advanced-pdes

[astral vine]

which is one

[brisk granite]

Are you sure 0^0 is one?

[astral vine]

Bro

[brisk granite]

Pardon me because I've spent an entire day researching after 0^0 and it's considered to be undefined

[astral vine]

no

this is well defined since it is taken as a convention

which is consistent with other notations, and makes since easier to write stuff as sum, integral etc.

this is like 0! to be 1

that's the same idea

[river path]

Well you have to be careful in analysis where you could be taking any kind of limit.

But yeah for combinatorial purposes it's 1.

[astral vine]

it consistent with few properties of real functions

here

[brisk granite]

Oh so we can consider that here

[astral vine]

there is no limit ryc

here this is just x{k-k}

[river path]

Yeah I agree with it here for sure

Yeah

[brisk granite]

Okay then that's okay otherwise it's troubling

[river path]

But this is a delicate channel for such claims.

[brisk granite]

I tried thinking the same but then others disagreed

Anyways so I do consider it 1 and then i get my solution

[astral vine]

I see what you mean

[river path]

Sorry I didnt even see the context beforehand. Just making sure.

[astral vine]

but 0^0=1, but a limit of the form 0^0 is almost never 1

which is the difficult thing to catch

[brisk granite]

So we can actually apply that to get the solution? And it would be correct?

[river path]

Ok yeah I definitely agree with this convention

[astral vine]

yes

[brisk granite]

Okay I'm taking your word for it.

[astral vine]

Does anyone knows a good reference for existence of Schwartz Kernel for general Heat Semigroups on almost arbitrary domains ?

[remote matrix]

Hello ! I am looking for help to solve the following two equations:
$\frac{\partial u}{\partial t} = D\frac{\partial ^2u}{\partial x^2} - du$ on one hand (diffusion + degradation)
and
$\frac{\partial u}{\partial t} = K\frac{\partial u}{\partial x} - du$ on the other hand (transport + degradation)?
on $[0, L]$ with mixed boundary conditions $u(t, 0) = \alpha$ and $\partial_t u(t, L) = -\varepsilon$
I have looked up for methods to solve this kind of equations but I have only found a way to solve diffusion only with Fourier transforms, and even that I am not totally comfortable with. I will be vvery grateful to anyone who takes the time to answer !

[untold delta]

Jeanne

[astral vine]

#odes-and-pdes

[remote matrix]

What is it for here then ?

[astral vine]

For higher PDE involving sharp regularity, existence, uniqueness of solutions in various function spaces

Harmonic Functions from Geometric and Complex Analysis, and Several Variables Complex Analysis are also allowed

undergrad/Physics/Enginerring PDEs are in #odes-and-pdes

(solving explicit PDEs with Fourier series, Separation of Variable etc...)

[onyx heart]

anatole keeping the advanced PDE channel straight

[remote matrix]

Okay thanks, sorry 🙂

[astral vine]

No problem :)

[verbal nebula]

Anyone have much experience with Eskin's Linear PDE book?

If so, is it any good? Worth the buy? Or should I just keep using a PDF?

[quaint herald]

It's okay. There's a lot in it but somehow I felt like it was too terse to be an introduction to many of the topics in it, but doesn't dive deep enough to be the ideal reference for mastering technical machinery like FIOs. @verbal nebula

Probably depends quite a lot on what you were hoping to learn from it, and how much it costs.

[verbal nebula]

It's the recommended text from Krupchyk

Here's the syllabus of the course

[quaint herald]

I think you'll find the other suggested text grigis-sjostrand a lot more enlightening for (II) and there are a million sources for (I).

It's decent for (III) though.

Idk, if it's as expensive as I am guessing it is, I wouldn't buy it for a course unless I was planning on making regular use of it in the future.

[verbal nebula]

I just ordered a copy for like $30

I took a look at the topics and it seems to be good enough for the purposes

One of my profs was a student of Eskin, and so some of the material is exactly how he taught us things like distributional calculus

Which may have influenced my decision

[quaint herald]

Fair enough, yeah I don't think it's too consequential a decision for the topics in that syllabus.

[verbal nebula]

It's strange how many approaches PDE has to me I suppose

Every person I've learned from had an exceedingly different point of view on them

[junior bloom]

PDEs as a field is large enough to be its own department and everyone involved wants to kill themselves because nothing ever gets solved.

I took PDEs as a course 3 times now and every time it covered new stuff.

3 levels, 2 unis.

[verbal nebula]

Am I choosing the wrong field?

It's just so wild to me - I recently got an article published in a good PDE journal but I feel like I know nothing

[junior bloom]

Have you met those people who work in Hodge theory? They're pushing PDEs but the tools are totally alien to me. Alg top.

Or Hairer and his stochastic renormalization, with geometric regularity structures.

Even within PDEs there are people who delve into hardcore harmonic analysis and others who do weird qualitative stuff to suss out solutions.

[quaint herald]

I don't think the fact that the field is so vast/broad is a bad thing. It means it is rich, with lots of niches to find and many creative ways to combine tools.

[verbal nebula]

Yeah - it seems like I'll be learning from the following triade: a qualitative PDE person, a probabilistic/discrete harmonic person, and a person that does microlocal stuff provided I get into my target school for PDEs

[junior bloom]

I don't think the fact that the field is so vast/broad is a bad thing. It means it is rich, with lots of niches to find and many creative ways to combine tools.

This is what Stockholm's syndrome looks like

But yeah, PDEs is incredibly rich and its applications beautiful.

[astral vine]

You know who I am thinking about

[verbal nebula]

I guess I was used to there being like a "canonical" introduction to a subject and I figured Evans PDE book would be that canonical introduction

Then there's a complete 180 from that

[quaint herald]

Nah the subject is just way too big for that. Evans is still a good first book to read though, or Taylor.

[forest cradle]

I regret trying to read Gilbarg Trudinger

Convinced me to switch one my quals to probability instead of pde 😅

[astral vine]

I didn't read the whole book but few parts I read of Gilbarg Trudinger are great

[forest cradle]

Yeah I think it would work much better as a reference than going through it page by page

[astral vine]

They contain very important proof and results, and proofs are clear

Grisvard : Elliptic PDEs in non-smooth domains is bit wilder, but is a nice "functional analytic" complement (but is older iirc)

[forest cradle]

From what I remember I just kept reading like the same theorem with very slight boundary differences over and over again, and I didn't really have much motivation to learn pde other than quals so I found it really tedious.

Maybe with a course it would have been better

[astral vine]

The idea behind this kind of presentation is to give the "why" when you change the assumptions

If the results still holds with weaker regularity, this will be probably due to a much deeper pov

etc...

This gives the insights about what will work "straightforward", what won't

Otherwise you won't be able to tell the difference between, e.g., the elliptic regularity of the Laplacian in the C² boundary case, the C1,alpha case, the Lipschitz case, etc...

[forest cradle]

Hmm maybe I will have to give it another chance if I need to learn more PDE some day

[heavy turtle]

if a functional has zero differential(variation) at every point of the function space, is it a constant?

this feels so obvious but I can't prove it

[heavy turtle]

Hm I found the answer

[strong nimbus]

Any hints on how to solve this? $y-f(x)+ \frac{\lambda}{\sqrt{1+(y')^2}}=C$, where $f(x)$ is given

[untold delta]

ImHackingXD

[river path]

Are you sure y' and y arent switched? This looks ridiculous

Uhh

[strong nimbus]

Euler-Lagrange equation from $I(y)=\int_{x_{0}}^{x_{1}}(y-f) d x$, with the restriction $\int_{x_{0}}^{x_{1}} \sqrt{1+\left(y^{\prime}\right)^{2}} d x=L$

[untold delta]

ImHackingXD

[river path]

hmm

i need to remember how to do these restricted calculus of variations problems

i'll remind myself

this equation is very weird to me

it's not what i would expect to show up

[strong nimbus]

We also know that $y\left(x_{0}\right)=f\left(x_{0}\right)=y_{0}$ and $y\left(x_{1}\right)=f\left(x_{1}\right)=y_{1}$

[untold delta]

ImHackingXD

[strong nimbus]

But I don't think it's important for what I'm asking

[river path]

so it's basically lagrange multipliers, right? when we take a directional derivative of the optimizing functional, it needs to be a scalar multiple of the directional derivative of the constraint.
consider $v$, and let's take $\frac{d}{dt} I(y + tv)\Big|{t = 0} = \lambda \frac{d}{dt} J(y + tv)\Big|{t = 0}$ where $J$ is your constraint.
notice that the integral of $f$ is a constant and does not change under variations. we find that the left hand side is $\int_{x_0}^{x_1} v, dx$ and that the right hand side is nastier, $\int_{x_0}^{x_1} \frac{y'v'}{\sqrt{1 + (y')^{2}}} , dx$.

[untold delta]

ryc

[river path]

(the right hand side should have a lambda)

now we need to integrate the right hand side by parts (yuck) and here your initial conditions become relevant. you can assume that v(x_0) = v(x_1) = 0. that actually cancels out the boundary term.

so differentiating y'/(sqrt(1 + (y')^2)) you'll find that the derivative of this should be equal to a constant (after we vary over all v's). ok, we can re-integrate and get that y'/(sqrt(1 + (y')^2)) is equal to Cx + D, how do you solve this? idk, but this should be the right equation I think.

i should check again after i finish breakfast

[strong nimbus]

No one ever really explained me that, so it's a lot of information

Could you check if my Euler-Lagrange derivation is correct? Because yeah that differential equation seems insane

[river path]

uhhh

yeah sorry

idk how you learned to do constrained optimization with this

in any case, the main thing i want to mention here is that i'm not sure why f(x) would show up in your E-L equations when it doesn't affect which y is the minimal y in your I(y) minimization

[strong nimbus]

No problem, appreciate your help

[untold delta]

ryc

[strong nimbus]

Wait, I think I figured it out, I used the formula $h-y^{\prime} \frac{\partial h}{\partial y^{\prime}}=C$, where $h=y-f+\lambda \sqrt{1 + (y')^2}$, because I thought $h$ didn't depend explicitly of $x$

[untold delta]

ImHackingXD

[strong nimbus]

If I am allowed to do this, I still don't get it

[winter dock]

what I do is write $F=y-f+\lambda(\sqrt{1+y'^2}-L)$ then using E-L eqn $\pdv{F}{y}=\dv{x}\pdv{F}{y'}$ I end up with $$1=\lambda \dv{x}\frac{y'}{2\sqrt{1+y'^2}}$$ and then I integrate wrt x immediately to get $$x+C=\lambda \frac{y'}{2\sqrt{1+y'^2}}$$ now you can solve explicitly for $y'$ and then integrate, then you gotta go back to your constraint, kind of looking gross now

[untold delta]

Merosity

[strong nimbus]

hmmm, so I'm not allowed to use the formula that I used actually

[winter dock]

yeah I think you can use the beltrami identity that might clean it up from what I did

cause it doesn't depend on x, only y and y'

so you're allowed yeah

[strong nimbus]

But the f doesn't disappear if I use that formula

[winter dock]

I think my first step of immediately integrating wrt x is basically the same thing

oh true

f(x) definitely depends on x

so yeah not allowed lol

[strong nimbus]

but it doesn't depend explicitly right? you could say y depends on x as well but x isn't written there

[winter dock]

F(x,y,y') depends on x

at least how I've defined it

yeah it depends on it explicitly

because if you were to write out f, it'd be something like x^3-sin(x^7)+... or whatever

if f(x)=x^3 and we explicitly wrote x^3 there instead, you'd say it depends on x

[strong nimbus]

but y will also depend on x, so we could write in terms of x explicitly as well no?

