I am learning PDE, and I encountered an equation.
\begin{cases}u_{t t}=a^2 u_{x x}+f(x, t), & 0<x<+\infty, t>0, \\ u(x, 0)=\varphi(x), u_t(x, 0)=\psi(x), & 0 \leq x<+\infty \\ u(0, t)=0, & t \geq 0\end{cases}
One way to solve it is by using odd reflection.
$\Phi(x)= \begin{cases}\varphi(x), & x \geq 0 \\ -\varphi(-x), & x<0\end{cases}$ $\Psi(x)= \begin{cases}\psi(x), & x \geq 0 \\ -\psi(-x), & x<0\end{cases}$ $F(x, t)= \begin{cases}f(x, t), & x > 0 \\ 0, & x=0 \\ -f(-x, t), & x<0\end{cases}$
Then we can solve this equation by the Duhamel principle(https://en.wikipedia.org/wiki/Duhamel's_principle) and confine the solution to $[0,\infty] \times [0,\infty]$. \begin{cases}U_{t t}=a^2 U_{x x}+F(x, t), & -\infty<x<+\infty, t>0 \\ U(x, 0)=\Phi(x), U_t(x, 0)=\Psi(x), & -\infty<x<+\infty\end{cases}
My question is that we can easily check that $\Phi$ and $\Psi$ are continuous on $R$, but $F$ cannot be continuous at 0, I think that violates the requirements to use the Duhamel principle. So we can not solve it this way.
In fact, we can solve this equation by odd reflection and the Duhamel principle. \begin{cases}u_{t t}=a^2 u_{x x}+f(x, t)-f(0,t), & 0<x<+\infty, t>0, \\ u(x, 0)=\varphi(x), u_t(x, 0)=\psi(x), & 0 \leq x<+\infty \\ u(0, t)=0, & t \geq 0\end{cases}
