02 - SN Partial Differential Equations and Applications Webinar -Thomas Bartsch
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Join Thomas Bartsch of Universität Gießen as he discusses the existence of L2 normalized solutions to nonlinear elliptic equations and systems. Don’t want to miss another webinar? Register here: https://go.sn.pub/lXgEaxHave an article on nonlinear elliptic equations? Submit Your Article: https://go.sn.pub/GT525VLearn more about publishing with SN Partial Differential Equations and Applications: https://go.sn.pub/UFulhi
SHOW LESS
00:05
I've never told us for more than 30 years and
00:10
so please allow me to introduce briefly his professional background
00:17
and Thomas obtained his PhD from Munich in 1987 and
00:23
then
00:23
he went to hidable for junior position and he was
00:29
awarded Heisenberg Fellow in 1995.
00:32
Then he spent a year in it hard Zurich.
00:38
There and he was hired as a full professor in
00:43
Giessen University in 1996.
00:46
Ever since he has been a cheer in analysis in
00:51
dissent.
00:52
And the Thomas has been a leading expert in topological
00:57
method.
00:58
But original method in nonlinear analysis,
01:02
developing new ideas techniques, in particular tord applications in nonlinear,
01:09
elliptic PDS and Hamiltonian Systems,
01:13
Voltex Dynamics and bifurcation theory,
01:17
among others. And more personal node and Thomas has been
01:22
a long time.
01:23
Collaborator, very good. Friend of mine and I think just
01:28
how and several others in the audience would share the
01:32
same feeling and myself.
01:34
I have learned so much from Thomas and we have
01:38
some very great moment and lately Thomas has been working
01:43
on so called normalized solutions and so he's going to
01:47
talk about the normalized solutions of nonlinear elliptic problems today.
01:54
Thomas, if I ready, please start your tag.
01:58
OK, thank you very much young for this very kind
02:03
introduction.
02:04
I will hope you can see.
02:07
And my screen by now.
02:17
I don't know it should work OK.
02:21
That is OK now.
02:23
Then adjust it. OK so you see OK.
02:30
OK, anyway thank you. She jumped for for your very
02:34
kind introduction.
02:36
Was always really pleasure to work with you and my
02:40
many other colleagues,
02:42
and some of them's already also attend this talk.
02:46
And also I would like to thank the organizers.
02:49
Of course for this webinar I think it's a very
02:53
nice idea and I'm very glad that I'm able invited
02:57
to give a talk here.
02:59
So the topic is normalized solutions of nonlinear elliptic problems,
03:04
equations and systems. Let me just get right into the.
03:10
Problem is here. OK so the start start have given
03:14
a domain in our end which you may think is
03:18
a bounded domain or all of our end whatever at
03:22
the moment completely open we have a potential function V
03:27
from Omega into arm.
03:28
A nonlinearity G from Omega Times R2R and positive #8.
03:33
So these are the data given.
03:35
Then we would like to solve this classical symbol India
03:40
elliptic PD 1 minus slash U + 3 of Xu
03:44
Plus Lambda U equals G of Xu.
03:46
But in addition, you want to add a mass constraint
03:51
will be require that the L2 Norm of you is
03:54
equal to a.
03:55
So. Requiring this additional constraint,
04:00
we have Lambda here as an additional unknown of the
04:04
problem,
04:05
so we look for a long time R and you
04:08
and H10 Omega to satisfy this equation.
04:11
Of course, most of you know that problem one has
04:15
been investigated actually since ages.
04:19
Since the 1960s, for sure,
04:21
there and all kinds of techniques have been developed.
04:24
You know, of course, variational methods,
04:27
topological methods, PD methods, sub super solution techniques,
04:31
whatever under all kinds of conditions on Omega or V
04:35
or G,
04:35
compared with this amount of work,
04:37
there's surprisingly little relatively little known about the problem.
04:42
When you have this additional mass constraint,
04:45
and so this I would like to present today.
04:48
It has been become an active area of research,
04:52
in particular during the last five or so years.
04:56
Some motivation maybe here here.
04:59
The next slide when you look at the nonlinear Schrodinger
05:03
equation here,
05:04
just you may think of F just the pure power.
05:08
Normalcy to some power. And then you make a standing
05:12
wave and that's your letter.
05:14
Of course, soon equation like like this one here.
05:17
But you see in the UNS arts there slam down
05:20
and you as an unknown.
05:22
Now most of the work as I mentioned,
05:25
deal with this problem for fixed Lambda.
05:28
But the two norm of Psy is independent of T.
05:31
This preserved as it is a conserved quantity,
05:35
so actually doesn't make sense.
05:37
Also from a physical point of view to prescribe the
05:41
L2 Norm and leave Lambda as an unknown in the
05:45
variable and.
05:46
Impact with this situation where you prescribe Lambda and test
05:50
their two norm open.
05:52
OK, so this is a problem but we would like
05:55
to deal with today.
05:57
So to get some idea,
05:58
we look at the very classical focusing nonlinear Schrodinger equation.
06:04
So if you have as the nonlinearity we just take
06:08
you to the P -
06:09
1, we have P in subcritical so it is bigger
06:12
than two but less than the Super left critical exponent
06:17
and for the moment or let's just look for positive
06:20
solutions. And then it is well known that this equation
06:25
here has a unique solution.
06:28
Positive solutions unique up to translations.
06:31
It is Radial, so you can assume as it has
06:35
a maximum at 0.
06:37
And you obtain these solutions,
06:39
for instance by the mountain pass theorem.
06:42
Very classical, nothing. You now?
06:44
What about the mass constrained?
06:46
That is easy to see here.
