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Join Kanishka Perera as he presents an abstract critical point theorem based on a cohomological index theory that produces pairs of nontrivial critical points with nontrivial higher critical groups.

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00:05

the prototype problem

00:07

that that we are going to look at today

00:09

so we have this uh

00:10

problem involving the uh laplacian

00:14

operator uh and we on the right hand

00:16

side we have

00:18

uh three terms the first term uh

00:21

has the same homogeneity as the p lab

00:24

launches so it's of the order p

00:26

p minus one and then we have a

00:30

a term uh of order less than p minus one

00:34

uh this sigma is uh less than p

00:38

and and and then we have a term of order

00:42

higher than p minus 1 this q is bigger

00:44

than

00:45

p so problems like these are generally

00:48

referred to as

00:49

problems with combined non-linearities

00:53

and the uh the

00:56

the term q can be up to the

00:59

uh the critical exponent npo n minus

01:03

p if p is less than n and no

01:06

restrictions on q

01:07

if p is greater than or equal to n

01:11

and lambda and mu are basically

01:15

two parameters to positive parameters so

01:18

so the the basic results that we want to

01:21

prove here is that for all positive

01:24

lambda

01:25

this problem has at least two

01:28

non-trivial solutions

01:30

if mu is small enough

01:40

so uh so let me denote by lambda 1

01:43

the the first directly eigen value of

01:46

the

01:46

the p laplacian on omega so this is

01:49

positive

01:50

and and if lambda is between 0 and

01:53

lambda 1

01:54

and mu is positive then you can

01:57

produce a pair of non-trivial solutions

02:00

by using a a local minimization argument

02:04

and the mountain pass theorem so what

02:06

you do is you produce

02:08

a an ordered pair of positive sub and

02:11

super solutions and then you can use get

02:14

a

02:15

local minimizer between those

02:18

sub and super solutions and then you

02:21

apply the the

02:22

mountain pass theorem to get a second

02:24

non-trivial solution so the

02:26

local minimizer will be at a negative

02:29

energy level

02:30

and the mountain pass point will be at a

02:32

positive energy level so this was an

02:35

argument that was uh first used by

02:39

uh ambrosetti brzezis and gerami

02:42

in their 1994 paper and after that

02:46

many people have extended this to

02:50

related problems like the the pilaf

02:52

plushie

02:53

the the original paper of uh adversity

02:56

brazilian

02:58

uh considered the the the blood function

03:01

so that's the case for p is equal to two

03:03

and it was followed by other people

03:05

extending this to

03:08

be different from two

03:12

and so here are some examples of uh

03:18

papers giving results on this uh problem

03:20

for

03:21

uh for the peel of fashion uh starting

03:24

with

03:24

uh asaroro and

03:28

as as alonso and alfred in 2000

03:36

so so the case that lambda is less than

03:38

lambda 1 is

03:39

more or less settled and and this talk

03:42

is mainly about the case where lambda is

03:44

bigger than or equal to lambda 1 so in

03:47

this case

03:48

the the variational functional no longer

03:51

has

03:51

the mountain fast geometry and

03:54

as far as i know there are no results

03:58

there are no multiplicity results for

04:00

this problem uh

04:02

available in the literature for that

04:03

case so so

04:05

i'm going to present a a linking theorem

04:08

uh

04:08

based on a co-homological index theory

04:11

that can be basically capture the

04:14

geometry of the functional when lambda

04:16

is

04:17

uh bigger than or equal to lambda 1

04:20

and and it still produces a pair of

04:22

non-trivial critical points

04:24

for all positive lambda provided that mu

04:27

is sufficiently small

04:33

but these critical points are

04:36

not going to be minimizers they're not

04:39

going to be local minimizers

04:41

or mountain pass points they are going

04:43

to be a

04:44

higher critical points in the sense that

04:47

they both have a a non-trivial

04:51

higher critical group so let me just

04:54

recall the definition of critical groups

04:58

quickly so suppose you have a c1

05:01

functional e

05:02

defined on a banana space and suppose u

05:05

0 is an isolated critical point of e

05:09

so the the critical groups of e at u0

05:12

are def

05:13

in dimension q are defined to be the the

05:15

relative

05:16

singular homology groups uh group of

05:20

the pair ec

05:23

and ec minus u0 where ec

05:27

is the sub level set where the

05:29

functional is

05:30

less than or equal to the the critical

05:32

value

05:33

corresponding to u0 so

05:37

so these critical groups basically

05:39

measure

05:40

the the change in the topology of the

05:42

sub-level sets

05:44

when you pass through the critical point

05:46

u0

05:51

okay so now let me describe the uh

05:55

the main result here which is the the

05:57

abstract critical point theorem

05:59

so suppose w is a banach space

06:03

and m is a bounded symmetric subset

06:07

of w minus 0 that's radially

06:10

homeomorphic to the unit sphere

06:12

so this means that if you take the

06:15

radial projection

06:16

onto the unit sphere and restrict

06:20

it uh to m that's a that's a

06:23

homeopathism

06:24

uh onto the unit sphere so in that case

06:28

we could

06:29

write down what the radial projection

06:33

onto m is in terms of the radial

06:36

projection onto

06:38

s the unit sphere so if you if you first

06:42

first if you first radially project onto

06:45

the unit sphere

06:47

and then pull it back using the inverse

06:51

of the restriction to m of the

06:53

projection onto the unit sphere

06:55

you get the radial projection

06:58

on to m so so we will make use of

07:02

this uh projection pi m

07:06

and then i will denote by i a the the z2

07:10

chordological index

07:11

of fertile and rabinovic for a symmetric

07:15

subset a contained in

07:17

w minus the origin

07:22

okay so so with that let me describe the

07:25

uh the abstract critical point theorem

07:28

so suppose you have a c1 functional on w

07:32

and and two disjoint

07:36

closed symmetric subsets of m

07:40

with the property that the index of

07:43

0 and the index of the complement of b 0

07:46

in

07:47

m are the same and and

07:50

their common value is finite

07:55

and then we are going to use these

07:58

the two sets a 0 and b 0 to construct

08:02

a linking structure so suppose there is

08:06

a

08:06

w 0 in m minus a 0

08:10

and numbers uh are little r

08:14

which can be zero uh or positive and

08:17

uh and then a row and r

08:20

uh in that order and uh and two

08:23

real numbers a and a less than b

08:27

and we are going to describe this sets a

08:30

1

08:30

a star b star a and b

08:35

so so let me show you what this sets

08:38

would look like

08:40

in a picture roughly speaking so

08:43

for example suppose m is is

08:47

looks like a sphere and

08:51

so so in this picture these two

08:54

shaded areas on

08:57

m denote the set a zero and this band

09:02

[Music]

