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A string moving in an elastic medium is governed by:

$u_{tt} = c^2u_{xx} − γ^2u$

where c and γ are constants. Solve this equation for a string of length L, fixed at the ends, subject to initial displacement f(x) and an initial velocity of zero.

I understand I need to find a separated solution in the form $u(x,t) = X(x)T(t)$.

After plugging that into the original equation, I got:

$X(x)T''(t) = c^2X''(x)T(t) - γ^2X(x)T(t)$

Then after dividing by $X(x)T(t)$,

$\frac{T''(t)}{T(t)} = c^2\frac{X''(x)}{X(x)} - γ^2 $

I know this is only true if each side is equal to a separation constant λ

$\frac{T''(t)}{T(t)} + γ^2 = λ $

$c^2\frac{X''(x)}{X(x)} = λ $

But I'm not sure if I'm heading in the right direction / where to go from here. I'm also not sure how to interpret the given "boundary conditions". I appreciate any help. Thanks in advance.

user114014
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1 Answers1

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Because you should be expecting oscillatory behavior in $x$, maybe it would be convenient to call the constant $-\lambda$. Then the $x$ equation becomes

$$X''+\frac{\lambda}{c^2} X=0$$

so that

$$X(x) = A \cos{\left ( \frac{\sqrt{\lambda}}{c} x\right )} + B \sin{\left ( \frac{\sqrt{\lambda}}{c} x\right )}$$

Because the string is fixed at the ends, the boundary conditions are

$$X(0) = X(L) = 0$$

$X(0)=0 \implies A=0$, and $X(L)=0$ means that

$$\sin{\left ( \frac{\sqrt{\lambda}}{c} L\right )}=0$$

This only happens for certain values of the argument, i.e., when

$$\frac{\sqrt{\lambda}}{c} L = n \pi \implies \lambda_n = \left (\frac{n \pi c}{L} \right )^2 $$

The $t$ equation becomes

$$T'' + \left [ \gamma^2 + \left (\frac{n \pi c}{L} \right )^2\right ] T = 0$$

which means that

$$T_n(t) = C_n \cos{\eta_n t} + D_n \sin{\eta_n t}$$

where

$$\eta_n = \sqrt{\gamma^2 + \left (\frac{n \pi c}{L} \right )^2}$$

The general solution is a linear combination over each $n$:

$$u(x,t) = \sum_{n=1}^{\infty} (C_n \cos{\eta_n t} + D_n \sin{\eta_n t}) \, \sin{\left ( \frac{n \pi}{L} x\right )} $$

You find the constants from the initial conditions. First, the condition that the initial string position is $f(x)$:

$$f(x) = \sum_{n=1}^{\infty} C_n \, \sin{\left ( \frac{n \pi}{L} x\right )} \implies C_n = \frac{2}{L} \int_0^L dx \, f(x) \sin{\left ( \frac{n \pi}{L} x\right )} $$

Next, the initial string velocity $u_t(x,0)=0$ implies that

$$0 = -\sum_{n=1}^{\infty} \eta_n D_n \, \sin{\left ( \frac{n \pi}{L} x\right )} \implies D_n=0$$

Thus the solution to your problem is

$$u(x,t) = \sum_{n=1}^{\infty} C_n \cos{\eta_n t} \, \sin{\left ( \frac{n \pi}{L} x\right )} $$

Ron Gordon
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    Very nicely done indeed, +1! I think it's also worth noting, from a physical point of view, that the effect of the $\gamma^2 u$ term is to shift the frequency of each mode upward. – Robert Lewis Feb 10 '14 at 20:58
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    @RobertLewis: thanks and yes, the $\gamma^2 u$ term shifts the temporal frequencies up. – Ron Gordon Feb 10 '14 at 21:03
  • Great explanation, thanks so much! Quick question - why isn't f(x)sin(...) included INSIDE the integral (before the dx) in the third last line? – user114014 Feb 10 '14 at 21:28
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    @user114014: Oh, it is. That's just my quirky way of writing integrals, with the differential first. It has been the topic of discussion so often here I've lost track. It's a physics thing, I guess: we like to think of integrals as operators on functions. – Ron Gordon Feb 10 '14 at 21:31
  • @ Ron Gordon: right, the temporal frequencies. – Robert Lewis Feb 10 '14 at 21:39