[winter dock]

we're not taking the total derivative here

[strong nimbus]

or is it because it's unknown for now

[winter dock]

you can think of it as if we're taking just the partial derivatives with respect to the slots of F(a,b,c)

in deriving the euler lagrange equation you do take the full derivative and the chain rule takes care of that but it gets whisked away when you separate it out

that's why it seems a bit awkward

$$\dv{x}F(x, y(x), y'(x)) =\pdv{F}{x}+\pdv{F}{y} \dv{y}{x}+\pdv{F}{y'}\dv{y'}{x}$$

[untold delta]

Merosity

[winter dock]

this is what occurs within the integral when deriving the E-L eqn

initially, but then you do a little integration by parts trick that flips the terms around a little, etc etc

the partial derivatives with respect to y and y' are basically placeholders for the second and third slots in F(a,b,c)

it might take a bit of thinking to see what I mean, I remember being really confused by this too when I first learned it haha

[prisma shadow]

Hey guys, I've been watching everything, cause I'm at the same class as @strong nimbus , and belive us, we truly apreciate your help !
Btw, i think you missed a 2 that should had cut, but aside from that I agree with you

[winter dock]

oh yeah good catch

from the derivative of the sqrt in the denominator there

[winter dock]

you're welcome, glad to hear it

[untold delta]

emme

[sweet epoch]

sorry I just saw #odes-and-pdes and reposted there, it seems that my question fits better there than here.

[tired hollow]

fourier? or green's functions

[sweet epoch]

The boundary conditions were wrong but with the right conditions I solved it

[tired hollow]

how did you get rid of infinite number of solutions?

due to All Neumann Conditions

[tired hollow]

bdry are right, compatibility conditions met

you only will have infinite number of solutions that differ from each other by constant

[sweet epoch]

No, that are BCS on u_x and u_y, if you integrate the equation you get an absurd

[tired hollow]

could you show the calcs?

[sweet epoch]

Sorry, got busy. So from $\laplacian u = 0$ we get $\int_R \laplacian u = 0$ now $\int_R \laplacian u = \int_R u_{xx} + \int_R u_{yy}$

[untold delta]

emme

[sweet epoch]

$= \int_0^2 \int_0^1 u_{xx}dx dy+ \int_0^1 \int_0^2 u_{yy} dy dx = \int_0^2 u_x(1, y) - u_x(0, y) dy + \int_0^1 u_y(x, 2) - u_y(x, 0) dx$ applying BCS we have that $u_x(1, y) = u_x(0, y) = 0$ while $u_y(x, 2) - u_y(x, 0) = 2\sin(2\pix) - sin(\pix)$ thus

[untold delta]

emme
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sweet epoch]

$= 0 + \int_0^1 2\sin(2\pi x) - \sin(\pi x) dx = -\frac{2}{\pi}$(or $\frac{2}{\pi}$

[untold delta]

emme

[sweet epoch]

so from $\int_R \laplacian u = 0$ we arrive to $-\frac{2}{\pi} = 0$

so those are incorrect BCS

[untold delta]

emme

[sweet epoch]

the correct BCs are

$u(0, y) = u(1, y) =0$ and $u_y(x, 0) = 2\sin(2\pi x)$, $u_y(x, 2) = \sin(\pi x)$

[untold delta]

emme

[sweet epoch]

with this conditions you can find a solution with a series expansion method

[tired hollow]

starting here makes no sense to me :\ sry

its ok :)

[sweet epoch]

why?

I was checking if the conditions where coherent with the equations since there where 4 conditions on derivatives

[tired hollow]

That can mean the solution exists but not unique
Difference in constant
I can code it but idk how to solve analytically
Perhaps via Greens function in bounded domains

Similar to differential equation with no IC

[sweet epoch]

If I integrate a null function over a domain, I get zero no matter what

Since Laplacian is zero that double integral must be zer9

But with that BCs you get 2/pi, that will lead 2/pi = 0
😬

With that integration I am not trying to find a primitive or u(X, y), also because the integration is definite over the rectangle R!

[solid flint]

Hello! If $f,g \in L^1(\mathbb{R})$ after the Young's Theorem we have $$|f*g|{L^1(\mathbb{R})} \leq |f|{L^1(\mathbb{R})} |g|_{L^1(\mathbb{R})}$$ in this case can we say that $f(y)g(x-y)$ belongs to $L^1( \mathbb{R \times R})$??

[untold delta]

Mikahopff

[river path]

Yes, you can (you can do that last thing automatically by fubini, actually, if you change variables with (y, z) = (y, x - y) after pushing the x integral inside, and in fact this is part of how the inequality is proved).

[solid flint]

Thank you, but that's confusing because somewhere I found that if f,g are in L^1 this doesn't imply that y-----> f(y)g(x-y) is in L^1 for a fixed x 1?

[quaint herald]

As usual with Tonelli's this is no problem though. It will be in L^1 for a.e. x, and the resulting a.e. defined function of x will be integrable.

[solid flint]

Thank you
Isn't f(y)g(x-y) defined for every x? it is just a matter of translating the function g I think
So if f and g are in L^1 (R) we can say that y----> f(y)g(x-y) is defined for every x and it is in L^1?

[quaint herald]

"the resulting a.e. defined function of x" refers to the integral of |f(y)g(x-y)| in y.

[solid flint]

Thank you, I got you now

so it is ok to say that if f and g are in L^1 (R) we can say that y----> f(y)g(x-y) is defined for a.e x and it is in L^1?

moreover, the resulting a.e. defined function of x is in L^1 too?

[quaint herald]

Well (assuming f and g are given to you as actual functions, rather than L^1 equivalence classes), y -> f(y)g(x-y) will actually be defined for any x, but the main point is that you can only be sure that a.e. value of x makes this function integrable.

and so taking this y-integral (when you can), gives you an a.e. defined function of x

if you extend this function of x to a totally defined function of x (by zero or whatever you like really because you are modifying it on a null set), this resulting function is L^1.

[solid flint]

Thank you so much for this explanation

[quaint herald]

no worries

[solid flint]

Do you know or can help me to find an example of functions where this does't work for some specific x's?

[quaint herald]

Well forgetting the specific context of convolution here for now, this is just a feature of how Tonelli works, so you should first think about integrals over squares.

Eg/ consider integrating f:[0,1]^2 -> R given by f(x,y)=0 for x > 0, f(0,y)=1/y for y>0, f(0,0)=0. Integrals in x define a function of y in [0,1], and this function is identically 0 (since f is supported at the left endpoint of any horizontal slice).
Similarly, for any positive x, integrating in y gives zero because the function is identically zero there, but on the vertical slice x=0, f is the non-integrable function 1/y, and so your "inner integral" when integrating in y then x would only be defined on (0,1].

(Or really, it would be "inf" working with the extended non-negative reals.)

But when things are no longer non-negative, then you can't make sense of an integral over such a slice at all.

[solid flint]

Thank you so much for this, it is much clearer in my mind now

[solid flint]

to make sure that I have understood this. Do you mean that this idea doesn't work if things aren't non-negative? Or do you mean something else?

[quaint herald]

No I just mean that when the integrand is non-negative there are two possibilities for the inner integral (integral of f(x,y) in y say).

Either
a) The integral is a finite non-negative real.
b) The integral is the extended non negative real number "inf".

In either case, one can say that the inner integral IS "defined" for any x, it might just happen to be infinite.

When f is instead real/complex, then there is no such dichotomy and at x s.t. f(x,y) is not integrable in y, there is simply no real/complex number we can associate to the y-integral of f(x,y).

But since in Fubini/Tonelli this happens on a x-set of measure zero, it doesn't actually matter and we can extend this function arbitrarily and the outer integral will be the same for any such choice.

Another way of thinking about it is that the inner integral defines an element of L^1 in the equivalence class sense, rather than a measurable function on your full measure space having finite integral.

This distinction is often blurred a little in mathematical writing because it can be tedious to state.

[solid flint]

I can't thank you enough.
Can you recommend any readings about these please?

[quaint herald]

Probably just like the product measures / fubini-tonelli sections of books like big rudin or royden or folland should be good enough, can't really remember their relative virtues on this particular topic but I know big rudin is pretty thorough.

[solid flint]

Thank you so much, I will check them

[solid flint]

Hello,
If $f=g$ in $\mathcal{D}(\Omega)'$ and $g \in L^2(\Omega)$ can we say that $f\in L^2(\Omega)$? any references for these kinds of regularity please?

[untold delta]

Mikahopff

[river path]

By "$f = g$ in $\mathcal{D}(\Omega)'$" do you mean $\langle f, \phi \rangle = \int g \phi d\mu$ for every test function $\phi$?

I'm assuming that's what you mean.

[untold delta]

ryc

[river path]

This has to do with density of the test functions in L^2.

Indeed, to check that f is in L^2, it suffices to show that f extends to a continuous linear functional on L^2 (since the dual of L^2 is itself). But this is true just because g does.

of course, f is really only an element of L^2 after you apply the riesz representation theorem to pick an element of L^2 which represents the linear functional given by f. they're different objects, the point is that f is represented by an element of L^2 (and that element is g)

[astral vine]

In general we don't make distinction when dealing withequality in D' , the distribution can be represented by integral of smooth compactly supported function against a nice representation in the desired function space, hence we identify every involved objects.

To be honest, @solid flint this is hard to find references where this kind of stuff is explicitly explained, every books in higher Maths dealing about Distribution Theory and PDE's generally omits it. The reason why you can make such identification, in the L² case, or even Lp, is as Ryc said, but you can generalize it to (fractional) Sobolev spaces and much more complicated function spaces.

That's also why we defined usual Sobolev spaces to be

[untold delta]

Anatole

[astral vine]

Which can be understood as "Distribution on Omega that can be identified (i.e. represented) as an Lp function, and its distributional derivatives too (up to the appropriate order)"

[solid flint]

Thank you @river path @astral vine

[twilit rover]

If two L^1 loc functions are equal as distributions, then they are equal almost everywhere

Proof is given here: http://www.math.chalmers.se/~hasse/distributioner_eng.pdf

theorem 1.7

@solid flint

[solid flint]

Thank you very much!

[solid flint]

Hello everyone, if we have a bounded subset $\Omega \subset \mathbb{R}^n$ with a sufficiently smooth boundary, what's the difference between $L^2(\partial \Omega)$ and $H^{1/2}(\partial \Omega)$?

[untold delta]

Mikahopff

[astral vine]

Check Zhonghai Ding's paper

For those who are interested, this is a legit link https://www.readcube.com/articles/10.1090/s0002-9939-96-03132-2

[junior bloom]

What's a good toy example for a simple PDE with a weak distributional solution that provably has no exact solution.

[astral vine]

There is many definition of weak solutions

[astral vine]

You have to be more precise

[junior bloom]

Solutions that exist in the Sobolev sense

[astral vine]

https://en.wikipedia.org/wiki/P-Laplacian

This is a bit overkill

[junior bloom]

Well thanks, it does look a bit overkill ahaha

but good to know

[crisp heart]

what is an "evolution problem" in terms of ODE or PDE?

[river path]

A problem with time dependence and initial data.

Or final data I guess, that doesnt really matter.

Time dependence is a perspective switch: we decide that one variable is time and the others are space instead of them all being space variables. But most common time dependent equations have some kind of symmetry between all the space variables which is broken just for the time variable. So this is why the distinction is useful.