09:04

is the set b zero and then again the

09:07

requirement is that the

09:09

index of a 0 and the index of the

09:12

complement of b

09:13

0 are the same and then we take this

09:17

other point w 0 on

09:20

m and and we define this set a1

09:24

so the a1 is the set of so here's how

09:27

you construct this at a1

09:28

you take all line segments joining

09:31

points on a0 to w0

09:35

and and you project those line segments

09:39

uh on to the the set m radial using the

09:43

radar projection

09:45

pi m so using this set a1

09:49

now we are going to construct through

09:51

two sets a and b

09:53

uh that link in a sense that i will

09:56

describe

09:56

uh soon so so what you do is

10:00

uh the the the the set b

10:03

is basically a scaled uh

10:07

copy of b0 so so so this b0 is basically

10:12

uh this

10:15

set b is basically set b zero scaled

10:18

dilated by row

10:22

and then the construction of the set a

10:24

is a little more complicated

10:26

so so here's what you do you take the

10:29

set a1

10:30

and you scale it by the

10:34

by little r so you get this this little

10:37

scaled copy of a1

10:41

and and little r is less than rho so

10:43

that will be inside

10:46

the band b and then you also take a

10:50

scaled copy

10:51

of a1 scaled by

10:54

capital r so that's this set

10:57

in the front

11:01

and and this set will be outside of b

11:04

because capital r is bigger than rho

11:07

and then you look at the the two copies

11:10

of a0

11:11

on the on these two sets the one in the

11:15

rear and the one in the the front and so

11:18

you have these two scale

11:19

copies of the set a0

11:24

on those two sets and you you join the

11:27

corresponding cop

11:29

points on those two copies by line

11:31

segments so so this part

11:35

and that part are made up of the

11:38

line segments joining points on the the

11:41

two copies of

11:43

a0 that are that are on uh

11:47

the front arc and the the and the real

11:49

rock

11:50

so so this is the uh this is the setting

11:54

so so we is also so i have described

11:57

those sets here

12:00

and and with that

12:04

here's the uh so here are the main

12:06

assumptions on e

12:08

so so and and also uh

12:12

we have two cells uh these two sets a

12:15

star and b

12:16

star so a star and b star are basically

12:19

uh filling in the two sets a and b

12:24

so b star is just the the set of all

12:26

line segments

12:28

joining points on b to the origin

12:32

and a star is is the set that

12:36

fills in between this this set a

12:40

so basically what you do is you look at

12:42

this real copy of

12:44

a0 and the front copy of

12:47

the real copy of a1 and the front copy

12:50

of a1

12:51

look at their corresponding points and

12:54

join the corresponding points uh by line

12:57

segments so

12:58

so a star is the is the the set filling

13:02

in this set a by those line

13:06

segments okay so with that

13:10

here are the assumptions on the

13:12

functional

13:13

we assume that on the set b stop

13:18

e is bigger than a so on the set that's

13:21

filling in b uh the function is bigger

13:24

than a

13:25

and then we assume that on a and b the

13:28

values of

13:29

the functional are bounded away from

13:32

each other

13:33

so meaning that the soup of e on

13:36

a is strictly smaller than the

13:39

inf of e on b

13:44

and finally we assume that on the set a

13:46

star

13:48

which fills in the set a

13:52

the function is is strictly smaller than

13:56

b so the soup of e on a star is less

14:00

than b

14:01

and then we also assume that the

14:03

functional

14:04

satisfies the palestinian condition

14:08

at all levels between a and b

14:12

then the conclusion is that he has a

14:14

pair of critical points

14:16

uh u1 and u2 with the

14:19

with the value the critical value of u1

14:23

between the the inf of e on v

14:26

star and the soup of e on a so

14:30

so this will be a a critical point with

14:32

uh with energy

14:34

less than the the supremum of e on a

14:37

and the second critical point will have

14:39

its uh critical value between the

14:42

infimum of e on b and the supremum of

14:45

e on this star so it will have energy

14:48

bigger than the

14:49

bigger the equal to the in freedom of e

14:51

on b

14:53

and then if and if in addition if you

14:56

assume that he has only a

14:58

finite number of critical points with

15:00

critical values in the

15:02

interval from a to b then you can choose

15:05

u1 and u2 in such a way that the

15:08

kth critical group of u1 is not trivial

15:12

and the k plus first critical group of

15:13

u2 is non-trivial

15:15

and then remember this k is the the the

15:18

common chromological index of

15:21

a0 and the complement of b0 in there

15:26

okay so this is the the abstract

15:28

critical point theorem

15:29

and and the the proof of this is based

15:33

on the the notion of uh

15:35

homological linking so let me quickly uh

15:38

recall the definition of chromological

15:40

linking

15:41

so suppose you have uh two disjoint

15:44

