Cut the recording at this point.

OK, so today we will finish our discussion of Poisson-Boltzmann

with an alternative proof of existence.

So as usual, we consider the Poisson-Boltzmann equation

with some right-hand side f, some Dirichlet boundary

conditions, and some Neumann boundary

conditions that are homogeneous on the remaining

part of the boundary.

And as I announced last time, we're

going to look at some continuation approach.

So we actually look for pi of xs with some kind

of artificial time variable s in 0, 1.

And we construct the problem so that for s equals 1,

we recover the original Poisson-Boltzmann.

And for s equals 0, we have a problem that we can solve easily.

What we're going to solve is the following.

We keep the nonlinear part, but then we change the right-hand side.

Simply take s times s, and we take minus 1 minus s sum ci zi.

And we take phi equals s times phi d

on the Dirichlet part of the boundary,

and we keep the homogeneous Neumann boundary condition.

OK.

The main advantage of this form is we can immediately

write down explicit solutions for s equals 0,

namely phi is equal to 0.

So the problem at s equals 0, let's call this phi 0.

The problem itself is the plus phi 0 is equal to sum from 1

to m ci times charges.

And then we have e to the minus charge phi minus 1.

And we see phi equals 0.

That drops out.

The Laplacian is also 0.

And we have homogeneous Neumann and Dirichlet boundary values.

So we have phi 0, 0 on gamma d.

And d phi 0 dn is 0 on gamma a.

OK, so we see there is a solution that is very smooth.

Obviously, we have phi 0 is in H1.

And it's also, of course, in L infinity.

We call this now V. More precisely, we can also write.

We want to have phi s.

We want to have phi s if we write this briefly like this

in s times phi d plus H1 d.

And here we have s equals 0.

So we just take H1, or if you want, H1 d.

OK, but for the moment, we keep this as our space.

OK, so we can start with, we still

have, since it's a semi-linear elliptic equation

with a monotone right inside, we still

have immediately the uniqueness for this problem.

So phi equals 0 is really the unique solution.