Julie Clutterbuck (Monash University)
PDE from the south
66 subscribers
For more information go to http://mat.ufcg.edu.br/pdefromthesouth/
00:00
to this
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um and so the picture you should be
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thinking of in your mind
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is if omega is convex
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um so here i'm i'm gonna
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pretty much work in r2 just because it's
00:23
easiest for us to do calculations but
00:25
most of these things work hard enough
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well some of these things work hard in
00:29
rn
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and i'll try and give pointers to where
00:32
everything goes through
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so here's uh well i'll draw a desire in
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here
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um and so our domain is some kind of
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omega down here and the graph of our
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function
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looks like this
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and we consider the level set so these
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are the
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the sets um whether our solution all has
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the same height
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so we have something like this
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so i'm going to write the level set as
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um let me back up a little bit
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let me consider a super level set
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so this is going to be given by omega c
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and this will be all the points in the
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domain
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which uh so take on a value that's
01:40
greater than c
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and the question is is that also convex
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so if um
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if you think about this little picture
01:58
here we can sort of project that back
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down to the plane
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and we get another little set in here
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and that's our super level set
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um and we could imagine if
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if we just drew omega itself
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um so here's some said omega then the
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boundaries of all those level sets
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might be sort of these concentric
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sets um so the question is
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you know when when is this true
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and the sort of answer is
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is kind of infuriating because the
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answer is
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this is almost always true um
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sorry i should actually mention when we
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go back to our problem
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uh another condition here is that f
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is positive or non-negative
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um
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yeah so so we expect our solutions to
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look something like this
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and people have looked at this a fair
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bit
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for different values of the right-hand
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side f
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and they've found that you know for
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almost all the f's that they look at
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this is always this pretty much always
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holds
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so i'm just going to mention some
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classic results
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um so for example in the case that
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f of u is equal to lambda u
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then the problem's just the um
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eigenvalue problem and
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and use the first eigen function
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and the property holds
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and this is due to breast camp and lib
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in 76 and this turns out to be a really
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important property
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um particularly in the theory of
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probability you'll find that a lot of
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the
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results in this area are sort of done in
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a very probabilistic language
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um
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there's sort of a more general result if
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uh say f of naught
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is equal to naught and
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f is non-decreasing
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as a function of one variable
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um this is due to caferelli
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and spruck
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i think in 82
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um and there's kind of a whole industry
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of
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finding these kind of results um
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and that would sort of be nice
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if if it was always true but the reason
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why it's actually intriguing
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is there's also a counter example
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um so this is quite recent and this is
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sort of what made me a little bit more
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interested in this problem in itself
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rather than as a tool for
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finding other pro finding other results
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um
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and this is due to
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hamel nadarish philly
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and sierra
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and they actually just constructed a
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very explicit domain
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and a right hand side g
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and their domain was some kind of very
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long
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stadium shape
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and what they found is on this
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kind of shape um
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the the level sets because it's such a
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long and narrow shape they're kind of
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almost
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flat but they're able to choose this g
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in such a way that
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um kind of had a little
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you know a little non-con convex part
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um so out towards the boundaries these
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are kind of
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quite regular but there was just one
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sort of
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very small part where there was a a
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little
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dent in the in the boundary of the level
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set
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so
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um so that's why this
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this problem is quite intriguing because
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it still hasn't been proven where this
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this dividing line is between um
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problems that have this property and
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problems that don't and so
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a plausible conjecture which um
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they mentioned in this paper is that
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is that the property holds
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when u is a semi-stable solution
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um and this means that
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for all test functions
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um we have
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the dirichlet energy
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minus f prime u phi squared is
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positive is greater than or equal to
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zero
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um
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and so
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the idea is that when this is kind of a
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variational solution it sort of
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minimizes
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um an energy then you should get the
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property and
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um this counter example doesn't satisfy
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that
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um so
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yeah that's that's sort of the the
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motivation and the interest there
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um and
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where where i'm coming from um
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so feel free to ask me any questions
10:08
here
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um and i should also say that i wrote
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some notes for this but they're kind of
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in a disgraceful
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condition at the moment so then i'm not
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quite ready to share them but i
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should be able to tomorrow um
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okay so this
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this is going to be what my plan is
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is i'm going to
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talk about
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just the sort of background and the
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notation um
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so maybe notions and notation
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um then
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i'm going to talk about the
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um
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[Music]
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the eigenvalue problem this is kind of
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the simplest
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case whoa
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um and
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then i'm going to talk about the general
