A sample problem for an exam is as follows:

Consider the wave equation $U_{tt} = 4U_{xx}, 0 < x < 1$ with $U(0,t)= U(l,t)= 0$ and $U(x,0)= x(1-x)$, $U_t(x,0)= \pi$.

Find $U(1/4,1/4)$ and $U(1/2,1/2)$ using the reflection principle.

We went over the solution in lecture but I lost my notes and I'm unsure how to proceed. If someone could provide a solution with some explanation I would greatly appreciate it.

Note: $U_x$ and $U_t$ denote the partial derivative with respect to $x$ or $t$ respectively.


2 Answers 2


The reflection method consists in extending the problem to $\mathbb R\times [0,\infty)$ in such a way that the given boundary condition $u(0,t)=u(1,t)=0$ becomes a consequence of some symmetry of initial conditions $u(\cdot, 0)=g$ and $u_t(\cdot, 0)=h$. In order to make sure that there is no motion at $x=0$, we need $g(-x)=-g(x)$ and $h(-x)=-h(x)$. Similarly, to fix the point at $x=1$ we need $g(2-x)=-g(x)$ and $h(2-x)=-h(x)$. Thus, both $g$ and $h$ are $2$-periodic functions. Their period from $-1$ to $1$ is shown below: red is the initial position $g$, blue is the velocity $h$.

Initial data

Now it remains to apply d'Alembert's formula with propagation speed $c=2$. $$u(x,t) = \frac12 (g(x-2t) + g(x+2t)) + \frac{1}{4} \int_{x-2t}^{x+2t} h(s)\,ds$$ The piecewise nature of $g$ and $h$ would make it awkward to write out these integrals in general, but it's not hard to evaluate them at particular points $(x,t)$, which is what you asked to do.


Here is a reference where the method is explained in detailes. Starting on page $19$.

  • $\begingroup$ That reference isn't helpful, because while the spatial domain of the problem stated by rmh52 is $x \in [0,\ell]$, the spatial domain in the reference is $x \in [0,\infty)$. $\endgroup$
    – Ron Gordon
    Dec 31, 2012 at 16:27
  • $\begingroup$ @rlgordonma The reference is at least somewhat helpful, because the method can be adapted to bounded intervals. I posted an answer. $\endgroup$
    – user53153
    Dec 31, 2012 at 18:12
  • $\begingroup$ Perhaps. But it just seems that the problem is more naturally solved by Fourier decomposition, which is more or less equivalent to your formulation. $\endgroup$
    – Ron Gordon
    Dec 31, 2012 at 18:48
  • $\begingroup$ @rlgordonma If you wanted a general formula for $U(x,t)$, the Fourier series is a natural tool indeed. Here we just want a numeric value of $U(1/4,1/4)$, which according to my answer is $\frac{1}{2}(g(-1/4)+g(3/4))+\frac14 \int_{-1/4}^{3/4} h = 0 +\frac14 (\pi/2)=\pi/8$. Compare this effort with expanding $x(1-x)$ into a Fourier series, and then trying to sum the infinite series back to find what $U(1/4,1/4)$ actually is. $\endgroup$
    – user53153
    Dec 31, 2012 at 19:04
  • $\begingroup$ @PavelM: Thanks for replying to his comment. $\endgroup$ Dec 31, 2012 at 22:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .