Simple 2-D Wave Equation Problem

[comwiz0]

Homework Statement

Solve the boundary value problem

\frac{\partial ^2 u}{\partial t^2} = c^2 (\frac{\partial ^2 u}{\partial x^2}+\frac{\partial ^2 u}{\partial y^2}),
0<x<a,
0<y<b, and
t>0

for the boundary conditions

u(0,y,t) = 0 and u(a,y,t) = 0 for 0 \leq y \leq b and t\geq0 and
u(x,0,t) = 0 and u(x,b,t) = 0 for 0 \leq x \leq a and t\geq0

and the initial conditions

u(x,y,0) = f(x,y) and
\frac{\partial u}{\partial t}(x,y,0) = g(x,y)

with a=b=1, c=1/\pi and the given functions f(x,y) = sin(\pi x)sin(\pi y) and g(x,y) = sin(\pi x).

Homework Equations

The derived solution to the two dimensional wave equation with the above boundary and initial conditions is

u(x,y,t) = \sum_{n=1}^\infty \sum_{m=1}^\infty (B_{mn} cos \lambda_{mn} t + B_{mn}^* sin \lambda_{mn} t) sin \frac{m \pi x}{a} cos \frac {n \pi y}{b}

where

\lambda_{mn} = c \pi \sqrt{\frac{m^2}{a^2} + \frac{n^2}{b^2}}

and

B_{mn} = \frac{4}{a b} \int_0^b \int_0^a f(x,y) sin \frac{m \pi x}{a} sin \frac{n \pi y}{b} dx dy

and

B_{mn}^* = \frac{4}{a b \lambda_{mn}} \int_0^b \int_0^a g(x,y) sin \frac{m \pi x}{a} sin \frac{n \pi y}{b} dx dy

The Attempt at a Solution

Alright, with all of that out of the way, this should be a pretty simple problem of just plugging in numbers. However, for some reason, I seem to keep getting B_{mn} and B_{mn}^* to be zero, which, in turn, means the whole thing is zero, unless I'm misunderstanding something or missing something - which, for as long as I've been staring at this problem, could very well be the case.

Starting with \lambda_{mn},

\lambda_{mn} = c \pi \sqrt{\frac{m^2}{a^2} + \frac{n^2}{b^2}} = \frac{\pi}{\pi} \sqrt{\frac{m^2}{1^2} + \frac{n^2}{1^2}} = \sqrt{m^2 + n^2}.

Next, B_{mn},

B_{mn} = \frac{4}{a b} \int_0^b \int_0^a f(x,y) sin \frac{m \pi x}{a} sin \frac{n \pi y}{b} dx dy
B_{mn} = 4 \int_0^1 \int_0^1 sin(\pi x)sin(\pi y) sin(m \pi x)sin(n \pi y)dx dy
B_{mn} = 4 \int_0^1 sin(\pi x) sin(m \pi x) dx \int_0^1 sin(\pi y) sin(n \pi y) dy
B_{mn} = 4 [\frac{sin(\pi m)}{2(m+1)\pi} - \frac{sin(\pi m)}{2(m-1)\pi}][\frac{sin(\pi n)}{2(n+1)\pi} - \frac{sin(\pi n)}{2(n-1)\pi}]

For all m and n where m, n = 1, 2, ..., B_{mn} = 0 because of this sin(\pi i) term. I might be missing something obvious here, but I think this is correct. This, in and of itself, doesn't bother me. However, repeating for B_{mn}^*,

B_{mn}^* = \frac{4}{a b \lambda_{mn}} \int_0^b \int_0^a g(x,y) sin \frac{m \pi x}{a} sin \frac{n \pi y}{b} dx dy
B_{mn}^* = \frac{4}{\sqrt{m^2+n^2}} \int_0^1 \int_0^1 sin(\pi x) sin(m \pi x)sin(n \pi y)dx dy
B_{mn}^* = \frac{4}{\sqrt{m^2+n^2}} \int_0^1 sin(\pi x) sin(m \pi x) dx \int_0^1 sin(n \pi y) dy
B_{mn}^* = \frac{4}{\sqrt{m^2+n^2}} [\frac{sin(\pi m)}{2(m+1)\pi} - \frac{sin(\pi m)}{2(m-1)\pi}][\frac{1}{n \pi} - \frac{cos(n \pi)}{n \pi}]

As before, for all m and n where m, n = 1, 2, ..., B_{mn}^* = 0 because of this sin(\pi i) term. So, given that what I have here seems to indicate that I'm doing something wrong, I figured I would ask for some assistance. I'm sure I'm missing something really simple, but I've been staring at this problem for too long to see it. Any thoughts?