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case
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um sort of for a general right hand side
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and then
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i'm going to take
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a step sideways and
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look at this in a geometric setting on a
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manifold
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and talk about some recent work that
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we've been doing in
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negatively curved spaces
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um
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so that's where i'm going with this i
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should mention that also that
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this has been you know fairly well
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studied
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um and maybe the nicest
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treatment of it but it's very old is the
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book by kavole
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on um rearrangements and
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level sets for elliptic equations in pde
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i think it's called that
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rearrangements and convexity of level
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sets in pde um
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it's just a short section at the end of
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the book where he he talks about these
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things
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but also this um hamel
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nader schwili sai paper it's got a
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really good introduction and it
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outlines some of the issues okay
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so let me
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move on um
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okay so i'm gonna be talking about
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two different kinds of convexity or
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concavity
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here so we all know that a
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convex set
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is one in which when we have
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two points
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um in the set the line between them
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is also contained in that set
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so
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um
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and we can also extend this
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to a convex function
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um so
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this is when the the epigraph
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um so the set that lies above the
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of the graph is a convex set
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so you know for a
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function in
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if one variable little b
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like that um but i'm going to be
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really not paying very much attention to
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regularity here and assuming that
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everything is
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um is quite smooth
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so for a c2 function
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we say
15:49
[Music]
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this is convex when the the hessian's
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positive
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semi-definite
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okay and
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we also have concave functions
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now if we um
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go back to looking at our original
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problem
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in general the solutions this and not
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concave
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um you know if
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if we're on a ball for example this is
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not true then then things really are
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concave but
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in general concavity is much too strong
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um and so we've got some weaker notions
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of concavity here
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that um will
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will will do for us um
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so let me just write that down so
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and i didn't give my main problem a
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label so let me go back and fix that up
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let's call let's call our problem
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oh star
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okay so um
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let's write
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some other notions of concavity
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which are
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ranked so firstly this is the strongest
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so f is concave
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okay so if it's c2 then the second
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derivative is a
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negative semi-definite um this is a
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little bit weaker
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uh and this is called alpha concavity
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so f to the alpha is concave
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so for some alpha in zero one
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um sort of the zero case of this
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is when
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log f is concave
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so this is called
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log concavity
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and the final notion is the one that
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i'm it's the weakest one and the one
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we're
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going to be looking at and this is that
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f
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is um
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quasi concave and this is exactly the
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property
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that the level sets
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uh convex or super level sets
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and yes so these are
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completely um ordered
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so this one
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um implies all of those
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and you can see that this
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quasi-concavity
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property is really really weak so for
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example if we have
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any radially symmetric
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domain
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and a radially symmetric solution
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then all the level sets are going to be
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balls
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or spheres
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well let's just think about two
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dimensions and say all the levels that's
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a circle
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but we could imagine that um this
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might not actually be concave um
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so we might have you of our
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might look something like you know like
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this
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um
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um
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yes so so this is this this
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quasi-concavity notion is
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really very weak
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um
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right
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um so what i'm going to do now is
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just talk a little bit about the
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geometry so this is really um
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sort of well-known things so
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feel free to you know zone out a bit
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um so
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but i should just just mention what's
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going on here
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um so
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so when we have such a solution you
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um clearly the shape of the level sets
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is completely given by you
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um but we kind of want to deal with the
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level sets as
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objects in themselves um
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so uh
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let's
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consider our domain and our function so
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this is
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you know what we what we're expecting
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our function to look like
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um some kind of
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turtle shell um
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so if we consider a level set
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up here um
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we can draw this down in the domain
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as some kind of
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level set um
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so this here is supposed to be
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the graph of u
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and this here is supposed to be the
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level set
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and what the implicit function theorem
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tells us is that we can always find
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a a little parametrization
24:03
of this level set at any point where the
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um
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the derivative is non-zero
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okay
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and what i'm going to do is i'm going to
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write the
24:50
um the level set
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the parametrization of the level set as
24:55
as some curve gamma
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so we're going to write this as gamma of
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x and so this will be from
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a little interval in r
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to r 2.
