07 - The Maximum Principle for the Laplacian

The Maximum Principle for the Laplacian

Suppose $u$ satisfies

$$\Delta u:=u_{xx}+u_{yy}\ge 0\text{ for }(x,y)\in D$$

everywhere within a bounded domain $D$ then

$$u\text{ attains maximum value on }\partial D$$

If $\Delta u\le 0$ on $D$ then $u$ attain minimum value on $\partial D$ (substitute $u$ with $-u$)
If $\Delta u=0$ then $u$ attains both maximum and minimum values on $\partial D$

Proof

Denote boundary of $D$ as $\partial D$ so $D\cup \partial D$ is a closed boundary set so compact
Hence $u$ must attain the maximum either on $D$ or on the boundary $\partial D$

Suppose that $u_{xx}+u_{yy}>0$ in $D$
If $u$ has an interior maximum at some point $(x_0,y_0)$ inside $D$ then we have the following

$$u_x=u_y=0,u_{xx}\le 0,u_{yy}\le 0$$

As we assumed that $u_{xx}+u_{yy}>0$ in all od $F$ then there is a contradiction
Hence $u$ cannot have an interior maximum within $D$ so it must attain it on boundary $\partial D$

Suppose now we have $u_{xx}+u_{yy}\ge 0$ in $D$
Consider

$$v(x,y)=u(x,y)+\frac{ϵ}{4}(x^2+u^2)$$

for some positive constant $ϵ$

Then

$$v_{xx}+v_{yy}=u_{xx}+u_{yy}+ϵ>0$$

which is in $D$, so $v$ attains maximum value on $\partial D$

So suppose now that

1. maximum value of $u$ on $\partial D$ is $M$
2. maximum value of $(x^2+y^2)$ on $\partial D$ is $R^2$

then maximum value of $v$ on $\partial D$ (and throughout $D$) is

$$M+\left(\frac{ϵ}{4}\right)R^2$$

In other words we have

$$u+\frac{ϵ}{4}(x^2+y^2)=v\le M+\frac{ϵ}{4}{R}^2$$

holding for all $(x,y)\in D$
Then if $ϵ\to 0$ we have that $u\le M$ throughout $D$ hence $u$ attains maximum value on $\partial D$