

The Fourier transform .1"[/ (t)] of f (t) is defined as

F(w) = .1"[/(t)] = .

!::27r Joo f(t)e-iwtdt,

V

(8.13)

-co

and the inverse Fourier transform can be written as

f(t) = .r-l [F(w)] = .

!::27r Joe-oo F(w)eiwtliv,

(8.14)

V

where exp[iwt] = cos(wt) + i sin(wt). The Fourier transform is a linear operator, and it has most of the properties of the differential operator and the integral operator. Hence, it is straightforward to prove that it has the following properties:

F[f(t)

+ g(t)] = .1"[/(t)] + .F(g(t)], F[of(t)]

= o.1"[f(t)],

.1"[( -it)n f(t)]

(8.15) (8.16)

dnF(w)

= dwn '

(8.17)

and

= (iwt F(w), ±oo) = ... = f(n-l)(t

.1"[/(n)(t)]

(8.18)

if f(t ---+ ±oo) = f'(t ---+ ---+ ±oo) ---+ 0. The transform can have different variations such as the Fourier sine transforn1 and the Fourier cosine transform. The Fourier transforms of some common functions are listed in the following table 8.1. D Example 8.2: For the triangle function f(t) and f (t) = 0 for It I > 1, its Fourier transform is

= 1-ltl for (It I ~

\1'21rF(w) = /_: (1- ltl)e-iwtdt

129

1)

8.1 Fourier Ti·ansform

f

=

0

{1

Integral Transforms

+ t)e-iwtdt +

-t

r (1- t)e-iwtdt. lo 1

Integrating by parts and using

. 2(W /2) = 1 - cosw , 2

Sill

we have

rn=F( ) = iw+ 1- eiw _ v £.1f w w2

e-tW

+ iw -1 = 2{1- cosw) w

2

w2

.

Hence, we have 1 sin 2 (w/2) 1 . 2 F(w) = .j2; (w/ 2)2 = .j2;smc (w/2). 0

Table 8.1: Fourier Transforms

F(w) = F[f(t)] F(w)e-twto F(w- wo)

f(t)

f(t- to) j(t)e-iw0 t 8(t)

1/v"Fi v"Fi8(w) .1. tw

1 sign(t)

2o

e-oltl e-(at)2

a2+w2 w2

(a > 0)

1 -7:'2" ""'7'f!r"e 4a v:.::o

! F(~)

f(at)

vt·

@e-alwl

1 a2+t2

J1[8(w- wo) + 8(w + wo)]

cos(wot) sin(wo) sinxax

(a > 0)

a

;[8(w + wo)- 8(w- wo)]

i ~'

(lwl < a); 0, (lwl > a)

The most useful Fourier transform for engineering and computational science is probably the discrete form, especially in digital signal processing. The discrete Fourier transform (DFT), 130

Integral Transforms

8.2 Laplace Transforms

for a periodic discrete function or signal x( n) with a period N, is defined by N-1

X[k]

=

L

x[n]e-i ~;", 2

(8.19)

n=O

and the inverse transform, also called the signal reconstruction, is defined by 1 N-1 x[n] = X[k]ei~. (8.20) N k=O

L

A periodic signal x( n + N) = x( n) has a periodic spectrum X[k + N] = X[k]. The discrete Fourier transform consists of N multiplications and N- 1 additions for each X[k], thus for N values of k, the computational complexity is of O(N 2 ). However, if N =2m (mE JV), many of the DFT calculations are not necessary. In fact, by rearranging the formula, one can get the complexity of O(N log2 N). This type of algorithms is called Fast Fourier Transform (FFT). Vast literature exists on the signal processing such as FFT, filter design and signal reconstruction.

8.2

Laplace Transforms

The Laplace transform .C [f (t)] of a function f (t) is defined as

F(8)

=.C[f(t)] =loco f(t)e-stdt,

(8.21)

where 8 > 0. The inverse Laplace transform .c- 1 (F(8)] is f(t) or f(t) = .c- 1 [F(8)]. The Laplace transforms of most simple functions can be obtained by direct integration. For simple functions t and eo:t, we have 1 t l + [--e-st lo co te-stdt = loco -e-stdt

00

.C(t] =

o

o

8

8

o

1 = -. 2 8

.C(eo:t] = {oo eo:te-stdt = [--1-e-<s-o:)tl oo = _I__ lo 8 -a o 8- a 131


Integral Transforms

Conversely' .c- 1 [ 1 ] = t .c- 1 [s-o -1- ] = eat. For the Dirac ST ' <5'-function, we have its Laplace transform

.C[d'(t)] =

rX) d'(t)e-stdt = e-stl t=O = 1.

lo

(8.22)

However, the inverse of a Laplace transfonn is usually more complicated. It often involves the partial fractions of polynomials and usage of different rules of Laplace transforn1s. From the basic definition, it is straightforward to prove that the Laplace transform has the following properties:

.C[af(t) + fjg(t)] .C[e0 t f(t)]

=

a.C[f(t)]

+ /3.C[g(t)],

(8.23)

F(s- a), s >a,

(8.24)

= e-os .C[f(t)],

(8.25)

.C[f' (t)] = s.C[f(t)] - f(O),

(8.26)

=

.C[f(t- a)]

.C[ ft f( r)dr]

lo

=

.!..c[j],

s

(8.27)

The Laplace transform pairs of common functions are listed below in table 8.2. 0 Example 8.9: In order to obtain the Laplace transform of f(t) = coswt, we shall first write f(t)

1 . t . t = coswt = '2(e w + e-zw ). 1

Then, we have

~[ [ e<-s+iw)tdt + foe e-s-iw)tdt) 2 Jo lo 00

=

1 =

1

1

2[s- iw + s + iw]

s =

s2 +w2 0

132

Integral Transforms


Table 8.2: Laplace Transform Function f(t) 1 8(t) tn,n > 0 cos( at) sin( at) eat

Laplace Transform F( s)

-s 1 n!

sn+T

s s:l+a:l a s:l+a:l 1

s-a

t1/2

1(

11" )

'2'S3'

t-1/2

tn f(t)

( s

cos( at+ !3) sinh(ot) cosh( at)

1/2

~

-1)ndnF(s) dsn

cos(~)-a sin(~)

s2+a2

a s1-a:l

s s2-a2

le-avs

erfc( 2 ~)

s

a2

_1_e--:rr

..;:;;

sinoVt 1-~-at (a> 0) a~ J3 (eat _ ePt) (0 =f= /3)

Both Fourier and Laplace transforms follow the convolution theorem. For two functions f and g, their convolution f * g obeys (8.28) and their Laplace transforms follow

.C[f(t) * g(t)] C 1 [F(s)G(s)] =

=

F(s)G(s),

fo' f(t- a)g(a)do:. 133

(8.29) (8.30)

8.3 lVa.velet

Integral Transforms

The Fourier transform has the similar properties

f(t)

* g(t) =

/_:

F[j(t) * g(t)]

8.3

f(t)g(t- u)du,

=

F(w)G(w).

(8.31) (8.32)

Wavelet

The Fourier transform is an ideal tool for studying the stationary time signal whose properties are statistically invariant over time. In the Fourier transform, the stationary signal is decomposed into linear combinations of sine and cosine waves

. ~'

v2~

~ cos(nt), ~sin(nt),

v~

v~

(n

=

1, 2, ... ).

(8.33)

The Fourier transform is very useful to analyse stationary signals where a stationary signal means that the frequencies of the signal do not change with time. For non-stationary signals whose frequencies f = w/2~ vary with time (see Figure 8.3), the Fourier transform does not work well. In addition, in the Fourier transform there is a tradeoff between frequency resolution and time resolution, ~w~t ~

1

2,

(8.34)

which is similar to the Heisenberg uncertainty principle for spatial and velocity intervals. The wavelet transform is an alternative approach to the Fourier transform to overcome the resolution problen1 using the Ivlother wavelet 1/J or prototype for generating the other windows functions, and all the used windows are in the form of either dilated/compressed or shifted. As a result, the wavelet transform is very powerful in dealing with non-stationary signals because the Fourier transforn1 is not suitable for such signals. In the wavelet transform, a transient signal is decomposed into elementary components of wavelets 134

Integral Transforms

...as>-

8.3 Wavelet

1

s0 = u; as

-1

0

10

2

4

6

f:ro/21t varies with time (t)

8

10

Figure 8.3: A stationary signal with two fixed frequencies (ft = 0.5 Hz and /2 =2Hz) and a non-stationary signal whose frequency varies with time. or wavelet packets. There are three major type of wavelets: Grossmann-l\1orlet wavelets, Daubechies wavelets and GaborIvlalvar wavelets. Wavelets are defined as a real-valued function 'ljJ(t) (t E 'R) in of the generator wavelet or n1other wavelet. The function 'l/J is both well localized, decreasing rapidly as t ---.. oo and oscillating in a wavery manner. To generate other wavelets, 'l/J( a, (3, t) is used by translating in time and change of scales. Grossmann-Ivlorlet wavelets are of the form 1

t-(3

Q

Q

-'l/1(--),

Q

> 0,

a,b E 'R,

(8.35)

where 'l/J a generator wavelet. The Daubechies wavelets have the form m,nEZ. (8.36) 135

8.3 lVa.velet

Integral Transforms

The Gabor-lVIalvar wavelets are in the form 1

w(t- m) cos[1r(n + '2 )(t- m)),

mE Z,n EN.

(8.37)

The continuous wavelet transform can be defined by \II J(T, s) =

1 JJST

J

t-T f(t) · 1/J(-)dt, 8

(8.38)

where T is the translation of the location of the window and s is the scale where s = 1 is for the most compressed wavelet. Wavelet analysis has vast literature and it is still an active research area in signal processing. Readers can search the latest research journals to follow the latest developments.

136

Chapter 9

Partial Differential Equations Partial differential equations are much more complicated compared with the ordinary differential equations. There is no universal solution technique for nonlinear equations, even the numerical simulations are usually not straightforward. Thus, we will mainly focus on the linear partial differential equations and the equations of special interests in engineering and computational sciences. A partial differential equation (PDE) is a relationship containing one or more partial derivatives. Similar to the ordinary differential equation, the highest nth partial derivative is referred to as the order n of the partial differential equation. The general form of a partial differential equation can be written as 8u 8u 8 2u 8 2u 8 2u (9 · 1) 1/J(x, y, ... , 8x' ay' 8x 2 ' ay2' 8xay' ... ) = O. where u is the dependent variable and x, y, ... are the independent variables. A simple example of partial differential equations is the linear first order partial differential equation, which can be written as 8u 8u (9.2) a(x, y) 8x + b(x, y) ay = f(x, y). 137

9.1 First Order PDE

Partial Differential Equations

for two independent variables and one dependent variable u. If the right hand side is zero or simply f(x, y) = 0, then the equation is said to be homogeneous. The equation is said to be linear if a, b and f are functions of x, y only, not u itself. For simplicity in notations in the studies of partial differential equations, compact subscript forms are often used in the literature. They are

{) 2 u

Uyy

={)y2 '

{]2u

Uxy

=8x8y'

(9.3)

and thus we can write (9.2) as

aux +buy= f.

(9.4)

In the rest of the chapters in this book, we will use these notations whenever no confusion occurs.

9.1

First Order PDE

The first order partial differential equation of linear type can be written as

a(x, y)ux

+ b(x, y)uy =

f(x, y),

(9.5)

which can be solved using the method of characteristics

(9.6) This is equivalent to the following equation in of parameter s

(9.7) which essentially forms a system of first-order ordinary differential equations. 138


9.2 Classification

The simplest example of a first order linear partial differential equation is the first order hyperbolic equation Ut +cux =

(9.8)

0,

where cis a constant. It has a general solution of

u

= ,P(x- ct),

(9.9)

which is a travelling wave along x-axis with a constant speed c. If the initial shape is u(x,O) = 1/J(x), then u(x, t) = 1/J(x-ct) at timet, therefore the shape of the wave does not change though its position is constantly changing.

9.2

Classification

A linear second-order partial differential equation can be written in the generic form in of two independent variables x andy, auxx

+ buxy + CUyy + gux + huy + ku =

/,

(9.10)

where a, b, c, g, h, k and f are functions of x and y only. If f (x, y, u) is also a function of u, then we say that this equation is quasi-linear. If ~ = b2 - 4ac < 0, the equation is elliptic. One famous example is the Laplace equation Uxx + Uyy = 0. If~ > 0, it is hyperbolic. One example is the wave equation Utt = c2Uxx·

If ~ = 0, it is parabolic. Diffusion and heat conduction equations are of the parabolic type Ut = KUxx·

9.3

Classic PDEs

Iviany physical processes in engineering are governed by three classic partial differential equations so they are widely used in a vast range of applications. 139

9.3 Classic PDEs


Laplace's and Poisson's Equation In heat transfer problems, the steady state of heat conduction with a source is governed by the Poison equation

(x,y)En,

(9.11)

or Uxx

+ Uyy

q(x, y, t),

=

(9.12)

for two independent variables x and y. Here k is the thermal diffusivity and f(x, y, t) is the heat source. If there is no heat source (q = 0), this becomes the Laplace equation. The solution or a function is said to be harmonic if it satisfies the Laplace equation. Heat Conduction Equation Time-dependent problems, such as diffusion and transient heat conduction, are governed by parabolic equations. The heat conduction equation (9.13) Ut = kuxx, is a famous example. For diffusion problem, k is replaced by the diffusion coefficient D. Wave Equation The vibrations of strings and travelling sound waves are governed by the hyperbolic wave equation. The 1-D wave equation in its simplest form is Utt

=

2

C Uxx,

(9.14)

where cis the speed of the wave. There are other equations that occur frequently in mathematical physics, engineering and computational sciences. We will give a brief description of some of these equations in later chapters. 140

Chapter 10

Techniques for Solving

PDEs Different types of equations usually require different solution techniques. However, there are some methods that work for most of the linearly partial differential equations with appropriate boundary conditions on a regular domain. These methods include separation of variables, series expansions, similarity solutions, hybrid methods, and integral transform methods.

10.1

Separation of Variables

The separation of variables attempts a solution of the forn1

u = X(x)Y(y)T(t),

(10.1)

where X(x), Y(y), T(t) are functions of x, y, t, respectively. In order to determine these functions, they have to satisfy the partial differential equation and the required boundary conditions. As a result, the partial differential equation is usually transformed into two or three ordinary differential equations (ODEs), and these ordinary differential equations often appear as eigenvalue problems. The final solution is then obtained by solving these ODEs. As a classic example, we now try to solve 141

10.1 Separation of Variables

Techniques for Solving PDEs

the 1-D heat conduction equation in the domain x E [0, L] and

t>O (10.2)

with the initial value and boundary conditions

u(O, t) = u(L, t) = 0,

u(x, 0)

=

1/J.

(10.3)

Letting u(x, t) = X(x)T(t), we have

X"(x)

---y- =

T'(t) kT.

(10.4)

As the left hand side depends only on x and the right hand side only depends on t, therefore, both sides must be equal to the same constant, and the constant can taken to be as -A 2 . The negative sign is just for convenience because we will see below that the finiteness of the solution T(t) requires that eigenvalues A2 > 0 or A are real. Hence, we now get two ordinary differential equations

X"(x) + A2 X(x) = 0,

T'(t) + kA 2T(t) = 0,

(10.5)

where A is the eigenvalue. The solution for T(t) is (10.6)

The solution for X(x) is in a generic form

X(t) =a cos Ax+ ,BsinAx.

(10. 7)

From the boundary condition u(O, t) = 0, we have a = 0. From

u(L, t) = 0, we have sinAL= 0,

(10.8)

which requires that AL = n1r. Please note that n f 0 because the solution is trivial if n = 0. Therefore, A cannot be continuous, and it only takes an infinite number of discrete values, called eigenvalues. Each eigenvalue A = An = n1r / L, (n = 142


10.2 Transform }.!fethods

1, 2, ... )has a corresponding eigenfunction Xn = sin(.Xnx). Substituting into the solution for T(t), we have (10.9)

By superimposing Un = XnTn and expanding the initial condition into a Fourier series so as to determine the coefficients, we have 00

u(x, t) =

L:: an sin( n~x )e-< nz > kt, 2

n=l

2 [L . n1rd~ an = L Jo 1/;(x) s1n( --y- )~.

10.2

(10.10)

Transform Methods

Laplace Transform The basic idea of the integral transform method is to reduce the number of the independent variables. For the 1-D timedependent heat conduction, it transforms the partial differential equation into an ordinary differential equation. By solving the ordinary differential equation, the solution to the original problem is obtained by inverting back from the Laplace transform. As an example, we now solve the 1-D heat conduction in semi-infinite interval [0, oo ), (10.11)

with the boundary conditions T(x, 0)

= 0, T(O, t) = T 0 .

(10.12)

Let T(x, s) = f 000 T(x, t)e-stdt be the Laplace transform of T(x, t), the equation then becon1es (10.13) 143

10.2 TI·ansform 1\tlethods


and the boundary condition at x = 0 becomes f'x=O = To/ s. The general solution to the ordinary differential equation can be written as IBx · 1'Fx T=Ae-v"f· +BeV"f·. The finiteness of the solution as x --+ oo requires that B = 0, and the boundary conditions lead to To T = -e s

lEx

V"k'.

By the inversion of the Laplace transform, we have X

T = Toerfc( . /U), 2vkt where erfc( x) is the complementary error function.

Fourier Transform Fourier transform works in the similar manner as the Laplace transform. The famous example is the classical wave equation Utt

=

2

V Uxx'

(10.14)

with the initial conditions u(x, 0) = 1/;(x) = exp[-(x - a) 2 ], and Ut(x, 0) = 0. Let u(w, t) = f~oo u(x, t)eiwxdx be the Fourier transform of u(x, t). This transforms the PDE problem into an ODE d2-u 2 2(10.15) dt2 = -v w u,

Jk:-

with

__ ,.T.( )

du(w, 0) _ dt - 0.

U-tpW,

The general solution in tenns of the parameter w is

u(w, t) = ijj(w) cos(vwt). By using the inverse Fourier transform, we finally have 1 u(x, t) = . !7C V 27r

!00 1/;(w) - cos(vwt)e-zwxdw . -oo

144

(10.16)


10.3 Similarity Solution

1

= [1P(x + vt) + 'l/1(x- vt)]

2

-_

1 [e -(x-a+vt)2

2

+ e -(x-a-vt)2] ,

(10.17)

which implies two travelling waves: one travels along the x-axis and the other along the negative x-axis direction.

10.3

Similarity Solution

The essence of similarity solution is to use the so-called sinlilarity variable ~ = xjtf3 so as to reduce the partial differential equation to an ordinary differential equation. For example, the diffusion equation {10.18) can be solved by using the similarity method by defining a similar variable {10.19)

or X

(=-

#"

(10.20)

In general, we can assume that the solution to the equation has the form 0 (10.21) U = {Kt) j[ {K~)f3]. By substituting it into the diffusion equation, the coefficients a and f3 can be detern1ined. For most applications, one can assume a= 0 so that u = f(() (see the following example for details). In this case, we have {10.22)

or

f"

=

-(f',B(Kt) 2/3-l, 145

(10.23)

10.3 Similarity Solution


where !' = df I d(. In deriving this equation, we have used the chain rules of differentiations fx = ~ and = ~ so that

*:

au = - {3kf'(()x = 8t

(~t)f3t

£ &

-{3(J'(()(~t)-1

,

(10.24)

and (10.25)

Since the original equation does not have time-dependent explicitly, this means that all the exponents for any t must be zero. Therefore, we have 2{3 = 1, or {3 = Now, the diffusion equation becomes

!.

!"(() Using (ln !')' = we get

=

-~f',

(10.26)

!"If' and integrating the above equation once, or

(10.27)

Integrating it again and using the ( 2 = 4{2 , we obtain

u =A

1

~

X

·2

e-t; ~ = Cerf( . .1'7'::7) + D, v4~t

t;o

(10.28)

where C and D are constants that can be determined from appropriate boundary conditions. For the same problem as (10.12), the boundary condition as x---+ oo implies that C+D = 0, while u(O, t) =To means that D = -C =To. Therefore, we finally have X

X

u = T0 [1- erf( .;;w)] = T0 erfc( .;;w)·

(10.29)

0 Example 10.1: For the similarity solution of the diffusion equation Ut = KUxx' we assume u = (Kt)o: !(,.}) 13 • Now we want to know why a = 0. Since

146


10.4 Travelling lVave Solution

the diffusion equation becomes

The requirement that no explicit (time t) appears in the equation leads to Q' = 0, Q' - 2,8 + 1 = 0. Thus, Q'

= 0,

You may think that why not divide both sides of the above equation by (Kt )o:, then we do not impose any requirement on a, thus f3 = ~- This non-zero a indeed appears in some nonlinear diffusions equations where the diffusion coefficient is not a constant and K{ u) may depend on u itself.

10.4

Travelling Wave Solution

The travelling wave technique can be demonstrated using the famous Korteweg-de Vries (KdV) equation (10.30) or (10.31) which is a third-order nonlinear partial differential equation. The interesting feature of this equation is that it has a solitary wave solution or soliton. The soliton phenomenon was first observed by John Russell in 1834 when he travelled along the Union canal in Scotland. Nowadays, telecommunications use solitons to carry signals in optical fibres. Now we seek the travelling wave solution in the form

u = (x- vt), 147

(10.32)

10.5 Green's Function


where v is the speed of the travelling wave. By substituting into the KdV equation, we have

¢"' + 6c/>c/>1 - v¢ = 0. Using (¢ 2 /2)' we have

(10.33)

= c/>c/>1 and integrating the above equation once, ¢" + 3¢2

-

vel> = A,

(10.34)

where A is an integration constant. The requirement of c/>, cl>', ¢" --+ 0 at far field x --+ ±oo leads to A = 0. Let '1/J = cl>', we get d'lj;

'1/J d¢

2

+ 3¢ - v¢ = 0.

(10.35)

Integrating with respect to¢, we get

~'l/J2 =

-¢3

+ ~v¢2.

(10.36)

Integrating it again and substituting back to u, we have the travelling wave solution

u = v sech 2[~ T (x - vt - <5) ],

2

(10.37)

where <5 is a constant and v /2 is the amplitude of the wave. We can see that the speed of the wave depends on the amplitude or height of the wave. That is to say, big waves travel faster than smaller waves. For linear wave equations, waves can travel in both directions, but here it is only possible for the soliton to travel in one direction, that is along x-axis direction in this scenario.

10.5

Green's Function

The method of Green's function is very powerful in solving elliptic equations. A Green's function inside the domain n is defined as (10.38) 148


10.6 Hybrid J.1etbod

where 8(x) is the Dirac delta function. It usually requires that G = 0 on the boundary surface r. Generally speaking, G =

G(x, y, z; ~, 17, (). For the hyperbolic equations such as the wave equation, we can define the Green's function as

Gtt - c2 ~G = 8(x- ~)8(y- 17)8(z- ()8(t- r).

(10.39)

The fundamental Green's function for this case is G(x,y,z,t;~,1J,(,T) = -

1

41rc2 r

r

8[(t- r)- -]. c

(10.40)

The Green's function method is very complicated, but it can be very neat in obtaining solutions. Readers can refer to the literature listed at the end of the book.

10.6

Hybrid Method

Some differential equations can be solved using one of the methods described above, but often a single method simply does not work. In this case, a hybrid method that combines several methods is needed. For example, the Crank's diffusion problem in an infinite cylinder is governed by the following equation:

8u {)t

18

8u

=; 8r [rll: 8r],

(10.41)

where ll: is the diffusion coefficient, and r is the distance in the polar coordinates. u(r, t) can be concentration or any other quantity. Now we want to solve this equation with the following boundary conditions:

u(r, t) = 0,

r =a,

(10.42)

r E (O,a).

(10.43)

and

u(r, 0)

=

1/J(r),

First, we use the separation of variables, we have (10.44) 149

10.6 Hybrid Method


then we have the equation for v( r) 1

v" + -v' + A2 v = 0, r

(10.45)

where v' = dv / dr and A is a parameter to be determined. This equation is essentially the Bessel equation of the zero-order v = 0. Thus, we can now assume that the general solution for u(r, t) has the following form 00

u(r, t) =

L DiJo(Air)e--\rKt,

(10.46)

i=l

where Di are undetermined coefficients. The boundary condition [u(r =a) = 0] requires that (10.47)

which means that paratneter Ai are the roots of the Bessel function J0 ( Aia) = 0. The initial condition gives 00

1/J(r)

=

L DiJo(Air).

(10.48)

i=l

Using the basic properties of the Bessel functions (10.49)

and (10.50)

loa rJo(Air)dr = ;i Jt(aAi),

(10.51)

the general solution can be written as

u(r, t)

2 ~ Jo(rAi) --\~Kt

= 2 L-

a

i=l

J 2 ( A·) e • 1 a z

150

Ina r'l/J(r)Jo(rAi)dr. o

(10.52)

10.6 Hybrid P.1ethod


The solution procedure shows that it requires a combination of separation of variables, Bessel functions, and power series. 0 Example 10.2: If we now want to solve the same diffusion equation in the cylindrical coordinates with slight different boundary conditions:

u(r =a, t) = uo = const,

u(r, 0) = ¢(r),

we have to make a transformation u = w + uo. Both u and w satisfy the same diffusion equation, but now the boundary conditions for w become

w(r =a, t) = 0,

w(r, 0) = ¢(r) - uo = 1/J(r),

which is the problem we have just solved. By substituting u = w + uo and 1/J(r) = ¢(r) - uo and using the properties of Bessel functions, we finally obtain

where hi = foa r¢(r)Jo(r>..i)dr. For a very special case when (r) = 0, we have

There are other important methods for solving partial differential equations. These include series methods, asymptotic methods, approximate methods, perturbation n1ethods and naturally the numerical methods.

151

Chapter 11

Integral Equations The calculus of variations is important in many optin1ization problems and computational sciences, especially the formulation of the finite element methods. On the other hand, integral equations are a different type of equation and they frequently occur in applied mathematics and natural sciences. In this chapter, we will briefly touch these topics.

