Upper and Lower Bound of Energy of vibrating String as $t→∞$

Question

I'm working on a question that involves a string satisfying $u_{tt}=c^2u_{xx}$ with the boundary conditions $u(0,t)=0$ and $u_x(L,t)=−γu(L,t)$ with $γ>0$.

So I'm given the energy term that the vibrating string satisfies, $E(t)=\frac{1}{2}\int_0^L [u_t^2(x,t)+c^2u_x^2(x,t)]dx.$

I am asked to figure out what happens to the energy term as $t→∞$ and find a possible lower and upper bound for the energy of the string. The steps I am told to take are to try to get rid of the integrals from $E(t)$ by first differentiating and then integrating with respect to $t$. This will involve integration by parts, the boundary conditions, and the PDE itself. I'm not sure how to go about doing the differentiation. Does this involve Leibniz integration since this is a PDE? Any help would be apppreciated.

Answer 1

As a starting point, probably, it is meant to differentiate by $\partial_t$ and partial integrate wrt. $dx$ under the time integral, and use the wave equation and the boundary values in order to reduce the kinetic term $u_{t,t}$ to $u_{x,x}$. I use subscripts for partial differentiaton and symbolically $\int_0^L f d1$ as the sum of $f$ at the boundaries $0,L$:

$$\int_0^L u_t^2 \, dx=\int_0^t \left(\int_0^L 2 u_t u_{t,t} \, dx\right) \, dt=2 c^2 \int_0^t \left(\int_0^L 2 u_t u_{x,x} \, dx\right) \, dt$$ $$=2 c^2 \int_0^t \left(\int_0^L u_t u_x \, d1-\int_0^L u_x u_{t,x} \, dx\right) \, dt=2 c^2 \int_0^t \left(-\int_0^L u u_{t,x} \, d1+\int_0^L u u_{x,x,t} \, dx+\int_0^L u_t u_x \, d1\right) \, dt$$

Answer 2

Let $u(x,t) = f(x) g(t)$.

$\partial_{tt} u = c^2 \partial_{xx} u$

$f(x) g''(t) = c^2 f''(x) g(t)$

$\frac{g''(t)}{g(t)} = c^2\frac{f''(x)}{f(x)} = -\lambda^2 = constant$

$f(x) = \sin(rx)$

$g(t) = \cos(st)$

Hence we have $s^2 = c^2 r^2 = \lambda^2$

Hence we have, $u(x,t) = \sin(rx) \cos(st)$

Boundary condition:

$u(0,t) = 0$

$u_x(L,t) = r \cos(rL) \cos(st) = -\gamma \sin(rL) \cos(st)$

So $r \cos(rL) = -\gamma \sin(rL)$.

Find $r$ such that, $r = -\gamma \tan(rL)$

So solution: $$u(x,t) = \sin(-\gamma \tan(rL) x) \times \cos(\gamma \tan(rL) c t)$$

Now calculate energy.

I presume energy behaviour does not depend on the particular solution. So use this solution and integrate and find your answer.