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Consider the PDE

\begin{align} u_{tt} - \nabla \cdot (c^2 \nabla u) + qu &= 0 \\ u(0,x) &= g(x) \\ u_t(0,x) &= h(x) \end{align}

where $c, q \geq 0$ depend only on $x$ and $0 < c_1 \leq c(x) < c_2$ for all $x \in \mathbb{R}^n$.

a) Fix $x_0 \in \mathbb{R}^n$ and $R_0 > 0$ and $t < R_0/c_2$. Let

\begin{equation} E(t)=\int_{B(x_0;R_0-c_2t)} \left(u_t^2 + c^2(x) \lvert \nabla u \rvert^2 + q(x) u^2 \right) d^nx \end{equation}

Show that $E(t)$ is decreasing.

b) Suppose that $\text{supp}(g), \text{supp}(h) \subset \{\lvert x \rvert < R \}$. Show that $u(t,x)=0 $ if $t > 0$ and $\lvert x \rvert > R + c_2 t$.

Showing a) seems easy as it is clear that $E(t_1) \geq E(t_2)$ for $0 \leq t_1 \leq t_2$ as increasing $t$ only decreases the radius of the ball over which we are integrating and the fact that the integrand is positive.

However, I am not sure how to deal with b). The approach that seems reasonable is to show that the energy is zero (which would imply $u$ is zero) for some initial moment and then use the fact that the energy is decreasing and non-negative to show that this holds for $t>0$ but still not sure.

In short, I'm looking for a hint in solving b).

Matthew Cassell
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    Careful about your argument for part a. The integrand may always be positive, but it's not constant in time. In particular, it could be increasing near the center of the sphere. The PDE ensures it can't be increasing fast enough to make $E'(t)$ positive, but you need to show that. – eyeballfrog Nov 16 '22 at 12:57

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