I am interested in the modes of a rectangular drum of size WxL that is fixed on 4 edges. The differential equation that governs the vibrations of a 2D membrane is $\nabla^2 \phi - \frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} = 0 $. Coordinate origin is on one of the corners of the rectangle. The modes are proportional to

$$\phi_{mn}(x,y)=\sin(\frac{\pi x m}{W})\sin(\frac{\pi y m}{L})$$

For mathematical simplicity, I will assume time dependency to be a complex exponential.

$$\phi_{mn}(x,y,t)=\sin(\frac{\pi x m}{W})\sin(\frac{\pi y m}{L})e^{i \omega_{mn} t}$$

$$(\frac{\omega_{mn}}{c})^2=(\frac{\pi m}W)^2+(\frac{\pi n}L)^2$$

I think following summation should be able to account for all physically meaningful solutions to the problem. (Please let me know if I am wrong, or if there is more to know.)

$$\phi(x,y,t)=\sum_m\sum_n c_{mn}\sin(\frac{\pi x m}{W})\sin(\frac{\pi y m}{L})e^{i \omega_{mn} t}$$

I am confused about what to do if we force an excitation $f(t)$ at the point $(x',y')$. I think the following summation should give the expansion

$$f(t)=\sum_m \sum_n c_{mn}\sin(\frac{\pi x' m}{W})\sin(\frac{\pi y' m}{L})e^{i \omega_{mn} t}$$

However, I can't wrap my head around how this expression can account for all possible $f(t)$. What I recall from Fourier transform is that an arbitrary function $f(t)$ requires continuous frequency components (ie $\omega \in \mathbb R$). Since the values of $\omega_{mn}$ are discrete, I don't see how this can be the general solution. Yet, it doesn't feel like it would violate any law of physics to force an arbitrary $f(t)$ at the point $(x',y')$.

One possibility is that instead of enforcing the solution equals $f(t)$ at $(x',y')$, I could put $f(t)$ as a forcing term in the equation $\nabla^2 \phi - \frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} = \delta(x-x')\delta(y-y')f(t) $. The more I think about it, the more reasonable this approach seems. I am not sure how to solve this either.

In summary, the question is: Solve $\nabla^2 \phi - \frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} = 0 $ for a rectangular region of size WxL that is fixed ($\phi=0$) at 4 edges and is "excited" by the arbitrary function $f(t)$ at the point $(x',y')$.

Also please let me know if I likely have misunderstood something about the subject and which detailed source I should read. Thanks.


1 Answer 1


There is no problem. Yes, Fourier analysis tells you that, but you forgot that it tells us that relative to a certain boundary condition. Here, you have the boundary conditions of the drum, and so it is no longer a continuous Fourier integral, but rather a discrete Fourier series, that is the solution.

  • $\begingroup$ If $\omega_{min}=\omega_{01}=2 \text{rad/s}$, how can I make these discrete frequency components add up to $f(t)=e^{i \omega_{forced} t}$ where $\omega_{forced}=\sqrt2 \text{rad/s}$? $e^{i\omega t}$ are orthogonal for different $\omega$. I don't think I can add up bunch discrete of $e^{i\omega t}$ and end up with an $f(t)=e^{i \omega_{forced} t}$ that is orthogonal to every frequency component we just added up. $\endgroup$ Commented Jun 6, 2023 at 20:34
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    $\begingroup$ I think the answer is that you cannot force arbitrary excitations on systems constrained that way. $\endgroup$ Commented Jun 6, 2023 at 20:41
  • $\begingroup$ What if I took a thin object and started vibrating a point on the drum at a frequency that don't correspond to any of the eigenfrequencies? Does the universe implode? Doesn't seem like it would be a problem. $\endgroup$ Commented Jun 6, 2023 at 20:48
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    $\begingroup$ It will be damped away rather quickly. i.e. it cannot resonate into a huge wave. $\endgroup$ Commented Jun 6, 2023 at 21:03
  • $\begingroup$ How will it be dampened? Where is the energy going? Presumably, it will still have a well defined solution, dampened or not. I am willing to reconsider the equations I came up with; as I explained, I am not totally sure it is the correct approach. But at the end of the day, I can always shake a drum at a frequency different from its eigenfrequencies. $\endgroup$ Commented Jun 6, 2023 at 22:26

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