06:47
Namely, you can obtain the solution.
06:50
Use a blunt down from the solution.
06:52
Use of one which corresponds to the value Lambda equals
06:55
one by a simple scaling you see here the scaling
06:58
you sublimed of X is obtained from use of 1
07:01
by the scaling.
07:03
And having this scaling, you can compute easily there two
07:07
norm of you,
07:07
some blunt. Use a plan to an compared with the
07:11
L2 Norm of use of 1.
07:12
Now the L2 Norm of use of one is just
07:15
some positive number and then you ask which L2 norms
07:19
can you achieve if you allow Lambda to vary and
07:22
you see you simply let Lambda very if Lambda varies
07:25
from zero to Infinity,
07:27
then this varies here from zero to Infinity.
07:31
Except. If the exponent of Lambda is equal to 0.
07:36
So that is a somewhat degenerate situation.
07:39
You see the exponent lumper of Lambda equal to 0
07:43
is just means just that P is equal to 2
07:46
4 /
07:46
N, so this exponent plays a special role here in
07:50
this problem of normalized solutions.
07:53
Let me just what I just said.
07:56
Again, now for this problem.
07:59
Here we have our PD together with the norm constraint
08:03
A2.
08:03
Normal view is equal to a.
08:06
A is prescribed. You and Lambda are looked for.
08:10
If P is different from 2 + 4 /
08:12
N then one can say that this problem has a
08:16
solution for all.
08:18
A bigger than 0. You simply solve this equation,
08:22
which determines the two norm for in terms of Lambda,
08:26
you can lump right lumped as a function of A
08:29
and then you have the solution.
08:31
Use of Lambda Lambda of this problem and the solution
08:36
is unique to translations.
08:38
On the other hand, when P is equal to 2,
08:41
two 2 + 4 over and then you can solve
08:43
the problem star only for one value a name.
08:46
If the value of the solution for Lambda equals 1
08:50
minus Laplacian new plus U equals U to the P
08:53
-
08:53
- Then you can also scale this U and obtain
08:56
the family of solutions you sub Lambda.
08:59
But they all lie on the same atmosphere.
09:03
So that means you see here this critical exponent 2
09:06
4 over in is kind of a very degenerate
09:09
situation.
09:09
In this setting, if P is equal to 2 +
09:12
4 over North,
09:13
you have a solution only for one value of A
09:15
and then you have an unbounded one parameter family of
09:19
solutions plus there translate so this exponent is therefore sometimes
09:24
called the mass critical Exponent.
09:27
And it actually, it also plays a role for the
09:30
time dependent analysts,
09:31
nonlinear Schrodinger equation. Just look at the problem as I
09:35
have this nonlinear classical nonlinear Schrodinger Equation.
09:38
The most classical cases of course equal to three and
09:41
P equal to four,
09:42
so that you have a cubic nonlinearity.
09:45
Then if P is mass subcritical,
09:48
so less than 2 + 4 over in,
09:50
then this problem has a solution for all size of
09:54
zero in H1.
09:55
The solution is glow exists globally and is bounded.
10:01
If P is equal to 2 + 4 /
10:02
N, you still have global existence,
10:05
but you need to require that there are two norm
10:08
of the initial value.
10:09
Size of 0 is small and then for other values
10:12
of initial initial conditions you may have blowup.
10:16
In fact for many.
10:18
And if peace mass super critical,
10:20
so P is bigger than 2 + 4 when you
10:23
have to require that the H1 Norm of the initial
10:26
condition is small,
10:27
then you have global existence and for many other initial
10:31
conditions you will have grow up.
10:33
So you see in this whole problem this exponent 2
10:37
4 /
10:37
N plays in the very essential role.
10:40
And here you see also it makes a difference whether
10:44
piece less than 2 + 4 /
10:46
N over the T is bigger than 2 + 4
10:48
/
10:48
N.
10:49
OK, now here. Let's get back to this general problem
10:55
here.
10:55
The variational approach I would actually I will present several
10:59
approaches to attack the problem.
11:01
One is the variational approach and I will give you
11:03
some details.
11:04
Occasionally you will see it gets a bit technical,
11:07
but I hope not too much.
11:08
But let me present the variational approach first because it
11:12
has been used in almost all papers of this subject.
11:15
You look at the energy functional written here J of
11:20
you because 1/2 integral gradient you squared plus V of
11:24
X squared minus the integral of capital G of Xu
11:28
Capital G is of course a primitive of Little G.
11:33
And you look at it and you don't find critical
11:36
points of J budgeted critical points of J defined on
11:40
the constraint to the edge to sphere.
11:42
And then Lambda does not appear in J.
11:45
It it appears as a LaGrange multiplier so,
11:48
but that's just very easy approach or not very natural
11:52
approach.
11:53
You have this variational functional.
11:55
Of course you have to require that the integrals are
11:59
well defined.
12:00
So for instance is Omega is equal to RN&V of
12:04
X becomes unbounded like view of X towards Infinity as
12:08
X Plus Infinity.
12:09
Then you need to require that this integral view of
12:13
XQ squared must be finite.
12:14
And of course, you also need to need to have
12:17
conditions on G so that this is defined in finite.
12:20
OK, but this is the variational approach and let me
12:23
show you now how it works in this setting.
12:25
Again, let's start with the sheep of the easiest case.
12:29
So we look at the case of the nonlinear Schrodinger
12:33
equation without any potential.
12:35
The domain is are in,
12:36
we just have the homogeneous nonlinearity,
12:39
the pure power due to the power P.
12:43
Then an important role place is played by the gagliardini
12:47
and work in inequality.
