% Problem P1



EGR 599 (Spring 2005) _____________________

LAST NAME, FIRST

Problem set #9

1. The Laplace’s equation

[pic] = 0

is given with the following boundary conditions

u(0,y) = u(a,y) = 0 0 < y < b

u(x,0) = 0 0 < x < a

and u(x,b) = f2(x) 0 < x < a

Use the method of separation of variables to obtain the solution (show all your work)

u(x,y) = [pic]Bnsin[pic]sinh[pic], where

Bn = [pic][pic]

Apply this solution to a 1(1 square plate where one side is at 100 and the other three sides are held at 0. Use Matlab pdetool to solve this problem numerically and provide the solution in a 3-D graph.

Solution

We assume that u(x,t) can be separated into X(x), a function of x alone, and Y(y), a function of y alone.

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

From the following relations

[pic] = T[pic] ( [pic] = T[pic]

[pic] = T[pic] ( [pic] = T[pic]

The original PDE becomes

[pic] = 0

Divide the above equation by XY to obtain

[pic][pic] = ( [pic][pic] = ( (2 = constant

Since the LHS depends on x only, and the RHS depends on y only, they must equal to a constant ( (2. The constant must be negative for non-trivial solution. The separation constant should be negative for the direction with both homogeneous boundary conditions. For this problem, x-direction has the homogeneous boundary condition at both ends.

The boundary conditions on u(x,t) are changed to the boundary condition on X(x)

u(0,y) = 0 ( X(0) Y(y) = 0 ( X(0) = 0

u(a,y) = 0 ( X(a) Y(y) = 0 ( X(a) = 0

The ODE with respect to x is

[pic]= ( (2X ( X = C1cos((x) + C2sin((x)

The constant C1 can be determined from the boundary condition at x = 0

At x = 0, X = 0 = C1(1) + C2(0) ( C1 = 0

At x = a, X = 0 = C2sin((a)

To avoid the trivial solution, C2 ( 0 and

sin((a) = 0 ( (a = n( ( (n = [pic]

The solution with respect to x is

X = Xn = C2sin[pic], n = 1, 2, 3, …

The ODE with respect to y is

[pic][pic] = (2 ( Y = An’cosh((ny) + Bn’sinh((ny)

The constant An’ can be determined from the boundary condition at y = 0,

Y(0) = 0 = An’ ( Y = Bn’sinh((ny)

The product solution is then

un(x,t) = Bnsin[pic]sinh[pic]

The general solution for the steady two dimensional heat equation is

u(x,y) = [pic]Bnsin[pic]sinh[pic]

The constants Bn can be determined from the boundary condition at y = b

u(x,b) = f2(x) = [pic]Bnsin[pic]sinh[pic]

The coefficients Bn sinh[pic] are then obtained from the Fourier sine series expansion of f2(x)

Bn = [pic][pic]

For a 1(1 square plate where one side is at 100 and the other three sides are held at 0, a = 1, b = 1, f2(x) = 100.

u(x,y) = [pic]Bnsin[pic]sinh[pic]

Bn = [pic][pic]= [pic](1- cos(n())

For n even Bn = 0, for n odd Bn = [pic]

u(x,y) = [pic][pic][pic] [pic]

[pic]

Matlab solution for a 1(1 square plate where one side is at 100.

2. Use the method of separation of variables to obtain the solution of the wave equation

[pic] = c2[pic], where c = 1/(

where u(x,y, t) = 0 on the boundary of a 1(1 square membrane, u(x,y, 0) = x(x – 1)y(y – 1), and the initial velocity is zero. Use Matlab pdetool to solve this problem numerically and provide the solution in a 3-D graph at t = 2, 4, 6, 8, and 10.

