Then, as I said, for the last exercises, you need the Lax-Meagre and Theorem.
Let's recall it.
It's a very general result of Lax-Meagre and Theorem.
It is not necessarily related to the existence of some second order partial differential
equation with some ellipticity condition on the coefficients of the second order derivatives.
No, no.
It's a general result on bilinear maps from cross-products of an Hilbert space with itself
into R. It states in particular that if H is in Hilbert space, B from H times H into
R is continuous and coercive bilinear map, which means it's a bilinear map such that
there exists constant C and chi positive, such that the absolute value of B of uv is
controlled by constant C times the H-norm of u times the H-norm of v for all u and
v in H. And B of uu is larger equal than small k, the H-norm of u squared for all u in H.
Then if we take L from H into R to be a continuous linear function on H, then there exists a
unique u in H such that B of uv is equal to L of v. The proof of this theorem heavily
relies on this representation theorem for Hilbert spaces, of course. And the coercivity
ensures the uniqueness. So this I say because I already corrected your homework, and I noticed
that many of you felt somehow compelled to give me the proof of the uniqueness of solution
separately. But no. If you can apply Lax-Milgram theorem, I mean, that's not a mistake, but
if you can apply Lax-Milgram theorem to prove the existence of the weak solution, you obtain
immediately the uniqueness. It's not like you have to prove something else. It's actually
immediately coming from the coercivity. So it's part of the statement. So it was not
necessary. Very well. So exercise two. We have u is an open bounded set, f is an actual.
We say that the function u in H, choose zero. So all the weak derivatives up to the second
order exist and they are true. And they have zero boundary values up to the first order.
We say this function is a weak solution of the boundary value problem for the B-harmonic
equation, which means a double appletion of u equal to f. So the appletion of the appletion
of u equal to f in u. And u equal to the u in the nu equal to zero on the u. If the appletion
of u against the appletion of v equal to integral over u f v for all v in H to zero, you have
to prove that there exists a unique weak solution. This is really the problem on which
you want to apply Lax-Meigren theorem. Why? Because clearly the left, this is, by the
way, okay, I'll give it to this. Sorry. The left hand side of this equation is clearly
your bilinear form on H to zero. And the right hand side is clearly the linear form on H
to zero. And proving that there exists a unique solution means there exists a unique u such
that B of u v is equal to L of v for all v in H to zero. Now, I noticed that some of
you somehow showed me that this definition of the weak solution is compatible with the
classical statement in the classical definition of solution, but this was not part of the
exercise. I mean, I'm not, I don't want to complain about, I want you to understand.
The point is that the first sentence, the second sentence here from we say to the period
here is not a result you have to prove is a definition. We say that the function in
H to zero is a weak solution of this boundary value problem if you have this integral equation
for all this function v in H to zero. Given this, prove me that there exists a weak solution.
Not check that indeed if u is C4 and the domain u is nice enough, then these two notions are
compatible. These, you know, it's the theory. I mean, you know, it was shown during the lecture
and it is the basic intuitive idea behind the very notion of weak solution. That is, if I would be
able to integrate by part testing against a nice function, test function, then I would transform my
point-wise partial differential equation into an integral identity on which I could employ some
results from functional analysis or and especially which would make sense even for functions which
are not classically differentiable, but only weakly. That is functions in H to zero, for instance.
This is important. Then again, it's not that it was wrong. It's just it's not answering the question.
It's like tell me another thing and I fear you miss the point. It's not something you have to prove.
It's a definition. Then it's an important statement that sometimes under some conditions
it's possible to show that weak solutions are indeed also strong solutions. This is true for
instance for the Dirichlet boundary value problem for the Laplace equation on domains,
bounded domains with C1 boundaries. That's true because Laplace equation is very beautiful.
It's not only elliptic, but I mean with constant coefficients. So it's very nice.
Actually, all the weak solutions are automatically analytic. It's infinity. So actually analytic.
So that's another thing. You're starting from the weaker definition. So the definition of weak
solution, you're able to show that you get actually strong solution, but that was not required here.
Here you had only to take the definition of weak solution. This is integral identity and show me
that exists a unique u satisfying it for every v in the Hilbert space you're given.
Now, sorry for all this. Let's come to the solution. So
first of all, I mean, here I put some observation. Notice that, I mean, it's a preliminary if you
want. The integral over u of the Laplacian of u squared for u in H2, 0 is equal to the actual
norm of the weak Hessian matrix of u squared. Why? Well, formally, because one says, well,
here there's written a sum over i and j from 1 to n of u x i, sorry, x j times u x i x i.
Here I wrote i i, sorry. You can bring out the sum, of course, because it's finite. And then
one could say, well, if I integrate by parts, I move twice, I move x i here, and then I move x j
here, and they get u x i x j squared. Wonderful, because this is exactly the actual norm of the
Hessian matrix squared. Point is, how do you justify this step? Because these functions are not
classically three times differentiable, and also the boundary of u can be ugly.
