Consider a square membrane with edge-length of L, and a circular membrane with radius R. The two membranes are fixed at the edges and are made of the same material and the fixation force at the edges is equal. Their eigen-angular frequencies are described respectively as: $0<\omega^¤_1<\omega^¤_2,...$ for the quadratic membrane and $0<\omega^\bullet_1<\omega^\bullet_2,...$ for the circular membrane.

What is the ratio $R/L$ such that the membranes have the same lowest eigenfrequencies, that is $\omega^¤_1=\omega^\bullet_1$ ? What is the ratio $\omega^\bullet_2/\omega^¤_{2}$ for the next-lowest frequencies the value of R/L square membrane in respect to the circular membrane?

I set up the PDE problem

$$\Delta u=-\lambda u$$


$$\Delta=\bigg(\frac{\partial^2 }{\partial x^2}+\frac{\partial^2}{\partial y^2}\bigg)$$ for the square membrane and

$$\Delta=\bigg(\frac{\partial^2 }{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}\bigg)$$ for the circular membrane

Then, for the first I get $\lambda=\frac{n\pi}{L}$ and for the second $\lambda=\frac{\alpha_{n,k}}{J_vR}^2$, ($\alpha$ are the Bessel zeros) considering the Laplace equation with Dirichlet conditions and the Bessel equation with solution for $\lambda>0$.

But from here, I am not sure how to find the answer.

Any ideas?


  • $\begingroup$ What do you mean by eigen-angular? eigen-functions that only depend on the radius? but what does it mean for the square? $\endgroup$
    – themaker
    Commented Dec 8, 2022 at 14:46
  • $\begingroup$ Yes, the angular frequencies of the two operators. I have no idea what it means for the square. $\endgroup$ Commented Dec 8, 2022 at 15:29

1 Answer 1


You are using the wrong equation. We need the wave equation, not the Laplace equation here.

Let $A$ be the amplitude of oscillation of your membrane. It obeys the wave equation (WLOG taking the wave speed $c=1$),

$$\partial_t^2A-\Delta A=0 $$ We look for a normal (pure frequency) mode of oscillation, and try separation of variables. For the square membrane this is $$A(t,x,y)=\mathrm e^{\mathrm i\omega t}X(x)Y(y)$$ Which yields $$-\omega^2\mathrm e^{\mathrm i\omega t}X(x)Y(y)+\mathrm e^{\mathrm i\omega t}X''(x)Y(y)+\mathrm e^{\mathrm i\omega t}X(x)Y''(y)=0 \\ \omega^2=\frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}$$ So let $$\frac{X''(x)}{X(x)}={\lambda}^2 \\ \frac{Y''(y)}{Y(y)}={\lambda}^2 \\ \omega^2=2\lambda^2$$ And having BCs $X(0)=Y(0)=X(L)=Y(L)=0$. Clearly we get eigenfunctions and eigenvalues of the form $$\lambda_n=\frac{n\pi}{L} \\ X_m(x)=\sin(m\pi x/L)~~,~~Y_n(y)=\sin(n\pi y/L)$$ Which gives normal frequencies for the square membrane as $$\omega^{\text{square}}_n(L)=\sqrt{2}\frac{n\pi}{L}$$

Applying the same approach to the circular membrane ($A(t,r,\theta)=\mathrm e^{\mathrm i\omega t}P(r)Q(\theta)$)gives us $$-\omega^2\mathrm e^{\mathrm i\omega t}P(r)Q(\theta)-\mathrm e^{\mathrm i\omega t}\bigg(P''(r)Q(\theta)+\frac{1}{r}P'(r)Q(\theta)+\frac{1}{r^2}P(r)Q''(\theta)\bigg)=0 \\ r^2\frac{P''(r)}{P(r)}+r\frac{P'(r)}{P(r)}+\omega^2r^2=-\frac{Q''(\theta)}{Q(\theta)}=-\lambda^2$$ I think we have to assume no dependency on $\theta$ here, otherwise the eigenvalues can't be ordered in a simple list $\omega_1,\omega_2,\dots$. Instead they will need two indices like $\omega_{n,k}$. So I assume $Q'(\theta)=0$. That leaves us with the Bessel equation $$P''(r)+\frac{1}{r}P'(r)+\omega^2 P(r)=0$$ Using the condition $P(0)<\infty$ we get $$P(r)=J_0(\omega r)$$ And applying the BC $P(R)=0$ we get the normal frequencies and eigenfunctions $$\omega^{\text{circle}}_n(R)=\frac{\alpha_{0,n}}{R} \\ P_n(r)=J_0\left(\alpha_{0,n}\frac{r}{R}\right)$$

So, $\omega^{\text{square}}_1(L)=\omega^{\text{circle}}_1(R)$ exactly when $$\frac{\alpha_{0,1}}{R}=\sqrt{2}\frac{\pi}{L} \\ \boxed{\frac{R}{L}=\frac{\alpha_{0,1}}{\sqrt{2}~\pi}\approx 0.541276}$$ The ratio of the next frequencies is $$\frac{\omega^{\text{circle}}_2(R)}{\omega_2^{\text{square}}(L)}=\frac{\frac{\alpha_{0,2}}{R}}{\frac{2\sqrt{2}\pi}{L}}=\frac{L}{R}\frac{\alpha_{0,2}}{2\sqrt{2}~\pi}=\frac{\sqrt{2}~\pi}{\alpha_{0,1}}\frac{\alpha_{0,2}}{2\sqrt{2}~\pi}=\frac{1}{2}\frac{\alpha_{0,2}}{\alpha_{0,1}}\approx \boxed{1.14771}$$

  • $\begingroup$ Superb, thanks! $\endgroup$ Commented Dec 8, 2022 at 16:03

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