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Show that a wave equation $\rho u_{tt}=Tu_{xx}$ has solution $u=0$ if $u(x,0)=u_t(x,0)=0$.

Thoughts: This is easy using the general solution to wave equations

$$u(x,t)=\frac12(\phi(x+ct)+\phi(x-ct))+\frac1{2c}\int_{x-ct}^{x+ct}\psi(s)\,ds$$

with $c=\sqrt{T/\rho}$ and $\phi$ and $\psi$ as initial displacement and velocity respectively. Obviously the form above will be $0$. However, I'm trying to show that the only solution is $0$ using conservation laws.

Conservation Laws: We have $E=KE+PE$ (energy is kinetic plus potential) and $dE/dt=0$. In particular,

$$KE=\frac12\int_D\rho u_t^2\,dx$$ $$PE=\frac12\int_D Tu_x^2\,dx$$

Question: I can't think of a good strategy to show that the conservation of energy implies that my PDE has only the trivial solution, even though intuitively it makes a lot of sense.

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    $\begingroup$ Why do you think you need the integral term? It is the sum of two functions, one a function of $x+ct$, the other a function of $x-ct$. $\endgroup$ – Mark Viola Sep 21 '15 at 19:22
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You know that $E(t)' =0$ where $$ E(t) = \frac{1}{2}\int (\rho u_t^2 + T u_x^2) dx = \frac{1}{2}\int (\rho u_t^2 - T u u_{xx}) dx $$ via integration by parts (assuming there is no wave at infinity). Clearly $$ E(0) = 0 $$ by the initial conditions. Thus we see $$ \frac{1}{2}\int (\rho u_t^2 + T u_x^2) dx = 0 \quad \forall t \in \mathbb{R}_+ $$ Since the integrand is non-negative, we see that $$ u_t = u_x = 0 \quad \forall x,t \in \mathbb{R} \times \mathbb{R}_{+}$$ along with the initial conditions, we conclude $$ u \equiv 0 $$

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The other way is using that Cauchy problem can have at most one solution, and noting that $u = 0$ is a solution, it is perfectly correct.

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