Bernatz R. Fourier Series and Numerical Methods for Partial Differential Equations

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172 HEAT TRANSFER IN 2D AND 3D

Introducing the separation constant —λ, the following ODEs, with transformed BCs,

result.

Y"(y) + XY(y) = 0 y(0) = 0 and Y(d) = 0 (6.115)

X"(y)-

(| + λ)χ(χ)=0 Χ(0) = 0 (6.116)

Equation (6.115) implies Y

(y) — sin(a

?/), with a

— ^, and Equation (6.116)

implies X

{x) = sinh(y/| + a^x). The solution will be complete once the trans-

formed boundary condition at x = c is satisfied. Consequently, it must be that

W*(c,y,s)

- ¿ A^inh (Jl+

) sin(a

y) = £[1] = 1 (6.117)

where

1 Γ* 1

sinhíy/f +

o%c)

With the coefficients determined, the series solution for W*(x, y, s) is given by

*(x,2/,s) = -^Cnsinhi W- + a^x) sm(a

y) (6.118)

where

/■<*

sinn^l + a¿c) Jo

The solution w*(x, y, t) to the constant BC version of IBVP (6.112) is the inverse

Laplace transform of

W*(x,y,s).

That is, w*(x,y,t) = C~

\W * (x,y,s)].

For the case of the time-dependent

BC,

as shown in IBVP

(6.112),

the solution pro-

cess through Laplace transformations outlined above remains identical until Equation

(6.117),

where the Laplace transformation £[l] is replaced with G(s)

—

C[g(t)\. Be-

cause G(s) is independent of

the resulting Fourier series solution for C[w(x, y,

t)]

C[w(x,y,t)] = G(s) ^ C

sinh IJ- + a^x\ sm{a

y) (6.119)

Instead of finding the inverse transformation at this time, consider the following

relationship:

s-C[w(x,y,t)} = G(s)sW(x,y,s) = G(s)sC[w*(x,y,t)} (6.120)

Using the fact that s£[w*(x, y, t)] = C[w^(x, y, t)], Equation (6.120) may be written

C[w(x,y,t)] = C[g(t)}Ciw*

(x,y,t)} (6.121)

Because the Laplace transform of a finite convolution is the product of the Laplace

transforms, it follows from Equation (6.121) that

w(x,y,t) — I g(t

—

r)w*(x,t,T)dr (6.122)

HOMOGENEOUS 3D IBVP 173

Integration by parts allows the integral formula in Equation (6.123) to be rewritten as

x,y,t)=

\ g

(t-T)w*(x,t,r)dT + g(0)w*(x,y,t)-g(t)f(x,y). (6.123)

The solution v(x,y, t) of the homogeneous version of IBVP (6.111) is found using

methods of Section 6.2. The solution u(x, y, t) of IBVP (6.111) is the sum of the

solutions v(x, y, t) and w(x, y, t).

6.7 HOMOGENEOUS 3D IBVP

The methods of separation of variables and Fourier series are easily extended to

IBVPs in three spatial dimensions. As usual, the methods are first described for an

IBVP having a homogeneous PDE and homogeneous BCs. The BCs given in IBVP

(6.124) are Dirichlet in each instance. The methods described for this case are valid

for each of the other 728 possible homogeneous BC cases.

IBVP<

= kSJ

u(x,y,z,0) = f{x,y,z)

u(0,y,z,t)

—

Wa,y, ζ,τ) = 0

i¿(x,0,

ζ,τ) = 0

u¡x,b,z,t) = 0

ulx, y,0, t) = 0

w(x,i/,c, t) = 0

0<y <b, 0<z<c

0<y <

0<z<c

0<x<a,

0<z<c

0<x<a,

0<z<c

0<x<a,

0<y <b

0<x<a,

0<y <b

(PDE)

(IC)

(BC1)

(BC2)

(BC3)

(BC4)

(BC5)

(BC6)

(6.124)

Assuming a separable solution of the form u(x,y,z,t) — X(x)Y(y)Z(z)T(t),

the PDE of IBVP (6.124) becomes

1 T'(t) _ X"{x)

Q/)

Z»(z)

kT(t) X(x) Y{y) Z(z)

Introduce the separation constant λ, and split Equation (6.125) in two, giving

1 T'(t)

(6.125)

and

k T(t)

X»(x) . Y»(y) , Z"(z)

= -λ.

