Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Caveat: I don't really know anything about PDEs, so this question might not make sense.

In complex analysis class we've been learning about the solution to Dirichlet's problem for the Laplace equation on bounded domains with nice (smooth) boundary. My sketchy understanding of the history of this problem (gleaned from Wikipedia) is that in the 19th century everybody "knew" that the problem had to have a unique solution, because of physics. Specifically, if I give you a distribution of charge along the boundary, it has to determine an electric potential in the domain, which turns out to be harmonic. But Dirichlet's proof was wrong, and it wasn't until around 1900 that Hilbert found a correct argument for the existence and uniqueness of the solution, given reasonable conditions (the boundary function must be continuous, and the boundary really has to be sufficiently smooth).

Is the physical heuristic really totally meaningless from a mathematical point of view? Or is there some way to translate it into an actual proof?

share|improve this question
    
Why is it physically obvious that the electric potential in the domain can't vary with time? –  Qiaochu Yuan Oct 28 '09 at 18:19
1  
Because Coulomb's law is time-independent. I think if it were not time-independent you could violate conservation of energy by exploiting the time dependence. –  Dinakar Muthiah Oct 28 '09 at 18:33
2  
Here's a more expanded version of what I mean, although perhaps your answer suffices to address this issue as well. To physically set up a Dirichlet problem you must bring some charges in from infinity and place them on the boundary. While you are doing this, the potential inside the boundary is changing. The "physical intuition" is that once the charges stop moving the potential should stabilize given infinite time. What physical principle prevents an oscillating potential that doesn't stabilize? –  Qiaochu Yuan Oct 28 '09 at 19:15
    
You make a good point. I guess you could bring in the walls for a waveguide in just the right way so that a standing wave remains inside the waveguide for all time. Of course in reality, the walls of the waveguide are not perfect conductors and the standing wave will be dissipate. But that's not really a fair argument. Maybe the correct argument is that the charges must be brought in very slowly so that we can model the system electrostatically for all time, and that any standing wave would have a frequency so low that we declare it to be static for our time scale. –  Dinakar Muthiah Oct 29 '09 at 19:20
2  
By the way, the physics suggests that a solution exists, but the uniqueness (which is easy mathematically) is not at all obvious. Nothing prevents the solution to depend on the way you brought the charges on the boundary. –  Andrea Ferretti Mar 5 '10 at 16:49
show 1 more comment

6 Answers

up vote 6 down vote accepted

Well, I don't understand the electrostatics, but here is another physical heuristic:

Impose a temperature distribution at the exterior, and measure (after some time has passed) the temperature in the interior. This gives a harmonic function extending the exterior temperature. [What's the electrostatic analogue? Formerly I had written "charge density", but now I am not sure if that's right.]

I think this strongly suggests a mathematically rigorous argument: We are naturally led to model the time-dependence of temperature in the interior. This satisfies a diffusion (or heat) equation, but in words:

"After a time \delta, the new temperature is obtained by averaging the old temperature along a circle of radius \sqrt{\delta}."

This process converges under reasonable conditions, as time goes to infinity, to the solution of the Dirichlet problem. Anyway, we are led to the Brownian-motion proof of the existence, which I personally find rather satisfying. Another personal comment: I think one should always take "physical heuristics" rather seriously.

[In response to Q.Y.'s comments below, which were responses to previous confused remarks that I made: neither the electric field nor the Columb potential is a multiple of the charge density on the boundary: the former is a vector, and in either case imagine the charge on the boundary to be concentrated in a sub-region; neither the electric field nor the potential will be constant outside that sub-region.]

share|improve this answer
    
If I recall correctly, as you approach the boundary, the potential approaches a constant multiple of the charge density. –  Qiaochu Yuan Oct 29 '09 at 5:37
    
Sorry, that's wrong. The electric field approaches a constant multiple of the charge density, so one should take a line integral of the target boundary value. –  Qiaochu Yuan Oct 29 '09 at 6:09
    
Indeed. Start a regular Brownian motion from some point x in the interior. Take the expectation of the boundary value at the first boundary point it hits. This gives a rigorous proof, and any open bounded domain will work (and bounded & measurable boundary values). –  George Lowther Oct 29 '09 at 23:11
add comment

The physical concept that "prescribing the potential at the boundary should determine the potential inside" does not even suggest a proof. (Btw, I think you meant "potential at the boundary" rather than "distribution of charge along the boundary.") At most it suggests a theorem. However, the physical principle that nature tries to minimize a quantity (in this case, the energy) does suggest an idea for a proof. This is precisely what Dirichlet (and others) tried to do in the 19th century, as well as what was eventually carried out by Hilbert.

To add to what Aguirre wrote, the basic problem is that just because you have an inf does not mean that you have a min: the problem is a lack of compactness. The technical fix is to weaken your topology (while also completing it) to get a larger space of functions in which you can successfully extract a min. The cost is that you no longer know if the min is an honest-to-goodness solution in the usual sense, so you have to work a little to prove that it is. (In other words, you've made the existence question easier at the expense of introducing a regularity question.)

A reference for this method of solving the Dirichlet problem (sometimes called the "direct method" in the calculus of variations) is Rauch's PDE book, which covers this material with a minimal amount of fuss.

share|improve this answer
add comment

The electrostatic intuition does lead to a correct mathematical formulation of the Dirichlet problem.