[crisp heart]

@river path thx)

[astral vine]

Usual PDE evolution problems with initial contidition at time 0, can be rewritten as an ODE with value in nice Function space, thinks about finite dimentional ODEs, $A$ being a matrix, $$\frac{\dd }{\dd t}u = Au$$.

Then this is vector valued ODE in $\mathbb{R}^n$, then you can change $A$ to e.g. $A=\Delta=\partial_x^2$ the 1-D Laplacian, then it becomes a vector valued ODE in time, with value in, say, $\mathrm{L}^p(\mathbb{R})$, and this the good way to think the Heat equation.

[untold delta]

Anatole

[crisp heart]

Thank you!

[late moth]

I can't really find a way to conclude that if a mollified $u_\epsilon(x)$ is harmonic, then so is $u(x)$ where the only assumption i have is that $u$ is continuous (I'm supposed to show that it's $C^2$)? Like,
[u_\epsilon(x) \to u(x)\quad \text{in compact subsets of $U$}]
so I'm a bit confused how to continue, like can I say something like $\nabla u_\epsilon \to \nabla u$ on these compact subsets of $U$? I can't say this for $\Delta u_\epsilon$ though because $u$ is only assumed to be continuous?\newline
Furthermore, I have that $u_\epsilon(x)\to u(x)$ pointwise for at least one $x\in U$. Is this sufficient to conclude that $u$ is $C^2$? I'm a bit stumped

[untold delta]

kirby

[astral vine]

Can you give more context, the precise statment you want to prove, what are assumptions on the function u etc.

[late moth]

Oh yeah, so u is weakly harmonic (where the L2 inner product of u and the laplacian of any smooth compactly supported function is 0) and continuous

the goal is to show that u is C2 and harmonic

I’ve gotten as far to show that the function u_eps (denoted as f^\eps in evans) is harmonic

[astral vine]

Notice that since u_eps is mollified in particular u is C² in the interior of U, hence you can apply the the mean value formula then dominated convergence theroem

(up to choose a susbsequence for the LHS)

In particular u coincides almost everywhere with a locally continuous function in U

in particular for any precompact openset V included in U

you get that u is C² on V

I am not very comfortable with interior strong regularity stuff, since this not really a relevant area in modern PDEs, so tell me if you think I'm wrong somewhere

(strong max. principle an mean value stuff doesnot admit nice genralization for complex valued elliptic operators, hence this kind of techniques and of interior regularity is very specific to Laplacian like elliptic operators (real coefficient etc.), hence this not relevant to study general Ellitptic PDEs)

[astral vine]

I assumed that at least u is in L²_loc

[late moth]

Yes, u is in L^2_loc, and I’ll look at this again in a little bit, and if I have any concerns, I’ll ping you

Thanks :)

[astral vine]

I just checked my Evans book, the idea is the same as in Theorem 6 p.28 at the beginning

[verbal nebula]

Shouldn't you have one of them frenchie books

[astral vine]

What do you mean ?

[verbal nebula]

I just wasn't expecting you to cite evans lol

[astral vine]

I don't like it but I bought it during my 4th year at uni thinking it would be useful, finally it is not as useful as intended.

But all my research books, those I use in my daily life, are English ones, but they are not as mainstream as Evans' book

[late moth]

Oh great, I realized this is how I did it after asking the question, thanks!

[empty terrace]

$\Delta = \sum_{i=1}^{D}\partial_{ii}, \quad \Delta^2 = \sum_{i=1}^{D} \sum_{j=1}^D\partial_{ii}\partial_{jj}$

[untold delta]

criver

[empty terrace]

I have the above simple expressions for the harmonic and biharmonic operator

Is there such a simple expression in the spatial domain (not involving the Fourier domain) for

$(-\Delta)^{\frac{3}{2}}$

[untold delta]

criver

[empty terrace]

In 1D this is fairly simple as it is simply $-\frac{d^3}{dx^3}$

[untold delta]

criver

[empty terrace]

Is there a simple generalization to higher dimensions (in the sense of powers of the Laplacian)

Does $\sum_i\partial_i\sum_j \partial_{jj}$ make sense for this?

[untold delta]

criver

[river path]

sadly, no, i don't believe there is a nice expression for it.

i'm not sure how to prove this

but

your hint should be to look at the symbol for this operator and to notice it's not a polynomial

differential operators correspond to polynomials on the fourier domain, but |k| and |k|^3 (which are the symbols for (-Laplacian)^(1/2) and (-Laplacian)^(3/2)) are not polynomials.

[empty terrace]

(4pi^2|w|)^p?

[river path]

so you're not going to find a nice representation for these I believe, unless you're ok with singular integral fuckery.

[empty terrace]

is there an expansion in terms of a series of partial derivatives that approximates it at least?

[river path]

hmmm, curious question

[empty terrace]

I know that for 0.5 the inverse decays as 1/(1 + |x|^2)^(3/2) in the spatial domain for instance, so typically one can truncate this in numerical schemes

[river path]

i think i've seen something like this with asymptotic expansions of pseudodifferential operators, but i'm not the person to answer this because i don't know what i'm talking about with that specifically

i see

[empty terrace]

This already helps, thanks. At least I won't waste time trying to prove that the schemes I came up with match it.

[tranquil steppe]

For the record, this is not true (for the reason already mentioned by ryc).

[empty terrace]

I reread this and noticed that I didn't exactly understand what he meant. Does it mean that |k| and |k|^3 are not polynomials but |k|^2 and |k|^4 are?

[tranquil steppe]

Yes.

In particular, |k| is also not a polynomial, hence even in one dimension (-Δ)^(1/2) cannot coincide with the usual derivative.

[empty terrace]

What do you mean by |k| is not a polynomial, you mean in terms of the real and imaginary components?

[tranquil steppe]

I don't understand the confusion. Some functions (like k) are polynomials, and some aren't.

[empty terrace]

you mean because of the modulus I guess

[tranquil steppe]

Yes, of course.

[empty terrace]

ok, this clarifies things, I had misunderstood it as poly p(|k|)

But yes, now it makes sense,

[astral vine]

@river path @empty terrace You can write it as a singular convolution operator applied to the biLaplacian

[river path]

Yes, definitely

This is probably the right way to do it, but I guess criver wanted to see some kind of pure differential operator, or some kind of series-like limit of them.

[astral vine]

$$(-\Delta)^{-\frac{1}{2}}f = \frac{2^\frac{d-1}{2}\Gamma\left(\frac{d-1}{2}\right)}{(2\pi)^\frac{d}{2}2^\frac{1}{2}\Gamma\left(\frac{1}{2}\right)}\int_{\mathbb{R}^d} \frac{1}{\left\lvert{x-y}\right\rvert^{d-1}}f(y)\mathrm{~d}y$$

[untold delta]

Anatole

[astral vine]

You can also use the semigroup representation of Fractional powers of sectorial operators

[river path]

I see

[astral vine]

$$(-\Delta)^{-\frac{1}{2}}f = \frac{1}{\Gamma(\frac{1}{2})}\int_{0}^{\infty} e^{t\Delta}f \frac{\dd t}{\sqrt{t}}$$

[untold delta]

Anatole

[astral vine]

(to see it use the Fourier transform and use expression of the Euler Gamma Function)

[river path]

wow, that's cool

[astral vine]

There is also all the fuckin' collection of Balakrishnan type formulas, involving Complement Formula for the Euler Gamma Function, but this one is the easiest to figure out and use

Balakrishnan formulas are useful but aweful to use

[astral vine]

and this is false

$$\mathcal{F} (-\partial_x^2)^\frac{3}{2}f = |\xi|^3 \mathcal{F}f(\xi) = i\mathrm{sgn}(\xi) \cdot[ - i\xi^3] \mathcal{F}f(\xi) = -\mathcal{F}[H\partial_x^3 f]{\xi}$$

Where H is the Hilbert Transform, hence

$$(-\partial_x^2)^\frac{3}{2}f =- H \partial_x^3 f$$

[untold delta]

Anatole

Anatole

[astral vine]

You can do similar treatment in higher dimension using Riesz Transforms instead of Hilbert Transform

(notice that the Hilbert transform is an automorphism on Lp(R) for all 1<p<+oo )

[empty terrace]

Any idea whether the n/2 powers in 1D correspond to polynomials of even degree similar to how \Delta^m u = 0 can be related to interpolation with piecewise polynomials of degree 2m-1?

[astral vine]

Can you reformulate your sentence please, I'm not sure about what I understood from it

[astral vine]

In 1D, you have

[empty terrace]

In 1d

[astral vine]

$$(-\partial_x^2)^\frac{m}{2}f = H^m \partial_x^m f$$

[untold delta]

Anatole

[empty terrace]

One can write

[astral vine]

H^m is m-th iteration of the Hilber Transform

H² = - I

[empty terrace]

$E(u) = \int_{\mathbb{R}}|D^m u|^2, , u(x_i) = f(x_i)$

[untold delta]

criver

[astral vine]

I forget everything about pointwise interpolation of functions

So I can't answer, sorry

[empty terrace]

it's in Duchon's paper

He argues that the solution u

Is the interpolant of degree 2m-1

piecewise polynomial interpolant

e.g. the bilaplacian would give you a natural cubic spline

Let me find a paper on this

[astral vine]

My Mathematical interests are far from this kind of stuff

[empty terrace]

https://www.geometrictools.com/Documentation/ThinPlateSplines.pdf

That's fine, ignore the link. Thank you for the detailed answers.

[empty terrace]

I found a paper including a lot of what you had mentioned: https://arxiv.org/abs/1801.09767

[astral vine]

This paper is famous, but my knowledge is more a consequence of what I learned during my Harmonic Analysis lectures

(and some papers)

[empty terrace]

The thing that really threw me off was that the kernel g that produces the standard finite difference discretisation is the standard forward finite difference. That is [1 -1] * [1 -1] = [1 -2 1]. So my guess was that it is supposed to be an analogue of sqrt(\Delta) for the discrete approximation of Delta (works also for higher dimensions through the tensor product). Although I am not so sure anymore after reading all of the above.

I am guessing this has something to do with the fact that \Delta = \nabla \cdot \nabla, although the above is a purely discrete story, and I do not know its order of approximation.

[late moth]

I’m not sure if this is adv PDEs, but I’m tasked to show that for $u$ harmonic,not constant, and $u>0$ in $U$ and there is $x_0$ in the boundary s.t. $u(x_0)=0$, and there exists an interior tangent ball $B$ at $x_0$ centered at a point $x_1$ with radius $r$. [\frac{\partial u}{\partial v}<0] where $v$ is the unit outward normal vector to $B$ at $x_0$

I typed this on my phone so I can clarify once I get on my laptop

[untold delta]

kirby

[late moth]

Okay, the hint I'm given is to show that $u(x) > c(|x-x_1|^{2-n} - r^{2-n})$ on $B(x_1,r)\setminus B(x_1,r/2)$ where $n$ is the dimension [I should've mentioned that $U \subset \R^n$]. I haven't really been able to make much headway, but it makes sense to me that from there, I'll be able to reduce the directional derivative to something negative using the fact that it'll be something like $\lim_{h\to 0} -u(x_0-hv)/h$

[untold delta]

kirby

[late moth]

Okay I got it in a different way, I just used Harnack’s inequality on the unit ball, then followed similarly to hopf’s lemma

[river path]

wait

why is this different from hopf's lemma

[late moth]

It's only different because I need to show the directional derivative < 0

Okay it's not "different", but u is strictly positive in B(x_1,r) (well, in U), so u(x) > u(x_0) for any x in B(x_1,r)

The proof would only be different because you don't need to worry about like -u(x) >= -c u(x_1) because you already have that u(x) >= cu(x_1)

[river path]

ah ok

mhm

[late moth]

Okay another question [from villani because I want to understand his fluids motivation section, and my PDEs knowledge is lackluster at best]. He says that in the Lagrangian formulation of Euler's equation you have that, where $m(t,x_0)$ is the flow of a particle in $\Omega$,
[\frac{d^2}{dt^2}m(t,x_0) = \left\frac{\partial v}{\partial t} + (v\cdot \nabla v)\right.]
Then, in order for incompressibility to be true, of course we need $\nabla \cdot v = 0$, but he recasts this into saying that $\text{det}\left(\frac{\partial m}{\partial x_0}\right) = 1$. I can work through this, he gives sufficient details.\newline
He searches for $m(t,x_0)$ as a family of diffeomorphisms from $\Omega$ to $\Omega$ (from the fact that $m$ is made injective to have uniqueness and surjective to avoid 'vacuums' as he puts it, which contradict incompressibility). Where I get confused is where he is able to say that this group with unit determinant produces trajectory maps from $m(t,x_0)$ that are measure preserving. This is in section 3.2 to motivate a physical interpretation of Brenier's polar factorization theorem.

