Dear students, welcome to the 5th exercise class. As customer we are going to see the
solutions to the previous exercise sheet. As is usually we recall the notation on the
Laplacian and for the balls, we recall that the surface measure for the ball BXR is given
by N omega N R to the N minus 1. And by the way, from now on we are going to use DS to
the noted surface measure as it is done in Evans book. I'm sorry I used the Hausdorff
measure before but it was because I do like this usually for work and so it was my habit.
By the way, according to the topic of the notation, it is fine if you use your own notation
in the solutions. I mean, you could follow one of the course but it's not mandatory.
What would be required to you is that in the case you use a different notation, you specify
the meaning. I mean, for me this is the standard symbol for the Laplacian but some of you use
a different one. That's fine. Just write it as it is Laplacian, not leave to my imagination
what is it? That's it. So the first exercise was simply an application of the weak maximum
principle. So if you have a U and V, C2 and continues up to the boundary of an open bounded
set U such that they are equal on the boundary and inside the Laplacian are equal, then U
is equal to V up to the boundary. Indeed, it's enough to apply the weak maximum principle
to the function U minus V and V minus U. Because W equal to U minus V is clearly solution
of the D-shaped problem and Laplacian of W equal to 0 on the boundary of U. Therefore,
the maximum of W over the closure of U is equal to a maximum on the boundary, but on
the boundary is 0. So W is less equal than 0 in U or in the closure of U. Therefore,
U is less equal than V in U. And so, repeating this argument for the opposite, so for V minus
U, we get the opposite inequality, V less equal than U inside the open set, so U and
V are equal. And that's it. You see, this is a quite strong result which does not actually
require that the set is connected, for instance. It's enough the weak maximum principle.