non-empty subsets a and b

15:47

of a banach space w then we say that uh

15:50

a

15:50

links b cohort logically in dimension

15:54

k if if the inclusion of a

15:57

in the complement of b induces

16:00

a non-trivial homomorphism between the

16:04

the homological homology

16:07

in in dimension k so

16:11

so so this inclusion induces this this

16:14

homomorphism

16:15

uh between the the reduced homology

16:18

groups

16:18

in all dimensions it will be from

16:23

the reduced co-homology of w minus b to

16:25

the reduced chromology of a

16:28

and the assumption is that this

16:31

homomorphism

16:32

is non-trivial in dimension k and if

16:36

that's the case then we say that these

16:37

sets

16:38

uh link homologically in dimension k

16:43

so so the proof of the theorem is based

16:45

on uh

16:47

on this new linking structure so so

16:50

basically you show that

16:52

uh this this set a links the set

16:56

b uh chromologically in in dimension key

17:00

and this is

17:01

and this is the the main step in the

17:04

proof of the theorem

17:05

and once you have that then you can use

17:09

the

17:10

the assumptions that that we made about

17:12

the function the values of the

17:14

functional

17:15

on these sets to produce

17:18

two pairs of regular values

17:22

of the functional with non-trivial

17:26

relative co-homology so you produce

17:30

a pair of regular values of

17:33

of e with the

17:36

reduced the relative co-homology uh

17:39

non-trivial in dimension k and then you

17:42

produce another pair

17:44

with the with the sub-level sets having

17:47

relative

17:48

homology non-trivial and dimension k

17:50

plus one and then this

17:52

these two give you the existence of uh

17:56

u1 and u2 and they also give you this

17:59

additional information about the

18:01

critical groups

18:02

uh when you have uh isolated critical

18:05

points

18:06

when you only have isolated critical

18:07

points

18:09

okay so so so we so so with that

18:13

let me uh go back to our

18:17

the p lab partial problem and then

18:19

indicate how to

18:21

uh apply this abstract critical point

18:23

theorem

18:24

to get a pair of non-trivial solutions

18:27

of this problem but rather than doing

18:30

this just for the pilar production i

18:31

will

18:32

do this for a uh a general class of

18:36

problems that model the the p laplacian

18:39

problem that

18:41

that that i mentioned at the beginning

18:44

because

18:44

uh this type of result applies not not

18:47

only to the pilaf passion but

18:49

to to any problem

18:52

any operator that has the same structure

18:55

as the pillar plaster for example

18:57

the fractional peel of plastino or

19:01

kirchhoff type non-broken problems so

19:04

so let me prove a

19:08

an abstract multiplicity result uh so

19:11

so our setting will be a will be in a

19:14

uniformly convex banach space w with

19:17

dual w star

19:20

and and i will prove a multiplicity

19:22

result

19:23

to a class of potential equations

19:26

so let me recall that a an operator

19:31

a continuous operator from w to its dual

19:34

is a potential operator if it's the

19:37

fresher derivative

19:38

of some c one functional on w

19:42

so this this c one function of capital h

19:46

is called the potential for uh for h

19:49

so capital h prime is h

19:53

so so we will consider this uh

19:55

non-linear operator equation

19:57

you have uh you have two operators a p

20:00

and b p ah so now first

20:03

example this a piece is the p laplacian

20:07

the b p is the u to the power p minus

20:11

one and then you have

20:12

uh two other potential operators

20:15

the first one will be uh

20:19

but of lower order than p minus one

20:23

then the second one will be of higher

20:25

order than p minus one

20:28

okay so these are all potential

20:29

operators uh

20:31

and and here and lambda and mu are

20:33

positive so

20:34

so here are our assumptions on all these

20:37

operators

20:38

so first of all a p is assumed to be p

20:41

minus one homogeneous

20:43

and odd meaning that this relation holds

20:47

for all u in w and t in r

20:52

and then we assume uh an inequality

20:55

of the holder type so apub

20:59

is assumed to be less than or equal to u

21:01

to the p minus 1

21:02

v and if the quantity holds

21:05

you can only if u and v are scalar

21:09

multiples of each other

21:10

so so in particular a p u u is the norm

21:14

of u to the power p

21:18

and again the the both the the the p

21:21

laplacian and then the fractional pillar

21:23

function

21:24

would satisfy these assumptions

21:28

and then bp is also assumed to be

21:31

p minus one homogeneous and odd and is

21:34

assumed to be

21:35

uh positive or u positive

21:39

and also assumed to satisfy a whole type

21:42

in your quantity

21:44

this this whole time inequality

21:47

and and then of course u to the power p

21:50

minus 1

21:50

does satisfy this and finally it's

21:53

assumed to be a

21:54

compact operator

21:58

okay so those are the operators a and b

22:00

and then so here are the assumptions