11.1

Calculus of Variations

The main aim of the calculus of variations is to find a function that makes the integral stationary, making the value of the integral a local maximum or minin1um. For example, in mechanics we may want to find the shape y( x) of a rope or chain when suspended under its own weight from two fixed points. In this case, the calculus of variations provides a method for finding the function y(x) so that the curve y(x) minimizes the gravitational potential energy of the hanging rope system.

11.1.1

Curvature

Before we proceed to the calculus of variations, let us first discuss an important concept, namely the curvature of a curve. In general, a curve y(x) can be described in a parametric form

153

11.1 Calculus of Variations

Integral Equations

in of a vector r( s) with a parameter s which is the arc length along the curve measured from a fixed point. The curvature K of a curve is defined as the rate at which the unit tangent t changes with respect to s. The change of arc length is (11.1)

We have the curvature

dt

-

ds

1 p ·

=Kn= -n.

(11.2)

where p is the radius of the curvature, and n is the principal normal. As the direction of the tangent is defined by the angle 8 made with the x-axis by t, we have tan() = y'. Hence, the curvature becomes

d() ds

dO dx dx ds ·

K=-=--

(11.3)

Fron1 () = tan- 1 y'(x), we have

dO [ _ 1 ( ')]' y" dx = tan y = (1 + yl2

r

(11.4)

Using the expression for dsjdx, the curvature can be written in of y(x), and we get (11.5)

11.1.2

Euler-Lagrange Equation

Since the calculus of variations is always related to some minimization or maximization, we can in general assume that the integrand 1/J of the integral is a function of the shape or curve y( x) (shown in Figure 11.1), its derivative y' (x) and the spatial 154

Integral Equations


coordinate x (or time t, depending on the context). For the integral I

=

l

'1/;(x, y, y')dx,

(11.6)

where a and bare fixed, the aim is to find the solution of the curve y( x) such that it makes the value of I stationary. In this sense, I[y(x)] is a function of the function y(x), and thus it is referred to as the functional. y

a '-----------__,. X

Figure 11.1: Variations in the path y(x). Here, stationary n1eans that the small change of the first order in y( x) will only lead to the second-order changes in values of I[y(x)], and subsequently, the change 8I of I should be virtually zero due to the small variation in the function y(x). 'Iranslating this into the 1nathematical language, we suppose that y(x) has a small change of magnitude off so that

y(x)

---+

y(x) + f((x),

(11.7)

where ( (x) is an arbitrary function. The requirement of I to be stationary means that

8I

= 0,

(11.8)

or more accurately,

di

df 'E=0 = 0,

for all ((x).

155

(11.9)


Thus I becomes

I(y, <) =

t

Integral Equations

,P(x, y + <(, y1 + <(')dx

= [ 1/J(x,y,y')dx+ [[<((: +(':Jdx+0(<2 ). (11.10) The first derivative of I should be zero, and we have (11.11)

which is exactly what we mean that the change ~I (or the first order variation) in the value of I should be zero. Integrating this equation by parts, we have (11.12) If we require that y(a) and y(b) are known at the fixed points x = a and x = b, then these requirements naturally lead to ((a) =((b)= 0. This means that the above right hand side of the equation is zero. That is,

81/J] b [( 8y' a= O,

(11.13)

which gives

1 [81/J- ~dx 81/J](dx b

a

8y

8y'

=

0.

(11.14)

As this equation holds for all ( (x), the integrand must be zero. Therefore, we have the well-known Euler-Lagrange equation (11.15)

It is worth pointing out that this equation is very special in the sense that 1/J is known and the unknown is y(x). It has many applications in mathematics, natural sciences and engineering. 156

Integral Equations


The simplest and classical example is to find the shortest path on a plane ing two points, say, (0,0) and (1, 1). We know that the total length along a curve y( x) is L=

l

(11.16)

J!+y'2dx.

Since '1/J = J1 + y'2 does not contain y, thus~= 0. Fron1 the Euler-Lagrange equation, we have (11.17)

its integration is 8'1/J

ay'

y'

v1 +

=

(11.18)

yn = A.

Rearranging it as '2

y

A2 A2 '

= 1-

A

I

or

y = v'1- A 2 '

(11.19)

and integrating again, we have y

= kx +c,

k-

A

- v'1- A 2 •

(11.20)

This is a straight line. That is exactly what we expect from the plane geometry. D Example 11.1: The Euler- Lagrange equation is very general and includes many physical laws if the appropriate form of 1/J is used. For a point mass m following under the Earth's gravity g, the action (see below) is defined as

1/J =

~mv 2 - mgy = ~m(y) 2 -

mgy,

where y(t) is the path, and now x is replaced by t. v = velocity. The Euler-Lagrange equation becomes

81/J d 81/J 8y = dt ( 8v ), 157

y is

the

11.1 Calculus of Variations or

Integral Equations

d -mg = dt (mv),

which is essentially the Newton's second law F = ma because the right hand side is the rate of change of the momentum mv, and the left hand side is the force.

Well, you may say, this is trivial and there is nothing new about it. This example is indeed too simple. Let us now study a more complicated case so as to demonstrate the wide applications of the Euler-Lagrange equation. In mechanics, there is a Hamilton's principle which states that the configuration of a mechanical system is such that the action integral I of the Lagrangian .C = T - V is stationary with respect to the variations in the path. That is to say that the configuration can be uniquely defined by its coordinates qi and timet~ when moving from one configuration at time t 0 to another time t = t* I

=

fo'• .C(t, q;, tj;)dt,

i =

1,2~

... ,N,

(11.21)

where T is the total kinetic energy (usually, a function of Qi), and V is the potential energy (usually, a function of q). Here Qi means . 8qi (11.22) qi =

m·

In analytical mechanics and engineering, the Lagrangian .C (=Kinetic energy - Potential energy) is often called the action, thus this principle is also called the principle of least action. The physical configuration or the path of movement follows such a path that makes the action integral stationary. In the special case, x -+ t, the Euler-Lagrange equation becomes (11.23) which is the well-known Lagrange's equation. This seems too abstract. Now let us look at a classic example.

158

Integral Equations


Figure 11.2: A simple pendulum.

0 Example 11.2: For a simple pendulum shown in Figure 11.2, we now try to derive its equation of oscillations. We know the kinetic energy T and the potential energy V are

V = mgh = mgl(l -cos 8).

Using C, = T - V, q = 8 and q =

which becomes

iJ, we have

d

.

-mgl sin 0 - dt (ml 2 0) = 0.

Therefore, we have the pendulum equation

This is a nonlinear equation. If the angle is very small (8 << 1), sin 8 ~ 8, we then have the standard equation for the linear harmonic motion

0

159


11.1.3

Integral Equations

Variations with Constraints

Although the stationary requirement in the calculus of variations leads to the minimization of the integral itself, there is no additional constraint. In this sense, the calculus of variation discussed up to now is unconstrained. However, sometimes these variations have certain additional constraints, for example, the sliding of a bead on a hanging string. Now we want to make the integral I stationary under another constraint integral Q that is constant. We have

I=

t t

1/J(x,y,y')dx,

(11.24)

cf>(x, y, y')dx.

(11.25)

subjected to the constraint

Q=

As for most optimization problems under additional constraints, the method of Lagrange multipliers can transform the constrained problem into an unconstrained one by using a combined functional J = I + >..Q or J

=

t[,P + >.¢]dx,

(11.26)

where ).. is the undetermined Lagrange multiplier. Replacing 1/J by [1/J + >..

same procedure of the derivations, we have

(11.27) Now we can come back to our example of the hanging rope problem with two fixed points. The total length of the rope is L, and it hangs frmn two fixed points ( -d, 0) and (d, 0). From the geometric consideration, it requires that 2d < L. In order to find the shape of the hanging rope under gravity, we now define its gravitational potential energy EP as

Ep

= j_:[pgy(x)ds] = pg j_:yJl + y12dx. 160

(11.28)

Integral Equations


The additional constraint is that the total length of the rope is a constant (L). Thus,

Jd

Q=

-d

J1 +y12 dx = L.

By using the Lagrange multiplier A, we have J = Ep J =

Jd [pgy +-X] VI+ y'2dx.

(11.29)

+ ,\Q, or (11.30)

-d

vi

Since \II = [pgy + -X] + y'2 does not contain X explicitly, or ~! = 0, then the Euler-Lagrange equation can be reduced into a simpler form in this special case. Using

d\11

8\11

ow dy

ow dy'

=8x - +8y -dx - +oy'-dx dx ,ow

,ow

= 0 + y oy + y oy''

(11.31)

and the Euler-Lagrange equation~= fx(~ ), we have (11.32)

which can again be written as

d ,8\11 dx[w- Y 8y'] =

o.

(11.33)

The integration of this equation gives \II - y

,8\11

8

y' = A = const.

(11.34)

Substituting the expression of \II into the above equation, the stationary values of J requires (11.35) 161


Integral Equations

Multiplying both sides by vf1 A cosh(= pgy +A, we have

+ y''l and using the substitution (11.36)

whose solution is cosh

-t[PYY

+A]= xpg + K. A

(11.37)

A

Using the boundary conditions at x = ±d and the constraint

Q = L, we have K = 0 and in1plicit equation for A . h(pgd) _ pgL SlllA-2A.

(11.38)

Finally, the curve for the hanging rope becomes the following catenary

A pgx pgd y(x) = pg[cosh(A)- cosh(-y)].

(11.39)

0 Example 11.9: For the hanging rope problem, what happens if we only fix one end at (a, 0), while allowing the free end of the hanging rope to slide on a vertical pole? lVell, this forms a variation problem with variable end-point(s). We assume that free end is at (0, y) where y acts like a free parameter to be determined. Now the boundazy condition at the free end is different. Since the variation of 81 = 0, we have

8J =

l

a

b

8\J! 8y

d 8\J! dx 8y'

[ - - -(-)](dx

8\J! oy'

+ [(-]b

a

= 0.

As the variation ( is now non-zero at the free end point, we then have

8\J! oy' From J = Ep

+ >..Q,

= o.

we have \J! = (pgy + >..) J 1 + y'2 • Thus, we get

a

8y' [(pgy + >..)vfl

162

+ y'2 ] = 0,

Integral Equations


or y'(pgy + >..)/

Jt + y'

2

= 0,

or

y' = 0.

In other words, the slope is zero at the free end.

0

Such a boundary condition of y' = 0 has the real physical meaning because any non-zero gradient at the free end would have a non-zero vertical component, thus causing the vertical slip along the rope due to the tension in the rope. The zerogradient leads to the static equilibrium. Thus, the whole curve of the hanging rope with one free end forms half the catenary. 0 Example 11.4: Dido's problem concerns the strategy to enclose a maximum area with a fixed length circumference. Legend says that Dido was promised a piece of land on the condition that it was enclosed by an oxl1ide. She had to cover as much as land as possible using the given oxhide. She cut the oxhide into narrow strips with ends ed, and a whole region of a hill was enclosed. Suppose the total length of the oxhide strip is L. The enclosed area A to be maximized is xb

A=

1

y(x)dx,

Xa

where X a and xb are two end points (of course they can be the same points). \rVe also ha.ve the additional constraint Xb

1 Jl +

y 12dx = L =

const.

Xa

This forms an isoperimetric variation problem. As L is fixed, thus the maximization of A is equivalent to make I = A + >..L stationary. That is

I= A+ AL = { ' [y + A.jl +ifl)dx. Using the Euler-Lagrange equation, we have

or

163


Integral Equations

which becomes

1 - >.. .!!._ ( dx

y'

Jt + y'2

) = 0

·

Integrating it once, we get

>..y'

--====x+K,

Jt + y'2

where K is the integration constant. By rearranging, we have y' =

±

X+ K

J>..2- (x + K)2

Integrating this equation again, we get

y(x) = =F)>.. 2

-

(x

+ K)2 + B,

where B is another integration constant. This is equivalent to

which is essentially the standard equation for a circle with the centre at (- K, B) and a radius >... Therefore, the most area that can be enclosed by a fixed length is a circle. 0

An interesting application is the design of the slides in playgrounds. Suppose we want to design a sn1ooth (frictionless) slide, what is the best curve/shape the slide should take so that a child can slide down in a quickest way? This problem is related to the brachistochrone problem, also called the shortest time problem or steepest descent problem, which initiated the development of the calculus of variations. In 1696, Johann Bernoulli posed a problem to find the curve that n1inimizes the time for a bead attached to a wire to slide from a point (0, h) to a lower point (a, 0). It was believed that Newton solved it within a few hours after receiving it. From the conservation of energy, we can determine the speed of the bead from the equation ~mv 2 + mgy = mgh, and we have

v = J2g(h- y). 164

(11.40)

Integral Equations


So the total time taken to travel from (0, h) to (a, 0) is t

=

lo

0

a-ds 1 = loa v

0

)1 + y'2 dx. J2g(h - y)

(11.41)

Using the simplified Euler-Lagrange equation (11.34) because the integrand \ll = )1 + yrl / J2g(h- y) does not contain x explicitly, we have

(1

+ y 12 )

2g(h- y) -

y

I

8 8y'

(1

[

+ y12)

] -A

2g(h- y) -

.

(11.42)

By differentiation and some rearrangements, we have y

'2

=

1 B = 2gA2.

B-h+y h-y '

(11.43)

By changing of variables TJ = h- y = ~(1 -cos 0) and integrating, we have x=

~ [0 -

sin 0] + k,

(11.44)

where 0 < 1r and k is an integration constant. As the curve must the point (0, h), we get k = 0. So the parametric equations for the curve become x

= ~ (0 - sin 0),

B y = h- "2(1- cosO).

(11.45)

This is a cycloid, not a straight line, which seems a bit surprising, or at least it is rather counter-intuitive. The bead travels a longer distance, thus has a higher average velocity and subsequently falls quicker than traveling in a straight line.

11.1.4

Variations for Multiple Variables

What we have discussed so far mainly concerns the variations in 2-D, and subsequently the variations are in y(x) or curves only. What happens if we want to study a surface in the full 3D configuration? The principle in the previous sections can be 165


Integral Equations

extended to any dimensions with multiple variables, however, we will focus on the minimization of a surface here. Suppose we want to study the shape of a soap bubble, the principle of least action leads to the minimal surface problem. The surface integral of a soap bubble should be stationary. Now we assume that the shape of the bubble is u( x, y), then the total surface area is

where

W=

{)u 1 + (-)2

ax

au = + (-)2 {)y

v+ 1

u2X

+ u2.y

(11.47)

In this case, the extended Euler-Lagrangian equation for two variables x and y becomes (11.48)

Substituting winto the above equation and using ~! since w does not contain u explicitly, we get

=

wu = 0 (11.49)

or

(1

+ u~)Uxx- 2uxUy + (1 + u;)uyy = 0.

(11.50)

This is a nonlinear equation and its solution is out of the scope of this book. This nonlinear equation has been one of the active research topics for more than a century. It has been proved that the fundamental solution to this equation is a sphere, and in fact we know that all bubbles are spherical. For some problems, we can approximately assume that Ux and uy are small, thus the above equation becomes Laplace's equation Uxx

+ Uyy 166

= 0.

(11.51)

Integral Equations

11.2 Integral Equations

The calculus of variations has many applications. The other classical examples include Fermat's principle in optics, SturmLiouville problem, surface shape minimization, the action principle, and of course the finite element analysis.

11.2

Integral Equations

From the calculus of variations, we know that the unknown y( x) to be optimized is inside the integrand of I. In certain sense, this is an integral equation. In fact, many physical processes and laws of conservation are expressed in of integral forms rather than their differentiation counterparts. Naturally, one of the ways of constructing an integral equation is to integrate from a differential equation. Integral equations are much more complicated compared with the differential equations. There is no universal solution technique for nonlinear equations, even the numerical simulations are usually not straightforward. Thus, we will mainly focus on the simplest types of integral equations.

11.2.1

Linear Integral Equations

Fredholm Integral Equations A linear integral equation for y( x) can be written in the following generic fonn

u(x) +A

l

K(x, TJ)Y(TJ)dTJ = v(x)y(x),

(11.52)

where K(x, rJ) is referred to as the kernel of the integral equation. The parameter A is a known constant. If the function u( x) = 0, the equation is then called homogeneous. If u( x) =/= 0, the equation is inhomogeneous. If the function v( x) = 0, then the unknown y( x) appears only once in the integral equation, and it is under the integral

167

11.2 Integral Equations

Integral Equations

sign only. This is called the linear integral equation of the first kind u(x)

+At

K(x, 1/)Y('I)d'l = 0.

(11.53)

On the other hand, if v(x) = 1, equation (11.52) becomes the integral equation of the second kind u(x) +A

l

K(x, 1/)Y('I)d'l

= y(x).

(11.54)

An integral equation with the fixed integration limits a and b, is called a Fredholm equation. If the upper integration limit b is not fixed, then the equation becomes a Volterra equation. The integral equation becomes singular and at least one of its integration limits approaches infinite. Volterra Integral Equation In general, the Volterra integral equation can be written as u(x) +A

lx

K(x, TJ)Y(TJ)dTJ = v(x)y(x).

(11.55)

The first kind [or v(x) = 0] and second kind [or v(x) = 1] are defined in the similar manner. The kernel is said to be separable or degenerate if it can be written in the finite sum form N

K(x, TJ) =

L, fi(x)gi(TJ),

(11.56)

i=l

where fi(x) and gi(TJ) are functions of x and TJ, respectively. A kernel is called a displacen1ent kernel if it can be written as a function of the difference (x - TJ) of its two arguments K(x, TJ) = K(x- TJ). 168

(11.57)

Integral Equations

11.3

11.3 Solution of Integral Equations

Solution of Integral Equations

Iviost integral equations do not have closed-form solutions. For linear integral equations, the closed-form solutions are only possible for the special cases of separable and displacement kernels.

11.3.1

Separable Kernels

For a Fredholm integral equation of the second kind with separable kernels, we can substitute the kernel (11.56) into the equation and we have

J. tr b N

u(x) +A

J;(x)g;(17)d1J

= y(x),

(11.58)

y(x).

(11.59)

which becomes

tr J. N

u(x) +A

/;(x)

b

g;(17)d17

=

Because the integration limits are fixed, the integrals over 1J should be constants that are to be determined. By defining <>;

=

t

g;(1J)Y(1J)d1J,

(11.60)

we now have the solution in the form N

y(x)

=

u(x) +A

L aifi(x),

(11.61)

i=l

where the N coefficients

Oi

are determined by (11.62)

fori= 1, 2, ... , N. Only for a few special cases, these coefficients can be written as simple explicit expressions. 169


11.3.2

Integral Equations

Displacement Kernels

For a singular integral equation with a displacement kernel, the equation can be solved by Fourier transforms if both integration limits of the integral are infinite. In this case, we have

u(x)

+A/_: K(x -1J)Y(1J)d1J

=

Y(1J).

(11.63)

Using the Fourier transforms and the convolution theorem, we have U(w) + Av'2?fK(w)Y(w) = Y(w), (11.64) which is an algebraic equation for Y(w). Its solution is simply

Y(w)

=

U(w)

.

1 - Av'2-iK(w)

(11.65)

The solution y( x) can be obtained using the inverse Fourier transform

y(x)

11.3.3

=

roo

V(w) eiwxdw. y'2-i 1-oo [1- Av'2-iK(w)]

_1_

(11.66)

Volterra Equation

A Volterra equation with separable kernels may be solved by transforming into a differential equation via direct differentiation. In the case of a simple degenerate kernel

K(x, 17) = f(x)g(1J),

(11.67)

we have

fox f(x)g(1J)Y(1J)d1J,

(11.68)

+ Aj(x) fox g(1J)Y(1J)d1J.

(11.69)

y(x) = u(x) +A which becomes

y(x) = u(x)

170

Integral Equations


If f (x) -=/= 0, it can be written as

y(x) f(x)

=

u(x) rx f(x) +A lo g(TJ)Y(TJ)dTJ.

(11.70)

Putting ¢(x) = u(x)/ f(x) and differentiating it, we have

~~~:~]' = 4>'(x) + Ag(x)y(x).

(11.71)

By letting W = y(x)j f(x), we have

w'(x)- .Xf(x)g(x)\ll(x) = ¢'(x),

(11.72)

which is a first-order ordinary differential equation for W(x). This is equivalent to the standard form

w' + P(x)\ll =

Q(x),

(11.73)

and

P(x) = -.Xf(x)g(x),

Q(x)

=

~~~:~]'.

(11.74)

We can use the standard technique by n1ultiplying the integrating factor exp[J P(x)dx] to obtain the solution. We get

y(x)

=

f(x)[e- J P(x)dx]{j [Q(x)ef P(x)dx]dx }.

(11.75)

With appropriate boundary conditions, the exact form of the solution can be obtained. 0 Example 11.5:

Let us tzy to solve the integral equa.tion of

Volterra type y(x) =ex+ lax ex sin(()y(()d(.

First, we divide both sides by ex, we get y(x)

ex

= 1 + rx sin(()y(()d(,

Jo

171

Integral Equations

11.3 Solution of Integral Equations whose differentiation with respect to x leads to y(x) '

[&]

. = y(x)sm(xL

or ..!..y'(x)- y(x)e-x = y(x) sin(x). ex Divide both sides by y(x) and using [Iny(x)]' = y'(x)fy(x), we have [In y(x)]' =ex sin x

+ 1.

By direct integration, we have lny(x) = x-

~ex cosx +~ex sinx.

Thus, we finally obtain y(x)

ex

= exp[x- 2

(cosx- sinx)]. 0

There are other n1ethods and techniques of solving integral equations such as the operator method, series method and the Fredholm theory. However, most integral equations do not have closed-form solutions. In this case, numerical methods are the best alternative.

172

Chapter 12

Tensor Analysis Ivlany physical quantities such as stresses and strains are tensors. Vectors are essentially first-order tensors. Tensors are the extension of vectors, and they can have any number of dimensions and any orders, though most commonly used tensors are second order tensors.

12.1

Notations

In tensor analysis, the Einstein sun1mation convention or Einstein notations 1 and notations for subscripts are widely used. Any lowercase subscript that appears exactly twice in any term of an expression means that sum is over its all possible values of the subscript. For example, in the three-dimensional case, we have 3

GiXi

=L

aiXi = GtXt

+ a2x2 + G3X3.

(12.1)

i=l 3

~jBjk

=L

AijBjk

= AitBlk + Ai2B2k + Ai3B3k·

(12.2)

j=l 1

This notation convention was introduced by Albert Einstein in 1916 when formulating the theory of General Relativity.

173

12.2 Tensors

Tensor Analysis

OUi = \7 . u = OUt 8xi OXt

+ OU2 + OU3 8x2

(12.3)

8x3 ·

The Kronecker delta 8ij which is a unity tensor (like the unity matrix I in matrix analysis), is defined as 8

ij =

1 (if i { 0 (if i

= j),

i= j).

(12.4)

For a tensor with three subscripts sin1ilar to 8ij, the Levi-Civita symbol or tensor is defined as Eijk =

+1 if (i,j, k) is an even permutation of (1, 2, 3), { -1 if (i,j, k) is an odd permutation of (1, 2, 3), 0 (otherwise). (12.5)

The tensors

8ij

and

Eijk

are related by

fijkfkpq = 8ip8jq -

8iq8jp·

(12.6)

Using the summation conventions, the matrix equation Ax=b,

(12.7)

can alternatively be written as (i=1,2, ... ,n).

12.2

(12.8)

Tensors

When changing the bases from the standard Cartesian e 1 = i, e2 = j, e3 = k to a new set of bases ei, e~, e~, a position vector x = (x 1, x2, x 3 ) in the old bases is related to the new vector x' = ( x!, x~, x~) in the new bases by a coefficient matrix Sij. Sij can be the rotation, translation, enlargement or any of their combinations. For example, the matrix for a simple rotation, with an angle of () around a fixed axis, Sij becomes

Sij= (

~:o ~: 174

D·

(12.9)

Tensor Analysis

12.3 Tensor Analysis

The orthogonality of Sij requires that

ssT= sTs =I or (12.10)

If the components ui of any variable u are transformed to the components u~ in the new bases in the same manner as

(12.11)

then ui(i = 1, 2, 3) are said to form a first-order Cartesian tensor (or vector in this case). If components of a variable such as strains aij are transformed as (12.12)

we say these components form a second-order tensor. The order of a tensor is also called its rank. Scalars have rank zero, vectors have rank 1, and second-order tensors have rank 2. In engineering and computing, the rank is associated with the number of indices to describe a tensor in of a multidimensional array. In this sense, a second-order tensor is equivalent to a two-dimensional array or a matrix. In a similar fashion, higher-order tensors can be defined, and for each order increase, then there is one Sij extra in the product for transforming, but no subscripts are allowed to appear more than twice (12.13)

and (12.14)

12.3

Tensor Analysis

Tensors in Cartesian Coordinates One of the main advantages of tensors is that a tensor is independent of any chosen frame of reference. Therefore, any 175


Tensor Analysis

physical laws or equations that are formulated in of tensors should be independent of frame of reference. For example, the stress -strain relation in linear elasticity is independent of frame of reference (for details see next chapter) (12.15)

where aij and Cij are the stress tensor and strain tensor, r& spectively. J.l and A are Lame constants. In the similar way as n1ulti-dimensional arrays or matrices, two tensors can be added or subtracted component-bycomponent if and only if they are the tensors of the same order. For second-order tensors, a tensor Tij is said to be symmetric if Tij = Tji, and antisymmetric if Tij = -Tji· An interesting property of a tensor Tij is that it can always be written as a sum of a symmetric tensor and an antisymmetric tensor Tij

= ~(Tij + Tji)[sym.] + ~(Tij- Tji)[antisym.].

(12.16)

All the formulas in vector analysis can be rewritten in the tensor forms using the summation convention and notations (12.17) \1 2 1/J =

{)21/J

= Oij

8xi8Xi

\7

X

(\7

X

u)i

{)1/J 8xi8Xj'

8uq 8XjXp

= fijkfkpq-- - .

