Solution to Goursat Problem

Question

I finished my maths degree a few months ago and one of the modules I did was methods in PDE’s. Question $1$ on the exam paper surrounded the goursat problem which I thought I’d revised throughly. However in lectures we studied the goursat problem in the first canonical form of the wave equation:

The problem that we will consider has the form:

\begin{align} u_{xy}&=f(x,y)\quad\quad(x\gt0,y\gt0),\\ u(x,0)&=\phi_1(x),\\ u(0,y)&=\phi_2(y),\\ \end{align}

However, on the exam it appeared in the form:

Find the solution to the following Goursat problem:

\begin{align} u_{tt}&=u_{xx},\quad\quad(-t\lt x\lt t,t\gt0),\\ \\ u&=\sin(x)\quad\text{on}\ x+t=0,x\lt0,\\ u&=\frac{x}{x+1}\quad\text{on}\ x-t=0,x\ge0.\\ \end{align}

I had absolutely no clue how to do this in the exam but since then I’ve thought long and hard about it as it has been bugging me ever since. I know I need to make a change of independent variables going from variables in which the problem is formulated to those in which the characteristics on the boundary conditions are given as coordinate lines but I’m so out of practice now that I’m struggling with even the change of variables. Could someone have a go at this so I can finally see how to do it?

Thank you in advance.

Answer 1

A drawing of the domain $\{(t,x)\,|\,-t<x<t\}$ of this function shows that it is the standard quarter-plane of the Goursat problem tilted by 45° clockwise. This suggests the change of variables $y=t-x$ (so that $y$ runs southeast) and $z=t+x$ (so that $z$ runs northeast).

Now we check that this makes our problem into a Goursat problem :

• for the boundary conditions, $y=0,z\geq0$ gives $x-t=0$ and $2x=t+x\geq0$, and $y\geq0,z=0$ gives $x+t=0$ and $-2x=t-x\geq0$ : check !
• for the equation, our change of variables gives $t=(y+z)/2$ and $x=(z-y)/2$, so we define $v(y,z)=u(\frac{z+y}{2},\frac{z-y}{2}）$. Now $v_y=\frac{1}{2}u_t-\frac{1}{2}u_x$ so $v_{yz}=\frac{1}{4}((u_{tt}+u_{tx})-(u_{xt}+u_{xx})) = \frac{1}{4}(u_{tt}-u_{xx})$ and the equation becomes $v_{yz}=0$ : check !

Answer 2

In fact the exam problem is $u_{tt}=u_{xx}$ with conditions $u(x,-x)=\sin x$ and $u(x,x)=\dfrac{x}{x+1}$ .

The general solution is $u(x,t)=f(x+t)+g(x-t)$ .

$u(x,x)=\dfrac{x}{x+1}$ :

$f(2x)+g(0)=\dfrac{x}{x+1}$

$f(2x)=\dfrac{x}{x+1}-g(0)$

$f(x)=\dfrac{\dfrac{x}{2}}{\dfrac{x}{2}+1}-g(0)=\dfrac{x}{x+2}-g(0)$

$u(x,-x)=\sin x$ :

$f(0)+g(2x)=\sin x$

$g(2x)=\sin x-f(0)$

$g(x)=\sin\dfrac{x}{2}-f(0)$

$\therefore u(x,t)=\dfrac{x+t}{x+t+2}+\sin\dfrac{x-t}{2}-f(0)-g(0)=\dfrac{x+t}{x+t+2}+\sin\dfrac{x-t}{2}+h(0)$

$u(0,0)=0$ :

$h(0)=0$

$\therefore u(x,t)=\dfrac{x+t}{x+t+2}+\sin\dfrac{x-t}{2}$