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.

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