% 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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