(6.126)

(6.127)

X(x) Y(y) Z(z)

Equation 6.127 may be split three ways by letting —λ = —v

—

η and pairing

corresponding terms on the two side of the equation. The result is three ODEs given

in Equations (6.128).

(6.128)

X"(x)

+ vX(x) =

Y"(y) +

Y(y) --

Ζ"(ζ)+ηΖ(ζ) =

= 0

174 HEAT TRANSFER IN 2D AND 3D

These ODEs, paired with the homogeneous BCs, define regular Sturm-Liouville

problems. From previous ID and 2D instances of similar problems, the resulting

eigenfunctions and eigenvalues are

ΤΙΊΤ

(x)

= sin(a

x) where a

= —, n = 1,2,3...

ΤΤ17Γ

(y) = sm(ß

y) where ß

= —, m = 1,2,3... (6.129)

Ιπ

Ζι{χ) = sin(7/x) where 7/ = —, I = 1,2,3...

Now that the eigenvalues a

, ß

, and 7/ are known, it follows that — λ =

-(al + β^ + 7/

) and the solution to the ODE in T(t) is

T{t) =

+/3

(6.130)

Combining the solution for each variable results in the general solution for

u{x,y,z,t) given in Equation (6.131)

00 00 00

u{x,y,z,t) = 5Z ΣΖ X^^nm/sin(a

x)sin(/5

?/)sin(7/2;)e~

(a;

+/3

+7i)t

n=l m=l /—1

(6.131)

The solution to IBVP (6.124) will be given by a particular solution of form shown

in Equation 6.131 provided coefficients

can be found so that

00 00 00

u(x,

ν,ζ,0)

= ΣΣΣ

Anrnl sin

)

sin

(/?m2/) sin(7/2;) = /(x, y, z)

n=l ra=l 1=1

(6.132)

The orthonormal ity of the eigenfunctions allow the coefficients to be determined in

the usual way forming the integral inner product of both sides of the equation of the

triple sum and /(x, y, z). The resulting triple integral formula is

pc pb pa

Anmi= / / f(x,y,z)sm(a

x)sm(ß

y)sin('y

z)dxdydz (6.133)

Jo Jo Jo

As an example, suppose a = b = c=l, k = 0.01, and the initial condition

/(x,

y, z) in IBVP (6.124) isgivenby /(x, y, z) = /i(x, y, z)*f

(x,

z)*f

(x, y, z)

where

_ Γ 2x 0.3 < x < 0.7

ji[x,y,z) -|

otherwise

r ( x f 2y 0.3 < y < 0.7

f ,

_ Γ 2z 0.3 < z < 0.7

}3{x,y,z) - I 0 otherwise

Figure 6.22(a) shows the /(x, y,z) — 1 surface, and Figure 6.22(b) depicts the same

surface for the triple Fourier series representation of /(x, y, z) given by u(x, y, z, 0).

HOMOGENEOUS 3D IBVP 175

The limit on each index in the sum is 15, so the total number of terms in the partial

sum is 3375. The Fourier series u = 1 surface is rather "stair-stepped" compared to

the / = 1 surface due to the low-order approximation. Figure

6.22(c)

is a plot of the

u(x, y, z, 0.1) = 1 surface. As expected, the volume within this surface decreases as

time increases due to the negative exponential term in the expression for u(x,

y,z,t)

(c)

u(x,y,

0.1) = 1

Figure 6.22 Surface plots for (a) f(x,y,z) = 1, (b) u(x,y, z,0) — 1, and (c)

u(x,y,z,0.1) = 1.