I'll be able to provide more details if needed, I just hastily wrote this before my meeting in 5 minutes lmao

[untold delta]

kirby

[astral vine]

What Villani reference ?

[late moth]

Topics in Optimal Transportation

[astral vine]

There is too much material I do not know, so I cannot help you, I hope someone can around here

[late moth]

It's okay, I think I'm just asking about how diffeomorphisms with unit determinant produce maps that are measure preserving. It's good to mention that $\Omega \subset \R^n$, so they'd send lebesgue to lebesgue. Which would actually probably mean that this isn't as much of a PDEs question, but it's perhaps relevant

[untold delta]

kirby

[astral vine]

That's because of the change of variable formula

if the the determinant is 1, it is the same as asking that the determinant of the Jacobian and its inverse is 1

[late moth]

Ah yes, I see now

Yes, I just checked Folland, and I see what you're saying. I just saw diffeomorphism and got a little

Thanks :)

[astral vine]

np

[junior bloom]

Is it more standard for double inner product $:$ to denote $A^{ij}B_{ji}$ or $A^{ij}B_{ij}$ in PDE lit

[untold delta]

teafortwo

[river path]

Ive only ever seen the latter

But idk

[astral vine]

to me it is the standard matrix one : $$A,:,B = tr(A^{T}B)$$

[untold delta]

Anatole

[solid flint]

Hello, Any good references for the spectral mapping theorem for Semigroups?

[astral vine]

Pruss and Pazy's papers

[astral vine]

Check Pruss Gearhart Theorem and related stuff

"ON THE SPECTRUM OF C0-SEMIGROUPS" , J.Pruss 1984

[solid flint]

Thank you @astral vine actually I have it but I completely forgot to check on it

by the way, which paper do you mean by Pazy?

[astral vine]

Forget about it, I thought that Pazy did something on spectral properties of semigroups, but I can't find it : I probably said shit

[solid flint]

it is ok, in case I find something I will let you know

[solid flint]

Hello, Are there any examples of operators that have only point spectrum ( eigenvalues) but it is continuous and not discrete??

[astral vine]

If The Operator is a normal one on Hilbert space this is impossible

[astral vine]

due to decomposition property of the spectral measure

This would imply there is uncountable many dirac masses in the decomposition of the spectral measure

which makes infinite integral on some continuous compactly supported function such that the support intersect the support of theses Dirac masses

See a Guide to Spectral Theory Chapter 8

[solid flint]

Thank you so much I will take a look

[junior bloom]

In Tao's notes on the Navier-Stokes https://terrytao.wordpress.com/2018/09/16/254a-notes-1-local-well-posedness-of-the-navier-stokes-equations/

he uses notation for the space $\dot{H}^{d/2-1}(\mathbb{R}^d)$. What is this.

[untold delta]

teafortwo

[river path]

the dot means you just require $|\xi|^{d/2 - 1}\hat{f}(\xi)$ to be in $L^2$, not $\langle \xi \rangle^{d/2 - 1} \hat{f}(\xi)$.

[untold delta]

ryc

[river path]

assuming you know what sobolev spaces are.

[astral vine]

The homogeneous Sobolev space $\dot{H}^{s,p}(\mathbb{R}^d)$ is defined to be the tempered distribution modulo polynomilas such that $$\lVert (-\Delta)^{\frac{s}{2}}f\rVert_{\mathrm{L}^p(\mathbb{R}^d)}<+\infty$$

[river path]

exactly

[untold delta]

Anatole

[river path]

it's the space defined by the highest derivative seminorm

[astral vine]

Since you consider d/2-1=s<d/2and p=2, you can assume that this is not tempered distribution modulo polynomials but only tempered distributions

(due to the Hardy-Littlewood-Sobolev Inequality, or the Sobolev Embedding Equivalently)

generally you can assume that iff s<d/p for any 1<p<+infty

[junior bloom]

What are you denoting by $\langle\xi\rangle$?

[untold delta]

teafortwo

[junior bloom]

I recognize the Sobolev space part, but I'm not sure what the dot is doing.

[untold delta]

Anatole

[junior bloom]

Hmm. Alright, got any reading to motivate this for me?

[astral vine]

You dont have control the 0 order terms

[junior bloom]

Ah.

[astral vine]

when you deal with unbounded domain

there is no Poincare or Sobolev-Wirtinger inequality to ensure control of lower terms from higher ones

hence you want to build function spaces that carry over the exact regularity you want

E.g. he homogeneous Sobolev space $\dot{H}^{1,p}(\mathbb{R}^d)$ is defined to be the tempered distribution modulo constant such that $$\lVert\nabla f\rVert_{\mathrm{L}^p(\mathbb{R}^d)}<+\infty$$

f may not be in any Lq space

[untold delta]

Anatole

[river path]

yeah sobolev spaces normally need ALL the derivatives up to order k to be in L^p

but homogeneous ones just need the kth derivatives

(so for example, you can add constants / polynomials up to order k-1)

[junior bloom]

I'm staring at the definition that has homogeneous Sobolev spaces as completions of Schwartz space.

[astral vine]

This is not true

[junior bloom]

Under the funny norm

https://math.stackexchange.com/questions/3483263/how-to-think-of-homogeneous-sobolev-spaces

[astral vine]

Because this space will contain stuff that are not even distributions

I have a counter example

Check Proposition 1.34 and related comments in Fourier Analysis and Nonlinear Partial Differential Equations

[junior bloom]

Cheers @river path @astral vine

It's really funny how I can be undertaking a lengthy graduate education in PDEs and still miss these basic definitions.

[astral vine]

They are not basic

Homogeneous functions spaces are quite recent

something like less than 15 years

they exists before of course but their real interest comes in quite recently

So this stuff is not as standard as he could be

This is a PDF available online which is free

the last part will answer all your question

[junior bloom]

Thank you, this looks good.

[astral vine]

(Chapter 6)

I am currently building a theory of those space on unbounded low regularity domain and gosh

I want to die

[junior bloom]

It's great that I can apparently make a career of just naively applying hard theorems to easy problems, because that's all of my publications so far.

[astral vine]

That's what I thought for me

unless the tools werent created at all

[junior bloom]

I ran into that issue before while doing renormalization in physics, had to deep dive for weird results. So much low-hanging fruit in mathematical physics where PDEs is concerned, y'all wouldn't believe.

I have no confidence that I can apply anything beyond the first 2 chapters of Taylor's PDEs volume 1 off the top of my head

Here's a problem that wasn't solved until a handful of years ago. What happens when a turbulent vortex hits a wall.

Suppose the wall just likes to reflect normal inwards.

3 decades ago? We were still answering what happens when Navier-Stokes reflects off walls.

[astral vine]

Free boundary problems are in the same fashion

solving it require really hard tools

Check Danchin, Hierber, Mucha, Tolksdorf

This is a 120 pages lenghty paper

[junior bloom]

120 pages? Oof. I guess I won't be checking it out

[astral vine]

the 40th first page are only about functional analysis and functional calculus

Then they make computations

and everything is still ugly

[junior bloom]

The way these boundary problems are often solved in physics are via stochastic renormalization, stuff with similar flavor (though usually easier) than what Hairer got the Fields Medal for.

I was staring at a really simple example recently of vorticity deformation (a symplectic 2-form under Lie transport) hitting a reflecting boundary. It seemed only possible to do by converting the problem to an SPDE and renormalizing it back to the Navier Stokes.

Which doesn't really make a ton of sense to me because there's no reason why injecting probability makes something more tractable, it seems to me there's purely deterministic tools that can handle this with the same flavor.

And there is, I think it's called geometric measure theory.

[astral vine]

I can't read geometric measure theory

Nobody really can

[junior bloom]

Neither can I

[astral vine]

if they say so

they are liars

[junior bloom]

Not even geometric measure theorists can read geometric measure theory

[astral vine]

exactly

[junior bloom]

My naive feeling is the process of solving deterministic problems through stochastic analogues is founded on the probabilities "mollifying" singular behavior the way Schwartz distributions do.

[astral vine]

So it acts like a "damping" term ?

[junior bloom]

The GMT correspondence works because its core consideration is "mollifying" geometric structures the same way.

Yeah, I get that feeling when I work with SPDEs.

[astral vine]

I'm not into SPDEs

but it seems we are not that far from each other

[junior bloom]

I mean I regard myself as an analyst, but realistically when you work in mathematical physics you won't be as good a mathematician or physicist as a pure person in either field, and you generally won't be working on pure problems, unless your name is Witten or Deligne or Atiyah or etc etc.

My career will be built on picking up low-hanging fruit by arbitraging what people in other fields don't know. 😌

You have no idea how many engineering professors have a career just by taking a well-known result in pure math and writing it out for the simple case of e.g. a nice manifold in R^3

If engineers could read Nash he'd have like 10x more citations

[astral vine]

lol

In Maths fields you are directly spotted, and some reviewer will say "there is nothing new, look at this russian paper written 20 years ago"

[junior bloom]

If you want to lament at the injustice of the world, in machine learning conferences there are top papers that just implement other people's work as algorithms with no further contributions.

Math has the highest bar, and I've published in enough fields to know this.

[astral vine]

Wow

[winter dock]

that's probably the most useful thing someone can do in math imo

[empty terrace]

Are the equations:

$\frac{du}{dx}(x) = 0, , u(x_i) = f(x_i), , i=1,\ldots,n$

[untold delta]

criver

[empty terrace]

$-\frac{du}{dx}(x) = 0, , u(x_i) = f(x_i), , i=1,\ldots,n$

[untold delta]

criver

[empty terrace]

$|\frac{du}{dx}(x)| = 0, , u(x_i) = f(x_i), , i=1,\ldots,n$

[untold delta]

criver

[empty terrace]

supposed to have different steady states. My first guess was no since I can multiply the second by -1 and so on, but I did some numerical simulations and the 3 different steady states I get seem to make perfect sense intuitively.

Specifically all yield piecewise constant interpolation of the Dirichlet data (x_i, f(x_i))

the difference is in the regions

For example the evolution $u_t=u_x$ leads to a steady state $u(x) = f(x_i), , x\in[x_i, x_{i+1})$, while $u_t=-u_x$ leads to $u(x) = f(x_i), , x\in (x_{i-1}, x_i]$, and $u_t = |u_x|$ leads to $u(x) = f(x_i), , x\in \left (\frac{x_i+x_{i-1}}{2}, \frac{x_i+x_{i+1}}{2} \right)$.