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um and
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this has a normal
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and a tangent
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um and it also has a curvature
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um
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which usually for parametrization would
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calculate this
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as
25:48
[Music]
25:49
the derivative of the
25:53
unit tangent
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but we don't need to have a
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arc length parameterization here we
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could have
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um just
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in terms of whatever our
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parameterization up here is
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and so this would be dd
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x of t
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in the direction of the normal times dx
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ds and here dxds would turn out to be um
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1 over the
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um
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the absolute value of the derivative of
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our
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curve um
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so the reason i'm doing this is because
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i want to relate
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what's going on with you to the
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curvature
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so i'm going to calculate
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some derivatives of u
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okay so since
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gamma is a level set
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so we have um u
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of gamma of x is equal to
27:25
a constant and so we have
27:29
d u um so taking the derivative of both
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sides with respect to x
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gives us d u
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gamma x is equal to zero so
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this is um d u is
27:45
in the direction of the normal
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and gamma x
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is the tangent
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so when we take the dot product of this
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it gives us zero
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um and we can also take second
28:08
derivatives here
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so we get
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d2u now i'm going to write this as the
28:16
hessian so this is going to be a
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um d2u
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of acting on gamma x gamma x
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um so in coordinates
28:30
um if you prefer to write it that way
28:35
this would be like uh
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d 2 u
28:43
dx i d x j
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um
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gamma prime or gamma x
28:55
i gamma x j
28:58
so this would be dt yeah this would this
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would be the matrix
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um entries here
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uh what am i doing here oh i'm taking
29:12
the derivative of this guy
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um but we also need to take the
29:15
derivative
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of this this tangent vector so we get
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um this
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and
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so now let me make my screen a bit
29:36
smaller so i can fit a little bit more
29:37
in
29:39
um now i'm going to sort of go back to
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my
29:43
curvature here and just
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sort of follow the calculation through a
29:49
little bit more so this is going to be
29:51
ddx
29:52
the tangent is
29:56
like this
30:05
and because i'm dotting this with the