12:49
Namely, you can estimate the LP norm of you,
12:51
which appears here to the power P.
12:54
I the L2 Norm of the gradient of you to
12:58
some power gamma times the L2 Norm of you to
13:01
the power,
13:02
1 minus, in Gamma is precisely this value.
13:05
Any times 1 / 2 -
13:07
1 / P.
13:09
OK, if you plug this estimate into the definition of
13:13
Jay,
13:14
so that means you replace this term by the two
13:18
norm of the gradient and there to normal view.
13:21
Then you get an estimate of J from below the
13:25
L2 Norm of you now appears just as a constant
13:29
because we fix the L2 Norm of U by a
13:32
ton of you should be a CSR.
13:34
A is the set of all functions you say in
13:38
H 1 /
13:38
N such that the two norm is equal to a.
13:42
So this simply appears here.
13:43
Then you have here the two norm of the gradient
13:46
of you to the power,
13:48
P.
13:50
And seeing this, you see the question whether,
13:53
for instance J is bounded from below depends of course
13:57
on the exponent gamma over 2.
13:59
If this is less than one then J is bounded
14:03
from below and coercive on incurs in cursive on this
14:07
a toothier.
14:08
In this, P / 2 less than one is again
14:12
equal or equivalent to that P is less than 2
14:16
4 /
14:16
N. So if P is Mars subcritical,
14:20
you can try to solve this problem by Minimization to
14:25
find the critical point of J on this to see
14:29
recalls a by Minimization.
14:31
This also works, by the way,
14:34
if P is equal to 2 + 4 /
14:36
N becausw, then this exponent is equal to 1.
14:40
And you just have to require that this coefficient is
14:44
small.
14:45
But this coefficient is small if a is small because
14:48
a is just there to normal view and this is
14:52
just a factor in this coefficient.
14:54
So if P is equal to 2 + 4 /
14:56
N and a small, you may also try to minimize.
15:00
OK, this holds in general also for more for quite
15:05
general nonlinearity,
15:07
so if you replace the home with the power you
15:11
to the power 3 by nonlinearity then if this is
15:15
mass subcritical or it is mass critical and the two
15:20
norm of the solution is small then the function is
15:26
bounded from below.
15:28
Discursive on the sphere and you can try to find
15:31
solutions of our problem simply simply in quotation marks by
15:36
Minimization.
15:37
So you take a minimizing sequence and then look for
15:41
convergent subsequence.
15:43
The model non linearity you can see here is simply
15:47
a sum of homogeneous nonlinearities due to the power PJ
15:51
minus 2 /
15:52
U and the PJS are all less than 2 +
15:55
4 over anything of the BJ of X here.
15:58
Being positive and bounded.
16:02
So that is the nonlinearity.
16:03
You may think of. Then I said This is minimization
16:08
is of course not just that easy.
16:11
Use the work on that started with Plyos.
16:17
Yours and Charles Stewart in the 1980s,
16:20
and who considered in particular yours,
16:23
also considered more general and more complicated functionals,
16:27
but it actually the work continued,
16:30
and even a quote here,
16:32
last paper from 2020 from this year by Roman Villa
16:36
model,
16:36
where they also minimize use minimization to obtain a solution
16:41
of this normalized solution problem.
16:44
They consider the case where Omega is equal to RN.
16:48
The nonlinearity G does not depend on X.
16:51
In the found conditions on V.
16:54
So that this information is achieved.
16:57
So you see this work on minimization is still active.
17:02
Now for my talk today and also actually a lot
17:06
of my work is not concerned with the mass subcritical
17:09
case,
17:10
but with the mass supercritical case.
17:13
So and here. So now we're looking,
17:15
looking at the most super critical case.
17:18
I look at the case where Omega is equal to
17:22
RN.
17:22
The equation does not depend on exit is autonomous.
17:25
There is no potential V.
17:28
They now want to require that she does not depend
17:31
on X.
17:31
So here I just ate some conditions,
17:33
continued support and growth conditions.
17:36
So this is a kind of.
17:38
Was it the rabinovitz type condition?
17:41
But the model in linearity again is this G it
17:45
being a power a sum of powers,
17:47
but now the exponents are all bigger than 2 +
17:51
4 /
17:52
N.
17:53
OK.
17:56
Let's look at this case and state.
17:58
I will state you 2 results which are by now
18:02
quite well known.
18:03
So the 1st result is by Lujon.
18:05
Maybe I think it might,
18:07
as far as I know,
18:08
he was really the first consider this mass supercritical case
18:12
for this problem,
18:14
and he had precisely these conditions at one at A
18:17
and H2.
18:18
Under these conditions he proved that this problem three has
18:23
a radius solution for every value of A.
18:26
And more than or about 15 or 16 years later,
18:30
in a joint paper with Sebastian Divalerio la,
18:33
we proved it under the same conditions that XO required
18:38
the existing unbounded sequence of failure solutions.
18:43
Which are now which are pair slumped in yuan for
18:46
every 8 positive,
18:47
and I would like to give you a in the
18:50
next on the next slide.
18:52
A short idea of how this proof works.
18:54
These ideas they this work by variational methods.
18:58
Both results have been proved proved using variational methods and
19:03
these methods have been extended to many other and more
19:07
general equations,
19:08
at least here a few quasilinear Schrodinger Schrodinger fossil show
19:13
com.
19:13
Fractional equations, systems of equations,
19:16
and it's of course impossible to mention too many of
19:20
these go into any details.
19:21
Actually, at the end of this,
19:23
the last three pages of these slides are there,
19:27
you will find a list of references where you can
19:30
dive into the literature,
19:32
and this will be made the made available on the
19:35
home page of the Journal.