[untold delta]

criver

[empty terrace]

This makes perfect sense to me but I do not understand how this can be reconciled with the fact that I can literally turn equation 1 into 2 by multiplying by -1 both sides. Is it something to do with the trace operator at the boundary?

That would be my guess since the steady states are discontinuous there, I am basically solving the same equation, but with different derivative operators at the boundary.

Seems to match taking a derivative from the right, from the left, and then some kind of 2-sided thing.

[tranquil steppe]

The main "reconciliation" should be that the whole problem is ill-defined - there are no functions satisfying these equations and constraints.

Your three approaches correspond to different perturbations of the starting problem. It doesn't have to be surprising that they yield three different limits. One could imagine other perturbations with other limits.

If you want to get to the heart of this, I'd recommend starting with answering the question "what we mean by a steady state". With some approaches you can encounter exotic things like "F(u) > 0 in viscosity sense" is not equivalent to "-F(u)< 0 in viscosity sense".

[astral vine]

This remind me few PTSD

Conservation laws through parabolic regularization etc.

Entropy solutions

[empty terrace]

I think I can "fix" this. If for $u_t+u_x = 0$ I set $u^{-}_x(x_i) = 0$, for $u_t-u_x = 0$ I set $u^{+}_x(x_i) = 0$, and for the eikonal equation $u^{+}_x(x_i) = 0 = u^{-}_x(x_i)$.

[untold delta]

criver

[empty terrace]

If the points are x_0, ..., x_n I would thus have well defined problems in (x_i,x_{i+1}) for each of the equations

so what I got corresponds to different boundary conditions I guess

so u_x = 0 is the same as -u_x = 0, but the numerical discretisation should depend on the boundary conditions, otherwise I guess it gets stuck in a local attractor or something

correction - the eikonal equation should have neither of the above neumann constraints

[late moth]

I'm a little bit confused on subquadratic growth of harmonic functions. More precisely, I'm given that $u$ is harmonic and
[ \lim_{R\to\infty}\frac{1}{R^2} \sup_{B(0,R)}|u| =0]
and I'm tasked to show that $u$ is linear on $\R^n$ (i.e. $u(x)=ax+b$ for real $a,b$). I initially thought to use the mean value property to generate a bound, but that hasn't gotten me anywhere, so I'm not really sure where to start with this. It makes sense to me that u would be about linear (or constant), but I can't really motivate it for myself

[untold delta]

kirby

[white hazel]

hmm if you pick a and b carefully, then u - (ax + b) is harmonic with a second order 0 at the origin

from there i think the mean value property should do something good

and also try apply the mean value property centered at arbitrary points

[late moth]

Oh yeah, that feels like a good way to do it, I'll try that for a second and I'll lyk how it goes

[quaint herald]

@late moth Do you know the Poisson kernel yet? This property comes out pretty immediately from that, and it is essentially identical to how you prove the analogous statement for holomorphic functions in C using the Cauchy integral formula.

[late moth]

Okay, so does this follow after allowing $w = u - (ax+b)$ as young_smasher suggested?
[w(x) = \frac{1}{\alpha(n)}\int_{B(0,R)}\frac{R^2 -|x|^2}{R|x-y|^n} w(y)dy?]
I see that I can take out a $1/R^2$, but I'm not sure if I can conclude that $w=0$ from that, I can see that $w(y) \le \sup{|u|} - ax+b$, so maybe that's sufficient. Unless this is the wrong direction of course

[untold delta]

kirby

[quaint herald]

No subtraction is needed @late moth , I don't see how it would really help anyway. The point is if you differentiate the Poisson kernel twice and use the estimate given, you can show that all second order derivatives vanish identically.

[late moth]

Ohhh, okay
I got the approximation idea in my head from some of my earlier problems so I was thinking about approximating that way

[quaint herald]

also, important to note that the poisson kernel expression is an integral over the sphere, not the ball.

gtg now but hopefully you can figure it out from here.

[late moth]

Ah yes, you mean that it's an integral over \partial B(0,R)?
But thanks for your help, I think I can go from here
Yeah, I was able to get it, thanks

[onyx heart]

that sucks

i am sure there are some meme conferences for sure

[patent raft]

😐

boundary value problem

[empty terrace]

Are there PDEs whose steady state results in a discontinous solution?

E.g. $u(x) = u(a), , x \in[a,(a+b)/2)$ and $u(x) = u(b), , x\in((a+b)/2, b]$

[untold delta]

criver

[empty terrace]

I know that the advection equation can propagate discontinuities for finite time, but I want when two of those meet in the middle to stop.

E.g. prescribing $u(t,a)=y_a$, $u(t,b) = y_b$, $u(0,x) = 0, x\in(a,b)$.

[untold delta]

criver

[empty terrace]

Basically I want $u_t + u_x = 0$ evolution from the left until (a+b)/2, and $u_t - u_x = 0$ evolution from the right, but without explicitly prescribing any conditions on (a+b)/2

[untold delta]

criver

[empty terrace]

One idea was to set the initial state to -infinity and have $u_t + H(|u_x|-m)|u_x| = 0$

[untold delta]

criver

[empty terrace]

Where $m>|u(a)-u(b)|$ and $H$ is the heaviside function.

[untold delta]

criver

[empty terrace]

Does this look reasonable?

[tranquil steppe]

There are too many messages, and in the end I don't know what your aim is. You mentioned singularities in finite time for some PDEs (Burgers' equation is one example), so what is it exactly that you need from these singularities?

[empty terrace]

For Dirichlet conditions $u(t,a) = y_a, u(t,b) = y_b$ the pde should produce a steady state $u(\infty, x) = y_a, , x\in[a,(a+b)/2)$ and $u(\infty, x) = y_b, , x\in ((a+b)/2, b]$.

[untold delta]

criver

[empty terrace]

As a generalisation in n-D the discontinuities must form along the interfaces of the Voronoi diagram induced by the Dirichlet data.

My formulation for achieving this in 1D is: $u(0,x) = min(y_a, y_b)-2m, , m = |y_a-y_b|, , x\in (a,b)$, $u_t = H(|u_x|-m)|u_x|$.

[untold delta]

criver

[empty terrace]

Although this $m$ and dependence on the initial state feels a little artificial. The H in the above is the heaviside function.

In n-D it's the same as above but with |u_x| being replaced with magnitude of the gradient I guess.

[untold delta]

criver

[tranquil steppe]

Again, putting it into one message would be helpful for anyone willing to read (mostly becasuse of the tex bot).
Your use of steady state implies that you just mean the limit at t->infty, which is not standard. For a first order PDE like the one you consider, there's no reason to expect that this limit solves the corresponding stationary PDE.

In the formulation you gave, the initial condition is constant, and it looks like the constant function solves the PDE. Am I missing something?

[empty terrace]

The initial condition is constant, except at the boundaries a and b, where it is respectively y_a and y_b. So the constant function does not solve the pde.

As I wrote it, it should act like a standard advection equation from a to the right: u_t + u_x = 0, propagating y_a to the right.

And it will propagate y_b from b to the left with the standard advection equation: u_t -u_x = 0

as I set up the initial state the step function will be 1 until those two fronts meet, at which point it will become zero and the transport should come to a halt at a discontinuity in (a+b)/2 (midway). In n-D "midway" corresponds to the faces of the Voronoi diagram induced by the Dirchlet boundary.

This dependence on the initial state and the m parameter in H feels quite artificial however, so I was wondering whether there's a way to set up a pde (even removing time dependence) that results in the same steady state solution without needing to manually prescribe the jump.

That is the PDE solution should reproduce jumps at Voronoi faces and have a constant value within each Voronoi cell equal to the Dirichlet point within that cell.

Advection equations of the above kind are transport equations, in this case it transports the Dirichlet data values. The absolute value makes it similar to the eikonal equation.

[tranquil steppe]

Hm, so it looks like there's a contradiction (or at least, a discontinuity) in the boundary conditions? It's not a bad thing necessarily, but it prompts the question: in what sense do you want the PDE to be satisfied?

[empty terrace]

I believe (I could be wrong though) that the above formulation satisfies these conditions. Since H(|u_x|-m)|u_x| vanishes at the discontinuity (the step basically ignores any jumps with a magnitude less or equal to m, it's like prescribing u_x = 0 when such a jump is encountered in the time evolution). But this feels rather artificial, depending on both the initial state and this parameter m, and also being a time evolution when the only thing I am interested in really is the steady state. For reference this arose from trying to model piecewise constant interpolation. And I am trying to figure out a robust formulation that would generalize to higher odd orders where the interpretation is not as easy. For example something involving d^3u/dx^3 which in the 1D case should yield piecewise quadratic C1 interpolation.

I don't know whether this is useful, but the formal solution (not depending on time) of the above relies on a distance field:

$u(x) = \begin{cases} y_a & |x-a| <|x-b| \ y_b & |x-b| < |x-a| \end{cases}$

[untold delta]

criver

[empty terrace]

I don't really care what the solution is at (a+b)/2 but (y_a+y_b)/2 would make sense

[empty terrace]

Ok, I figured out my problem. The reasonable thing to do (in order to match bspline interpolation) is to prescribe Neumann conditions explicitly after all.

then for the two subintervals I just solve u_t+u_x= 0 (left subinterval) and u_t-u_x= 0 (right subinterval)

[untold delta]

DvaNapasa

[junior bloom]

Is there a definition for the trace of a measure?

Weird operation I'm seeing in Young measure lit for dissipative solutions and I have no idea what it's saying.

[astral vine]

Depends on your measure

Can you give me the reference you are actually reading

[junior bloom]

Sure

[astral vine]

Thanks

[junior bloom]

Check out Wiedmann (2017), top of p. 11.

About the Weak-Strong uniqueness of the Navier-Stokes limiting solution.

Actually scratch, I figured this one out lol

[astral vine]

Okay lol

[wooden sundial]

Hi, how do I solve dirichlet problem on unit square?

[astral vine]

Fourier Series

(or equivalently Fredholm alternative which gives existence of eigen vector with eigenvalues)

[junior bloom]

your two tools for solving pdes:

1. fourier series
2. give up

youre welcome

[astral vine]

I assumed @wooden sundial wanted to solve it explicitly

There is also an unusable Integral formula which is used For regularity and Functional Analytic purpose

And for fully abstract problems (for arbitrary data): Lax-Milgram, or Harmonic Analysis(related to above method with unusable integral formula)

[empty terrace]

There's a third option: numerics.

[junior bloom]

Numerics is basically giving up

[empty terrace]

I reserve giving up only for when there are also no working numerics.

[tired hollow]

ended up with this after reducing from PDE, prof says statement correct but askd to find approxiimation

idk how :|, there is sqrt that messes up the computaitons

[astral vine]

s is a non negative number and you want asymptotic as s goes to infty

[astral vine]

?

or is it as w goes to 0/ +\infty

[tired hollow]

it is for large s w check behavior

the integral itself is divergent

s is nonnegative ye

g also nonnegative

[tired hollow]

didnt get this

[astral vine]

The integral is convergent, but not absolutely convergent

[tired hollow]

wait why convergent

Bessel function is $J_0(xs)=\sqrt{\frac{2}{\pi xs}}cos(xs-\frac{\pi}{4})$

[untold delta]

DvaNapasa

[astral vine]

I know but you cna make sense of this integral

[tired hollow]

ye

[astral vine]

So this is covnergent

It makes sense $\int_0^A$ as A goes to infty

[untold delta]

Ryc for mod²

[astral vine]

this involve Fresnel integrals

[tired hollow]

okay so subst $\infty$ with $\lambda$

[astral vine]

(in this precise case for J_0)

[untold delta]

DvaNapasa

[tired hollow]

and then dance?

