advanced-pdes

[untold delta]

Rotta

[astral vine]

This question has been solved several weeks ago by Gomez

This cannot hold asking only w in H^1

[patent fjord]

it doesnt hold in general case, there is additional assumptions

[astral vine]

Being in a subdomain doesnot change the problem unless you cna prove additional inner regularity, inside the domain away from the boundary

and even there

Lw in L² and w in H^1 gives you a twisted normal derivative in H^{-1/2}, not the actual normal derivative

[patent fjord]

so the smoothness of $h$ cant help

[untold delta]

Rotta

[astral vine]

like, for Lu=-div(A D u)

you obtain ADu . n in H^{-1/2}

but this tells you nothing about Du . n

[patent fjord]

yeah, im strugling with the $\nabla \cdot A\nabla w$

[untold delta]

Rotta

[astral vine]

Why do you want to reach such trace theorem ?

Is it an exercise, an intermediate step for your research problem ?

[patent fjord]

it looklikes that such inequality is used in a paper

trying to rework the steps

[astral vine]

Give me the paper

gonna check

[patent fjord]

proof of the theorem 1.2 page 6226 bottom inequalities

did u download the paper?

[astral vine]

Yes

Still do not know how they get rid of the matrix

WAIT

their definition is fucked

check the bottom of ^page 6219

the coefficients of A are involved in their definition

so everthing is fine

problem solved.

[patent fjord]

so the estimate works then? or what they do?

the conormal

[astral vine]

Yes and no, you didn't take advantage of Lw=0

but otherwise everything is fine

[patent fjord]

w should be smooth in D_2 since Lw = h in D_1 and Lh = 0, hence h is smooth and Lw = 0 in D_2-D_1, so w is smooth in D_2-D_1

i mean can it give me an estimate $|h|{L^2(\Omega')} \leq C|w|{H^1(\Omega)}$ ?

[untold delta]

Rotta

[void flame]

Could I get some help with this? I know I'mprobably supposed to derive this when integrating over an epsilon ball excluding the origin, but I struggle to make much progress

[rotund jetty]

since $2 < d = 3$, $1/|x|^2$ is integrable on any ball containing origin, hence is locally integrable

[untold delta]

L

[void flame]

yeah, that part i understand

i struggle to find what is the laplacian of 1/|x|^2 as a distribution

[rotund jetty]

Do $\int u(x)\Delta\phi(x),dx = \lim_{r \to 0} \int _{\mathbb{R}^3 \setminus B_r(0)}u(x)\Delta\phi(x),dx$ then integrate by parts and use $\Delta u(x) = 0$.

[untold delta]

L

[void flame]

yes, that is precisely what i'm sturggling with ;_;

this is what I got

[rotund jetty]

keep going? The integral over boundary of Omega_eps can be taken over boundary of B_eps(0).

[lilac barn]

Try to express the laplacian in polar coordinates and then calculate the quantity after converting into polar

[void flame]

ok, will try again tomorrow, thx :3

[patent fjord]

So we can find an estimate for The equation $|h|{L^2} \leq C|w|{H^1}$?

[untold delta]

Rotta

[patent fjord]

i mean $|h|{L^2(\Omega')} = <h, L^*w>{L^2(\Omega')} = <Lh, w>{L^2(\Omega')} = 0$ which implies the estimate, but $|h|{L^2(\Omega')} = 0$ seems very odd

[untold delta]

Rotta

[astral vine]

No, check pages 6219-6220

Moreover, the conventions of this paper states that L* is only the FORMAL adjoint not the adjoint it self.

so you only have <Lu,w>=<u,L*w> + <ADu.n,v>-<u, ADv.n>.

The biggst difficulty of the paper comes from the choice on conventions, which are a huge mess with regard to the standard ones.

[patent fjord]

ahhh i see my mistake, thanks

[patent fjord]

ok from this it is easy to see that $|h|{L^2(\Omega')} \leq C|w|{H^1(\Omega)}$ from $<Lu,w>=<u,L*w> + <ADu.n,v>-<u, ADv.n>.$, still the proof feels bit weird since the process is very similiar compared to $|\partial_v w|{H^{-1/2}} \leq C|w|{H^1} + |h|_{L^2}$

[untold delta]

Rotta

[astral vine]

No

I was talking about the whole proof for the trace estimate

And I think you are missing assumptions to use

I was just saying that you did have missing terms, but one them cancels out

due to h =0 in the annulus

[patent fjord]

if we suppose that $Lh = 0$ in D_1 and L* = h in D_1 and otherwise 0, then <Lu,w>=<u,Lw> + <ADu.n,v>-<u, ADv.n> is 0=<h,Lw> -<h, ADw.n>

[untold delta]

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

[astral vine]

What are you doing ?

I don't get it

[patent fjord]

trying to show $|h|{L^2} \leq c|w|{H^1}$

[untold delta]

Rotta

[patent fjord]

cause i have $|\partial_v w|{H^{-1/2}} \leq C|w|{H^1} + C|h|_{L^2}$

[untold delta]

Rotta

[astral vine]

Your gal is till about to prove the end of page 6226 right ?

[patent fjord]

yes

[astral vine]

Then this is not how to do it

Okay here is the proof

[patent fjord]

idk any other way than using the definition of the conormal derivative in the paper, but cool if you are willing share ur thoughts

[untold delta]

Functionanatolysis

[astral vine]

The w_a term in the proof can be obtained the same way

but using cnacelation in the annuli

and the fact that the boundary of D_2 \ D_1 is the reunion of d D_1 and d D_2.

The first line, the first inequality, this is H^{-1/2} not H^{1/2}, my bad

[patent fjord]

shouldn't $(Lh, \Phi){D_1}$ be $(Lh -ch , \Phi){D_1}$ ? according to definition

[untold delta]

Rotta

[astral vine]

Does not change that much the thing

Since Lh is 0

this adds a term c<h,Phi>_D

L^2 norm of Phi will be aborsbed in the H^{1} norm

the same for h

[patent fjord]

yes, but in case of w it becomes $(h -cw, \Phi)_{D_1}$ which causes all the problems, i was able to prove the case of $h$ similarly

[untold delta]

Rotta

[astral vine]

No

What is w ?

I mean

[patent fjord]

the problem L*w = h in D_1 otherwise 0

[astral vine]

You have to prove the same estimate

conormal derivative in H^{-1/2}( d D_1) bounded by H^1 norm of w in D_2\D_1

You domain here becomes D_2\D_1 not D_1 !

so L*w = 0 in D_2\D_1 by definition/construction

[patent fjord]

ahhhh

the issue have been that ive been trying to prove H^1 norm D_1

lol

[astral vine]

If you read the estimate end of page 6226

[patent fjord]

yes, i thought it for very weird way

[patent fjord]

ok, i think i can do it now, big thanks for your help

[astral vine]

you are welcome

[wintry delta]

Hello everyone, I was looking for a solution to my problem and found the lecture notes https://people.math.osu.edu/tanveer.1/m6451/week2.pdf. The problem is as follows: Prove that, if u is a classical (strong) C1 solution for Cauchy problem for scalar conservation laws, then u is also a weak solution of it. In there lecture notes it is at the end of Lemma 2, page 7. Can anyone point out for me a definition of strong solution, ideally how do you get def of strong solution from the weak one?

[unborn quiver]

A strong solution is just a solution where derivatives can be evaluated directly.

