OK, so last time we started with the analysis of the Poisson-Boltzmann equation.

Just to remind you, we have the electric potential phi and species with charge Ci.

The problem we look at is minus Laplacian of phi is sum i from 1 to n, some positive constant Ci, times z i e to the minus Ci phi.

OK, so we could rewrite this in a more abstract form as a nonlinear, and then we have the permanent charge of course, the right hand side.

We could write this as a semi-linear elliptic equation in some domain omega with boundary condition phi equal phi d.

And some Dirichlet part of the boundary and homogeneous Neumann boundary conditions on the remaining part of the boundary.

OK, then we have introduced an appropriate space for the weak formulation.

We look for phi d plus what we call H1d, and H1d was a subspace of H1 with zero trace on the Dirichlet part of the boundary.

OK.

For this we derived first the weak formulation with a test function psi for all psi in this space H1d.

And since F is monotone, we have seen that F is the derivative of some function G, where G is convex.

And we have seen that this is at least formally the optimality condition of a variational problem of minimizing the functional J,

which is Dirichlet energy 1 half gradient phi squared minus integral F, sorry, nonlinear part integral G of phi minus integral F times phi.

OK, we have shown with the direct method of calculus of variations the existence of a minimizer phi in phi d plus H1d.

Another remaining step for today to show really that it's a weak solution is to verify that it's, or somehow we can really compute the first order optimality condition.

So we can compute a variational derivative on the one hand, and also that this is really well defined.

And this is the major problem in the beginning.

If phi is only in H1, or in some L P space, it's not clear that F of phi is really bounded.

So in general, if we have phi in some L P space,

then this does not imply that F of phi is in some L Q space.

Unless we have one of the following additional properties.

One option is that F satisfies a growth condition.

And the second option is that F is only continuous, but P is really infinitive.

If we have phi in L infinity, it's a bounded function almost everywhere.

So if you apply something continuous to something bounded, it stays bounded.

So then it will be true for Q equals infinity.

So let's check what we have. If you have a growth condition.

On F, this would mean that F of some argument, say, S is lesser equal than some constant times the absolute value of S

to some power alpha.

Particularly, we need this at infinity.

Then we see immediately if we have integral F of phi to the power Q, so the L Q norm of this thing to the power Q.

Then this is lesser equal than C1 to the power Q integral omega phi to the power Q times alpha.

So this means if Q is P divided by alpha, everything is fine.

At least we could define this in some L Q space.

The problem, particularly for Poisson-Boltzmann, is that F of phi is the sum of Ci times some exponentials.

And we see that some exponential function of phi does not satisfy any growth condition.

So no matter what P and Q we want to use, it will never work.

Because for large arguments, for every alpha, this will be violated.

We will always be the opposite inequality.

So what's the only option we can do?

Well, the only chance is we need to go for P equals infinity.

So we need to refine our theory of the energy minimizer and of the weak solution

to have phi not just in this Sobolev space H1, but also in L infinity.

And this is non-trivial if the dimension D is greater than 1.

So D equals 1, we know H1 is continuously embedded into L infinity, even into the space of continuous functions.

So there we know immediately that the minimizer in this space is also bounded.

So it's in L infinity, of course, only if you have the additional property that phi D is in L infinity.

And this can be guaranteed if phi D is bounded on the Dirichlet boundary,

and then also the H1 extension can always be chosen so that it's bounded with the same bound.

It's a simple property you can prove.