[astral vine]

??

what is lambda

[tired hollow]

and evaluate $A(\lambda)$

[untold delta]

DvaNapasa

[tired hollow]

upper bound of integral

[astral vine]

You don't need that

A is the constant that goes to infty

[tired hollow]

i dont get :(

[astral vine]

$$\mathcal{A}(s,\xi)= \Re \int_{0}^{+\infty} J_0(xs), e^{i\xi \sqrt{gx}},\dd x = \lim_{A\rightarrow + \infty} \Re \int_{0}^{A} J_0(xs), e^{i\xi \sqrt{gx}},\dd x $$

[untold delta]

Ryc for mod²

[astral vine]

First you should consider $s^2\mathcal{A}(s,\xi)$

[untold delta]

Ryc for mod²

[tired hollow]

@astral vine and use final value theorem?

[astral vine]

no

there is several transformation to achieve before

[tired hollow]

what kind of im confused

[astral vine]

change of variable

[tired hollow]

we have covered stationery phase, fourier laplace integrals and some steepest descents :|

[astral vine]

use the definition of J0

[tired hollow]

oh

$$\mathcal{A}(s,\xi)= \Re \int_{0}^{+\infty} J0(xs), e^{i\xi \sqrt{gx}},\dd x = \lim{A\rightarrow + \infty} \Re \int_{0}^{A} J_0(xs), e^{i\xi \sqrt{gx}},\dd x $$

[astral vine]

The achieve a Stationary Phase Arg

[untold delta]

DvaNapasa

[tired hollow]

here you missed x

$$\mathcal{A}(s,\xi)= \Re \int_{0}^{+\infty} J0(xs), e^{i\xi \sqrt{gx}}, x \dd x = \lim{A\rightarrow + \infty} \Re \int_{0}^{A} J_0(xs), e^{i\xi \sqrt{gx}},x \dd x $$

[untold delta]

DvaNapasa

[tired hollow]

this blows the stuff out :\

[astral vine]

it does not

[tired hollow]

but whenever i try it goes divergent

[astral vine]

Again you can make sense of it

[tired hollow]

or give up because cant even get what the problem statement means xd

[tacit pendant]

Probably the wrong thread to ask, but what is the dP called for P in a stochastic process? I think it's something like a rate of change? I'm not familiar with stochastics and differential equations

[tranquil steppe]

You could call it a rate of change, or a differential, I guess. But the precise meaning of the stochastic ODE you wrote here is given by equality of two stochastic integrals.

[tacit pendant]

Thank you!

[tranquil steppe]

Or maybe not. Since you mentioned stochastic processes, I assumed that z is a Wiener process, but it's a silly assumption on my part. Maybe it's just a usual ODE.

[solid flint]

Hello, in this MSE tread (https://math.stackexchange.com/q/2117107/311112), in a comment and in the answer, it is stated that if the operator $A$ isn't closed then $A-\lambda I$ isn't surjective. How is this? What's the relation between the closed-ness of the operator and its surjectivity?
Also, are there a relation between the closed-ness of an operator and the closed-ness of its range?
Thank you in advance.

[untold delta]

Mikahopff

[astral vine]

That's nto what he means @solid flint

He said : "The operator is ill-defined, hence you have to take its closure to make it well defined"

"So assumed that you have considered its closure, this is not a surjective operator"

this is not a consequence of closedness

he means that "if the operator have a domain such that it is well defined, THIS particular operator is not surjective"

[solid flint]

Thank you very much @astral vine

there still a lot of work to do for my English!

[astral vine]

This was not clear to me at my first read too

But I get what the post's author means considered an example where it holds

[junior bloom]

@astral vine @quaint herald You guys know of any results that do something like:

1. It's a weak solution on a Riemannian manifold
2. The solution preserves some group of invariances that's geometrically relevant to the metric (or symmetries, as physicists call them)

Weak can be in any sense. Sobolev, Young, etc.

[astral vine]

I'm not sure what are you asking exactly, but it reminds me stuff like solutions of Hamiltonian systems

Maybe @river path

[river path]

It reminds me of that too but idk anything specific about this

[astral vine]

I have checked the Shallow Water wave Lecture I had during my Master degree

[junior bloom]

Yep! KdV shallow water dynamics has Young measure solutions that allow for singular behavior.

One of things that motivated my current search.

[untold delta]

Functionanatolysis

[astral vine]

Is that the sort of thing you are looking for ?

[junior bloom]

Yeah but replacing f with something like a distribution or random variable or measure.

"Weak" solutions

[astral vine]

[astral vine]

Maybe checking those (it talks about Sobolev solutions but I don't if it is what you are looking for) :

Existence and Uniqueness Solution of Euler-Lagrange Equation in Sobolev Space W1,p(Ω) with Gateaux Derivative
Ratna Dwi Christyanti, Ratno Bagus Edy Wibowo, Abdul Rouf Alghofari

Composition functionals in higher order calculus of variations and Noether's theorem
Gastão S. F. Frederico, J. Vanterler da C. Sousa & Ricardo Almeida

[junior bloom]

Close but I'm actually looking for something simpler. Something as simple as here's a weak solution to a PDE on a manifold that preserves e.g. special relativity/some symplectic form/etc under its evolution.

[astral vine]

It's kinda far from my speciality domain, so I can't tell

I just checked my former Lecture and its related references

[junior bloom]

I've checked fluid dynamics lit and there's nothing here in GR Navier-Stokes. My first thought was to ask physicists because, you know, GR has wave and diffusive characters and lots of singularities that force weak solutions.

Was a bit of a hail mary. Definitely a new topic.

[astral vine]

GR Navier-Stokes is probably not investigate due to pseudo-riemannian metric instead of a pure riemannian one

which make lose parabolic caracter of the Heat operator

[junior bloom]

Ooh good point

So singularities are embedded into it by default.

[astral vine]

The question is maybe interesting but all usual technics are no longer available

[junior bloom]

I've seen relativists do Ricci diffusion with singularities so it's still probably possible?

IDK this is out of my depth now

[astral vine]

You have to check under what assumptions on the pseudo-riemannian manifold is this done

(bounded by below curvature etc.)

[river path]

i'm able to do energy estimates to show global existence in C(R_t, H^N) for this equation. is there a way to do energy estimates on this if a_kj also depend on time?

(k and j are summed over)

the way i'm doing it currently is differentiating the equation in time to get L^2 estimates on del_t^n u, and then using the equation to convert these into estimates on L^2 norms of space derivatives.

(assume whatever you want on a: uniformly elliptic, smooth in x and t, even some decay is fine)

so when you introduce time dependence, then whether you differentiate in x or in t you get these extra product rule terms with derivatives of a_kj. on the RHS, i get one term with an extra derivative of u and one without, so when i multiply by either u* or a derivative of u*, take imaginary parts and integrate, i always end up with an extra bad term (pairing u with one of its derivatives)

it's possible you just don't get energy estimates. i'm not sure. you still get the usual L^2 conservation.

[astral vine]

If your matrix is time dependent, you need some additional assumptions

[river path]

i'm okay with assumptions

i'm more just trying to figure out how to handle the lower order terms in the energy estimates

things can be as nice as wanted

[astral vine]

Your operator is a divA grad one

oh wait you want Higher order estimates

[astral vine]

assuming smoothness and uniform ellipticity in (t,x) and by induction considering $$ \partial_t\lVert \partial_{x_k}^m u(t)\rVert_{\mathrm{L}^2(\mathbb{R}^d)}^2$$ and using Gronwall estimates ?

[untold delta]

Functionanatolysis

[astral vine]

(diagonal derivatives estimates is sufficient on Rn)

[river path]

hmm

so when i do this one time, i get $\partial_t \int |\partial_{\ell} u|^2 = \mathrm{Im} \int (\partial_k \partial_\ell \overline{u})\partial_{\ell} a_{kj} \partial_j u$

[untold delta]

ryc

[river path]

The term on the right hand side comes from applying del_ell to a (we then multiply the equation by del_ell u* in order to get rid of the leading order derivatives)

i'm not sure how to estimate this

or if i should be testing the equation with something else maybe

[astral vine]

Maybe it was not clear but I also assumed uniform ellipticity of

[untold delta]

Functionanatolysis

[astral vine]

the same goes for mixed derivative in time-space for A

[river path]

yes that's ok

uhhh

am i being silly

how does this help when you're pairing del_l grad u with grad u

ummm

somehow i need to get rid of the grads in order to apply gronwall

[astral vine]

Uniform ellipticity, A and all its mixed derivatives with high enough smoothness, implies square root property of any order

say $$ \lVert [\partial_t^\ell \mathcal{L}^\alpha(t)] f\rVert_{\mathrm{L}^2(\mathbb{R}^n)} \simeq \lVert \nabla^2 f\rVert_{\mathrm{L}^2(\mathbb{R}^n)} $$

[untold delta]

Functionanatolysis

[astral vine]

(here f is independent of time

and the operator $\partial_t^\ell \mathcal{L}^\alpha(t)$ stand for $$ \partial_t^\ell \mathcal{L}^\alpha(t) f = -\mathrm{div}( [\partial_t^\ell \partial^\alpha_x A(t,x)]\nabla f)$$

[untold delta]

Functionanatolysis

[river path]

aahhhh

ok

i forgot about this

thank you

[astral vine]

np

I recall that is true since you assumed some very strong "super ellipticity" assumption

[river path]

i mostly wanted to see if things could hold up under any sort of time dependence - i don't need any kind of super strong ellipticity for time independent coefficients which is what's important

[river path]

(using this "differentiate in time instead of space" trick)

[astral vine]

Intereseting things comes in when you look at non autonomous elliptic divergence form operators with rough coefficients

[river path]

I see

I haven't thought about nonautonomous operators before at all

[astral vine]

moreover you added troubles since you were dealing with the Schrodinger Equation is kinda Hyperbolic

(the parabolic case is already sufficiently difficult)

[river path]

Yes, that's a but annoying to deal with here

I was kind of expecting it to fail

[astral vine]

Theorem : If everything is smooth enough, and your PDE is on the whole Rn then it works except if you are talking about NSE.

[river path]

Lol

Yeah I just have no idea what to expect for well posedness of nonautonomous equations. But I suppose for ODEs nothing changes.

[astral vine]

In general you cannot expect gain of regularity, and only expect some loss in the case of strong solutions

Check out some Monniaux/Auscher/Portal works

aslo some El Maati Ouhabaz work

those are only in the case of Parabolic

and hell

I wanna kill myself

[river path]

Hmm

Thank you for the recommendations

I will do my best to avoid needing to think about this

But that might fail

[astral vine]

There is few definitions of bifurcation, chat is yours, and in which context you want to apply it ?

[astral vine]

I get that but involved techniques depend on the kind of problem you want to investigate, classical finite dimensional ODEs, PDEs (Reaction Diffusion, etc.)

I guess it's about finite dimensional ODEs

[late moth]

Okay, I'm kind of stuck on starting this problem, I know it’s probably fairly simple. Let $g:\R\to\R$ be a continuous function s.t. $\lim_{x\to-\infty}g(x) = a$ and $\lim_{x\to\infty}g(x)=b$. Let $u(x,t)$ be the solution of the heat equation on $\R\times (0,\infty)$ given by the Green's function. Show that for every $R > 0$,
[\sup_{|x|\le R} |u(x,t) - \frac{a+b}{2}| \to 0]
as $t\to\infty$.

