Okay, let's continue with exercise number five, which is the second exercise on exercise

sheet two.

The first part of this exercise isn't really an inverse problem exercise, but a PDE, so

a partial differential equation exercise.

So either you've seen that before or you haven't, well, it will be beneficial to see this in

both cases.

So this is a heat equation in one dimension.

It describes how an initial heat condition, so it's just, I don't know, maybe this might

be an initial heat condition on a one dimensional rod, changes over time if you have some boundary

conditions.

So what happens here?

Let's say, so this is the heat, which is the function u.

This is the domain, the one dimensional one.

So you have heat attached to each point on this domain.

So let's think about a rod which is unevenly warm.

So it's quite hot at this point and the heat drops off to the boundaries in this manner.

So quadratically here and linearly here, right?

So this might be the initial condition u0.

And the boundary conditions are zero, so u of t and point zero here and u of t at point

one here is always zero.

So we can think of this rod being cooled from both sides by zero degrees Kelvin or something

like that.

And this differential equation up here, so this describes how the heat dissipates here.

And you can probably guess that the whole structure will cool down.

So you will probably see something like that after some time.

Sorry, that's not quite exact.

It has to be equal to zero at the boundaries.

So when time progresses, you get a different heat distribution.

So this is for some time t here.

Well, let's remove this x because it isn't really an x dependency here.

So this is the function.

How do we compute this function u of tx from this partial differential equation?

And the trick here is to make the ansatz that u of tx can be decomposed in its two variables.

So there's a t dependent function, capital T of t times a space dependent function, capital

X of x.

So there's no reason why this should work.

This doesn't work for most PDEs at all, but it works for the heat equation.

So just trust me that this is a smart thing to do.

And then the PDE becomes, well, what is dt u of tx?

This time derivative operator only works on this.

So this is t prime of t times X of x, and differentiating with respect to space twice

only applies to the x operator here, this x function.

So the PDE becomes t prime of t times X of x is equal to t of t times x prime of x.

Sorry, x double prime.

And well, now we can put this in this form, write for all t, x in 0, 1.

So I have to remove the equality here, just to be exact.

And if this is to hold for all t and all x, then this means that this equality has actually

has to be a constant.

T prime of t divided by t of t has to be equal to c, some constant in R.

This has to be equal to x prime of x divided by X of x for all t.