In weak solutions, you look at the distributional derivatives of a function $u$. I.e. for every test function $\phi\in C_c^\infty(\bR;\bR)$,
$\int_{\bR} u_x(x)\phi(x)dx=-\int_{\bR}u(x)\phi_x(x)dx$ (this generalizes for higher dimensions, I’m just on phone rn)

When you have a strong solution, $u_x(x)$ is continuous, but using integration by parts, the identity above still holds for each test function.

[untold delta]

kirby

[unborn quiver]

I hope that helped without giving you the answer directly

[azure frigate]

these are definitions of sobolev norm from 2 different books. does anyone know about why you would choose one definition over the other? Does it even matter in terms of what you can prove about sobolev spaces?

just let M=1 in the first definition to compare

[sand echo]

They are equivalent norms (a consequence of the fact that all norms of R^n are equivalent). So it doesn't really matter

[azure frigate]

oh im stupid

but it does slightly change how you would prove certain things, no?

[sand echo]

Basically not really lol

[meager dune]

Well if you know the norms are all equivalent then that is the optimal situation for proving things

[lament patrol]

I am having trouble with these general energy-type estimate problems while studying for my phd's quals. \mathbb{R}^3 means begin with Kirchoff formula then take \partial x_i and get to work, but I can't seem to understand the point of the problem and how to avoid doing the wrong "trick" at the wrong time (backwards integration, Holder, CS, etc) any help would be super appreciated 🙏

[lilac barn]

Try Fourier transform

(take FT solve the ode and then try to get the Lp control on the required quantities by Lp' control on the transform)

[lament patrol]

Good idea, but each problem “should” be designed to not require FT according to the committee, since it’s a timed exam. Although, I have a huge distrust for the exam committee because they very often have typos and don’t communicate expected materials…
So perhaps I should try FT anyway and just see what happens

[ocean ether]

PDE finals tomorrow

does this look about right

• euler lagrange states the minimizer of a lagrangian (PDE) satisfies a nice property about its derivative wrt x(t) and x'(t)
• hodograph transformation converts a PDE with respect to the variable
• legendre transformation is some convex analysis thing
• fourier/laplace transformation transforms PDEs into nicer ones where we can use to obtain the OG PDE solution
• cole-hopf transformation turns a quadratic PDE into a nice linear PDE
• separation of variables separates a potential solution into two functions to possibly obtain a nicer solution

[buoyant pike]

Legendre transform moves between a Lagrangian and Hamiltonian

[ocean ether]

does the rest look about right

[autumn stratus]

how does one seperate the PDE Au_xx+Bu_xy +Cu_yy=0

I know theres some trick with differentiating

[astral vine]

What are A,B and C ? Constants ? functions ? Positive ? Negative ? etc.

[patent fjord]

do we have weighted-cacciopolli type estimates $\int_{B(0,r)}|x|^\alpha|\nabla u|^2dx \leq 2M\int_{B(0,2r)}|x|^\alpha|u|^2dx+2/r^4\int_{B(0,2r)}|x|^\alpha|u|^2dx$, where $\alpha \in \mathbb{R}$ ?

[untold delta]

Rotta

[ocean ether]

bombed it

[flat crow]

I don't know where to put this question because I'm beginning to studying these from the POV of existence and uniqueness but this seems like a question that can be answered with vector calculus and I'm blanking

The wikipedia page for MHD lists these equations (in 3D), but theres 7 unknowns (3 from u and B, and one from p) and 8 equations. Can we remove the condition that divB = 0 and deduce it from the others?

[lilac barn]

The pressure is usually recoverable from velocity via inverting a nonlocal operator. Start by hitting the divergence on the equation for u and then invert the laplacian for p.

[naive breach]

Dear all, I having trouble with the discontinuous Garlerkin method (DG). My question is the following , by applying the DG method could I deduce a upwind/downwind discretization scheme for my PDE (a PDE to the convection diffusion type) ?

[buoyant pike]

#numerical-analysis

[naive breach]

Oups... wrong channel

[astral vine]

This is true only on the whole space, this approach is wrong on domains as such.

From the mathematical point of view One should see the gradient of the pressure as a correction term to ensure that the incompressibility condition (and the boundary condition when one considers the domain with a boundary) is preserved for the solution, but this correction term is always uniquely determined and well-defined. On the whole space Cocat's procedure gives you how to recover the pressure.

Furthermore, notice that the equation div(B) =0 is not a part of the system

otherwise one should have similarly to the equation for u with pressure p an additional term as gradient of some electric charge to ensure the divergence free condition for the magnetic field is met. In this case, w have more equations than unknowns... And the system becomes overdetermined.

So the point is to not impose free divergence condition for B (except at time 0) and to only show that it remains through the evolution.

The momentum equation contains 3 sub equations, the magnetic equation contains it self again 3 sub equations, div(u)=0 is another one. (u,b,p) yields 7 unknowns, Hence you have 7 equations and 7 unknowns. This is fine. If you add div(B) =0 you obtain more equations than unknowns, which might be problematic.

[lilac barn]

How will you recover the divergence-free condition on B?

[astral vine]

First you need to assume that B(0) is divergence free. The standard method to prove it is to show that the solution B of the a priori non-divergence free system, is such that div(B) satisfies the heat equation with homogeneous boundary and initial condition and 0 force term, by uniqueness of weak solution for the heat equation (up to an appropriate framework), this imposes div(B)(t) = 0 for all t>0.

There is some algebra involving the divergence free condition for u (which one is enforced and included in the system)

[lilac barn]

Yes, you can do the same thing with velocity as well. And get rid of pressure or vice-versa the divergence-free condition on velocity. Of course you need to take care of boundary conditions but as they weren't mentioned here I gave the full space approach

[astral vine]

This is not the same as for the pressure

Because you method only works on the whole space

[lilac barn]

You can solve for the laplacian with boundary conditions.

[astral vine]

No it is wrong

[lilac barn]

How?

[astral vine]

This will give you an other solution

unless I mistaken what you meant

[lilac barn]

What you're saying doesn't make sense as you're suggesting the pressure is a fully unknown quantity for a Navier-Stokes in bounded regime

[astral vine]

No

but it cannot be deduced exactly the same way

On domains, say with dirichlet B.C

[untold delta]

Functionanatolysis

[lilac barn]

That doesn't matter

[astral vine]

Yeah but this what is usually done on the whole space to deduce it

otherwise

can actually solve a Laplace equation for p with Neumann boundary condition depending on u

But this is not the approach used for the magnetic field

since this is a standard heat equation and "no pressure term is involved"

[lilac barn]

I mean formally speaking it has to satisfy the Laplace equation. By uniqueness it has to coincide with the method I said. Of course on whole space, you can invert by Fouriee transform but it also works in bounded domain as well as long as you can invert the laplacian

[astral vine]

Not really due to the non-zero B.C.

This is not exactly about inverting it

But I get what you mean

[lilac barn]

For me inverting is being able to solve the Laplacian with the boundary condition specified if present.

[astral vine]

Then yes.

[lilac barn]

I think that's what people also use on the literature as well.

[astral vine]

My point was to say this is not the method one can/should use for the magnetic field.

(while being spiritually close)

[lilac barn]

Well when you hit the Leray-operator on the velocity equation it's action on nabla pressure is to kill it. So from this pov, the operator doesn't know if there was a magnetic-field related pressure or not. All that approach tells you is how to propagate divergence-free conditions.