I'm just not really sure on how to leverage the properties of the heat equation for this problem in particular.

[untold delta]

kirby

[astral vine]

Whaat is is g ?

The initial data ?

The source term ?

[late moth]

The initial data

[astral vine]

Then write down the explicit solution

[untold delta]

Functionanatolysis

[astral vine]

replace a and b by -1 and 1 first

[astral vine]

(the idea is to make a change of variables to put everything in g, then cut the integral into parts and make it again one whole sided integral on (0,+ìnfty), then apply DCT using continuity of g)

[late moth]

Sorry for the late response, I haven't gotten around to working on this. So we want to show
[\left|\int_{\R}e^{\frac{-|x-y|^2}{4t}}g(y)dy - \frac{a+b}{2}\right| \to 0.]
Replacing $a,b$ with $-1$ and $1$ respectively (this is just a scaling of $g$ right?), and rewriting the above as
[\int_0^\infty e^{\frac{-(x-y)^2}{4t}} g(y)dy + \int_{-\infty}^0 e^{\frac{-(x-y)^2}{4t}}g(y)dy = \int_0^\infty e^{\frac{-(x-y)^2}{4t}} g(y)dy - e^{\frac{-(x+y)^2}{4t}}g(-y)dy ]
This is where I get stuck with change of variables. I presume that we want to write g(y) as a function of $t$ and $x$ (and $-t$ and $x$ for the second term) through this, then by continuity, we have boundedness, and hence by DCT, we can conclude convergence to $\int_0^\infty \text{something}\cdot(a + b)dy = 0$.

I think this is a call for me to review multivariable calc again.

[untold delta]

kirby

[astral vine]

I said to put everything in g

[untold delta]

Functionanatolysis

[late moth]

Ah yes, I switched the order of the steps you mentioned. I did the change of variables as above, and it agrees with what you wrote. Thanks, I think I have it from here :)

[mellow flicker]

can someone please tell me if it's possible to use Maclaurin series on an equation like this one

[astral vine]

probably teh wrong channel

[mellow flicker]

where should i ask?

[astral vine]

I don't know, but I dont get how you want to solve this using MacLaurin series ?

[mellow flicker]

so the equation up is the green curve and i want to get it as close as i can to the red curve which I'm trying to model it

so i was willing to use MacLaurin series to get it as close to the red one as i could

[astral vine]

weird stuff

[valid edge]

Hey yall

Is «find the first integral of the corresponding Euler equation for the following functional» a question that make sense to yall?

Its one that i am attempting to grasp the meaning of for an assignment ive got

My attempts at this is to convert the limit to dx instead of ds and plug into Euler equation

[valid edge]

I think i figured out what they want

[tired hollow]

does anyone know where "epsilon regularity" started? Is there an easy example of it working that I can see? Hopefully something geometric (on smooth manifolds or something)?

I saw a theorem that was like: If (energy) on a ball is less than epsilon, and f is in L^{1,p} for (some appropriate p), and solves (a nonlinear elliptic equation we are trying to solve), then f is smooth in the half ball / interior

Is this better or different from elliptic regularity, if that even makes sense to ask? (I kind of just want to ask - why can't we use elliptic regularity, but maybe that is too specific to this question I'm looking at). Sorry I want to know the details but most of the time people just say "just use elliptic regularity" or "this is epsilon regularity" and then I don't understand how it's used exactly

[astral vine]

what is epsilon regularity ?

[tired hollow]

that's kind of my question haha, but really the theorem I semi-quoted was called epsilon regularity

[astral vine]

But I don't what do you mean by it ?

[tired hollow]

The people who use it make it seem like it's a well known thing so it might not be

[astral vine]

Give me full explicit statement with all assumptions (with something like a Precise ref)

[tired hollow]

oh well this is in a youtube lecture

hold on

https://www.youtube.com/watch?v=BvDSeTbtAKg theorem is written down at 1h 15min or so - it is about the Yang-Mills equation

but the point is I've seen the term "epsilon regularity" come up in places, so I'm curious as to how people think about it / what it is

theorem (called epsilon regularity) statement is : If the Yang-Mills energy over a ball is less than epsilon, and if A is in L^{1,n/2} and solves the Yang-Mills eq, then A is smooth in the interior (half ball)

[tranquil steppe]

"Elliptic regularity" refers to theorems of the form "if u solves the PDE in the unit ball, then it's smooth in the ball of size 1/2". Typically, this can be proved for linear (or otherwise well-behaved) PDEs. Or for nonlinear PDEs, if we assume a priori that u has good regularity. For example, if the PDE involves du to some high power, but u is assumed to be Lipschitz, then this assumption somehow kills the nonlinearity.

"Epsilon regularity" refers to theorems of the form "if u solves the PDE and its energy is small in the unit ball, then it's smooth in the ball of size 1/2". Here "energy" can be some L^p norm of du, for example (depends on the context). As you can see, the statement is strictly weaker than "elliptic regularity". Typically, this is the best one can hope for in case of (some) nonlinear PDEs.

If it's known that solutions of your PDE can have singularities (which I guess is the case for Yang-Mills, for example?), then you can be sure that standard "elliptic regularity" theorems are not valid in your case. So you try at least to get "epsilon regularity", which is still useful - it usually implies that the solution is smooth outside some small singular set (typically, of a lower Hausdorff dimension).

[tired hollow]

cool that clears up a bunch thanks!

[tranquil steppe]

I see that a similar question has been asked on MathOverflow:
https://mathoverflow.net/questions/39227/epsilon-regularity-what-does-it-say-and-where-does-it-come-from
One of the answers is by Terry Tao, by the way.

[tired hollow]

oooh nice

thanks

[junior bloom]

Potentially silly question. How should I apply the trace theorem to boundary conditions of the form $(Tu)|{\partial\Omega} = 0$ as opposed to $u|{\partial\Omega}=f$? Where $u\in W^{k,p}(\mathbb{R}^d)$.

[untold delta]

teafortwo

[astral vine]

what is T ?

The most general trace theorem say that you can make sens of the restriction of a function $u\in W^{k,p}(\mathbb{R}^d)$ on any closed subset of $F\subset\mathbb{R}^d$ , $F$ being a $d-1$-set (in particular $C^k$ or Lipschitz boundary of an openset)

[untold delta]

Functionanatolysis

[astral vine]

You wan also make sense of some boundary conditions like $\nu \cdot A\nabla u$, $A$ to be an $\mathrm{L}^\infty$ matrix, using sesquilinear form

[untold delta]

Functionanatolysis

[astral vine]

hence the trace is well defined but in weak sense

Let $\mathcal{L}u:=-\mathrm{div}(A\nabla u)$ with Neumann boundary condition $\nu \cdot A\nabla u_{|_{\partial\Omega}} =0$, $A\in\mathrm{L}^\infty(\Omega,\mathcal{M}_n(\mathbb{C}))$ uniformly elliptic, is defined by the sesquilinear form $\mathfrak{a}$, with domain $\mathrm{D}_2(\mathfrak{a})=\mathrm{H}^1(\Omega)$ on $\mathrm{L}^2(\Omega)$,

$$\mathfrak{a}(u,v) = \langle A\nabla u, \nabla v\rangle$$

[untold delta]

Functionanatolysis

[astral vine]

The same goes for higher order system of order 2m (in above example m=1$)

If everything is smooth (the matrix A for instance) you can make sense of it in the strong usual sense of trace

(Generally using pseudo-differential operators, check out Lopatinskii-Shapiro boundary conditions)

[errant gust]

Can anyone clarify what is meant by (ii)? I am not sure what boundary conditions they are talking about here

[twilit rover]

maybe periodic

[junior bloom]

Let's say $T$ is a linear functional.

[untold delta]

teafortwo

[junior bloom]

Can the $\nu$ in your generalization be another vector field?

[untold delta]

teafortwo

[astral vine]

nu the exterior normal can be given by a 1 form acting by the wedge or interior product on a differential form, but it always represent the outward normal unit vector

[astral vine]

For any linear functional you cannot generally makes sense of it

[ancient chasm]

Hi everyone
I am trying to derive the adjoint Poisson equation for the following problem to find the sensitivity of an objective function with respect to a decision variable, but I get stuck in the middle of the way. Here is my derivation.

This is the point that I do not know how to proceed further. I do not know how I should set the boundary condition of 𝜆 on ∂Ω2. Basically, I do not know if I should eliminate the third or fourth integral in the last equation. Besides, after setting the boundary condition of 𝜆 on ∂Ω2 and eliminating either of terms, I do not know how to treat with the other integral term in calculation of 𝛿𝐽/𝛿𝑎.

I appreciate any help or resources.

Thanks a lot

Just in case you need a PDF file of the derivation, here it is:

[late moth]

Let $g: \R^n\to \R$ is bounded and h"older continuous with $\alpha\in(0,1]$. Consider $u$ to be the solution to the heat equation with initial data $g$. I am tasked to show that there is a $C$ such that
[\sup_{x,t}|\partial_t u| + \sup_{x,t}\left{\sum_{i,j}|\partial^2_{x_ix_j}u|\right}\le C\sup_{x\ne y}\frac{|g(x)-g(y)|}{|x-y|^\alpha} t^{(\alpha/2) - 1}.]
I know that because $\int_{\R^n} \Phi(x-y,t)dy = 1$ for all $t$,
\begin{align*}
\partial^2_{x_ix_j}\int_{\R^n} \Phi(x-y,t)dy = 0 \
\partial_{t}\int_{\R^n} \Phi(x-y,t)dy = 0
\end{align*}
I also know that
[\int_{\R^n} \Phi(y,t)|x-y|^{\alpha}dy \le Dt^{\alpha/2}]
where $D$ is some other constant depending on $n$ and $\alpha$.\newline
Setting this up I have
[\sup_{x,t}\left|\partial_t \frac{1}{(4\pi t)^{n/2}}\int_{\R^n}e^{-|x-y|^2/4t}g(y)dy\right| + \sup_{x,t}\left{\sum_{i,j}\left|\partial^2_{x_ix_j}\frac{1}{(4\pi t)^{n/2}}\int_{\R^n}e^{-|x-y|^2/4t}g(y)dy\right|\right}]
Now rearranging we get that this quantity is less than or equal to
[\sup_{x,t}\int_{\R^n}\left|\partial_t \frac{1}{(4\pi t)^{n/2}}e^{-|x-y|^2/4t}g(y)\right|dy + \sup_{x,t}\left{\sum_{i,j}\int_{\R^n}\left|\partial^2_{x_ix_j}\frac{1}{(4\pi t)^{n/2}}e^{-|x-y|^2/4t}g(y)\right|dy\right}]
I'm not sure from here how to continue to get estimates and used the above results I found.

Then, after this, I need to justify how this can be used for proving Duhamel's formula to reduce the regularity requirements on $f$ in the inhomogeneous case (this is just confusing because we're applying a bound in $g$ for a homogeneous heat equation)

[untold delta]

Kirby

[late moth]

I can also carry the sup into the sum so then I only have to worry about the integral terms

[astral vine]

The result you are looking for is called Maximal Regularity Estimates

Here you want to check out Maximal Regularity in Holder spaces/Continuous bounded functions

Doing it for the heat semigroup "by hand" is not that nice

but doable

The problem is that you cannot expect to reproduce it for more general regular elliptic operators

To check out this Kind of estimate check last Chapters of 'Interpolation Theory' by Alessandra Lunardi

[late moth]

Thanks, I'll look at it, I knew that the calculation was probably going to be messy, but thanks for some broader context to it as well

[astral vine]

If you still want to do it

You should check Homogeneous besov spaces as a realization of Holder spaces for non integer index

The descirption of besov norms through the Heat Semigroup should give you the desired result in few lines

[late moth]

I'll check that out too, thanks :)

[astral vine]

Lemma 2.34 in "Fourier Analysis and Nonlinear PDEs" by Bahouri, Chemin and Danchin.

