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

292 FINITE ANALYTIC METHOD

y

+

BC4

BC3

V

2

W(x,y)

= 0

BC1

BC2

Figure 11.9 Laplace problem with various types of

BCs.

e) c = 4,d= 1,BC1: -W

y

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

f)

c

=l,d=l,BCl: -W

y

-W = 0,BC2: W = 2/(l-y),BC3:

W

y

+W

= 0,

BC4:

W = 0

11.5 Figure 11.10 shows the problem domain for a general 2D Poisson BVP on a

rectangular domain. The four boundary conditions of various types are prescribed

in each of the exercises below. Solve each of the examples by the FA method. Use

y+

BC4

BC3

V

2

W(x,y)

= f(x,y)

BC1

BC2

Figure 11.10 Poisson problem with various types of

BCs.

an appropriate software application to accommodate variations in the given problem

from an example in the textbook. Your 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(l - x), BC4:

W =

0,f(x,y)

= 10xy

EXERCISES 293

b) c

=

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

x

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

jy

x

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

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

x

= 0, BC3: W = 0, BC4:

-W

x

+ W = 0,/(x,y) = 10xy

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

x

= y - |, BC3: VF = 0, BC4:

W

x

=0,/(x,y) = x(2-x)

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

x

= 0,

/(x,y) = 10(x-±)(±-y)

11.6 The general 2D IBVP for heat transfer is shown below. Unless stated oth-

erwise, assume a(x,y,¿), /(x,y), a

l9

a

2

, .·., di, d

2

, gi(x,t),...,g

4

(y,t) are all

zero.

Use an appropriate software application to solve the following IBVP by the FA

method. Include a 3D plot of u(x, y, 0), u(x, y, 2), and the limiting (steady-state)

temperature field if it exists.

u

t

(x, y, t) = kV

2

u(x, y, t) + g(x, y, t)

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

IBVP<

(PDE)

(IC)

ait¿(x,0, t) + a

2

u

y

(x,0,t) = gi(x,t) 0<x<c (BC1)

biu(c, y, t) + b

2

u

x

(c, y, t) = g

2

(y, t) 0 < y < d (BC2)

c\u(x,d,

t) +

c

2

u

y

(x,d,

t)

—

gs(x,t) 0<x<c (BC3)

diu(0,y,t)+d

2

Mx(0,y,í) = <7

4

(í/,í) 0<y<d (BC4)

a) c = 1, d = 2, k = 0.1, f(x,y) = x H- y, ai = &i = ci = di = 1,

#2(2/,

¿) = 2/(2 - y), 03(z, ¿) = z(l - a).

b) c = 1, d = 1, k = 0.2, g(x, y, t) = 100#

[0

.

4

,ο.6]

χ [0.4,0.6]

0,2/), /(s,

2/)

= 0,

ai = &ι = ci =di = l,g

2

(y,t) = 2/(1 -y),gz{x,t) = x(l -x).

c) c= 1, d= 1,

fc

= 0.2, ai = 1, a2 =

&2

=

C2

= d2 = 1 92(y,t)

—

sin t.

d) c = 1, d = 1, fc = 0.01, /(x,y) = 2, -a

x

= &i = ci = -di = 1,

¿2 =

¿>2

= c

2

= d

2

= 0.2 g

2

(y, t) = 1.

e) c= 1, d= 1, k = 0.2, q(x,y,t) = ¿ï[o.4,o.6]x[o.4A6](s,y)sin(§t), a

2

=

&i = C2 = di = 1

11.7 Use the FA method to solve the following 2D transport problem. Include a

3D plot of

</>(x,

y, 0), φ(χ, y, 2), and the limiting (steady-state) temperature field if it

exists.

Take each parameter below to be zero unless otherwise specified.

IBVP<

Φί

+

ν>φ

χ

+ νφ

υ

= W

2

</> + g(x, y, t)

0(x,y,O) = /(x,y)

αι^(χ,Ο,ί) + a

2

</>

2/

(x,0,¿) = yi(x,¿)

bi<¡)(c,

y, t) +

&2<Mc,

2/,

¿) =

02(2/,

i)

ci0(x, d, t) +

ο

2

φ

υ

(χ,

d, t) = y

3

(x, t)

di0(O,y,¿) +d

2

0x(O,y,t) = ^4(2/,*)

(PDE)

(IC)

0 < x < c (BC1)

0 < y < d (BC2)

0 <x < c

0<y <d

(BC3)

(BC4)

a) c = 1, d = 2, t¿=0.2, fc = 0.1, g(x, y,

£)

= x + y, ai = b\

—

c\

—

d\ — 1.

b) c = 1, d = 2, u=0.2, v = -0.2, fc = 0.05, /(x, y, t) - (x - x

2

)(y - y

2

),

«1 = &2 = ci = d2 = 1.

