APM346-2012 > Final Exam

problem 4

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Djirar:
Let $u$ solve the initial value problem for the wave equation in one dimension
\begin{equation*}
\left\{\begin{aligned}
& u_{tt}-  u_{xx}= 0 ,\qquad&& ~{\mbox{in}} ~\mathbb{R} \times (0,\infty),\\[3pt]
&u (0,x) = f(x), \qquad&& ~{\mbox{on}}~ \mathbb{R} \times \{t=0\} ,\\[3pt]
&u_t(0,x)= g(x),  \qquad&& ~{\mbox{on}}~ \mathbb{R} \times \{t=0\} .
\end{aligned}\right.
\end{equation*}
Suppose $f(x)=g(x)=0$ for all $|x|>1000.$ The  kinetic energy is
$$
k(t)= \frac{1}{2}\int_{-\infty}^{+\infty} u_t^2 (t,x) dx
$$
and the potential energy is
$$
p(t)= \frac{1}{2}\int_{-\infty}^{+\infty} u_x^2 (t,x) dx.
$$
 Prove

* $k(t)+ p(t)$ is constant with $t$ (so does not change as $t$ changes),
* $k(t)=p(t)$ for all large enough times $t$.
problem 4 part (a)

Chen Ge Qu:
I thought we were supposed to wait until Prof. Ivrii posted the problems...?

In any case, my solution to Problem 4a is attached.

Pei Zhou:
My answer to question 4.a

Danny Dinh:
My solution to part b.

Victor Ivrii:
I will grade problem 4 Dec 24. There is a solution for both (a) and (b) basically coinciding with one of Danny (without an obvious misprint): solution $u(x,t)=\phi (x-ct)+ \psi (x+ct)$ (thus does not working in higher dimensions where one based on energy integral works). Then
\begin{equation}
c^{-2}u_t^2 + u_x^2= \bigl( \phi'(x-ct)+\psi'(x+ct)\bigr)^2 + \bigl( \phi '(x-ct)-\psi'(x+ct)\bigr)^2=2 \bigl( \phi'(x-ct)\bigr)^2 + \bigl( \psi'(x+ct)\bigr)^2
\end{equation}
and integrating with respect to $x$ and making change of variables $x_{new}=x-ct$, $x_{new}=x+ct$ in the integrals containing $\phi'$ and $\psi'$ respectively we get
\begin{equation}
\int\bigl(c^{-2}u_t^2 + u_x^2\bigr)\,dx= 2 \int \bigl( \phi'^2 (x)+\psi'^2(x)\bigr)\,dx
\end{equation}
where r.h.e. does not depend on $t$. Here $\phi,\psi$ are related to $f,g$ in conditions by $f(x)= \phi(x)+\psi(x)$, $c^{-1}g(x)=-\phi'(x)+\psi'(x)$. 

Meanwhile
\begin{equation}
c^{-2}u_t^2 - u_x^2= \bigl(- \phi '(x-ct)+\psi '(x+ct)\bigr)^2 + \bigl( \phi '(x-ct)+\psi'(x+ct)\bigr)^2=-4  \phi '(x-ct) \psi'(x+ct)
\end{equation}
and the r.h.e. is $0$ unless both factors are not $0$ which can happen only as $|x-ct|\le R, |x+ct|\le R\implies  c|t|\le 2R$.  Therefore as $|t|\ge 2c^{-1}R$ the r.h.e. is identically $0$ and integrating with respect to $x$ we get (b)

Remark.  Obviously as either $u=\phi(x-ct)$ or $u=\psi (x+ct)$ we have $u_t^2=c^2u_x^2$ and $k(t)=p(t)$ for all $t$. The above proof is based on the observation that any solution with initial data supported in $\{|x|\le M\}$ after time $T=2c^{-1}M$ breaks into two waves $u_1=\phi(x-ct)$ and $u_2=\psi (x+ct)$which do not overlap. This is definitely not true for equation on the finite interval with energy preserving boundary conditions.

Remark. In higher dimensions (a) still holds for $u_{tt}-c^2\Delta u=0$
\begin{equation}
\int \bigl( |u_t|^2 + c^{-2}|\nabla u|^2\bigr)\,dV = \text{const}
\end{equation}
and (b) is replaced by
\begin{equation}
\int \bigl( |u_t|^2 - c^{-2}|\nabla u|^2\bigr)\,dV \to 0 \qquad \text{as}\quad  t\to \pm \infty.
\end{equation}
 While (a) is proven easily by an energy integral method, (b) becomes more subtle.

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