[astral vine]

You need to be able to exchange the Leray projector and then the Laplacian you use to convlude

P D² = D² P

(you want to prove B= PB)

[lilac barn]

Yes I am assuming whole-space setting. My point is to argue that projection argument doesn't tell anything about the existence or nonexistence of pressure. Merely the propagation of divergence-free conditions.

[astral vine]

This is equivalent, except on bounded domaisn they might be some tedious regularity issues

But on the whole space

this is easy to see through the heat flow

You are perfectly right

but again this tells us that the divergence free condition for B not need to be stated in the system of equations since it emerges from it

(and always should actually)

[arctic whale]

If we apply harnack inequality on different annuli we get different constants, how is that helpful? Is there some change of coordinate trick involved?

[arctic whale]

Figured it out: Harnack constant is invariant under scaling of domains

[gentle mesa]

And thats on word 💯

[exotic void]

I am struggling to understand why q is in L^(3/2) C^2 and how does that imply ∂tv is in L^(3/2) L∞ and v is Holder. I couldn't find the correct version of regularity theory used here. Any help would be greatly appreciated. (Here b is just a constant)

[lilac barn]

What's Q?

[exotic void]

Parabolic cylinder, Qr = B(0,r) × (-r^2, 0) ⊂ R^3 × R

[ocean ether]

whats the difference between gateaux and frechet differentiability

[waxen bobcat]

similar to the difference between directional differentiability and total differentiability

[ocean ether]

ah

ok

👍

[lilac barn]

So the laplacian of q is 0 so it's a harmonic function. So it's C2

Now you can look at the equation for partv and make similar assertions

[exotic void]

I think I am not entirely sure what you mean by making similar assertions on part v. Would you mind elaborating?

[trim ledge]

How do you guys solve for v the following PDE (which uses vector calculus)?

$grad(div(\bold{v}))=\bold{0}$

[untold delta]

Kardashiana7

[buoyant pike]

What spatial domain

[trim ledge]

A general spatial domain, in other words R^n.

Let's start with a smaller case, such as R^2.

[buoyant pike]

Do you have any thoughts about this

[lilac barn]

Fourier transform

[trim ledge]

You mean like this?:

$\tilde{f}(k)=\int^{\infty}_{-\infty} f(x)e^{-ikx} dx$

[untold delta]

Kardashiana7

[buoyant pike]

k and x should be vectors

[trim ledge]

$\tilde{f}(\bold{k})=\int^{\infty}_{-\infty} f(\bold{x})e^{-i\bold{k}\cdot\bold{x}} dx$

Like this?

[buoyant pike]

Modulo some latex errors yes this is one solution path

[trim ledge]

[Oh, I forgot the bold the x.]

[untold delta]

Kardashiana7

[trim ledge]

I think I fixed the Latex errors. Actual errors, not so much.

It's been nearly four years since I've gotten my Master's.

So, if we take x = v, and f(x) = div(grad(v)) = 0, then we get:

[buoyant pike]

No don't take x=v

[trim ledge]

Then what do you recommend I take x as? We've only got v in the original PDE.

[buoyant pike]

v is a function of x

[trim ledge]

OK. Let f(x) = v.

Then:

$\tilde{f}(\bold{k})=\int^{\infty}_{-\infty} \bold{v}e^{-i\bold{k}\cdot\bold{x}} d\bold{x}$

[untold delta]

Kardashiana7

[trim ledge]

I still don't see how that would answer the question:

$grad(div(\bold{v}))=\bold{0}$

[untold delta]

Kardashiana7

[trim ledge]

This means that:

$\forall i, \frac{\partial}{\partial x_i} div(\bold{v}) = 0$

[untold delta]

Kardashiana7

[buoyant pike]

Tbh I don't think the fourier transform is that helpful

Have you heard of the helmholtz decomposition

[trim ledge]

I have not, but I have heard of Helmholz' Equation, which is:

$\nabla^2 \phi - k^2 \phi = 0$

[untold delta]

Kardashiana7

[buoyant pike]

No

Unrelated

[trim ledge]

Then no, I have not heard of the Helmholz Deocmposition at all.

[buoyant pike]

The helmholtz decomposition says that for a vector field $\mathbf{v}$, you can write this as the sum of an irrotational part and an incompressible part $\mathbf{v}=\nabla f+\nabla\cross\mathbf{g}$

[untold delta]

Angetenar

[buoyant pike]

Then $\nabla\cdot\mathbf{v}=\Delta f$ because the divergence of a curl is 0

[untold delta]

Angetenar

[buoyant pike]

And $\nabla(\nabla\cdot\mathbf{v})=\nabla\Delta f$

[untold delta]

Angetenar

[buoyant pike]

This lets you characterize solutions to your original pde

Namely, you need your solution to be $\mathbf{v}=\nabla f+\nabla\cross\mathbf{g}$ for $f$ satisfying $\nabla\Delta f=0$ and $\mathbf{g}$ is arbitrary

[untold delta]

Angetenar

[trim ledge]

Thank you very much. If only I knew about Helmholz' Decomposition while I was still at uni... Alas, that never happened.

[buoyant pike]

$\nabla\Delta f=0$ means that $\Delta f$ is constant so $\Delta f=C$ and then you can solve a Poisson equation to get $f$

[untold delta]

Angetenar

[trim ledge]

Thank you very much. I'll try to look up Poisson equations sometime in the near future.

I'm glad I asked a smart question about PDEs.

[quaint herald]

You can also note grad(div(V))=0 => div(V)=C.
If the components of V are y_1,...,y_d, then for arbitrary smooth y_2,...,y_d, we get a unique smooth solution on R^d for specified values of y_1 on the hyperplane {x_1=0} just by integrating the first order ODE for y_1 in x_1.

[pure fulcrum]

hey

looking a bit for pointers

Studying some stochastic control, i've come across situations where the HJB pde can be solved through convex duality using Fenchel transform of the value function (for context)
It feels sightly adhoc but in the process, they were introducing a new semimartingale that they were calling "dual process", I wonder if there's a more general framework of duality for semimartingales that was at play and could shed some light on the solutions

https://www.imperial.ac.uk/media/imperial-college/faculty-of-natural-sciences/department-of-mathematics/math-finance/LIN_HAOYIN_01788182.pdf this is here in 1.2

[unborn gyro]

Just to clarify: what does this notation mean?

also this notation:

[buoyant pike]

C0 in time, Lipschitz or Hs in space

[unborn gyro]

I guess the real question I had was what norm is it equipped with?

[kind olive]

(Spectral theory question, please let me know if it needs to be moved)
Does anybody knows where I can find a reference (book) introducing the Spectral shift function ?

[tepid sleet]

Not a book, but this paper is a decent reference with examples in geometry: https://arxiv.org/html/2404.18422v1

[kind olive]

Thanks !

[kind olive]

I like the introduction, it's easy to read

[distant current]

hello ! I have a question about an explicit computation of the adjoint. take the gradient operator $\nabla : H^1 \rightarrow L^2$ which is continuous of norm 1. it admits an adjoint, but what would its adjoint look like ?

[untold delta]

benjamin noez

[distant current]

maybe the answer is that by Reisz theorem, we don't have much but the existence (and uniqueness but well it's not what i'm looking for) of this adjoint. but sometimes we can compute it so I'm wondering if anyone knows for this example

oh well $\nabla$ should take value in $(\L^2)^n$ where n is the dimension of the space

[untold delta]

benjamin noez

[shell jackal]

Things depend a bit on boundary conditions but assuming that you're working on a compact domain where there are none you're essentially asking the solution to the following problem: given g in L^2 in the domain of the adjoint, meaning that f |-> <nabla f, g> is continuous, how can we characterize the element h of L^2 representing this linear functional? so by definition thats h such that <nabla f, g> = <f, h>, so again by definition h is the negative of weak gradient of g and so g in W^1,2 and (nabla)^* g = -nabla g

(So essentially, the adjoint of the gradient on a compact domain is its negative, because of integration by parts)

[distant current]

the domain of the adjoint is not the whole L^2 ?

I don't see what would be the domain then. why would nabla g make sense ?

[shell jackal]

It kind of depends on what you want by an adjoint I guess

[distant current]

? my definition of adjoint gives $\langle \nabla f, g \rangle = \langle f, \nabla^* g\rangle$ for $f\in H^1$, $g\in L^2$

[shell jackal]

Like, if you have a bounded map from a Hilbert space to itself, T: H -> H, then the adjoint is given by taking the dual map H^* -> H^* and then identifying H with its dual using the Riesz representation right

[distant current]

yes yes I agree

[untold delta]

benjamin noez

[shell jackal]

So when you have a map T: H_1 -> H_2 between distinct Hilbert spaces and you take the dual to get H_2* -> H_1* the riesz isomorphisms on the two spaces are different and arent necessarily like, easily compatible

[distant current]

but if T:H ->K is bounded, then it's adjoint is a map T^*:K->H

[shell jackal]

Like you have to unwind more isomorphisms I guess to think of the adjoint as just going from K to H

basically when you're working with one Hilbert space H you can freely identify it with its dual but when you're working with multiple Hilbert spaces this starts to become kind of nasty

[distant current]

it's also by Riesz theorem that we can. define the map from K to H

I dont see the issue if it's doable

[shell jackal]

well its certainly doable its just going to be problematic for you if you dont know what the dual of H^1 (usually this is denoted by H^-1) looks like in its own right

[distant current]

oh

ok I have a definition for H^-1 actually

you mean that I should define the adjoint of nabla as a map from L^2 to H^-1 ?

[distant current]

it's in terms of Fourier transforms for me

[shell jackal]

Like theres an obvious inclusion of H^1 into L^2 right? so if you identify L^2 with its dual you get an inclusion of L^2 into H^-1. so we have H^1 subset L^2 subset H^-1, but obviously taking H^-1 = H^1 in the naive sense leads to the absurd conclusion that H^1 = L^2 so clearly this is not a good way to think about things

[shell jackal]

Yeah this is one option

(technically (L^2)^n)

[distant current]

let's juste take one dimensional space

n=1

[shell jackal]

the other option is to think of the gradient as an unbounded operator from L^2 -> L^2, with dense domain H^1. then you can talk about the adjoint map L^2 -> L^2, which will be densely defined on some subspace of L^2

[distant current]

I see what you mean thanks

[shell jackal]

If you think of things this way, the adjoint map you will get is going to be defined on H^1, and will be given by the negative of the gradient

for the reason i outlined above

so thats gonna be nabla*: H^1 -> L^2, and it will extend to a map L^2 -> H^-1, which is the adjoint in the other sense

does that make some amount of sense

[distant current]

so if I understand properly you'd define the adjoint from H1 to H^-1, extend the domain by density of H1 in L2 and then use the identification H^-1≈H^1 ?

or am I completely mistaken ?

[shell jackal]

Umm im not sure what you mean exactly

[distant current]

ok Ill re write it betterly

$\nabla: H^1 \rightarrow L^2$ as an adjoint $L^2\rightarrow H^-1$ that we can define on H1. we extend this definition by density of $H^1 \subset L^2$

[untold delta]

benjamin noez

[shell jackal]

Yeah this sounds right

[distant current]

then use the identification $\H^1 =H^{-1}$ to get an element of H1

[shell jackal]

Oh

I would not use the identification personally

I mean you hypothetically could

[untold delta]

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

[shell jackal]

but its not really the "right" way to think about it

[distant current]

you have a particular reason for that ?

it's a preference ?

I mean I have a definition for H^-1 so I know what's these éléments look like, and I know how to identify H1 and H^-1. since I don't like the space H^-1 I'd préfère use the identification to work with a nicer space

[shell jackal]

Well the problem is that if you have T: H^1 -> L^2 and you take T*: L^2 -> H^-1, and you have like, T*y right? after identifying this with some z in H^-1, then <x, z> is NOT going to equal <x, T*y>

[distant current]

ho

right

[shell jackal]

and so its not gonna equal <Tx, y> either

[distant current]

yup

so right

I see

[shell jackal]

so if you try to define the "adjoint" from L^2 back to H^1 then the map has no nice formal properties at all

its just a map

you cant say <Tx, y> = <x T*y> if you do that

[distant current]

yes I see

thank you very much !

for your patience

and explanations

[shell jackal]

unfortunately this is one of those situations where for practical purposes you kinda have to deal with H^-1 on its own terms and not just take the isomorphism with H^1

you really need to think of it as a space of distributions

[distant current]

yeah ok it's unpleasant but I understand why it has to be so

[shell jackal]

bc differential operators usually go between different Hilbert/Banach spaces

(they lower the order)

[shell jackal]

no problem, hope it helped

[distant current]

a lot

[quaint herald]

When ironing out these details for the first time it can be useful to be really explicit, like subscript your inner products with the space the IP is taken in, introduce explicit notation for the Riesz "identifications" and the "inclusions" of higher reg spaces in lower. Working instead with embeddings rather than set inclusions clears up the confusions that can occur when you have several identifications floating around at once.

It's like a notoriously confusing point.

[mellow cloud]

anyone got good intuition for sobolev capacity?

what i know so far is that it’s an outer measure finer than lebesgue measure and is somehow induced by sobolev norms

[flat crow]

I'm reading about the existence of solutions to 3D Euler, and I don't understand how to show the inequality
\begin{equation}
|\langle (u \cdot \nabla) u, u\rangle_{H^s}| \leq C |u|{H^s}^2 |\nabla u|{L^\infty}
\end{equation}
for $s > \frac 52$. From what I understand, the proof of this inequality should hinge on the fact that $H^s$ is an algebra for $s > \frac 32$

[untold delta]

jamiecjx

[lilac barn]

Did they not provide you a proof for this? The proof requires some handling of the commutator at the Fourier side and an application of Galgiardo Nirenberg iirc

Most likely the inequality was proven earlier anf they're referencing it

[flat crow]

This question came from this section of a survey on general results about 3d euler

so I haven't seen a proof of it

[lilac barn]

Okay check Theorem 2.5 of Vlad's book

[flat crow]

ok I understand why it's true now, thanks for the reference

that was a lot more involved than I expected

where is my beloved viscosity term :((

[lilac barn]

Removed in recent patch because too OP

[flat crow]

after some more searching, I found this proof in another piece of literature (reproduced in condensed form) which in my head makes clear why such an inequality can hold

the only potentially troublesome term when beta=0 vanishes magically

[lilac barn]

It's L2 though at the end. That's much easier to get than Linfty

A much more simpler idea is this: The worst two extremes of this term is when the s-derivatives fully hit on u or nabla u. When they fully hit on u, you immediately get the desired inequality. When they fully hit on nabla u, then your divergence free condition let's you kill this term.
So the remaining terms are just interpolation of these two and thus just the required bound.

