In this note, a method introduced by David Maxwell was considered. More precisely, we proved that
on a compact manifold without boundary where , , and admits at most one solution.
In this note, we consider the case when has boundary and together with our PDE above, we have the following Neumann boundary condition
where is the outward normal vector field on .
As before, we assume that and are solutions of the above PDE. Setting . We wish to prove that at least is constant. Recall that the following holds
where is our conformal change. For the detailed calculation, we refer the reader to the previous notes.
Following the previous notes, we first have
Using the equation for , we conclude that
Note that under the conformal change , there holds . Therefore,
Thus, we have proved that
We now use the test functions . Indeed, we first have
provided along the boundary . Clearly, if then the right hand side of the preceding equality is non-positive. A similar argument using as a test function gives us the same thing. Hence is constant since
To prove , we need to take care the sign of coefficients. An easy way to handle this is to assume that on .
To conclude our notes, let us consider the following Neumann boundary condition
where is a scalar function. We then have
In other words,
Using our new conformal change, we then have
provided along the boundary .
On the other hand, there holds
Using the test function , we also obtain
We eventually obtain that is constant. If we assume that on , then it must hold that .