# Non-Homogeneous 1-Dimensional Wave Equation with arbitrary initial/boundary conditions

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).

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 $v(x,t)=h(x+t) + g'(x+t) + \displaystyle\int_{-t}^0 f(x-s,t+s)ds$.

Is this correct?

Let $\begin{cases}p=x+t\\q=x-t\end{cases}$ ,

Then $u_x=u_pp_x+u_qq_x=u_p+u_q$

$u_{xx}=(u_p+u_q)_x=(u_p+u_q)_pp_x+(u_p+u_q)_qq_x=u_{pp}+u_{pq}+u_{pq}+u_{qq}=u_{pp}+2u_{pq}+u_{qq}$

$u_t=u_pp_t+u_qq_t=u_p-u_q$

$u_{tt}=(u_p-u_q)_t=(u_p-u_q)_pp_t+(u_p-u_q)_qq_t=u_{pp}-u_{pq}-u_{pq}+u_{qq}=u_{pp}-2u_{pq}+u_{qq}$

$\therefore u_{pp}-2u_{pq}+u_{qq}-(u_{pp}+2u_{pq}+u_{qq})=f\left(\dfrac{p+q}{2},\dfrac{p-q}{2}\right)$

$-4u_{pq}=f\left(\dfrac{p+q}{2},\dfrac{p-q}{2}\right)$

$u_{pq}=-\dfrac{1}{4}f\left(\dfrac{p+q}{2},\dfrac{p-q}{2}\right)$

$u(p,q)=F(p)+G(q)-\dfrac{1}{4}\int_0^q\int_0^pf\left(\dfrac{r+s}{2},\dfrac{r-s}{2}\right)~dr~ds$

$u(x,t)=F(x+t)+G(x-t)-\dfrac{1}{4}\int_0^{x-t}\int_0^{x+t}f\left(\dfrac{r+s}{2},\dfrac{r-s}{2}\right)~dr~ds$