[flat crow]

Yeah this makes sense to me

and oops I should have mentioned later on I found out I only needed $(u\cdot\nabla u, u) \leq C |u|_{H^3}^3$ for existence/uniqueness for euler and not L inf

[untold delta]

jamiecjx

[lilac barn]

Linfty is usually used to get a more precise maximal time existence criteria

But yes, it's not needed

The book I mentioned by Vlad does everything you need

[flat crow]

funnily enough the next question on my exercise sheet is just "prove that at the time of maximal existence $T$, $\int_0^T | \nabla u|_{L^\infty} \dd s = \infty$ and that the integral is finite for any $t < T$ or something like that

[untold delta]

jamiecjx

[sinful herald]

I was reading Galdi's Book on Navier Stokes. I am having doubt in proof of following theorem (attached relevant part of proof ) can someone please help me, here Omega is an Exterior domain and u is in L1_loc with all derivatives of order 1 in Lq and 1<=q <n, I did not understand how they prove existence of r_m in above, also grad*u is a projection of gradient on sphere, and I think there should not be n-q-1 in the second inequality of II.6.16, but I am not sure

[wintry delta]

Hello guys, I was preparing to exam and I have troubles understanding solutions of last year exercises. Here is first part of solution for 1.a. I pointed out 3 questions:

1. Why we have shock at x=-1 with ro left and ro right both equal to 0? Shouldn't it be 0 from ro left where x is less than -1, so it's not in (-1, 0) and 1 for ro right, where x is inside (-1, 0)?
2. Why we replaced y/8 by x/8t? Where did y = x/t came from?
3. What is happening at (3)? Why we make x_2 equal to x_1 and what do we get from it?

[wintry delta]

I understand even less for (b), almost nothing is clear to me

[serene mica]

I was reading this paper https://arxiv.org/pdf/2402.07534 and was confused about the portion I have screenshotted below. This can be found on page 4 of the paper. Anyway, I think everything makes sense up until the very final sentence. u clearly is in L^{\infty} and H^1 since it is smooth on a compact set, the compact set being T^d, but Im not sure how they conclude u = 0. We also have the condition that \int u = 0, but even with this, I have no clue. Was hoping someone else could see what I am missing here

[astral vine]

I am not sure what you mean but this is deduced from elliptic /Parabolic (maximal-) regularity theory

he used the claim "There exists a unique mild solution u in C(0,T;L^p(T^d))"

This is not something that could be directly deduced by basic/elementary considerations.

[lilac barn]

To get u=0, write down the energy balance for this equation or equivalently, steady Navier-Stokes (don't remove the pressure). You will see that the condition boils down to the gradient vanishing everywhere and then combined with average being 0, it implies u identically vanishes.

[astral vine]

To do this you need to show that the solution is H^1

[lilac barn]

They said in their post they have that u is H1

[astral vine]

The claim v is smooth is not clear to me

I cna prove it lies in H^1nL infty

But not the smoothness

[hidden coral]

cross-post from #book-recommendations, but where does one go to learn about fractional sobolev spaces?

[frank tide]

a first course in fractional sobolev spaces is a good book

giovanni leoni

[hidden coral]

thanks

[sand echo]

Another good reference is https://arxiv.org/pdf/1104.4345

[flat crow]

https://en.wikipedia.org/wiki/Lions–Magenes_lemma

In this article the Lions-Magenes lemma is stated in terms of three spaces $X_0 \subset X \subset X_1$ where $X_1 = X_0^*$ but I have only seen this lemma for spaces $V \subset H \subset V'$ where $H = L_{0,\sigma}^2$ and $V = H_{0,\sigma}^1$

Is $X_0 \subset X \subset X_1$ supposed to be interpreted as a Gelfand triple? I don't see how this statement can make sense if it isnt e.g. $H^1 \subset H^{\frac 12} \subset H^{-1}$

[untold delta]

jamiecjx

[lilac barn]

Yes they're supposed to a Gelfand triple

[untold delta]

Zanarcane

Zanarcane

Zanarcane

Zanarcane

[shadow topaz]

Could anyone help me with the very 1st pde please ? I tried variable séparations but I wasn't succesful...

[native smelt]

I'm working through Evans (and it's my first time learning PDEs so sorry if it's a stupid question :) ), and now I'm reading about solutions to the Poisson problem in the unit ball and half plane using Green's functions. Here are a few questions:

1. He introduces Green's functions as:
   $$G(x,y) = \Phi(y-x) - \phi^x(y)$$
   where $\phi^x$ is an auxiliary function such that
   $$
   \begin{cases}
   \Delta \phi^x = 0 & \text{ in } U \
   \phi^x = \Phi(y-x) & \text{ on } \partial U.
   \end{cases}
   $$
   And my question is, whether any such $\phi^x$ that is defined everywhere in $\bar U$ works?

2. Second, (and maybe this will be answered later in the book), is that so far this has only been used to solve Poisson problems, but I read that this can be used in many other linear non-homogenous problems $L u = f$. Do Green's functions have the same form in those cases?

thanks

[untold delta]

{Wilhelm}

[buoyant pike]

1. What do you mean by "any"? You'd be hard pressed to find multiple phi^x that work

2. With appropriate modifications yes

[native smelt]

I mean, suppose you find a $\phi^x$ that satisfies the BVP and is defined on all points of the domain of interest, then does it necessarily (alongside the fundamental solution) define a valid Green's function in the sense that $L G = \delta_x$?

[untold delta]

{Wilhelm}

[buoyant pike]

Well, what do you think

[native smelt]

I'll re-read that section of Evans, but from what I remember I feel like it would as long as the inverse differential operator $L^{-1}$ (if it exists), is of a certain form. Namely an integral of the form $L^{-1} f = \int G f dy$

[untold delta]

{Wilhelm}

[native smelt]

Which I think holds at least for the Poisson problem? But maybe also for other types of BVPs...

The proof in the section I'm in uses the multidimensional integration by parts formula, and just assuming that $\phi^x$ satisfies the BVP above is enough for stuff to work out nicely to
$$
u(x) = -\int_{\partial U} g(y) \frac{\partial G}{\partial \nu} d S(y) + \int_U f(y) G(x,y)dy
$$

So yeah, I suppose any such solution would work for a Poisson problem. Is that correct?

But also this proof so far relies on the particular form of $L = \Delta$

[untold delta]

{Wilhelm}

[native smelt]

If anything I said is wrong I appreciate a correction :)

[mortal slate]

I know nothing of pdes. Is there a solution to <grad f, u>(f'' + f) = g where u is a vector and g is a fixed function?

Like is this a well known problem?

[sinful herald]

I want to show that $0$ is not an eigenvalue of $A_r : D(A_r) \to H_r(\Omega)$. I was trying it this way, Suppose $0$ where an eigenvalue then there exists $u \neq 0$ such that $A_r u = 0$ which corresponds to existence of $p \in L^r_{\text{loc}}(\Omega)$ with $\nabla p \in L^{r}(\Omega)$ which satisfies in $\Omega$
\begin{equation*}
\begin{split}
\Delta u + \nabla p = 0 \
\mathrm{div} , u = 0 \
u|{\partial \Omega} = 0
\end{split}
\end{equation*}
Then I took divergence to obtain in $\Omega$
\begin{equation*}
\begin{split}
\Delta p = 0 \
\frac{\partial p}{\partial n} = 0
\end{split}
\end{equation*}
Which gives that $p$ is constant and hence we obtain $u = 0$ Is this correct ? Where $A_r$ is stokes operator, $H{r}$ is divergence free space and $D(A_r) = D(\Delta) \cap H_{r}$

[untold delta]

Math&Tea

[lilac barn]

How do you obtain u is 0 from p is constant

[sinful herald]

p is constant then $\nabla p = 0$

[untold delta]

Math&Tea

[sinful herald]

p is constant then $\nabla p = 0$, so I used the stokes equation, which is now, $\Delta u = 0$ with $u|_{\partial \Omega} = 0$

also here $\Omega$ is exterior domain...