19:37
OK, now let me give you some idea of how
19:40
one.
19:40
How JA proceeded because these ideas have been used also
19:44
in any other papers following that.
19:46
So on simply looks at the various at the variation
19:51
functional.
19:51
So this is the functional as a set,
19:54
no potential involved. It's simply 1/2 trading you squared minus
19:59
integral of G of U.
20:01
We look at the two sphere in the space of
20:04
radial functions because we will also use the compact embedding
20:09
of H1 route into a four between two and two
20:13
star. And within this two sphere we haven't,
20:17
we introduce addition additional parameter,
20:19
namely, they're two norm of the gradient.
20:23
So this piece of row consists of all U in
20:27
each one.
20:28
Having the L2 Norm equal to a in the Altoona
20:31
Mauve,
20:32
the gradient equal to roll.
20:34
So be sobro separates the sphere into two components in
20:38
one component.
20:39
The item of the grade in this lesson role in
20:42
another component V,
20:44
L2 Norm is bigger than row.
20:47
And Jean observed that in this mass supercritical case,
20:51
the function J actually has amount infrastructure.
20:56
Namely, there exists over one and there exist two elements
21:02
you want you to in the sphere.
21:05
Such that.
21:07
That you wanted you to lie in the two different
21:10
components of the complement of Bureau one.
21:12
So the two norm of the gradient of you one
21:15
is less than rule one.
21:16
There two norm of the gradient of you too is
21:20
bigger than women.
21:21
And you can choose this in such a way that
21:24
the energy,
21:25
the functional J of you,
21:27
one in jail for two is less than the infamous
21:30
of J on this piece of Roman.
21:33
This is nothing more than saying You have a mountain
21:36
pass structure and you have a classical Min Max description.
21:40
You look at all at all paths connecting U-1 to
21:43
U2 on this L2 sphere,
21:44
and you make an influx,
21:45
the classical infamous. And then of course,
21:49
classical methods tell you there exists a police Mail sequence
21:54
for this mountain pass critical value of this functional J
21:59
constraint to the sphere.
22:01
But then the problem is,
22:02
is this police Mail sequence bounded?
22:05
And here you see there is already a difference compared
22:09
to the case when you look at J not constrained
22:12
to their two sphere button all on H1.
22:15
Cause in this case then when you want to prove
22:19
boundedness of Alice Mail sequence,
22:22
or do you always have to look at is at
22:25
J Prime of you ally to you?
22:28
But J Prime of you applied to you is not
22:31
allowed,
22:31
is not allowed in this setting,
22:34
because here we have just a constraint fit critical Palace
22:38
mayor sequence.
22:39
That means if you have you in this sphere is
22:43
a we can only test J prime of you with
22:46
elements of the tangent space of the sphere at you.
22:50
And you itself is not in the tangent space of
22:53
the sphere in the tangent space of the sphere.
22:56
Are all functions with satisfy integral U times v =
23:00
0.
23:00
So the classical way to obtain a bounded by a
23:04
similar sequence does not work here.
23:06
Now what works is this kind of scaling and when
23:10
you look at the scaling here you have you given
23:14
you in this tooth.
23:15
Here you look at the function as star go and
23:19
you look at defining this way.
23:21
Then of course, you see that here you will run
23:24
into trouble if your problem depends on X.
23:26
If there is an explicit X if there's a potential,
23:29
or if the nonlinearity depends on X.
23:32
But here we consider a case where it does not
23:35
depend on X and then you can compute J of
23:38
a star you easily you get the art.
23:41
The out of the argument of you simply by transformation
23:45
and so J of estar.
23:46
You simply is squared over two times the integral gradient
23:50
you squared minus 1 /
23:52
S Integral G of S to the over 2 times
23:55
you.
23:55
But the argument in you is simply X.
23:58
And so this has the effect that this function considered
24:02
as a function of S,
24:03
and you call this J sub Taylor is also the
24:07
River is also differentiable with respect to S.
24:10
Where is this object here is not differentiable with respect
24:14
to S and does not yield an element of each
24:17
one.
24:17
If you take the derivative with respect to S.
24:21
Then one also observes that there are two norm of
24:24
Esther.
24:24
You is the same as they are to normal view.
24:27
And the mountain pass structure for J which you find
24:31
here,
24:31
which address recall immediately translates into a mountain pass structure
24:37
for Jason to do.
24:38
So you see one star you if you put S
24:41
equal to 1 you just obtain you back.
24:44
So Jay tailed off on you want history of Urology?
24:47
Told of one U2 is jails you too.
24:49
And this infant woman which you see here is the
24:53
SMS in from here.
24:55
And so yeah, in this way amount in path structure
24:59
for this functional Jayson Tatum,
25:01
which is defined on R *
25:03
H one.
25:05
And you look at Jade Subtilior,
25:07
constrained to our times, this here and now.
25:09
You have a policy sequence for this.
25:13
So what does it mean?
25:14
So having a police Mail sequence for this reduced functional,
25:18
you now have two derivatives which converge towards zero,
25:21
namely the gradient with respect to you and the derivative
25:26
objective with respect to S.
25:28
So let us compute this for later.
25:30
First, compute the derivative of the gradient of Jay~ with
25:34
respect to you,
25:35
and then one sees is nothing else but the gradient
25:39
of Jay constrained to this fear.
25:41
Of SN star UN. And this means that SN Stallion
25:45
is the policies police made sequence for Jay on the
25:50
sphere.
