Let $u(x,t)\in{C^2}$ be a solution for the wave equation in $\mathbb{R}^n$.

\begin{align*} u_{tt}-\Delta{u}&=0 ,\ \ \ \ \ \ \ x\in{\mathbb{R}^n}, t>0\\ u(x,0)&=f(x), \ x\in{\mathbb{R}^n}\\ u_t(x,0)&=g(x), \ x\in{\mathbb{R}^n}. \end{align*}

Let $x_0\in{\mathbb{R}^n}$ and $T>0$, define for $0\leq t\leq T$ the energy integral: $$e(t):=\frac{1}{2}\int_{B(x_0,T-t)} (u_t^2+|\nabla_xu|^2))\mathrm dx.$$

A: Prove for all $0\le t\le T , e'(t)\le0$ (Hint: Use the polar coordinates: $$\int_{B(x_0,T-t)} (u_t^2+|\nabla_xu|^2)dx)=\int_{0}^{T-t}(\int_{\partial B(x_0,r)}(u_t^2+|\nabla_xu^2|)dSx)dr$$ then use Green's formula and finally use the formula $xy\le 1/2(x^2+y^2)$).

B: Prove that there is only one solution to the wave equation in $\mathbb{R}^n$.

I tried many many times to solve part A according to the tips and using the derivation rules for integrals, but every time I was stuck.

In part B,by assuming that there are two solutions to the given problem $u_1 and u_2$, defining $u:=u_1-u_2$. Then $u$ solves the homogeneous problem. using A, $e(t)$ is an decreasing function. Note that $e(0)=0$, so $e(t)\leq 0$. On the other hand, clearly $e(t)\geq 0$ by definition of $e(t)$, finally $e(t)=0$. Thus both $u_t=0$ and $\nabla {u}=0$ , then $u$ is a constant, $u(x,0)=0$ leading to $u=0$ and $u_1=u_2$. That proves that there's only one solution to the wave equation $\forall x \in \mathbb{R}^n$ and $t \in [0,T]$. Is that right?

Can someone help me in part A and give me tips for part B?

  • $\begingroup$ The hint of part a has several steps. Which parts are you stuck? $\endgroup$ – Arctic Char Jul 11 '20 at 10:50
  • $\begingroup$ In the derivation, I couldn't use the green formula after getting that integral on the boundary of the ball on $u_tu_{tt}+\nabla_xu\nabla_xu_t$ , knowing that i had to get integral of tha ball itself, (not its boundary) in order to use green formula, but maybe i was mistaken in the derviation.. $\endgroup$ – Usermat Jul 11 '20 at 11:31
  • $\begingroup$ @Arctic Char can you hep please. $\endgroup$ – Usermat Jul 12 '20 at 8:39

One has

$$2e(t):= \int_{B(x_0,T-t)} (u_t^2+|\nabla_xu|^2)\mathrm dx)=\int_{0}^{T-t}\left(\int_{\partial B(x_0,r)}(u_t^2+|\nabla_xu^2|)\mathrm dS_x \right)dr$$

which gives

\begin{align*} e'(t) &= -\frac{1}{2}\int_{\partial B(x_0,T-t)}(u_t^2+|\nabla_xu^2|)\mathrm dS_x + \int_0^{T-t} \int_{\partial B(x_0,r)}(u_tu_{tt}+\langle \nabla_xu, \nabla_x u_t\rangle)\mathrm dS_x \mathrm dt\\ &=-\frac{1}{2}\int_{\partial B(x_0,T-t)}(u_t^2+|\nabla_xu^2|)\mathrm dS_x + \int_{B(x_0, T-t)}(u_tu_{tt}+\langle \nabla_xu, \nabla_x u_t\rangle)\mathrm dx \end{align*}

The Green's theorem gives

\begin{align*} \int_{B(x_0,T-t)}\langle \nabla_xu, \nabla_x u_t\rangle\mathrm d x &= \int_{B(x_0,T-t)}(\nabla_x (u_t \nabla_x u) - u_t \Delta u)\mathrm d x \\ &= \int_{\partial B(x_0,T-t)} u_t \langle \nabla_x u, \vec n\rangle \ \mathrm dS_x-\int_{B(x_0, T-t)} u_t \Delta u\ \mathrm d x \end{align*}

where $\vec n$ is the unit normal vector of $\partial B(x_0, T-t)$. Using the equation $u_{tt} = \Delta u$, we have

\begin{align*} e'(t) = \frac{1}{2}\int_{\partial B(x_0,T-t)}(-u_t^2-|\nabla_xu|^2 + 2 u_t \langle \nabla_x u, \vec n\rangle )\mathrm dS_x \end{align*}


$$ |u_t \langle \nabla_x u, \vec n\rangle | \le |u_t| |\nabla _x u| | \vec n| = |u_t| |\nabla _x u| \le \frac{1}{2}(u_t^2 + |\nabla _x u|^2),$$

we obtain that $e'(t) \le 0$.

  • $\begingroup$ Arctic Char Thanks! I just did not understand why: $\int_0^{T-t} \int_{B(x_0,r)}(u_tu_{tt}+\langle \nabla_xu, \nabla_x u_t\rangle)\mathrm dS_x \mathrm dt\\=\int_{B(x_0, T-t)}(u_tu_{tt}+\langle \nabla_xu, \nabla_x u_t\rangle)\mathrm dx$. That was the reason i did not manage to solve it. Is what I did in part 2 right? $\endgroup$ – Usermat Jul 12 '20 at 12:32
  • $\begingroup$ It is just polar coordinates @Almaa your work to part B is correct. $\endgroup$ – Arctic Char Jul 12 '20 at 13:49

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