[untold delta]

Math&Tea

[lilac barn]

Yeah looks correct. You should also discuss why p = constant are the only solutions

[sinful herald]

Thanks. Yes, actually I remember seeing that only solution to that p equation are constant somewhere earlier...

that's why I assumed that...

[lilac barn]

Put the Dirichlet boundary condition and show that the solutions are unique

[astral vine]

Why the normal derivative of p is 0 ?

Shouldn't it be

[untold delta]

Functionanatolysis

[astral vine]

a priori nothing tells you that the normal component of the laplacian of u vanishes on the boundary

Furthermore, unless your domain is smooth, your definition of the Dirichlet-Stokes operator might be wrong, be careful.

(in general the (gradient of the) pressure term might not lie in any Lr space)

[sinful herald]

I understand my mistake here, thank you very much for pointing it out

[minor mulch]

for a nonlinear heat equation of the form $$\partial_t u = \partial_x^2u + u\phi$$ for some $\phi : \bR^2\to\bR$, is there any way to heuristically guess the regularity of $u$ in terms of that of $\phi$?

[untold delta]

memorylessfunctor

[minor mulch]

like how in the linear case, the solution gains two derivatives in space and one derivative in time compared to the forcing term

[glass jungle]

https://arxiv.org/abs/2501.08353 Is this real? I'm somewhat skeptical, but I don't know enough about PDEs to tell.

[minor mulch]

it references millenium prize in the abstract

so like… definitely not

[glass jungle]

Ok

Definitely some suspicious things going on with it

[minor mulch]

the first step of the proof of corollary 4.1 seems just wrong

they start with smooth initial data

and then just claim it is actually analytic

[lilac barn]

It is definitely wrong. A more immediate way to see this is the same Corollary 4.1 claims that there exists a solution with 0 energy at all times. In particular it is almost surely 0 at time 0, which is precisely the initial condition.

[glass jungle]

Ok

[lilac barn]

I dont think we can generally say much about regularity of phi without smallness on phi.. For example, I can show that if the L2 norm of phi remains small then u is L2(H2).
My guess is, you have the same regularity conditions as f, but u would need smallness on all derivatives of phi.
Alternatively, you can trade in small+regularity with extreme amounts of integrability.
Example: Using Fourier analysis, we can also show that if e^int phi is say L2 then the solution is analytic.

[minor mulch]

does f mean the initial data u(0,x)? or something else

[lilac barn]

By f I mean arbitrary forcing as opposed to uphi

[minor mulch]

ok makes sense

do you have any intuition about what should go wrong if phi is large, say in L2?

im guessing its just not possible to say anything in general

[untold delta]

cocat

[untold delta]

cocat

[untold delta]

cocat

[lilac barn]

@minor mulch

[minor mulch]

thanks, i’ll read and reply a bit later

[lilac barn]

Just the last image is sufficient

[minor mulch]

i dont understand the estimates in the 3rd or 5th display

shouldnt the 5th display have the H1 norm of u^2 on the RHS?

im assuming you use cauchy schwarz and then bound the L2 norm by H1

oh i see it should be the square of the H1 norm

and it follows from using the L\infty bound

[lilac barn]

Yes, it should be ^2 the rest is unaffected.

[minor mulch]

can you explain the 3rd display a bit more? i understand you are writing the LHS as <ut, u\phi>

is this just some basic inequality about inner products?

[lilac barn]

Wdym by third display?

[minor mulch]

like the 3rd math display here

[lilac barn]

We have <phiu,ut>, then I use Cauchy schwarz with epsilon

ab <= Ca^2 +eb^2 for large C and small e

[minor mulch]

ok sure

[minor mulch]

thanks @lilac barn this is illuminating

[lilac barn]

Yw!

[verbal nebula]

The first issue I see is the pre-print is explaining multi-index notation

Damn if only NS was solved, then I could finish my thesis faster

[flat crow]

theres also this gem in the paper

[meager dune]

lol this is what i was going to say

an actual solution would be very modest lol

[meager dune]

lmao

[minor mulch]

i think it depends a bit on the author, but it is definitely true that the focus would be on the problem rather than the prize money

the way the paper is structured is hilarious

referencing “millenium prize problem (A) and (B)” like its a homework submission

[meager dune]

yeah lol

[minor mulch]

but i think usually the quickest way to determine if these things are wrong is to check if the bibliography has more than 10 references

and not counting undergraduate textbooks

[dire fjord]

Hey has anyone participated in the reading group for Evan’s PDE last year?

I am interested with engaging with them so I would like to know if there are any talks to holding those sessions again this year or any documents containing required reading or suggested exercises.

Any advice or input would be greatly appreciated

[manic tinsel]

Sorry that this isn’t really a PDE question but there isn’t an advanced ODE channel. Is there a faster way to solve non-homogenous ordinary differential equations than via this method? I’ve got a qualifying exam coming up and I’m not sure I can work fast enough to do this kind of problem given the past example I’m looking at.

[fathom schooner]

Yeah theres faster methods

Depends on the problem

[manic tinsel]

What about in cases where the functions of x and y are linear, such as this problem I pulled from an old qualifying exam?

[fathom schooner]

Wdym

[manic tinsel]

Problems like this. Is there a method for finding solutions to problems of this type other than the method I posted before?

[fathom schooner]

Hmm

[sinful herald]

Dose any one know a reference where it's proved that $\Delta : W^{s+2, p} \to W^{s,p}$ is an isomorphism, (including negative sobolev spaces)?

[untold delta]

Math&Tea

[sinful herald]

I dont think its a channel to share this staff

[stable shale]

I'm not sure it proves that exact statement, but the book 'Functional Analysis, Sobolev spaces and Partial Differential Equations' by Haim Brezis covers a lot of stuff on Sobolev spaces. If you can't find the solution in there, you will likely find one of his references that does have a solution.

[sinful herald]

I dont think it deals with negative or fractional sobolev space, but I will take a look if it refers to anything of that sorts

[stable shale]

Ah no it probably doesn't. But yeah defo have a look at his references for more specialised texts

[sinful herald]

yes, I will go through the reference...

[astral vine]

On the whole space this is wrong

and on bounded domain this true only if the domain has some regularity

and that you have a boudnary value involved

such has Dirichlet, neumann, robin

On the Laplacian and its (non-)invertibility on the whole space : https://math.stackexchange.com/questions/4738465/properties-of-the-inverse-laplacian-operator?noredirect=1&lq=1

[astral vine]

The reason this fails for the whole space is somewhat not that hard to grasp but really not trivial, and relies on a lot of technicalities and concepts.

In the case of Bounded domains domains the exponents s for which this is true, will depend on the regularity of the boundary. For instance if the boundary is C², say with dir. BC. then this is true for all s between -1+1/p and 1/p for all 1<p<+oo.

If the domain is Lipchitz, this is true only for Sobolev space such that the Sobolev index is "not far from L²" (not true for all p and all s (for instance, this requires p to be s.t. 3/2-eps<p<3+eps, for some eps>0).

If the domain is C1,a, 0<a<1, then this is true for all 1<p<+oo, all -1+1/p<s<-1+a+1/p.

