This problem is from our recitation which I do not have solutions for, and I'm stuck on the very last part where I need to satisfy the $u_t(x,0)$ initial condition.
The problem is:
$$
\left\{
\begin{split}
u_{tt} &= c^2u_{xx}, \quad\qquad x>0,t>0\\
u(&0,t) = 0 \qquad\qquad\qquad t>0\\
u(&x,0) = \sin(x) = f(x)\;\quad x>0\\
u_t&(x,0) = e^{-x} = g(x)\quad\quad x>0
\end{split}
\right.
$$
Since we have Neumann conditions we use an even extension. Let
$$
u(x,0) =
\begin{cases}
f(x) &\text{if } x > 0 \\
f(-x) &\text{if } x < 0
\end{cases}
$$
and
$$
u_t(x,0) =
\begin{cases} g(x) &\text{if } x > 0 \\
g(-x) &\text{if } x < 0
\end{cases}
$$
Then D'Alembert's formula gives:
$$
\begin{split}
u(x,t) &= \frac{1}{2}\big(f(x-ct)+f(x+ct)\big) + \frac{1}{2c}\int_{x-ct}^{x+ct} g(s)\,ds\\
&= \frac{1}{2}\big(f(x)+f(x)\big) + \frac{1}{2c}\int_{x}^{x} g(s)\,ds \\
&= f(x)=\sin(x)
\end{split}
$$
thus the First initial condition is satisfied. For the second one we have
$$
u_t(x,t) = \frac{1}{2}\big(-cf(x-ct)+cf(x+ct)\big) + \frac{d}{dt}\bigg [ \frac{1}{2c} \int_{x-ct}^{x+ct} g(s)\,ds \bigg ]
$$
My problem is I'm not sure what to do with the
$$
\frac{d}{dt}\bigg [ \frac{1}{2c} \int_{x-ct}^{x+ct} g(s)\,ds \bigg ]
$$
term. Should I differentiate the integral and then plug in $t = 0$ for the whole expression? How do I get the term $e^{-x}$ from this?
Best Answer
FTC says $$\frac{d}{dt}\int_{a(t)}^{b(t)}f(s)ds = f(b(t))b'(t)-f(a(t))a'(t).$$ Substituting your functions, we get $$\frac{d}{dt}\frac{1}{2c}\int_{x-ct}^{x+ct}g(s)ds = \frac{1}{2c}(g(x)c+g(x)c) = g(x),$$ as desired.