Lol I tried

The result was given in three lines

I didn't expect it was that straight forward

[astral vine]

also be careful you didnt mentionned that above inequality was for finite time T, you used the same notation for intrinsic t, and the finite time

[late moth]

Oh yes, you're right. This helps a lot, I'll look at the reference you gave

[astral vine]

I think the real estimate you want to prove is

which is nothing but a decay estimate

[untold delta]

Functionanatolysis

[astral vine]

That's why quantify objects and introduce everything is improtant in a statement

[tranquil steppe]

I'm not sure this is the best advice for someone just starting to study the heat equation. And in any case, if you dived into all the steps required before the three line proof, I'm pretty sure you'd have to do one of these tedious calculations anyway.
I agree that the direct approach fails us in more general contexts, but it's a good start.

@late moth, I guess you're not very far from the goal. Expressing $u$ as $u(t,x) = \int \rho_t(x-y) g(y) dy$, and applying two derivatives, we have $\nabla^2 u (t,x) = \int \nabla^2 \rho_t(x-y) g(y) dy$, as you described. Then using your observation, we can turn it into $\int \nabla^2 \rho_t(x-y) (g(y)-g(x)) dy$ (since $g(x)$ is a constant). Using Holder regularity of $g$, we're left with bounding $\int |\nabla^2 \rho_t(x-y)| |x-y|^\alpha dy$, which shouldn't be hard.

[untold delta]

SingularityResearch

[astral vine]

Indeed the proof of above Lemma contains kinda tedious calculations, but there is also some easier version, just this is the most recent one and the book contain other interesting and fundamental results about the Heat Equations

(anf Functional Spaces)

But since what he needs to prove is probably the simplest version (there were some supremum in time that shouldn't be there in his above statement)
#advanced-pdes message
(replacing the Laplacian by the Hessian which is not a big deal in the whole space)

So I agree now, its and your approach are now the simplest one.

[late moth]

The supremum in time bounds shouldn't be there?
I've just written that from the problem I was solving.
I took a break and just saw this set of messages so sorry for the late response, I think I understand the process for bounding the laplacian (and the hessian), I just have to deal with finding a bound on these second derivatives, which appear to be multiples of the fundamental solution.

I do appreciate the sources you gave though

[late moth]

Yeah I see what you're going off of here (re: my last message), and as I mentioned, I already have a bound for moments of the heat equation, so once I'm able to deal with the multiplicative terms of the derivatives (I'm getting Phi(x-y,t) * [|x-y|^2/4t^2 - 1/2t] for the laplacian case), I think I'm done

[tranquil steppe]

Doing it the most direct way (as I suspect you're trying right now) is OK if you're good with computations. You just need to get it done.

[late moth]

Yeah, I was able to make some headway with it that was sufficient to get the bound

[tranquil steppe]

@late moth However, one can derive the right bound in a more comprehensible way. We're left with $\int |\nabla^2 \rho_t(x-y)| |x-y|^\alpha dy$, which is the same as $\int |\nabla^2 \rho_t(y)| |y|^\alpha dy$, so the only question is the $t$-dependence of this quantity. Substituting $y = z \sqrt{t}$ in the integral, and noting $\nabla^2 \rho_t(y) = t^{-n/2-1} \nabla^2 \rho_1(z)$ (you don't need to compute these derivatives at all!), the above integral is reduced to the form $t^{\frac{\alpha}{2}-1} \int |\nabla^2 \rho_1(z)| |z|^\alpha dz$. And now you don't care for the integral - as long as it's finite, it's just a constant.

[untold delta]

SingularityResearch

[late moth]

Thanks a lot for this guys, it helped a ton

[twilit rover]

Other than ODEs, what are important first topics one should study for PDEs?

[astral vine]

Most of the ODEs stuff is useless for PDEs

Measure Theory, Functional Analysis (Banach Spaces (Including Hilberts) and Operator theory), Lp spaces, Distributions (and Tempered ones), Euclidean Fourier Analysis (basics), Sobolev spaces.

Then depending on the kind of PDEs you want to look at

you will need some additional complements

Like Spectral Analysis, (Euclidean) Harmonic Analysis or Semigroup Theory

Interpolation Theory could also be useful

[junior bloom]

I'd love to see operator algebra in action in solving a PDE (rather than limited to the possibly unphysical case of algebraic QFT), or Hodge theory generalized to be more useful.

Practically all you need to know to study PDEs is everything in Taylor's 3-volume series. Good luck.

Don't @ me until you finish it. 🙂

[astral vine]

Again it depends on the way want to study PDEs

And, imho, Taylor is kinda rough and raw for a first approach, and deals almost only with L² setting

[junior bloom]

Taylor is more geometric than harmonic, but he does hit PDOs in Vol 2

[astral vine]

I know

[junior bloom]

Which I definitely have not read

[astral vine]

Just opened it, then closed it just after

[junior bloom]

The classic Taylor experience

[astral vine]

I still didn't finish to figure out how Mitrea, Mitrea,Taylor 2001 worked out

I need it to generalize some stuff

[verbal nebula]

Have you found representation theory to be useful at all in PDEs?

Teafortwo?

[astral vine]

Iirc, William Beckner did somework in this direction

Investigating Fourier Multipliers on Sln

The original question arised from applied Functional Analysis for PDEs

[junior bloom]

I've seen an application of Schur's lemma for representations in studying symmetries from Schrodinger's equation, which strictly speaking isn't even a PDE.

That's about it lol

Er, something something dynamical systems

We have dynamical people here and I'm not one of them so I won't say more

[astral vine]

There is also things about studying some PDEs on Heisenberg group but this is not general Representation Theory (linked to what teafortwo mentionned)

[carmine blaze]

not advanced but I got $u(x,t)=f^\prime(x-t)-g^\prime(x-t)$, where $f^\prime,g^\prime\in C^2(\mathbb{R},\mathbb{R})$ as the general solution to the parabolic pde $u_{xx}+2u_{xt}+u_{tt}=0$ can anyone verify/deny?

[untold delta]

Dpao

[carmine blaze]

I expected to get one of the functions as a multiple of t

[quaint herald]

Yes it should be something like A(x-t)+tB(x-t)

[carmine blaze]

do you mind taking a look at what I did and letting me know where I went wrong?

[carmine blaze]

nvm if i work backwards its clear why A(x-t)+tB(x-t) where A and B C^2 works so i think thats good enough

[twilit rover]

Use the mean value theorem or Fundamental thm of calculus

the main idea is to write $$F(x_1 + y_1, x_2 + y_2) = F(x_1, x_2) + F(x_1 + y_1, x_2) - F(x_1, x_2) + F(x_1 + y_1, x_2 + y_2) - F(x_1 + y_1, x_2)$$

[untold delta]

IlIIllIIIlllIIIIllll

[cerulean light]

thank you

[junior bloom]

OK this is a weird and general question, I'm trying to do something really stupid

what can I say about functions $f$ such that $(-\Delta)^sf\equiv0$, e.g. for the standard Laplacian $f$ is harmonic. harmonic things are very nice and appear everywhere. where do the fractional kernels appear?

[untold delta]

teafortwo

[junior bloom]

are they trivial?

actually what am i saying, of course they arent

[astral vine]

depends on your definition of fractional Laplacian

The Fractional Laplacian is not the same if you consider the Dirichlet, Neumann Laplacian

and both are different from the standard Fractional Laplacian on the whole Rn

For sectorial non negative operators the kernel of its fractional powers is Nothing but the Kernel of the Operator

For the Dirichlet Laplacian on a bounded smooth connexted domain (say on L²), its kernel is 0

For the Neumann Laplacian constants

for the Laplacian on the whole Rn, 0

[junior bloom]

how to show the last one

er, for fractional

[astral vine]

Are you familiar with tempered distribution theory ?

[junior bloom]

yes

[astral vine]

If you consider a tempered distribution

then its support should be reduced to 0

Fourier transform of the solution should some finite sum of delta dirac masses at 0 and higher order derivatives of it

taking back the Fourier transform

you solution must be a polynomial

but your solutions probably lies in some Lp or Sobolev space

[junior bloom]

ah

yeah

i see it

[astral vine]

hence it should be 0

if you don't have restriction on f other than tempered distribution, then you can't say more than "f must be a polynomial"

But this motivates, the "why" about people chosing to consider homogeneous Sobolev and Besov spaces lying in S'/polynomials

which sucks

I mean really sucks

because you won't be able to chose a canonical representation to make sense of a value a.e.

Hence considering Homogeneous Sobolev or Besov spaces on domains is nearly impossible since composition with functions does not make any sense (to talk about the description near the boundary)

This is why it sucks

[junior bloom]

That Fourier transform trick to get the kernel is really funny and direct. I should meditate of how I can interpret this in terms of frequencies and wavelengths

S'/poly sounds useless

even without boundaries

[astral vine]

If you do not do this, then Homogeneous Sobolev/Besov norms are no longer norms

And if you ask for other ambient distributions spaces that might works, then the Sobolev/Besov norms are indeed norms, but underlying spaces are no longer completes

So you have either to choose : loss of pointwise definition, and canonical representative, or loss of completness on a huge part of the scale

[junior bloom]

With spaces as weak as these what am I using completeness for again

If I can only produce a dense sequence that converges to something I want, well I basically just described the state of the Navier-Stokes problem

[astral vine]

If space are no longer complete then the Theorem of extensions for uniformly continuous (hence linear continuous) densely defined map doe snot hold any more

so you lose A LOT of properties In particular some density results cannot be used

[astral vine]

However if the space you are looking at is

[untold delta]

Functionanatolysis

[astral vine]

with norm

[untold delta]

Functionanatolysis

[astral vine]

Then in this case no matter what your currently doing (using S'h or S'/poly) this space can be uniquely determined as function spaces

i.e. you can choose a unique representation no matter what is your ambient "distribution space" and they are all complete

trouble arise for higher regularity , s ≥ n/2

(s ≥ n/p in th Lp case)

[rugged minnow]

@orchid reef Have you done dynamical systems' course?

[junior bloom]

Must appreciated @astral vine @orchid reef

So what explicitly are these other potential ambient distribution spaces that aren't complete?

I work in stochastic analysis too, its connection with renormalization in physics.

[junior bloom]

mama mia

[astral vine]

Th important thing is the first one

S'h

[junior bloom]

homogenous Besov

[astral vine]

It remebers me the reaction Ange had a year ago during my presentation

#discussion message

[junior bloom]

I'm missing lots of notation here. Where's this page from?

[astral vine]

my own work

A good reference

is Fourier Analysis and Nonlinear Partial Equations by Bahouri Danchin and Chemin, edited by Springer in 2011, Chapters 1-2

this is why BCD11

[junior bloom]

Too far down the rabbit hole for me

[astral vine]

Just the chapters 1 and 2 can be handle with elementary Fourier/Harmonic Analysis

everything is made from scratch

[junior bloom]

All my things have been Besov at worse

usually L^p