# Solving Wave Equations with different Boundary Conditions

Right now I'm studying the wave equation and how to solve it with different boundary conditions (i.e. $u(x,0);u(0,t);u_t(x,0);u_x(x,0);u(x,x);u_t(x,x)...$)

I know how to solve it when the boundary conditions are $u(x,0)=f(x)$ and $u_t(x,0)=g(x)$ with d'Alambert's formula. But I don't know how to solve it in any other case.

Does anyone know of any book, notes, etc. where I can find an explanation on how to solve the wave equation with many kinds of boundary conditions?

Examples:

$$u_{tt}-4u_{xx}=16$$ $$u(0,t)=t;u_x(0,t)=0$$

or

$$u_{tt}-4u_{xx}=2$$ $$u(x,x)=x^2;u_t(x,x)=x$$

I'm talking about the wave equation with many kinds of initial conditions, not just only the ones in d'Alambert's solution.

Thanks.

• More specific examples? Your question talks about what are usually called initial conditions, since they are specified at time $t=0$. – abnry Mar 30 '14 at 1:32
• Check the laplace method for solving the wave equation. It is well-suited for problems with different boundary conditions and more robust than the d'alembert solution – user32882 Aug 25 '17 at 13:10

I admitted for the current stiff PDE education system, it is hard to find the examples of solving PDEs with sightly innovative types of conditions.

So it is the time to use the brains ourselves.

For your first example, it gives the message that the roles of $x$ and $t$ can be interchanged.

Please use the general solution of the form $u(x,t)=f\left(t+\dfrac{x}{2}\right)+g\left(t-\dfrac{x}{2}\right)+4t^2-x^2$ instead for convenience.

$u(0,t)=t$ :

$f(t)+g(t)+4t^2=t$

$f(t)+g(t)=t-4t^2~......(1)$

$u_x(x,t)=f_x\left(t+\dfrac{x}{2}\right)+g_x\left(t-\dfrac{x}{2}\right)-2x=\dfrac{1}{2}f_t\left(t+\dfrac{x}{2}\right)-\dfrac{1}{2}g_t\left(t-\dfrac{x}{2}\right)-2x$

$u_x(0,t)=0$ :

$\dfrac{1}{2}f_t(t)-\dfrac{1}{2}g_t(t)=0$

$f_t(t)-g_t(t)=0$

$f(t)-g(t)=c~......(2)$

$\therefore f(t)=\dfrac{t}{2}-2t^2+\dfrac{c}{2},g(t)=\dfrac{t}{2}-2t^2-\dfrac{c}{2}$

$\therefore u(x,t)=\dfrac{t}{2}+\dfrac{x}{4}-2\left(t+\dfrac{x}{2}\right)^2+\dfrac{c}{2}+\dfrac{t}{2}-\dfrac{x}{4}-2\left(t-\dfrac{x}{2}\right)^2-\dfrac{c}{2}+4t^2-x^2=t-2x^2$

Please use the general solution of the form $u(x,t)=f(2t+x)+g(2t-x)+\dfrac{t^2}{2}-\dfrac{x^2}{8}$ instead for convenience.

$u(x,x)=x^2$ :

$f(3x)+g(x)+\dfrac{x^2}{2}-\dfrac{x^2}{8}=x^2$

$f(3x)+g(x)=\dfrac{5x^2}{8}~......(1)$

$u_t(x,t)=f_t(2t+x)+g_t(2t-x)+t=2f_x(2t+x)+2g_x(2t-x)+t$

$u_t(x,x)=x$ :

$2f_x(3x)+2g_x(x)+x=x$

$3f_x(3x)+3g_x(x)=0$

$f(3x)+3g(x)=c~......(2)$

$\therefore f(x)=\dfrac{5x^2}{48}-\dfrac{c}{2},g(x)=\dfrac{c}{2}-\dfrac{5x^2}{16}$

$\therefore u(x,t)=\dfrac{5(2t+x)^2}{48}-\dfrac{c}{2}+\dfrac{c}{2}-\dfrac{5(2t-x)^2}{16}+\dfrac{t^2}{2}-\dfrac{x^2}{8}=-\dfrac{x^2-5xt+t^2}{3}$

Your two examples are having the conditions in same positions. Note that for the cases of having the conditions in different positions, their PDEs may still have infinitely many solutions.

For example $u_{tt}=u_{xx}$ with $u(x,0)=0$ and $u(0,t)=0$ ,

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

$u(x,0)=0$ :

$f(x)+g(x)=0~......(1)$

$u(0,t)=0$ :

$f(t)+g(-t)=0~......(2)$

$\therefore f(t)-f(-t)=0$

$f(t)=\Phi(t)$ , where $\Phi(t)$ is any even function of $t$

$\therefore g(x)=-\Phi(x)$ , where $\Phi(x)$ is any even function of $x$

$\therefore u(x,t)=\Phi(x+t)-\Phi(x-t)$ , where $\Phi(x)$ is any even function of $x$

Another good example is Does the wave equation require an initial function for one of its derivative?.