(12.18) (12.19)

Similarly, the divergence theorem can be rewritten as the following form f 8 ui dV = J uinidS. (12.20) lv 8 xi !s The tensor forn1s are sometimes useful to the proof of the complex relationship among vectors and tensors. They also become handy for the implementation of numerical algorithms. 176


Tensor Analysis

Using the tensor notations, we have the identity oiiAi = Ai· The cross product Ax B can be expressed as (12.21)

Another way of denoting the derivatives of tensors a and v is to use following notations

aa

(12.22)

ai=~,

'

UXi

where the index of the spatial component Xi is denoted by a comma to avoid any potential confusion with other indices. With these notations, the important operators involving the V'-operator can be written as

and V'.v

grad¢ = ¢, i = V' ¢,

(12.23)

Y' 2 ¢ = ¢,ii = ~¢,

(12.24)

82 u1 82 u2 82 u3 = Vi,i = !i""""2 + '£l"2 + -8X3 2 · UXl UX2

(12.25)

A very special case is that (12.26) 0 Example 12.1: Let us now use the tensor notations to prove ax (b x c) = b(a ·c) - c(a ·b). Fl·om above expressions for cross

products, we know that

a

X

(b

X

c) = Eijkaj(Ekpqbpcq) = EijkEkpqajbpCq

= (8ip8jq- 8iq8jp)ajbpcq = 8ipbp(8iqajcq)- 8iqcq(8iPaibp)·

Using 8ipbp = b and 8iqcq = c, we have

a(b x c) = b(tSjqajcq) - c(tSiPaibp)· By renaming the indices (j - i and q - j) so that 8iqajcq 8ijaici = a • c and tSiPaibP = a • b, we finally obtain

ax (b x c)= b(a ·c)- c(a ·b). 0

177


Tensor Analysis

Tensors in Non-Cartesian Coordinates The tensors we have discussed so far are expressed in Cartesian coordinates. In non-Cartesian coordinates, they are more complicated. Tensor analysis is very important in theoretical physics and differential geometry where formal mathematical theory is required. In fact, many books on tensor analysis use a modern approach in of tensor duality, covariance and contravariance concepts. In the simplest term, a tensor such as a vector v can be expressed as the sum of its components multiplying by the basis vectors (12.27)

where vi(i = 1, 2, 3) are called the covariant components of v in the contravariant basis vectors ei, while vi(i = 1, 2, 3) are called the contravariant components of v in the covariant basis vectors ei. For the curvilinear coordinates (Ql , Q2, q3) at any point P on a position vector r(q1 , q2 , q3 ), the basic vectors are given by (i=1,2,3),

(12.28)

where ei and ei are reciprocal systems of vectors, and ei · ei = 8j, where 8j acts in the similar way as 8ii· In many books on tensor analysis, ei is also written as €i = ei. Furthermore, the tensor product (also called outer product) of two tensors is rather complicated. For example, the tensor product u Q9 v = uvT of two vectors u and vis given by

U@V = (

~~ )@(Vt

V2 V3)

