So, we need to check that the claim is true. As before, we can, without loss of generality, take B positive, because B equal to zero is trivial, because there is no transport, so the function is constant in time, there is no, the B is zero.
Okay, it's likely like before. B negative can be done in a analogous way, just being careful, or the sign changes. So, as before, let's rewrite what is written over here.
So, if x, so, first of all, we integrate in x, and then in time, because we are considering u against phi t.
And if x is negative, x will be always less than tb, because b is positive and t is time is positive.
So, when x goes from minus infinity to zero, and t goes from zero to plus infinity, x minus tb is always negative, and so u is always zero.
So, with respect to u, I mean zero, against phi t. And you must admit, this is much simpler than a previous case, because this is automatically zero without calculations.
Plus, now, when x is positive instead, then, when x is positive, we have two cases. When t is less than x over b, we have one. And otherwise, zero.
So, to integral, when t is less than x over b, so, t goes from zero to x over b, we have one against phi t.
See, I know it's simple, I'm writing everything to be precise. Plus, the integral from x over b to plus infinity of zero against phi t.
Now, clearly, this is zero. Clearly, this is zero. What remains is this, the integral from zero to plus infinity of, well, phi of x x over b minus phi of x.
Zero, in dx. And this is this first term. And now, we need to consider u against b phi x.
So now, as before, first integrating time, so from zero to plus infinity, and then in space, if x is less than tb, so if x goes to no minus infinity to tb, there is zero.
So, zero b phi x. And this is trivial. Plus, the integral from tb to plus infinity of one against b phi x in dx in dt.
Clearly, this is zero. And so, what remains is the integral from zero to plus infinity of b, and now phi as compact support in r, so phi of plus infinity is zero.
So, there is only the evaluation in tb, minus b phi tb t in dt. But now, notice that, thanks to a trivial change of variables, so x equal to tb, this becomes minus the integral from zero to plus infinity, because b is not negative,
of b phi x x over b dx, sorry, dx over b. b simplifies, and this term is equal to this term. Here, there is a minus, so they cancel, because remember, we need to sign this second part.
Okay? So, this, the integral of u phi t plus the integral of u b phi x is equal to summing up these two terms over here. These two simplify, and it remains only this. So, what we end up with, let me delete here, the parts with zero, because anyway, I mean, this calculation is clear.
So, all in all, we conclude that this integral over here, now we write it to what is equal.
So, let's keep it for safety. So, these sum of the integrals will be equal to what? The integral from zero to plus infinity of phi x x over b dx minus the integral from zero to plus infinity phi x zero.
In dx minus the integral from zero to plus infinity. So, this is the first term where we integrated by parts in t. Now, integration by parts in x gives us the integral from zero to plus infinity of phi x x over b dx, and then there is this term, the initial data.
But the initial data means the every side function. So, g2 is equal to 1 when x is positive and zero when x is negative. So, here there is victim plus integral from zero to plus infinity phi of x zero in dx.
This simplifies with this, and this one here simplifies with this one here. So, all in all, we prove that the sum of these three integrals must be zero for all phi c1 with compact support in r times zero plus infinity.
And this shows that h of x minus tb is a weak solution of the linear homogeneous transport equation. By the way, all of this works also in a random, just here it's much simpler to write it down.
Maybe let me just add one thing. Those of you who know a little of sublet functions, distributions, weak derivatives and so forth, might notice that h is not a problem. The weak derivative of this function is not a function.
I mean, the distribution of the derivative of that function is well known. It's a Dirac delta centered in the origin. And why so? Because if you integrate h of x against phi prime, it's the integral from zero to plus infinity only, so you get minus phi of zero.
This is the dual definition. So, and this is not a function of Dirac delta. This Dirac delta is a distribution of order zero. And if some of you doesn't know, it's not a problem. We will come back to this. Just to say, it's not important. That's interesting.
It's not even extremely important that the function admits weak derivatives in the sense of measurable functions. It may be enough that it has weak derivatives as distributions. Actually, this is a rather measurable.
Okay, doesn't matter. So that the condition is dissatisfied. I mean, because it's a very particular condition. I'm not asking that this function has derivatives in some LP spaces because it doesn't help.
But it is satisfied, this relation. Maybe we should consider this method again after we introduce some of the spaces. Again, again, just to be clear, no, you were not supposed to give me all this detailed explanation. The purpose of the exercise was to make you a little thing with your own mind on, well, the rule.
Let's call it like this. To solve the transport equation seems straightforward. Now, take the initial data, compose, take the composition with the map which sends x and t into g of x minus tb. Voila. This is the solution.
But one can say, well, but to write down g of x minus tb, it's enough that the function g is well defined everywhere. No. And this function is well defined everywhere. You need zero. I set it equal to one.
It is discontinuous, sure. But to write down the composition, you don't need any particular other condition. And for every fixed x, when t goes to zero, I'm fine. It's really no. But it is discontinuity also here. But anyway, almost.
So the point is, when does this abstract, I'm sorry, this is a formal, which means not regular, but when only manipulating the rule without checking really the conditions still give you a meaningful answer?
Well, the point is, up to lowering the requirements on a solution, the answer is, yeah, pretty much in many general settings.
The point is this. In order that the composition g of x minus tb is the solution in the classical sense, so point-wise everywhere, g must be c1, because we are going to take the derivatives in t and in x everywhere.
There is no way to avoid the c1 condition. But in order to get, for instance, that this equation is satisfied for almost every x. But you don't even need Lipschitz as one would have imagined. This function is discontinuous in the origin. And still, almost everywhere the solutions work.
Even weaker, if you want to be a solution in a weak sense, after testing against a smooth, you see, one functional compact support distribution sense still works.
This rule, the composition with x minus tb to get the solution to the transport equation. That was the point of this exercise, to make you a little thoughts on this fact that it's true that it's not enough to apply the formula to get the solution.
But if we don't require the solution to be differentiable everywhere, to solve the equation everywhere, or even to solve it point-wise, but only in average, with respect to integration by parts, it still works. That's it.