Induction on the n-dimensional, radially symmetric wave equation

[tjackson3]

Homework Statement

Consider the radially symmetric wave equation in n dimensions

$$u_{tt} = u_{rr} + \frac{n-1}{r}u_r$$

Use induction to show that the solution is

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

for n odd and

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

n even.

Homework Equations

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

$$u=\left(\frac{1}{r}\frac{\partial}{\partial r}\right)^m \frac{f(t-r)}{r}$$

solves

$$u_{tt} = u_{rr}+ \frac{2m}{r}u_r$$

We want to use this to show that

$$u=\left(\frac{1}{r}\frac{\partial}{\partial r}\right)^{m+1} \frac{f(t-r)}{r}$$

solves

$$u_{tt} = u_{rr}+ \frac{2m+1}{r}u_r$$

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 $L_m = \partial_t^2 - \partial_r^2 - (2m/r)\partial_r$ and $G = \left(\frac{1}{r}\frac{\partial}{\partial r}\right)$. Then our solution for m would be $G^m f(t-r)/r$; i.e. $L_mG^m f(t-r)/r = 0$. We want to show that $L_{m+1}G^{m+1} f(t-r)/r = 0$

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

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

which you could almost factor if $\partial_r^2$ and G^m 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!

 

[mighty2000]

Thank you for your post. I understand that you are struggling with the inductive step in proving the solution for the radially symmetric wave equation in n dimensions. I will try to provide some guidance and suggestions to help you with your proof.

Firstly, I would suggest trying to simplify the problem by considering the case where n=3 (since the solution is known for 2 dimensions, as mentioned in your post). This will give you a better understanding of the problem and may help you see the general pattern in the solution.

Next, let's focus on the odd case (n=2m+1). As you have already mentioned, the first step of the induction is pretty straightforward. Now, for the inductive step, we want to show that

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

solves

u_{tt} = u_{rr}+ \frac{2m+1}{r}u_r

To do this, we can use the fact that we know the solution for n=3 (i.e. m=1). This means that

u=\left(\frac{1}{r}\frac{\partial}{\partial r}\right) \frac{f(t-r)}{r}

solves

u_{tt} = u_{rr}+ \frac{3}{r}u_r

Now, let's look at the term

\left(\frac{1}{r}\frac{\partial}{\partial r}\right)^{m+1} \frac{f(t-r)}{r}

We can rewrite this as

\left(\frac{1}{r}\frac{\partial}{\partial r}\right)^{m} \left(\frac{1}{r}\frac{\partial}{\partial r}\right) \frac{f(t-r)}{r}

Using the product rule for derivatives, we can expand this as

\left(\frac{1}{r}\frac{\partial}{\partial r}\right)^{m} \left(\frac{1}{r}\frac{\partial f(t-r)}{\partial r} - \frac{f(t-r)}{r^2}\right)

Now, we can use the fact that

u=\left(\frac{1}{r}\frac{\partial}{\partial r}\right