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Simple 2-D Wave Equation Problem

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data

    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].

    2. Relevant 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]



    3. 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?
     
    Last edited: Oct 8, 2012
  2. jcsd
  3. Oct 8, 2012 #2
    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.
     
  4. Oct 8, 2012 #3
    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: [itex]a=1[/itex] and [itex]b=1[/itex]. [itex]f(x,y)[/itex], 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.

    What do you mean here? If m or n is set to any whole number value, [itex]sin(\pi m)[/itex] or [itex]sin(\pi n)[/itex] are equal to 0.
     
  5. Oct 8, 2012 #4

    LCKurtz

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    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.
     
  6. Oct 8, 2012 #5
    Silly me, I missed that! Please ignore that part.

    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.
     
  7. Oct 9, 2012 #6
    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?
     
  8. Oct 9, 2012 #7
    Between this: [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] and this:
    [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], what was the intermediate equation?

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