Okay, so we start with some words on the exercises.

So in exercise four, I think it's rather straightforward to do.

So the idea is to implement a scheme for the nonlinear diffusion equation and test it for different values of d tilde.

So basically most schemes should work well. What my suggestion is for the discrete scheme is to take into account

that minus the divergence is somehow the adjoint of the gradient operator.

So if you have DL2 scalar product of gradient U with some vector field phi,

and you have say phi 0 on the boundary or something like this, then this would be minus integral U divergence of phi.

So in the discrete setting, this would mean if you have A matrix that computes grad U,

then you would somehow have for a vector say phi transpose AU is of course the same as U transpose A transpose phi.

So this is somehow the same operation as here. So then minus A transpose should be the matrix for the divergence,

or A transpose is the matrix for minus divergence.

So in other words, you want to have in this discrete setting still this property,

and the advantage is then you can kind of really preserve the energy dissipation.

So if you have a discrete energy would be probably something like this F.

So this is then sum over the pixels F of your say I call it gradient T.

So this would be a discrete gradient that you do in the pixels.

For example, you do forward differences or backward differences, or if you want you can also do finite elements.

Then you can compute the directional derivative.

Okay, you have some scaling, some constant that depends on the volume.

So then this would be F prime is the same computation by the chain rule as we did in the continuum setting.

And then we would have here the discrete gradient of U in that pixel with the discrete gradient of phi.

So this was basically D tilde.

So then you see why you want to have this edge property now if you have a discrete,

you would like to have this to be equal to D tilde of the discrete gradient squared.

So you want to indicate by parts in a sense that this should be minus the divergence, the discrete divergence of this gradient U times phi.

So here you have a discrete scalar product on the pixels and then you want to have kind of if you have the discrete transpose here,

it's like having integration by parts.

So ideally what you could also do is just to say, okay, I choose any kind of discrete gradient operator here.

That's all I have to do, then I can compute everything by kind of this formula.

Okay, so I would define minus divergence just by this equality here.

Then you will get automatically a scheme that decreases the energy also in the discrete scheme.

And to rather hints that you may take into account in the tests.

So for Perona-Malek, you have to be a bit careful with the scaling of your image and maybe also of the domain.

If you compute one over one plus gradient U squared and you do, for example, rescaling of the image.

Okay, so for example, if you want to go from gray values from 0 to 255 to 01, you will divide by 255 or something like this.

Essentially, you would go to U tilde is a constant times U.

And then you see what happens here is you get one divided by one plus constant squared gradient U tilde.

So you can put this constant in front is like one over C squared.

Divided by one over C squared plus gradient U tilde squared.

So if you plug this into the PDE, then DTU tilde is C times DTU.

So this is C times divergence of D tilde gradient U gradient U tilde.

I see this constant is kind of nice because it gives you here.

So C times gradient of U is again gradient U tilde.

And here we insert the formula from before.

So it's one over C squared divergence of one divided by one over C squared plus gradient U tilde squared gradient U tilde.

So you see you change effectively two things.

One is the constant here is not equal to one anymore.

And the other thing you change is somehow the time scale.

So if C is very small, one over C squared is very large, so it will become very fast.

If C is very large, then this is small, so the evolution will become very slow.