# Why don't elliptic PDE's have a time coordinate?

Usually second-order linear PDE's are classified as elliptic, parabolic, or hyperbolic (or ultrahyperbolic) depending on the eigenvalues of the coefficient matrix. The three cases correspond to the three most famous second-order PDE's:

In the general study of such equations, it is common to refer to one of the coordinates as time in the parabolic and hyperbolic case, but in the elliptic case all of the coordinates are usually thought of as spatial (at least in the treatments I have seen).

My question is -- is there a good theoretical reason for this? Or is this just a tradition, based on the fact that the main application of Laplace's equation in physics are spatial? Is the equation $$u_{tt} = -\nabla^2 u$$ useful for modeling any physical situations?

• Isn't that just $-\Delta u=0$ in the space $(t,x)$? – Siminore Jan 31 '14 at 17:11
• Can it be related to the fact that elliptic PDEs have complex characteristics? I have no idea, sincerely... – Dmoreno Jan 31 '14 at 17:24
• @Siminore I've reworded the last part to make it more clear what I'm asking. – Jim Belk Jan 31 '14 at 17:24
• The reason for this is that in physics we usually have to deal not only with the equation, but also with initial-boundary problem. And some of the initial and boundary conditions lead to a well posed problem, whereas others do not. When we talk about the "time" variable, it usually means for mathematicians that natural initial conditions can be set. However, if you consider your example for the "Laplace" equation with a "time" variable, your initial condition+equation will not constitute a well-posed problem. – Artem Jan 31 '14 at 17:50
• @Artem If you could expand on your comment a bit, I would consider it an answer. For example, is it true that Laplace's equation does not have a unique solution on $[0,1]\times[0,\infty)$ if I specify the value of the function on the boundary? (My intuition is that it would have a unique solution, but I can't say for sure.) – Jim Belk Jan 31 '14 at 18:25

## 2 Answers

The reason for this is that in physics we usually have to deal not only with the equation, but also with initial-boundary problem. And some of the initial and boundary conditions lead to a well posed problem, whereas others do not. When we talk about the "time" variable, it usually means for mathematicians that natural initial conditions can be set. However, if you consider your example for the "Laplace" equation with a "time" variable, your initial condition+equation will not constitute a well-posed problem.

For example for the Laplace equation the initial value problem is not well-posed, since small perturbation in the "initial" conditions leads to very large difference for finite t. You should google Hadamard's example of an ill posed problem. In general, Laplace equation does not go good with "initial-value" problems.

About comment by @mkl314:

This question is right to the money and very deep mathematically. Note that distinction between "initial" and "boundary" comes from the physical interpretation of our equations. Nothing prohibits to set the initial conditions for an arbitrary manifold in the space. However, for some initial conditions the problems will be well-posed, for others --- ill posed. It is a very complicated question in general: How to set the initial conditions for a given system of PDE of order $n$ to guarantee the existence, uniqueness, and continuous dependence on the initial conditions. The well known examples of heat, wave, and Laplace equations how we treat them in undergraduate courses come from physical intuition, not from carefully chosen mathematical conditions.

One of the reasons is that in some physics problem, stationary solutions (equilibria) satisfy an elliptic equation. The stationary solutions of the heat equation and the wave equation satisfy Laplace's equation.

Also, the method of separation of variables when applied to the heat or the wave equation leads to elliptic equations.

• This is a quite interesting point of view. Let me add some more, if you have no inconvenient. You can see this clearly, for example, in the case of the 1-D heat equation without addition of heat, so $u_t = u_{xx}$. Then, if you choose $t_c$ as a characteristic time of the problem you are considering, then $\partial u / \partial t |_{t = t_c} \approx 0$ if $t_c \gg 1$, so it yields $0 = u_{xx}$. This characteristic time is related to the thermal (or viscous, or masic, ...) diffusivity coefficient (and also to the initial conditions). Cheers! – Dmoreno Jan 31 '14 at 17:47
• This gives a good reason beyond physics that elliptic equations are thought of as spatial, but it still doesn't entirely satisfy me. Does it somehow not make sense (or is somehow not useful) to think of one of the variables in Laplace's equation as time, or is this just a convention based on the most common case? – Jim Belk Jan 31 '14 at 19:05