Solution

u(x,y, t) = [pic][pic] sin[pic]sin[pic][pic]

[pic]

[pic]

[pic]

[pic]

3. Three edges of a thin 1- by 2-m plate are held at 0oC, while the fourth edge, at y = 1 m, is held at 100oC. The steady state temperature distribution in the plate is

T(x, y) = [pic]

where An = ________________ Bn = _______________ Cn = __________________

Solution

Solution: (From Don Phillippi)

Hand calculation:

Based on conditions noted above the following is concluded:

[pic] and [pic] and [pic] equation should be used

(page 5-53 class notes)

Basic equation is: [pic]

modify the basic equation to fit the problem statement yielding:

[pic] where [pic] equals [pic]

note that [pic] therefore [pic] equals [pic]

therefore [pic] (from page 5-53) in basic equation equals[pic] in problem statement

Begin by solving the basic equation, then put it is the specified form to find the requested unknowns:

[pic]= [pic]

[pic]= [pic]

[pic] ( [pic]

3. (continued):

T(x, y) = [pic]

replace sinh terms with e terms to match problem statement:

T(x, y) = [pic]

T(x, y) = [pic]

since subscript implies “n” terms, leave in terms of “n”

[pic] where [pic]

Note: could input [pic], but it looks nicer this way

and from above:

[pic] and [pic]

%Prob9_1 MATLAB program to plot u(x,y,t)

x=[0:.02:2];

y=[0:.02:1];

[X,Y]=meshgrid(x,y);

n1=length(X);n2=51;

uxy=zeros(n2,n1);

%t=infinity, steady state problem

for n=1:200

Cn=n*pi./2;

uxy=(1-(-1)^n)*sin(n*pi./2*X).*sinh(Cn*Y)./n/(sinh(Cn))+uxy;

end

uxy=200*uxy/pi;

mesh(X,Y,uxy)

%End M-file

3. (continued):

[pic]

4. Use d’Alembert’s formula to solve the following boundary value problem

[pic]= c2[pic] c = 1, L= 1

B.C. : u(0,t) = 0 and u(L,t) = 0, for t ( 0

I.C. : u(x,0) = f(x) = [pic] and

[pic](x,0) = g(x) = x, for 0 ( x ( L

Solution: (From Don Phillippi)

Hand calculation: (from pages 5-23 through 5-25 notes)

Initial conditions: [pic] [pic]

and: [pic] [pic]

therefore: [pic]

for: [pic]

[pic]

[pic] and [pic] ( [pic] and [pic]

[pic] where [pic]

[pic]

dividing through by “c” and integrating both sides with respect to x yields:

[pic] where “s” is a dummy variable

[pic]

let: [pic]

4. (continued):

[pic]

using: [pic] or [pic]

therefore: [pic]

yielding: [pic]

[pic]

[pic] putting in above yields:

[pic]

or: [pic]

replace x by x + ct for F(x) and x by x – ct for G(x), yielding:

[pic]

[pic]

inverting integral:

[pic]

adding F(x+ct) and G(x-ct):

[pic] or:

[pic]

since c = 1 (given in problem statement):

[pic]

4. (continued):

[pic]

where [pic] an odd 2-periodic extension and G is the anti-derivative of [pic]

let: [pic]

[pic]

consolidating:

[pic]

where: [pic]= [pic]

and: G(x) = [pic]

5. Determine the first time the strings returns to its initial shape for the string of unit length with boundary conditions u(0,t) = 0 and u(L,t) = 0, and initial conditions:

a) f(x) = sin (x, g(x) = 0, c = [pic] Ans: t =2(

b) f(x) = [pic], g(x) = x, c = 1 Ans: t =2

6. Consider the following boundary value problem

[pic]= c2[pic] c = 1, L= 1

B.C. : u(0,t) = 0 and u(L,t) = 0, for t ( 0

I.C. : u(x,0) = f(x) = [pic], [pic](x,0) = g(x) = 0

a) Use d’Alembert’s method to plot the string at time t = 0, ¼, ½.

[pic]

b) For t = ¼, identify the points on the string that are still in rest position.

Ans: .75 < x ( 1

c) Take a point x on the string with zero initial displacement (1/2 ( x ( 1). How long does it take before the point x starts to vibrate?

Ans: t > x ( [pic]

d) What is the answer in (c) for an arbitrary string constant c > 0?

Ans: t > [pic]

7. Solve [pic]= c2[pic] c = 1, L= (

B.C. : u(0,t) = 0 and u(L,t) = 0, for t ( 0

I.C. : u(x,0) = f(x) = [pic]

Ans: u(x,t) = [pic][pic]

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