-The case of bounded C² domains can be done by hand from the behavior on the flat half-space (which share issues similar to what happens on the whole space), you can prove the result for -1+1/p<s< eps, for some eps>0. (To reach eps = 1/p is super deep and technical actually)
-The case of bounded Lipschitz domains is super hard and Technical : see Fabes Mendez Mitrea 1999, and Jerison and Kenig 1995.
-The case of C1,a domains is a bit less hard than the previous one but requires more technology, and is no where to be seen except in Maz'ya and Shaposhnikova's book, butonly for the Dirichlet Laplacian (actually their assumption on the boundayr is sharper and the result is hidden in the book.)

[sinful herald]

Thanks a lot for the references, I will look through these

[mortal slate]

This isa stupid question but is there a name for an ode of the form x'(t)^2 + x''(t)x(t) = g(t)

Basically where we have terms with derivatives being multiplied to each other

I just want existence of something like this but I don't know where to start. I guess applying Picard Lindelof would be nice but this is about all I know for odes.

This is a PDE's chat, but I asked odes and never got a response.

[flat crow]

why not integrate the equation wrt t twice

[lilac barn]

(x'x)'=g and thus (x^2)''=2g.

[manic tinsel]

Another question from an old qualifying exam. Can anybody point me to a source or an example solving this sort of problem? I can't find one in the textbook I am using to study

[sweet osprey]

what does it mean for a domain to be "regular for the dirichlet problem"? just that the dirichlet problem has a unique solution on that domain?

[prisma marsh]

gilbarg and trudinger ch 2.8 is your friend here

[arctic whale]

Gilbarg Trudinger seconded.

[sweet osprey]

might read later, ty :)

[astral vine]

There can be multiple definition. One is just about existence and uniqueness. One is about existence uniqueness, and the solution gains actually two derivatives compared to the forced term.

[reef tartan]

I am not sure in which subcategory to out this so I ll just put it here:
I have this ode:
y'' + (2+tanh(x/epsi)y =0
where epsi is really small. I want to study the effect of epsi on the solution. Some intuition about this:
At - inf this equation behaves as exp(-ix) + exp(ix) and at + infinity it behaves as exp(-ix/3) + exp(ix/3). I am not interested in the exact solution but only in how does epsi effects the low orders. Not looking for answers but for techniques to solve similar problems. I have been looking at series for a few hours now but it gets messy very quickly.

[harsh veldt]

Hello. I have a basic question. Is it immediate that $D = {u\in H^{2m}(\Omega)\vert u\vert_{\partial\Omega} = g}$ is dense in $H^{2m}(\Omega)$? I feel like it is but I want to make sure.

[untold delta]

emphatic_wax

[exotic lava]

you mean the trace operator?

[harsh veldt]

I want to argue that $D$ is dense in $H^{2m}$

[untold delta]

emphatic_wax

[exotic lava]

well it isn't

it's a closed hyperplane affine subspace

[harsh veldt]

So this means that $H^{2m}_0(\Omega)$ is also not dense in $H^{2m}(\Omega)$?

[untold delta]

emphatic_wax

[exotic lava]

yes

wait I said hyperplane

[exotic lava]

it isn't a hyperplane it's just a closed affine subspace

H_0 is a closed subspace

[harsh veldt]

okay okay. I might need to revisit this problem haha I'm trying to prove that an operator generates a C0 semigroup lol

[buoyant pike]

Ask in #modeling and I will answer there

@reef tartan

[astral vine]

If you have a boundary say lipschitz, smooth compactly supported function (away from the boundary) , are NEVER dense in H^s, as long as s>1/2..

[astral vine]

With non-homogeneous boundary condition, in general Elliptic equations does not induce an operator, due to the lack of linearity. What kind of Elliptic operator are you dealing with, I think I may be of help if needed.

[arctic whale]

Does this follow from Green's representation formula? I am trying to derive it using the representation formula but I am not getting there for some reason.

[prisma marsh]

id mimic the proof of theorem 2.1 directly instead

[arctic whale]

Do you start with computing $\int_{\partial B} \frac{\partial }{\partial \nu}(u - N(f)) dS$?

[untold delta]

Co-aerA

[arctic whale]

where N(f) is the newtonian potential of f

[prisma marsh]

start with the fact that $\int_{\partial B_{\rho}}\frac{\partial u}{\partial \nu} ds = \int_{B_{\rho}}\Delta u = (\geq, \leq)\int_{B_{\rho}}f$

[untold delta]

razor

[arctic whale]

But that doesn't do the trick no?

[prisma marsh]

look at p. 14 in G-T

[arctic whale]

the trick in the original proof is that the laplacian is >, <, =, zero so you get that some function of the radius is either decreasing, increasing, or constant wrt the radius

[prisma marsh]

integrate from rho from 0 to R

[arctic whale]

or atlest that's how I understand the proof in GT

[prisma marsh]

change limits in the integral of f

you get something like $$R^{1-n}\int_{\partial B_{R}}u dS - n\omega_{n}u(y) = (\geq, \leq)\int ^{R}{0}\rho^{1-n}\int{B_{\rho}(y)}fdx d\rho$$ then change limits in that last integral

[untold delta]

razor

[prisma marsh]

you then need to integrate one more time to get the solid ball version

[arctic whale]

yeah that also seems that it might work; I will give it a try. Thanks a lot

[wintry delta]

I don't get it. Why is it zero condition if trace in interval 0, 1/2?

[buoyant pike]

Not entirely sure but maybe 5 and 6 should be switched in the sentence?

[wintry delta]

Not sure

[celest fiber]

i wonder if someone here might be able to help?

[buoyant pike]

What is the difference between hydrodynamic turbulence and wave turbulence

[lilac barn]

The main difference would be in energy cascade: Hydrodynamic/Kolmogorov is via vortex stretching+dissipation whereas in wave turbulence, it is via nonlinear interactions of different wave modes.

[buoyant pike]

Hmmmm I see

[candid token]

,tex how come the $W^{k,\infty}$ norm is defined as such? all the other $W^{k,p}$ norms are the $p$-norms on the $|D^\alpha u|{L^p}$, so wouldn't it make sense to say $|u|{W^{k,\infty}(U)}= \sup_{|\alpha| \leq k} \operatornamewithlimits{ess,sup}_U |D^\alpha u|$?

[untold delta]

minitarrasque

[wind mortar]

what is ess

oh, i see

i mean i just don't see why it would be the supremum over the supremums

that would sorta only involve one derivative, right?

when for p < infinity all of those cases involve multiple derivatives?

i guess it comes down to...

you have a sum because it's multidimensional

so since it's still multidimensional even when p is infinity, you should still have a sum instead of just a supremum

[candid token]

not if |alpha| > 1

[wind mortar]

no, but it would only be one term?

no sum?

[buoyant pike]

I don't really understand what your question is, all the W^k,p norms involve sums over derivatives of various orders

[restive frost]

it doesn't matter which definition you use because they only differ by the choice of norm on a finite dimensional space which are all equivalent

for example the wikipedia page uses the definition you suggested

[candid token]

sure, i kind of figured that it didn't really matter

it just seemed sort of an odd choice to me

[ornate peak]

I’ve never studied PDEs and my only background here is a first course in ODEs, and some Fourier analysis, but I’ve recommended some physics problems lately and I’m wondering if anyone here knows of like a standard reference/book for learning how to solve these types of problems (this is an interest, I’m not taking a course).

I’m not even sure how one solves a boundary value problem. These are the sorts of problems I’ve been recommended

And if this is the wrong channel for this please let me know.