22:02

on uh f and g so let's denote by capital

22:06

f

22:06

the the potential of little f with f

22:09

zero equals zero

22:11

you can always achieve this by just

22:13

subtracting

22:14

f 0 from f that will give you another

22:18

potential with

22:20

values 0 and 0. and this is assumed to

22:23

satisfy

22:25

this limit so f of t u divided by t to

22:28

the power

22:28

p goes to zero uh obviously goes to

22:32

positive infinity as t goes to zero

22:34

so this basically means that f the the

22:36

potential f

22:37

was like a power that's uh smaller than

22:40

b

22:41

in our example we had sigma uh less than

22:44

p so so u to the power sigma minus 1

22:47

will satisfy this

22:50

and the limit is assumed to be uniform

22:52

on compact subsets of w

22:54

minus zero and it's assumed to be

22:56

positive

22:57

uh except at the origin and f is assumed

23:01

to be bounded on bounded subsets of w

23:06

and then here are the assumptions on g

23:08

so again denote by capital g

23:10

the the potential with capital g of 0

23:13

equals 0

23:14

and this is assumed to go to 0 when

23:17

divided by

23:18

the normal q to the power p when u goes

23:20

to 0

23:21

so this is basically saying that g is of

23:24

a higher power than

23:26

p near the origin

23:32

so so in our example this is something

23:35

like u to the power q where q is bigger

23:37

than p

23:38

and g is also assumed to be positive uh

23:41

except at the origin

23:42

and bounded on boundary subsets of w

23:46

and uh it's and this is the assam main

23:49

assumption on on g

23:51

at infinity so g of t u over t to the

23:54

power

23:54

p is assumed to go to positive infinity

23:57

when t

23:58

goes to positive infinity uh again

24:00

uniformly on

24:02

compact subsets of w minus zero

24:06

so so again in in our example uh

24:09

this is a higher power than uh

24:13

p so uh when at

24:16

infinity it will go to infinity

24:21

so so the the solutions of this equation

24:24

uh are the critical points of this c1

24:28

functional

24:29

the first term is the the potential of

24:33

the operator a p and the second term is

24:38

lambda times the potential of the

24:40

operator v p and then you have the

24:43

the two potentials of f and g and you

24:45

see that if you

24:46

if you differentiate this e prime is

24:49

exactly what you get when you

24:52

bring all these terms of the equation to

24:54

the left-hand side

24:57

okay so with that here's the the main

24:59

multiplicity result

25:01

for this abstract equation assume that

25:06

a1 through g4 hold and you have the the

25:09

psa condition

25:10

for all c and then

25:13

if lambda is positive depending on

25:16

lambda

25:17

there is a mu zero such that this

25:20

equation has

25:21

two non-trivial solutions u1 and u2

25:24

for all mu between 0 and mu 0

25:29

the first solution has negative energy

25:32

and the second one has

25:33

positive energy

25:38

and then let's say let me just indicate

25:40

how

25:41

the abstract critical point theorem is

25:43

applied to

25:44

get this multiplicity result so the the

25:47

proof is going to involve

25:49

the non-linear eigenvalue problem uh apu

25:52

equals

25:52

lambda vpu so so here's the the

25:56

variation of formulation of this

25:58

eigenvalue problem let's denote by

26:02

uh m the the

26:05

the sphere of radius p to the power 1

26:07

over p

26:08

in w uh so this is a

26:11

it's basically a bounded complete uh

26:14

symmetric

26:15

c one finster manifold it's it's really

26:18

homomorphic

26:18

unit sphere so it fits into the

26:21

uh the setting of the the abstract

26:24

critical point theorem

26:27

and then the uh the eigenvalues of this

26:29

problem

26:30

coincide with the critical values of

26:32

this functional

26:34

psi given by p over

26:39

basically this is the one over the

26:41

potential

26:42

of the operator vp

26:48

and then you can construct a sequence of

26:52

eigenvalues using a minimax scheme

26:56

based on on some index theory

26:59

the the standard way to do this is to

27:00

use the

27:02

the the genus of the of christmas

27:06

elsie as the index but if you use the

27:10

the critical values based on the genus

27:13

you

27:13

you don't get in any you don't get

27:15

enough information

27:17

about the the the indexes of the

27:20

sub level sets to apply the the abstract

27:24

critical point theorem so so what we are

27:25

going to do here is to

27:27

construct the the sequence of

27:28

eigenvalues based on the

27:30

the co homological index so so that's

27:33

denoted by capital f

27:35

the class of symmetric subsets of m and

27:38

by f k the the sets with index

27:42

at least k and you and you set lambda k

27:45

to be the

27:46

the min max based on this this class of

27:50

sets

27:50

fk and then one can show that this is an

27:54

unbounded

27:55

and increasing sequence of eigenvalues

27:58

the first eigenvalue is positive

28:00

etc and then and this sequence of

28:03

eigenvalues

28:04

uh based on the chromological index was

28:07

uh first introduced in this

28:09

2003 paper in topological methods in

28:14

nonlinear analysis

28:17

okay so now uh let me denote by

28:20

size super array the the sub level set

28:24

where psi is less than or equal to a and

28:26

by

28:27

size sub a the the super level set drags

28:30

bigger than your equal to a then

28:34

so here's what you get from using the

28:36

the cohomological index

28:38

uh if lambda k is less than lambda k

28:40

plus 1

28:42

then the index of the sub level set

28:44

where

28:45

size less than or equal to lambda k is

28:47

equal to k

28:49

and also the index of the complement

28:52

of the super level set where psi is

28:55

bigger than or equal to

28:56

lambda k plus 1 is also k and you see

28:59

that this is the type of

29:01

assumption that we had for the sets a 0

29:04

and b

29:05

0 in the abstract critical point 0

29:08

so so so that that assumption was

29:12

based on this observation

29:16

and and but to apply the critical point

29:19

theorem

29:20

then we also need to uh

29:23

replace this lambda with cavity is set

29:26

that's a little more regular

29:29

so here's a key lemma in the in the

29:31

application

29:33

so the the dilemma is that if lambda k

29:35

is less than lambda k plus one

29:37

then the uh and then some level said

29:40

silent

29:40

k has a compact symmetric subset a zero

29:44

of index k so and so this cylinder k

29:48

in general need not be compact uh but

29:53

but you can construct a subset a compact

29:56

symmetric subset of the same index

29:58

as cylinder key and that this will be

30:01

useful

30:03

so so here's how the the abstract

30:05

critical point theorem is applied

30:08

if if lambda is between 0 and lambda 1

30:11

you take the set s0 to be the empty set

30:13

and b0 to be the whole sphere m so in

30:16

this case

30:17

the index of a0 is 0 so is the index of

30:20

the

30:22

complement of b0 and if lambda is bigger

30:25

than or equal to lambda 1

30:27

then lambda is between lambda k and

30:29

lambda k

30:30

plus 1 for some k bigger than or equal

30:33

to 1

30:34

and then you take a zero to be the

30:36

compact symmetric subset of

30:38

psi lambda k with index k and b zero to

30:41

be the

30:42

super level set where size bigger than

30:44

equal to lambda k plus one

30:46

and then so we have this this relation

30:50

which is the main assumption in the

30:52

theorem and you can then just apply the

30:54

abstract critical point theorem to get

30:56

the multiplicity result

30:59

okay so so let's uh let's see how this

31:03

is applied

31:04

in our prototype example so let's go

31:08

back to this

31:09

this original problem and i will replace

31:12

the

31:12

the the term u to the power sigma minus

31:15

1

31:16

with a more general non-linearity

31:19

uh so so the

31:22

so this f is something like u to the

31:25

power p minus

31:26

sigma q to the power sigma minus one

31:28

with sigma less than one

31:30

but it's not assumed to be odd in

31:33

in you if it's already new then you can

31:36

get

31:36

multiple results using the symmetry

31:40

with the functional will be even and

31:42

then you have

31:43

you can use other other results for

31:48

critical points of symmetric functionals

31:50

so so here we are not assuming that so

31:53

so these results uh does not involve

31:56

when c matrices symmetry does not play

31:58

any any role in these results so this

32:02

okay so this little f is uh

32:05

it goes like t to the sigma minus one

32:08

uh near zero and is

32:12

and it's primitive uh is assumed to be

32:14

positive

32:16

uh except at t equal to zero

32:19

and they may assume a growth condition

32:21

on f

32:22

uh so so we assume that it's its growth

32:26

is uh

32:26

so critical is lower than

32:30

the the power q

32:33

and and then q can be

32:37

up to the the critical exponent

32:41

p o n minus p if p is

32:44

less than n

32:47

and then the the the solutions uh of

32:50

this problem uh

32:52

coincide with the critical points of

32:54

this functional so this is the

32:56

variational function

32:57

associated with with our problem

33:00

so so the first result here is for the

33:03

subcritical case

33:05

so suppose uh q is

33:08

less than p star if p is less than

33:11

n more restrictions if p is bigger than

33:13

or equal to n

33:15

and and and lambda is positive

33:19

then applying the the the multiplicity

33:22

result for the abstract

33:24

operator equation we get uh two

33:27

non-trivial solutions

33:28

one with negative energy and one with

33:30

positive energy

33:32

for all all small enough mu

33:35

and and so basically you apply the

33:37

abstract result

33:39

uh with these choices of of operators

33:43

so so so ap in this case will be

33:46

uh will be the operator from

33:50

w1p0 to its dual uh

33:53

given by this equation so apu acting on

33:56

v is

33:57

this this integral and the the other

34:00

operators are defined uh

34:02

similarly and then the

34:05

so basically the operator equation uh

34:07

coin

34:08

[Music]

34:10

is the weak formulation of the uh the p

34:12

laplacian equation

34:17

and also the for the linear case p equal

34:20

to two

34:21

this result was uh actually recently

34:24

obtained in in this paper with

34:26

uh candido and morano to appear in

34:30

combinations with contemporary

34:32

mathematics

34:34

but the but the method used there was uh

34:37

very different so they are

34:40

the the operator is linear the laplacian

34:43

is linear operator so it has

34:44

linear eigen spaces so you can do uh

34:49

linking uh arguments based on

34:52

linear subspaces but for p not equal to

34:55

two this

34:56

just doesn't work because it does not

34:59

the p log partial for p different from

35:00

two

35:01

does not have linear eigen spaces so you

35:04

need

35:06

something based on the the sub level

35:08

sets like in the

35:09

get start critical point theorem here

35:15

okay so so here's a another application

35:18

of the

35:18

the abstract multiplicity result so

35:22

suppose we consider this the kirchhoff

35:24

type

35:25

non-local problem so here we are looking

35:30

at a

35:33

sub critical problem so this omega is a

35:35

bounded domain in rn

35:38

n is 1 2 or 3

35:41

and this q is between 4 and 6

35:46

so in dimension 3

35:48

[Music]

35:50

6 is the critical exponent so it's

35:53

assumed to be less than 6 if

35:54

n is 3 and no restrictions

35:58

uh a bow if n is one or two

36:02

and again f is the same type of uh

36:07

function as in in the early example

36:10

so it's uh so f

36:13

goes like t to the power sigma minus 1

36:17

where sigma is less than 4

36:20

and and q is

36:24

is bigger than 3

36:27

times q is bigger than 4 so

36:31

so so here p is basically 4

36:34

and this is the the p minus 1 term which

36:37

is the

36:39

term that's a lower power and this is

36:41

determines of a higher power

36:45

so so so here and then by the way this

36:48

is a non-local problem because we have

36:49

this uh

36:50

this integral in front of the laptop

36:54

so so so at each point this equation

36:58

depends on the values of u uh on the

37:01

whole domain

37:02

making it a non-local problem

37:07

but the this problem has pretty much the

37:09

same structure as the

37:11

the p laplacian equation that we looked

37:13

at so

37:15

uh so a really a similar result holds

37:17

for this

37:18

so so here's the uh the variation of

37:20

functional

37:23

and and the result is that uh

37:25

[Music]

37:26

if you assume or what we just assumed

37:29

about

37:30

p and about q and r

37:33

uh then in some critical case again you

37:36

can find two solutions

37:38

uh one at a negative energy level one at

37:40

a positive

37:41

energy level if the the parameter mu is

37:45

more

37:45

and again there are no restrictions uh

37:48

on

37:49

on lambda

37:52

so i mean so if you compare this

37:55

functional with the

37:56

the functional for the feel of function

37:58

you you see that

37:59

the first two terms are of the of the

38:01

order four so this is

38:04

something like the laplacian equation

38:05

with p equal to four

38:09

and and again here are the the choices

38:12

of the operators a

38:13

p b p f and g so in this case the

38:17

operator

38:18

a p will be a non local operator

38:21

it's action on the action of apu and v

38:24

is

38:25

given by this this right hand side

38:28

and and the rest of the operators are

38:31

just like in the

38:32

bill of national case

38:37

okay so so that's the uh

38:40

the some critical case so here you see

38:43

is an extension of the abstract

38:45

multiplicity result

38:47

uh that that you can apply even to the

38:49

the critical case

38:51

so suppose that uh

38:55

the so we are going back to the abstract

38:57

operator equation

38:58

so assume the assumptions a1 through g4

39:02

on the operators a b b f and g

39:06

and assume that for each mu

39:10

there is some c new a positive number c

39:13

mu such that

39:14

e satisfies the psa condition below the

39:18

level c

39:18

mu then we can still get

39:22

uh two critical points to non-trivial

39:25

critical points

39:26

if lambda is between 0 and lambda 1

39:29

uh assume that there is a w0 in

39:32

m such that the the functional

39:36

on the ray joining the

39:39

the the ray going through the the

39:43

point w zero uh is

39:46

strictly smaller than c mu

39:49

the supremum of the function on the rays

39:51

uh strictly smaller than c mu

39:54

for small mu and if lambda is between

39:57

lambda k and lambda k plus 1 for some k

40:00

assume there is a compact symmetric

40:02

subset

40:04

of the sub level sets psi lambda

40:07

with index k and a w

40:10

0 in in the complement of that set

40:14

such that if you if you look at

40:18

the lines joining

40:23

basically we if you take

40:26

if you take the if you take points in

40:30

c and form a linear combination like

40:33

sv plus t w 0 with s

40:36

and t non-negative then the supremum of

40:40

e

40:40

on that set is strictly smaller than c

40:43

mu

40:45

again for small mu then the conclusion

40:48

is that

40:48

uh if mu is small enough this the the

40:52

opponent equation will again

40:53

have two non-trivial solutions u1 and u2

40:58

with the the energy of u2

41:01

positive and strictly smaller than c mu

41:05

if mu is small enough

41:09

so so so this can now be applied to the

41:12

the critical peripheral problem so so

41:15

here's the

41:16

the critical peel of flashing problem

41:19

so say f is as before but now the

41:22

the q is replaced with the critical

41:24

exponent

41:26

p star and p is assumed to be

41:30

smaller than n

41:34

so to ah just to state the the

41:38

multiplicity result for this problem

41:40

let me denote by uh sigma of negative

41:43

delta p

41:44

the directly spectrum of the negative

41:46

laplacian in omega

41:48

so this is a set of lambda such that

41:51

this

41:51

eigenvalue problem has a non-trivial

41:54

solution

41:56

and let me denote my capital s the

41:59

the best subaru uh constant for the

42:01

embedding of

42:03

w1p0 omega in lp star

42:06

omega

42:10

so so again the the critical points

42:14

of this functional are the the solutions

42:16

of our problem

42:17

and here's the the the multiplicity

42:21

result for the

42:22

critical case so in the sub-critical

42:25

case we didn't have any

42:27

uh restrictions on n or p

42:32

and we also have any restrictions on

42:35

lambda it was just assumed to be

42:37

positive

42:38

but here we do have some restrictions

42:42

so suppose n is bigger than or equal to

42:44

p

42:45

squared and and suppose lambda is not an

42:48

eigen value of the

42:50

uh the negative pillar cluster then

42:53

you can find a a small enough mu

42:57

and a kappa such that the problem has uh

43:00

two non-trivial solutions uh one with

43:03

negative energy one with positive energy

43:05

and the energy would be less than uh

43:08

this right hand side so this this one

43:10

over n s to the n of p

43:12

minus kapha mu is basically the

43:15

threshold for compactness

43:17

uh in this functional e so the

43:19

functional will

43:20

uh satisfy the the psc condition

43:24

if you are below this level with the

43:27

kappa uh sufficiently small

43:31

and then you can apply the the abstract

43:34

multiplicity result

43:35

uh and show that this problem has

43:39

two non-trivial solutions

43:42

and again the for the semilinear case uh

43:45

p equal to this this problem was that

43:47

this theorem was proved

43:49

for the critical case in the paper that

43:51

i just mentioned

43:52

but again the the method in that paper

43:55

does not apply here because uh

43:57

p is different from you know

44:00

okay okay so let me

44:04

also just uh state the the result for

44:07

the the fractional pillar platinum

44:09

uh so so here's the the corresponding

44:12

critical problem

44:14

for the fractional pillar caution so

44:16

this negative delta

44:17

uh sp is the

44:21

p fraction plus in and then you have the

44:23

first two terms

44:24

on the right hand side that are the same

44:26

and

44:27

and then the last term will be uh

44:29

critical with respect to

44:31

the fractional p laplacian so this

44:33

fractional exponent

44:35

is n p over n minus s p

44:38

so p is assumed to be less than n over s

44:41

this is between 0 and 1

44:44

and the assumptions on f

44:48

are the same as before okay so i'm

44:53

almost out of time so let me just uh

44:55

skip the

44:58

the setting of the problem and just take

45:00

the

45:01

the excess the multiplicity result so so

45:04

here is the

45:05

uh the functional that corresponds to

45:08

uh our fractional uh pilab caution

45:11

problem

45:12

uh so so you have this

45:16

so so this is defined on the the

45:17

fractional solar space wsv

45:19

0 and again there are

45:22

restrictions on n p and lambda so

45:26

n is assumed to be s bigger than s p

45:29

squared

45:30

and we assume that lambda is not an

45:32

eigenvalue

45:33

of the fractional field of platinum

45:37

and then the the same conclusion as

45:39

before holes maybe that you get

45:41

two non-trivial solutions one with

45:43

negative energy one with positive energy

45:45

and

45:46

and this right hand side is now the uh

45:49

the threshold for compactness in the

45:52

functional e and and you you get two

45:56

nautical oscillations for

45:57

uh such really small museum

46:01

okay so thank you very attention

46:08

thank you for the uh peer rather very

46:12

nice talk very interesting talk

46:14

thank you very much any

46:17

questions or comments

46:24

you can give a

46:27

questions in the dialogue that says the

46:30

chat you can also

46:33

you you want some someone else attached

46:36

to some

46:40

listeners

46:49

uh equations frequently pilar plus

46:53

and uh what are the main difficulties

46:58

uh between the fractional and the no

47:02

fraction

47:04

yeah so the main difficulties in uh

47:07

estimating uh

47:08

some of the integrals because if you

47:10

look at the first term

47:13

[Music]

47:14

of the functional that you have this

47:16

this double integral

47:18

uh over r and cross rn and and

47:23

when you try to show that uh you can

47:26

construct these sets

47:28

uh c uh using this w0

47:32

where the function is strictly smaller

47:33

than the threshold for compactness

47:36

to to get that estimate you have to do

47:39

more work

47:39

to estimate that that first non-broker

47:42

turn

47:43

so that's that's one place where the uh

47:46

the additional difficulty comes in

47:48

and now logo non-local operator yes yes

47:50

right

47:51

and then and also the the talented

47:54

functions for the fractional pillar

47:56

function are still not known

47:57

so in for the peanut partial you know

48:00

the

48:00

the minimizers in the circle of

48:02

inequality in rn

48:04

but for the fraction of health fashion

48:06

you you don't know what they are

48:07

but but but you could still get around

48:09

this difficulty by using some estimates

48:12

for the for the for the talented

48:15

functions this this was done in a paper

48:17

with

48:18

marcus cosina sun ramos kony and yang

48:21

young

48:22

yeah so that that technique applies here

48:27

i also uh super professor

48:30

ask you please give me a static pde

48:34

to solve values for content

48:38

algebra in x y c t for

48:41

three dimension problem give us basic

48:44

equation for boundary values asks you

48:51

oh i'm sorry hear it clear can you

48:53

repeat that please you can see the

48:55

you can see what is right at the

48:58

chat chat with that says there's a

49:02

on the church yeah yeah yeah yeah you

49:04

can you can see that

49:08

a question that you can see

49:15

maybe he's asking what are the boundary

49:17

conditions

49:18

specifically for the specific problem

49:21

one of the specific problems you look at

49:23

well the boundary values are all always

49:25

taken to be the

49:27

zero boundary values so

49:30

those are the directory problems but but

49:32

you can you can apply this to 911

49:34

problems as well

49:35

the only thing that changes is that

49:37

instead of working in w1p0 you'll be

49:39

working in

49:40

some other space so w1p for example for

49:42

the for the neumann problem

49:44

but again you can define the the

49:46

operators apbp

49:48

in the same way so this abstract

49:49

multiplication result

49:51

applies even with other types of

49:53

boundary conditions

49:54

so the only thing you'll need is that

49:56

the boundary conditions will also

49:58

have to be homogeneous of order p minus

50:01

one

50:05

yes

50:09

and a questions or comments

50:17

so actually i had one uh just uh

50:20

we're speaking about also along the

50:22

lines of the fractional

50:23

uh equation i was just wondering

50:27

uh if instead of fractional p laplacian

50:30

you had something with a

50:31

with an integral kernel possibly

50:33

non-translation invariant like some k of

50:36

x y

50:37

would this follow from from a similar

50:39

kind of technique or is that a much more

50:41

complicated

50:42

uh problem i uh i haven't thought about

50:45

it but

50:47

so so the question is does this fit into

50:49

the abstract operator equation

50:51

setting so uh if you're

50:55

or if the operating defined in terms of

50:57

this this kernel satisfies the

50:59

the assumptions a1 through a4 then then

51:02

all this builds apply

51:03

so just have to check those four

51:05

conditions to make sure that they're

51:06

satisfied

51:08

okay so it's not some so it's not like

51:09

translation lack of translation and

51:11

variance

51:11

immediately rules out the whole thing

51:15

probably not though that's interesting

51:20

so uh i can solve non-linear

51:23

uh partial differential equations and

51:27

i would like for each talk uh

51:30

for the person to give a specific

51:33

equation

51:35

you know and specific boundary values

51:37

all zero

51:38

is fine for boundary values uh

51:41

because i would like then to try my code

51:45

and see if i get the correct solution

51:48

for it but you give me those general

51:51

equations

51:53

uh you know i don't have something i can

51:57

put into the computer to get a numerical

51:59

solution of

52:01

yeah yeah well i mean for so if i may go

52:05

back to my

52:06

uh slides again for a second

52:09

so you see so this first problem i don't

52:12

know if you can

52:13

see my screen still so this is not yeah

52:16

okay so let me just share this screen

52:18

again

52:21

you can share that okay so so this so

52:24

this is the

52:25

uh the model problem so so you can for

52:28

example

52:29

take a p that's different from two so

52:32

for example let's say take p equal to

52:34

uh four and then pick a sigma that's uh

52:38

smaller than four so let's say you take

52:39

sigma to be three

52:41

and uh can add q to be uh

52:45

let's say five so also suppose you're

52:47

working in in three dimensions

52:49

so so that uh this is this is a the the

52:52

critical case

52:53

when q is equal to the matrix this is

52:55

sub critical when q is less than

52:57

six so so you could try to do that

53:00

problem so you have specific values for

53:02

p sigma and q and you can take

53:06

omega to be a wall for example and

53:09

and then you have this this problem with

53:11

specific values and you can see

53:14

if it has two solutions for

53:17

a small enough mu

53:21

and then no restrictions on lambda for

53:23

example you can take lambda to be one

53:27

is u like the sine of x times the cosine

53:30

of y

53:32

and the arc tangent of z

53:35

uh you know i want a specific equation

53:43

a specific equation not a general

53:46

equation

53:47

so so the specific equation is negative

53:49

for laplacian of u

53:51

equals uh u to the power

53:54

power 4 plus mu u2 uq

53:58

plus u to the power phi what value

54:01

for mu well okay yeah so you need to

54:05

plug in let's say small values of mu

54:07

and see if you do get two solutions i

54:09

know i don't know how small it has to be

54:14

have you not worked a numerical solution

54:17

no i don't no i don't do numerical

54:20

solutions

54:20

oh okay sorry your day your lady

54:24

we uh here configure the

54:28

you see things not

54:29

[Music]

54:31

yes yes yes this is some difference

54:34

maybe

54:35

okay so as we talk about

54:38

very very very

54:41

okay so thank you the speaker and all

54:43

the listeners again

54:45

uh we talked about give a very nice talk

54:48

very interesting

54:50

so thank you you're again thank you

54:54

okay see you bye

54:58

everyone okay very very good thanks

55:01

everybody

55:02

[Laughter]

55:04

thank you