 

[voko]

B_{mn} = \frac{4}{a b} \int_0^b \int_0^a f(x,y) sin \frac{m \pi x}{a} sin \frac{n \pi y}{b} dx dy
B_{mn} = 4 \int_0^1 \int_0^1 sin(\pi x)sin(\pi y) sin(m \pi x)sin(n \pi y)dx dy
B_{mn} = 4 \int_0^1 sin(\pi x) sin(m \pi x) dx \int_0^1 sin(\pi y) sin(n \pi y) dy
B_{mn} = 4 [\frac{sin(\pi m)}{2(m+1)\pi} - \frac{sin(\pi m)}{2(m-1)\pi}][\frac{sin(\pi n)}{2(n+1)\pi} - \frac{sin(\pi n)}{2(n-1)\pi}]

There are at least two mistakes here. The transition from the first line to the second is incorrect. If you change the variables so that x: (0, a) -> (0, 1) nd y: (0, b) -> (0,1), that change should affect x and y everywhere, including f(x, y).

The second mistake is ignoring m = 1 and n = 1.

 

[comwiz0]

There are at least two mistakes here. The transition from the first line to the second is incorrect. If you change the variables so that x: (0, a) -> (0, 1) nd y: (0, b) -> (0,1), that change should affect x and y everywhere, including f(x, y).

Can you clarify what you mean by the transition from the first line to the second line being incorrect? All I am doing here is substituting in the given values into the first line: a=1 and b=1. f(x,y), as given in the problem statement, does not include an a or b term. It is just simply inserted in place in the equation. The third and fourth terms inside the integrals on the second line included an a and b, respectively, and they were set equal to one. The values of x and y are not being worked with at that stage.

As an aside, the solution given on the last line I was able to symbolically confirm as well with my TI-89.

The second mistake is ignoring m = 1 and n = 1.

What do you mean here? If m or n is set to any whole number value, sin(\pi m) or sin(\pi n) are equal to 0.

 

[LCKurtz]

Next, B_{mn},

B_{mn} = \frac{4}{a b} \int_0^b \int_0^a f(x,y) sin \frac{m \pi x}{a} sin \frac{n \pi y}{b} dx dy
B_{mn} = 4 \int_0^1 \int_0^1 sin(\pi x)sin(\pi y) sin(m \pi x)sin(n \pi y)dx dy
B_{mn} = 4 \int_0^1 sin(\pi x) sin(m \pi x) dx \int_0^1 sin(\pi y) sin(n \pi y) dy
B_{mn} = 4 [\frac{sin(\pi m)}{2(m+1)\pi} - \frac{sin(\pi m)}{2(m-1)\pi}][\frac{sin(\pi n)}{2(n+1)\pi} - \frac{sin(\pi n)}{2(n-1)\pi}]

The second mistake is ignoring m = 1 and n = 1.

What do you mean here? If m or n is set to any whole number value, sin(\pi m) or sin(\pi n) are equal to 0.

What he means is that the integral formula you used for those integrals does not work when $m=1$ or $n=1$. They have to be done separately.

 

[voko]

All I am doing here is substituting in the given values into the first line: a=1 and b=1.

Silly me, I missed that! Please ignore that part.

What do you mean here? If m or n is set to any whole number value, sin(\pi m) or sin(\pi n) are equal to 0.

Do you see that in the formulae you got after integration, you have (m - 1) and (n - 1) in denominators? When m = 1 and n = 1, those terms blow up. This is an indication of a problem with your integration. The problem is that before integration those terms were $\cos (m - 1)\pi x = \cos (1 - 1) \pi x = 1$ and $\cos (n - 1) \pi y = \cos (1 - 1) \pi y = 1$, i.e., constants, so their integrals cannot be zero.

 

[comwiz0]

I believe I understand what you two are saying, but I'm not quite sure I know how to implement it. I've been working on this assignment for a bit too long to really comprehend anything at this point unfortunately. Can you be a bit more specific about how I might start to set something like that up?

 

[voko]

Between this: B_{mn} = 4 \int_0^1 sin(\pi x) sin(m \pi x) dx \int_0^1 sin(\pi y) sin(n \pi y) dy and this:
B_{mn} = 4 [\frac{sin(\pi m)}{2(m+1)\pi} - \frac{sin(\pi m)}{2(m-1)\pi}][\frac{sin(\pi n)}{2(n+1)\pi} - \frac{sin(\pi n)}{2(n-1)\pi}], what was the intermediate equation?

Evaluate this intermediate equation for m = 1 and n = 1. The same applies to $B_{mn}^*$.