30:07
normal
30:08
i don't need to worry about taking the
30:09
derivative of the length here
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um
30:15
and what i'm going to get is a
30:18
1 over
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uh gamma x squared gamma
30:25
x x dot um the normal here
30:29
is again d u
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uh divided by the length of du
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um and the reason why i write this is
30:38
because i wanted to sort of
30:40
recognize this part here and this part
30:42
here are
30:43
the same so here we've kind of managed
30:47
to pick up our
30:48
curvature in terms of u
30:51
so this gives us that d2u
30:59
is equal to minus um
31:03
kappa and
31:08
got some whoop
31:16
uh and
31:19
d u there
31:23
and maybe we write everything in terms
31:25
of the unit tangent
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um
31:34
okay um so this is kind of
31:41
a fairly natural identity
31:45
so if we think about our
31:48
problems this is this telling us the
31:50
curvature of the level set
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um
32:02
and the sign of the curvature of the
32:04
level set is given by
32:06
um the
32:11
sort of one component of the hessian and
32:12
the component of the hessian that's
32:14
purely in the direction of the tangent
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um and
32:20
this is kind of natural when we think
32:22
about our picture
32:24
up here um
32:28
if we consider all our level sets then
32:31
we can see that um the level sets are
32:34
not containing all the information about
32:36
you they're just containing
32:38
um yeah
32:41
they're just containing part of the
32:42
information because you could have
32:44
a different function
32:47
for you here that had exactly the same
32:49
level sets um
32:51
if it was changing direction in the term
32:53
in the direction of the normal
32:54
at a different speed um
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so so
33:01
um yes so that part of the part of the
33:05
hessian of
33:07
mu is given giving us the curvature
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um and
33:13
this also gives us sort of another way
33:16
to think about
33:17
the problem so original problem which
33:21
you've probably all forgotten it so
33:24
write it down again if we think about
33:29
our original equation then if you
33:31
satisfy this equation
33:33
then this on the
33:39
one hand is saying that the trace
33:45
of the hessian is
33:49
negative um
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so the hessian is this matrix that has
33:56
negative trace
33:57
and what we want to know is
34:01
is one particular oh
34:04
i just saw your message it said the
34:06
sharing stopped
34:08
no but that was fine that's just i think
34:11
one person had a problem
34:13
oh okay it's okay
34:16
great um so the so the traces the trace
34:19
is negative
34:20
um but we really want to know about sort
34:23
of one specific
34:24
um component whether this is
34:29
this is also negative
34:33
um so
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yeah it's kind of you know it's almost
34:38
there it's it's not impossible that the
34:40
curvature should be positive but it's
34:42
also not directly given
34:44
um so what we need to do is use that
34:48
extra information that's given by this
34:50
being a solution and
34:51
a domain being convex and
34:54
get some kind of global properties going
34:56
on there that will give us that
34:58
sort of extra um
35:02
extra bit of information that will go
35:03
from saying okay this trace is
35:05
um negative two actually this component
35:09
is negative um
35:13
so hopefully that makes sense um
35:18
and i did say that i would say what's
35:20
going on with
35:21
higher dimensions you can of course do
35:22
this in higher dimensions
35:24
and you get um you can do this for the
35:28
the mean curvature um
35:32
and the question there is is are all the
35:34
curvatures of
35:36
your level sets positive
35:39
um
35:43
right so i should maybe connect up a few
35:46
dots a bit
35:48
more um if
35:54
if gamma is the boundary
36:01
of a convex set
36:07
then kappa is
36:10
positive or non-negative
36:14
[Music]
36:16
if kappa
36:19
is positive then we say gamma
36:22
is locally convex
36:28
um but this this doesn't imply that it's
36:31
the boundary of a convex set because we
36:33
have
36:34
um you know curves that
36:39
might look like this
36:42
which are locally convex but they're not
36:44
the boundary of a convex set
36:48
um
36:52
good
36:58
and now that we've made this
37:02
calculation here we can
37:05
observe that log concavity will
37:08
always imply quasi concavity
37:12
um so
37:23
so if u is
37:29
log concave
37:34
then u
37:37
is quasi concave
37:46
um and we can just do that
37:49
so suppose
38:01
so this means that log of you
38:12
but we can
38:17
we can calculate what's going on with
38:19
this um
38:21
f equals log u because we have
38:26