25:50
So far we haven't gained anything because he know already
25:54
that Jay that they existed police Mail sequence or mountain
25:58
pass corresponding to the mountain possibility for J on the
26:01
sphere. But now let's go to the second relation,
26:04
namely that the derivative of Jay Taylor with respect to
26:08
S converges to zero.
26:09
You compute this and you see this is then the
26:12
function P of N star.
26:13
Ewan appears. And this is just this function here.
26:17
He'll fuse, integrate into squared plus N times gradient of
26:21
capital G of you minus in over 2 integrati or
26:25
few times you.
26:26
So this is just an elementary computation.
26:29
Now we have two things.
26:30
Every police Mail sequence for J Tilden,
26:33
or equivalently a Polish meal sequence for J plus this
26:38
relation P of installing an converges to 0.
26:41
And in fact, this relation can be shown.
26:44
It can be used to show that SN star UN
26:47
is really a bounded palacial sequence.
26:50
OK. By the way, this relation P of U equals
26:54
to 0 is the poziv identity,
26:57
which has to be satisfied by any solution of the
27:02
PD.
27:02
So in a sense we find the Palace mayor sequence
27:05
so it almost satisfies the PD and it almost satisfies
27:08
the poziv identity.
27:10
And then this type of Alice Mail sequence is bounded.
27:14
OK, but that's now I will be shortly going over
27:17
the rest of the proof and without any details.
27:21
Having a Palace male sequence for the constraint functional means
27:25
we get first of all LaGrange multipliers.
27:28
These come from the fact that we constrained to their
27:32
two sphere and the couple Lambda N&VN,
27:35
which is in style you and does not satisfy the
27:39
equation,
27:39
but it almost satisfies the equation because after all.
27:43
It's a Lambda in Veganism Palace Mayor Sequence and not
27:48
a critical point.
27:49
OK, then we know that we have just a just
27:51
said we have no that Viennas bounded,
27:54
so it converges weakly and so we have a solution
27:57
of our problem but.
27:59
Being a week limit you a week limit of the
28:01
solutions means that there are two norm of useless for
28:05
equal to a,
28:05
so this is the next problem.
28:07
The first problem was to get a bounded police Mail
28:10
sequence.
28:11
Now we need to control the L2 norm.
28:13
And for this one first needs to show that the
28:16
LaGrange multipliers are bounded and bounded away from zero,
28:21
and once one has done this,
28:22
one can actually show that VN is strongly convergent.
28:26
The Alice Miller sequence converges strongly,
28:29
and then one has a limit which lies on the
28:32
correct sphere and solves the problem.
28:35
OK, that would be the sketch of the proof of
28:38
this result of Lujon,
28:40
which has been generalized as I mentioned to many other
28:45
problems.
28:45
For the infinitely many solutions,
28:48
one in this paper with Sebastian,
28:51
we used a version of the fountain theorem.
28:54
You know, the fountain theorem is critical point,
28:58
theorem 4. Symmetric functionals defined on banner spaces.
29:03
Here we have a version for the functional defined on
29:07
the sphere.
29:08
The proof was a little bit in this paper with
29:12
the Valula was,
29:13
let's say, use quite advanced algebraic topology,
29:16
namely, logical index. In a recent paper with young and
29:21
you and Charlie you be found very also treated.
29:25
The problem now for Chicago equation,
29:28
there's a proof which has a more elementary touch.
29:32
It just depends on the possible and theorem.
29:36
OK, so that's the result for our for the proof
29:41
of these two results.
29:43
A variation is which I like a lot,
29:46
is to use the poziv identity as an additional constraint.
29:52
And we did this also.
29:53
So you take the two sphere.
29:57
This is the sphere of a two normal view equal
30:00
to A and you simply add the constraint of you
30:04
equal to 0.
30:05
This is not quite a manifold in general,
30:08
but in the paper with Nicholas one of 'em we
30:11
were still able to,
30:12
and actually I should say that on this with this
30:16
constraint the functional is bounded below in cursive,
30:20
so that is a very nice situation,
30:22
so you can minimize and make a applied to standing
30:26
steel,
30:26
Anon type arguments and the extra we were able to
30:30
do this so we were able to prove that the
30:33
influence is achieved.
30:35
And also that this min Max values which come from
30:38
Lucia Nation German Thierry remember the capital G wasn't isn't
30:43
even function,
30:44
so this is invariant J of minus U is equal
30:47
to JFUP of minus use equal to be of you
30:50
using list and she'll run arguments.
30:52
We were also able to find that these see subpar
30:55
achieved in the papers.
30:57
We established this, but actually not not for an equation
31:01
but for system.
31:02
We just mentioned how it works for an equation.
31:06
But that's also very nice approach.
31:09
Mentioning system I would now like to proceed to a
31:14
system.
31:14
And I want to present this for two reasons.
31:18
One is that the system has been investigated since about
31:24
2005.
31:24
Because by many many people and it still offers very.
31:31
Interesting questions and problems at the second reason is that
31:35
I want to present a different way of proof naming,
31:39
not the variational proof to approach this problem.
31:42
I will have one slide just stating the variational proof
31:47
too so that you can see it works,
31:49
but one gets much better results with a different proof
31:53
approach,
31:54
which I will also, which I would like to sketch
31:58
here.
31:58
So many of you know this.
32:00
This is as I said,
32:01
it has been first there's the 1st paper by Lynn
32:04
in vain since 2005.
32:05
Since this first paper there were many people.
32:08
Many people's papers have appeared.
32:11
I mentioned only these few because I stopped all the
32:14
papers which appeared.
32:15
Even the papers which appeared after 2008.
32:18
There are too many, and so I just mentioned the
32:21
first.
32:22
Really, the first papers on the on the topic.
