Ngô Quốc Anh

February 6, 2012

A note on the maximum principle

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 18:18

Today, let us discuss a very interesting stuff. Let say M is a compact manifold without boundary of dimention n \geqslant 3.

On M, we consider the following simple PDE

\displaystyle -\Delta u + hu = a

where h and a are smooth functions with h>0 and a \geqslant 0. Since h>0, it is well-known that the operator -\Delta +h is coercive, see here. A standard variation method tells us that there exists a weak solution u to the above PDE. By regularity theorem, u is at least a C^2 function, thus, a strong solution (in the classical sense).

Next, we claim that u \geqslant 0. To this purpose, assume that the solution u achieves its minimum at some point x_0 \in M. In particular, there holds

-\Delta u (x_0) \leqslant 0.

This, together with the fact that h>0 and a \geqslant 0, implies that u(x_0) \geqslant 0. Thus, we have shown that u \geqslant 0 in M.

Once we have the non-negativity of u, in view of the strong maximum principle, either u \equiv 0 in M or u>0. In other words, the solution u cannot achieves its minimum inside the manifold. Since the manifold has no boundary, it is natural to think that the solution u cannot achieve its minimum although u>0. This is clear a contradiction to the fact that the manifold is compact and u is of class C^2(M).

So something went wrong but what and why?

In fact, we have made a small mistake. In view of the strong maximum principle, we can only claim that the solution can only achieve its non-positive minimum value on the boundary. Therefore, there are cases so that u may achieve its positive minimum inside M. Thus, there is no contradiction here.

In order to see this, let us go back to a proof of the strong maximum principle. Roughly speaking, it starts with the following simple one.

Lemma 1. If -\Delta u +hu >0 at any point in M, then u cannot have non-positive minimum value in M.

Proof. The proof is standard. Assume that at x_0 \in M, the function u realizes its minimum, besides, u(x_0) \leqslant 0. In particular, -\Delta u(x_0) \leqslant 0 and h(x_0)u(x_0) \leqslant 0. These force -\Delta u + hu \leqslant 0 at x_0. A contradiction.

Form the proof above, if u(x_0)>0, we cannot get any contradiction. This is why we can claim either u \equiv 0 or u>0 since u can achieve its positive minimum in M.

Notice that, if we don’t have any h (in the operator), the non-positivity can be dropped.


1 Comment »

  1. I have a question.

    If h is not smooth but only bounded, say h \in L^{\infty}(M). Then does the result still hold?

    Comment by baotangquoc — September 1, 2016 @ 16:27

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