MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I recently learned about Dirichlet problems and was wondering if there were similar solutions in the case where only few temperature points are known instead of a continuous temperature boundary.

For instance, say the temperature is known at the three points of an equilateral triangle, and is assumed to be at a steady state. Is it possible to derive a differential (and of closed form) equation that describes the temperature within the triangle?

Additionally, what are these type of problems called?

share|cite|improve this question
Why would you expect the temperature at three points to uniquely describe the temperature within the triangle? – Qiaochu Yuan Jul 5 '10 at 9:37
@Qiaochu: why not? He ever specified that he is working with the Laplace operator, or what the physical laws are. Or even what the function spaces are. It is perfectly conceivable to me to be able to pose a question whose answer is a three-parameter family depending on the values at three prescribed points. – Willie Wong Jul 5 '10 at 9:45
@Qiachu My intuition is that this problem is ill posed and cannot be answered exactly. However, I believe there is an optimal answer to the question even if there is not an unique answer. For instance one could marginalize over all possible solutions (assuming a maximum temperature perhaps). Or maybe if the constraints could result in some dominating configuration due to multiplicity. Maybe one could use the points to estimate a boundary that could be used to answer the question? Not really sure, hopefully someone else is :) – Jonathan Fischoff Jul 5 '10 at 9:59
@Qiaochu (this is a duplicate of my earlier comment. I so badly mistyped your name that I felt bad): for example, take the equation $\partial^2 f = 0$ on $\mathbb{R}^2$. The general solution is $f(x) = c + v\cdot x$ and depends, indeed, on three real parameters, which can be fixed at three points. Whether these equations are meaningful is a different question. – Willie Wong Jul 5 '10 at 11:03
up vote 3 down vote accepted

I think you may have gotten things backwards.

The point of differential equations is to describe macroscopic (global) phenomena via microscopic (local) physical laws, as differential operators are strictly local objects. Solving a differential equation one often finds families of solutions, which can live in various different function spaces of different regularity. The study of well-posedness of a differential equation often becomes the study of under what conditions can we obtain the existence of a unique solution. It is often found that the "degrees of freedoms" in the families of solutions can be restricted by prescribing boundary data of sufficient regularity. This is the case of elliptic operators and leads to the Dirichlet problem.

Now, the amount of data to be prescribed at the "boundary" is not the same for every problem. For elliptic type problems one only need to give a Dirichlet or Neumann type condition, but in general giving both conditions may lead to non-existence of a solution (over-determined problem). But for a different type of boundary and a different type of equation (say hyperbolic/wave equation in the initial value problem formulation), it is necessary to prescribe both the "Dirichlet" and the "Neumann" conditions for the question to be well-posed.

What you are asking is sort of an opposite problem: you are asking that given the value of a function on some subset of, say, the plane, what differential equations are well-posed for this data. This problem has too many solutions. Just to give a few

  • As I mentioned in the comments, the equation $\partial^2f = 0$ (vanishing Hessian)
  • Or, let $v,w$ be arbitrary unit vectors not parallel to the sides of the triangle, then you can set $v(f) = w(w(w(f))) = 0$. The solution is constant in the $v$ direction, and along integral lines of $w$ must be quadratic, which is determined by three constants.
  • Or, let $v$ be the unit vector going from point 1 to point 2, and let $w$ be the unit vector that connects point 3 to the line formed by point 1 and point 2 perpendicularly. Let $a$ be a number that is not a multiple of the distance between points 1 and 2, and $b$ be a number that is not a multiple of the distance between point 3 and the line, then take $v(v(f)) = - a^2 f$ and $w(w(f)) = - b^2 f$. The general solution is a trigonometric function depending on one translation coordinate and one scaling parameter.

But if you ask that also the function represents the steady state of a solution to the classical heat equation (in other words a solution to the Laplace equation), then the answer is no: you can extend the temperature profile on the boundary of your triangle arbitrarily from the three fixed data points you gave. For every extension (say differentiable) there is a corresponding solution to the Laplace equation. In other words, there are many, many steady-states whose temperature are as given at those three points on the triangle. So the map from your data to admissible steady-state temperature distributions is necessarily non-unique. So unless you prescribe additional conditions to pick out which of the many possible solutions you want, it is in general impossible (by definition) to write down a well-posed differential equation doing what you want it to do.

Edit: I just saw your answer to Qiaochu's comment

Yes, the devil is in the details on how you insert the constraint though. The Dirichlet problem is well posed. The three-point version isn't. By counting dimensions your constraint needs to be strong enough to mod-out a infinite dimensional set. For example, my first example of an equation $\partial^2 f = 0$, is one possibility. A solution to that equation most certainly solves the Laplace equation. It is equivalent to extending the data to be linear along the boundary of the triangle, and solving the Laplace equation. It also happens to be the one that also minimizes the $H^2$ norm among all solutions to the Laplace equation. Is it a meaningful one? I dunno, what do you think? But it certainly is optimal when considering one metric.

In any case, any conditions you can impose that leads to a unique solution most certainly will be equivalent to one that leads to a unique set of compatible boundaries. Then you can impose differential conditions (ask that $f$ solves a second order ODE along each segment of the boundary) or algebraic conditions. Unless you have a physical justification, optimality really is in the eye of the beholder.

share|cite|improve this answer
I like your answer, but I feel compelled to completely understand the specifics before I accept it. In the mean time I'm now interested in assumptions that could make this well posed. I completely agree that this requires two steps, first estimate a boundary, then solve the ODE. Perhaps one could assume the points are sampled from a circular boundary. Then solve the 1d heat problem on a circle? – Jonathan Fischoff Jul 5 '10 at 15:15
That won't work. The circle is a closed smooth manifold, so the only steady states for the heat equation (ie harmonic functions) are constants. In general the restriction of a harmonic function to a submanifold is not harmonic, so surely that is not the right way to get boundary data. Now, if you assume that the points are sampled from a circular boundary, then you can probably say something about the three point correlation function of the boundary value, which will put some, but not complete control on what the boundary value is. – Willie Wong Jul 5 '10 at 18:04
One problem with "optimality" is that a set of three points is of measure zero on the circular boundary. So it is hard to imagine a probablistic way of characterizing what the allowed boundary function is. Perhaps one may ask for a function $g$ on the boundary circle, such that on average, taking measurements along three points on a equilateral triangle, the three measured temperatures are $T_1,T_2,T_3$. But to consider this even as a conditional probability requires putting a probability measure on allowed functions. – Willie Wong Jul 5 '10 at 18:15
As per your point about the circle. Awesome. That clears up some initial confusion I had about how steady states could have a non zero temperature gradient at all. – Jonathan Fischoff Jul 5 '10 at 20:03
Getting back to optimality. Here's my take on what you are suggesting. Assume that whatever function the boundary has, the average temperature that results in the boundary must be the average given by the three points. Now imagine quantizing the boundary in space and values. By a similar argument to one Boltzmann made, you can arrive at a distribution for the boundary. Of course this would mean you have infinite number of boundary functions. So maybe assume the boundary is continuous? That would reduce allowable functions. I'm just guessing here. – Jonathan Fischoff Jul 5 '10 at 20:44

Your Answer


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.