= (

U3

~~~~ ~~~~ ~~~: ) . (12.29) U3V1

U3V2

U3V3

Using these basis vectors, we can write a second-order tensor in tern1s of covariant components aii and contravariant components aii as (12.30) 178

Tensor Analysis


In a given frame of reference, the fundamental metric tensor is defined by (12.31) This tensor is always symmetric 9ii = 9ji, its determinant g = l9ij I = det(gij) is related to the Jacobian J = Vff· The derivatives in non-Cartesian coordinates are far more complicated, and they usually involve the Christoffel coefficients rt = ek · ~. For example, the divergence of a vector is defined by the covariant differentiation u1i or V · u = u~i = ~u~ +r~.ui = ui.+r1 .ui. It is worth pointing out that ei and ei vq1 l] ,] •1 become identical and r~j = 0 in Cartesian coordinates, and it is therefore not necessary to distinguish the contravariant and covariant vectors and components. Mathematically speaking, the formal approach is preferred. In engineering mathematics, however, the simple formulation in of multidimensional arrays in Cartesian coordinates is a more convenient approach, especially from the computational point of view. That is why we have used an over-simplified approach here. In the next chapter, we will study the theory of linear elasticity as an application of tensor analysis.

179

Chapter 13

Elasticity 13.1

Hooke's Law and Elasticity

The basic Hooke's law of elasticity concerns an elastic body such as a spring, and it states that the extension x is proportional to the load F, that is

F=kx,

(13.1)

where k the spring constant. However, this equation only works for 1-D deformations. For a bar of uniform cross-section with a length L and a cross section area A, it is n1ore convenient to use strain e and stress u. The stress and strain are defined by

u

F

e

= A,

~L

=

L'

(13.2)

where ~Lis the extension. The unit of stress is N jm 2 , while the strain is dimensionless, though it is conventionally expressed in mjm or% (percentage) in engineering. For the elastic bar, the stress-strain relationship is

u=Ec,

(13.3)

where E is the Young's modulus of elasticity. Written in F and x = ~L, we have EA k= EA (13.4) F= L~L=kx, L' 181

13.1 Hooke's Law and Elasticity

Elasticity

where k is the equivalent spring constant for the bar. This equation is still only valid for any unidirectional compression or extension. For the 2-D and 3-D deformation, we need to generalize Hooke's law. For the general stress tensor (also called Cauchy stress tensor)

(13.5)

and strain tensor e

= ( ::: ::: ::: ) = ( c::;x

c::;y

:~~ :~~ :~: ) ' c31

c::;::;

c32

(13.6)

c33

it can be proved later that these tensors are symmetric, that is u = uT and e = eT, which leads to O"xy

= O"yx'

O"xz

= O":;x,

O"yz

= cyx,

cxz

= c::;x,

cy:::

= O"::;y'

(13.7)

= c:::y·

(13.8)

and cxy

Therefore, we only have 6 independent components or unknowns for stresses and 6 unknown strain cmnponents. The strain tensor is defined by the displacement u T = 1 8ui (2 8Xj

Cij = -

where write

Xt

= x, x2 =

y, and x3

+ -8uj ), 8 Xi

(13.9)

= z. Sometimes, it is useful to

1 e = 2'(V'u + V'uT).

(13.10)

The generalized Hooke's law can be written as 1

Cxx

= E[axx- v(ayy

cyy

= E[ayy- v(axx

1

182

+ O"::;z)],

(13.11)

+ O"zz)],

(13.12)

Elasticity


1 c::::: = E [u::::-

v(uxx

+ O"yy)],

1+v

cxy = ----g-O"xy,

1+v

ex;::

= ----g-O"x;:;,

(13.13) (13.14) (13.15) (13.16)

where v is the Poisson's ratio, and it measures the tendency of extension in transverse directions (say, x and y) when the elastic body is stretched in one direction (say, z). It can be defined as the ratio of the transverse contract strain (normal to the applied load) to the axial strain in a stretched cylindrical bar in the direction of the applied force. For a perfectly incompressible material, v = 0.5, and v = 0 rv 0.5 for most common materials. For example, steels have v = 0.25 rv 0.3. Some auxetic material such as polymer foams or anti-rubbers have a negative Poisson's ratio v < 0. This generalized Hooke's law can concisely be written as (13.17) where we have used the Einstein's summation convention O"kk = O"xx+O"yy+O"::::· Another related quantity is the pressure, which is defined by p

=

1 3

--O"kk

=

(13.18)

The negative sign comes from the conventions that a positive normal stress results in tension, and negative one in compression, while the positive pressure acts in compression. Sometimes, it is more convenient to express the stress tensor in of pressure and devitoric stress tensor Sij (13.19) 183


Elasticity

If we want to invert equation (13.17), we have first express in of Ckk so that the right hand side of the new expression does not contain the stress akk· By contraction using j ---+ i, we have

akk

(13.20)

where we have used 8ii = 811 + 822 + 833 = 1 + 1 + 1 = 3 and In engineering, the quantity

aii = akk·

Ckk

= Cxx

8 2 u2

fPut

82 u3

+ Cyy + Cz::: = Bx2 + By2 + Bz 2

= V · u, (13.21)

means the fractional change in volume, known as the dilation. This gives that E (13.22) CJii = CJkk = _ v 1 2 Ckk· Substituting it into equation (13.17), we have 1 +v

Cij

v

= ----g-aii - E (

1

E _

(13.23)

v Ckk)8ij,

2

or after some rearrangement 1+v ----g-aii

= Cij

v

+ 1 _ 2v Ckk8ij,

(13.24)

which can be written as (13.25)

where J.L and A are La1ne constants. They are E

G = J.l = 2(1 + v)'

A=

vE

(1

+ v)(1 - 2v)

.

(13.26)

This stress-strain relationship can also be written as u = 2Ge + A(V · u)l5.

(13.27)

In engineering, G = J.L is called the shear modulus, while K = 3 ( 1 ~2 ,) is called the bulk modulus which is the ratio of pressure -p to the volume change ~ V. 184

Elasticity

13.2

13.2 P.tfaxwell's Reciprocal Theorem

Maxwell's Reciprocal Theorem

For an elastostatic problem, the balance of force leads to

V ·U+ b

= 0,

(13.28)

where b is the body force or force per unit volume. For a small cube volun1e element, the total body force dfi along the xi-axis is 8aii dfi = -dbi = dV, (13.29)

0x·J

here we have used the index summation conventions. Similarly, the total force along the j-axis is d/j

=

oaiidV. OXi

(13.30)

Since there is no relative rotation of the cube element because the cube element must be at rotational equilibrium, thus the result moment must be zero. Taking the moment of the two force components about any point (say, a corner of the cube). This leads to (13.31) which means the stress tensor is symmetric. This is in fact the compatibility condition for stresses. Alternatively, consider a cube element at rotational equilibrium with a volume dV = 8x8y8z, and the dimensions of the elements are 8x, 8y and 8z, respectively. We consider the forces along the four faces that are parallel to z-axis (shown in Figure 13.1). If we take the moment about a line which is parallel to z-axis and goes through one corner point, we only have to consider the two faces that are far from this point because the two faces through this point do not contribute to the moment. The forces on the two faces are O'xydAtdY and ayxdA2dx where the surface areas are dAt = 8x8z and dA2 = 8y8z. The total moment about point D is (13.32) 185

13.2 !viaxwell's Reciprocal Theorem

Elasticity

Thus, axy = ayx· Similar arguments for other faces along xand y-directions about any other points and we have aii = Uji·

z y

8y Figure 13.1: A cubic element in an elastostatic body. Using the stress-strain relationship, it is straightforward to prove that the strain tensor is also symmetric (cij = cji)· In engineering, this tensor is often written as

(13.33)

so as to emphasize that the non-diagonal elements are for shear components Txy etc. From the matrix algebra, we know that a square matrix can always be expressed in of eigenvalues and eigenvectors. For the stress tensor, the eigenvalue problen1 (u- ail)ft = 0,

(13.34)

provides the the principal stresses ai(i = 1, 2, 3) (eigenvalues) and their principal directions ftT = n1, n2, n3 (eigenvectors). In the coordinate system formed by the three eigenvectors, the stress is expressed by the three principal stresses along three principal directions, and there are no shear stress components. The non-trivial solutions require the determinant of the co186

Elasticity

13.2 P.tfaxwell's Reciprocal Theorem

efficient matrix must be zero. That is

(Jx- (J Txy Tx:; ay- a Txy Ty:; (J:; - (J Ty:; Tx::;

= 0.

(13.35)

It can be expanded into a cubic equation

a3- (ax+ ay

+ a::;)a2

+[axay + aya:;

+ a::;ax- (r;y + r;::; + r;:;)]a -[axaya::; + 2TxyTy:;Tx::;- (axr;::; + ayr;::; + a::;T;y)] = 0.

(13.36)

There are three invariants (It, h and I3) for the second-order symmetric stress tensor aii, and these invariants satisfy the characteristic equation

(13.37) where I1

h

=

= tr( a ii) = ax + a y + a::;,

axay + aya:: + a:;ax- (r;y + r;:; + r;:;;),

(13.38) (13.39)

and

Under appropriate transformations, this tensor can be transformed into a diagonal form

(13.41) where a1, a2, a3 are principal stresses. Written in of principal stresses, the three invariants become

(13.42) (13.43) 187

13.2 !viaxwell's Reciprocal Theorem

Elasticity

and (13.44)

Now consider an elastic body when n concentrated load f 1 , f 2 , ... , fn acted upon the body at n different points. The displacements at each point in the direction of the corresponding force are Ql, Q2, ... , Qn. For a linear elastic body, the principle of superposition applies and we have (13.45) (13.46)

(13.47)

where Cij are the influence coefficients or flexibility matrix. The total work done due to this set of loads is (13.48)

There is a very useful theorem concerning these coefficients. It is called Maxwell's reciprocal theorem or Maxwell-Betti theorem, which states that the influence coefficients (or flexibility) matrix is symmetric Cii = Cji· That is to say, the displacement at point i due to a unit load at another point j is equal to the displacement at j due to a unit load at point i. This theorem is essentially equivalent to say the displacements are path-independent and independent of the order of the loads applied upon the elastostatic body. For two forces f i and f i, the final displacements are the same where fi is applied first, then fi, or fi is applied first then f i, or even both are applied at the same time. In other words, the system has no memory of the load history. In the case of only two forces, we first apply /i slowly (so as to reduce the dynan1ical effect) with fi = 0, the displacement of point i is Cii/ i and the displacement of point j is Ciif i. The work 188

Elasticity

13.3 Equations of !vlotion

done is !Ciifl where Jl = fi · fi. Now with /i kept fixed, we apply I i slowly, the additional displacement at point i is Cij f i and the additional displacement at point j is Cjj f j. In this case, the extra work done is Cijfi · fj + !Cii!J. The total work done at the final state is

W

1

2

1

= 2ciifi + 2ciifi2 + Cii(fi · fj).

(13.49)

If follow the same procedure but slowly apply I i first, then

f i, the total work done is now (13.50)

As the total work done should be independent of the order in which the loads are applied, this requires W = W, which leads to (13.51) Since fi · fj is a dot product and thus a scalar, we now get (13.52)

This completes the proof of the :rviaxwell's reciprocal theorem. This theorem is the important basis for boundary element analysis and virtual work method in computational engineering.

13.3

Equations of Motion

For a general solid where the inertia is not negligible, we have (13.53)

where p is the density of the elastic body. In some books, the following form of body force b = pf is used, in this case, the force f means the force per unit mass. Together with the generalized Hooke's law and relationship with displacement u, 189

13.3 Equations of Motion

Elasticity

we have the following set of equations of motion for an elastic body. (13.54) (13.55) (13.56)

There are 15 equations (6 for stresses, 6 for strains, 3 for displacement) and we have 15 unknowns (6 stress con1ponents, 6 strain components and 3 displacements). Therefore, the elastic field should be uniquely determined if appropriate boundary conditions are given. There are other compatibility equations as well, and we will briefly discuss them later. If we write the equations of motion using the bold font notations, we have {)2u

V·u+b=p8t 2 ,,

(13.57)

u = 2Ge + .X(V · u)d,

(13.58)

1 e = 2'(Vu + VuT).

(13.59)

If we substitute the generalized Hooke's law and displacement into the first equation (13.57), we have {)2u

V · [2Ge + .X(V · u )«5] + b = P 8t2

(13.60)

,

or {)2u

V. [G(Vu + VuT) + .X(V · u)«5] + b = P {)t 2

,

(13.61)

which leads to 2

[)2u

(G + .X)V(V · u) + GV u + b = P {)t 2 190

•

(13.62)

Elasticity

13.3 Equations of !vlotion

Using G +A= G/(1- 2v) and after some rearrangements, we have (13.63)

which is the well-known Cauchy-Navier equation. This equation s both longitudinal wave (P wave) and transverse wave (S wave). In the simplest 1-D case without any (external) body force b, we can take V · u = 0 for S-wave, the equation is simplified as (13.64)

thus its wave speed is

vs=~·

(13.65)

For the P-wave in 1-D, the displacement field is non-rotational, i.e., V x (V xu)= 0. From the identity V(V · u) = V x (V x u) + V 2 u, the 1-D Cauchy-Navier equation becomes P

82u1

fJt2 = (A + 2G)

a2ut

ax2 .

(13.66)

Then, the speed of P- wave is Vp

= J(.>..+p2G).

(13.67)

Since A+ 2G > G, therefore, P-waves always travel faster than S-waves. Furthermore, from the definitions of the strain components in of displacements uT = (ut, u2, u3) = (u, v, w), we have

au

Cxx

=ax'

Cyy =

8v oy'

By assuming the displacements are continuous and differentiable functions of positions, we differentiate Cxx with respect

191

13.4 Airy Stress Functions

Elasticity

to y twice, we have (13.69)

Similarly, differentiate

cyy

with respect to x twice, we have (13.70)

Now differentiate cxy with respect toy once, and with respect to x once, we have (13.71)

where we have used the interchangeability of partial derivatives fPvj8x8y = 8 2vj8y8x. This can be rearranged as

2 8 cxx 8y2

+ 82cyy 8x 2

=

28

2

cxy

8x8y '

(13.72)

which is the compatibility equation. In the same fashion, we can derive other compatibility equations (13. 73)

(13.74)

13.4

Airy Stress Functions

For certain engineering problems, the solutions in a plane is of concern. In this case, we are dealing with plane strain and plane stress problems. For a plane stress problem, we assume that a:::: = 0 (but c:;:; =I= 0), then the plane stress problem involves no stress components depending on z. That is to say Uxz = Uyz = Uzz = 0. We have only three independent stress 192

Elasticity


components axx' aYY' and axy· The generalized Hooke's law reduces to 1 (13.75) Cxx = E(axx- vayy), 1

Cyy = E(ayy- vaxx),

1+v

Cxy = ----g-axy·

(13.76) (13.77)

However,

-v 1 -v

C::;::; = - -(Cxx

+ Cyy),

(13.78)

which is not zero in general. For plane strain problems, it is assumed that C::;::; = 0. Thus, there are only three independent strain components Cxx' Cyy' and Cxy, however, the stress a::;::; = v(axx + ayy) is not zero. The compatibility equation becomes

82Cxx + fPcyy 8y 2 8x 2

= 2 fPcxy

8x8y ·

(13.79)

For plane strain problems with no body forces, the equilibrium equations are automatically satisfied if the stress components are related to a scalar function q,, called Airy's stress function. The Airy's stress function is defined by (13.80)

In this case, the compatibility equation becomes (13.81)

which is a biharmonic equation and can be written as (13.82)

In cylindrical polar coordinates (r, 8, z), it becomes

82 [8r2

18 + ; 8r

+

182 2 r2 882] q, - O.

193

(13.83)


Elasticity

Figure 13.2: A crack in an elastic plate. Now consider a semi-infinite crack in an infinite solid as shown in Figure 13.2, assuming the elastic body deforms in plane strain. The far field is subjected to bounded stress at infinity. The surfaces of the crack shall be stress free, which leads to the following boundary conditions U()()

=

[)2 {)r2

= 0,

Ur()

=-

{) 1 {) {)r (-; {)O)

= 0, at 0 = ±7r. (13.84)

Let us try a solution of the form = rn+l f(O),

(13.85)

and substitute it into the governing biharmonic equation, we get

d?-

[d02

+ (n + 1) 2 ][dd?-02 + (n-

2

1) ]f(O) = 0.

(13.86)

As the second-order equation y" + >.. 2 y = 0 has a general solution y = A sin >..O + B cos >..O, we can here use this method twice, the general solution takes the following form

f(O)

=

Acos(n + 1)0 + Bsin(n + 1)0

+C cos(n- 1)0 + D sin(n- 1)0.

(13.87)

The boundary conditions become

aeo

=

rn- 1n(n + 1)/(0), 194

(13.88)

Elasticity


and Ure

= rn-ln{(n + 1)[Asin(n + 1)8- Bcos(n + 1)8]

+(n- 1)[C sin(n- 1)8- D cos(n- 1)8]},

(13.89)

at 8 = ±1r. We know n = 0 is trivial. From the first equation, we have 1 3 n = ±2, ±1, ±2, ... ,

sin(2n7r) = 0,

(13.90)

and rn(n ~ 1) does not converge, therefore, they are not suitable solutions. The constraint now becomes n ::; 0, but the solutions has singularity as r-----+ 0. This is however acceptable in the crack propagation as the stress concentrations do physically exist. Substituting the general solution into the boundary conditions with n = 1/2 and 8 = ±1r, we get B- D = 0.

3A+C= 0,

(13.91)

By defining the stress intensity factor K1 for the crack,

K _ 3A-/2i 4

I-

(13.92)

'

which is for the opening (model I) of the crack. It is a limit of stress at 8 = 0 (13.93) K1 =lim uee(r, 8)1 .

r-o

6=0

Finally, the solution of stresses can be written as }(/

Urr

= . M=:(1

v 21rr

. 2

+ Slll

K1 U()()

= v'27fTCOS }(/

Ure

2

8

8

2

2

-)COS-,

3

8

2'

8 . 8

= v'21fT cos 2 s1n 2· 195

(13.94)

(13.95) (13.96)

13.5 Euler-Bernoulli Beam Theory

Elasticity

Once we have the stress distribution, we can get the strains. Then the displacements are the integration of the strains, and we have (13.97) and ue =

13.5

K1(1 - v) fiT . 30 E V-;r[sin

. 0

2 - (5- 4v) sm 2].

(13.98)

Euler-Bernoulli Beam Theory

The Euler-Bernoulli beam theory is a simplified theory for calculating the deflection of beams under a distribution of load force using the linear isotropic theory of elasticity. The basic assumptions for the beam theory are: 1) the beam is isotropic and elastic; 2) the beam deformation is dominated by bending, and distortion and rotation are negligible; 3) the beam is long and slender with a constant cross section along the axis. Under these assumptions, we can now derive the governing equations. Let u(x, t) be the deflection of the beam (shown in Figure 13.3), A be the area of the cross section, and f(x, t) be the force per unit length. The first assumption implies that the bending moment M is proportional to the curvature K of the bending. That is M=EIK,

(13.99)

where E is the Young's modulus and I is the area moment of the beam's cross section. In mechanics, I is also called the second moment of area or the area moment of inertia. It is worth pointing out that the area moment about a horizontal axis through the centroid is defined by (13.100) 196

Elasticity


which has a unit of [m] 4 , and it should not be confused with the mass moment of inertia J (also often denoted as I, but we use J here) about an axis, which is defined by (13.101) with a unit of [Kg] [m] 2 . Both E and I do not change along the x-axis. For a cylindrical rod with a radius of R, we have I= 7rR4 j4. For a rectangular beam with a base width of band a depth of h, we have I = bh3 /12.

V(x)/ X

r---I ~

/v(x+dx)

f(x, t)dx

...-r------=-=--=-:..:-:.:-~-

- - -- ---- -_____ - -- - ,_J

Tu(x,t) Figure 13.3: Beam bending. The second assumption means that the shear V (x) is related to the bending moment 8M 8x = V(x),

and the third assumption means

tx << 1. Therefore, we have

82u M::::::: EI 8x2' or

V:::::::

(13.102)

~(EI82u).

(13.103)

(13.104) 8x 8x 2 For a small volume element (also shown in Figure 13.3), the mass of the element is pAdx where p is the density, and the

197


Elasticity

fff.

acceleration is The shear force variation V(x V (x) + ~dx, and the total force is

V(x)- V(x

+ dx) + f(x, t)dx =

8V

[f(x, t)- ax ]dx.

+ dx) (13.105)

Using the Newton's second law of motion, we have

8V

82 u

f(x, t) - ax = pA at2 .

(13.106)

Substituting the above expression for V, we have

82 u pA at2

82

82 u

+ ax2 [EI ax2] = f(x),

(13.107)

which is the Euler-Bernoulli equation. If there is no force f(x, t) = 0, the equation becomes a homogeneous form

{)2u pA fJt2

{)2

a2u

+ {)x2 [EI {)x2] =

0,

(13.108)

which is a fourth-order wave equation. It governs the waves that travel along a beam, a rod or any slender column. This equation can essentially explain why spaghetti and dry pasta almost always break into more than two fragments. You can try in your kitchen to break a slender spaghetti by holding its two ends and gradually form an arc and bend beyond its curvature limit. When a spaghetti rod snaps, it will generally break three to pieces. This phenomenon is very interesting and once puzzled the famous physicist Richard Feynman for quite a while, and it was recently studied by two scientists that the brittle fragmentation process is virtually governed by this Cauchy-Navier equation. They found that the sudden relaxation of the curvature by first breaking will lead to a burst of flexural waves along the spaghetti rod, and these waves locally increase the curvature in the rod, resulting in more fragmented pieces. ()2 For the elastostatic problem, 7Jfi : : : : 0, we have

82 {)2u ax2 [EI {)x2] = q(x), 198

(13.109)

Elasticity


where q(x) = f(x) is the applied force per unit length. This equation will be used to determine the deflection of a bean1. 0 Example 19.1: Let us now use the Euler-Bernoulli theory to calculate the shape of a heavy cantilever with a uniform cross section under its own gravity. For a beam under its own gravity, the force is constant q(x) = pgA per unit length where g is the acceleration due to the Earth's gravity. If the length of the cantilever is L, then the total weight is W = pgAL, thus q = ~. Therefore, we have d4 u W EI dx4 = q = T

where we have use EI =canst. Integrating it twice, we have

u

EI dx 2 =

q

2x

2

+Ax+B.

At the free end, the beam cannot bending moment and/or shear, which implies that M(L) = 0 and V(L) = q*L+A = 0. These conditions lead to A = -qL and B = qL 2 /2. Integrating the above equation again, we have du q EI- = -x3 dx 6

-

qL - x2 2

qL 2

+ -x+C 2 ·

As the beam is fixed at x = 0, we have u = 0 and :: = 0 at x = 0. Thus we have C = 0 from u = 0. Integrating once again, we have

As Ux = 0 at x curve becomes

= 0, we have D = 0.

Therefore, the final deflection

The end deflection 8 = u(x = L) is

0

199

Chapter 14

Mathematic al Models What we have discussed so far in of differential equations is very limited. In engineering and natural sciences, there are so many different kinds of phenomena that require both mathematical modelling and computer simulations as well as experimental studies. In most cases, the classical models (using the heat conduction equation and the wave equation and others) are simply not adequate to described these phenomena. Therefore, we have to broaden our view to study other kinds of partial differential equations. In fact, mathematical modelling per se is a subject with vast literature, and subsequently we have to focus on the relevant equations and to introduce them very briefly.

14.1

Classic Models

Before we introduce more complicated partial differential equations, let us first remind us the three types of classic partial differential equations because they are widely used and occur in a vast range of applications. To a certain extent, almost all books or studies on the partial differential equations will have to deal with these three types of basic partial differential equations.

201

14.1 Classic Models

14.1.1

l.!fathematical l.!fodels

Laplace's and Poisson's Equation

In heat transfer problems, the steady state of heat conduction with a source is governed by the Poison equation

(x,y) En,

(14.1)

or Uxx

+ Uyy

=

q(x, y, t),

(14.2)

for two independent variables x and y. Here k is thermal diffusivity and I (x, y, t) is the heat source. If there is no heat source (q = 0), this becomes the Laplace equation. The solution or a function is said to be harmonic if it satisfies Laplace's equation. In order to determine the temperature u completely, the appropriate boundary conditions are needed. A simple boundary condition is to specify the temperature u = uo on the boundary an. This type of problem is the Dirichlet problem. On the other hand, if the temperature is not known, but the gradient 8uf 8n is known on the boundary where n is the outwardpointing unit normal, and this forms the Neumann problem. FUrthermore, some problems may have a mixed type of boundary conditions in the combination of au + (3 g~ = 1', which naturally occur as a radiation or cooling boundary condition.

14.1.2

Parabolic Equation

Time-dependent problems, such as diffusion and transient heat conduction, are governed by the parabolic equation (14.3)

Written in then-dimensional case Xt = x, x2 = y, X3 = z, ... , it can be extended to the reaction-diffusion equation (14.4)

where

I is the reaction rate. 202

1\t/athematical 1\t/odels

14.1.3

14.2 Other PDEs

Wave Equation

The vibration of strings and travelling sound waves are governed by the hyperbolic wave equation. The 1-D wave equation in its simplest form is -

2

Utt- C Uxx'

(14.5)

where cis the velocity of the wave. Using a transformation of the pair of independent variables ~ =

fort> 0 and -oo as

x+ct,

'f/ = x- ct,

(14.6)

< x < oo, the wave equation can be written u~ 11 =

0.

(14.7)

Integrating twice and substituting back in of x and t, we have

+ ct) + g(x- ct), arbitrary functions of x + ct

u(x, t)

=

f(x

(14.8)

where f and g are and x - ct, respectively. We can see that there are two directions that the wave can travel. One wave moves to the right and one travels to the left at a constant speed c.

14.2

Other PDEs

We have shown examples of the three major equations of secondorder linear partial differential equations. There are other equations that occur frequently in mathematical physics, engineering and computational sciences. We will give a brief description of some of these equations.

14.2.1

Elastic Wave Equation

As we have seen in the linear elasticity, the wave in an elastic isotropic homogeneous solid is governed by the following 203

14.2 Other PDEs

!Ylathematical A1odels

equation in of displacement u, p

82 u 8t2

J.L\7 2 u +(A+ J.L)\7(\7 · u)

=

+ b,

(14.9)

where p is density. A and J.l are Lame constants. b is body force. Such an equation can have two types of wave: transverse wave (S-wave) and longitudinal or dilatational wave (P-wave).

14.2.2

Maxwell's Equations

The scientific essence of modern wireless communications is governed by the :Maxwell's equations for electromagnetic waves (14.10)

8B

VxE=- at'

(14.11)

\7 · B = 0,

(14.12)

\7 x B = J.LoJ

1 8E

+ c2 7K'

(14.13)

where Pe is the charge density, E is the electric field, B is the magnetic field and J is the current density. eo and J.lo are the permittivity and permeability of the free space, respectively. Finally, c is the speed of light. The first equation is the electrostatic equation, the second one is the Faraday's law and the last equation is the Ampere-Iviaxwell's law.

14.2.3

Reaction-Diffusion Equation

The reaction-diffusion equation is an extension of heat conduction with a source Ut

=

Dtl.u + f(x, y, z, u),

(14.14)

where D is the diffusion coefficient and f is the reaction rate. One example is the combustion equation Ut

= D Uxx

+ Que ->..ju ,

204

(14.15)


14.2 Other PDEs

where D, Q and A are constants. The other example is the Fitz-Hugh-N agumo equations for transport of a nerve signal =

Ut

Uxx

+ u( 1 -

U) ( U -

Q) -

V,

Vt = f(U- /V),

(14.16) (14.17)

where /' > 0, 0 < a < 1 and E << 1. These equations are sometimes also called the equations in excitable media.

14.2.4

Fokker-Plank Equation

The time evolution of the probability density function of position and velocity of a particle system is described by the FokkerPlank equation 8p

n

aD!ll

8t

i=l

axi

n

n

-=[-L:-1 +L:L: i=l i=l

82 n!~l lJ

]p,

(14.18)

8xi8xi

which can be generalized as N

8p at

="" "" (- 1)k_!__ _!__{D(kJ L- . L-. 8Xii ... 8Xik

ZI , ••••

k=lt}, ....lk

. }

1kp '

(14.19)

.

where D[k] (x 1 , x 2 , ... , xn) are tensors. In the special case of N = 2, nl 11(xt' ... , Xn) is the drift vector, while Dl21 is the diffusion tensor.

14.2.5

Black-Scholes Equation

In the option pricing model, the value of an option u( 8, t) at time t is governed by the well-known Nobel-winning BlackScholes equation

au

au

{)t = ru - r 8 a8

-

1

2a

2 28

2

u

8 882 '

(14.20)

where 8 is the current stock price of the underlying stock and r is the risk-free interest rate. t is the time until option expiration. a is the stock volatility or the standard deviation of 205

14.2 Other PDEs

!Ylathematical A1odels

stock returns. The Black-Scholes equation is very similar to the extended version of the diffusion equation. The interesting feature for a call option with an exercise price E and expiry time T, the change of variables T =

(T- t)a 2 /2,

V = uer(T-t)

and x

2r

= ln(S/E) + (2

(J

-1)r,

(14.21)

(14.22)

can transform it into a standard diffusion equation (14.23)

Then, we can use the standard methods such as integral transforms to solve this equation.

14.2.6

Schrodinger Equation

The famous Schrodinger equation is the revolutionary equation in quantum mechanics and molecular dynamics ·i:_

aw -

tt2

t"'72 ,y,

ZH----v

8t

2m

':l'+

u·T· ':l'

'

(14.24)

where tt is a Planck constant. This equation can be obtained from the energy formE=::+ U (where pis the momentum) using differential operator mapping E ---+ itt£ and p ---+ -itt V. W is the probability wave function and U = V (r) - E is the energy potential. This is a complex partial differential equation.

14.2. 7

Navier-Stokes Equations

The Navier-Stokes equations for an incompressible flow can be written as V·u=O, (14.25) Ut

+ (u · V)u =

1 2 ReV u - Vp,

206

(14.26)


14.2 Other PDEs

where Re = pUL/J.l is the Reynolds number. U is the typical velocity and L is the length scale. p and J.l are the density of the fluid and its viscosity, respectively. In computational fluid dynamics, most simulations are mainly related to these equations. In the limit of Re << 1, we have the Stokes flow (slow flow) governed by (14.27) In another limit Re

>> 1, we have the inviscous flow

V'. u = 0,

Ut

+ (u · V')u =

-V'p.

(14.28)

We can see that the equations are still nonlinear even in this simplified case.

14.2.8

Sine-Gordon Equation

Another important equation that appears in a wide range of applications in physics and many other fields is the Sine-Gordon equation Utt- Uxx + sin(u) = 0, (14.29) which can generally be written as Utt

=

Uxx

+ asin(wu).

(14.30)

This is a nonlinear hyperbolic equation. Almost all these equations are very difficult for mathematical analysis, and most of them do not have closed-form solutions under most common boundary conditions. Therefore, the numerical methods are a good alternative in this case. In the next few chapters, we will introduce various numerical methods in details.

207

Chapter 15

Finite Difference Method The finite difference method is one of the n1ost popular methods that are used commonly in computer sin1ulations. It has the advantage of simplicity and clarity, especially in 1-D configuration and other cases with regular geon1etry. The finite difference method essentially transforms an ordinary differential equation into a set of algebraic equations by replacing the continuous derivatives with finite difference approximations on a grid of mesh or node points that spans the domain of interest based on the Taylor expansions. In general, the boundary conditions and boundary nodes need special treatment.

15.1

Integration of ODEs

The second-order or higher order ordinary differential equations can be written as a first-order systen1 of ODEs. Since the technique for solving a system is essentially the same as that for solving a single equation

dy dx = f(x,y), 209

(15.1)

15.1 Integration of ODEs

Finite Difference Metbod

then we shall focus on the first-order equation in the rest of this section. In principle, the solution can be obtained by direct integration,

y(x) =Yo+ {x f(x, y(x))dx,

lxo

(15.2)

but in practice it is usually impossible to do the integration analytically as it requires the solution of y( x) to evaluate the right-hand side. Thus, some approximations shall be utilized. Numerical integration is the most common technique to obtain approximate solutions. There are various integration schemes with different orders of accuracy and convergent rates. These schemes include the simple Euler scheme, Runge-Kutta method, Relaxation method, and many others.

15.1.1

Euler Scheme

Using the notations h = ~x = Xn+l - Xn, Yn = y(xn), Xn = x 0 + n~x (n = 0, 1, 2, ... , N), and ' = djdx for convenience, then the explicit Euler scheme can simply be written as

Yn+l = Yn

+

Xn+l

1

f(x, y)dx ~ Yn

+ hf(xn, Yn)·

(15.3)

Xn

This is a forward difference method as it is equivalent to the approximation of the first derivative 1

Yn =

Yn+l- Yn ~X

(15.4)

The order of accuracy can be estimated using the Taylor expansion

h2 "I n + ··· Yn+l = Yn + hY'I n + TY ~ Yn

+ hf(xn, Yn) + O(h2 ).

(15.5)

Thus, the Euler method is first order accurate. For any numerical algorithms, the algorithm must be stable in order to reach convergent solutions. Thus, stability is an important issue in numerical analysis. Defining 8y as the 210

Finite Difference Method


discrepancy between the actual numerical solution and the true solution of the Euler finite difference equation, we have (15.6) In order to avoid the discrepancy to grow, it requires the following stability condition 1~1 ::; 1. The stability restricts the size of interval h, which is usually small. One alternative that can use larger h is the implicit Euler scheme, and this scheme approximates the derivative by a backward difference y~ = (Yn- Yn-d/h and the right-hand side of equation (15.2) is evaluated at the new Yn+l location. Now the scheme can be written as (15.7) The stability condition becomes (15.8)

U ::;

which is always stable if f'(y) = 0. This means that any step size is acceptable. However, the step size cannot be too large as the accuracy reduces as the step size increases. Another practical issue is that, for most problems such as nonlinear ODEs, the evaluation of y' and f'(y) requires the value of Yn+l which is unknown. Thus, an iteration procedure is needed to march to a new value Yn+l, and the iteration starts with a guess value which is usually taken to be zero for most cases. The implicit scheme generally gives better stability. 0 Example 15.1:

To solve the equation dy

dx = f(y) = e-y- y,

we

use

the explicit Euler scheme, and we have

211



Suppose the discrepancy between real solution y~ and the numerical Yn is 8yn so that y~ = Yn + 8yn, then the real solution satisfies y~+l

= y~ + hf(y~).

Since f(y~) = f(Yn)

Yn+l

+ :~8yn, the above equation becomes + 8yn+l = Yn + 8yn + h[f(Yn) + J'(Yn)8yn]·

Together with the Euler scheme, we have

8yn+l

= 8yn + J'8Yn·

Suppose that 8yn ex en, then we have en+l =en+ h/'en'

e= 1 + hf'.

or

In order for the scheme to be stable (or en- 0), it requires that or

- 1 ~ 1 + hf'

=1-

h(e-Yn

+ 1) ~

1.

The stability condition becomes 2 O <-- e-y, ~ +1

0

15.1.2

Leap-Frog Method

The Leap-frog scheme is the central difference 1

Yn+1 - Yn-1

2Ax

Yn =

'

(15.9)

which leads to Yn+1 = Yn-1

+ 2hf(xn, Yn)·

(15.10)

The central difference method is second order accurate. In a similar way as equation (15.6), the leap frog method becomes 8yn+1 = 8Yn-1

+ 2hf'(y)8yn,

(15.11)

or 2

8yn+1 = ~ 8Yn-1'

(15.12)

where ~ = 1 + 2hf'(y)~. This scheme is stable only if 1~1 ::; 1, and a special case is 1~1 = 1 when f'(y) is purely in1aginary. Therefore, the central scheme is not necessarily a better scheme than the forward scheme. 2

212


15.1.3

15.2 Hyperbolic Equations

Runge-Kutta Method

We have so far seen that stability of the Euler method and the central difference method is limited. The Runge-Kutta method uses a trial step to the midpoint of the interval by central difference and combines with the forward difference at two steps Yn+l/2

= Yn + .;. f(xn, Yn),

(15.13) (15.14)

This scheme is second order accurate with higher stability compared with previous shnple schemes. One can view this scheme as a predictor-corrector method. In fact, we can use multisteps to devise higher order methods if the right combinations are used to eliminate the error order by order. The popular classical Runge-Kutta method can be written as

a= hf(xn, Yn), b = hf(xn + h/2, Yn C

=

+ a/2), hf(xn + h, Yn + b/2),

d = hf(xn + h,yn +c),

a+ 2(b +c)+ d

(15.15) ' 6 which is fourth order accurate. Generally speaking, the higherorder scheme is better than the lower scheme, but not always.

Yn+l = Yn +

15.2

Hyperbolic Equations

Nun1erical solutions of partial differential equations are more complicated than that of ODEs because it involves time and space variables and the geometry of the domain of interest. Usually, boundary conditions are more complex. In addition, nonlinear problems are very common in engineering applications. Now we start with the simplest first order equations and then move onto more complicated cases. 213


15.2.1

Finite Difference Metbod

First-Order Hyperbolic Equation

For simplicity, we start with the one-dimensional scalar equation of hyperbolic type, (15.16)

where c is a constant or the velocity of advection. By using the forward Euler scheme for time and centered-spaced scheme, we have u~+ 1 - u"!-

~t

J

+ c[

J

un

un

i+1; j-1] = 0,

(15.17)

where t = n~t, n = 0, 1, 2, ... , x = x 0 + jh,j = 0, 1, 2, ... , and h = ~x. In order to see how this method behaves numerically, we use the von Neumann stability analysis. Assun1ing the independent solutions or eigenmodes (also called Fourier modes) in spatial coordinate x in the form of uj = ~neikhi, and substituting into equation (15.17), we have ~n+ 1 ikhj - ~n ikhj

~

e

~e

= ~n c~t eikh(j+1) ~

h

ihk(j-1)

2

e

Dividing both sides of the above equation by using sinx = (eix- e-ix)/2i, we get

~=

. ~n

1- ic~t sin(kh).

(15.18)

exp( ikhj) and

(15.19)

The stability criteria 1~1 ~ 1 require . 2 kh 0 ( c~t h )2 Sill ~ .

(15.20)

However, this inequality is impossible to satisfy and this scheme is thus unconditionally unstable. To avoid the difficulty of instability, we can use other schemes such as the upwind scheme and Lax scheme. For the upwind scheme, the equation becomes u~+ 1 - u"!1

1

~t

+ c[

u"!- - u"!1

214

h

1

-

1

]

= 0,

(15.21)



whose stability condition is

1~1 =II- c~t [1- cos(kh) + isin(kh)]l ~

1,

(15.22)

which is equivalent to c~t

0<--,;:~l.

(15.23)

This the well-known Courant-Friedrichs-Lewy stability condition, often referred to as the Courant stability condition. Thus, the upwind scheme is conditionally stable.

15.2.2

Second-Order Wave Equation

Higher order equations such as the second-order wave equation can be written as a system of hyperbolic equations and then be solved using numerical integration. They can also be solved by direct discretization using a finite difference scheme. The wave equation (15.24)

consists of second derivatives. If we approximate the first derivatives at each time step n using

,

u'- uft-1- uf i-

ui-1

~X

=

uf- uf-1 ~x

(15.25)

then we can use the following approximation for the second derivative u~- u~-1 U·" =......;....-....;........;. t

~X

uf+1- 2uf + uf-1 (~x)2

(15.26)

This is in fact a central difference scheme of second order accuracy. If we use the similar scheme for time-stepping, then we get a central difference scheme in both time and space. 215

15.3 Parabolic Equation


Thus, the numerical scheme for this equation becomes

ui+1- 2uf + ui-1 2 uf+1- 2uf + uf-1 (~t)2 =c (~x)2

(15.27)

This is a two-level scheme with a second order accuracy. The idea of solving this difference equation is to express (or to solve) ui+ 1 at time step t = n+ 1 in of the known values or data uf and ui- 1 at two previous time steps t = n and t = n - 1.

15.3

Parabolic Equation

For the parabolic equation such as the diffusion or heat conduction equation 8u = ~(D8u)

8t

8x

(15.28)

8x '

a simple Euler method for the time derivative and centered second-order approximations for space derivatives lead to (15.29) The stability requirement ~ ::; 1 leads to the constraint on the time step (see the example), (15.30) This scheme is shown in Figure 15.1 and it is conditionally stable. 0 Example 15.2: From equation (15.29), we can apply the von Neumann stability analysis by assuming uj = f.neikhj, we have

Dividing both sides by

f.neikhj,

Dt::..t

we have

f. = 1 + """h2"[e

1·kh

216

+ e-z·kh -

2].


15.3 Parabolic Equation

t

t=n+1

Q I I I

0

j-1

6

Ot=n

j

j+1

x·J

Figure 15.1: Central difference in space and explicit Euler timestepping. Using cosx = (eix + e-ix)/2 and sin2 (x/2) = (1- cosx)/2, we obtain 4DD.t .

2

kh

e=1-~sm(2).

Since sin( X)

~

1' thus e~ 1 requires 4DD.t -1 -< 1 - h2 - -< 1. 0

or 0

A typical feature of a solution to the diffusive system is that the profile is gradually smoothed out as time increases. The time-stepping scheme we used limits the step size of time as larger time steps will make the scheme unstable. There are 1nany ways to improve this, and one of most widely used schemes is the implicit scheme. To avoid the difficulty caused by very small timesteps, we now use an implicit scheme for time derivative differencing, and thus we have DL).t

u~+l - uf!- = --(u~+l J

J

h2

J+l

1 + 2u~+l + u~+ ) J J-1 .

(15.31)

Applying the stability analysis, we have

1

~ = 1 + fiT""sm 4D.6.t · 2 kh' 2 217

(15.32)

15.4 Elliptical Equation

Finite Difference !v!etbod

whose norm is always less than unity (1~1 ~ 1). This means the implicit scheme is unconditionally stable for any size of time steps. That is why in1plicit methods are more desired in simulations. However. there is one disadvantage of this method, which requires more programming skills because the inverse of a large matrix is usually needed in implicit schen1es.

15.4

Elliptical Equation

In the parabolic equation, if the time derivative is zero or u does not change with time Ut = 0, then it becomes a steadystate problem that is governed by the elliptic equation. For the steady state heat conduction problem, we generally have the Poisson problem,

V · [K(u, x, y, t)Vu]

= f,

(15.33)

If K is a constant, this becomes

f

q=-. K

(15.34)

There are many methods available to solve this problems such as the boundary integral method, the relaxation method, and the multigrid method. Two major ones are the long-time approximation of the transient parabolic diffusion equations, the other includes the iteration method. The long time approximation method is essentially based on fact that the parabolic equation (15.35)

evolves with a typical scale of ../'Kl. If VKJ, >> 1, the system is approaching its steady state. Assuming t ---+ oo and K >> 1, we then have 2 f 1 V U = - - -Ut ---+ 0. (15.36) K

218

K


15.4 Elliptical Equation

The long-time approximation is based on the fact that the parabolic equation in the case of K = const degenerates into the above steady-state equation (15.33) because Ut -----+ 0 as t -----+ oo. This approximation becomes better if K >> 1. Thus, the usual numerical methods for solving parabolic equations are valid. However, other methods may obtain the results more quickly. The iteration method uses the second-order scheme for space derivatives, and equation (15.34) in the 2-D case becomes Ui+l,j - 2Ui,j

(Ax)2

If we use Ax comes

+ Ui-l,j

+

Ui,j+l - 2ui,j

+ Ui,j-1

(Ay) 2

= q.

(15.37)

= Ay = h, then the above equation simply be(15.38)

which can be written as

Au=b.

(15.39)

In principle, one can solve this equation using Gauss elimination; however, this becomes impractical as the matrix becomes large such as 1000 x 1000. The Gauss-Seidel iteration provides a more efficient way to solve this equation by splitting A as

A=L+D+U,

(15.40)

where L, D~ U are the lower triangle, diagonal and upper triangle matrices of A~ respectively. The iteration is updated in the following way: (15.41) This procedure stops until a prescribed error or precision is reached.

219

Chapter 16

Finite Volume Method 16.1

Introduction

The finite difference method discussed in the previous chapter approximates the ordinary differential equations and partial differential equations using Taylor series, resulting in a system of algebraic equations. The finite volume method resembles the finite difference method in certain ways but the starting point is the integral formulation of the problem. It uses the integral form of the partial differential equations in of conservation laws, then approximates the surface and boundary integrals in the control volumes. This becomes convenient for problems involving flow or flux boundaries. For a hyperbolic equation that is valid in the domain n with boundary an' (16.1)

or written in of flux function F have

8u

at + v · F =

=

F(u)

=

-K.'\Ju, we

(16.2)

q.

The integral forn1 of this equation becomes

In ';;: + In dQ

V' . F

221

=

In

qdf!.

(16.3)

16.2 Elliptic Equations

Finite Volume !v!etbod

If the integral form is decomposed into many small control volumes, or finite volumes, n = U~ 1 ni and ni ni = 0. By defining the control volume cell average or mean value

n

(16.4) where \ti = IOil is the volume of the small control volume the above equation can be written as

ni,

(16.5) By using the divergence theorem

fv \7 · F = £F · ndA,

(16.6)

we have

(16. 7) where dS = ndA is the surface element and n is the outward pointing unit vector on the surface ri enclosing the finite volume ni. The integration can be approximated using various nun1erical integration schemes. In the sin1plest 1-D case with h = ~x, the integration 11(i+l/2)h

ui = h

udx,

(16.8)

(i-1/2)h

is a vertex-centred finite volume scheme. In the following sections, we will discuss the three major types of partial differential equations (elliptic, parabolic and hyperbolic) and their finite volume discretizations.

16.2

Elliptic Equations

Laplace's equation is one of the most studied elliptic equations \7 2 u(x, y) = 0, 222

(x, y) E 0,

(16.9)

Finite Volume l'vlethod

16.3 Parabolic Equations

its integral form becomes

1 n

'\7 2 udO. =

1-8n8u · r

dS

= 0.

(16.10)

For the simple regular grid points (i~x,j~y), the control volume in this case is a cell centred at (i~x, j ~y) with a size of ~x (along x-axis) and ~y (along y-axis), and the boundary integral on any cell consists of four parts integrated on each of the four sides. By using the simple approximation ~ with g~ = (ui+l,j- ui.j)/~x and = (ui,j+l- ui,j)j~y, we have

t

1

f2·. t,]

8u - dO.= 8n

~y A

uX

(ui+l.i + Ui-t,j-

2ui,j)

~X

+ ~y (ui,j+l + ui,j-1 Dividing both sides with obtain (ui+l,j

~x~y,

2ui,j) =

and letting

~x

+ Ui.j+l + Ui-l,j + Ui,j-t)- 4Ui,j

(16.11)

0.

=

= 0,

~y

= h, we (16.12)

which resembles finite difference methods in many ways. In fact, this is exactly the Laplace operator for a 5-point differencing scheme.

16.3

Parabolic Equations

For the case of heat conduction (16.13)

we have its integral form [ [

{)u

lt ln ( 8t -

82 u

k {)x2 - q)dxdt = 0. 223

(16.14)


Finite Volume !v!etbod

If we use the control volume from (i - 1/2)h to (i + 1/2)h where h = D.x, and with time from step n to n + 1, we have

{(n+1)At 1(i+1/2)h {Ju fJ2u lnAt (i-1/2)h ( {)t - k 8x 2

q)dxdt

-

= O.

(16.15)

By using the mid-point approximation

l

'1/J(x)dx = '1/J[(a; b)](b- a),

(16.16)

and the DuFort-Franke! scheme where we first approximate the gradient [)2u uf+1 - 2uf + uf-1 (16.17) {)x2 = h2 then replace -2uf with -(uj+ 1 + uj- 1), we have u~+1- u~-1 t

t

=

2D.t

(16.18)

where we have used the central scheme for time as well. The finite volume scheme is more versatile in dealing with irregular geometry and more natural in applying boundary conditions. Following the stability analysis, we get 1~1 < 1 is always true and thus the Dufort-Franke! scheme is unconditionally stable for all .Dot and D.x.

16.4

Hyperbolic Equations

For the hyperbolic equation of the conservation law in the onedimensional case {)u {Jt

8w(u) _

+ ---ax- -

0

'

we have its integral form in the fixed domain

1

Xb -{)ud x ={)- 1Xb udx

Xa

{Jt

{Jt

224

Xa

(16.19)

Finite Volume l'vlethod

=


-{w[u(xb)]- w[u(xa)]} = 0.

(16.20)

If we use the mid-point rule u* to approximate the integral, we have

(16.21) If we choose the control volume [(i -1/2)~x, (i + 1/2)~x] centred at the mesh point Xi = i~x = ih with the approximation ui ~ ui in each interval, and using the forward differencing scheme for the time derivative, we have

(16.22)

By further approximation of the flux \ll(xi+l/2) have the upward scheme

~

\ll(xi), we

(16.23)

which is conditionally stable as we know this in the finite difference method. For the simplest flux w(u) = cu, we have

u

n+l

n

= ui -

hc~t (uin

-

n ) ui-1 '

(16.24)

and its stability requires that 0

<1 < c~t h - .

225

(16.25)

Chapter 17

Finite Element Method In the finite difference method, we approximate the equations at a finite number of discrete points, and there are many limitations in finite difference methods. One of such disadvantages is that it is not straightforward to deal with irregular geometry. lviore versatile and efficient n1ethods are highly needed. In fact, the finite element method is one class of the most successful methods in engineering and have a wide range of applications. The basic aim of the finite element method is to formulate the numerical method in such a way that the partial differential equation will be transformed into algebraic equations in of matrices. For time-dependent problems involving partial differential equations, the equations can be converted into an ordinary differential equation, which will in turn be discretized and converted into algebraic equations by time-stepping or some iteration techniques. For example, a linear elastic problem can be formulated in such a way that it is equivalent to the equation of the following type Ku=f, (17.1) where K is the stiffness matrix, and f is a vector corresponding to nodal forces and some contribution from boundary conditions. u is the unknown vector to be solved and it corresponds to the nodal degree of freedon1 such as the displacement. 227

17.1 Concept of Elements

Finite Element Metbod

17.1

Concept of Elements

17.1.1

Simple Spring Systems

The basic idea of the finite element analysis is to divide a model (such as a bridge and an airplane) into many pieces or elements with discrete nodes. These elements form an approximate system to the whole structures in the domain of interest, so that the physical quantities such as displacements can be evaluated at these discrete nodes. Other quantities such as stresses, strains can then be evaluated at certain points (usually Gaussian integration points) inside elements. The simplest elements are the element with two nodes in 1-D, the triangular element with three nodes in 2-D, and tetrahedral elements with four nodes in 3-D. In order to show the basic concept, we now focus on the simplest 1-D spring element with two nodes (see Figure 17.1). The spring has a stiffness constant k (Njm) with two nodes i and j. At nodes i and j, the displacements (in metres) are Ui and Uj, respectively. /i and /j are nodal forces.

k

Figure 17.1: Finite element concept. From Hooke's law, we know the displacement is related to f, or

f

= k(~u).

~u = Uj - Ui

(17.2)

At node i, we have (17.3) and at node j, we get (17.4) 228

Finite Element 1\lethod


These two equations can be combined into a matrix equation or

Ku = f.

(17.5)

Here K is the stiffness matrix, u and f are the displacement vector and force vector, respectively. This is the basic spring element, and let us see how it works in a spring system such as shown in Figure 17.2 where three different springs are connected in series.

Figure 17.2: A simple spring system. For a simple spring system shown in Figure 17.2, we now try to determine the displacements of Ui ( i = 1, 2, 3, 4). In order to do so, we have to assemble the whole system into a single equation in of global stiffness matrix K and forcing f. As these three elements are connected in series, the assembly of the system can be done element by element. For element E1, its contribution to the overall global matrix is

k1 ( -k1

-k1 ) ( k1

Ul )

U2

= (

fi ) ' /2

(17.6)

which is equivalent to (17.7) where

(17.8)

229


and

!'f

1


(ft, /2, 0, 0). Similarly, for element E2, we have

=

(

k2

-k2 ) ( U2 )

-k2

k2

U3

-/2 ) '

= (

(17.9)

/3

or

KFU

0 k2 -k2 0

!}

0 -k2 k2 0

(17.10)

where we have used the balance at node 2. For element E3, we have (17.11)

or (17.12)

where /4 = /. has been used. We can now add the three sets of equations together to obtain a single equation (

-t kl

-k2 kt + k2 -k2 0

-k2 0 k2 + k3 -k3

0 ) ( U2 UJ -k3 U3 k3 U4

)

(

=

-/2+/2 !J ) - j3 + /3 ' f*

or Ku=f,

(17.13)

where K

(

=

T kl

= Kt +K2 +K3

-kl kt + k2 -k2 0

0 -k2 k2 +k3 -k3 230

_t} k3

(17.14)



and (17.15)

In general, the matrix K is singular or its rank is less than the total number of degrees of freedom, which is four in this case. This means that the equation has no unique solution. Thus, we need the boundary conditions to ensure a unique solution. In this spring system, if no boundary condition is applied at any nodes, then the applied force at the node 4 will make the spring system fly to the right. If we add a constraint by fixing the left node 1, then the system can stretch, and a unique configuration is formed. In our case where there are no applied forces at nodes 2 and 3, we have (17.16) 0 Example 11.1: For k 1 = 100 Njm, k2 = 200 N/m, and k3 = 50N/m, and J. = 20 N, the boundary at node 1 is fixed (u 1 = 0). Then, the stiffness matrix is

-100 300 -200 0

0 -200 250 -50

_t )' 50

and the force column vector fT = (0, 0, 0, 20).

The rank of K is 3, therefore, we need at least one boundary condition. By applying u 1 = 0, we now have only three unknown displacements u 2 , u 3 , u 4 • Since u 1 = 0 is already known, the first equation for Ut becomes redundant and we can now delete it so that the reduced stiffness matrix A is a 3 x 3 matrix. Therefore, we have 300

A=

(

-~00

-200 250 -50

and the reduced forcing vector is gT = (0, 0, 20).

231

~

50

),


Finite Element Method

The solution is u =A -lg =

(

0.2) 0.3 . 0.7

Therefore, the displacements are u 2 = 0.2 m, u 3 = 0.3 m, and u 4 = 0.7 m.

Theoretically speaking, the force should be 20N eve1ywhere in the spring systems since the mass of the springs is negligible. Let us calculate the force at nodes 2 and 3 to see if this is the case. At the node 2, the extension in element E 1 is ~u = u 2 - u 1 = 0.2 m, thus the force at node 2 is

f2 =

kt~U

= 100

X

0.2 = 20N.

Similarly, at node 3 of element E2, we have

which is the same at node 3 of element E 3

So the force is 20 N everywhere.

17.1.2

0

Bar and Beam Elements

The spring system we discussed earlier is limited in many ways as a spring does not have any mass and its cross section is not explicitly included. A more complicated but realistic elen1ent is the bar element as shown in Figure 17.3, which is a uniform rod with a cross section area A, Young's elastic modulus E, and a length L. A bar element can only tension and compression, it cannot bending. For this reason, it is also called a truss element. The displacements at nodes i and j are Ui and Uj, respectively. The forces at the corresponding nodes are fi and fi. Now we have to derive its stiffness matrix. Assuming the bar is linearly elastic, the stress a is thus related to strain f via

232

Finite Element .Atfethod


a= EE. Since f = (uj- ui)/L and a= f /A where F is the force in the bar element, we have

f =

EA

y(~u)

=

k(~u),

(17.17)

where ~u = Uj - Ui is the extension or elongation of the bar element. Now the equivalent spring stiffness constant is

k

= EA

(17.18)

L"

Therefore, the stiffness matrix K for this bar becomes K = (

k -k ) = EA ( 1 -k k L -1

-1 ) . 1

A,E,L

(17.19)

j

u·J

u·t

Figure 17.3: Bar element. We have up to now only discussed 1-D systems where all displacements Ui or Uj are along the bar direction, and each node has only one displacement (one degree of freedom). We now extend to study 2-D syste1ns. In 2-D, each node i has two displacements ui (along the bar direction) and vi (perpendicular to the bar direction). Thus, each node has two degrees of freedom. If we rotate the bar elen1ent by an angle (} as shown in Figure 17.4, we cannot use the standard addition to assemble the system. A transformation is needed between the global coordinates (x, y) to the local coordinates (x', y'). From the geometrical consideration, the global displacements ui and vi 233



at node i are related to the local displacement

u~

and (usually)

v~ = 0.

(

~~ )

= (

~::o

:: )(~; ).

(17.20)

y

Figure 17.4: 2-D transformation of coordinates. Using the similar transformation for Uj and transformation for the two-node bar ele1nent

I

U=

(~) -ro (

~o

sinO 0 cosO 0 0 cosO -sinO 0

Vj,

we get the

si~O) (:}

COS

0

Vj

which can be written as

u'

= Ru,

(17.21)

where

si~O}

cosO sinO 0 -sinO cosO 0 R== ( 0 cosO 0 0 -sinO cosO 0

(17.22)

The same applies to transform the force, f' = Rf,

234

(17.23)


17.2 Finite Element Formulation

and the stiffness matrix in local coordinates is

K'u'

=

f'.

(17.24)

As the calculation is n1ainly based on the global coordinates, and the assembly should be done by transforming the local systems to the global coordinates, by combining the above two equations, we have K'Ru=Rf, (17.25) or

R- 1 K'Ru = Ku = f,

(17.26)

which is equivalent to a global stiffness matrix (17.27)

The stiffness matrix K is a 4 x 4 matrix in 2-D. Bar elements can only elongate or shrink, they do not bending or deflection. For bending, we need the beam elements which include a rotation around the end nodes ()i and ()i. In this case, each node has three degrees of freedom (Ui, Vi, Oi), and the stiffness matrix is therefore a 6 x 6 matrix in 2-D. For more complicated elements, it is necessary to use a formal approach in of shape functions and weak formulations.

17.2

Finite Element Formulation

17.2.1

Weak Formulation

l\llany problems are modelled in of partial differential equations, which can generally be written as .C(u)

= 0,

X

E !l,

(17.28)

where .C is a differential operator, often linear. This problem is usually completed with the essential boundary condition £(u) = (u- u) = 0 for X E {)!lE and natural boundary conditions B(u) = 0 for X E anN. Assuming that the solution can 235


Finite Element !v!etbod

be approximated by uh over a finite element mesh with an averaged element size or mean distance h between two adjacent nodes, the above equation can be approximated as (17.29)

Multiplying both sides of the equation by a test function or a proper weighting function, integrating over the domain and using associated boundary conditions, we can write the general weak formulation of Zienkiewicz-type as

fn C(

uh)widn

+ f B(uh)widr + f &(uh)widrE ~ o, lanN lanE

(17.30)

where (i = 1, 2, ... , M). If we can approximate the solution uh by the expansion M

uh(u, t)

=L

M

ui(t)Ni(x)

i=l

=L

UjlVj,

(17.31)

j=l

it requires that Ni = 0 on 8f2E so that Wi = 0 on {}O.E. Thus, only the natural boundary conditions are included since the essential boundary conditions are automatically satisfied. In addition, there is no much limitation on the choice of Wi and wi. If we choose wi = -wi so as to simplify the formulation, we have (17.32)

17 .2.2

Galerkin Method

There are many different ways to choose the test functions Wi and shape functions Ni. One of the most popular methods is the Galerkin method where the test functions are the same as the shape functions, or Wi = Ni. In this special case, the formulation simply becomes (17.33) 236



The discretization of this equation will usually lead to an algebraic matrix equation. On the other hand, if we use the Dirac delta function as the test functions Wi = 8(x - ~), the method is called the collocation method which uses the interesting properties of the Dirac function

fo

f(x)8(x- Xi)dn = f(xi)·

together with 8(x- Xi)

=

1 at x

(17.34)

= Xi and 8(x- Xi) = 0 at

xfXi·

17.2.3

Shape Functions

The n1ain aim of the finite elen1ent method is to find an approxinlate solution uh(x, t) for the exact solution u on some nodal points, M

uh(x, t) = L Ui(t)Ni(x)

(17.35)

i=l

where ui are unknown coefficients or the value of u at the discrete nodal point i. Functions Ni (i = 1, 2, ... , M) are linearly independent functions that vanish on the part of the essential boundary. At any node i, we have Ni = 1, and Ni = 0 at any other nodes, or M

LNi = 1,

Ni(Xj) = 8ij·

(17.36)

i=l

The functions Ni (x) are referred to as basis functions, trial functions or more often shape functions in the literature of finite element methods. For the simplest 1-D element with two nodes i and j, the linear shape functions can be written as (17.37) where L =

lxi- xil, which is shown in Figure 17.5. 237


Finite Element !v!etbod

X

j ~ = xflxi-

xil

Figure 17.5: The 1-D linear shape functions. Using the assumptions that ui(t) does not depend on space and lVi(x) does not depend on time, the derivatives of u can be approximated as M

8u

8uh """' 1 a~ fJx = ~ Ui(t)N (x), X

i=l

M .

OUh

"""'

vt

i=l

. u~~=~ui

N( x, )

(17.38)

3t.

where we have used the notations: 1 = dj dx and· = Higher order derivatives are then calculated in a similar way. The ultimate goal is to construct a method of computing ui such that the error uh - u is minimized. Generally speaking, the residual R varies with space and time, so we have (17.39) There are several methods to minimize R. Depending on the scheme of minimization and the choice of shape functions, various methods can be formulated. These include the weighted residual method, the method of least squares, the Galerkin method and others. 238


17.3 Elasticity

17.3

Elasticity

17.3.1

Plane Stress and Plane Strain

The stress tensor u and strain tensor e are not written as tensor forms but vector forms u = (axx, ayy, a:;::, axy, ay::, a::x)T, and € = (f.xx' f.yy, f.:;:;, "fxy, 'Yyz, 'Yzx)T. The strain tensor is usually defined as 1 oui 8u· cij = - ( - + _ J ), (17.40) 2 8xj 8xi where one applies the engineering shear strain f.xy = 2cxy· Hooke's elasticity can be expressed as (17.41)

where Dis a 6 x 6 symmetrical matrix as functions of Young's modulus E and Poisson's ratio v. Two special cases that are commonly found in many applications are the plane stress (a zz = 0, but f.z:; =/= 0) and plane strain (f.:::: = 0, but azz =/= 0). The commonly used formulation is the displacement-based formulation or u-based formulation. In the 2-D case, the displacen1ent u = (u, v)T and the strain f. and stress a are defined as (17.42)

Now the stress-strain relationship becomes a=

D(f.- f.o),

(17.43)

where D is a 3 x 3 matrix. The strains are given by

au

8v

ox

8y '

f.x=-, f.y=-

au

8v OX

l'xy=-+-,

{)y

(17.44)

where f.o is the initial strain due to temperature change or thermal loading. If there is no such change, then the initial strain can be taken to be zero in most applications. 239

17.3 Elasticity


The equilibrium of force in elasticity leads to

V' · u+ b = 0. where b = [fx stress, we have

(I7.45)

/y]T is the body force. In the case of plane

E ( I D=-v 2 I- v 0

vI 0

0 ) 0 . (I - v)/2

(I7.46)

In the case of plane strain, we have

