2
$\begingroup$

This is a follow-up question of What would happen if the boundary value for $u_{tt}=a^2u_{xx}$ is that $u|_{x=0}=0$ and $u|_{x=l}=\sin\frac{n\pi a}lt$.

In the following one-dimensional wave equation system:

$$ u=u(x,t),\;\frac{\partial ^2u}{\partial t^2}=a^2 \frac{\partial ^2u}{\partial x^2} \\ u(0,t) = 0, \; u(l,t) = \sin \omega x \\ u(x,0) = 0, \; \frac{\partial u}{\partial t}(x, 0) = 0 $$

It is known that resonance will occur if $\omega$ approaches any of the values $\cfrac {n\pi a}l,\;n=1,2,\ldots$, which are the natural frequencies for the system.

However, my friend worked on the problem and gave a surprising conclusion: When $n$ is an even number, the amplitude of the system will not go up infinitely. Here's part of his proof:

The solution to the above partial differential equation system is

$$ u(t,x) = \cfrac {\sin\cfrac{\omega x}a}{\sin \cfrac{\omega l}a}\sin\omega t + 2\omega al\sum_{k=1}^{+\infty}\cfrac{(-1)^{k+1}}{(\omega l)^2-(k\pi a)^2}\sin\cfrac{k\pi at}l\sin\cfrac{k\pi x}l $$

with a little rewriting:

$$ \begin{align} u(t,x) = & \cfrac {\sin\cfrac{\omega x}a}{\sin \cfrac{\omega l}a}\sin\omega t \\&+ \cfrac{(-1)^{n+1}}{(\omega l)^2-(n\pi a)^2}\sin\cfrac{n\pi at}l\sin\cfrac{n\pi x}l \\&+ 2\omega al\sum_{k=1\\k\ne n}^{+\infty}\cfrac{(-1)^{k+1}}{(\omega l)^2-(k\pi a)^2}\sin\cfrac{k\pi at}l\sin\cfrac{k\pi x}l \end{align} $$

And when $\omega=\cfrac{n\pi a}t$ is taken into the equation:

$$ u(t,x) = \lim_{y\to n\pi}\left(\cfrac 1{\sin y} + \cfrac{(-1)^{n+1}}{y-n\pi}\right)\sin\cfrac{n\pi x}l\sin\cfrac{n\pi at}l \\ + \cfrac{2n}\pi\sum_{k=1\\k\ne n}^{+\infty}\cfrac{(-1)^{k+1}}{n^2-k^2}\sin\cfrac{k\pi at}l\sin\cfrac{k\pi x}l$$

When $n$ is an even number, there is (Taylor expansion):

$$ \lim_{y\to n\pi}\left(\cfrac 1{\sin y} + \cfrac{(-1)^{n+1}}{y-n\pi}\right) = \lim_{y\to n\pi}\cfrac{1+(-1)^{n+1}+o(1)}{y-n\pi+o(y-n\pi)} = 0 \;\textrm{(or }\cfrac{1}{n\pi}\textrm{)} $$

and the infinite part of the sum

$$ \cfrac{2n}\pi\sum_{k=n+1}^{+\infty}\cfrac{(-1)^{k+1}}{n^2-k^2}\sin\cfrac{k\pi at}l\sin\cfrac{k\pi x}l $$

is bounded.

Therefore the amplitude will not go all the way to infinity.

I have not been able to reproduce his result:

image

Can anyone tell me what I am missing here, or if my friend's proof is flawed?

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.