176 HEAT TRANSFER IN 2D AND

EXERCISES

6.1

Explain

why the

solution

to the

Neumann problem

is,

generally,

not

unique.

6.2

Justify

the

general solvability statement given

Equation (6.81).

6.3

Figure

6.23

shows

the

problem domain

for a

general

Laplace's problem

with Dirichlet boundary conditions

on a

rectangular domain. What follows

is a

list

W(0,y)

(y)

W(x,d)=g

(x)

W(x,y)=0

W{x,0)=

(x)

W(c,y)=g

(y)

Figure 6.23 Laplace problem with Dirichlet BCs on a rectangular domain.

of Laplace BVPs with Dirichlet

BCs.

Each boundary function

gi has a

default

value

zero unless specified

the given exercise. Solve each

the

examples using

appropriate

methods.

Your solution must include (i)

specification of all subproblems

whose

sum

equals

the

original BVP (ii) a contour plot

the

steady-state temperature

surface,

(iii)

plots verifying

the

solution satisfies

the

boundary conditions,

and (iv)

the value

W(c/2,d/2).

c= l,d= 1,02(2/) =

2/(1

~y)

c = 5, d

1,^2(2/) =

2/(1 -2/)» ^4(2/)

= 1

c= 1, d=

gi(x) = 4x(l

—

x), gs(x) =

sin(27rx)

d) c=l,d

2,0

(2/H#[o.8,i.2](2/)

10,

d = 1,02(2/) =

^9s(x)

= 1

6.4

Figure

6.24

shows

the

problem domain

for a

general

Laplace's problem

with Neumann boundary conditions

on a

rectangular domain. What follows

a list

Laplace BVPs with Neumann

BCs.

Each boundary function

#¿ has a

default value

zero unless specified

in the

given exercise. Solve each

of the

examples using appropriate methods. Some examples may not require series solution

techniques. Your solution must include (i) consideration of the solvability criteria

(ii)

a specification

all subproblems whose

sum

equals

the

original BVP,

(iii) a

contour

plot

the steady-state temperature surface,

(iv) a

plot

the normal derivative

the

temperature field

for

each boundary that has

non-homogeneous boundary condition,

and

(v) the

value W(c/2,

d/2).

2, 03 (x)

—

EXERCISES 177

-W

(0,y) = g

(y)

{c,y)

= g

(y)

Figure 6.24 Laplace problem with Neumann BCs on a rectangular domain.

b) c= l,d = l,gs(x) = \ -x,94(y) = cosny

c) c = 2, d= \,g

(y) = y,94(y) = -\

d) c= l,d = \,gi{x) = 2,g

{y) = -3, g

(x) = -2, g

(y) = 3

e) c = 1, d = 1, </i(x) = -2x, 0

(y) = 2 - 3y

f) c = 2, d= l,flfi(x) = x,g

(y) = y- §

6.5 Figure 6.25 shows the problem domain for a general 2D Laplace BVP on

a rectangular domain. The four BCs of various types are prescribed in each of

the exercises below. Solve each of the examples using appropriate methods. Your

BC4

Figure 6.25 Laplace problem with various types of

BCs.

solution must include (i) a specification of all subproblems whose sum equals the

original BVP, (ii) a contour plot of the steady-state temperature surface, (iii) plots

that verify each of the four BCs are satisfied, and (iv) the value W(c/2, d/2).

a) c

1, d = 1, BC1: W = 0, BC2: W

= 0, BC3: W = x(l - x), BC4:

178 HEAT TRANSFER IN 2D AND 3D

x-\,

BC2: W = 0, BC3: W

= O, BC4:

b) c

1, d

1, BC1: -Wj, + W = 0, BC2: W = 0, BC3: W = x(l - x),

BC4:

W = 0

c) c=l,d=l,BCl: -Wj, + W = 0,BC2: W = 0,BC3: W

+ W = x,BC4:

d) c= 1, d = 1,BC1: -W,

W* = 0

e) c = 4, d = 1,

BC1:

-W

= 0, BC2: W = y(l - y), BC3: W = x, BC4:

Wx-0

f) c=l,d=l,BCl: -W

-W = 0,BC2: W = t/(l-2/),BC3: Vt^+W = 0,

BC4:

W = 0

6.6 Figure 6.26 shows the problem domain for a general 2D Poisson BVP on a

rectangular domain. The four BCs of various types are prescribed in each of the

exercises below. Solve each of the examples using appropriate methods. Your

2/f

BC4

BC3

W(x,y)

= f(x,y)

BC1

BC2

Figure 6.26 Poisson problem with various types of

BCs.

solution must include (i) a specification of all subproblems whose sum equals the

original BVP, (ii) a contour plot of the steady-state temperature surface, (iii) plots

verifying each of the four BCs are satisfied, and (iv) the value W(c/2, d/2).

a) c = 1, d = 1, BC1: W = 0, BC2: W = 0, BC3: W = x{\ - x), BC4:

W =

0,f(x,y)

= 10xy

b) c = 1, d = 1, BC1: W = 0, BC2: W* = 0, BC3: W = x(l - x), BC4:

= 0,/(x,y)-5x + 5^

c) c = 1, d = 1, BC1: W = a, BC2: W

= 0, BC3: W = 0, BC4:

-Μ^ + ^ =

0,/(Χ,Ϊ/)

= 10ΧΪ/

d) c = 2, d = 1, BC1: W = 0, BC2: W

= y - ±, BC3: VF = 0, BC4:

W^ = 0,/(x,2/)=x(2-x)

e) c=l,d= 1,BC1: W = 0, BC2: W* = 0, BC3: W* = 0, BC4: W

= 0,

/(x,

/) = 10(x-|)(|-2/)

EXERCISES 179

6.7 Solve the given nonhomogeneous IBVP using methods outlined in this chapter.

Your solution must include (i) a contour plot of u(x,y, 1), (ii) plots verifying that

your solution satisfies the given BCs, and (iii) the value u(c/2, d/2,1).

{x,y,t)

= 0.1

ti(x,S/,0) =0

t{x,y,t) + u

{x,y,t)\

IBVP<

u(x,0,t)

uh,y,t)

u(x, 1,£)

u(0,y,t)

(x,y,t)

u(x,y,0)

-x(l

■x)

0<x< 1

0 < y < 1

0 <x< 1

0 < y < 1

(PDE)

(IC)

(BCl)

(BC2)

(BC3)

(BC4)

0.1 [u

(x, y, t) +

(x,

y, t))

x + y

u(x, 0,i) = 0

uh,y,t) = y(2-2/)

u(x,2,¿) = x(l

—

u(0,2/,

¿)

—

{x,y,t)=

0.05V

IBVP<

u(x,2/,0) = #[0.4,0.6]

x [0.4,0.6]

0>2/)

u(x, 0,t) = 0

u(l,2/,í) = ?/(l -y)

u(x, Ι,η = x(l

—

i¿(0,2/,í) = 0

{x,y,t)

0.2V

w(x,2/,0) = 0

(PDE)

(IC)

0 < x < 1 (BCl)

0 < y < 2 (BC2)

0 < x < 1 (BC3)

0 < y < 2 (BC4)

(PDE)

(IC)

0<x < 1

0<y< 1

0 <x < 1

0<2/< 1

(BCl)

(BC2)

(BC3)

(BC4)

IBVP<

(x,0,t)=0

0<x<l

uJl,2/,tj+u(l,y,t) = l 0<y<l

%(χ,1,ί) =0 0<χ<1

(0,y,t)=0 0<2/<l

(PDE)