df is equal to d u
38:29
divided by u
38:33
and here we recall that
38:37
use positive on the interior
38:43
so this is okay um and in fact we can
38:49
for solutions to these these elliptic
38:51
equations we can
38:52
show that this is well defined up to the
38:54
boundary um
38:57
well it's it's not it's well defined up
39:00
to the boundary on the boundary it
39:01
becomes
39:02
um unbounded um
39:06
anyway we can calculate the second
39:08
derivatives
39:10
of f and this is given by the second
39:12
derivatives of u
39:16
minus 1 over
39:20
u squared
39:26
so vector product of d u cross d u
39:30
um and
39:37
now when if we know
39:41
u is log concave then
39:44
we know that um
39:46
[Music]
39:50
d2f is uh
39:54
negative semi-definite
39:59
and now
40:04
let's go back and remember what our
40:07
our curvature was
40:10
so we have
40:15
d2u
40:18
in tangent directions um
40:22
is going to be given by u
40:26
d2 f
40:30
in tangent directions plus uh
40:36
um
40:43
plus 1 over
40:46
u d u
40:50
d u also in tangent directions
40:53
um but now we
40:56
realize that d u here is a normal
40:59
direction to the level set
41:01
so all of this
41:04
part here will
41:08
cancel out
41:12
and
41:17
this is equal to minus kappa
41:21
d u
41:25
so what we've found is that
41:28
if this d2f
41:32
is
41:36
a negative semi-definite
41:40
then this means that our curvature must
41:43
be
41:46
non-negative
41:48
um
41:52
so that's that's sort of a
41:56
short little calculation that lets you
41:58
see that this
42:00
um log concavity is certainly sufficient
42:04
to show
42:07
the quasi convexity the convexity of the
42:09
level sets
42:16
okay so any questions so far
42:34
okay so let's
42:39
move on
42:42
um
42:45
okay so
42:49
let's start talking about some specific
42:51
equations
42:53
um and one that's really nice for me
42:59
is the eigenvalue equation um
43:02
so our equation now becomes
43:08
like this we're going to take laplacian
43:11
of u
43:12
but now i'm also going to consider a
43:14
potential
43:17
um
43:21
so here v will be a convex potential
43:27
so it's just a convex function that's
43:30
defined
43:33
on our domain
43:39
um and
43:46
uh here
43:49
we still got our
43:53
zero
43:56
boundary value and
44:00
our positive function on the interior
44:03
and once
44:06
we've set up our problem like this the
44:09
general theory of
44:10
eigenvalues tells us that there's
44:13
exactly one lambda that satisfies this
44:18
and that lambda is the eigenvalue
44:25
okay so for general so if you think of
44:27
some random lambda here it
44:29
doesn't necessarily have solution but
44:32
there's
44:33
exactly one lambda where it does have a
44:35
solution
44:38
and i guess u
44:42
is the eigen function
44:46
um and because this is the the positive
44:48
one
44:49
and there's many eigenfunctions some of
44:50
which most of which will change
44:52
sign but because this is the one that
44:54
doesn't change sign this is also called
44:56
the ground state
45:03
um for the physicists
45:06
um
45:09
and so this this problem comes up
45:14
a lot um
45:18
and it's a physical meaning
45:21
um and it's also
45:24
used a lot in probability
45:27
um
45:31
and
45:34
the result here is is due to breast
45:37
cancer lab
45:50
um so their result is that you
45:54
well if omega is convex
45:59
then u is quasi convex
46:07
grassy yeah quasi context i always get
46:10
confused when i write quasi convex so
46:11
i'm not sure whether it's quasi convex
46:13
or quasi concave
46:17
because it's a concavity property
46:20
but it's talking about the convexity of
46:22
the level sets
46:31
and this is quite an old um
46:34
result and
46:37
there's a lot of really very nice proofs
46:40
of this
46:44
and i'm going to
46:48
sort of present one that's due to um
46:52
to yao it's in a paper by singer wong
46:57
uh yao and yao
47:01
um so
47:08
yeah that's that's what i'll do maybe
47:09
maybe i'll give you some examples
47:11
first
47:16
um
47:21
oh this is this is a really easy one the
47:24
nice thing about this problem
47:26
is that it's one you can give to your
47:28
engineering students and tell them to
47:30
solve it by separation of variables for
47:32
some
47:32
really simple domains um
47:37
so if we take a box
47:42
as our domain
47:45
and a
47:48
zero potential then we can explicitly
47:51
write down the solutions
47:54
this is given by u equals
47:57
sine of x
48:01
pi on a sine of
48:06
y pi and b
48:10
and you can check that
48:15
um u is
48:18
log concave
48:23
um but not
48:28
concave
48:32
and just now that i've written that i
48:34
realize that kind of somewhat undersold
48:36
breast cancer libs result maybe i write
48:39
the stronger thing here u is
48:41
log concave
48:46
and hence quasi concave
49:00
and
49:02
another similar
49:06
example here is um the case when
49:10
omega is a ball
49:14
and you can even
49:18
um
49:22
give it a potential here so let's write
49:24
v of x
49:26
is x squared um
49:31
and because this potential is purely
49:33
radial
49:35
we can solve in polar coordinates
49:43
by separation of variables
49:54
