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comwiz0
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Homework Statement
Solve the boundary value problem
[itex]\frac{\partial ^2 u}{\partial t^2} = c^2 (\frac{\partial ^2 u}{\partial x^2}+\frac{\partial ^2 u}{\partial y^2})[/itex],
[itex]0<x<a[/itex],
[itex]0<y<b[/itex], and
[itex]t>0[/itex]
for the boundary conditions
[itex]u(0,y,t) = 0[/itex] and [itex]u(a,y,t) = 0[/itex] for [itex]0 \leq y \leq b[/itex] and [itex]t\geq0[/itex] and
[itex]u(x,0,t) = 0[/itex] and [itex]u(x,b,t) = 0[/itex] for [itex]0 \leq x \leq a[/itex] and [itex]t\geq0[/itex]
and the initial conditions
[itex]u(x,y,0) = f(x,y)[/itex] and
[itex]\frac{\partial u}{\partial t}(x,y,0) = g(x,y)[/itex]
with [itex]a=b=1[/itex], [itex]c=1/\pi[/itex] and the given functions [itex]f(x,y) = sin(\pi x)sin(\pi y)[/itex] and [itex]g(x,y) = sin(\pi x)[/itex].
Homework Equations
The derived solution to the two dimensional wave equation with the above boundary and initial conditions is
[itex]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}[/itex]
where
[itex]\lambda_{mn} = c \pi \sqrt{\frac{m^2}{a^2} + \frac{n^2}{b^2}}[/itex]
and
[itex]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[/itex]
and
[itex]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[/itex]
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 [itex]B_{mn}[/itex] and [itex]B_{mn}^*[/itex] 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 [itex]\lambda_{mn}[/itex],
[itex]\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}[/itex].
Next, [itex]B_{mn}[/itex],
[itex]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[/itex]
[itex]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[/itex]
[itex]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 [/itex]
[itex]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}][/itex]
For all [itex]m[/itex] and [itex]n[/itex] where [itex]m, n = 1, 2, ...[/itex], [itex]B_{mn} = 0[/itex] because of this [itex]sin(\pi i)[/itex] 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 [itex]B_{mn}^*[/itex],
[itex]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[/itex]
[itex]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[/itex]
[itex]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 [/itex]
[itex]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}][/itex]
As before, for all [itex]m[/itex] and [itex]n[/itex] where [itex]m, n = 1, 2, ...[/itex], [itex]B_{mn}^* = 0[/itex] because of this [itex]sin(\pi i)[/itex] 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?
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