(IC)

(BCl)

(BC2)

(BC3)

(BC4)

IBVP<

ut{x,y,t) =

0.01V

(PDE)

u(x,y,0) = 2 (IC)

-0.1ι^(χ,

0, t) +

0.2u(x,

0, t) = 0 0 < x < 1 (BCl)

0.W1,2/, t) +

0.2w(l,

2/,

í) = 1 0 < y < 1 (BC2)

O.lu^x,

1, t) +

0.2u(x,

1,t) = 0 0 < x < 1 (BC3)

-0.1^(0,2/,í) + 0.2u(0, y, t) = 0 0 < y < 1 (BC4)

180 HEAT TRANSFER IN 2D AND 3D

6.8 Solve the given nonhomogeneous IBVP, with a time-dependent BC, using

methods outlined in this chapter and Duhamel's theorem. Your solution must include

(i) a 10-contour plot of u(x, y, 0.1) and a 10-contour plot of u(x, y, 0.5) and (ii) the

value u(c/2,d/2,l).

(x,y,t) = 0.1V

w (PDE)

IBVP<

u(x,y,0) = 0

u(x,0,t) =0

u(l,y,t) = sint

uix, l,t) = 0

u{0,y,t)=0

0<x< 1

0 <y< 1

0<x < 1

0 <y < 1

0.1V

IBVP^

(x,y,t)

u(x,y,0) = y*x

Uy(X,0,t) = 0

u\l,y,i) = sint

(x, l,t) =0

u(0,y,t)=0

f u

(x,y,t) =0AV

u(x,y,0) = 0

IBVP<

^(x,0,t) = 0

u(l,2/,t) = 2/(1

t¿

(x, l,t) = 0

M(0,Í/,Í)

= 0

0<x < 1

0 < ?/ < 1

0<x< 1

0 < y < 1

(IC)

(BCl)

(BC2)

(BC3)

(BC4)

(PDE)

(IC)

(BCl)

(BG2)

(BC3)

(BC4)

y) sin t

(PDE)

(IC)

0 < x < 1 (BCl)

0 < y < 1 (BC2)

0 < x < 1 (BC3)

0 < y < 1 (BC4)

6.9 Suppose a cubed-shaped solid (unit length on each edge) has homogenous

Dirichlet BCs on each of its six faces. The initial condition on the temperature is

u(x,y,z,0) = 64x(l - x)y(l - y)z{\ - z). Given k = 0.01, solve IBVP and

generate a 3D implicit plot of the u

—

1 surface at (i) t = 0 and (ii) t = 2.

6.10 Redo exercise 6.9 except let the boundary condition at z — 0 be such that

u(x, y, 0,

£)

= 2. The initial condition on the temperature is u(x, y, z, 0) = 64x(l

—

x)y(l

—

y)z{\

—

z). Given k = 0.01, solve IBVP and generate a 3D implicit plot of

the u = 1 surface at (i) t = 0 and (ii) t = 2.

CHAPTER 7

WAVE EQUATION

This chapter

devoted to the development and solution techniques for

the

hyperbolic-

type wave equations in ID and 2D. In both cases, the assumption is the medium is

elastic, such as an elastic string in ID or an elastic membrane in 2D.

7.1 WAVE EQUATION IN 1D

The ID wave equation is derived for the case of a vibrating string. The following

assumptions are made:

The mass of the string per unit length is constant.

The tension per unit length is constant in position and time.

The deflection of the string is small relative to the length of the string.

Figure 7.1 depicts an infinitesimal section of the string with the tension vector at

each end. Because the deflection of the string at any point is in the vertical direction

only, the dependent variable is y(x, t). The governing PDE results from Newton's

second law of motion. The motion is assumed to be in the vertical direction, so an

accounting of the forces in the vertical must be made. These forces are

Fourier Series and Numerical Methods for Partial Differential Equations, 181

First Edition. By Richard Bernatz