and when we do this we get
49:57
um an ode
50:02
for u
50:08
just as a function of of r
50:12
um and we can
50:16
use ode techniques then to show that
50:19
it's um
50:24
that it's concave
50:38
so we actually get cavity for for this
50:40
one
50:42
um
50:47
and yeah use
50:51
solves an equation that looks a little
50:54
bit like the um
50:55
the bessel load here
50:59
um
51:20
so
51:23
okay um
51:26
so this
51:31
let's write this
51:35
let's call this one the l
51:43
and this is the proof of bl
51:50
um
51:55
and this is using a method of
51:59
singer wong
52:03
yeah so this as i mentioned there's
52:05
there's many different proofs of this
52:08
um
52:12
this is from a paper in the early 80s um
52:17
there's also some
52:18
some nice proofs of this due to korva
52:22
um and others
52:27
so what we're going to do here
52:31
is um
52:34
sorry i'm just finding my notes
52:56
there we go
53:01
okay so let's
53:05
um
53:11
let's just consider f
53:15
which is going to be
53:18
log of u
53:21
and use the eigen function
53:24
as a positive function we can define the
53:28
log of it um but we should expect it to
53:31
look
53:31
like if we took a slice through our
53:33
domain
53:36
um like this
53:39
and we have u
53:43
looking like this then
53:46
um
53:51
f would would look something like
53:56
um
54:04
well too closely to the edge there
54:13
so
54:23
so f is going to be unbounded towards
54:25
the um
54:26
the boundaries um and it's going to have
54:29
unbounded
54:30
um derivative as well
54:36
so
54:38
we've already looked at this but i'll
54:41
write it down again
54:43
so the derivative df is given by d u
54:46
divided by u and the second derivatives
54:50
of f are given by the second derivatives
54:54
of u um
54:56
[Music]
54:58
minus u squared
55:01
d u cross d u
55:06
um and
55:11
in fact we can write this
55:15
as um one on u
55:19
d2u minus df
55:24
cross df
55:29
and so since u satisfies
55:36
the eigenvalue equation
55:47
f will satisfy
55:51
something that looks pretty similar um
55:56
laplacian of f is equal to
56:00
lambda minus v plus
56:08
df squared
56:16
okay
56:21
and we also have the boundary conditions
56:24
um
56:28
f of x going to minus infinity
56:31
as we approach the boundary
56:35
and um
56:40
df also
56:45
becomes unbounded
56:58
okay so what we're going to do is we're
57:01
going to
57:01
try and show that um firstly that
57:06
uh the hessian d2f
57:09
is negative towards the boundary and
57:11
then we'll show that it can't have an
57:12
interior positive maximum
57:16
so a kind of a maximum principle type
57:18
idea
57:21
so this is what we're claiming
57:25
is that
57:38
well
57:42
so we're claiming that
57:45
um f is is concave near the boundary
57:49
um and we'll just do this by
57:52
taking a orthogonal basis
58:03
and
58:07
what we are going to observe is that on
58:09
the boundary
58:12
um u is constant
58:17
is equal to zero and so
58:22
um
58:33
and so the tangential derivatives are
58:36
always zero
58:44
and d u
58:47
is in the direction of
58:51
the normal okay so that's kind of a long
58:54
way of saying that the
58:55
boundary is the level set at height zero
59:08
okay so what we're finding here is that
59:15
the tangential part
59:20
is given by um
59:28
uh let's go back to up here
59:32
is one on u d2u
59:35
minus well tangential part
59:47
okay this df is
59:51
in the direction of the normal and so
59:54
these guys are going to be zero
59:57
and this one here is a curvature term
60:02
okay so this is
60:16
and here kappa is the curvature of the
60:19
level set
60:27
which of course is just the curvature
60:31
of the boundary
60:35
okay so this is positive
60:45
okay we should also check out what the
60:47
um the hessian is doing in the
60:49
normal directions
60:54
um
61:09
and this is given by
61:22
the um
61:27
so the same thing but in
61:31
normal directions here um
61:34
so this one
61:38
is just the
61:45
length of the derivative of f
61:50
this one here now we can think of we can
61:54
use our equation
61:55
for um u here and
61:58
write this one the
62:01
the normal part of the hessian as
62:05
a
62:11
as a laplacian
62:16
minus the tangential part
62:25
[Music]
62:32
um
62:35
then we can use our equation
62:44
so this is minus lambda u
62:48
plus v u
62:51
uh minus
63:04
so here we've got our eigenvalue
63:08
um
63:14
and well we can get rid of our u that's
63:17
potential
63:28
um
63:32
and here we've got
63:36
a kappa term
63:46
and the point of all this is
63:50
that um oh sorry i should have
63:59
collected these guys
64:02
um
64:09
that's right this is
64:12
minus df
64:30
okay the point is that
64:34
this one's these are all positive
64:39
these are maybe not positive but these
64:42
are bounded
64:45
and this df is becoming huge near the
64:52
boundary
64:56
so near the boundary this is all going
64:59
to be
65:00
[Music]
65:02
less than or equal to zero um
65:09
so so we don't
65:13
we don't need to worry about what's
65:15
going on with f near the boundary
65:17
um so f is certainly concave
65:34
up
65:40
um
65:44
okay so that's that step um
65:49