Let's consider an electric charge distribution of two thin layers (one layer is positive and the other is negative) located along a closed surface $S\subset\mathbb R^3$. Assume that $d>0$ is the distance between charges along the normal $n_p$ to the surface at point $p$. Let $\rho\in C(S,\mathbb R)$ denote the distribution's density.

A pair of two opposing charges $+Q=\rho/d$ and $-Q=-\rho/d$ creates an electric field. The limit of the field when $d\to 0$ is known as the dipole. For any $x\in \mathbb R^3$, the dipole potential at the point $p\in S$ has the form $$\frac{\rho}{d}\Phi(x-(p+n_pd/2))-\frac{\rho}{d}\Phi(x-(p-n_pd/2))=\rho\frac{\partial \Phi(x-p)}{\partial n_p} +o(1)\quad{\rm as\ \ } d\to0,\qquad(1)$$ where $\Phi(x)=-(4\pi|x|)^{-1},\quad x\in \mathbb R^3,$ is the fundamental solution of Laplace's equation. (1) gives the dipole potential of a single dipole at the point $p\in S$ and the integral $$u(x)=\int_{S}\rho(p) \frac{\partial \Phi(x-p)}{\partial n_p} dp,\quad x\in\mathbb R^3,$$ is the potential of the whole distribution.

Now, a simple computation shows that $u(x)$ is a harmonic function when $x$ is not on the surface. It has a jump when $x$ passes through $S$: $$u_{-}(x_0)=u(x_0)-2\pi \rho(x_0),\quad u_{+}(x_0)=u(x_0)+2\pi \rho(x_0),\quad x_0\in S,\qquad\qquad\qquad\qquad (2)$$ where $u_{-}(x_0)$ ($u_{+}(x_0)$) is the limit from the interior (exterior) of the surface.

Relations (2) are integral equations w.r.t. the unknown density (assuming that the potential on the surface is known). The equations can be solved using the Fredholm approach. The function $u(x)$ then gives a solution to the Dirichlet problem.

Edit. See a nice little textbook by Arnold where he shows how to make the physical intuition rigorous in this problem.

share|improve this answer
add comment

I am not an expert on PDEs, but I know that often the existence and uniqueness of solutions to partial differential equations are obtained by Banach's fixed point theorem or similar results. Essentially, the idea is that if solutions to a differential equation are "more stable" then the boundary conditions, you should be able to construct a unique solution by a limit process.

If you look into this problem, it very much looks like one of the problems solvable this way. What reinforces my opinion is that you're actually asked to find something in a zero space for a Laplace operator and this is a very good operator, it's elliptic I think, so, e.g. the exp (-t\Delta) expression behaves well and has projection to the zero space as limit.

Hope that gives some direction and keywords to search :)

share|improve this answer
add comment

Physical intuition tells us that the potential determined by a distribution of charge on the boundary will minimize the energy. If D is a bounded domain in R^d (d=3 in real applications), then the energy of a potential u on D is given by Dirichlet's integral:

DI(u) = \int_D |grad(u)|^2 dx

Thus, if we minimize DI(u) among all functions u whose restriction to the boundary of D is equal to a certain function f, we will have solved Dirichlet's problem: find u: D -> R such that

Laplacian(u) = 0 in D, u restricted to the boundary of D = f.

The Dirichlet integral is bounded below (by zero), and so has an infimum when computed on functions whose restriction to the boundary of D is equal to f. What is not obvious at all is that it attains its minimum. The proof of this fact had to wait until the development of what we now call Hilbert space methods. In particular, one of the problems was to determine the right class of functions to work with: hence the emphasis on all functions above.

Once the existence of a solution is proved, uniqueness follows from the maximum principle.

share|improve this answer
    
It is my understanding that this was Dirichlet's original argument, but he had no rigorous way to prove that a limit of a minimizing sequence exists. Only much later, could existence, uniqueness, and regularity (which is also important) could be proved. –  Deane Yang Oct 30 '09 at 1:59
add comment

Do you mean that you want to rigorously prove a mathematical statement on the basis of physics, or that you want a proof based on the mathematical theory of a particular physical phenomenon? I assume you have to mean the latter.

The Dirichlet problem can be solved by the mathematical theory of Brownian motion, which describes a physical phenomenon discovered in the nineteenth century.

But I don't know any proof based on electromagnetic theory (except tautologically, by incorporating the theory of harmonic functions into electromagnetism).

share|improve this answer
    
I would assume the question means, can you prove the Dirichlet problem has a solution by assuming that certain laws of physics are valid (and in particular, that a certain abstract mathematical model of them is well-defined and consistent)? –  Eric Wofsey Oct 28 '09 at 19:40
    
I think he's specifically referring to "translate it into an actual proof", which should mean actual mathematical proof. –  Ilya Nikokoshev Oct 28 '09 at 23:46
    
This seems like a reasonable response to the question. Because it relates to a physical problem doesn't prove the result mathematically, or help to prove it, unless the physical situation can be handled with rigorous mathematics. The mathematical theory of Brownian motion does this (although it wasn't understood mathematically in the 19th century). –  George Lowther Oct 29 '09 at 23:21
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.