Solve by direct methods the $1$-dimensional non-homogeneous wave equation
$$u_{tt} – u_{xx} = f(x,t), \hspace{1cm} u(x,0) = g(x), \hspace{1cm} u_t (x,0) = h(x).$$
Our solution will be of the form $u(x,t) = p(x,t) + o(x,t)$, where $p(x,t)$ is a particular solution satisfying homogeneous boundary conditions:
\begin{align*}
p_{tt} – p_{xx} &= f(x,t) \\
p(x,0) &= 0 \\
p_t (x,0) &= 0
\end{align*}
and $o(x,t)$ is a solution of the corresponding to the homogeneous PDE:
\begin{align*}
o_{tt} – o_{xx} &= 0 \\
o(x,0) &= g(x) \\
o_t (x,0) &= h(x)
\end{align*}
By D'Almbert's formula we know that
$$o(x,t) = \dfrac{1}{2}\left( g(x + t) + g(x – t) + \int_{x-t}^{x+t} h(y)dy\right)$$
So our goal is to find the solution to the particular solution $p$. That is we need to solve
\begin{align*}
p_{tt} – p_{xx} &= f(x,t) \\
p(x,0) &= 0 \\
p_t(x,0) &= 0
\end{align*}
I know that the solution should be
$$p(x,t) = \dfrac{1}{2}\int_0^t \int_{x-(t-s)}^{x+(t-s)} f(r,s)drds.$$
where Duhamel's principle should be utilized somewhere.
Question: How do I solve the PDE involving $p$? My lecture notes only have the solution involving $o$ (i.e. derivation of D'Almbert's formula).
Best Answer
Here is an attempt:
Let $v(x,t) := u_t + u_x$. Then, we have that $$v_t - v_x = u_{tt} + u_{xt} - u_{xt} - u_{xx} = f(x,t)$$ where \begin{align*} v(x,0) &= u_t (x,0) + u_x (x,0) \\ &= h(x) + g'(x) \end{align*} Now we let $w(s) = v(x-s,t+s)$, so that \begin{align*} w'(s) &= -v_x(x-s,t+s) + v_t(x-s,t+s) \\ &=f(x-s,t+s) \end{align*} Hence, $$w(\tau) - w(-t) = \int_{-t}^{\tau} f(x-s,t+s)ds$$ so that, \begin{align*} w(\tau) &= w(-t) + \int_{-t}^{\tau} f(x-s,t+s)ds \\ &= v(x+t,0) + \int_{-t}^{\tau} f(x-s,t+s)ds \end{align*} Therefore, $$v(x,t) = w(0) = h(x+t) + g'(x+t) + \int_{-t}^0 f(x-s,t+s)ds$$ So that we're left with: $$\begin{cases} u_t + u_x &= v(x,t) \\ u_x (x,0) &= g(x) \\ u_t (x,0) &= h(x) \end{cases}$$ Now if we let $z(s):=u(x+s,t+s)$, then: \begin{align*} z'(s) &= u_x(x+s,t+s) + u_t(x+s,t+s) \\ &= v(x+s,t+s) \end{align*} Therefore, $$z(\tau) - z(-t) = \int_{-t}^{\tau} v(x+s,t+s)ds$$ \begin{align*} z(0) = u(x,t) &= u(x-t,0) + \int_{-t}^0 v(x+s,t+s)ds \\ &= g(x-t) + \int_{-t}^0 v(x+s,t+s)ds \end{align*} where $v(x,t)=h(x+t) + g'(x+t) + \displaystyle\int_{-t}^0 f(x-s,t+s)ds$.