~~~ i~v v 1-v

E

D

=

(I - 2v) (

l~v l~v

0 0 (1-2v) 2(1+v)

) .

(I7.47)

Clearly, for the I-D case plane stress when v = 0, /y = 0 and = Txy = 0, the equation of force balance simply becomes

uy

E 82u -I--2 8 2 + fx = 0,

-v

(I7.48)

x

where we have used the stress-strain relationship. This I-D equation is essentially the same as the I-D heat transfer equation u" + Q = 0 that will be discussed in detail later, so the solution technique for the I-D heat transfer shall equally apply. Therefore, we shall focus on the 2-D case in the rest of this section. Displacements (u, v) in a plane eletnent can be interpolated from nodal displacements. For example, using a triangular element (i,j, m) with three nodal points as shown in Figure I7.6 (xi, Yi), (xj, Yi), and (xm, Ym), we have

u

= (:) = (N;I,N;I,Nml]

240

=Nd,

(I7.49)


17.3 Elasticity y

X

Figure 17.6: Schematic triangular mesh and the layout of a triangular element. where I is a 2 x 2 unitary matrix, i.e., (17.50)

and N = ( Ni 0

0 Ni

N·J 0

(17.51)

By defining a differential operator

Ld=

fx0 -iu, 0 ) ( {) w i)y

(17.52)

'ljX

we can rewrite the above forn1ulation as (17.53)

where B =LdN.

(17.54)

Now the equation (17.45) becomes

where K =

Ku=f,

(17.55)

In

(17.56)

BTDBdV, 241

17.3 Elasticity


and (17.57)

where r is the surface traction (force per unit area).

17.3.2

Implementation

In the case of 2-D elastic problems, the simplest elen1ents are linear triangular elements, thus we have Bi = LdNi, or

B

= LdN = 21~

c;

0

b·J

0

bm

0

Ci

0

c·J

0

c·l

b·l

c·J

b·J

Cm

~). bm

(17.58)

where 1 Xi 1 ~=- 1 x·J 2 1 Xm

Yi Yi Ym

(17.59)

is the area of the triangular element. The linear elen1ent implies that the strain is constant in the element. The stiffness matrix K~j) (i,j = 1, 2, ... , 6) of each triangular element can be expressed as (17.60)

For a triangular element with three nodes (i,j, m), each node has two degrees of freedom (ui, Vi), then the stiffness matrix K(e) for each element is a 6 x 6 matrix. In general, if each element has r nodes, and each node has n degrees of freedom, then the local stiffness matrix is a rn x rn matrix. If the region has M nodes in total, then M n equations are needed for this problem. For the present case (n = 2), we need 2/v/ equations for plane stress and plane strain. In order to calculate the contribution of each element to the overall (global) equation, each node should be identified in some way, and most often in of the index matrix. The 242


17.3 Elasticity

nodal index matrix for three nodes (i, j, m) can be written as

IDnode =

(i, i)

(i,j)

(i, m) )

(j, i)

(j,j)

(j, m)

(m,j)

(m, m)

( (m, i)

(17.61)

.

The nodal index matrix identifies the related nodes in the global node-numbering system. However, the main assembly of the stiffness matrix is about the corresponding equations and the application of the boundary conditions, thus we need to transfer the nodal index matrix to the equation index matrix in of the global numbering of equations with two degrees of freedom for each node (e.g., equations 2i -1 and 2i for nodal i, equation 2j - 1 and 2j for node j, etc). Thus, for each entry in the stiffness matrix, say (i,j), in the nodal index matrix, we now have four entries, i.e., (

.. ) Z,J

~ ( (2i -1,2j -1) (2i -1,2j)) (2i, 2j - 1)

(2i, 2j)

'

etc

(17.62)

The equation index matrix now has 6 x 6 entries, and each entry is a pair such as (2i- 1, 2j- 1), ... , (2m, 2m), etc. By writing it as two index matrices (ID = JDT), we now have

IDequ = JD~u,

JDequ =

2i -1 2i -1 2i -1 2i -1 2i -1 2i -1

2i 2i 2i 2i 2i 2i

2j 2j 2j 2j 2j 2j

-1 -1 -1 -1 -1 -1

2j 2j 2j 2j 2j 2j

2m-1 2m 2m-1 2m 2m-1 2m 2m-1 2m 2m-1 2m 2m-1 2m

So that the contribution of K~) to the global matrix sin1ply K

(J Dequ(/.J),J Dequ(/,J)] -

I,J

K

(J Dequ(I,J),JDequ(/.J)]

= 1,2, ... ,6.

243

(17.63)

Kij

is

+ K(e) (/.J)' (17.64)

17.4 Heat Conduction


Similarly, the contribution of the body force and external force can be computed

J(l) = f(l) + j(e)(l) where l = 2i- 1, 2i, 2j- 1, 2j, 2m- 1, 2m etc.

17.4

Heat Conduction

Heat transfer problems are very common in engineering and the geometry in most applications is irregular. Thus, finite elen1ent methods are especially useful in this case.

17 .4.1

Basic Formulation

The steady-state heat transfer is governed by the heat conduction equation or Poisson's equation V · (kVu)

+ Q = 0,

(17.65)

with the essential boundary condition u

=u,

x

E

anE,

(17.66)

and the natural boundary condition (17.67)

Using the formulation similar to the formulation (17.33) in of u ~ uh, we have

f [V · (kVu) + Q]1VidO.- f [k 8u- q]lVidr = ln lanN 8n

0.

(17.68)

Integrating by parts and using Green's theorem, we have

244

Finite Element !vletbod

+ f

ln

Since

lVi =

Q1Vidn-


f [k 8uh lanN 8n

-

q]JVidr =

o.

(17.69)

0 on {)O.E, thus we have

f [ ]Nidr = f [ ]Nidr. lan lanN

(17.70)

Therefore, the above weak formulation becomes

Substituting uh = Ej\1~ 1 UjlVj(x) into the equation, we have

-

qNidr = 0.

[

lanN

(17.72)

This can be written in the compact matrix form M

L KijUj =

KU = f,

fi,

(17.73)

j=l

= [Kij], (i,j = 1, 2, ... , M), uT = (ut, U2, ... , UM ), and (ft, !2, ... , !M ). That is,

where K

cT =

Kii =

fi

=

In

k"V Ni "V lVjdO.,

f QNidn + f qNidr. ln lanN

(17.74) (17.75)

0 Example 17.2(a): As a simple example, we consider the 1-D steady-state heat conduction problem,

u"(x)

+ Q(x) = 0,

witb boundazy conditions

u(O) =

/3,

u'(l) = q. 245



For a special case Q(x) = rexp(-x), we have the analytical solution u(x) = ({3- r)

+ (re- 1 + q)x +re-x.

(17.76)

Then equation (17. 75) becomes

For the purpose of demonstrating the implementation procedure, let us solve this problem by dividing the interval into 4 elements and 5 nodes. This will be discussed later in more details. D

17.4.2

Element-by-Element Assembly

The assembly of the linear matrix system is the popular elementby-element method. The stiffness matrix Kin equations (17.73) and ( 17.75) is the summation of the integral over the whole solution don1ain, and the don1ain is now divided into m elements with each element on a subdomain ne (e = 1, 2~ ... , m). Each element contributes to the whole stiffness matrix, and in fact~ its contribution is a pure number. Thus, assembly of the stiffness matrix can be done in an elen1ent-by-element manner. Furthermore, Ki,j =I= 0 if and only if (or iff) nodes i and j belong to the same elements. In the 1-D case~ Ki,j =I= 0 only for j = i - 1, i, i + 1. In finite element analysis, the shape functions Nj are typically localized functions~ thus the 1natrix K is usually sparse in most cases. The element-by-element formulation can be written as

(17. 77) and m

fi =

L fi(e), e=l

246



In addition, since the contribution of each element is a sinlple number, the integration of each elen1ent can be done using the local coordinates and local node numbers or any coordinate system for the convenience of integration over an element. Then, the nonzero contribution of each element to the global system matrix K is simply assembled by direct addition to the corresponding global entry (of the stiffness n1atrix) of the corresponding nodes or related equations. In reality, this can be easily done using an index matrix to trace the element contribution to the global system matrix. 0 Example 17.2(b):

The assembly of the global system matrix for the example with 4 elements and five nodes is shown below. For each element with i and j nodes, we have L= he,

Kljl =

[f

kN;Njdx] = :. (

fz~ e) =

! -1) 1 ' 1

Q he ( 1 ) 2 1 '

so that, for example in elements 1 and 2, these can extend to all nodes (with hi = xi+l -xi, i = 1, 2, 3, 4),

k/ht -k/ht

K<'l = (

~

-k/ht k/ht 0 0 0

0

0

0 0 0 0

0 0 0 0

and so on. Now the global system matrix becomes

K= 247


k/ht -k/hl 0 0 0

(

-k/ht k/hl + k/h2 -k/h2 0 0

U=U~}


0 -k/h2 k/h2 + k/h3 -k/h3 0

I=

0 0 -k/h3 k/h3 + k/h4 -k/h4

0 0 0 -k/h4 k/h4

}

( Q(hl Qh./2 + h2)/2 ) Q(h2 + h3)/2 , Q(h3 + h4)/2 Qh4/2 + q

where the last row of I has already included the natural bounda1y D condition at u' ( 1) = q.

17.4.3

Application of Boundary Conditions

Boundary conditions can be essential, natural or mixed. The essential boundary conditions are automatically satisfied in the finite elen1ent formulation by the approxin1ate solution. These include the displacement, rotation, and known value of the solution. Sometimes, they are also called the geometric boundary conditions. In our example, it is u(O) = /3. Natural boundary conditions often involve the first derivatives such as strains, heat flux, force, and moment. Thus, they are also referred to as force boundary conditions. In our example, it is u'(1) = q. The natural boundary conditions are included in the integration in the finite elen1ent equations such as (17.75). Thus no further in1position is necessary. On the other hand, although the essential boundary conditions are automatically satisfied in the finite element formulations, they still need to be implemented in the assembled finite element equations to ensure unique solutions. The imposition of the essential boundary conditions can be done in main several ways: a) direct application; b) Lagrangian n1ultiplier and c) penalty method. To show how these methods work, we use the 1-D poisson equation on the distinct points xi(i = 1, 2, ... , 1\11) E (0, 1] to aid our discussion. 248



Direct Application In this method, we simply use the expansion uh = Et! 1 uilVi, and apply directly the essential boundary conditions at point i to replace the corresponding ith equation with ui = ui so that ith row of the stiffness matrix K in equation (17. 73) becon1es (0, 0, ... , 1, ... , 0) and the corresponding /i = f(i) = ui. All other points will be done in the similar manner. For example, the boundary conditions u(O) = a and u(M) = (3 in the 1D case mean that the first and last equations are replaced by u1 =a and UM = {3, respectively. Thus, Ku = 1, It =a (all other coefficients are set to be zeros: K12 = ... = K1M = 0, and KMM = 1, !M = /3 with Ki\111 = ... = KM,M-1 = 0). Then, the equations can be solved for (ub u 2 , .•• , uM )T. This method is widely used due to its simplicity and the advantage of time-stepping because it allows bigger time steps. Lagrangian Multiplier This method is often used in the structure and solid mechanics to enforce the constraints (Ui = Ui). The variation is added by the extra term .X(ui- ui) where A is the Lagrange multiplier. Now we have (17.79)

whose variation 8II = 0 leads to (17.80)

Because 8u and 8-X are arbitrary, we have

where ei = (0, 0, ... , 1, 0, ... , O)T (its ith entry is equal to one). This method can be extended tom Lagrangian multipliers. 249



Penalty Method One of the most widely used methods of enforcing the essential boundary conditions is the so-called penalty method in of a very large coefficient I' Ui = ft at Xi E an E' so that /Ui = '")'U can be directly added onto Ku =f. In the 1-D example, it simply leads to Ku = Ku +!', KMM = KMM+/, and It =It+ /O::, fA,/ = fM + '"'(/3. The common rule for choosing I' is that '"'f >> max IKiil· Usually, '"Y ~ lOOOmax IKiil should be adequate. The penalty method is widely used in steadystate problems. However, it may affect the efficiency of timestepping since it increases the maximum eigenvalue of the stiffness matrix, and thus very small time steps are required for convergence. The advantage of the penalty method is that the handling of the essential boundary conditions becomes simpler from the implementation point of view. The disadvantage is that the conditions are only satisfied approximately. 0 Example 17.2(c): Following the same example of the 1-D steady state beat conduction discussed earlier, we now use the direct application method for the essential boundary conditions. lVe can replace the first equation E;=l K 1iui = f 1 with u 1 = (3, so that the first row becomes K 1j = (1 0 0 0 0) and f 1 = (3. Thus, we have

K=

(

1 -k/ht 0 0 0

0

k/ht + k/h2 -k/h2 0 0

0 -k/h2 k/h2 + k/h3 -k/h3 0

~~} ( I=

U= (

0 0 -k/h3 k/h3 + k/h4 -k/h4

~

Q(ht + h2)/2 Q(h2 + hJ)/2 Q(h3 + h4)/2 Qh4/2 + q

0 0 0 -k/h4 k/h4

) ,

For the case of k = 1,Q = -1,h 1 = ... = h 4 = 0.25,

250

}

f3 = 1 and


17.5 Time-Dependent Problems

q = -0.25, we bave

( K=

0

0

-4

8

-4

0 0

~

-4

8

-4

0 0

-4

1

0

~}

f

= (

-4 -4 4 8

-0.25 -0.25 1 ) -0.25 -0.375

Hence, tbe solution is

U = K- 1 J =

1.00 ) 0.72 0.50 . ( 0.34 0.25 D

17.5

Time-Dependent Problems

The problems we have discussed so far in our finite element analysis are not time-dependent, and the solutions obtained are the steady-state solutions. However, most realistic problems involve time, and thus we will now discuss the tin1e-dependent problems.

