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