Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A string of length $2L$ is fixed at both ends. The displacements on it satisfy the equation ${\partial^2 y\over \partial t^2}=\nu{\partial^2 y\over \partial x^2}$ . Also, $y(x,0)=0$ and for $t<0$, it oscillates in its fundamental mode. At $t=0$, the change in ${\partial y \over \partial t}=c\delta(x-L)$. How do I find $y(x,t)$ for $t>0$?

Thanks.

I know the form of the general solution to the (free) wave equation, I just don't know how to apply boundary conditions. It would be great if someone would kindly elaborate.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Fixed at both ends means that $y(0,t)=y(L,t)=0$ for all $t$. This will result in solutions that are sine waves with periods that are integer fractions of $2L$. With no dissipation, each mode will oscillate forever. If you find the set of modes, you can find the velocity at each point when the string passes through $x=0$. The initial condition gives you the velocity at each point at $t=0$ and at that time $x=0$. So you need to find a superposition of the modes that have the given velocity at $t=0$.

share|improve this answer
    
Thanks, Ross. So I know that the general solution is $y(x,t)=\sum_{n=1}^\infty \sin({n\pi x \over 2L})(a_n \cos({n\pi \nu t\over 2L})+b_n \sin({n\pi \nu t\over 2L}))$. But I am still confused about the boundary condition. Isn't $\delta(x-L)=\infty$ at $x=L$? –  Alex Nov 1 '11 at 21:09
    
@Alex: yes. Usually you would just be given ${\partial y \over \partial t}$ at $t=0^+$ and I would take it that way: $y'(L,0^+)=c, y'(x,0^+)=0 \text{for} x \ne L$, but that is not the statement you have. –  Ross Millikan Nov 1 '11 at 21:54
    
Thanks, Ross, so how could I impose the BC to find the wave function? Incidentally if I found the form of the wave for $t<0$ (fundamental mode) and I am given that $y(x,0)=0$. Does this mean that I can find coefficients of the $t<0$ wave using this? Or does the wave no longer hold? Thanks again! :) –  Alex Nov 1 '11 at 22:05
    
@Alex: You are given it oscillates in its fundamental mode for $t<0$ and $y(x,0)=0$, which says $a_n=0$ and $b_n=0 \text{for} n>1$. I don't see how you find $b_0$ for $t<0$. The fixed BC were used to find the sine expansion in your earlier comment. For $t>0$ you Fourier analyze the motion at $t=0^+$ to find the $b_n$. The $a_n$ are all $0$ because $y(x,0)=0$ –  Ross Millikan Nov 1 '11 at 22:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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