# Wave equation on a disk (circular membrane)

Solve wave equation in a disk, axisymmetric case

$$\begin{cases} \frac{\partial^2u}{\partial t^2}=\frac{c^2}{r}\frac{\partial}{\partial r}(r\frac{\partial u}{\partial r}) \,\,\, \,,00\\ u(r,0)=f(r),\quad\frac{\partial u}{\partial t}(r,0)=g(r),\quad u(a,t)=0 \end{cases}$$

My attempt:

Note the function $$u}$$ does not depend on the angle $$\theta ,}$$ because we have axisymmetric case of a circular membrane.

Let $$u(r,t)=R(r)P(t)$$ and replacing in the PDE we have: $$R(r)P''(t)=c^2[\frac{R'(r)}{r}+R''(r)]P(t)\tag1$$

Dividing $$(1)$$ for $$R(r)P(t)$$ we have:

$$\frac{P''(t)}{c^2P(t)}=[\frac{1}{r}\frac{R'(r)}{R(r)}+\frac{R''(r)}{R(r)}]$$

The left-hand side of this equality does not depend on $$r,}$$ and the right-hand side does not depend on $$t,}$$ it follows that both sides must be equal to some constant $$\lambda.}$$

Then $$\lambda=\frac{P''(t)}{c^2P(t)}=[\frac{1}{r}\frac{R'(r)}{R(r)}+\frac{R''(r)}{R(r)}]$$

Of this we have two equations

$$\begin{cases} P''(t)-c^2\lambda P(t)=0\\ rR''(r)+R'(r)-r\lambda R(r)\tag2 \end{cases}$$

We're going to solve $$P''(t)-c^2\lambda P(t)=0$$

If $$\lambda=0$$ then the solution is of the form: $$P(t)=c_1+c_2t$$

If $$\lambda>0$$ then the solution is of the form

$$P(t)c_1e^{ckt}+c_2e^{-ckt}$$

If $$\lambda<0$$ then the solution is of the form

$$P(t)=c_1\cos(ckt)+c_2\sin(ckt)$$

Here i'm stuck.

• I guess you mean $rR’’+R’-\lambda rR=0$ for the second equation – MPW Feb 20 at 23:29
• Yepp, true @MPW THANKS – Bvss12 Feb 20 at 23:30
• And so now you just have two 2nd order ODEs to solve. That gives you $R$ and $P$, which is what you want. There will be corresponding bd/init conditions too. – MPW Feb 20 at 23:32
• But i think i need found the Sturm-Liouville base but i'm vry stuck :( @MPW – Bvss12 Feb 20 at 23:35
• These are very basic ODEs, any text will show you how to solve them. I’m not going to type in that lecture here. Even Google will show you – MPW Feb 20 at 23:42

The BVP

$$r^2R'' + rR' - \lambda r^2 R = 0, \quad R(0) < \infty, R(a) = 0$$

only has a non-trivial solution when $$\lambda < 0$$.

You can check that $$\lambda = 0$$ returns a general solution of $$A+B\ln r$$, and $$\lambda > 0$$ returns modified Bessel functions, neither of which will satisfy the boundary conditions.

The substitution $$\rho = \sqrt{-\lambda}r$$ results in Bessel's equation (check this for yourself), so we have a general solution

$$R(r) = AJ_0(\sqrt{-\lambda}r) + BY_0(\sqrt{-\lambda}r)$$

Note that $$Y_0$$ blows up at $$r=0$$ so we need to set $$B=0$$.

This leaves the boundary condition $$J_0(\sqrt{-\lambda}a)=0$$. Let $$\alpha_n$$ be the zeroes of $$J_0(x)$$ with $$n=1,2,3,\dots$$ then we can rewrite the solution as

$$R_n(r) = J_0\left(\frac{\alpha_n}{a}r\right)$$

up to a multiplicative constant.

Therefore

$$u(r,t) = \sum_{n=1}^\infty \left[C_n \cos\left(\frac{\alpha_n}{a}ct\right) + D_n \sin\left(\frac{\alpha_n}{a}ct\right)\right]J_0\left(\frac{\alpha_n}{a}r\right)$$

The initial conditions give

\begin{align} u(r,0) &= f(r) = \sum_{n=1}^\infty C_n J_0\left(\frac{\alpha_n}{a}r\right) \\ u_t(r,0) &= g(r) = \sum_{n=1}^\infty \frac{c\alpha_n}{a}D_n J_0\left(\frac{\alpha_n}{a}r\right) \end{align}

To determine the remaining constants, you must find the Fourier-Bessel series of $$f(r)$$ and $$g(r)$$. The process is very similar to the Fourier series, since Bessel functions are also mutually orthogonal.