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Math Help - Finding the solution to the wave equation on a closed interval (non-zero boundaries)

  1. #1
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    Finding the solution to the wave equation on a closed interval (non-zero boundaries)

    Hey guys, apologies if this is in the wrong section. I know wave equations are a type of differential equation, but for this particular question I'm not really solving the DE as such. My course is called Applied Analysis so I thought this would be the place to put the question.

    We were given the following wave equation with initial- and boundary-conditions:



    We were then asked to compute u(3/2, 3/2) using an "appropriate characteristic parallelogram". I know that the characteristics are x-t = 0 and x+t = 0 and that the solution to this problem is u(3/2, 3/2) = u(A) + u(B) - u(C), but finding the points for my parallelogram is proving to me quite confusing as my geometry is seriously shocking. I know that this is a simple question with a simple solution.

    The solution has the points (1,1), (2,1) and (3/2, 1/2) all in region III, so if I could get some help as to how these points were found, that would be great.
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  2. #2
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    I'm not sure what you are doing here. The point (3/2, 3/2) is obviously on the characteristic x- t= 0 which passes through (0, 0). u(3/2, 3/2)= u(0, 0) which we are told is 0.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    I'm not sure what you are doing here. The point (3/2, 3/2) is obviously on the characteristic x- t= 0 which passes through (0, 0). u(3/2, 3/2)= u(0, 0) which we are told is 0.
    The wave equation has two characteristic lines, so is not determined by its values in only one of them.

    For the question, I'm not going to compute them explicitly (I'm terrible at calculations...and lazy), but the idea is to take your point and draw the characteristics passing through it, this characteristics intersect either the boundary or the line t=0, then between your original point and these intersections you pick points that will guarantee that you can compute the function there, in your case they pick two of them to be in the characteristic x+t=2 (this is the one emanating from the right endpoint). This guarantees the third point will be on the right boundary.

    Notice the choice of points is somewhat arbitrary, personally I would have chosen one point to be (1,0) instead of (3/2,1/2) and (1/2,1/2) instead of (1,1). You can check these work too.

    PS. Don't do the calculations: Draw a picture and remember the characteristic lines have the form x-t=c then solve.
    Last edited by Jose27; June 9th 2011 at 10:05 PM.
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