32:25
All these papers deal with the case that Lambda One
32:28
and Lambda two are prescribed.
32:30
So that means no information on their two norms of
32:33
you and Lee.
32:34
Now here the normalized probe solutions problems mean means we
32:38
want to prescribe their two norms of you and of
32:42
V,
32:42
and we take Lambda One and Lambda.
32:44
Two is our notes. OK.
32:48
So here's the result which I stadium which we achieved
32:52
recently.
32:53
That is, you may forget about the details.
32:56
I will just like to later to show you the
32:58
idea which can be understood without knowing any details.
33:02
But here is the result.
33:04
You take the unique positive radio solution of this problem,
33:08
minus lashing you. Plus you go to you to the
33:12
three in R3.
33:13
Then you use this solution capital U.
33:16
You see the PS Here in this Rayleigh quotient as
33:20
a weight function in the denominator.
33:23
You look at this Rayleigh quotient and then it is
33:26
not difficult to see that this in film is achieved.
33:28
It is positive and it is less than one.
33:32
So this this constant is needed in the hour result,
33:36
namely, maybe I just go back for a moment to
33:39
the system so that those of you have not worked
33:43
on this system C.
33:44
Again, how it looks like we have you take me
33:47
one and you two fixed this will they will just
33:50
be taking fixed in positive Lambda one and Lambda two
33:54
unknowns. But better is an additional parameter in the solution
34:00
will depend on the Perimetre.
34:02
For in the next result,
34:04
the only considered the case.
34:05
Better positive. You may of course also look at the
34:09
case of better negative,
34:10
as many of the papers also do,
34:12
but we look at the case better positive and the
34:15
following result describes parameter ranges of better so that we
34:20
can solve this solution for all values of A&B.
34:23
So here are the two intervals,
34:25
so you see either better is less than 1000.
34:27
Zero times the minimum of these two numbers.
34:30
We want you to. Or it is bigger than tall
34:33
sub zero temps.
34:34
The maximum of new material.
34:37
In either case, we see that the result states that
34:43
for any.
34:44
Prescribed values of the two norms of you and their
34:47
two normal V.
34:48
This problem. The system has a solution.
34:52
Lambda one Lambda 2UV. That is,
34:55
I think a very strong result.
34:57
In fact, I tend to believe it's it is optimal.
35:01
We have examples of better lying in between these two
35:05
values,
35:06
where the existence of the solutions depends on this quotient,
35:10
a / B. And I should also say that this
35:14
theorem improves significantly and earlier result of myself with Jean
35:20
equals father.
35:22
There we also found ranges of beta where is solution
35:26
exists,
35:26
but there the better depends also on not only on
35:30
this tour 0 and U1U2,
35:32
but also on a B itself.
35:34
In fact, in this paper with blue Jean Nicolas rather,
35:38
we did not have a single value of beta so
35:41
that the system has a solution for all values of
35:45
A&B.
35:46
OK, now the result with Jean.
35:48
So how that has been achieved by variational methods and
35:54
we can just be very short.
35:56
There's a function associated to it.
35:58
We looked at this functional constraint now to the product
36:02
of the two spheres.
36:04
First component has economical, a second component is Toonami equals
36:08
B,
36:08
Lambda one, Lambda two appears like wash multipliers.
36:12
In these papers, he saves the one with Jean.
36:15
So when we had several more papers in this.
36:19
Then jaw continued. Nicholas have also continued to work on
36:23
this system.
36:24
We use the approach which I sketched in quite some
36:28
detail for the system in a presented result of Jean
36:33
and myself with the Viola.
36:35
Benedetta Norris, so this be I should.
36:37
This is a bit confusing.
36:39
This is Benedetta I'm not involved in this research.
36:42
Benedetta Norma San. Who would have ours in January of
36:46
that CD.
36:46
Also have some very nice results at the end I
36:49
would just short dimension their work.
36:51
They consider this system but with potentials.
36:54
So there's a V1 of X times you squared and
36:57
the two of X times the squared entering,
36:59
but they're tracking potentials, meaning we one of X convergence
37:03
thoughts,
37:04
Infinity as extras, Infinity. And the same work for V2.
37:08
So this gives you also some compactness in the problem,
37:12
but it does not allow you to do scaling arguments
37:16
easily.
37:17
So now as I said,
37:18
I want to prove proof to look at the system
37:21
because I would like to present a completely different approach,
37:25
not the variational approach. What is the different approach?
37:30
So we take the system here.
37:32
You see the system, then this is the one which
37:35
we want to solve.
37:36
You see that I did not write the constraint on
37:39
the L2 norms,
37:40
but what I do is the following.
37:42
So we solve this for all various Lambda one and
37:46
Lambda Two and then we simply check which values of
37:50
L2 norms appear.
37:52
Very simple or innocence. Solve it for Lambda one and
37:55
Lambda two.
37:56
Let Lambda One and Lambda two run through all values
38:00
of all positive numbers.
38:01
You see they should be positive because otherwise we are
38:05
here in the essential spectrums along the one in Lambda
38:08
2 positive they go from zero to Infinity and then
38:11
simply collect which values for the solutions appear.
38:14
Witcher 2 norms of appear for the solutions.
38:17
Now here very to use variational methods to solve.
38:21
This seems not to be very successful or promising 'cause
38:26
it's difficult to control the two norms,
38:29
but what we do is we want to find global
38:33
connected continue of solutions along which these L2 norms vary
38:39
continuously.
38:40
And if we know that for the such a continuum
38:43
of solutions,
38:44
the two norm of you converges to zero and at
38:47
some other end of the flow of this continuum it
38:50
converges,
38:51
stores Infinity, then along this continuum all positive values of
38:56
the two norms must be achieved.
