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Hey I'm trying to understand what kind of boundary data I can pose for the wave equation. Let's work in one dimension for now. It appears I should be able to pose any Neumann, Dirichlet or Robin boundary conditions.

I've heard that you need your boundary data to be 'consistent with the Cauchy data'. I would like to understand this better.

If I think about the infinite string problem so solve $u_{tt} - u_{xx} = 0$ on $[0,\infty)$ and I consider a characteristic $x+t = x_0$ I must have $(u_t(x_0 - t, t) - u_x(x_0-t,t))$ is constant (I'm thinking of a point (x=0,t) on the boundary x=0 and a point from my the x-axis (x_0,0) with a characteristic connecting them). This tells me that I can only choose one of $u(x=0,t)$ or $u_x(x=0,t)$ but not both since the must agree on the bounary.

Is this what it means for "cauchy data to be consistent with boundary data"? Is there an analagous statement for the heat equation?

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You may want to look up the literature for "initial boundary value problems". For example, the book Initial-boundary value problems and the Navier-Stokes equation by H.-O. Kreiss and J. Lorenz has a treatment for the heat equation case in Chapter 7. – Willie Wong Aug 24 '10 at 15:14
Basically the idea of consistency is that sometimes overspecification of boundary data leads to non-existence of solutions. A trivial example is the transport equation $\partial_t u + c\partial_x u = 0$ on $[0,\infty)\times [0,\infty)$. Let $u(0,x) = \phi(x)$ have compact support, and prescribe the boundary condition $u(t,0) = \psi(t)$. When $c \geq 0$ this IBVP is wellposed, the Cauchy and Boundary data are consistent. But in general when $c < 0$ it is not, as the characteristics intersect both boundaries and the solution is over-prescribed. – Willie Wong Aug 24 '10 at 15:22
Thanks Willie. However I have a question about this example. Wouldn't you in this case propogate your characterstics from the x=0 and t=0 axes seperately and then define a shock wave where they collide? Your argument seems to suggest that the data propogated from the t=0 axis (if c < 0 so it may hit the x=0 axis) needs to be consistent with the boundary data on x=0. However with the introduction of a shock wave wouldn't this remedy the situation? – Dorian Aug 24 '10 at 16:41
Well, yes and no. The point is that if you allow your solutions to be sufficiently discontinuous, you enlarge the set of possible solutions, and thus also the set of boundary conditions admissible. If you restrict to classical (or at least continuous) solutions, then shocks are not allowed. That said, the use of shocks for this problem is rather unnatural. How do you determine where the shock front sits? Also you'd have to specify some sort of entropy condition for the jump etc. Allowing shocks in this case really is more trouble than it is worth. – Willie Wong Aug 24 '10 at 20:14
up vote 2 down vote accepted

I am just going to answer here the problem actually posed in the title.

The one dimensional wave equation can be re-written as $\partial_u\partial_v \phi = 0$, where $u$ and $v$ are the null variables $x + t$ and $x - t$. The initial data then is prescribed on $u+v \geq 0, u-v = 0$, while the boundary is $u+v = 0, u-v \geq 0$.

So we see that the wave equation implies that the function $\psi(x,t) = \partial_u \phi(x,t)$ solve a transport equation with negative velocity (cf. my second comment above). Thus if your boundary condition is given such that $\partial_u \phi(x,t)$ is well-determined along the boundary by just the data given there, you will reach an inconsistency. This is one of the ways of seeing why you cannot prescribe simultaneously Dirichlet and Neumann conditions at the same time.

(On the other hand, to make the IBVP well-posed, you need to specify $\partial_v \phi(x,t)$ along the boundary, since it satisfies a transport equation with positive velocity. That just one of Dirichlet or Neumann conditions suffice follows from the fact that $\partial_v \phi(0,t)$ can be solved from $\partial_u \phi(0,t)$ [transported from Cauchy data] and the boundary data.)

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