09 - SN Partial Differential Equations and Applications Webinars
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Join Yanyan Li of Rutgers University as he examines the question of blow up rates in higher dimensions for the insulated conductivity problem. Don’t want to miss another webinar? Register here: https://go.sn.pub/lXgEaxHave an article on the insulated conductivity problem? Submit Your Article: https://go.sn.pub/GT525VLearn more about publishing with SN Partial Differential Equations and Applications: https://go.sn.pub/UFulhi
00:06
so
00:06
it's my great pleasure to introduce
00:09
today's speaker
00:10
i'll be brief because we all know
00:14
him very well so okay
00:17
uh received his phd at the quran long
00:21
time ago
00:22
and he joined the ruckus from
00:25
1990 and he he has been
00:29
distinguished professor there from 2000
00:33
and he hold
00:36
many guest positions including
00:40
the one at the beijing normal university
00:45
he received many honors and among those
00:48
are
00:48
simon's fellow this year and the
00:51
inaugural class of fellows
00:54
from american mass society and he is on
00:57
the editorial boards of many journals
01:00
including
01:01
advances in mass
01:06
as we all know the union made tremendous
01:09
contributions to
01:10
advancement of the
01:13
pd's geometry and the theory
01:17
i mean analysis of pdes lately he has
01:20
been
01:21
producing some breakthrough results on
01:24
pde
01:26
arising from mechanics especially
01:29
the theory of composite materials today
01:32
he will talk
01:33
on this which i'm eager to
01:37
hear and then all yours
01:41
thank you humber and thank you very much
01:43
for the
01:45
kind introduction and
01:48
it's a great pleasure for me to be here
01:51
to see you
01:52
and to see many friends and
01:56
so i will share the screen
02:00
now so today uh i will talk about
02:04
this greeting estimates
02:07
for the insulated conductivity problem i
02:10
will explain
02:12
uh what it is so i will
02:15
i will mention uh some earlier results
02:19
and then we'll we'll be
02:23
uh talking about a recent result with uh
02:27
with my students drawing young on this
02:30
problem so first i will restrict to
02:35
mainly just the greeting estimates so uh
02:38
we know that for harmonic functions
02:42
if you you are if we are in the domain
02:44
and if we are given any control
02:47
in low norms for example it's one norm
02:50
l2 norm
02:51
and then when we go to the interior we
02:54
will have
02:55
higher derivative estimates so for
02:58
instance the gradient estimates
03:02
and so
03:05
in in a bounty domain
03:08
and uh if we look at the
03:12
divergence form elliptic equations of
03:16
second order
03:19
so the coefficient is elliptic
03:23
so this matrix is symmetric and have a
03:25
positive and
03:27
lower bounded upper bound and we give
03:30
some
03:30
regularity not only our infinity gives
03:33
some
03:33
continuity for instance gradient bounded
03:38
then if we have h1 norm
03:42
bounded by one when we go to interior
03:46
we will have and we will have a
03:50
gradient estimate
03:54
so so the cur if the coefficient
03:58
is not so smooth so
04:01
so still in abundant domain
04:04
still elliptic we have uniform
04:07
ellipticity
04:10
and second order divergence form
04:13
equation
04:15
and give control of h1 funded
04:19
then the georgie nash theory
04:25
will tell us that there is a
04:29
holder estimate the holder exponent
04:32
depends on
04:34
the electricity
04:37
so the
04:40
this alpha in general cannot
04:44
reach one so it's a it
04:47
we know that so so it will depend on
04:49
this ellipticity
04:53
so there's one situation which
04:58
which comes from at at least coming from
05:01
uh the study of uh composite material
05:05
so it's like this so we are
05:09
in a domain so this omega is the bigger
05:12
one
05:13
so this is a unit size it's like a take
05:15
a ball
05:16
of radius 10 for example and
05:20
there are two inclusions they are unit
05:23
size they are not very very tiny
05:25
so they are unit size boundary are
05:27
smooth
05:29
and this is another one
05:32
boundary smooth unit size but they are
05:36
they can be getting close so this is the
05:39
epsilon is small so the distance between
05:42
them
05:42
are small so and
05:46
in the limit situation they can't even
05:48
touch
05:50
so the coefficients we are given
05:54
are piecewise constant
05:57
so let's say inside here
06:01
we have a k and here we have a k
06:04
and outside we have a one so
06:07
the k is greater than zero
06:11
and less than infinity so they are
06:13
elliptic so this is elliptic
06:16
and uh but the coefficient is
06:20
l infinity coefficient of l infinity
06:23
in general so well it's better than our
06:26
infinitive
06:29
but it's not yeah so
06:33
so so if you control
06:37
if you are given h1 control now the
06:40
question
06:40
is do we know that
06:44
uh l infinity is under control so
06:48
the main point is this you
06:51
the question is you want this constant
06:54
to be independent of epsilon you want
06:57
this constant to be independent of
06:59
epsilon
07:00
so that's the question so this
07:03
does not follows from the georgia
07:06
national theory
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the general theoretical theory because
07:11
uh
07:11
uh there will only get older
07:15
but but but there the coefficients are
07:18
allowed to be in all l
07:21
infinite coefficients so here the
07:24
discontinuity is rather special
07:26
it occurs across a smooth surface
07:31
so if allowing this epsilon
07:35
depending on epsilon
07:38
allowing this constant depending on
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epsilon so this result was known
07:46
much earlier so it was
07:49
this is the rather classical
07:52
it's called transmission problem so
07:55
so the the transmission problem
07:59
uh the transmission problem
08:04
will give you a constant which depends
08:06
on
08:07
epsilon so the for example it is uh uh
08:12
i so i i think this goes back to at
08:15
least
08:16
1960s at least i think there
08:19
this was well understood already
08:23
so the solutions were known
08:26
back then already known that the
08:29
solution
08:30
will be piecewise
08:33
in this case will be piecewise smooth
08:36
actually
08:37
after the boundary and in here piecewise
08:40
moves
08:40
up to the boundary and another the other
08:43
wall piecewise moves up to the boundary
08:45
and across the normal uh
08:49
across here the normal derivative
08:53
has a gap
09:00
so then
09:04
so evo babushka and collaborators that
09:09
there were
09:10
uh two papers i think and uh
09:13
in 1999 quite uh some time ago
09:17
so the in the paper the paper they have
09:21
uh
09:22
is about uh so the title of the paper is
09:26
it's called damage analysis of fiber
09:28
composites
09:30
part one statistical analysis and fiber
09:32
scale
09:33
so there they used
09:36
the lamest system to make a model
09:40
so this is a homogeneous
09:43
isotropic linear system of
09:47
elasticity which i will write down the
09:50
equation
09:51
so they use this as a model
09:54
and they observed numerically
09:57
that for linear system of
10:01
elasticity l infinity norm
10:06
of the gradient is actually
10:10
independent of epsilon so it's
10:12
independent of epsilon
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so uh
10:19
my colleague michael vogelis
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and uh bernie taye they have a
10:27
work on the first work
10:30
on such situation to to see
10:34
whether one can prove uh this
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rigorously so
10:41
actually they were both former student
10:43
of
10:44
evo babushka and
10:49
so they proved they started with two
10:51
dimension
10:52
in two dimension if we take
10:55
two unit ball and let it touch
10:59
each other so
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and you take a unit size to man uh yeah
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you take a domain okay
11:07
so then you take ellipticity k inside
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k here it's okay to take
11:15
three different positive constants so i
11:17
just take simple
11:18
simple one to take one
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here so we are having a piecewise
11:25
constant
11:26
coefficient and this is the equation
11:30
and assume that you have a h1 solution
11:35
and they prove that the gradient is
11:38
l infinity so namely so the main thing
11:42
is here
11:43
so the main thing is here so l infinity
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so means
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mainly is here so gradient u is
11:50
l infinity because as we said by the
11:54
classical transmission problem we
11:56
know that away from this starting point
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everything is fine everything is fine
12:07
so since the proof was using
12:12
complex variable because in
12:14
two-dimension
12:15
a harmonic function so this will be
12:17
harmonic
12:19
here a harmonic function it can be
12:21
linked to
12:22
a holomorphic function the real part
12:25
of homomorphic function so so
12:28
the so michael
12:32
uh uh told me this problem
12:36
yeah so showed me the results
12:39
and uh well actually at the time
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so i was quite surprised to see this
12:47
result
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and uh so wondering you know whether
12:51
there's a pde theory of it so i've never
12:53
seen such a result before
12:55
so and so then michael
12:58
and i uh worked on that
13:01
and and have a pde result
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so namely if we look at
13:08
second order divergence form elliptic
13:11
equations
13:12
in all dimensions with piecewise smooth
13:15
coefficients
13:16
like the the domain
13:19
geometry are like what we described if
13:22
before
13:23
like this so piecewise smooth
13:26
coefficients up to the boundary
13:28
piecewise most
13:30
and they are elliptic diverse form and
13:33
in that case
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indeed gradient u is bounded
13:38
is bounded independent of the distance
13:41
epsilon
13:41
and when they are touching greeting u is
13:44
in
13:45
l infinity
13:48
and uh so that because when we
13:51
through that our starting point actually
13:53
is the is
13:55
the georgie nash so it does it did uses
13:58
the
14:00
elliptic equation so it didn't uh uh
14:03
one need for for linear system
14:06
of elasticity uh like
14:09
the main system so one needs to
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uh to do some
14:16
different things so so then with uh
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louis nuremberg
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so we proved uh such a result for
14:23
divergence form
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elliptic systems with piecewise smooth
14:27
coefficients so which include
14:29
linear system of elasticity
14:34
so about linear system of elasticity
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so so it it is such equation
14:42
so so so the number of
14:45
unknowns is x1 through xn
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and the number of equations is u1 and un
14:52
so the number of equations uh
14:56
and the number of unknowns are the same
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so
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so here is uh um by
15:04
it is a so so here
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running through and by and and this is
15:09
running through and backend
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so we have n unknowns and equations so
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this corresponds to
15:15
any equations and equations any
15:18
equations and unknowns
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so so
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so the so the ellipticity
15:26
is given by a i j alpha beta you
15:31
test against symmetric
15:35
matrices so we have positivity
15:39
so that that is called
15:42
a linear system of elasticity the
15:45
elliptic one so
15:47
so for system there's one type which is
15:50
strong stronger is called strongly
15:54
uh uniformly elliptic so
15:58
which is tested for all
16:02
matrices and here you don't impose this
16:05
symmetry this symmetry is you see that
16:08
you switch
16:09
i j alpha beta you you see i j
16:12
and alpha beta you it's the same yeah
16:15
and here you can switch
16:16
i and alpha so this corresponds to
16:20
the invariance when you set up the
16:22
problem uh
16:23
model the problem uh one has translation
16:26
in variance and rotation environs
16:28
so so this invariance will give your
16:31
equation some
16:32
uh environments
16:36
the main system is a is a
16:39
it's an important uh case
16:43
and uh so it is like this this is you
16:46
you still have n you have n variable
16:48
this is laplacian u
16:50
as a vector to do divergence and take
16:53
gradient and ellipticity is
16:56
this is positive and lambda n is the
16:58
dimension and this is
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this is ellipticity so when we try to so
17:04
when we
17:04
try to prove some results we first look
17:07
at the
17:08
equation case like a conductivity case
17:11
so that's a scalar equation and when we
17:13
go to system
17:15
and we we certainly want to look at this
17:18
system first
17:20
like like in babushka's uh
17:26
they use this first okay
17:31
so after so let's go back to the uh
17:35
conductivity problem first the scalar
17:37
equation
17:38
so we come back to the geometry
17:42
so we have a domain which is a smooth
17:46
nice one
17:47
unit size and d1 is a
17:50
domain d2 is a domain unit size
17:55
and epsilon is the distance so
17:58
here the epsilon will be small
18:02
so epsilon positive will be small so you
18:05
are interested in
18:06
obtaining estimates um
18:09
either abstract independent or pin down
18:13
the dependence on epsilon well
18:18
so
18:26
so let's for simplicities for from now
18:29
on let's just look at
18:31
uh d1 d2 for simplicity let's just
18:35
think of it them as a strictly convex
18:41
so things you know can be done for more
18:45
general for some
18:46
some results but let's just take it for
18:50
simplest case so then
18:54
this is the setup we have k inside
18:57
ellipticity
18:58
k here and the one so it's a
19:01
transmission problem
19:02
and we would like we are interested in
19:04
the quantitative control
19:07
of the gradient mainly say in terms of
19:11
epsilon and k for instance because when
19:14
k
19:14
goes to infinity uh ellipticity
19:17
deteriorates
19:18
and also when k goes to zero electricity
19:21
deteriorates
19:23
so well for the equation the maximum
19:27
principle immediately gives
19:30
that's an independent l infinity norm
19:33
and the gradient as we mentioned
19:36
gradient l
19:37
infinity is bounded
19:40
by a constant depending on k
19:44
so one asks naturally
19:48
say what happens for k goes to infinity
19:51
and
19:52
in particular for k equal to infinity
19:56
or for k goes to 0 in particular for k
20:00
equal to 0 so
20:03
so for the conductivity this is the
20:06
perfect conducted conductor case is the
20:08
insulated conductor case
20:10
so for k go equal to infinity
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so for k equal to infinity
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and uh we
20:23
here actually the results here stated
20:25
does require
20:27
strictly convex otherwise the rate will
20:29
be different but one can also capture
20:31
the rate
20:32
so so let's first start from k equal to
20:35
infinity
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so for k equal to infinity so inside
20:41
we have infinity infinitives infinity
20:44
inside two balls
20:46
one outside so one infinity
20:50
infinity so of course uh
20:55
this does not
20:59
make much sense it looks like but you
21:01
can
21:02
you you you can define it in the
21:04
following way you take a k
21:07
you take a k and put a boundary
21:09
condition like that you solve it
21:10
uniquely
21:11
and you send k to infinity
21:14
and this uk will goes to you so uniquely
21:18
so it you can take it as definition or
21:23
you can define it by inside a constant
21:27
inside here some constant the constant
21:31
the two constants are determined by
21:34
two non load line local boundary
21:37
conditions the normal
21:39
integrating and here the flux you
21:42
integrate it it should be zero
21:44
so these two conditions pin down
21:47
uniquely
21:48
the two constants inside so that's the
21:50
solution so two equivalent definition
21:54
there are some there are some uh
22:01
these are more from engineering
22:03
direction one of them
22:05
is a book perhaps this one i think this
22:08
one is the book
22:08
is a book and this one is a paper
22:12
so there they they produce
22:15
special solution in the case n
22:18
equal to two they produce solution
22:24
that
22:26
the rate is like this when f1 goes to 0
22:29
the gradient u will be
22:30
like this so then
22:34
so the first work is by amari khan
22:38
and lim so they look at the
22:42
in r2 and a global situation and they
22:46
pin down that
22:47
the rate in this both
22:50
i think so they pin down the rate is
22:54
exactly that yeah
22:57
so then uh
23:00
then when eventually the
23:05
the result is in
23:08
go to all dimensions and can be
23:10
localized
23:11
and this is ellen
23:21
so so the the result is
23:25
in two dimension it is that and
23:29
this is epsilon log epsilon in three
23:31
dimension
23:32
and in four dimension and up it
23:36
is 1 0 epsilon
23:39
so the the rate is getting worse here
23:44
so and this is actually optimal in the
23:49
sense that
23:50
you are able to find boundary value fee
23:54
so that you actually reach
23:58
this rate so
24:05
so so so these rates are known
24:09
now to be optimal
24:12
well it is also oh
24:15
let's uh for linear system of elasticity
24:19
so this is the system oh
24:21
as we said yeah so as we said
24:24
if we take three elliptic constants
24:27
so finite case so like uh three
24:32
sets of numbers so inside d1 you have
24:35
one set of mu and
24:37
lambda the main constants inside d2 you
24:40
have a second
24:41
set of learning constants instead
24:44
outside the two you have the third set
24:47
so
24:48
then you know that gradient you
24:51
are bounded
24:54
depending on the ellipticity so this
24:57
this was
24:59
and
25:03
for the infinity so if you if we make
25:08
them to infinity inside
25:11
so one can also define it by taking mu
25:15
and n lambda plus 2 mu you you have a
25:18
unique solution
25:19
and then you send
25:23
this ellipticity go to infinity
25:27
it will pin down one's unique solution
25:30
so in that case
25:32
if we look at the lame system
25:35
and uh so it was proved
25:39
that jiguan bao and haigan li and
25:42
so the rates are
25:45
similar are the same as in the uh for
25:48
the conductivity
25:50
scalar case still angle 2 is this rate
25:56
so how about the
25:59
for k goes to infinity and for k goes to
26:03
zero
26:04
so
26:08
in that case so if k goes to
26:11
infinity for k goes to infinity
26:15
and in dimension 2 well the right rate
26:18
is actually here so c is independent of
26:21
k
26:21
epsilon of course uh i i think here
26:25
should be
26:26
this so uh okay
26:29
so really i think it's one okay so
26:33
yeah so that's certainly should be
26:35
square epsilon
26:36
so and uh
26:39
so this was first uh
26:43
identified uh
26:46
by amari khan lim and amari kang lee lee
26:49
and lim in a two-dimensional case
26:52
in a r2 and minus two balls
26:57
this was identified and and they they
27:00
could
27:01
so the proof is much more than that so
27:04
they can
27:05
pin down much more different radius and
27:07
so on
27:08
different k 1 k 2 and so on
27:11
so then a proof
27:14
also localized and proof with general
27:17
domain
27:18
in two dimension was made by hunji
27:21
doom and hai gandhi and
27:24
[Music]
27:27
sorry and
27:30
so this 2d this was settled
27:35
so for an equal greater equal than 3
27:37
this
27:38
is still not not
27:41
no so what are the what should be the
27:44
right
27:44
rate and when we have both
27:48
epsilon k so that's not known
27:53
for example if we keep k equal to
27:56
positive constants we see that
28:00
this rate the greeting is bounded
28:03
because this will take over yeah so it's
28:06
bounded
28:07
if we take k equal to infinity this term
28:10
is not there
28:11
so epsilon we see
28:14
gradient epsilon square root rate
28:17
so so in general when we had results
28:20
earlier for k
28:22
for fixed constant and for k equal to
28:25
infinity
28:26
it is desirable to have a uniform
28:29
rate with both epsilon k
28:32
dependence so that like this
28:35
yeah but that's uh not known
28:39
yet and it's certainly interesting
28:43
to pursue
28:46
so for k equal to 0
28:50
this is a insulated case
28:55
so for n equal to 2
28:59
so in r2 so we take a harmonic function
29:04
fix any harmonic function and at
29:06
infinity
29:08
we assume they go to they have the same
29:12
ux goes to h
29:16
then in that case it was proved by
29:19
in the same paper by
29:24
these authors and uh
29:27
the same paper mentioned earlier and we
29:29
have this rate
29:31
so actually this problem in this case is
29:34
dual
29:35
to the k equal to infinity case it's a
29:37
dual problem so therefore they have the
29:39
same
29:40
so in two dimensions it's a dual so
29:43
for k equal to 0
29:48
and for insulated case
29:52
so for n greater equal than two so
29:57
there yes
30:00
this is zero k equal to zero so you have
30:02
zero zero one
30:04
so this case so a rate was
30:08
given so by the in this
30:11
paper so the rate was given
30:17
n equal to 2 is sharp and i said here
30:20
this is sharp
30:23
so but in for a while
30:26
it is not known whether n greater or
30:29
equal than three
30:30
is sharp so
30:50
oh this is for k goes to 0.
30:54
so so for k goes to 0
30:58
so this rate
31:01
is like this so this is due actually
31:06
and uh so
31:09
here and uh this is c
31:12
over that and so and
31:16
and this is proved by but this is in the
31:19
met
31:20
mentioned earlier that those papers they
31:24
they they they treated both k goes to
31:26
infinity
31:27
and k goes to 0.
31:31
and this is a separate
31:37
so for k equal to 0
31:41
and insulated case for n greater equal
31:45
than three
31:46
is that sharp so this is the wrong date
31:48
so this should be zero
31:51
so then
31:55
so this this has not been known for a
31:57
while
31:58
and in 2016
32:02
yun examined a situation in
32:07
r3 so
32:11
in r3 we take two balls
32:15
and the epsilon apart and we take
32:19
zero zero inside therefore
32:22
what one has is a neumann condition
32:26
outside on the boundary this is on the
32:29
boundary
32:30
and laplacian u equal to 0
32:34
outside exterior domain harmonic
32:36
function
32:37
and at infinity the rate that you
32:40
approach
32:41
and why are close together why is a
32:43
harmonic function
32:44
so they are close together and in that
32:47
case
32:50
showed both sides both sides
32:54
the rate is not c over root epsilon
32:58
the rate is actually this number
33:01
so the rate is 2 minus
33:05
root 2 divided by 2. so yeah
33:12
so
33:16
so this number is less than one half so
33:19
so it's it's better it doesn't uh
33:21
it's not as bad as square root
33:25
so so recently um
33:29
young
33:34
he's my current student and uh
33:38
so we proved that
33:41
in n greater or equal than three indeed
33:44
the problem the estimate
33:48
is not optimal so this estimate can be
33:50
higher be
33:51
improved
33:55
so the improvement is by a number of
33:58
beta
33:58
depending on the dimension and the
34:01
domain
34:03
so one can improve a beta since this
34:05
result works not only for laplacian
34:08
we can have piecewise uh we can have
34:11
second order divergence form
34:14
elliptical equations here uh we smooth
34:18
coefficient after the boundary so it's
34:19
still
34:20
like this so one can improve
34:24
but we our proof as i will show
34:28
does not give a good
34:32
bomb for example it's not like this for
34:35
example in this case
34:37
we are not able to say it is this bomb
34:40
or or so our bond is not not
34:44
as good yeah so we can write down a bond
34:47
it's it's not
34:48
a the method you know will give an
34:50
explicit bound
34:52
but that bound is is certainly far from
34:54
uh
35:00
optimal
35:02
okay so so me i'll say about
35:06
a proof
35:09
so so let's just take a simpler
35:14
uh everywhere i will keep simplifying
35:17
things
35:18
so let's say uh we take a set up this is
35:22
a dividing surface so the blue one
35:25
uh enlarged picture of these two so
35:29
what what i draw is here
35:32
the blue one here and blue one here
35:36
and we have some epsilon distance so we
35:39
just
35:40
enlarge there so to draw it bigger and
35:44
so this part is epsilon so this part is
35:47
like
35:48
epsilon and we just take say
35:52
a paraboloid a paraboloid
35:57
and this is minus paraboloid so we take
36:02
we take any point here we are going to
36:05
make an estimate for greeting there
36:08
so we are going to estimate the gradient
36:09
there so
36:11
we will prove this there's a beta so
36:14
in fact we prove the estimates give this
36:17
this is the more natural quantity
36:19
so this is smaller than this epsilon
36:22
plus so
36:22
when we go out this number will become
36:26
us a better better
36:29
which is natural
36:32
so so let's take a point
36:36
we take a point here so this height
36:41
will be proportional we take this height
36:46
basically it's like this height more or
36:47
less like this height
36:49
proportional to
36:53
yeah proportional to that
36:57
yeah proportional to that so so this is
37:00
proportional to that
37:01
so we take this height so we are
37:05
in the in the volume we are going to
37:07
take a gamma so we are going to work
37:09
with a small gamma so this gamma
37:12
the choice of the gamma will enter into
37:15
uh
37:15
the this beta value also yeah
37:19
so so we are going to take a smaller
37:22
gamma
37:24
and we are going to work in order to
37:27
estimate this point we are going to work
37:30
with uh
37:36
we are going to work with uh
37:40
like a cylinder type domain like like
37:42
this
37:43
yeah like this region we are going to
37:45
work inside that
37:46
a shape like that and the r we are going
37:50
to take
37:50
r is some with the right scale so
37:54
we take the we work with this r
37:59
so then we
38:09
then we prove a hanak so
38:12
so without loss of generality we assume
38:14
u is between 0 and 1
38:16
so it's positive so we can
38:19
uh yeah so this
38:23
this is the shape so this is the shape
38:26
2r
38:26
so so the shape is 2r is is this
38:30
shape is this shape with
38:35
so so this this lens
38:38
this radius is like a r yeah 2r and then
38:42
minus
38:43
of r over 4. so it's like a annulus
38:46
ring type shape so on that shape
38:51
if you assume the top has a
38:55
normal derivative equal to 0 so when i
39:00
have hanukkah i have this harness
39:07
well uh the way i say
39:11
okay yeah yes yeah so we have this
39:14
hanukkah
39:15
so this constant depends on epsilon
39:17
independent epsilon r and u
39:19
so this c goes into that gain
39:23
of beta so so so yeah
39:26
so that's why our beta is not so uh
39:29
so so good so so
39:32
not not not optimal not optimal yeah
39:36
so so for simplicity let's simply take
39:40
this x zero or two behind the center
39:43
to be at the center so take take this so
39:45
this third time should be the square
39:47
root
39:49
of epsilon yeah so
39:54
so so then the shape
39:57
here so
40:00
we have a once we have this harness
40:05
maximum principle will tell us that
40:10
the soup will be bounded by inf
40:13
for the whole thing for the whole not
40:16
only the
40:16
cell i mean for the whole thing you have
40:19
so once you have this hanak
40:21
for the whole thing oscillation will
40:24
decay
40:25
we will get a beta prime gamma prime
40:28
beta prime
40:30
so so this number also
40:34
goes into the game
40:39
so without any work this oscillation
40:42
is bounded by a constant so with that
40:46
we are going what one obtained
40:50
a square one over square root so
40:54
by working with the domain
40:57
with actual height and square root
41:00
epsilon
41:02
radius so it's kind of flat region in
41:04
that region so
41:06
so all the transformations are not
41:08
singular so it worked well
41:10
so here
41:13
here we if we want to gain we want we
41:15
need to go out
41:17
larger so to get control
41:21
so once we have this we are getting
41:25
the control what we want
41:28
so we can make a transformation
41:32
and flatten that part so we can we can
41:35
we can we can take a region like
41:39
this is epsilon and
41:43
we can take a region
41:47
to be root epsilon should be okay so we
41:49
take root epsilon
41:50
and this the kind of fat uh like that
41:54
and uh and and then we can transform
41:58
to here so it will be become such
42:01
equations
42:02
so so that
42:05
so the bottom is a unit size
42:09
because you see this ratio is root
42:11
epsilon
42:12
so this is this height is root epsilon
42:15
so
42:16
so it's a unit disk and the height is
42:20
root epsilon
42:22
and then on this here you have we have
42:25
normal derivative equal to 0
42:27
and we have a elliptic equation of
42:29
divergence form
42:30
and this aij is good is icon holder
42:36
when you make this transformation in
42:37
this scale you can keep
42:40
aij holder so in this situation
42:44
then one has estimates
42:48
like this so we are able to control
42:51
so for example in our case we can take
42:54
this as zero actually
42:55
so i think in this case then you don't
42:58
we don't even have this part
43:00
those we can take as zero so therefore
43:04
we can control the gradient uh like this
43:08
or or we can take here if we normalize
43:12
properly uh between
43:16
yeah so yeah we will see this we are
43:19
going
43:19
so when we you we normalize we can
43:22
normalize property we can
43:23
take gradient estimates you can always
43:25
insert a constant
43:27
so we are going to see at the top
43:30
we see this l infinity this
43:33
so so we gain so we get so here we will
43:37
get
43:38
epsilon to some beta
43:41
so the whole thing so when we scale back
43:44
we are having
43:45
when we scale back so we have
43:49
another root epsilon and we get our
43:51
result so
43:52
this so then so so this estimate
43:56
is by some
43:59
results uh in our earlier uh
44:02
work with uh um mongalis and nuremberg
44:06
so yeah able to
44:09
control obtain such estimates so
44:14
so that gives the proof here
44:17
so that's just the beginning somehow
44:21
a step and
44:24
uh so for instance natural question is
44:27
whether or not
44:28
one can obtain some result like i
44:31
you know for example in such in this
44:33
case
44:35
is it possible to actually get this rate
44:38
or to look at linear system of
44:42
elasticity
44:43
can one get such rate
44:48
so for linear system of elasticity uh
44:51
uh one
44:54
one we do not know uh
44:58
uh such result so because
45:01
because we have used uh we have used
45:05
maximum principle it
45:08
it for for linear system of elasticity
45:11
it was not the same method as before uh
45:15
for the equation case earlier
45:18
we can get we can get something like
45:20
this so
45:22
we can get the gradient ul infinity
45:25
and will be controlled by something like
45:28
root epsilon and say ul infinity
45:31
so this this one can obtain and
45:34
uh one question is whether or not one
45:36
can gain this bound
45:38
uh cane this beta power and uh
45:41
we we haven't seen that yet so we are
45:44
trying to
45:46
replace this hanak part
45:49
so we replace this part so we have not
45:53
succeeded yet and
45:58
so i think i will stop here
46:02
okay thank you for a wonderful lecture
46:05
and then and congratulations and good
46:07
results
46:08
so you improved the the the
46:12
estimates one of a square root
46:13
absolutely in three dimensions
46:15
right so let's take a few questions uh
46:19
anybody
46:21
have a question
46:27
i i i do yeah all right please go ahead
46:31
beautiful talk thank you as always thank
46:34
you
46:34
so i was wondering in in that part where
46:37
you use harneck neck quality
46:39
so instead of using harlequin equality
46:42
uh
46:42
you were able to prove a helder estimate
46:46
uh a sharp helder estimate like
46:49
solutions are
46:50
heather continues with some exponent
46:52
alpha which could be sharp
46:55
that respect the junk would would you be
46:58
able to improve this beta
47:00
so uh you uh to
47:04
to improve the beta so
47:07
to have a better beta
47:11