17.5.1

The Time Dimension

As the weak formulation uses the Green theorem that involves the spatial derivatives, the time derivatives can be considered as the source term. Thus, one simple and yet instructive way to extend the finite element forn1ulation to include the time dimension is to replace Q in equation (17.65) with Q- GUt/3Utt = Q - GU - /3il so that we have

V' · (kV'u)

+ (Q- GUt- f3utt)

= 0.

(17.81)

The boundary conditions and initial conditions are u(x, 0) = 4J(x), u = u,x E {)O.E, and kg~- q = O,x E {)O.N. Using 251



integration by parts and the expansion uh = EJ~ 1 uiNi, we have

'Lrj (k'VN/vlVj)dn] + i=l L J[(NiaNi)ui + (Ni/3Ni)uj]dn i=l n n M

M

- f NiQdn - f Niqdr ln lanN

=

o,

(17.82)

which can be written in a compact form as

Mii+ Cu+Ku = f,

(17.83)

Kii =In [(k'V Ni 'V Ni )]dO.,

(17.84)

where

fi

=

[ NiQdn

ln

+ [ Niqdr, lanN

(17.85)

and (17.86)

The matrices K, M, C are symmetric, that is to say, Kij = Kji, Mij = Mji, Cij = Cji due to the interchangeability of the orders in the product of the integrand k, Ni and Nj (i.e., 'V Ni · k · 'VNi = k'VNi'VNi, Nia1Vi = 1ViaNi =aNiNi etc). The matrix C = [Cij) is the damping matrix similar to the damping coefficient of damped oscillations. M = [Mii] is the general mass matrix due to a similar role acting as an equivalent mass in dynamics. In addition, before the boundary conditions are imposed, the matrix is usually singular, which may imply many solutions. Only after the proper boundary conditions have been enforced, the stiffness matrix will be nonsingular, thus unique solutions may be obtained. On the other hand, M and C will be always non-singular if they are not zero. For example, for the 1-D elen1ents (with nodes i and j),

-1) 1

'

252

(17.87)

Finite Element .Atfethod

but


M;~) = /3~e ( ~ ~ ) , C~~) = t}

o.he ( 2 1 ) 6 1 2 '

det[M(e)] # 0,

det[C(e)]

# 0,

(17.88)

Clearly, if M = 0, it reduces to the linear heat conduction. If C = 0, it becomes the wave equation with the source term.

17 .5.2

Time-Stepping

Fron1 the general governing equation

Mii+ Cu+Ku = f,

(17.89)

we see that it is an ordinary differential equation in of time and matrices. Thus, in principle, all the time-stepping methods developed in the standard finite difference method can be used for this purpose. For a simple center difference scheme, we have .. (un+1 _ 2un + un-1) u= (~t)2

(17.90)

so that equation (17.89) becomes (un+1 _ 2un + un-1)

M

(~t)2

(17.91)

Now the aim is to express un+ 1 in tenns of un and un- 1.

17.5.3

1-D Transient Heat Transfer

In the case of heat conduction (M = 0), we have

Cu+Ku=f,

253

(17.92)



or

. c- 1 (f -

U=

K U. )

(17.93)

Using the explicit time-stepping method, we can write it as u

n+l

- u

n

tl.t

= c-t (f- K n) u

(17.94)

'

so that we have (17.95) 0 Example 17.9: have

For a transient heat conduction problem, we OUt

= kuxx

+ Q,

and

u(x, 0) = 0, u(O, t) = 1, u'(l) = q. The formulation with 5 nodes and 4 elements leads to

C=

( 2h, ht ~ 0 6 0 0

0

ht 2(ht + h2)

h2

0 0

h2

2(h2 + ha)

ha

0 0

ha

2(ha + h4)

h4

0

h4

2h4

0 0 ) 0

For the case of a= 6,k = 1,Q = -1,h 1 = ... = h 4 = 0.25, we have 0.25 0 0 ( 0.25 0.5 0 0.25 0 1 0 ) 0.25 0.25 C= 1 0 . 0 0 0.25 0 1 0.25 0 0 0.25 0.5 0 0

17.5.4

Wave Equation

For the wave equation (C = 0), we have

Mii+Ku=f. 254

(17.96)


Using


un+1 - 2un + un-1 u = (~t)2 ..

(17.97)

we have

where I is an identity or unit matrix. For example, the 1-D wave equation 82u 82u (17.99) 8t2 = c {)x2' with the boundary conditions

u(O) = u(1) = 0,

u(x, 0) =

e-(x- 1/ 2 )

2 ,

(17.100)

= 0,

(17.101)

can be written as

fo NiNjdx, 1

Mii

=

Kii

=

l

cNfNjdx, f

and u 0 is derived from the u(x, 0) = exp[-(x- 1/2) 2 ]. The finite element methods in this book are mainly for linear partial differential equations. Although these methods can in principle be extended to nonlinear problems, however, some degrees of approximations and linearization are needed. In addition, an iterative procedure is required to solve the resultant nonlinear n1atrix equations. The interested readers can refer to many excellent books on these topics.

255

Chapter 18

Reaction Diffusion System The partial differential equations we solved in the previous chapters using the three major numerical methods are linear equations. We know that the generalized forn1s of parabolic equations are nonlinear reaction-diffusion equations. l\1athematically speaking, nonlinear equations are far more difficult to analyse if it is not impossible. From the numerical point of view, some extra linearization and approximations should be used for the nonlinear . However, the finite difference scheme should still be useful for most nonlinear equations though they should be implemented more carefully. Before we proceed to study the nonlinear system, let us review the fundamental characteristics of linear parabolic equations by solving the linear heat conduction equations.

18.1

Heat Conduction Equation

18.1.1

Fundamental Solutions

From the similarity solution in section 10.3, we know that both diffusion equation and heat conduction equation may mathe257

18.1 Heat Conduction Equation

Reaction Diffusion System

matically have a similarity variable defined by X

(=

(18.1)

ym'

where k is either the thermal diffusivity or diffusion coefficient. In engineering, the coefficient k has the unit [n1] 2 /[s], thus the unit of kt is [1n] 2, which means the variable ( is dimensionless. Any two combinations of x and kt giving the same ( will have similar solutions. Using the similarity variable (, the heat conduction equation can be transformed into an ordinary differential equation

!"(()

=

-2(/',

(ln !')' = -2(.

or

(18.2)

Integrating it once, we have (18.3)

where C is an integration constant. Integrating it again, we have xo 2 u = C x e-< d( + D, (18.4)

1

which is the general solution of the heat conduction equation. If the domain is semi-infinite or infinite so that x 0 ~ 0, then we get X (18.5) =Aerf( ~)+B. v4kt 0 Example 18.1: domain, we have

For the heat conduction in a semi-infinite

au= k8 2 u 8t with the boundazy condition U=

UQ,

8x 2 '

(X

~

0),

and an initial condition

u = 0,

(x > 0) 258

at t = 0.


18.2 Nonlinear Equations

The general solution is X

u = Aerf( /"A"U) v4kt

+ B.

For x ~ oo, erf(x/ ../4kl) ~ 1, we have

A+B

= 0.

At x = 0, erf(O) = 0, we get B = uo. The solution is X

X

u = uo[l - erf(. 17'i:!')] = uoerfc r:rr:r· v4kt v4kt However, if u is constant (u = uo) in the initial region x E [-h, h), then we have uo h-x h+x u = -[erf-- + erf--]. 2 J4kt J4kt 0

The solutions of heat conduction do not always involve the error function because error functions only occur when the integration involves semi-infinite or infinite domains. If the domain has a finite length, then the solutions often consist of power series or even special functions. For example in heat conduction through a plane sheet with zero initial temperature, its two surfaces are held at constant temperatures with the boundary conditions u = u 0 at x = 0 for (t ~ 0), and u = 0 at x = L for (t ~ 0). The general solution can be written as u

~ uo . n1rx -kn21r2t;L2 = uo (1 - -x) +-2 L...., - sm --e ,

L

1T' n=l

n

L

(18.6)

which is a slowly convergent series.

18.2

Nonlinear Equations

18.2.1

Travelling Wave

The nonlinear reaction-diffusion equation

8u 8t

82 u

= D {)x 2

259

+ f(u),

(18. 7)



can have the travelling wave solution under appropriate conditions of f(O) = f(1) = 0, f(u) > 0 for u E [0, 1], and f'(O) > 0. For example, f(u) = ')'u(1 - u) satisfies these conditions, and the equation in this special case is called the KolmogorovPetrovskii-Piskunov (KPP) equation. By assuming that the travelling wave solution has the form u( () and ( = x - vt, and substituting into the above equation, we have

Du"(()

+ vu'(() + f(u(())

= 0.

(18.8)

This is a second-order ordinary differential equation that can be solved with the appropriate boundary conditions

u( -oo)

--+

1,

u( oo)

--+

0.

(18.9)

The KPP theory suggests that the limit of the speed of the travelling wave satisfies v 2?:

18.2.2

2JDf'(O).

(18.10)

Pattern Formation

One of the most studied nonlinear reaction-diffusion equations in the 2-D case is the Kolmogorov-Petrovskii-Piskunov (KPP) equation {)u fPu 8 2 u (18.11) {)t = D( ax2 + {)y2) + '"YQ(u), and

q(u) = u(1 - u).

(18.12)

The KPP equation can describe a huge number of physical, chemical and biological phenomena. The most interesting feature of this nonlinear equation is its ability of generating beautiful patterns. We can solve it using the finite difference scheme by applying the periodic boundary conditions and using a random initial condition u = random( n, n) where n is the size of the grid. 260



1

0.5 0

200 0

0 200

Figure 18.1: 2-D pattern formation forD= 0.2 and 'Y = 0.5. Figure 18.1 shows the pattern formation of the above equation on a 100 x 100 grid for D = 0.2 and 'Y = 0.5. We can see that rings and thin curves are formed, arising from the random initial condition. The landscape surface shows the variations in the location and values of the field u(x, y) can be easily demonstrated. The following simple 15-line Matlab program can be used to solve this nonlinear system.

% Pattern formation: a 15 line matlab program % POE form: u_t=O*(u_{xx}+u_{yy})+gamma*q(u) %where q(u)='u.*(1-u)'; % The solution of this POE is obtained by the % finite difference method, assuming dx=dy=dt=1 % Written by X S Yang (Cambridge University) % Usage: pattern(100) or simply >pattern function pattern(time) % Input number of time steps if nargin<1, time=100; end 261

% line 1 % line 2



% Initialize parameters n=100; % 0=0.2; gamma=0.5; n=200; 0=0.2; gamma=0.5;

% line 3

%Set initial values of u randomly u=rand(n,n); grad=u*O;

% line 4

% Index for u(i,j) and the loop I = 2:n-1; J = 2:n-1;

% line 5

% Time stepping for step=1:time, % line 6 % Laplace gradient of the equation % line 7 grad(I,J)= u(I,J-1)+u(I,J+1)+u(I-1, J)+u(I+1,J); u ={1-4*0)*u+O*grad+gamma*u.*(1-u); %line 8 % Show results pcolor(u); shading interp; % line 9 % Coloring and colorbar colorbar; colormap jet; % line 10 drawnow; % line 11 end % line 12 % plot as a surface surf(u); shading interp; view( [ -25 70]); % ------------- End of Pragram

% line 13 % line 14 % line 15

-----------------

If you use this program to do the simulations, you will see that the pattern emerges naturally from the initially random background. Once the pattern is formed, it evolves gradually with time, but the characteristics such as the shape and structure of the patterns do not change much with time. In this sense, one can see beautiful and stable patterns. 262


18.3

18.3 Reaction-Diffusion System

Reaction-Diffusion System

The pattern formation in the previous section arises naturally from a single equation of nonlinear reaction-diffusion type. In many applications, we often have to simulate a system of nonlinear reaction-diffusion equations, and the variables are coupled in a complicated manner. The pattern formation in the previous section comes from the instability of the nonlinear reaction diffusion system. In order to show this, let us use the following mathematical model for enzyme inhibition and cooperativity. For example, the following system consists of two nonlinear equations (18.13)

(18.14)

and f(u, v) = au(1 - u) g(u,v) =

bu- cv 1 +u+v

vd 1+u+v'

,

(18.15) (18.16)

where Du and Dv are diffusion coefficients, while a, b, c, d are all constants. This reaction diffusion system may have instability if certain conditions are met. The steady state solutions are obtained from f(uo, vo) = 0 and g(uo, vo) = 0. They are

uo

b~ = 2a [V1 + 4/;2 - 1],

vo = 0.

(18.17)

Let 1/J = (u- uo, v - vo) be the small perturbation, then 1/J satisfies 81/; = D\12,.,, + M"'' (18.18) ~ ~,

at

263



where (18.19)

and

1

M =

(1

( -(2au0 +b) a(1 -duo)+ c ) . 0

+ uo)

(18.20)

Writing 'l/J in the form of (18.21)

where the sumn1ation is over all the wavenumbers k, we have (18.22)

where I is a 2 x 2 unity matrix. This eigenvalue equation has two roots. Since ~(A) > 0 implies that instability, this requires that Dv d - < ~-----:~ (18.23) Du (2auo +b) The range of unstable wavenumbers between the two roots of k2 at the bifurcation point is given by k 2 = dDu - Dv(2auo +b) [1 ± ± 2DuDv ( 1 + Uo)

with 8

J1

+4

D D 8] u v '

_ (2auo +b) - [dDu- Dv(2auo + b)]2 ·

(18.24)

(18.25)

If the unstable criteria are satisfied, any small random perturbation can generate complex patterns. Similar to the nonlinear KPP equation (18.12), beautiful patterns also arise naturally in the following nonlinear system

8u 8t 8v 8t

82 u

82 u

-

=

Da ( 8x2

+ 8y2 ) + 'Yf (u, v)'

(18.26)

=

82v 82v _ Db( 8x2 + 8y2) + f3g(u, v),

(18.27)

264



and

](u, v) = u(1- u),

g(u, v) = u- u 2 v,

(18.28)

for the values Da = 0.2, Db = 0.1, "( = 0.5 and {3 = 0.2. With different functions j (u, v) and g( u, v), these equations can be used to sin1ulate the pattern formation in a wide range of applications where nonlinear reaction-diffusion equations are concerned.

265

Chapter 19

Probability and Statistics All the mathematical models and differential equations we have discussed so far are deterministic systems in the sense that given accurate initial and boundary conditions, the solutions of the system can be determined (the only exception is the chaotic system to a certain degree). There is no intrinsic randomness in the differential equations. In reality, randomness occurs everywhere, and not all models are deterministic. In fact, it is necessary to use stochastic models and sometimes the only sensible models are stochastic descriptions. In these cases, we have to deal with probability and statistics.

19.1

Probability

19.1.1

Randomness and Probability

Randomness such as roulette-rolling and noise arises from the lack of information, or incomplete knowledge of reality. It can also come from the intrinsic complexity, diversity and perturbations of the system. The theory of probability is mainly the studies of random phenomena so as to find non-random regularity. 267

19.1 Probability

Probability and Statistics

For an experiment or trial such as rolling dices whose outcome depends on chance, the sample space n of the experiment is the set of all possible outcomes. The sample space can be either finite or infinite. For exan1ple, rolling a six-sided die will have six different outcomes, thus the san1ple space is n = {1, 2, 3, 4, 5, 6}. The elements of a sample space are the outcomes, and each subset of a sample space is called an event. For example, the event S = {2, 4, 6} is a subset of n. In a sample space n, the outcomes of an experiment is represented as numbers ( 1 for heads and 0 for tails for tossing coins). A real-valued variable that is defined for all the possible outcomes is referred to as a random variable, which is a function that associates a unique numerical value with every outcome of an experiment, and its actual value varies from trial to trial as the experiment is repeated. The values of a random variable can be discrete (such as 1 to 6 in rolling a single die) or continuous (such as the level of noise). If a random variable only takes discrete values, it is called a discrete random variable. If its values are continuous, then it is called a continuous random variable.

Figure 19.1: Venn Diagrams: An Band AU B. Two events A and B can have various relationships and these can be represented by Venn diagrams as shown in Figure 19.1. The intersection An B of two events means the outcome of the random experiments belongs to both A and B, and it is the case of 'A AND B'. If no event or outcome belongs to the intersection, that is An B = 0, we say these two events are 268


19.1 Probability

mutually exclusive or dist. The union A U B denotes the outcome belongs to either A or B or both, and this means the case of 'A ORB'. The complement A = n - A (or not A) of the event A is the set of outcomes that do not belong to A but in the sample space n (see Figure 19.2). The A- B means the outcomes in A only.

Figure 19.2: Venn Diagrams:

A = n - A and A - B.

Probability P is a number or an expected frequency assigned to an event A that indicates how likely the event will occur when a random experiment is performed. This probability is often written as P( A) to show that the probability P is associated with event A. For a large number of fair trials, the probability can be calculated by P(A) = NA(number of outcomes in the event A). N n (total number of outcomes)

(19.1)

0 Example 19.1: If you tossed a coin 1000 times, the head (H) occurs 511 times and the tail (T) occurs 489 times. The probability P(H) and P(T) are

P(H) =

511 = 0.511, 100

P(T) =

489 = 0.489. 1000

and

There are three axioms of probability, and they are: Axiom I : 0

~

P(A)

~

1. 269

19.1 Probability


Axiom II: P(f!) = 1. Axiom III: P(A U B)= P(A)

+ P(B), if

An B = 0.

The first axiom says that the probability is a number between 0 and 1 inclusive. P(A) = 0 corresponds to impossibility while P(A) = 1 corresponds to absolute certainty. The second axiom simply means that an event must occur somewhere inside the sample space. The third axiom is often called the addition rule. Since A and A are mutually exclusive (An A= 0), we have

P(A)

+ P(A) = P(A u A) = P(fl) = 1,

(19.2)

P(A) = 1 - P(A),

(19.3)

or which is usually called the NOT rule. The third axiom can be further generalized to any two events A and B

P(A U B) = P(A)

+ P(B) -

P(A n B).

(19.4)

In a special case when events Ai (i = 1, 2, ... , n) exhaust the wholesamplespacesuch that A= Uf=tAi = AtUAiU ... UAn = nand Ai n Ai = 0(i =I= j), n

P(A n B)=

L P(Ai n B).

(19.5)

i=l

Since n n B = B, we also get n

P(n n B) = P(B) =

L

P(Ai

n B),

(19.6)

i=l

which are the useful properties of the total probability. For example, if you randomly draw a card from a standard pack of 52 cards, what is the probability of it being a red king or a diamond with a face value being a prime number (if its 270


19.1 Probability

face value is counted from 1 to 13). The prime numbers are 2, 3, 5, 7, 11, 13, therefore they are 6 cards that forms the primes. The possibility of event (A) of drawing a red king is P( A) = ~ = ,Js. The probability of event (B) of drawing a prime number is P(B) = 562 = 6 • As a diamond king (13) is also a prime, this means P( A n B) = ~. Therefore, the probability

:f

P(A U B) = P(A) 1 = 26

3

+ 26

+ P(B) -

P(A n B)

1 7 - 52 = 52.

Two events A and B are independent if the events has no influence on each other. That is to say, the occurrence of one of the events does not provide any information about whether or the other event will occur. In this case, the probability of both occuring is equal to the product of the probabilities of the two individual events P(A) and P(B) P(A n B)= P(A)

* P(B).

(19.7)

This can be easily extended to n mutually independent events Ai(i = 1, 2, ... , n). The probability of all these events happening is n

P(L nAi) = Ilf:: 1 P(Ai) = P(Al)P(A2) · · · P(An)·

(19.8)

i=l

0 Example 19.2: The probability of drawing a king from a pack of cards (Event A), and showing an even number of rolling a sixsided die (event B) is P(A n B). We know P(A) = 4/52 = 1/13, and P(B) = 3/6 = 1/2. Since these two events are independent, the probability that both events occur is

1 1 1 P(A n B) = P(A)P(B) = - · - = 13 2 26. 0

If the two events are not independent, then one may affect

the other event, in this case, we are dealing with the conditional probability which will be discussed later in the next section. 271

19.1 Probability


In calculating the probabilities, it is useful to know the possible combinations and permutations of certain objects. Suppose you have 5 pairs of shoes, 4 pairs of tros, 7 shirts and 2 hats. This is equivalent to the lineup problem from your feet to your head. In this case, as the event of selecting each thing to wear is in a similar manner of putting it into slots in successive stages, the total number of all possible ways is simply the multiplication of all the possible choices for each stages. All possible outfits you can wear form a permutation problem, and the total number is 5 x 4 x 7 x 5 = 700. In order to line 5 objects marked A, B, C, D, E, in the first place, there are 5 possible choices, the second place has only 4 options, the third place 3 choices, the fourth place has 2 choices, and there is only one left for the last place. Thus the number of all possible permutations is 5 x 4 x 3 x 2 x 1 =51. Following this line of reasoning, n objects can in general be pern1utated . I ways. Inn. Suppose there are n = 20 students in a class (named 81, 8 2, .... , 820 ), we want to select 5 students at random to fonn a 5-student team to work on a project. This is different from the lineup problem because once you have selected any five students (say) 8~, 81, 8w, 81s, 819, it does not n1atter what order you selected them, the final formed team is the same. There are 51 permutations within the same tea1n. Order does not count in this case. This is a combination problem (also called a comn1ittee problem). As before, there are 5 places to line up the students, and the total number of all permutations for selecting 5 students is 20 * 19 * 18 * 17 * 16. Therefore, the total number of combinations (of selecting 5 students) is

c5 =

20 * 19 * 18 * 11 * 15 = 201 = 15504 (19.9) 51 5!15! . In general, the total number of all possible combinations of selecting k objects from n is 20

nck

= ( ~ ) = kl(nn~ k)!" 272

(19.10)


19.1 Probability

The consistency requires OJ= 1. 0 Example 19.9: A club of 5 is chosen at random from 8 female students, 10 male students, and 7 teachers. What is the probability of the club consisting of 2 female students, 2 male students, a.nd 1 teacher? The total number of clubs is 25 Cs. If two female students are selected, we have 8 C2. Similarly, 10 C2 for selecting 2 male students, and 7 C 1 for selecting one teacher. Therefore, the total number N of forming the 5-member club is

N = sc2IOC27Cl = ~ ~ 0166

253

2SC5

·

0

There is an interesting 'birthday paradox' which is related to this context. The birthday paradox was first proposed in 1939 by Richard von l\1ises, which states that what is the probability of two people having the same birthday in a group of n people. For a group of 367 people, it is certain that there must be two people having the san1e birthday as there are only 365 (or 366 if someone was born in a leap year) possible birthdays. Ignoring 29 February and the year of birth and assuming that the birthdays are evenly distributed throughout the year, we only have 365 different birthdays (days and months only). If the event A denotes that all the n people will have different birthdays (no birthday matching), the first person can have any date as his or her birthday, 365/365. The second person must be in other 364 dates, which is 364/365, and the kth person has (365- k + 1)/365. Therefore, the probability of no two people having the same birthday is 365

P( A, n)

365 * (364)

364

= 365 x 365 x .. · x

* ... * (365- n + 1) 365n

(365- n 365

+ 1)

_ 365! (19.11) - (365- n)f365n ·

Now the probability of two people with the same birthday is P(A, n)

365!

= 1- P(A, n) = 1- (365- n)!365n. 273

(19.12)

19.1 Probability


The factorial 3651 is a large number, but you do not have to deal with such large numbers. You can use a simple calculator to estimate it. For five people, the probability of two people with the same birthday is

P(.A 5 ) '

= 1 _ 365

* 364 * 363 * 362 * 361 365 5

~ 0 027 .

'

( 19 . 13)

which seems insignificant. However, the interesting thing is that for n = 23, the probability becomes 365! P(A, 23) = 1 - (365 - 23)136523 ~ 0.507.

(19.14)

This means that you have slightly more than a 50-50 chance of finding two people sharing the same birthday. If you increase n, you get P(A, 30) ~ 0. 706 for n = 30, P(.A, 40) ~ 0.891 for n = 40, and P(A, 50) ~ 0.970 and P(A, 70) ~ 0.9992 (aln1ost certainty) for n = 70. It is worth noting that there is some difference in combinations when the member drawn is placed back or not. Suppose there are 10 red balls and 10 white balls in bag. If we draw a ball (say a red, event A) from the bag and then put it back (with replacement), then we draw another ball (event B). P(A) = 1/20 and P(B) = 1/20. The probability of getting two red balls are P(A n B) = P(A) * P(B) = 1/400. We call this easel. For a second case (Case II), if we do not put it back after we have drawn the first ball (without replacmnent), then the probability of event B is now different P(B) = 1/19 as there is now only 19 balls in the bag. The probability of getting two red balls now becomes P(A n B) = .Jrr x .fg = ~~ which is different from 1/400. The reason here is that the two events are not independent in the case of no-replacement. If we use notation 'BIA' which means that event B occurs given that event A has occurred, then we can use P(BIA) to denote the probability of event B when there is no replacement in event A in the scenario 274


19.1 Probability

described in Case II. Now P(B) becomes P(BIA). Hence, we have P(A n B) = P(A)P(BIA), (19.15) which is often called the multiplication rule in probability theory. Similarly, we can get P(A

n B) =

P(B)P(AIB).

