proof of weak maximum principle for real domains
First, we show that, if (where denotes the Laplacian on ) on , then cannot attain a maximum on the interior of . Assume, to the contrary, that did attain a maximum at a point located on the interior of . By the second derivative test, the matrix of second partial derivatives of at would have to be negative semi-definite. This would imply that the trace of the matrix is negative. But the trace of this matrix is the Laplacian, which was assumed to be strictly positive on , so it is impossible for to attain a maximum on the interior of .
Next, suppose that on but that does not attain its maximum on the boundary of . Since is compact, must attain its maximum somewhere, and hence there exists a point located in the interior of at which does attain its maximum. Since is compact, the boundary of is also compact, and hence the image of the boundary of under is also compact. Since every element of this image is strictly smaller than , there must exist a constant such that whenever lies on the boundary of . Furthermore Since is a compact subset of , it is bounded. Hence, there exists a constant so that for all .
Consider the function defined as
At any point ,
In particular, if lies on the boundary of , this implies that
Since this inequality implies that cannot attain a maximum on the boundary of .
This leads to a contradiction. Note that, since on ,
which implies that cannot attain a maximum on the interior of . However, since is compact, must attain a maximum somewhere on . Since we have ruled out both the possibility that this maximum occurs in the interior and the possibility that it occurs on the boundary, we have a contradiction. The only way out of this contradiction is to conclude that does attain its maximum on the boundary of .
|Title||proof of weak maximum principle for real domains|
|Date of creation||2013-03-22 14:35:21|
|Last modified on||2013-03-22 14:35:21|
|Last modified by||rspuzio (6075)|