This Page Intentionally Left Blank

APPENDIX A

FA 1D CASE

The finite analytic solution for the ID transport equation,

φ

χχ

= 2Αφ

χ

+ Βφ

ι

(A.l)

is detailed in this appendix. For the one-time step formulation, the initial condition

φι and boundary conditions, φ\γ and

ΦΕ

are

φ(χ, 0) = φ!(χ) = a

s

(e

2Ax

- 1) + b

s

x + c

s

(A.2)

φ(—h,t) = φνν(ί)

— a>w

+ bwt (A.3)

and

where

0(ft,

¿)

= <M¿) = α

£

(ί) = a

E

4- M (A.4)

as — O

4 sinrT Aft

cs = ΦΞΟ

t>SE

-

ΦΞνν

- coth AhtysE + <?W - 205c)

2ft

Fourier Series and Numerical Methods for Partial Differential Equations, 295

First Edition. By Richard Bernatz

bs

296 FA 1D CASE

r

&E = ΦβΕ

_ ΦΕΟ -

ΦβΕ

aw = Çsw

, 0W/C* -

4>SW

b

w

and

&£

τ

With the introduction of a change of variable

0 = we

Ax

~^

1

the convective transport Equation (A.l), Initial Condition (A.2), and Boundary Con-

ditions (A.3) and (A.4) are transformed to

w

xx

= Bw

t

(A.5)

w(x, 0) = a

s

e

Ax

+ b

s

xe~

Ax

+ (c

s

- a

5

)e"

A:E

= </>(x) (A.6)

w{-h,

t) = e-

AhJr

^

l

{aw + b

w

t) = </>w(¿) (Α.7)

w(h,

t) = e-

Ah

+^\a

E

+ b

E

t) = <M¿) (Α.8)

The solution of Equation (A.5) for w can be obtained by superposition of solutions

w

—

wl + w% where wl solves

(A.9)

(A. 10)

(A.11)

(A. 12)

and iu2 solves

(A. 13)

(A.

14)

(A. 15)

(A. 16)

The solution wl is obtained using the method of separation of variables. Assuming

wl = X(x)T(t), and substituting for wl in Equation (A.9), we may separate the

variables as

x"

τ'

— = B-— = constant = -λ

2

(A.

17)

X x

Two ordinary differential equations result for Equation

(A.

17),

X"+X

2

X

= 0

(A.

18)

\2

Γ +—Τ = 0 (A.19)

JD

Wlxx =

wl(x,0) =

wl(-M) =

wl(h,t) =

WZ

XX

-

w2(x,0) --

w2(-h,t) =

w2(ft,t) =

= ß.wlt

= ΦΙ(Χ)

= 0

= B

·

w2

t

= 0

= <l>w(t)

= <M¿)

FA 1D CASE

297

with boundary conditions on

X

given by

X(-h) = X{h) = 0 (A.20)

The boundary conditions specified in Equation (A.20) stipulate solution to Equation

(A.

18)

having form

X(x) = a

n

sin[A

n

(x 4- h)] (A.21)

where λ

η

= ^

(n

=

1,2, 3,... ). The solution to Equation

(A.

19)

has the form

T{t)=b

n

e-^

t

(A.22)

Using the principle of superposition, the general solution for wl can be written as

oo

0

wl(x,t)

=

2^a

n

e""ß"

i

sin[A

n

(x

+

/i)] (A.23)

n=l

where the coefficients a

n

can be determined by applying the initial conditions given

in Equation

(A.

10)

as

oo

wl(x,0)

=

Φ(χ) =

Σ

αη

sin

t

A

n(^

+

h)] (A.24)

n=l

Invoking the orthogonality condition of

the

sine function, this initial condition results

1

f

h

-

/

Φ(χ) sin[A

n

(x 4- h)]da

"

J-h

-h

asEon

+

b

s

hE

ln

4- (c

5

-

as)£2n (Α.25)

where

1

/

,/l

E

0n

= -

e

Ax

sm[X

n

(x + h)]dz

n

J-h

;[e-

Ah

-(-l)

n

e

Ah

] (A.26)

^nh

t.-Ah (

Λ

\η„ΑΗΛ

(Ah)

