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Induction on the n-dimensional, radially symmetric wave equation

  1. May 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider the radially symmetric wave equation in n dimensions

    [tex]u_{tt} = u_{rr} + \frac{n-1}{r}u_r[/tex]

    Use induction to show that the solution is

    [tex]u = \left(\frac{1}{r}\frac{\partial}{\partial r}\right)^{(n-3)/2} \frac{f(t-r)}{r}[/tex]

    for n odd and

    [tex]u = \left(\frac{1}{r}\frac{\partial}{\partial r}\right)^{(n-2)/2} \int_0^{t-r}\ \frac{f(\tau)}{\sqrt{(t-\tau)^2-r^2}}\ d\tau[/tex]

    n even.

    2. Relevant equations



    3. The attempt at a solution

    I'm stuck even on the odd case. I feel like the even case could be handled using the method of descent that is used to derive the 2D solution from the 3D solution, but I could be wrong. Also, I can prove the first step of the induction pretty easily in both cases; it's just a matter of proving the inductive step itself that's presenting trouble.

    I've tried this in a few different ways. First off, let's take n odd, and in particular, n=2m+1. Then we have that

    [tex]u=\left(\frac{1}{r}\frac{\partial}{\partial r}\right)^m \frac{f(t-r)}{r}[/tex]

    solves

    [tex]u_{tt} = u_{rr}+ \frac{2m}{r}u_r[/tex]

    We want to use this to show that

    [tex]u=\left(\frac{1}{r}\frac{\partial}{\partial r}\right)^{m+1} \frac{f(t-r)}{r}[/tex]

    solves

    [tex]u_{tt} = u_{rr}+ \frac{2m+1}{r}u_r[/tex]

    The trouble is, I don't know if you could write out an explicit formula for the solution that would be very helpful. You could apply the "binomial theorem" (for derivatives), but it doesn't seem helpful. My thought - which may be completely off, so if you have a better suggestion, please let me know - was that you could define two operators, L and G, as [itex]L_m = \partial_t^2 - \partial_r^2 - (2m/r)\partial_r[/itex] and [itex]G = \left(\frac{1}{r}\frac{\partial}{\partial r}\right)[/itex]. Then our solution for m would be [itex]G^m f(t-r)/r[/itex]; i.e. [itex]L_mG^m f(t-r)/r = 0[/itex]. We want to show that [itex]L_{m+1}G^{m+1} f(t-r)/r = 0[/itex]

    For that, we have to see how the m operators interact with the m+1 operators. Clearly [itex]G^{m+1} = G G^m[/itex]. Note that [itex]L_{m+1} = L_m - (1/r)\partial_r = L_m - G[/itex], so [itex]L_{m+1}G^{m+1} = (L_m - G)(GG^m) = L_m(GG^m) - G^{m+2}[/itex]. To handle the first term, we'd like to switch G and Lm, because then we'd just have zero. Unfortunately, they don't commute, but [itex](GL_m - L_mG) = (2/r^2)((1/r)\partial_r - \partial_r^2)[/itex], which means that if we want to switch the two, we have [itex](GL_m + (2/r^2)(\partial_r^2 - (1/r)\partial_r))G^m) = (2/r^2)(\partial_r^2 - (1/r)\partial_r)G^m[/itex]. The second term is just another G, which means that our operator equation becomes

    [itex](\partial_r^2G^m - (2/r^2)G^{m+1}-G^{m+2})(f(t-r)/r) = 0[/itex]

    which you could almost factor if [itex]\partial_r^2[/itex] and Gm commuted, which they don't, and I don't know how you'd work out the commutant for that.

    Anyway, that's my attempt at the solution. I'm sure there's a wildly easier way to do it, but I'm not seeing it. Any advice is very much appreciated!
     
  2. jcsd
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