Duhamel's Principle for the wave equation 
attempt
a) To prove Duhamel's principle, we must show that (3) satisfies (1) and (2). One has
$$ u_t = U(x,0,t)$$
and so $u_{tt} = U_t(x,0,t) = f(x,t) $. Also, is $u_x = 0$? Can someone clarify this? I believe we have to differentiate under the integral, there is some trick that I forgot.
for b) then we see that I think we have to let $v(x,t) = u (x,t) - h(x) $ and observe that $v(x,0) = 0 $ and $v_t(x,0) = u_t(x,0)=g(x)$ so this choice doesnt work. How do we make this work?
 A: To prove Duhamel's principle all you have to do is differentiate. Remember that if
$$F(t) = \int _{a(t)} ^{b(t)} f(x, t) \ \mathrm d x$$
then
$$F'(t) = \int _{a(t)} ^{b(t)} \frac {\partial f} {\partial t} (x, t) \ \mathrm d x + f(b(t), t) \ b'(t) - f(a(t), t) \ a'(t) \ .$$
In your case, this means
$$u_t (x,t) = \int _0 ^t U_t (x, t-s, s) \ \mathrm d s + \underbrace{ U(x, 0, t) } _{=0} - 0 = \int _0 ^t U_t (x, t-s, s) \ \mathrm d s$$
so that, when differentiating once more,
$$u_{tt} (x,t) = \int _0 ^t U_{tt} (x, t-s, s) \ \mathrm d s + U_t (x, 0, t) - 0 = c^2 \int _0 ^t U_{xx} (x, t-s, s) \ \mathrm d s + f(x,t) = \\
= c^2 \left( \frac \partial {\partial x} \right) ^2 \int _0 ^t U (x, t-s, s) \ \mathrm d s + f(x,t) = c^2 u_{xx} (x,t) + f(x,t)$$
which is exactly your equation.
To check the initial conditions,
$$u(x,0) = \int _0 ^0 U (x, 0-s, s) \ \mathrm d s = 0$$
and (using the calculation of $u_t$ performed above)
$$u_t (x,0) = \int _0 ^0 U_t (x, 0-s, s) \ \mathrm d s = 0 \ .$$

To solve the problem
$$\left\{ \begin{eqnarray}
& u_{tt} - c^2 u_{xx} &= f(x,t) \\
& u(x,0) &= g(x) \\
& u_t (x,0) &= h(x)
\end{eqnarray} \right.$$
consider the two auxiliary problems
$$\left\{ \begin{eqnarray}
& v_{tt} - c^2 v_{xx} &= 0 \\
& v(x,0) &= g(x) \\
& v_t (x,0) &= h(x)
\end{eqnarray} \right.$$
and
$$\left\{ \begin{eqnarray}
& w_{tt} - c^2 w_{xx} &= f(x,t) \\
& w(x,0) &= 0 \\
& w_t (x,0) &= 0 \ .
\end{eqnarray} \right.$$
The first one is homogeneous, so it can be solved with the classical approach. The second one is non-homogeneous, so you reduce it with Duhamel's principle to the homogeneous problem
$$\left\{ \begin{eqnarray}
& W_{tt} - c^2 W_{xx} &= 0 \\
& W(x,0,s) &= 0 \\
& W_t (x,0,s) &= f(x,s)
\end{eqnarray} \right.$$
that you solve with the same classical method as above, and then
$$w(t,x) = \int _0 ^t W(x, t-s, s) \ \mathrm d s \ .$$
Check for yourself (it is easy) that if $v$ and $w$ solve the problems above, then their sum $v + w$ solves the problem given initially.
Finally, you show that this solution is unique by using some uniqueness theorem given to you in the lecture notes.