2

+

(X

n

h)

2

'

1

/"

Ei

n

— T2 I

xe~

Ax

sm[\

n

(x + h)]dx

(_1)«

β

-

ΛΛ

]

2(Ah)(X

n

h

Ah t

Λλη

^

ΑΗλ

(Ah)

2

+

(X

n

h)

2[

[e

Ah

+

(-l)

n

e-

Ah

]

(A.27)

^n,h

,Ah , i

-¡\n-Ahi

(Ah)

2

+

(X

n

h)

2

1

f

h

E2n

= T /

e"

Ax

sin[A

n

(a: 4- ft)]cte

λτιΛ

[e

Ah

-

(-l)

n

e-

Ah

]

(A.28)

μΛ)2

+

(λ

η

Λ)

21

298 FA 1D CASE

To solve Equation

(A.

13),

note that the boundary conditions specified in Equations

(A.

15)

and

(A.

16)

are functions of

time.

The solution for this problem can be deduced

from the similar constant boundary conditions by the use of Duhamel's theorem. That

is,

w2 solves

w2 = [ -J^ (x, μ, t - μ)άμ (Α.29)

where w2 satisfies the zero initial condition and constant boundary conditions

w2

xx

= Bw2

t

(A.30)

w2(x,0) = 0 (A.31)

w2(-h,t) =

Φ\ν{μ)

(Α.32)

w2(h,t) = Φ

Ε

(μ) (Α.33)

The solution for w2 can be obtained by the superposition of the steady-state solu-

tion Φ and the transient solution v, which satisfy homogeneous boundary conditions.

That is,

w2 = Φ(χ, μ) + υ(χ, μ, t - μ) (Α.34)

where

0 (Α.35)

ά

2

Φ

dx

2

Φ(-Λ,μ) = Φνν(μ) (Α.36)

Φ(Λ,μ) = Φ

Ε

(μ) (Α.37)

and

Vxx = B'Vt (A.38)

ν(χ,μ,-μ) - -Φ(χ,μ) (Α.39)

ν(±ή,μ,ί-μ) = 0 (Α.40)

The steady-state solution Φ(χ, μ) for the ordinary differential equation (A.35) is

known to be

Φ(Χ,/Χ) =

¿

[Φ£;(μ)

~

Φ

^

(μ)] +

2

[ΦΕ(Μ)

+ Φ

^

(μ)] (Α,41)

which, in turn, is the initial condition for Equation (A.30). The solution for v is

similar to that for it;

1,

except that the initial condition Φ/(χ) is replaced by — Φ(χ, μ).

Therefore,

v =

Σ

bne

~~^

1 s[n

[^n(x + h)} (A.42)

n=l

where b

n

can be obtained from the initial condition given in Equation (A.31). That

is,

b

n

= / —Φ{χ,μ)8ΐη[\

η

(χΛ-}ι)]άχ

J-h

FA 1D CASE 299

Then,

X

n

w2 = v + Φ

00

1 A

2

= Σ TTK-ir^W - «M/*)]e-#* sin[A

n

(x 4- ft)]

Φ

Ε

(μ) - Φτν(μ) , Φ#(μ) +

Φ\γ(μ)

+

2ft *

+

2

ύ)2 is known, w2 can be determined using Duhamel's theorem and substi-

w2 in Equation (A.29). Thus,

Now that w2 is known, w2 can be detei

tuting for w2 in Equation (A.29). Thus

w2 = /

——

(χ,μ,ί-

μ)άμ

OO x »f

n=l

00

_/x)

sin[A

n

(x + ft)]d/i

00

λ Γ*

= - Σ M l [(-!)

η

Φ

Β

(μ) - <Μμ)]β-

^β-^*«η[λ„(χ + ft)] y* eT"[

$

„ - (-1)"Φ

Β

]φ

the FA element is given by

Σ

71=1

Now, the analytic solution on

= we

J

= (wl+w2)e

Ax

-^

00

= ]P

eAX AR l sin

[

A

nO + h){a

n

n=l

+

M f

e

^^

w

^ -

{-ν

η

*Ε{μ)]άμ}

(Α.43)

When the formula given in Equation (A.43) is evaluated at the node P(0, r), a

six-point formula for φρ is obtained.

φ

Ρ

= φ(0,τ)

OO 00

« > A

2

+\±

ηπ

= Ye

^ 2

Λ

'+Κ.

