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Consider the wave equation $u_{tt}-\sum a^{ij}u_{x_i}u_{x_j}=0$. Assume that for any $v\in \Bbb{R}^n$, we have $\sum\limits_{i,j} a^{ij}v_iv_j\leq c^2\sum\limits_{k=1}^n v_k^2$. Prove that if $u,v$ are two solutions with the intial data satisfying $u(0,x)=v(0,x)$ for $|x|\geq R$, then $u(t,x)=v(t,x)$ for all $t>0$, $|x|\geq R+ct$.

I don't know how to solve this question exactly. I know that I'm supposed to use the fact that if we have a cone such that at $t=0$, we have $u-v=0$, then $u=v$ throughout the cone. However, we don't have that here. Also, I don't know how to use the condition that $\sum\limits_{i,j} a^{ij}v_iv_j\leq c^2\sum\limits_{k=1}^n v_k^2$. Any hints will be helpful.

UPDATE: We haven't really done in Fourier analysis in that. So hints without assuming Fourier transforms would be very helpful.

Anju George
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1 Answers1

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Express $u(x,t)$ in terms of its Fourier transform, $$u(x,t) = \int dk_1 \dots dk_ndt \ \hat{u}(\vec{k},\omega)e^{i(\vec{k}\cdot \vec{x} - \omega t)}, \quad \textbf{(1)} $$

substitute this into the differential equation and obtain a relationship between $\vec{k}$ and $\omega$.

$$-\omega^2 + \sum a^{ij} k_ik_j=0,\quad \textbf{(2)}$$

use the inequality to demonstrate that,

$$ \omega^2 < c^2 \|\vec{k}\|^2,\quad \textbf{(3)}$$

$$ \frac{\omega^2}{\|\vec{k}\|^2} < c^2 .\quad \textbf{(4)}$$

Now we have that the phase velocity of our waves is slower than the speed $c$.

Spencer
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