(19.16)

This is essentially a conditional probability problem which forms the main topic of the next section.

19.1.2

Conditional Probability

In calculating the probabilities, we often assume that all possible outcomes of an experiment such as drawing a card are equally likely. Probabilities can change if additional information is known or some other event has already occurred and thus P(BIA) denotes the probability that event B will occur given that event A has already occurred. The conditional probability can be calculated by P(BIA) = P(B n A) P(A) .

(19.17)

P(AIB) P(A n B) P(B) .

(19.18)

Conversely, we have

Using equation (19.15), we can write the above formulation as P(AIB)

=

P(A)P(BIA) P(B)

P(A)P(BIA) P(A)P(BIA) + P(A)P(BIA)'

which is the Bayes' theorem. Here have used P(A) = 1 - P(A). 275

Au A

(19.19) =

0 and

19.1 Probability


As an example, we consider the drug test in sports. It is believed that the test is 99% accurate if athletes are taking drugs. For athletes not taking drugs, the positive test is only 0.5%. It is assumed that only one in 1000 athletes takes this kind of drug. Suppose an athlete is selected at random and the test shows positive for the drug. What is the probability that the athlete is really taking the drug? If event A denotes an athlete is taking the drug, and B denotes the event that the individual tests positive. Thus, P(A) = 1/1000, P(BIA) = 0.99 and P(BIA) = 0.005. The probability that the athlete is actually taking the drug is P(AIB) _ P(A)P(BIA) - P(A)P(BIA)+P(A)P(BIA)

0.001 * 0.99 ~ 0.165. 0.001 * 0.99 + 0.999 * 0.005 This is surprisingly low in probability.

(19.20)

0 Example 19.4: The classical problem of three cards consists of three cards: one blue card (B) is blue on both sides, one white card (W) is wbite on both sides, and one mixed card (A1) is wbite on one side and blue on the other. If you draw one card at random from a bag and place it on a table, suppose that the side you can see is blue, what is the probability of other side is also blue? This a conditional probability problem. There are 3 blue faces and 3 white faces, thus tbe total probability of showing a blue face (F) is P(BF) = 1/2, and probability of pull the blue-blue card out is P(BB) = 1/3, while the probability of showing a blue face is P(BFIBB) = 1 if the pulled card is blue-blue one. Then, the probability of other side being also blue is P(BBIBF)

= P(BFIBB)P(BB) = 1 X P(BF)

!

t =~

3·

!\lost people will intuitively guess that the probability is not correct.

(19.21)

t, which is 0

Another related problem is the so-called l\1onty Hall problem (or three door problem) in a classical game show. Suppose 276


19.1 Probability

that you are given the choice of three doors. There is an expensive car behind one door, behind other two doors are goats. If you choose one door (say, A) at random, then the host opens one of the other door (say, B), which he knows there is a goat behind it, to reveal a goat. Now you have a choice either to stick with your original choice or swap with the other door (say, C). What is your best strategy based on probability? Initially, the prize car behind any door (Y) has a priori probability P(any) = 1/3, so your initial choice P(A) = 1/3. As the host knows where the prize is, if the car is behind A, the host will open B or C so 1/2 each. If the car is behind B, the host will never open B, and if the car is behind C, the host will surely open B. Nlathematically, this gives P(OpenBIA) = 1/2, P(OpenBIB) = 0, and P(OpenBIC) =I. So the total probability of opening door B is P(OpenB) = P(A)P(OpenBIA) +P(C)P(OpenBIC)

1

= '3 x

1

+ P(B)P(OpenBIB)

1

1

1

2 + 3 x 0 + 3 x 1 = 2.

(19.22)

Now the probability of the car behind door Cis P(CIOpenB)

=

P(OpenBIC)P(C) P(OpenB)

= 1 x ~ = ~' ~

(19.23)

3

which is greater than your initial choice 1/3. Therefore, the best strategy is to switch your choice. This game has other variations such as the three envelope problem and others, but the analysis and strategy are the same.

19.1.3

Random Variables and Moments

Random Variables For a discrete random variable X with distinct values such as the number of cars ing through a junction, each value Xi may occur with a certain probability p(xi)· In other words,

277

19.1 Probability


the probability varies with the random variable. A probability function p(xi) is a function that defines probabilities to all the discrete values xi of the random variable X. As an event must occur inside a sample space, the requirement that all the probabilities must be summed to one leads to n

LP(Xi) = 1.

(19.24)

i=l

The cumulative probability function of X is defined by

P(X ~ x) =

L

p(xi)·

(19.25)

Xi<X

For a continuous random variable X that takes a continuous range of values (such as the level of noise), its distribution is continuous and the probability density function p( x) is defined for a range of values x E [a, b] for given limits a and b [or even over the whole real axis x E (-oo,oo)]. In this case, we always use the interval (x, x + dx] so that p(x) is the probability that the random variable X takes the value x < X ~ x + dx is ~(x)

= P(x

<X~

x + dx)

= p(x)dx.

(19.26)

As all the probabilities of the distribution shall be added to unity, we have

l

p(x)dx

= 1.

(19.27)

The cumulative probability function becomes

~(x)

=

P(X

~ x)

=

J.x p(x)dx,

(19.28)

which is the definite integral of the probability density function between the lower limit a up to the present value X= x.

Mean and Variance Two main measures for a random variable X with a given probability distribution p(x) are its mean and variance. The mean 278


19.1 Probability

J1. or the expectation value of E[X] is defined by J1.

=E[X] =<X>= j xp(x)dx,

(19.29)

for a continuous distribution and the integration is within the integration limits. If the random variable is discrete, then the integration becomes the sumn1ation

E[X]

=

L

XiP(Xi)·

(19.30)

The variance var[X] = a 2 is the expectation value of the deviation squared (X- J..L) 2 . That is (19.31)

The square root of the variance a = Jvar[X] is called the standard deviation, which is simply a. This simply becomes a sum (19.32)

for a discrete distribution. In addition, any other formulas for a continuous distribution can be converted to their counterpart for a discrete distribution if the integration is replaced by the sum. Therefore, we will mainly focus on the continuous distribution in the rest of the section. Other frequently used measures are the mode and median. The mode of a distribution is defined by the value at which the probability density function p( x) is maxin1um. For an even number of data sets, the mode may have two values. The median m of a distribution corresponds to the value at which the cumulative probability function (m) = 1/2. The upper and lower quartiles Qu and QL are defined by (Qu) = 3/4 and (QL) = 1/4. 279

19.1 Probability


Moments and Moment Generating Functions In fact, the mean is essentially the first moment if we define the kth moment of a random variable X by E[Xk]

=Jlk =I xkp(x)dx,

k = 1, 2, ... , N.

(19.33)

Similarly, the kth central moment is defined by E[(X- Ill]

= vk

=I

(x- p,)kp(x)dx,

k = 1, 2, ... , N. (19.34)

Obviously, the variance is the second central moment. From these definitions, it is straightforward to prove (19.35)

and var[ax + .6] = a 2 var[X].

(19.36)

where a and .6 are constants. Ivlost probability functions can be expressed in of moments and moment generating functions. The moment generating function is defined by (19.37)

where v E 'R is a real parameter. By expanding exp[vx] into power series and using the definition of various moments, it is straightforward to that (19.38)

and 2 _

a -

~Gx(O) _ [dGx(0)] 2 dv 2 dv ·

280

(19.39)


19.1.4

19.1 Probability

Binomial and Poisson Distributions

Binomial Distribution A discrete random variable is said to follow the binomial distribution B(n,p) if its probability distribution is given by

B(n,p) = nCxpx(l- Pt-x,

nc -

x-

n! xl(n-x)l'

(19.40)

where x = 0, 1, 2, ... , n are the values that the random variable X may take, n is the number of trials. There are only two possible outcomes: success or failure. p is the probability of a so-called 'success' of the outcome. Subsequently, the probability of the failure of a trial is q = 1 - p. Therefore, B(n,p) represents the probability of x successes and n- x failures in n trials. The coefficients come from the coefficients of the binomial expansions n

(p+qt = LnCxpxqn-x = 1,

(19.41)

x=O

which is exactly the requirement that all the probabilities should be summed to unity. 0 Example 19.5: Tossing a coin 10 times, the probability of getting 7 heads is B(n, 1/2). Since p = 1/2 and x = 7, then we have

10c7 (.!.)1(.!.) 3 = ~ ~ o 111 2 2 128 . . 0

It is straightforward to prove that J1. = E[X] = np and a 2 = npq = np( 1 - p) for a binomial distribution. Another related distribution is the geometric distribution whose probability function is defined by P(X

= n) = pqn-1 = p(1- p)n-1'

(19.42)

where n ~ 1. This distribution is used to calculate the first success, thus the first n- 1 trials must be in failure if n trials are needed to observe the first success. The mean and variance of this distribution are J1. = 1/p and a2 = (1 - p)jp2. 281

19.1 Probability


Poisson Distribution

The Poisson distribution can be thought as the limit of the binomial distribution when the number of trial is very large n --+ oo and the probability p --+ 0 (small probability) with the constraint that A = np is finite. For this reason, it is often called the distribution for small-probability events. Typically, it is concerned with the number of events that occur in a certain time interval (e.g., number of telephone calls in an hour) or spatial area. The Poisson distribution is

A> 0,

(19.43)

where x = 0, 1, 2, ... , n and A is the mean of the distribution. Using the definition of mean and variance, it is straightforward to prove that J.L = A and a 2 = A for the Poisson distribution. The parameter A is the location of the peak as shown in Figure 19.3. 0 Example 19.6: If you receive 3 calls per hour on your mobile phone on the average. If you do not switch your phone off, what is the probability that it begins to sound (one call is enough) during any one-hour class? \-'Ve know that .,\ = 3. The probability of no phone call at all is

Thus, the probability of sounding is P(X > 0) In fact, the probability of receiving one call is

~

1 - 0.0498

~

0.95.

31e-3 P(X = 1) = - ,-::::::: 0.149, 1

and the probability of receiving two calls is

0

282


19.1 Probability

0.05r------..-------..-----.-----;===::::::;, --- A=S - -· A=10 0.04 A=15

,-, I

\

I

0.03 'iC'

\

II

\

I

rt

\

:'

0.02

\

:

\

I

\

I

\

I

\

0.01 :'

,

I

25 X

Figure 19.3: Poisson distributions for different values of .X = 5, 10, 15. The moment generating function for the Poisson distribution is given by (19.44)

19.1.5

Gaussian Distribution

The Gaussian distribution or normal distribution is the most important continuous distribution in probability and it has a wide range of applications. For a continuous random variable X, the probability density function (PDF) of a Gaussian distribution is given by p(x) =

1

(x-,_.)2

. m=e-~,

ay27T

(19.45)

where a 2 = var[X] is the variance and J-l = E[X] is the mean of the Gaussian distribution. From the Gaussian integral, it is easy to that 00

p(x)dx = 1,

/ 00

283

(19.46)

19.1 Probability


0.1 r---.---~--.------.---;::::======il <J=5 - -· <J=7 --- <J=10

0.08

/.'

0.02

,':' I

I I

I

, ,' ,·'

____ ,. ......_..... ,

-20

-10

10

0

20

30

X

Figure 19.4: Gaussian distributions for a =

5~

7, 10.

and this is exactly the reason that the factor 1/ .;2-i comes from the normalization of the all probabilities. The probability function reaches a peak at x = J..l and the variance a 2 controls the width of the peak (see Figure 19.4). The cumulative probability function (F) for a normal distribution is the integral of p( x), which is defined by

4>(x) = P(X < x) =

1 21ra 2

~

!X

e-

(<-w2 2a

d(.

(19.47)

-oo

Using the error function defined by Chapter 1, we can write it as 1 X-J..t (19.48) 4>(x) = ro[1 + erf( ro )]. v2 v2a The moment generating function for the Gaussian distribution is given by (19.49) The Gaussian distribution can be considered as the limit of the Poisson distribution when A >> 1. Using the Sterling's approximation xl rv ..f2;iX(xje)x for x >> 1, and setting J..l =A and a 2 =A, it can be verified that the Poisson distribution can 284


19.1 Probability

be written as a Gaussian distribution (19.50) where J..t = A. In statistical applications, the normal distribution is often written as N (J..t, a) to emphasize that the probability density function depends on two parameters J.L and a. The standard normal distribution is a normal distribution N(J..t, a) with a mean of J.L = 0 and standard deviation a = 1, that is JV(O, 1) This is useful to normalize or standardize data for statistical analysis. If we define a normalized variable X-J.L

~=-,

(19.51)

a

it is equivalent to give a score so as to place the data above or below the mean in the unit of standard deviation. In of the area under the probability density function, ~ sorts where the data falls. It is worth pointing out that some books define z = ~ = (x - J.L) /a in this case, and call the standard normal distribution as the Z distribution. Table 19.1: Function¢> defined by equation (19.53). ~

¢>(~)

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

0.500 0.540 0.579 0.618 0.655 0.692 0.726 0.758 0.788 0.816

~

¢>

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9

0.841 0.864 0.885 0.903 0.919 0.933 0.945 0.955 0.964 0.971

285

19.1 Probability


Now the probability density function of standard normal distribution becomes (19.52) Its cumulative probability function is

¢(~) = ~ ~~ e-~: d~ = -21 [1 + erf(. ~)]. v27r

v2

-oo

(19.53)

As the calculations of ¢ and the error function involve the numerical integrations, it is usual practice to tabulate ¢ in a table (see Table 19.1) so that you do not have to calculate their values each time you use it.

19.1.6

Other Distributions

There are a dozen of other important distributions such as the exponential distribution, log-normal distribution, uniform distribution and the x2-distribution. The uniform distribution has a probability density function 1

p= - - , {3-a:

X=

(a:,b],

(19.54)

whose mean is E[X] = (a: + {3) /2 and variance is a 2 = (/3 a:) 2 /12. The exponential distribution has the following probability density function

f(x)

=

..\e-.Xx (x > 0),

(19.55)

and f(x) = 0 for x:::; 0. Its mean and variance are (19.56) The log-normal distribution has a probability density function 1 (lnx- JL) 2 f(x) = ~exp[ (19.57) ], 2a 2 x 21ra2 286


19.1 Probability

whose mean and variance are (19.58)

The x2-distribution, called chi-square or chi-squared distribution, is very useful in statistical inference and method of least squares. This distribution is for the quantity 2

_~(Xi- /-li)2

Xn-L

,

(19.59)

ai

i=l

where then-independent variables Xi are normally distributed with means J-li and variances a'f. The probability density function for x2-distribution is given by p(x)

= 2n/2:(n/2) x1'-le-xf2

(19.60)

where x 2:: 0, and n is called the degree of freedom. Its cumulative distribution function is ( )

= "'f(n/2, x/2)

X

f(n/2)

'

(19.61)

where "Y(n/2, x/2 is the incomplete gamma function. It can be verified that the mean of the distribution is n and its variance is 2n. For other distributions, readers can refer to any books that are devoted to probability theory and statistical analysis.

19.1.7

The Central Limit Theorem

The most important theorem in probability is the central limit theorem which concerns the large number of trials and explains why the normal distribution occurs so widely. This theorem is as follows: Let Xi(i = 1, 2, ... , n) ben independent random variables, each of which is defined by a probability density function Pi (x) with a corresponding mean J..Li and a variance a'f. The sum of all these random variables n

e=

Lxi =

Xt +X2 + ... +Xn,

i=l

287

(19.62)

19.1 Probability


0.2 >-0.15

C)

i:::s

tr

! 0.1

LL

0.05

2

1

3 4 Outcome

5

6

7

Figure 19.5: A uniform distribution. is also a random variable whose distribution approaches the Gaussian distribution as n ~ oo. Its mean E[8] and variance var[8] are given by E[8] =

n

n

i=l

i=l

L E[Xi] = L Jli,

and

n

var[8]

= L var[8] = i=l

(19.63)

n

La?.

(19.64)

i=l

The proof of this theorem is out of the scope of this book as it involves the n1oment generating functions, characteristics functions and other techniques. In engineering mathematics, we simply use these important results for statistical analysis. In the special case when all the variables Xi are described by the same probability density function with the same mean Jl and variance a 2 , these results become E[8] = np,,

var[8] = na 2 .

(19.65)

By defining a new variable {n

=

8- np, a ..jn n

288

'

(19.66)


19.2 Statistics

0.2

0.15

0.05

I

2

4

6

8

Outcome

10

I

12

14

Figure 19.6: A bilinear distribution. then the distribution of ~n converges towards the standard normal distribution N(O, 1) as n ~ oo. Let us see what the theorem means for a simple experiment of rolling a few dice. For a fair six-sided die, each side will appear equally likely with a probability of 1/6 ~ 0.1667, thus the probability function after rolling it 15000 times approaches a uniform distribution as shown in Figure 19.5. If we now roll two independent dice 15000 tin1es and count the sum (1-12) of the face values of both dice, then the sum obeys a bilinear distribution as shown in Figure 19.6. If we roll n = 15 independent dice, the sums of the face values vary from 1 to 90. After rolling the 15 dice 10,000 times, the distribution is shown in Figure 19.7 and it approaches to a normal distribution as n ~ oo.

19.2

Statistics

Statistics is the mathematics of data collection and interpretation, and the analysis and characterisation of numerical data by inference from sampling. Statistical methods involve reduc-

289

19.2 Statistics


0.1 r----.------.------.---r-----r-----.----, 0.08 >-

~0.06

CD ::I

117'

~0.04 0.02

~0

30

50

60

Outcome

70

80

Figure 19.7: An approximate Gaussian distribution (the outcomes of the sum of face values in rolling 15 dice). tion of data, estimates and significance tests, and relationship between two or more variables by analysis of variance, and the test of hypotheses.

19.2.1

Sample Mean and Variance

If a sample consists of n independent observations Xt, x2, ... , Xn on a random variable x such as the price of a cup of coffee, two important and commonly used parameters are sample mean and sample variance, which can easily be estimated from the sample. The san1ple mean is calculated by

1

X

=<x>= -(Xt + X2 + ... + Xn) n

1

n

= - LXi,

n

(19.67)

i=l

which is essentially the arithmetic average of the values Xi. Generally speaking, if u is a linear combination of n independent random variables Yt, Y2, ... , Yn and each random variable Yi has an individual mean J.li and a corresponding variance 290


a[,

19.2 Statistics

we have the linear combination n U

=

L O:iYi

+ G2Y2 + ··· + GnYn,

= O:tYl

(19.68)

i=l

where the parameters ai(i = 1, 2, ... , n) are the weighting coefficients. From the central limit theorem, we have the mean J.lu of the linear combination n

n

J.lu = E(u) = E(L GiYi) = L aE(yi) = L GiJ.li· i=l

(19.69)

i=l

Then, the variance u~ of the combination is n

u~ = E[(u- J.lu) 2] = E[L o:i(Yi- J.li) 2],

(19.70)

i=l

which can be expanded as n

a~ =La~ E[(Yi - J.li)2] i=l n

+

L

O:iGjE[(Yi- J.li)(Yj- J.lj)],

i,j=l;i#j

(19.71)

ar

where E[(Yi - J.Li) 2 ] = Since Yi and Yi are independent, we have E[(Yi - J.li)(Yi - J.li )] = E[(Yi - J.li)]E[(Yi - J.li )] = 0. Therefore, we get n

u~ =

L:a'faJ.

(19.72)

i=l

The sample mean defined in equation (19.67) can also be viewed as a linear combination of all the Xi assuming each of which has the same mean J.li = J.l and variance u'f = u 2 , and the same weighting coefficient Gi = 1/n. Hence, the sample mean is an unbiased estimate of the sample due to the fact J.lx = Ef=t J.L/n = J.l. In this case, however, we have the variance n

1

0"2

0"~ = ~ -0"2 = x

~ n2 t=l

291

n'

(19.73)

19.2 Statistics


which means the variance becomes smaller as the size n of the sample increases by a factor of 1ln. The sample variance 8 2 is defined by

S2

1 ~ n- 1 i=l

-2

= - - L...,(xi- x) .

(19.74)

It is worth pointing out that the factor is 1l(n -1) not 1ln because only 1I (n - 1) will give the correct and unbiased estimate of the variance. From the probability theory in the earlier sections, we know that E[x 2] = J..L 2 + a 2. The mean of the sample variance is J..ls2 =

1~ 2 E[-L...,(xi-x) ] n- 1 i=l

Using E[x 2] =

J..L

2

+ a 2In,

Jls2 =

1~ 2 2 E[(xi -nx )]. (19.75) 1 n- i=l

= --L

we get

n ~ 1 ~ { E[x~] - nE[x2 ]}

1 a2 n(J..l2 + -)} = a2. (19.76) n-1 n 0 bviously, if we use the factor 1In instead of 1I(n - 1), we would get J..l 82 = n~ 1 a 2 < a 2, which would underestimate the sample variance. The other way to think the factor 1I (n - 1) is that we need at least one value to estimate the mean, we need at least 2 values to estimate the variance. Thus, for n observations, only n - 1 different values of variance can be obtained to estimate the total sample variance.

= --{ n(J..l2 + a2) -

19.2.2

Method of Least Squares

Maximum Likelihood For a sample of n values Xt,X2, .•• ,Xn of a random variable X whose probability density function p( x) depends on a set of k parameters f3t, ... , f3k, the t probability is then

292


19.2 Statistics

= p(xt, f3t, ... , !3k)p(x2, f3t, ... , !3k) · · · p(xm f3t, ... , f3k)·

(19.77)

The essence of the maximum likelihood is to maximize 4> by choosing the parameters /3i. As the sample can be considered as given values, the n1aximum likelihood requires that

84>

0/3i = 0,

(i = 1, 2, ... , k),

(19.78)

whose solutions for /3i are the maximum likelihood estimates. Linear Regression

For experiments and observations, we usually plot one variable such as pressure or price y against another variable x such as time or spatial coordinates. We try to present the data in a way so that we can see some trend in the data. For n sets of data points (Xi, Yi), the usual practice is to try to draw a straight line y =a+ bx so that it represents the major trend. Such a line is often called the regression line or the best fit line as shown in Figure 19.8. y

Figure 19.8: Least square and the best fit line. The method of least squares is to try to determine the two parameters a (intercept) and b (slope) for the regression line from n data points. Assuming that Xi are known more precisely and Yi values obey a normal distribution around the potentially best fit line with a variance a 2 . Hence, we have the probability (19.79) 293

19.2 Statistics


where A is a constant, and f(x) is the function for the regression [f (x) = a + bx for the linear regression]. It is worth pointing out that the exponent L:i=dYi- f(xi)] 2 ja2 is similar to the quantity X~ defined in the x2-distribution. The essence of the method of least squares is to maximize the probability P by choosing the appropriate a and b. The maximization of P is equivalent to the minimization of the exponent 'l/J n

'l/J = L)Yi - f(xi)f ·

(19.80)

i=l

We see that 'l/J is the sum of the squares of the deviations f.r = (Yi- f(xi)) 2 where f(xi) =a+ bxi. The minimization means the least sum of the squares, thus the name of the method of least squares. In order to minimize 'l/J as a function of a and b, its derivatives should be zero. That is

8'1/J n 7) = -2L[Y- (a+ bxi)] = 0, a

(19.81)

i=l

and (19.82)

By expanding these equations, we have n

n

na+bLxi = LYi,

and

n

a LXi i=l

i=l

i=l

n

n

+ b L:x~ = LXiYi, i=l

(19.83)

(19.84)

i=l

which is a system of linear equations for a and b, and it is straightforward to obtain the solutions as (19.85) 294


19.2 Statistics

where

1

n

y = - LYi·

(19.87)

n i=l

If we use the following notations

(19.88) and

n

n

Kxx =LX~,

Kxy = LXiYi,

i=l

(19.89)

i=l

then the above equation for a and b becomes KxxKy -KxKy a= n K X X - (KX )2 '

(19.90)

The residual error is defined by fi

= Yi - (a + bxi),

(19.91)

whose sample mean is given by 1

J-te

n

L n i=l

=-

fi

1

1

n

n

n

i=l

= -Yi - a - b- L Xi

= y - a - bx = [y - bx] - a = 0.

(19.92)

The sample variance 8 2 is (19.93) where the factor 1/ (n - 2) comes from the fact that two constraints are need for the best fit, and the residuals therefore have a n - 2 degrees of freedom. 295


19.2 Statistics

Correlation Coefficient The correlation coefficient rx,y is a very useful parameter to find any potential relationship between two sets of data Xi and Yi for two random variables x and y, respectively. If x has a mean J..lx and a sample variances;, andy has a mean J..ly and a sample variance s~' the correlation coefficient is defined by (19.94)

where cov(x, y) = E[(x - J..lx](Y - J..ly) is the covariance. If the two variables are independent cov(x, y) = 0, there is no correlation between them (rx,y = 0). If ri,y = 1, then there is a linear relationship between these two variables. r x,y = 1 is an increasing linear relationship where the increase of one variable will lead to increase of another. r x,y = -1 is a decreasing relationship when one increases while the other decreases. For n sets of data points (Xi, Yi), the correlation coefficient can be calculated by

or

rx,y where Kyy

nKxy -KxKy

= --;.===::::::::::=======,

J(nKxx- K';)(nKyy- K~)

= Ei=l

(19.96)

yr

0 Example 19.7: Is there any relationship between shoe size and height among general population? By collecting data randomly among our friends, we have the following data:

Height (h): 162, 167, 168, 171, 174, 176, 183, 179 (em); Shoe size (s): 5.5, 6, 7.5, 7.5, 8.5, 10, 11, 12. From these data, we know the sample mean J.lh = 172.5, J.Ls = 8.5.

296


19.2 Statistics

0.15....--------.-----r----r-------.