38:58
With varying of course, this Lambda one and Lambda two,
39:01
of course, Derrick.
39:03
OK, so now let's the simple fact.
39:06
Now what we do is we look this simplify the
39:09
problem with it.
39:10
So the old problem for the new promise,
39:13
6 almost identical. There's just one small difference.
39:17
You see. Instead of Lambda one,
39:19
there's Lambda in instead of Lambda 2,
39:22
there's one, so we reduce the parameter by one.
39:26
This is possible since with the scaling and it depends
39:30
on the fact that he is no potential.
39:33
Appearing in the nonlinearities independent of X.
39:36
Attending this available so we can reduce one of the
39:40
parameters and just have one parameter Lambda and keep the
39:45
others fixed.
39:46
We want Mewtwo is completely fixed and we investigate the
39:50
existence of solutions for various data,
39:53
but for fixed beta we had take Lambda as parameter
39:57
and look for continual solutions.
40:00
So the relation between 6:00 and the problem with the
40:05
fixed none of the normal solutions problems is as follows.
40:09
If you have a solution,
40:11
Lambda UV of this system 6 here.
40:14
Then you can solve our original problem where you find
40:19
a solution with Lambda One and Lambda Two and UV
40:23
having prescribed norm a /
40:25
B. If the quotient a /
40:27
B is the same as the quotient of their two
40:32
norms of U&V.
40:33
That is simply by the scaling and it is easy
40:37
computation.
40:38
There's no no fact behind,
40:40
so the first idea is of course now we fix
40:43
better positive and look at the quotient of their terms
40:48
of U&V for all solutions,
40:50
Lambda UV.
40:51
So take the set of all solutions Lambda UV.
40:54
They solve 6. Look at the quotient of the L2
40:58
norms.
40:59
And see what kind of ratios appear.
41:05
And if the image of Robetta is all of our
41:08
plus.
41:09
Then our system four and five has a solution for
41:13
all values of A&B.
41:14
OK.
41:16
So next idea is the fight now.
41:18
As I mentioned, global connected continue of this of solutions.
41:24
Of our problem 6. And connected continue they will bifurcate
41:29
from trivial or semi trivial solutions.
41:32
So I make I modify six a little bit and
41:35
allow here that you and we are also equal to
41:38
0.
41:38
In that case I have trivial solutions,
41:40
so one trivial solution for each value of Lambda.
41:43
Of course you equal to 0 V equal to 0
41:46
is a trivial solution.
41:48
But you also have semi trivial solutions where the second
41:51
component is equal to 0 or the first component is
41:54
equal to 0.
41:54
So these are nice connected branches and the capital use
41:58
are explicitly unique positive radial solutions of this problem,
42:02
with the corresponding values of Lambda mu one and Tier
42:05
1 mutual that you see.
42:06
The second branch does not depend on Lambda,
42:09
because there is no longer appearing here.
42:13
OK, let me give you a picture of the situation.
42:16
Looks load now on the left picture you see the
42:19
Lambda X is here.
42:20
The UX is here. In the VX is here.
42:23
Here is the first semi trivial solutions where V is
42:26
equal to 0,
42:27
the second component and here's the branch where U is
42:31
equal to their zero.
42:32
And they're both you and we are equal to 0.
42:35
This would be the branch of trivial solutions here,
42:37
but there will be no bifurcation from this.
42:40
Now, in order to stay within time,
42:43
I will be rather short.
42:44
This picture on the right just illustrates where when he
42:48
can,
42:48
if possible bifurcation. This would be the Lambda access.
42:52
This is the better access that you see.
42:54
Two curves better depending on Lambda.
42:56
Better one of London, better to off long term and
43:00
if you take better here then you fix it.
43:02
You go to the right,
43:03
you hit this. Each of these curves precisely once and
43:07
for instance here you go down to the cross running
43:10
Lambda.
43:10
Yes, precisely one Lambda into this Lambda value.
43:13
There's bifurcation in this case from the second curve.
43:16
That means from this curve.
43:18
Fix better here. Hit this curve.
43:20
Go to the Lambda value.
43:22
Take Lambda here, go here,
43:23
then there's verification. This is what the picture means,
43:28
so we computed the bifurcation values.
43:30
And now I'm sitting where one has bifurcation from both
43:33
of these curves.
43:34
This depends on beta in some.
43:36
For better small there's no bifurcation from either of these
43:40
curves.
43:41
In case better is a little bit larger,
43:43
this verification from one in case better is large.
43:47
Then there's bifurcation from both cross.
43:51
So you have bifurcation. So we compute this bifurcation points
43:55
that can be done rather explicitly.
43:57
Then we formulate the problem as a fixed point equation.
44:01
Now as a set we use topological methods and not
44:04
variational methods.
44:05
We formulate the problem at 6 point equation.
44:08
So the solution of our systems are fixed points of
44:11
some nonlinear operator,
44:13
a Lambda better applied to UV,
44:15
but you only look for solutions in the cone UV
44:18
greater equal to 0.
44:20
Then he compute Local 6 point indices for those who
44:23
are not so well informed about this business,
44:26
then this is just a variation of the later show
44:30
degree and it it measures the fixed points inside the
44:34
Cone P.
44:35
So we compute these fixed point indices and you see
44:38
these numbers,
44:39
which I wrote here. These are the local fixed point
44:43
index of a semi trivial solution.
44:45
For this corresponding value of London and you see this
44:50
fixed point index changes for the curve teaser two.