so our next step is going to be trying
65:52
to show that f is concave on the
65:54
interior
65:54
we don't exactly do that um
65:59
so let me
66:02
tell you exactly what i am going to do
66:06
um
66:10
okay and the reason why i'm doing this
66:12
is because i think it's a really nice
66:14
technique
66:14
um which sort of uses maximum principle
66:18
ideas
66:19
and it
66:23
tells you well the idea is that you
66:25
assume that you're at
66:26
some maximum of a function and then you
66:30
can use
66:32
that assumed maximality to really say
66:34
quite a lot about
66:35
about your function so
66:39
what i'm going to do is consider
66:46
um the hessian d2f
66:52
has a function on the
66:55
tangent space so we're going to consider
66:58
it as a
67:02
function from omega cross
67:06
s1 cross s1 to
67:09
r so it takes
67:14
a point in omega so that's that's the
67:19
point at which we evaluate it
67:22
and to
67:26
um tangent directions
67:31
and it maps them to a real number
67:39
and we've just shown that um
67:44
that this is non-positive near the
67:48
boundary
67:49
um so if
67:52
this is
67:55
non-positive everywhere
68:03
then we're done
68:10
um let's suppose
68:14
that
68:20
um we have
68:28
um
68:33
a a point
68:38
x naught in our domain and
68:45
and
68:50
some direction
68:55
um which we'll call e1
69:03
um where
69:10
this has a positive maximum
69:26
um and in fact
69:30
what i'm going to do is i'm going to set
69:33
um g of x
69:38
is equal to the hessian
69:45
evaluated at x
69:48
but just in this direction e1
69:52
so
69:56
so what i'm doing here is i'm starting
69:58
off with the hessian and i'm saying okay
69:59
let's suppose it's got a maximum
70:03
in some particular direction and then
70:06
i stopped thinking about it as a um
70:09
as a function to find on the tangent
70:12
space and i go back to thinking about it
70:14
on the domain just to find on the domain
70:17
omega but restricting it to that
70:19
direction in the tangent space
70:23
so what we have is we have
70:28
g of x naught which is equal to
70:32
this maximum d2
70:37
f at x naught e one e one and
70:41
this is going to be greater than
70:44
or equal to
70:47
d2f at x
70:53
and because we're assuming that this is
70:55
the maximum
70:57
also over all um
71:02
other tangent directions it's also going
71:04
to be greater than
71:07
this so for all
71:11
x and
71:15
here
71:24
um
71:26
okay and we use
71:31
coordinates orthonormal coordinates
71:37
uh defined by this choice
71:48
and
71:52
because we've we've chosen this point to
71:56
be a maximum
71:57
um e1 is going to be
72:01
a a eigenvector
72:06
of the hessian at x naught
72:23
um and so what we have is that
72:26
d2 f um
72:30
at x naught of
72:33
e1 e2
72:38
is equal to zero okay so
72:41
so e1's kind of diagonalized the um the
72:44
hessian
72:48
um
72:54
and okay
72:58
so at this at this interior point of
73:00
this
73:01
interior maximum x naught of g
73:10
we have the gradient of g
73:16
is equal to zero and
73:19
we have the laplacian of g
73:24
is less than or equal to zero
73:31
and so we can calculate these and we
73:33
find that
73:34
0 is
73:41
greater than or equal to and now we just
73:44
take the derivative through
73:54
and so we get derivatives
74:00
of laplacian of f
74:07
and we know what laplacian of f is
74:09
because we've worked out the
74:10
elliptic equation satisfied by f and so
74:13
we get
74:26
um
74:29
and a bunch of things drop
74:32
out but what so we don't get any
74:35
derivatives of the
74:36
eigenvalue for example but we do get the
74:38
derivatives of the
74:40
um potential
74:43
v and we get derivatives of
74:47
df and so we get this is minus
74:51
two um
74:54
we get
74:58
here
75:05
we get d
75:08
e1 f d e k
75:12
f um
75:18
squared and we get
75:22
a third derivative d
75:26
e k d
75:29
e k or d one
75:33
d one of f
75:37
um
75:42
now this if we stare at it long enough
75:47
we can
75:47
realize that this is a
75:52
gradient of g term
75:59
and so it's equal to zero
76:05
and this this term is his um
76:10
well it has a sign
76:15
and but we're actually not going to
76:17
conclude anything here yet we're just
76:19
going to conclude
76:21
that
76:26
this term here
76:30
oh no sorry let me go back let me go
76:33
back
76:41
what's going on with this term here is
76:43
that we have
76:45
[Music]
76:46
this
76:49
um
76:54
this kind of diagonalization of the
76:56
matrix here
76:59
so in fact the only term that comes out
77:01
of this
77:02
is a d e1
77:06
f
77:11
sorry i've made a mistake here this
77:14
should just be a second derivative
77:32
this is due to um
77:42
this
77:49
okay so this has told us that at this
77:52
maximum point we've got this
77:53
um this inequality
77:58
for g
78:01
so this is g here so we've got
78:10
d2f at x naught
78:20
this is g squared
78:25
um this is always going to be greater
78:27
than
78:28
um ve1
78:37
so the second derivative of the
78:39
potential
78:41
and we can bound this by writing this as
78:43
the minimum
78:46
of the
79:06
okay
79:09
um
79:12
so so that doesn't look too bad um
79:16
and because we sort of assume that g was