~ 0.1 Q.

c::

.!!

~

c!o.os

Figure 19.9: Confidence interval 'Y = 1 -a. The covariance cov(h, s) = E[(h- J..lh)(s- J..ls)] = 13.2. \Ve also have the standard deviation of height sh = 6.422 and the standard deviation of shoe size Ss = 2.179. Therefore, the correlation coefficient r is given by 13.2 :::: _ _ r = cov(h, s) : : : : 0 94 ShSs 6.422*2.179 This is a relatively strong correlation indeed.

19.2.3

D

Hypothesis Testing

Confidence Interval The confidence interval is defined as the interval fh ~ X ~ 02 so that the probabilities at these two limits fh and fh are equal to a given probability 'Y = 1- a (say, 95% or 99%). That is (19.97) The predetermined parameter 1' is always near 1 so that it can be expressed as a small deviation a << 1 from 1 (see Figure 19.9). If we choose 1' = 95%, it means that we can expect that about 95% of the sample will fall in the confidence interval while 5% of the data will not. 297


19.2 Statistics

~ ~

If a

For the standard normal distribution, this means P( -8 ~ 8) = 1- a, so that a 4>(~ ~ 8) = 1 - 2" (19.98)

= 0.05, we have ( ~

~

8)

= 0.975 or 8 = 1.960. That is to x ~ J..L + 8a. We also know that

say, -8 ~ ~ ~ 8 or J..L - 8a ~ if you repeat an experiment n times, the variance will decrease from a 2 to a 2 jn, which is equivalent to say that the standard deviation becomes a/ .Jii for a sample size n. If a = 0.01, then 8 = 2.579, we have a a J..L- 2.579 Vn ~ X ~ J..L + 2.579 .;n· (19.99) On the other hand, for 8 = 1, we get J..L- a ~ x ~ J..L +a and 1' = 0.682. In other words, only 68.2% of the sample data will fall in the interval [J..L- a, J..L +a] or

x

= J..L

±a,

(19.100)

with a 68.2% confidence level. It is conventional to use 1 = 0.95 for probably significant, 0.99 for significant, and 0.999 for highly significant. 0 Example 19. 8: The sample data of the time taken for a quick lunch at a restaurant are as follows: 19, 15, 30, 20, 15, 23, 28, 22, 23 minutes. Suppose you want to attend a lecture at 12:30, at what time you should start your order if you want to take 5% chance of being late? The sample mean is J.L =

x=

1 g(19 + 15 + 30 + 20 + 15 + 23 + 28 + 22 + 23) = 21.67.

The sample variance is

which gives a standard deviation of a = 5.15 minutes. If you are willing to take 5% chance, then (~) = 0.95, it gives~ = 1.645. So you shall start x = J.L +~a = 30.15,

298


19.2 Statistics

which is about 30 minutes earlier or at about 12:00.

0

Student's t-distribution The Student's t-test is a very powerful method for testing the null hypothesis to see if the means of two normally distributed samples are equal. This method was designed by W. S. Gosset in 1908 and he had to use a pen name 'Student' because of his e1nployer's policy in publishing research results at that time. This is a powerful method for hypothesis testing using smallsize samples. This test can also be used to test if the slope of the regression line is significantly different from 0. It has become one of the most popular methods for hypothesis testing. The theoretical basis of the t-test is the Student's t-distribution for a sample population with the unknown standard deviation a, which of course can be estimated in of the sample variance 8 2 from the sample data. For n independent measurements/data Xt, x2, ... , Xn with an estimated sample mean x and a sample variance 8 2 as defined by equation (19.74), the t-variable is defined by X-J.L t = (Sf Vn).

(19.101)

The Student's t-distribution with k = n - 1 degrees of freedom is the distribution for the randon1 variable t, and the probability density function is (19.102) It can be verified that the mean is E[t] = 0. The variance is a 2 = kf(k- 2) fork> 2 and infinite for 0 < k ~ 2.

The corresponding cumulative probability function is

F(t)

=

r( k+t)

-r

v'f:rrr(k/2)

1t

(2

k

1

[1 + -]-+d(.

oo

299

k

(19.103)


19.2 Statistics

This integral leads to a hypergeometric function, and it is not straightforward to calculate, that is why they are tabulated in many statistical tables. For a confidence level of -y = 1 - a, the confidence interval is given by Q

F(O) = 1- 2"'

(19.104)

which is usually tabulated. For a = 0.05 and 0.01 (or 1- a/2 0.975 and 0.995), the values are tabulated in Table 19.2.

=

Table 19.2: Limits defined by F(O) = 1 - a/2 in equation (19.104). k

F(O)o.975

F(O)o.995

12.7 4.30 3.18 2.78 2.57 2.45 2.37 2.31 2.26 2.23 2.09 2.01 1.98 1.96

63.7 9.93 5.84 4.60 4.03 3.71 3.50 3.36 3.25 3.17 2.85 2.68 2.63 2.58

1 2 3 4 5 6 7 8 9 10 20 50 100 00

Suppose we are dealing with the 95% confidence interval, we have p( -0 ~ t ~ 0) = 1 - a = 0.95 or p(t ~ 0) = 1 a/2 = 0.975, we have 0 = ta,k = 12.70(k = 1), 4.30(k = 2), 3.18(k = 3), ... , 2.228(k = 10), ... , 1.959 for k--+ oo. Hence,

s

s

- < vn _,...,II+ 0-. ,II. ., - 0rn300

(19.105)


19.2 Statistics

This is much more complicated than its counterpart of the standard normal distribution.

Student's t-test There are quite a few variations of the Student's t-test, and most common are the one sample t-test and the two sample t-test. The one sample t-test is used for measurements that are randomly drawn from a population to compare the sample mean with a known number. In order to do statistical testing, we first have to pose precise questions or form a hypothesis, and such hypothesis is conventionally called the null hypothesis. The basic steps of a t-test are as follows: 1. The null hypothesis: Ho: J.L = J.Lo (often known value) for one sample, or H0 : f..Lt = f..L2 for two samples; 2. Calculate the t-test statistic t and find the critical value () for a given confidence level 1 = 1 - o by using F( t ::s; 0) = 1- of2; 3. If ltl > 0, reject the hypothesis. Otherwise, accept the hypothesis. 0 Example 19.9: A group of candidates (say, more than 100 students) have claimed to have an averaged IQ of 110 (or J..Lo = 110). Then, you randomly sample 11 students to do the IQ test and results are: x = IQ = 106,112, 103,108,108, 109,100, 106,106,99,101. Test the hypothesis:

Ho: J..L= J..Lo, at a confidence level of95%. From the data, we know that n Then, we have

= 11, x =

105.273, S

t = (x- J..L) = (105.273- 110) ~ _ _ _ 3 846 (S/.fii) 4.077/JIT

301

= 4.077.

19.2 Statistics


We only use the positive value if we look at the statistical tables. lVe also know for k = n- 1 = 10 degrees of freedom at a 95% confidence level, (} = 2.228. At a 95% confidence level, the probability oft>(} is 0.025 (or 2.5%) and the probability t < -8 is also 0.025. Thus, the hypothesis is not valid at a 95% confidence level. At the same level of confidence, the true mean J.Lo lies in in the range of x- 2.228 * Sj.J[f ~ J.Lo ~ x + 2.2288/.Jff. or

102.53

~ J.Lo ~

108.00. 0

Another important t-test is the two-sample paired test. Assuming that two pairs of n sample data sets Ui and Vi are independent and drawn from the same normal distribution, the paired t-test is used to determine whether they are significantly different from each other. The t- variable is defined by

n(n- 1) """~ (U-.t. _ iT.)2' 6t=l Vi where {fi = Ui - [J and 1% = Vi -

sd2 =

(19.106)

V. In addition,

1 ~ n- 1 i=l

- - L.)Ui-

-

Vi) 2 .

(19.107)

This is equivalent to apply the one-sample test to the difference ui - Vi data sequence. 0 Example 19.10: A novel teaching method of teaching children science was tried in a class (say class B), while a standard method was used in another class (say class A). At the end of the assessment, 8 students are randomly drawn from each class, and their science scores are as follows: Class A: Ui = 76, 77, 76, 81, 77, 76, 75, 82; Class B: Vi= 79, 81, 77, 86, 82, 81, 82, 80. At a 95% confidence level, can you say the new method is really better than the standard method? If we suppose that the two methods do not produce any difference in results, that is to say, their mean are the same. Tl1us the null

302


19.2 Statistics

hypothesis is:

Ho: J..L.-\ = JlB· lVe know that V = 77.5, Sd = 2.828. We now have

t

=

[J-

V = 81. The combined sample VBriance

v = 77.5-81 = -3 5

Sd/Vn

2.828/v'8

..

We know from the statistical table that the critical value 8 = 2.37 for F(8) = 1 - o:/2 and k = n - 1 = 7. As t < -8 or t > 8, we can reject the null hypothesis. That is to say, the new method does produce better results in teaching science. 0

The variance analysis and hypothesis testing are important topics in applied statistics, and there are many excellent books on these topics. Readers can refer to the relative books listed at the end of this book. It is worth pointing out that other important methods for hypothesis testing are Fisher's F-test, x2-test, and non-parametric tests. What we have discussed in this chapter is just a tip of an iceberg, however, it forms a solid basis for further studies.

303

References Abramowitz M. and Stegun I. A., Handbook of Mathematical Functions, Dover Publication, (1965). Arfken G., Mathematical Methods for Physicists, Academic Press, ( 1985). Ashby lvi. F. and Jones D. R., Engineering Materials, Pergamon Press, (1980). Atluri S. N., Methods of Computer Modeling in Engineering and the Sciences, Vol. I, Tech Science Press, 2005. Audoly B. and Neukirch S., Fragmentation of rods by cascading cracks: why spaghetti do not break in half, Phys. Rev. Lett., 95, 095505 (2005). Bathe K. J., Finite Element Procedures in Engineering Analysis, Prentice-Hall, (1982). Carrrier G. F. and Pearson C. E., Partial Differential Equations: Theory and Technique, 2nd Edition, Academic Press, 1988. Carslaw, H. S., Jaeger, Conduction of Heat in Solids, 2nd Edition, Oxford University Press, (1986). Chriss N., Black-Scholes and Beyond: Option Pricing Models, Irwin Professional Publishing, (1997). 305

19.2 Statistics


Courant R. and Hilbert, D., Methods of Mathematical Physics, 2 volumes, Wiley-lnterscience, New York, (1962). Crank J., Mathematics of Diffusion, Clarendon Press, Oxford, (1970). Devaney R. L., An Introduction to Chaotic Dynamical Systems, Redwood, (1989). Drew, D. A., Mathematical modelling of two-phase flow, A. Rev. Fluid Mech., 15, 261-291 (1983). Fenner R. T., Engineering Elasticity, Ellis Horwood Ltd, (1986). Fowler A. C., Mathematical Models in the Applied Sciences, Cambridge University Press, (1997). Farlow S. J., Partial Differential Equations for Scientists and Engineers, Dover Publications, (1993). Fletcher, C. A. J., Fletcher C. A., Computational Techniques for Fluid Dynamics, Vol. I, Springer-Verlag, GmbH, (1997). Gardiner C. W., Handbook of Stochastic Methods, Springer, (2004). Gershenfeld N., The Nature of Mathematical Modeling, Cambridge University Press, (1998). Goldreich P. and S. Tremaine, The dynamics of planetary rings, Ann. Rev. Astron. Astrophys., 20, 249-83 (1982). Goodman R., Teach Yourself Statistics, London, (1957). Gleick J., Chaos: Making a New Science, Penguin, (1988). Hinch E.J., Perturbation Methods, Cambridge Univ. Press, (1991). Hull J. C., Options, Futures and Other Derivatives, PrenticeHall, 3rd Edition, (1997). 306


19.2 Statistics

Jeffrey A., Advanced Engineering Mathematics, Academic Press, (2002). John F., Partial Differential Equations, Springer, New York, (1971 ). Jouini E., Cvitanic J. and l\.fusiela NL, Handbook in Mathematical Finance, Cambridge Univ Press, (2001). Kardestruncer H. and Norrie D. H., Finite Element Handbook, l\.fcGraw-Hill, (1987). Keener J., Sneyd J., A Mathematical Physiology, SpringerVerlag, New York, (2001). Korn G. A. and Korn T. M., Mathematical Handbook for Scientists and Engineers, Dover Publication, (1961). Kreyszig E., Advanced Engineering Mathematics, 6th Edition, Wiley & Sons, New York, (1988). Kant T., Finite Elements in Computational Mechanics. Vols. 1/11, Pergamon Press, Oxford, (1985). Langtangen, H P, Computational Partial Differential Equations: Numerical Methods and Diffpack Programming, Springer, (1999). LeVeque R. J., Finite Volume Methods for Hyperbolic Problems, Cambridge University Press, (2002). Lewis R. W., Nlorgan K., Thomas H., Seetharamu S. K., The Finite Element Method in Heat Transfer Analysis, Wiley & Sons, (1996). Mandelbrot B. B., The Fractal Geometry of Nature, W.H. Freeman ( 1982). Mitchell A. R. and Griffiths D. F., Finite Difference Method in Partial Differential Equations, Wiley & Sons, New York, (1980). 307

19.2 Statistics


Ivloler C. B., Numerical Computing with MATLAB, SIAM, (2004). I\1urray J. D., Mathematical Biology, Springer-Verlag, New York, (1998). Ockendon J., Howison S., Lacey A., and lVIovchan A., Applied Partial Differential Equations, Oxford University Press, (2003). Pallour J.D. and lVIeadows D. S., Complex Variables for Scientists and Engineers, lVIacmillan Publishing Co., (1990). Papoulis A., Probability and statistics, Englewood Cliffs, (1990). Pearson C. E., Handbook of Applied Mathematics, 2nd Ed, Van Nostrand Reinhold, New York, (1983). Press W. H., Teukolsky S. A., Vetterling W. T., Flannery B. P., Numerical Recipe in C++: The Art of Scientific Computing, 2nd Edition, Cambridge University Press, (2002). Puckett E. G., Colella, P., Finite Difference Methods for Computational Fluid Dynamics, Cambridge University Press, (2005). Riley K. F., Hobson M. P., and Bence S. J., Mathematical Methods for Physics and Engineering, 3rd Edition, Cambridge University Press (2006). Ross S., A first Course in Probability, 5th Edition, PrenticeHall, (1998). Selby S. I\1., Standard Mathematical Tables, CRC Press, (1974). Sicardy B., Dynamics of Planetary Rings, Lecture Notes in Physics, 682, 183-200 (2006). Strang G. and Fix G. J., An Analysis of the Finite Element Method, Prentice-Hall, Englewood Cliffs, NJ, (1973). 308


19.2 Statistics

Smith, G. D., Numerical Solutions of Partial Differential Equations: Finite Differerence Methods, 3rd ed., Clarendon Press, Oxford, (1985). Thomee V., Galerkin Finite Element Methods for Parabolic Problems, Springer-Verlag, Berlin, (1997). Terzaghi, K., Theorectical Soil Mechanics, New York, Wiley, (1943). Turcotte. D. L. & Schubert, G., Geodynamics: Application of Continuum Physics to Geological Problems, 1st. ed., John Wiley, New York, (1982). Weisstein E. W., http:/ jmathworld.wolfram.com Wikipedia, http:/ jen.wikipedia.com Wylie C. R., Advanced Engineering Mathematics, Tokyo, (1972). Versteeg H. K, Malalasekra W., 1vialalasekra W., An Introduction to Computational Fluid Dynamics: The Finite Volume Method, Prentice Hall, (1995). Yang X. S., Young Y., Cellular automata, PDEs and pattern formation (Chapter 18), in Handbook of Bioinspired Algorithms, edited by Olarius S. and Zomaya A., Chapman & Hall I CRC, (2005). Yang X. S., An Introduction to Computational Engineering with Matlab, Cambridge Int. Science Publishing, (2006). Zienkiewicz 0 C and Taylor R L, The Finite Element Method, vol. 1/11, IvicGraw-Hill, 4th Edition, (1991).

309

Appendix A

Mathematical Formulas This is a summary for the mathematical formulas that have appeared in various sections in this book. We list these formulas for your easy reference.

A.l

Differentiations and Integrations

Differentiation Rules:

+ uv

1

1

(uv) = u v

( ~)I= v 1

{f[g(x)]} =

I

u v- uv

1

I

v2

! [g(x)] · g1(x) 1

Leibnitz's Theorem:

d~n (uv) = u(n)v + nu
( n) r

=

)u(n-r)v(r)

n! r!(n- r)!

Integration by parts

1 b

dv u dx dx

b

= [uv] a +

l

311

1 b

du v dx dx

A .2 Vectors and !via trices

!v.Iathematical Formulas

Differentiation of an integral d 1b(x) db da 1b(x) {)u(x, y) -d u(x,y)dy = [u(x,b)-d -u(x,a)-d ]+ dy X

a(x)

X

0X

a(x)

X

Power Series ~

e-

zn

z2

= 1

+ z + -21. + ... + 1n. ...

z3

zs

sin z

= z - 3J + 5f - ... ,

sinh z

= z + 3J + 5f + ... ,

z3

cos z

=

(z E C)

z4

z2

1-

2f + 41 - ...

zs

Complex Numbers ei 0 =cosO+ isinO, z

[ei1r

+1=

0].

= x + iy = rei 0 = r(cosO + isinO)

De Jvloivre's formula:

[r( cos 0 + i sin O)t = rn( cos nO+ i sin nO)

A.2

Vectors and Matrices

Dot Product

Cross Product

a x b = nlallbl sin 0 = Eijkajbk =

312

i

j

a1

a2

bt

b2

k a3 b3

1\t/athematical Formulas

A .2 Vectors and lvlatrices

Vector Triple

[a, b, c]

= a · (b x c)

=

a1

a2

a3

bt

b2

b3

CJ

C2

C3

a· (b x c)= b · (c x a)= c ·(ax b)= -a· (c x b) ax (b x c)= (a· c)b- (a· b)c Divergence Theorem of Gauss

fffv V·udV= jfsu·dS Stokes's Theorem

Green's Theorems

f (udx + vdy)

=

If

8v au (ax - ay )dxdy

Identities

v. v

Xu= 0,

v

XV¢>= 0

V X (if>u) = (j>V Xu+ (V¢) xu V · (¢u) =

4>V · u + (V4>) · u

V x (V x u) = V(V · u) - V 2 u Inverse, Trace and Determinants

IAB .... ZI

= IAIIBI ... IZI,

A vi= Aivi,

IAI

= detA

eig(AB) = eig(BA) 313

A.3 Asymptotics

!v.Iathematical Formulas

tr(AB) = tr(BA),

tr(A +B) = tr(A) + tr(B)

1

det(A - 1) =

det(AB) = det(A)det(B)

det(A)'

Exponential Matrices

1

00

etA=

L

=L

oo

ln(IA)

~

(tAt = l+tA+ - A 2 + ... 1 2 n=O n.

(-1)n-1 nl

1 1 An= A- 2A2 + 3A3 + ...

n=1

eAeB = eA+B

~etA dt

(if AB = BA)

= AetA = etA A

(eA)-1 = e-A,

A.3

det(eA) = etrA

Asymptotics

Gaussian Distribution 1

p(x; J-t, a)= ~ exp[-

(x- J-t) 2 2a 2

27ra2

1

00

-oo

00

1

-oo

2n -ax2 x e dx

=

e -QX2dx=

]

---+

<5(x)

l1---+ 00.

If0:

(-1)n ·1· 3· · · (2n -1) 2n

r-:;r V7n+T'

(n > 0).

Binomial Distribution

B(k ; n,p ) = (n-n!k)tkt p k(1 - p )n-k ' (k = 0' 1' 2' ... , n ) 314

1\t/athematical Formulas

B(x; n ~ oo, p)l

np»l

A .4 Special Integrals

"'p(x; J.L, a), p, = np, a 2 = np(1 - p).

Poisson Distribution

e-.XAx f(x;A) = - - , 1 X.

f(x; A>> 1) : : : : p(x; p,, a), B(k

A.4

~

x; n

~

x

EN a 2 =A

p, = A,

oo,p) "'f(x; A= np),

(lim np =A) n-oo

Special Integrals

Gamma Function

r(x) = fooo tx-le-tdt = fooo e-t+(x-1) Intdt r(n + 1) = nl, r(-

1

2) = -2y'7r,

r(

e

Stirling's Formula

(~tV27ffi, e

Error Functions

315

2) = v'1r

(x ~ oo).

r(x + 1)::::::: (=)xJ27rX,

nl:::::::

1

n

>> 1

Index 1-D, 209, 254, 255 2-D, 261 Airy stress function, 192 algorithms, 210 analytic function, 67 analytical function, 67 assembly by element, 246 asymptotic, 151 error function, 314 Gamma function, 314 Stirling's formula, 315 Bessel equation, 88 Bessel function, 24, 151 bifurcation, 99 binomial distribution, 281 birthday paradox, 273 Black-Scholes equation, 205 boundary condition, 252 essential, 248 natural, 248

multiple variables, 165 pendulum, 159 shortest path, 156 central difference, 212 central limit theorem, 287 chaos, 99 complex integral, 70 residue, 70 complex variables, 62 coordinates cylindrical, 15 polar, 15 spherical, 15 correlation coefficient, 295 cross product, 30 cumulative probability function, 284 curl, 38 curvature, 153

determinant, 49 difference equation, 95 differential operator, 84 calculus of variations, 153 brachistochrone problem, 164lifferentiation, 1 implicit, 4 constraint, 160 partial, 9 curvature, 153 163 2 rule, problem, Dido's hanging rope problem, 163 vector, 32 316

INDEX

diffusion equation, 146 displacement, 239 divergence, 38 divergence theorem, 313 dot product, 28, 30 DuFort-Frankel scheme, 224 dynamic reconstruction, 102 dynamical system, 102

INDEX

gradient, 38 Green's function, 148 Green's identity, 41 Green's theorem, 313

harmonic motion, 109. 159 heat conduction, 139, 223, 253 hybrid method, 149 hyperbolic equation, 224 elastic wave, 203 first-order, 214 elasticity, 181, 240 second-order, 215 beam bending, 197 hyperbolic function, 65 Cauchy-Navier equation, 19J.hypothesis testing, 297 elastostatic, 185, 198 Euler-Bernoulli theory, 196 index matrix, 243 Hooke's law 181 inner product, 28 ' Maxwell-Betti theorem, 185 integral multiple, 12 strain tensor, 182 Bessel function, 88 stress tensor, 182 stress-strain relationship, 184 Cauchy's theorem, 71 differentiation, 12 elliptic equation, 218 Gaussian. 18 error function, 20 line, 38 Euler scheme, 210 residue theorem, 73 Euler-Lagrange equation, 154 special, 17 exponential distribution, 286 integral equation, 153, 167 displacement kernel, 169 finite difference method, 209 Fredholm equation, 167 finite element method, 227, 248 sparable kernel, 169 finite volume method, 221 Volterra equation, 168, 170 Fokker-Plank equation, 205 integral transform Galerkin method, 238 Fourier, 125 Gamma function, 22 Fourier transform, 144 Gauss's theorem, 41 Laplace, 131 Gauss-Seidel iteration, 219 Laplace transform, 143 Gaussian distribution, 283 wavelet, 134 317

INDEX

integration, 5 by parts, 6 iteration method, 219 Jacobian, 13 kinematics, 33 Lagrangian, 158 Lame constant, 184 Laplace's equation, 139, 201 Laurent series, 69 leap-frog scheme, 212 least-square, 238 Leibnitz theorem, 3 linear difference equation, 95 linear system, 56 log-normal distribution, 286 Lorenz attractor, 103

INDEX

ordinary differential equation complementary function, 81 general solution, 81 homogenous equation, 81 linear system, 85 particular integral, 81 oscillation damped, 112 forced, 109 natural frequency, 112 small amplitude, 119 undamped, 109 outer product, 30

parabolic equation, 202, 216 pattern formation, 260 bifurcation, 264 instability, 263 pattern formation , 261 mathematical model, 201 PDE, 138, 141, 203, 213 matrix, 47 Poisson distribution, 281 exponential, 52 Poisson's equation, 201, 244 Hermitian, 53 probability, 267 inverse, 50 axiom, 270 Maxwell's equations, 204 conditional, 271, 275 mean, 278 distribution, 279 method of least square, 292 independent events, 271 moment generating function, 280 moment, 280 l\1onty hall problem, 276 Navier-Stokes equation, 206 permutation, 272 Navier-Stokes equations, 206 random variable, 268, 277 node, 242 randomness, 267 normal distribution, 283 probability density function, 283 normal modes, 116 null hypothesis, 301 quadratic form, 55 ODE, 77, 80, 81

random variables, 277 318

INDEX

INDEX

reaction-diffusion, 257, 262 Sturm-Liouville equation, 86 reaction-diffusion equation, 204 Taylor series, 69 recurrence equation, 95 tensor, 173 Residue theorem, 73 analysis, 175 Riccati equation, 77 Cartesian, 175 Riemann (-function, 68 notations, 173 Rien1ann hypothesis, 69 rank, 175 Runge-Kutta method, 210, 213 vector, 176 Saturn's rings, 43 three card problem, 276 Schrodinger equation, 206 time-dependent, 251 self-similarity, 105 time-stepping, 217, 253 separation of variables, 141 explicit, 254 series implicit, 211 asympottic, 17 transient, 254 power, 8 Travelling wave, 147 Talor, 8 travelling wave, 259 shape functions, 238 triangular element, 240 similarity solution, 145 uniforn1 distribution, 286 Sine-Gordon equation, 207 upwind scheme, 214 soliton, 147 stability condition, 211, 215 standard normal distribution, 285 statistics, 289 confidence interval, 297 linear regression, 293 maximum likelihood, 292 sample mean, 289 sample variance, 289 steady state, 245 stiffness matrix, 242 Stokes's theorem, 41 stress intensity factor, 195 Student's t-distribution, 299 Student's t-test, 301

variance, 278 vector, 27, 29 triple, 31 vector calculus, 32 Venn diagram, 268 vibration, 109 wave equation, 139, 202, 203, 215, 255 weak forn1ulation, 236 Young's modulus, 181

319

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