The second exercise concerned a particular case of the wave equation.
Let's see, let's put it like this.
Because you notice in the wave equation we have a Laplacian here in the space coordinate.
We shall see this in more detail, and of course a second derivative in time.
So, let's see v non-zero, which will be the velocity actually, the normal velocity.
v, c2 function of r3, bounded, such that u, x, y, z, t is equal to v of x, y, z minus c, t.
And suppose that this is the solution to this wave equation in r3 times 0 plus infinity.
Assume that for z in r and t positive fixed z, u, x, y, z, t goes to 0 as rho goes to plus infinity,
where rho is the norm of the vector x, y, this sub-vector.
Then, u must be identically 0.
So this was an application of Liouville theorem.
Because the idea is, if you define tau to be z minus c, t, u, x, y, z, t is equal to v, x, y, tau.
Then, by the chain rule, the second derivative of u and z is equal to the second derivative of v in tau,
and the second derivative of u in time, in t, sorry, is equal to c squared, the second derivative of v in tau.
But therefore, since u satisfies, by assumption, is a solution of the wave equation,
and the second derivative of u and z is equal to...
This part simplifies, because the second derivative of u and z is equal to the second derivative of v in tau.
The second derivative of u in t is equal to the second derivative of v in tau, times c squared,
c squared, and 1 over c squared simplify, so this part cancels out.
So, the fact that u is a solution of the wave equation gives us that v is an harmonic function in r2 for every tau fixed.
This means, since v is bounded, this is one of the assumptions, so it was important yet to use it,
therefore, this theorem implies that v must be constant in x and y in r2.
And so, if it is constant in x and y, it's the only function of tau.
So, vx and y, tau, is equal to some g of tau, for some bounded function c2 over r.
But now, g of tau, which is g of z minus ct, is equal to v of x, y, z minus ct.
But this is equal to u x, y, z, t, by definition.
And this is going to zero, as the normal x, y, so square root of x squared plus y squared, goes to plus infinity,
for all z in r and t positive.
But here, there is no dependence on x and y, as we proved.
And therefore, the only conclusion is that g must be identically equal to zero.
Let us come now to the third exercise, which was a somewhat explicit formulation of Harnack's inequality.
In the case of a very nice open set, it is the ball centered in the origin with radius r positive.
So, if u is c2 in this ball, and it is harmonic and non-negative,
then u of x is controlled by the value of u in the origin, times this constant function, depending on r,
and the norm of x, for all x in B0.
The hint was, use Poisson formula for the ball.
If u is c2 in some ball B0s, and continues up to the boundary,
then u of x is equal to s squared minus norm of x squared over n omega n s, integral over the boundary of B0.
s u of y divided by norm of x minus y to the power n, in the s of y, for all x in B0s.
Now, the point is that, u is c2 in B0r, period.
You don't know that u is continuous up to the boundary of this ball.
So, you cannot directly use Poisson formula for s equal r.
This is a mistake, because if you don't have continuity up prior, you cannot say that this formula holds.
And for these continuous functions, there are counter-examples.
However, for all s small than r, it works.
And for all s small, because for all s small than r, u is c2 in B0r, so it's particular c2, B0s, and continues up to the boundary.
Therefore, fixed x in B0r, x will belong to B0s for some s between r and the norm of x.
Because norm of x is strictly less than r, so there is always some space here.
Therefore, for these values of s, we get u of x is equal to this, is less equal than this,
because we use the reverse triangle inequality, that is, the norm of x minus y is larger equal than the norm of y minus norm of x,
because norm of x is less than s, and on the sphere with radius s, the norm of y is s.
And now this term is constant, so we bring it outside.
We factorize s squared minus norm of x squared in s minus norm of x times s plus norm of x, simplify one of the two.
And we write here, we notice that it will be nice to have the average integral over here,
so we multiply and divide by s to the n minus 2, and we get also this factor.
And now here, again, it's important that we are working in a ball B0s strictly, smaller than, I mean,
strictly contained in B0r for s strictly less than r, because we are going to use the average property of harmonic functions on the spheres.
That is, this average integral is equal to a value of u in the center of this ball, which is the origin, and so we get u of 0.
And then one could say, ah, but s is less than r, so let's put r.
And here, however, you cannot do it.
But how to conclude?
Well, the first, here, the left-hand side does not depend on s.
This function is clearly continuous. Actually, you can also see here, it's continuous in s.
We can take any s as close to r as we want, so we push s to r minus, and we're done.
The lower bound can be got, can be obtained in a similar way.
Again, x in B0r fixed, so we take s here.
This time, in the denominator inside the integral, we get, we will use the real usual triangle inequality,
so to get s plus normal x, bring outside, de-factorize, etc., multiply by s to the n over 2 above and below to get the average integral, mean value property.
And again, this does not depend on s, so push s to r minus, we are done.
This last step was not really necessary, it's enough to stop here, I think this, but I put it for fun.
Now, you might wonder, okay, that's all very nice, but how do you actually get the Harnax inequality?
Because Harnax inequality says that the maximum is controlled by the infimum over the closure of any set compactly contained inside your domain.
Okay then, so let v be an open set compactly contained in B0r.
Let x1 and x2 in the closure of v such that u of x1 is equal to the maximum of u over the closure of v, and u of x2 is equal to the minimum of u over the closure of v.
Why is this theorem ensures that this x1 and x2 exist, because this is a compact in B0r?
Very well, but then, thanks to exercise 3, we have that the maximum over v of u is equal to u of x1.
Therefore, this is controlled by r to the i minus 2, r plus norm of x1 divided by r minus norm of x1 to the i minus 1, u of 0.
But now, this is telling us that u of x is controlled in terms of u of 0, but then, since the inequality is woven below, u of 0 is controlled by u of x.
And therefore, the trick is simply use the other inequality, this time with x2 to make here appear u of x2.
So we will multiply by r plus norm of x2 to the power m minus 1 divided by r minus norm of x2 times r to the n minus 2.
So this factor cancels out, we get this function of r, x1 and x2 times the minimum of u over the closure of v.
And now, this is a somewhat more explicit constant.
And notice, in the proof of general Harnack inequality, the point is, I want to connect with some segments, the first, the point where is the maximum and the point where is the minimum.
Here it's telling you, when you are inside the ball, it's enough to pass, from the point of the maximum to the center and then to the minimum.
Which makes sense somehow, because the symmetry, I mean, the ball as a center, no?
I mean, there is this nice inequality that tells you that, yeah, you can actually do it.
Now, you could tell me, but this requires that I actually find this point to have here a constant.
It doesn't seem so nice, true.
However, if you want somehow a nicer constant, maybe too less precise, of course,
since v is compactly contained in v0r, in particular, v must bear some distance from the boundary of v0r.
Which means, v will be contained in v0s for some s less than r.
Therefore, this function can be controlled upon both by r plus s divided by r minus s to the power n.
Because normal x1 and normal x2 are both less than s.
And so, to increase the numerator, simply put s and we are done.
Then the denominator, to decrease the denominator, because to get the larger fraction we need to decrease the denominator,
we need to take the largest possible value of x2 and x1, which is s.
And so, r minus s becomes as small as possible. And so, we are done.