Toronto Math Forum
APM3462012 => APM346 Math => Final Exam => Topic started by: Djirar on December 20, 2012, 01:33:07 PM

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)

I thought we were supposed to wait until Prof. Ivrii posted the problems...?
In any case, my solution to Problem 4a is attached.

My answer to question 4.a

My solution to part b.

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 $u(x,t)=\phi (xct)+ \psi (x+ct)$ (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'(xct)+\psi'(x+ct)\bigr)^2 + \bigl( \phi '(xct)\psi'(x+ct)\bigr)^2=2 \bigl( \phi'(xct)\bigr)^2 + \bigl( \psi'(x+ct)\bigr)^2
\end{equation}
and integrating with respect to $x$ and making change of variables $x_{new}=xct$, $x_{new}=x+ct$ 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 '(xct)+\psi '(x+ct)\bigr)^2 + \bigl( \phi '(xct)+\psi'(x+ct)\bigr)^2=4 \phi '(xct) \psi'(x+ct)
\end{equation}
and the r.h.e. is $0$ unless both factors are not $0$ which can happen only as $xct\le R, x+ct\le R\implies ct\le 2R$. 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 $u=\phi(xct)$ 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 $u_1=\phi(xct)$ 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.

I finished grading problem 4(a),(b). While (a) was mastered by many, only few managed with (b).
Also some tried to prove that $\lim_{t\to \infty} k(t)=\lim_{t\to \infty} p(t)=\frac{1}{2}E$ while in fact a stronger statement was required $p(t)=q(t)=\frac{1}{2}E$ for $t\ge T$.
The weaker statement could be proven then under relaxed condition $E<\infty$ instead of $f,g$ being supported in $x\le M$.