So you must justify it. The trick is, we know that since u is in H2, 0, there exists a sequence of
functions, phi k, in C infinity C of u, such that phi k converges to u in H2. What does it mean?
It converges with respect to the Wk2 norm, which means it converges in L2, the first, the gradient,
converges to the weak gradient of u in L2, and the Hessian converges to the Hessian of u in L2.
For this function, phi k, I can integrate by parts. Actually, it's not integration by parts.
It's the definition of weak derivatives. Yeah, sorry, it's a little ugly, but
the integral over u of u x j x j phi k x i x x x i is equal to, well, minus 1 squared,
the integral over u of, small u, against phi k x i x i x j x j. This is the definition of
second weak derivative with respect to x j. Simply, no? But then, again, the definition of
second weak derivatives, I can move and put another minus 1 squared, which is 1, of course.
I can move the derivative in x i and the derivative in x j over u. And what remains is phi k x i x j.
Now, remember that you have a convergence which implies the convergence of any partial derivatives
for any multi-index alpha with length less equal than 2, so in particular, the second partial
derivative. So it means the integral over u of u x i x i u x j x j, same mistake as before,
is equal to a limit as k goes to plus infinity integral over u of u x j x j phi k x i x i.
But this will show for every k is equal to the integral over u of u x x i x j phi k x i x j
convergence and we get the integral over u of u x i x j squared. Again, this is just to know,
to show how these things can be made rigorous, this notion of weak derivatives and some of the
spaces can be exploited to its full power without any more saying, yes, this could be
rigorous if the functions are regular. No, no, no, it is true, this kind of equality is true for
Sobolev functions in H2 with zero boundary values, of course, because I can approximate them in a nice
way with smooth functions with bounded support, with compact support. Indeed, and again, the
compact support allow me not to consider the boundary of u, so I don't care even if it's ugly.
Instead, I noticed that some of you were putting boundary terms, but no, the boundary of u, u is
only a bounded open set, so the boundary can be incredibly irregular, it can be, I don't know,
the Foncoc-Snowflake, which is a fractal set with fractal boundary. I don't care because phi k,
my approximated sequence, doesn't see it for every fixed k, so I can, this step works and then I pass
to a limit and exploit the two convergence of the second derivatives to the weak second derivatives.
Now, given this primacy, we can define our bilinear form, as I told you, so b from H2, 0 times H2, 0
into r is b of u, v, u, v, sorry, equal to the integral over u of Laplacian u against Laplacian v.
Continuity, take absolute value, let's equal them, bring absolute value inside, apply Cauchy's
vert, so we get the L2 norm of Laplacian u, L2 norm of Laplacian v. Laplacian is well defined
because when I choose, so it's L2, 0, so it's the weak Laplacian, of course. But we just proved that
the L2 norm of Laplacian is equal to the L2 norm of the Hessian, and so the L2 norm of the Hessian
is, of course, controlled by the H2, 0 norm, because we are putting more stuff, I mean,
also the H2 norm of u and the H2 norm of the weak gradient of u. Coercivity, well, b of u,
u is the integral over u of the Laplacian of u squared, but this is the L2 norm of the Hessian.
We claim this is larger equal than, see, this constant should depend also on u, sorry,
this is larger equal than a constant, the H2 norm squared of u, because of Poincare inequality.
Why? Well, let's write it explicitly. The H2 norm squared of u is equal to the L2 norm of u squared
plus the H2 norm of the u squared plus the H2 norm of the Hessian of u squared,
but Poincare inequality, the H2 norm of u is controlled by a constant, so let's put a square
plus one, the H2 norm of the gradient of u and put it together as two pieces.
And then there is the Hessian. Well, but u is in H1, 0 since u is in H2, 0. But since u is in H2, 0,
the gradient of u is in H1, 0 with values in Rn. Or if you prefer, each first week,
week first partial derivative of u is in H1, 0. So we can apply again vector with, I mean,
you can either write this as the sum of the second, of the L2 norm of the second partial,
of the first partial derivatives of u, or just like this, it's fine. We can apply again
Poincare inequality to the L2 norm of the u squared, so we get another c squared
and times the L2 norm of the Hessian. And so this small c will be 1 over this 1 plus c squared plus
c to the power 4. Now we apply Lax-Migner and Theorem since clearly the map which sends v
into the integral of u of f against v is linear continuous on H1, choose 0.
Linearity is obvious. Continuity, well, it's the same argument we showed before for exercise 1,
because take absolute value, bring it inside, the Cauchy-Schwarz, you get the L2 norm of f,
the L2 norm of v. But the L2 norm of v can be trivially controlled by the H2 norm of v,
so well done. Therefore, Lax-Migner theorem implies that there exists a unique u such that
Buv is equal to integral over u of fv. And so well done.