Solve wave equation in a disk, axisymmetric case

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

My attempt:

Note the function ${\displaystyle u}$ does not depend on the angle ${\displaystyle \theta ,}$ 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:


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

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


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


Here i'm stuck.

  • $\begingroup$ I guess you mean $rR’’+R’-\lambda rR=0$ for the second equation $\endgroup$ – MPW Feb 20 at 23:29
  • $\begingroup$ Yepp, true @MPW THANKS $\endgroup$ – Bvss12 Feb 20 at 23:30
  • $\begingroup$ 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. $\endgroup$ – MPW Feb 20 at 23:32
  • $\begingroup$ But i think i need found the Sturm-Liouville base but i'm vry stuck :( @MPW $\endgroup$ – Bvss12 Feb 20 at 23:35
  • 1
    $\begingroup$ 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 $\endgroup$ – MPW Feb 20 at 23:42


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

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.


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

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.


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