J. L. Davies says in his book,

"The basic principle in PDEs is that boundary value problems are associated with elliptic equations while initial value problems, mixed problems, and problems with radiation effects at boundaries are associated with hyperbolic and parabolic equations."

John Crank attests to this in his book saying,

"A free-boundary value problem requires the solution of an elliptic partial differential equation. For a moving boundary problem the equation is of parabolic type."

I might have unterstood these texts wrong, but it seems to imply that I cannot solve for a moving boundary problem in the case of a hyperbolic system. But why? For example, if I ask you to study the refraction of light through a moving slab of glass, then why shouldn't you be able to solve it?

In other words, what is the basic principle in PDE talking about? How do I overcome its limitations in case of hyperbolic boundary value problems?


2 Answers 2


There is no fundamental difficulty with hyperbolic equation on a domain with moving boundary, as long as the boundary moves slower than the characteristic speed. If the boundary is allowed to move at or faster than the characteristic speed, there will be complications.

  • $\begingroup$ Thank you for the answer. Could you kindly elaborate on why the quoted authors say the things they do (e.g. that moving BVP need to parabolic)? $\endgroup$ May 16, 2018 at 14:45
  • 2
    $\begingroup$ @NanashiNoGombe: My guess is that they are just describing what they chose to cover in their book, not that it is impossible. $\endgroup$
    – timur
    May 16, 2018 at 15:34

Let us consider the simplest hyperbolic equation: the linear advection equation $u_t + c u_x = 0$. The initial data is zero and the moving boundary condition is $u(v t,t)=f(t)$. The method of characteristics gives

  • $\frac{\text d t}{\text d s} = 1$, letting $t(0)=t_0$ gives $t=s+t_0$.
  • $\frac{\text d x}{\text d s} = c$, letting $x(0)=v t_0$ gives $x=cs+vt_0$.
  • $\frac{\text d u}{\text d s} = 0$, letting $u(0)=f(t_0)$ gives $u=f(t_0)$.

Therefore, $$ u = f\left(\frac{x-ct}{v-c}\right) \qquad\text{if}\qquad v\neq c \, . $$ One can observe that there is a problem if $v=c$, as suggested in the answer by @timur.


You must log in to answer this question.

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