exactly yes i mean of course depending
47:14
on
47:14
on how how sharp your
47:17
half exponent for for the other
47:19
continuity would be
47:21
inside of these uh so replacing the
47:25
harneck
47:25
by a sharper hilder estimate
47:30
on solutions and solution so so yeah
47:33
that that's that's where the beta
47:35
yeah so it will be good to have
47:38
different ways of
47:39
doing it to to improve the beta so the
47:41
beta if we write down a beta like this
47:44
so it's not so so so
47:47
so so good so our our way of proof
47:52
so even though i only describe this
47:54
three dimension
47:55
but but actually our results will go
47:57
with any
47:59
any dimension and greater or equal than
48:02
three
48:02
and then the question is why the method
48:05
doesn't work in 2d for example
48:07
okay so this meant so that's exactly in
48:10
2d
48:11
we are going to see shape they are
48:14
disconnected
48:15
so that's the only place in some sense
48:19
so the 2d will be like this two pieces
48:23
so our proof works for dimension and
48:25
greater or equal than three because the
48:27
shape
48:28
will be connected
48:31
so the hardness uh go that way otherwise
48:35
it become disconnected and that so
48:38
because in 2d
48:39
we are not supposed to improve that
48:41
that's sharp
48:42
so that's where the 2d makes the
48:46
difference
48:49
another quick question uh i understand
48:52
that
48:52
that all those problems involve
48:54
divergence form equations for physical
48:56
reasons
48:57
uh but have people look at the
49:01
number versions theory so for the
49:04
earlier one
49:05
i think jingang soon had a paper so the
49:08
first like uh for the earlier one for
49:11
the
49:12
uh for like i think uh a very good
49:15
question
49:16
yeah yes yes i i know ching dong soon
49:19
has a paper
49:20
uh for like uh uh
49:24
uh my joint work with
49:27
bargaining so one can look at uniformly
49:30
elliptic
49:31
case and there are natural uh numbers
49:34
once
49:35
nitro derivative order one should look
49:37
at
49:39
and uh and that was studied by
49:43
uh jinganjing in a paper i forgot the
49:46
ear so so it's a
49:50
non-divergence
49:53
so i think it's it's it's it's around uh
49:57
so around here so it's a non-divergence
50:00
non-divergence
50:04
equation uh
50:08
jingam so
50:11
i i forgot the ear so ear
50:15
i forgot yeah
50:18
so so that has been uh
50:22
there's a paper so uh force for
50:26
system i don't know so i i i do not know
50:29
whether there's a
50:31
uh system one that i i
50:34
don't know okay excellent thank you
50:38
thank you another and yeah
50:43
any other question or comments or
50:45
whatever
50:47
so let me ask you this uh
50:50
that your beta it depends on the include
50:53
shape of inclusions and what
50:55
can you tell us it's implicit so it's
50:57
maybe
50:58
quite uh difficult to classify i mean
51:00
clarify but uh
51:02
can you tell us a little bit more about
51:04
uh
51:05
how it depends on the what kind of
51:07
geometry it depends on
51:10
that's the one we are hoping to be able
51:13
to
51:13
to do that to do we are hoping to
51:17
to get that so you see our beta comes
51:20
from
51:20
mainly comes from this hanukkah that's
51:22
all yeah
51:23
yeah so we are going to have a region
51:26
like this
51:27
okay for appropriate geometry here
51:30
yes and in there we are going to have a
51:33
hanak
51:35
so that hanukk will be uh
51:38
so this cannot certainly depend on this
51:40
shape depending on this shape
51:43
so we have a hana but that that hannah
51:46
to get hannah
51:47
to the usual
51:51
it's usually quite quite far
51:54
from uh optimal
51:59
and also and also not only that we you
52:01
see
52:02
to go down we are going to use this hana
52:06
once you have this hanak so we are going
52:09
like
52:09
a a direct down to get a holder control
52:13
so
52:13
that's also not so good that's also far
52:17
from optimal
52:18
yeah i see
52:22
so there is some other problems to look
52:24
into i mean to investigate we
52:26
investigate this hana game quality and
52:27
things like that right
52:29
yes yeah interesting yeah very good
52:32
yes yes i couldn't hear you
52:35
[Laughter]
52:36
okay yes uh let's
52:40
wrap up and uh thank you very much for a
52:42
wonderful talk
52:44
and good to see you again thank you very
52:46
much good to see you homebabe yeah
52:48
yeah yeah yeah yeah yes yes i'll see you
52:52
next year again
52:53
okay bye bye everybody thank you thanks
52:56
a lot
52:57
thank you very much thank you very much
52:58
thank you for for for all the
53:00
participants
53:01
for your patience thanks a lot and
53:04
glad to see many friends yeah
53:08
thank you thank you everybody bye
53:11
thank you very much bye-bye