Any guidance, a question in PDE-one dimensional wave equation Let $u∈C^2(\mathbb{R}×[0,∞))$ be a solution of the following one-dimensional wave equation:
$$
\begin{align*}
&u_{tt}−u_{xx} = 0,& x∈\mathbb{R},\, t >0
\\
&u(0,x) = f(x) ,& x∈\mathbb{R}
\\
&u_t(0,x) = g(x) ,& x∈\mathbb{R}
\end{align*}
$$
$f$ , $g$ are derivable functions with support in $[-M,M]$.
Define $k(t):=\frac{1}{2}\int_{-\infty}^{\infty}u^2_t(x,t)\,\mathrm{d}x$ and $p(t) :=\frac{1}{2}\int_{-\infty}^{\infty} u^2_x(x,t)\,\mathrm{d}x$.
Prove:

*

*For every $t>=0$, $K(t)+P(t)=\text{constant}$.


*For every $t>M$, $K(t)=P(t)$.
I began solving part 1 by taking $K(t)+P(t)=e(t)$
and calculating $e'(t)$ to show that it equals $0$, using the wave equations and Dalimber formula but I got nothing, I did not manage to show this.
I will be grateful if someone can give me tips or any direction.
 A: You asked for tips so I won't write all details.
For part 1, you started in the right direction by calculating $e'(t)$, but you do not need the d'Alembert formula for that, just integrate one of the terms that you get, namely $\partial_t(u_x^2) = 2u_xu_{xt}$, integrate that by parts and you will get part 1. Actually you do have to get rid of the boundary term when you integrate by parts and the d'Alembert gives that is zero because $u$ has bounded support at all times.
For part 2, it is helpful to use the d'Alembert formula because it shows how $u$ is a sum of left and right traveling waves. After time $M$, these waves have disjoint support, i.e. they do not overlap after time $M$. You can use that in calculating $K$ and $P$ because the formula $(a+b)^2 = a^2+b^2$ that is usually false, is true when $a$ and $b$ are functions with disjoint support. I will leave that as the tip.
Edits in response to comment:
First the d'Alembert formula for the
solution to the problem
is
$$
 u(x,t) = \frac{1}{2}\Big(f(x+t)+f(x-t)+\int_{x-t}^{x+t}g(s)ds\Big).
$$
From this you can deduce that $u=0$ when $x> t+M$ or $x< -t-M$.
So also $u_t = u_x = 0$ in those regions.
That is enough information to prove part 1. Here is how.
$$
 e'(t) = \int(u_tu_{tt}+u_xu_{xt})dx
 $$ $$
       =\int(u_tu_{xx}-u_{xx}u_t)dx
       =
            0.
$$
That used the wave equation on the first term and integration by parts
on the second term. The integration by parts worked because $u_x$
and $u_t$ are zero for large $|x|$ as we saw above.
For part 2 we first
differentiate the solution formula to get
$$
 u_t(x,t) = \frac{1}{2}\Big(
                 f'(x+t)+g(x+t)-f'(x-t)+g(x-t)
                       \Big),
 \qquad
 u_x(x,t) = \frac{1}{2}\Big(
                 f'(x+t)+g(x+t)+f'(x-t)-g(x-t)
                       \Big).
$$
Define functions
$h(s) = \frac{1}{2}(f'(s)+g(s))$ and $k(s) = \frac{1}{2}(f'(s)-g(s))$.
Observe that
$$
 u_t(x,t) = h(x+t)-k(x-t),
 \qquad
 u_x(x,t) = h(x+t)+k(x-t).
$$
The main idea is that $h$ and $k$ travel to the left and right,
respectively, and since each has compact support the supports
eventually pull away from each other. In particular
$h(x+t)k(x-t) = 0$ when $t$ is sufficiently large. Therefore
$$
 u_t^2 = h^2(x+t)+k^2(x-t),
 \qquad
 u_x^2 = h^2(x+t)+k^2(x-t)
$$
when $t$ is sufficiently large, giving part 2.
But what does sufficiently mean? The problem says $t>M$.
Let's see why that is right. The supports of $f$ and of $g$
are in $[-M,M]$, so are the supports of $h$ and of $k$. Therefore
the support of $h(x+t)$ is in $[t-M,t+M]$, and
the support of $k(x-t)$ is in $[-t-M,-t+M]$.
When $t > M$ these are disjoint as required.
A: Another method to solve part 2:
Write the solution $u(x,t)$ by using the D'elamber formula, then derivate it and find $u_x , u_t$, now for every $t>M$ : we consider two cases, 1.$x>t+M , 2.x<t+M$, and noticing that some functions equal 0 considering that $f,  g$ are supported in $[-M,M]$.
