Okay, so last time we discussed that the continuous problem, our model problem, at least our model

problem, the Poisson equation with, but also more general elliptic equations, fulfill maximum

principles.

There are various ranges of maximum principle, there's the notion of the weak maximum principle

saying that the maximum under certain conditions is attained at the boundary, there's the strong

maximum principle saying if the maximum is attained, the maximum on the closure is attained

in an interior point, then the solution has to be constant.

In this sense, I was a little bit confused last time, in this sense this formulation is

correct.

We talk about the maximum, it's now in the discrete case about the maximum at the points,

grid points in the interior and at the boundary with the exception of those points which do

not interact with anything.

And in the situation that is the claim in the situation that the maximum is attained

at an interior node, an interior grid point, then the solution has to be constant.

And an immediate consequence of this is from the strong maximum principle is the weak maximum

principle also in the discrete setting saying that the maximum on the set of the interior

grid points is less or equal to the maximum on the set including the boundary grid points.

Okay, sorry, that was also correct, it's a trivial statement on the set of the boundary

grid points.

Okay, so let's go through the proof.

I have the proof here on the transparency.

I think we don't have to write it on the blackboard.

It's quite easy to comprehend.

And here we will see which properties of the discretization matrix A H will enter.

And this we will formalize in the sense that we will then in the next step speak about

sets of equations that is discretizations of differential equations, not necessarily

in two dimensions, not necessarily only with Dirichlet boundary conditions which have these

properties and see what we can do with these properties.

So in a second there will come a long list of conditions and it will be helpful for you

if you have some material where you can have a look at this list of conditions that you

can, if I then refer to number five or six or something, you can have a look at it.

That I'm not obliged to go back and forth all the time.

So let's start here with the proof still in our model case, rectangle, five point stencil,

Poisson equation.

So we have the value, the maximal value here attained at this point.

We call this value U bar.

And then we look at the corresponding stencil where this U bar is in the center.

As the right hand side, so the equation is then four, if I omit the one over H square,

so I multiply through by this factor, the equation is then four U star minus all the

neighbors, the values at all the neighbors, the four neighbors or maximal four neighbors

which appear in the stencil.

Equals the right hand side.

As the right hand side is less or equal to zero, that is a very important assumption

here.

We can write this in this inequality just by estimating the F by zero from above.

So we have this inequality here just by the assumption, just by the way how the stencil

looks and by the assumption F less or equal to zero.

And this is just the fact that we have at least four neighbors.

And of course, U H is the maximum and each of these individual values can be estimated

by U H.