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The equations are

\begin{equation*} \begin{cases} u_{tt} = (c^2(x)u_x)_x,&\quad 0<x<L,\ t >0\\ u(0,t) - u_x(0,t) = 0, &\quad t>0\\ u(L,t) + u_x(L,t) = 0,&\quad t>0\\ u(x,0) = f(x) , u_t(x,0) = g(x),&\quad 0<x<L, \end{cases} \end{equation*}

where $c(x) >0$ for $0<x<L$. Don't know really go about to show it's unique other than supposing $2$ solutions exits, and letting $w$ be the difference between the two the multiplying by $w_t$, and integrating with respect to $x$. However, I'm getting lost in the integration step. Please help? Thank you.

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  • $\begingroup$ For some basic information about writing math on MSE see here, here, here and here. $\endgroup$ – user409521 Apr 6 '17 at 3:33
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Here's my attempt at a proof. If you also want to consider consider weak solutions, there are a few extra considerations, but the argument is the same.

The general strategy is to show that if the energy of a solution is zero at any time, it must be the trivial solution. We can use this to show the difference between two solutions must be the trivial solution, implying the two solutions are identical.

For this problem, I'll use the following energy functional.

$$U=\frac 12\int_0^Lu_t^2+c^2u_x^2dx$$

Since non-constant wave speed is a somewhat uncommon case, if might be useful to explicitly show that $U$ has the properties we want. We can start by showing it is constant for a particular solution.

We can take the derivative of both sides w.r.t. $t$, propagate the time derivative into the terms of the integrand, noting that $c$ is still constant w.r.t. time.

$$U_t=\int_0^Lu_tu_{tt}+c^2u_xu_{xt}dx$$

$$U_t=\int_0^Lu_tu_{tt}dx+\int_0^Lc^2u_xu_{xt}dx$$

We can then use integration by parts on the 2nd term.

$$U_t=\int_0^Lu_tu_{tt}dx+\int_0^L\left[\frac\partial{\partial x}\left(c^2u_xu_t\right)-(c^2u_x)_xu_t\right]dx$$

$$U_t=\int_0^Lu_tu_{tt}dx+c^2u_xu_t|_0^L-\int_0^L(c^2u_x)_xu_tdx$$

Since $u(0,t)=u(L,t)=0$, $u_t(0,t)=u_t(L,t)=0$, so the 2nd term is always zero. We can then recombine the other two terms and factor.

$$U_t=\int_0^Lu_tu_{tt}-(c^2u_x)_xu_tdx$$

$$U_t=\int_0^Lu_t(u_{tt}-(c^2u_x)_x)dx$$

If $u$ is a solution to the PDE, $u_{tt}-(c^2u_x)_x$ will be zero everywhere, so we have $U_t=0$, and $U$ must be constant.

We also need to show that $U=0$ implies the trivial solution. To do this, note that $u_t^2$ and $u_x^2$ are nonnegative and $c^2$ is strictly positive. This tells us that $u_x=0$ almost everywhere. Since solutions must be twice differentiable, this is enough to conclude that $u(x)$ is constant for fixed $t$. The only way this satisfies the boundary conditions is $u(x,t)=0$.

From the above, we have shown that, for a particular solution, if $U=0$ at any time, $U=0$ for all $t$, and $u(x,t)=0$. This is the statement we use to prove uniqueness.

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Suppose there are two solutions, $\phi$ and $\psi$, that satisfy the differential equation, the initial conditions, and the boundary conditions. We can then consider their difference $\delta=\phi-\psi$.

Since the PDE is linear, we know $\delta$ is also a solution. Also, $\delta$ has the values of $\phi-\psi$ on the boundaries:

$$\delta_tt=(c^2\delta_x)_x$$

$$\delta(0,t)=\delta(L,t)=0$$

$$\delta(x,0)=0\ \ \implies\ \ \delta_x(x,0)=0$$

$$\delta_t(x,0)=0$$

$\delta$ still satisfies the boundary conditions and the PDE, so our energy functional still applies. From the initial conditions, it is clear that the energy of $\delta$ is $0$ at $t=0$. We can thus conclude that $\delta$ must be $0$ everywhere, which implies $\phi=\psi$. Thus, solutions to the problem must be unique.

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  • $\begingroup$ Thanks for writing down all details, however in the OP has curious boundary conditions (2nd and 3rd line) which I don't see why $\delta$ would satisfy $\endgroup$ – Noix07 May 4 at 11:27

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