τ

. ηπ

B

sin —

n=l

L + ^ f e^^M - (-1)

η

Φ

β

(μ)]φΙ (A.44)

300

FA 1D CASE

Because sin ^ =0 for even numbers n, Equation (A.44) can be further simplified

by letting n = 2ra

—

1, so that

]Γ(-1Γ

+Ι

β w^Uom

m=l ^

+ ^ £ β^[Φ

]ν

(μ) + Φ

Ε

(μ)]άμ^

(A.45)

where

./o

e^aw + er

Ah

a

E

) Γ ε

ρ

™

μ

άμ + (e

Ah

b

w

+ e'

Ah

b

E

) Γ μβ

ρ

™

μ

άμ

Jo Jo

= {e

Atl

a

w

+

e

-

Ah

a

E

)~(e

F

-

T

- 1) + (e

Ah

b

w

+ β~

Α

Η

Ε

)τ

■F

m.

„Ah,

F

n

,T

0

F

m

r

+ -ΕΓ-)

F mT -t

i

mT

with

and

2.

\2

A

2

+ A:

B

Λγη.

— ;

Therefore,

Σ(-ΐ)

BhFrr

m+1 -F

m

r

asEom + bshEi

m

4- (cs

—

as)E

2

m

+ e-

Ah

b

E

)

(e

Ah

a

w

+

e-

Ah

a

E

)

(e

F

™

T

- l) + (e

Ah

b

w

Fm.r i

T e

4-

m

'

x

m

i

Define

and

so that

*

=

Σ

^ (-ir+

l

\

m

he-

F

^

\ [(Ah)

2

+ (X

m

hy

i = 1, 2

oo

5](-ir+

1

e

i

'™^

m

= (e

Ah

+

e-

Ah

)P,

FA 1D CASE

301

ra=l

¿

(-l)

m+1

e

FmT

E

lm

=

2Ah(e

Ah

+

e~

Ah

)P

2

-

{e

Ah

-

e~

Ah

)P

1

OO

5^

(-l)

m+1

e

F

™

T

E

2m

=

(e

Afc

+

e-^JPx

771=1

5>i

r

+V^-=^(e*»'-l)

V Jzirí^Ln-e-

Qi-Pi

5>ι

Γ

+ν-

m=l

BhFr,

-(e

i

+

Γ mT

FmT

^ (-i)">+iA

m

fe

_

Bff_

y,

(-l)-+iA

m

/i(l

-

e-

F

-

T

)

2^

C4M2 -4-C\...M2

T Z>

^ (^Λ)2

+

(A

m

ft)2

T

JDU2

=

Qi

(Q2-P2)

r

ra=l

[(A/l)

2

+

(A

m

/l)2]2

With these relationships, the expression for φρ becomes

φ

Ρ

=

asie^ + e-^P^bshpAhie^ + e-^Pi-ie^-e-^Pi]

+(c

s

-

a

s

)(e

Ah

+

e~

Ah

)P

x

+ (e

Ah

a

w

+ e-

Ah

a

E

){Q

1

-

P

1

)

Bh

2

+ (e

Ah

b

w

+ e-

Ah

b

E

)r

1

Qi + —(P2-Q2)

[Φβε

-

Φβιν

-

coth(Ah)WsE +

Φβητ

-

2</><?c)](4Ahcosh(Ah)P

2

-2sinh(Ah)Pi)

+

^sc(e

A/l

+

e~

Ah

)P

x

+ (ε^φβιν

+e-

A

^

SE

)(Qi

-

Pi)

+

[e

Ah

Wwc

-

Φβ\ν) + ε~

ΑΗ

{φΕα

-Φβε)}

fíh

2

Qi + (P2-Q2)

T

(2cosh(i4ft)0

sc

-

e

Ah

<f>

sw

-

e-

Ah

φsE)(2Ahcoth(Ah)P2

-Pi)

+

2cosh(A/i)Piφ

8α

+ (e

A

^sw + e-

A

^

SE

)(Qi

-

Pi)

+[(e

A

^wc +

β~

ΑΗ

φ

Ε

ο)

-

(e

A

^sw +

e^

SE

)]

Qi +

—

(P2-Q2)

T

= (e

Ah

4>sw

+

e-

Ah

φsE){Pι

-

2Ahcoth(Ah)P

2

+ Q

Í

-P

1

-Q

1

-

—

(P

2

-

Q2) + (e

A

'^wc

+

e-

A

^Ec){Qi +

—

(P

2

-Q2)

[ +

<^sc{2cosh(A/i)(2A/icoth(yl/i)P

2

-

Pj

+

Pj)}

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

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