# Do the class of PDEs that lack initial conditions have a name?

I am trying to think of what this kind of problem is called. I illustrate it with a telegrapher's equation with (hopefully) standard notation.

Find $u:\Omega\times \mathbb{R} \to \mathbb{R}$ such that

\begin{align*} u_{tt}+u_t-\nabla^2 u &= 0& \mathrm{for}\, (x,t)\in \Omega\times \mathbb{R}\\ u&= u_{bc} \sin(\omega t)& \mathrm{for}\ (x,t)\in \partial\Omega \times \mathbb{R} \end{align*}

The main point is that I am interested in solving the PDE for all time, or perhaps more precisely I am interested in the limit cycle behavior of the PDE, (after any transients associated to the initial condition have died down).

Mathematically, I guess a valid problem would be the following. Fine $u: \Omega \times \mathbb{R}\to \mathbb{R}$ such that \begin{align*} u_{tt} +u_t - \nabla^2 u &= 0 &\mathrm{for} (x,t)\in \Omega \times \mathbb{R}\\ u &= u_{bc} \sin(\omega t)& \mathrm{for} (x,t)\in \partial\Omega \times \mathbb{R}\\ u(x,t+2\pi/\omega) &= u(x,t)& \mathrm{for} (x,t)\in \Omega \times \mathbb{R} \end{align*}

Does this problem have a name?

## 1 Answer

Such problems (sometimes called lateral Cauchy problems) are in general not well-posed (meaning they either lack a solution, or there are infinitely many of them, or the solution is unstable under perturbations of the boundary conditions).

For parabolic (or dissipative) equations, it makes sense to study the stationary limit (simply omit the term $u_t$ in the PDE together with the initial condition), but I don't think that's the case for wave-like hyperbolic problems, where information is transported without loss (meaning the influence of the initial conditions never disappears, unless you have absorbing boundary conditions, in which case the limit solution is the trivial solution $u\equiv 0$).

Your second problem doesn't really make sense mathematically, because the solution $u$ is either completely determined by the periodicity and boundary conditions (if they are compatible), or no solution exists (if they are not).

• I understand point about the wave equation and the influence of the initial conditions. I should have written used a telegrapher's equation, or a heat equation, something to add some dissipation. – fred Sep 26 '15 at 18:06
• I am going to change the question in one about the telegrapher's equation. I am not sure I understand this phrase "..., because the solution $u$ is either completely determined by the periodicity and the boundary conditions (if they are compatible),..." I agree with that phrase, but it seems to me that that means the second problem does, in fact, make sense mathematically. What do you mean? I thought I was careful to make sure that the boundary conditions and periodicity constraints are compatible. Is this not true? – fred Sep 26 '15 at 18:57
• @fred They are (were) compatible in your specific case; it just doesn't make sense as a PDE, since the differential equation is completely irrelevant. For a PDE to be well-posed means that it has a (unique, stable) solution for any choice of data (initial conditions, boundary conditions, and right-hand side) within a reasonably broad class of functions. – Christian Clason Sep 26 '15 at 21:50
• And I'm not sure the modified (linear) telegraph equation actually has a periodic solution... (At least not without a lower order term.) – Christian Clason Sep 26 '15 at 21:53