1
$\begingroup$

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:

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

The left-hand side of this equality does not depend on ${\displaystyle r,}$ 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

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

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

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

Here i'm stuck.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.