# Well-posedness of wave equation on finite domain

Consider the initial boundary value problem $$u_{tt} = u_{xx}$$ for $0 < x < 1, \; 0 < t$ with the initial data $$u(x,0) = \phi(x)$$ and $$u_t(x,0) = \psi(x).$$

Assuming the initial conditions are sufficiently smooth, for which constant values $a,b,c,d$ do the boundary conditions $$au_x + bu_t = 0 \; \text{ at } x = 0$$ $$cu_x + du_t = 0 \; \text{ at } x = 1$$ lead to a well posed problem?

My guess is that $\frac{b}{a} = 1$ and/or $\frac{d}{c} = 1$ might be important, or that $ad-bc \neq 0$ might be important. I am having a bit of trouble coming up with what these boundary conditions mean physically in the first place, as this usually hints at what is required for well posed-ness mathematically. Any and all help is appreciated, thanks in advance.

*This is the beginning of a problem focused on constructing a convergent finite difference scheme for these well-posed problems. Probably not of very much use at this stage.

• Hint: reformulate the problem using riemannian invariants – uranix Dec 17 '16 at 17:51
• I've had difficulty finding useful resources for "riemannian invariants and boundary conditions", could you suggest a place where I could learn about these? I am completely unfamiliar with these, but from what I can judge from wikipedia, it seems like its a push in the right direction. – Merkh Dec 17 '16 at 19:43

Hint. Reformulate the problem as a pair of first order equations.

First, let's rewrite the second order wave equation as a pair of two of first order: introduce $v = u_t, w = u_x$: $$v_t - w_x = 0\\ w_t - v_x = 0$$ or in matrix form $$\frac{\partial}{\partial t}\begin{pmatrix}v\\w \end{pmatrix} + \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} \frac{\partial}{\partial x}\begin{pmatrix}v\\w \end{pmatrix} = 0 \tag{1}$$ The matrix $A = \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix}$ has the following left eigenvectors: $$\omega_1^\top A = \lambda_1\omega_1^\top, \qquad \lambda_1 = -1, \; \omega_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\\ \omega_2^\top A = \lambda_2\omega_2^\top, \qquad \lambda_2 = +1, \; \omega_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$$ Riemmanian invariants are $r_{1,2} = \omega_{1,2}^\top \begin{pmatrix}v\\w\end{pmatrix}$ $$r_1 = v + w = u_t + u_x\\ r_2 = v - w = u_t - u_x$$

Multiplying the system by $\omega_{1,2}^\top$ from the left gives $$\frac{\partial r_1}{\partial t} - \frac{\partial r_1}{\partial x} = 0\\ \frac{\partial r_2}{\partial t} + \frac{\partial r_2}{\partial x} = 0.$$

So we've converted a hyperbolic system of first order $(1)$ to a pair of advection equations for $r_1, r_2$. Any hyperbolic system with constant coefficients can be converted to a set of advection equations this way.

The well-posed problem consists of one left boundary condition for $r_2$ and one right boundary condition for $r_1$.

I might be mistaken, but I suspect that the only requirement for the boundary condition at $x = 0$ is $$a u_x + b u_t = 0$$ should be expressable as a boundary condition for $r_2$ like $$r_2 = f(r_1).$$ Simlilarly, at $x = 1$ $$c u_x + d u_t = 0$$ should be expressable as $$r_1 = g(r_2).$$

• I see. The original problem I set out to solve was for $u_{tt} - u_{xx} = u,$ so perhaps that is why it seems like the result you found for the boundary condition is somewhat trivial. Thanks a bunch!! I am going to go try this for $u_{tt} - u_{xx} = u.$ – Merkh Dec 18 '16 at 13:17