79:19
going to be a positive maximum
79:22
and
79:25
you know here v is it has some minimum
79:28
so we know this has to be positive
79:30
it doesn't look so interesting and this
79:32
is kind of the crux of the proof here
79:34
the final step is to use a continuity
79:38
method
79:49
to find a counter example
79:56
or contradiction
80:04
um and you know this is
80:10
already gone on for some time um so i'll
80:13
just
80:13
briefly describe how we do this we
80:16
interpolate between
80:24
the problem on a known domain
80:35
and our given
80:40
problem that we're interested in
80:43
and so our known domain here will be
80:49
[Music]
80:51
the ball
80:55
with the potential given by x squared
81:00
so
81:04
and at t equals one so t is a parameter
81:08
here
81:09
is just add general domain
81:12
omega
81:16
and whatever potential we're interested
81:18
in
81:24
just call that v and so
81:30
we write down the problem
81:35
laplacian of u
81:39
t plus v t u t
81:42
is equal to lambda t
81:46
u t on
81:49
omega t so this will be omega one up
81:54
here
81:57
omega naught is equal to br
82:04
and so we can interpolate
82:07
the potentials in a pretty
82:12
explicit manner we can just say vt is
82:14
equal to
82:15
one minus t v naught
82:19
this is v naught it's x squared plus
82:22
t b1
82:28
and this is a convex
82:31
potential and
82:34
it's actually strictly convex
82:38
for
82:43
t less than one
82:48
and we can similarly um
82:53
interpolate the domains
83:04
such that it preserves this convexity
83:19
and there's actually a very explicit way
83:21
you can do this
83:23
by using the support function
83:30
okay and now there's just something
83:32
that's kind of a fact
83:37
[Music]
83:40
is solutions
83:43
to this problem
83:46
with the parameter uh really strongly
83:49
continuous
83:51
um
83:54
uh so
83:57
that's cool
84:02
let's call this one
84:05
blue star
84:29
and when i say they're continuous i mean
84:31
i think they're continuous in
84:33
c21 and probably maybe even something
84:37
better
84:38
um so they're continuous under
84:42
perturbations of the potential and also
84:44
continuous
84:45
under perturbations of the domain
84:51
and so
84:55
what we do here is we take our
84:58
known problem over here at t equals zero
85:02
we say okay um we
85:06
know that uh our property holds
85:09
t equals zero because this was the
85:12
example we had before
85:14
and by ode methods we can show that
85:18
our solution at t equals 0
85:22
is um
85:26
is not just log concave it's concave
85:29
um and
85:38
um
85:40
and then we then we go along and say
85:43
okay why don't we find the first point
85:46
as t goes along here um
85:50
at which this property fails to hold
85:53
okay
85:54
so let me
85:57
let me write this down um so
86:02
so at t equals naught we have d2f
86:07
um
86:10
in fact d2f is
86:14
less than zero um
86:22
so this is f t
86:25
f t equals log of u t
86:30
t equals zero
86:35
um so what we do is we suppose that
86:40
t naught is
86:44
the smallest
86:53
t naught uh
86:59
well actually
87:08
actually the first t naught
87:12
at which this fails to hold
87:22
um so actually i've got something
87:25
stronger here
87:38
um okay and then then this gives us the
87:42
contradiction because
87:43
we have
87:48
there exists some
87:51
sort of x naught
87:55
e1 such that
87:59
d2 f
88:03
t naught
88:06
at x naught in some direction e1
88:10
is equal to zero
88:14
um but the problem is
88:18
is we have our previous estimate here
88:26
um
88:28
which is
88:32
is actually that at such a if we assume
88:34
this is a maximum point
88:36
which it has to be if this is the first
88:38
one
88:39
then we have this inequality here
88:42
okay and so because our
88:45
interpolated potential here is actually
88:47
strictly
88:48
um convex
88:52
this is going to be strictly positive
88:56
let's call this
88:59
yellow star
89:07
okay so this contradicts
89:20
okay because
89:38
okay um so
89:44
let me fix up a few loose points here
89:50
suppose okay let's suppose that t naught
89:53
is
89:53
less than one here um
89:57
we might have just
90:01
uh not strict convexity at t equals 1
90:05
because we're not assuming
90:06
that v the v we're interested in is
90:08
strictly convex
90:12
and if we look at the way we defined v
90:14
we can see that
90:16
as long as as long as our the v we're
90:19
interested
90:20
in is convex and our original v naught
90:23
which was like x squared is is strictly
90:25
convex and the
90:26
strict convexity will go across for all
90:28
t
90:30
less than one okay
90:34
so then we're done
90:45
okay um
90:49
yeah so so that's that's the proof judo
90:53
to essentially to yao and his group
90:57
um and there's some more there's some
91:00
more proofs and more approaches there
91:01
and i'm hoping to say a little bit about
91:03
those in
91:04
the subsequent um days
91:09
um yeah i'm going to finish there but
91:13
you know if you have any questions then
91:16
please
91:18
please please ask them
91:27
okay i'm not sure if you can hear me
91:31
yeah i can hear you okay i have a
91:34
problem