44:54
It changes at this value of Lambda it Park change
44:58
from zero to minus one.
45:00
For the courtesy of one,
45:02
you also have a change of fixed bindings in it
45:04
from minus one to zero.
45:06
I think a change of fixed point index gives you
45:08
global bifurcation.
45:09
That's like the famous rabinovitz global bifurcation theory and you
45:14
get global connected Continuum.
45:16
Now you see here in case that these two bifurcation
45:19
points coincide,
45:20
you immediately get that role.
45:22
Better of this of this set of this continuum is
45:26
equal to our plus becausw.
45:28
What does this mean? That it bifurcates?
45:30
Here you have passed UV on the branch such that
45:34
you converges .0.
45:35
So the air to normal view converges to zero.
45:38
There two normal V converges with some value,
45:41
so the quotient of two normal view divided by two
45:45
normal V Converter,
45:46
Scott 0. He is just the other way around there
45:49
to know more few divided by their tied Poonam of
45:52
the convergence starts Infinity because here there to Novi converges
45:57
to 0 because this is a branch.
45:58
The same attribute on TV is equal to 0.
46:01
So along this connect connected continuum,
46:04
this role better passes from zero to Infinity,
46:07
so it shifts all real numbers,
46:09
positive real numbers and that means we have solutions for
46:14
problem for all values of a B.
46:16
OK, then this situation can also appear now for reasons
46:19
of time I will not go through it,
46:21
but you can show the same argument.
46:25
OK, then I would skip this.
46:27
This is not the topological methods and our 2nd result
46:31
now in the last minutes let me mention 2 problems
46:35
where no scaling works.
46:36
One is a problem on domains.
46:39
I take the basic problem but now on a domain
46:43
we prescribe the eternal.
46:45
Problem 7 can be solved immediately as you know,
46:48
if Lambda is bigger than the minus Lambda one where
46:52
Lambda One is the first eigenvalue of minus Laplacian,
46:55
then you simply solve this by the mountain pass theorem.
46:58
For instance, the solution is in fact unique is Omega
47:02
Zabal,
47:02
so this is results from the 1970s and 1980s.
47:07
And also from the 1970s one sees that minus Lambda
47:10
one is a bifurcation point,
47:11
the existing global connected set of solutions bifurcating from minus
47:16
number 10.
47:17
And that means, since they bifurcate from lump minus down
47:21
to 1 zeros,
47:21
that means you converges to zero.
47:24
You find solutions if the L2 Norm is small.
47:28
If their two norm is L2 Norm of,
47:30
use small enough that this problem must have a solution.
47:34
This is the bifurcation scenario,
47:36
so he flung down. He is there two norm of
47:39
the solution and you have bifurcation from here.
47:43
And then I would like to present a nice result
47:45
of several nice results from this group of people,
47:48
nor his property to bars had seen it 2 papers.
47:51
The first of all short.
47:52
If P is mass subcritical.
47:55
Then the curve of Lambda U.
47:59
Is looks more or less like this,
48:01
and in particular it becomes unbounded there to a big
48:05
also be a value there too long becomes unbounded means
48:08
therefore all positive values of a.
48:11
You have a solution of this problem of the normalized
48:15
solution problem.
48:16
If P is equal to 2 + 4 /
48:18
N, the situation is different.
48:21
Then you will not find solutions for all values of
48:27
A.
48:27
In fact, if Omega is a ball,
48:29
this is with which they studied in detail,
48:32
there exists a solution only up to some value.
48:35
A star in the curve looks like this.
48:39
And if P Supercritical, the situation is like this.
48:42
You have solutions up to a star including a star.
48:47
But again, on the ball this is,
48:49
and if it only if in case of domain they
48:52
also have some more results but not in this dead
48:55
sharpness.
48:56
Here, the method of proof is again little bit different.
49:00
I see I run out of time.
49:02
I will not go through that they replace the two
49:05
norm by 1 by introducing another parameter with this current
49:09
scaling they solve this problem by maximization.
49:12
Having here you see an additional constraint.
49:15
This gives you 2 LaGrange Multipliers,
49:18
Lambda and Mu, and then they investigate the range of
49:22
LaGrange Multiplier.
49:23
Mu's role varies. OK, that's one thing and the very
49:27
last result is now for the system on our end,
49:30
again with not now with the potential.
49:33
But again, a situation where you cannot do the same.
49:39
I mentioned already a result by Ecomony Yamamoto,
49:42
who consider the case where is less than 2 +
49:45
4 over and they had the case where V of
49:48
X convergent or zero.
49:49
This is a difficult case because then here this operator
49:53
1 slash UV as essential spectrum zero to Infinity.
49:56
I mentioned another result by North of ours.
49:59
That scene in which are also mentioned before they consider
50:03
trapping potential.
50:04
When I mentioned there result in case of a system
50:08
but this.
50:08
Lots of course works in case of RN.
50:11
And to conclude this is now the last slide together
50:15
with Ricardo model and material.
50:17
It's Ian generated Seeney be considered this problem?
50:23
When V of X converges to zero and this is
50:27
more supercritical.
50:29
The this theorem is infected.
50:31
Corollary over more stronger result,
50:34
but we obtain the result if either.
50:38
Movie or if P is close to the exponent 2
50:41
4 /
50:42
N and a is kind of small but independent of
50:45
him.
50:46
OK, let's stop with this.
50:48
We have asset is a special case of corollary over
50:52
more general result.
50:53
But I think my time is over and dystopia.
50:57
You see there are three more pages,
51:00
but these are just references to the literature and you
51:04
can look at these on the home page of the
51:08
Journal when the slides are made available.
51:11
So thank you very much.