Inhomogeneous Wave Equation Derivation

Question

This is an assignment question which I've been working on to solve the inhomogeneous wave equation $u_{tt} - c^{2}u_{xx} = f(x,t)$.

I separated the equation out into a system of two equations: $u_{t} + cu_{x} = v$ and $v_{t} - cv_{x} = f(x,t)$.

It says to solve the first differential equation to find that $u(x, t) = \int\limits_{0}^{t} v(x-ct+cs,s) ds$.

My idea is to use the linearity of the differential equation $u_{t} - cu_{x}$ to get that the solution to the homogenous equation is $f(x-ct)$ and then guessing that another solution is $g(x+ct)$ and using this to derive the solution given but didn't have any luck.

What's the general process to solving these kind of differential equations? I'm mostly attempting random ideas to solve it.

Answer 1

Similar to how to solve this PDE:

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

Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}$

$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial x}=\dfrac{\partial^2u}{\partial p^2}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}$

$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial t}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial t}=c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}$

$\dfrac{\partial^2u}{\partial t^2}=\dfrac{\partial}{\partial t}\left(c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial t}+\dfrac{\partial}{\partial q}\left(c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial t}=c^2\dfrac{\partial^2u}{\partial p^2}-c^2\dfrac{\partial^2u}{\partial pq}-c^2\dfrac{\partial^2u}{\partial pq}+c^2\dfrac{\partial^2u}{\partial q^2}=c^2\dfrac{\partial^2u}{\partial p^2}-2c^2\dfrac{\partial^2u}{\partial pq}+c^2\dfrac{\partial^2u}{\partial q^2}$

$\therefore c^2\dfrac{\partial^2u}{\partial p^2}-2c^2\dfrac{\partial^2u}{\partial pq}+c^2\dfrac{\partial^2u}{\partial q^2}-c^2\dfrac{\partial^2u}{\partial p^2}-2c^2\dfrac{\partial^2u}{\partial pq}-c^2\dfrac{\partial^2u}{\partial q^2}=f\left(\dfrac{p+q}{2},\dfrac{p-q}{2c}\right)$

$-4c^2\dfrac{\partial^2u}{\partial pq}=f\left(\dfrac{p+q}{2},\dfrac{p-q}{2c}\right)$

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

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

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

Answer 2

The one dimensional wave equation is easy because as you did you can factor it into (setting $c=1$) $(\partial_t-\partial_x)(u_t+u_x)=f$, or calling $v\equiv u_t+u_x$ we have $v_t-v_x=f$, or $\frac{dv(s,x+t-s)}{ds}=f(s,x+t-s)$, so that integrating we get $v(t,x)-v(0,x+t)=\int_0^t f(s',x+t-s')\,ds'$. Now as for the $u_t+u_x=v$ equation, this is the same as $\frac{du(s,x-t+s)}{ds}=v(s,x-t+s)$ and integrating gives $u(t,x)-u(0,x-t)=\int_0^tv(s',x-t+s')\,ds$. Then you can throw them together.

In response to the comment:

Well $u_t+u_x$ is just the derivative of $u$ along the line $x-t=$ const., or in other words it is just $\nabla u\cdot(1,1)$ if you prefer. So we have $v(t_0+s,x_0+s)=u_t(t_0+s,x_0+s)+u_x(t_0+s,x_0+s)=\frac{du(t_0+s,x_0+s)}{ds}\,,$ and presumably you are given an initial condition $u(0,x)=g(x)$, and since we know that $v(s,x_0+s)=\frac{du(s,x_0+s)}{ds}$ then to find $u(x,t)$ we want to integrate $s$ from $0$ to $t$ to get $u(t,x_0+t)-u(0,x_0)=\int_o^t v(s,x_0+s)\,ds\,.$ Now this is ok but since we are free to choose $x_0$ it is simplest if we choose $x_0=x-t$ because then we get $u(t,x)-u(0,x-t)=\int_o^t v(s,x-t+s)\,ds$.