I am learning a bit of fourier analysis, with an interest in physics as well. I originally posted this question on the math stack exchange, but perhaps you physicists have more experience in these kinds of things. The standard equation to solve is the steady state heat equation (Laplace equation) in the plane is

$$ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0 $$

Now I understand that, on functions with a fixed boundary, the solutions to this equation give the steady heat distribution, assuming that the heat at the boundary is a constant temperature. Thus the harmonic solutions make sense.

However, I am not convinced about the physical intuition of a solution on an open subset of the plane. Since heat should disperse equally, shouldn't the only steady state solutions be constant, if there is no boundary at a fixed temperature. Louiville's theorem for harmonic functions (a bounded entire function is constant) tell me that on the whole plane, heat must eventually disperse equally, but I can't figure out intuition for why this isn't the case on bounded open subsets of the plane. Is there a different equation I should be looking at? I can't seem to find any part of the derivation of the heat equation that uses the fact that a boundary must be fixed.


I am not entirely sure if I understand your question correctly, but here's my take on my interpretation of your question. Feel free to correct me if I misunderstood.

In physics, we are typically interested in solving problems that potentially have a physical realization, meaning we may be looking at boundary value problems that have unique solutions (well, at least in classical physics). For a PDE with an elliptic operator to have a unique solution, you need boundary conditions everywhere on the boundary. The "boundary-free" problem (on open bounded subsets of $\mathbb{R}^2$) you seem to be talking about above does not describe a definite physical problem, and thus does not have a direct physical interpretation.

In addition, the problem of heat transfer in the whole plane (I assume you mean all of $\mathbb{R}^2$) likewise is of no physical interest, since it does not correspond to any realistic physical situation. I suppose one could speculate, in principle, about the solution of a heat transfer problem in the "entire universe", meaning $\mathbb{R}^3$, and bounded solutions of the Laplace equation in this case are indeed constant (with an arbitrary value of the temperature). I would even go so far as to say that this makes some intuitive sense, even though, of course, such a problem is entirely academic, and has no relationship to the state of our real physical universe.


Disclaimer: I'm an engineer, and my knowledge of higher-level math is probably lacking here. But I'll try to guess what you mean by the question. It's not very clear as-is what boundary conditions you're using, since an open boundary on the plane doesn't really qualify as a boundary for the heat equation itself (I'm assuming here that by the words "open boundary" you're defining that section of the plane as a thermodynamic closed system).

It seems to me, however, that you're using the boundary condition of an infinite plane, with a heat point source at constant temperature. In that case, the only steady-state solution is indeed a constant temperature, where the entire plane adopts the same temperature as your point source. (Note that this is only reached as time tends to infinity!) This applies no matter what open boundary you surround your point source with. If you want to discuss how heat flows across the open boundary before the equilibrium of constant temperature is reached, then your answer is no longer governed by the steady-state heat equation, but rather the transient one, which in 2 dimensions and at constant thermal conductivity looks like this:

$\frac{\partial^2T}{\partial x^2}+\frac{\partial^2T}{\partial y^2} = \frac{1}{\alpha}\frac{\partial T}{\partial t}$

I've replaced $f$ with $T$ since that's the function you usually use the heat equation to solve (temperature as a function of position). Also, the parameter $\alpha$ is called the thermal diffusivity of whatever substance your plane is made out of.

The solutions to this equation, with an infinite plane and a heat point source, would show heat spreading equally in all directions through the plane, but if you draw an open boundary that isn't a circle with your point source at its center, then heat won't cross your boundary in equal quantities at all points, regardless of what time you look at. Perhaps that's what has you confused?

  • $\begingroup$ By an open subset of the plane I mean a topologically open subset - a shape which does not contain its boundary (en.wikipedia.org/wiki/Open_set). In this case, no boundary condition is needed. I am thinking of this subset of the plane as being a closed system (no heat can escape or enter), but there is no fixed temperatures, and heat is free to spread to any part of the open subset of the plane. In this case, my intuition is that the solutions to whatever `steady heat equation' (which may be different from the standard equation) defines the dynamics of the system should be constant. $\endgroup$ Jun 8 '16 at 21:50
  • $\begingroup$ Actually, what you're describing is an insulated boundary condition (dT/dx = 0 @ x = whatever), which is a special case of the Neumann boundary condition and your intuition is correct. The steady-state solution is that your entire subset of the plane is the same temperature as your source. BTW this is not a closed system, it's an isolated system. A closed system allows heat to cross the boundary, as described in my link above. $\endgroup$ Jun 8 '16 at 22:04
  • $\begingroup$ OK - I suppose that physically, I must assume my shape has boundary because its physically impossible not to. The insulated boundary condition then states that if $M$ is the shape on which we are doing the head equation, then on the boundary $(\partial M)$, the gradient of the function is perpendicular to the normal vector of the shape at that point. Do you know of a resource proving that the only harmonic functions with this property are constant? $\endgroup$ Jun 8 '16 at 23:57

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