[buoyant pike]

Strauss has a book on PDEs that is a good place to start

#odes-and-pdes is probably a better fit as well

[frank tide]

richard haberman - applied pde with fourier series and boundary values problems might be a good place to look depending on ur background

lots of examples

[ornate peak]

Thank you for the suggestions. I study abstract harmonic analysis, so this stuff is a bit foreign to me.

[untold delta]

<Andrew>

<Andrew>

<Andrew>

<Andrew>

<Andrew>

<Andrew>

<Andrew>

[astral vine]

a priori mu is arbitrarily large so chose it so that at least a+\mu >0.

Two comments : it seems you forgot derivatives in some place, and that you forgot some assumption on p(x) ? Am I right ?

[untold delta]

<Andrew>

<Andrew>

<Andrew>

[astral vine]

Then everything is fine

notice that coercivity might have multiplicative constants in your process that may depends on a, p, etc.

[untold delta]

<Andrew>

<Andrew>

<Andrew>

[astral vine]

yeah

Beta might depend on a, p_0 and the mu you did chose

What you have to prove is that B(u,u) + mu |u|^2 > beta |u|H^{1}

[untold delta]

<Andrew>

[astral vine]

choose mu = |a|+1, beta =p_0+1 should satisfy the estimate

[void flame]

Could I get some help with this Q (the last two, about the Fourier transform of sin(x^2) and |x|)? We haven't learned about distributions so I think we would just evaluate the integral and pass to the limit. So far ive tried a few factors but the integrals either diverge or is very difficult to evaluate.

[buoyant pike]

Well what have you tried

[quaint herald]

The factor mentioned leads to convergent integrals for eps > 0. Evaluation hints:
|x|: Break the integral into two pieces
sin(x^2): Cauchy's theorem

[celest fiber]

is there someone here that is familiar with Laplace PDE's to give me some feedback on the problems i've solved?
I've learned this alone and I've got a test tomorrow and i really need confirmation that I'm doing good

[sweet osprey]

no need to ask. just send them and see if anyone answers

[celest fiber]

i've got the following problem:
$$\begin{cases}
\Delta u\left(x,y\right)=0 & 0<x,y<\pi\
u_{x}\left(0,y\right)=u_{x}\left(\pi,y\right)=0 & 0\leq y\leq\pi\
u_{y}\left(x,0\right)=u_{y}\left(x,\pi\right)=x & 0\leq x\leq\pi
\end{cases}$$

using separation of variables we get that $$\frac{X''\left(x\right)}{X\left(x\right)}=-\frac{Y''\left(y\right)}{Y\left(y\right)}\triangleq-\lambda$$

then i find the eigenfunctions for $X_n = \cos(nx)$ with the eigenvalues of $\lamda _n = n^2 | n\in \mathbb{N} _0$

now solving the Y i get that:
$$Y\left(y\right)=\begin{cases}
A_{+}\cosh\left(ny\right)+B_{+}\sinh\left(ny\right) & n>0\
A_{0}y+B_{0} & n=0
\end{cases}$$

that means my general solution is of the form

$$u\left(x,y\right)=\frac{A_{0}y+B_{0}}{2}+\sum_{n=1}^{\infty}\cos\left(nx\right)\cdot\left[A_{+}\cosh\left(ny\right)+B_{+}\sinh\left(ny\right)\right]$$

now i use the initial values $u_{y}\left(x,0\right)=u_{y}\left(x,\pi\right)=x$ to get that $$A_{0}=\pi,\quad B_{+}=\frac{2\left[\left(-1\right)^{n}-1\right]}{\pi n^{3}}$$

doing the same with the second condition i get that $$A_{+}=\frac{2\left[\left(-1\right)^{n}-1\right]\left(1-\cosh\left(n\pi\right)\right)}{\pi n^{3}\sinh\left(n\pi\right)}$$

and plugging everything back in i get the solution:

$$u\left(x,y\right)=\frac{\pi y+B_{0}}{2}+\sum_{n=1}^{\infty}\frac{2\left[\left(-1\right)^{n}-1\right]}{\pi n^{3}}\cdot\cos\left(nx\right)\cdot\left[\frac{\left(1-\cosh\left(n\pi\right)\right)}{\sinh\left(n\pi\right)}\cdot\cosh\left(ny\right)+\sinh\left(ny\right)\right]$$

[untold delta]

Henry_quite_hungry

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<recently read> \lamda 
                       
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                                                   _n = n^2 | n\in \mathbb{N...
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misspelled it (e.g., `\hobx'), type `I' and the correct
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and I'll forget about whatever was undefined.```

[astral vine]

this better fits #odes-and-pdes

That being said I have doubts on A_0

When you produce a solution,especialy of this form, you should provide calculations

[celest fiber]

i've found the respective cosine series for x and got to this: $$x=\frac{\pi}{2}+\sum_{n=1}^{\infty}\frac{2\left[\left(-1\right)^{n}-1\right]}{\pi n^{2}}\cos\left(nx\right)$$

[untold delta]

Henry_quite_hungry

[astral vine]

ok so you just identify

fine

[celest fiber]

what? what do you mean by identify?

[astral vine]

no worries, why you didn't put a value for B_0 ?

[celest fiber]

since we only have conditions for the derivative it's impossble to get a value for the free term, we simply dont have information on that

[astral vine]

Oh I tought it was the Dirichlet problem not the Neumann pb

my bad

[celest fiber]

i've wrote in the other channel #odes-and-pdes what my problem was with my solution, if you think it's better to talk here thought i'll paste the rest of the question

[astral vine]

k

[celest fiber]

(at the end of my solution it's just sinh(ny) it didn't fit)
but the solution i found online for this problem was this one
while these are close they aren't exactly the same and i want to understand why and where might i've done a mistake

[mortal slate]

is anyone here well suited with Monge Ampere equations existence and regularity?

$$\det(\nabla^2 u + A(x)) = f(x, u, \nabla u)$$ I want to find existence of solutions to

[untold delta]

Brayden

[buoyant pike]

Is the standard existence theory not sufficient?

[mortal slate]

A(x) is a term I have

[buoyant pike]

Do you know anything about A(x)

[mortal slate]

here is the entire equation I wish to solve

$$\det (\nabla^2 u + \varepsilon \nabla^2 \varphi)(\nabla u + \varepsilon \nabla \varphi) = g.$$

[untold delta]

Brayden

[mortal slate]

$\varphi$ is given, so is $g$

[untold delta]

Brayden

[mortal slate]

everything here is C^\infty.

[buoyant pike]

Hmmm is this $f=\frac{g}{\nabla u+\eps\nabla\varphi}$?

[untold delta]

Angetenar

[mortal slate]

yes

move the epsilon to the other term, I guess. It doesn't matter.

am I tripping or can one just do a change of variables

[buoyant pike]

Ok if you write $v=u+\eps\varphi$ you get $\mathrm{det}(\Delta v)=f(x,v,\nabla v)$

[untold delta]

Angetenar

[buoyant pike]

To which you can apply the standard theory can't you

[mortal slate]

Even if we don't know that u exists apriori?

[mortal slate]

also why can you write this because grad u is a vector

[buoyant pike]

Is g a vector then?

[mortal slate]

yes

so we just do this coordinate wise then?

[buoyant pike]

Yeah

[mortal slate]

word

[buoyant pike]

I mean the det part is a scalar

So if the two sides are equal then they are equal componentwise

[mortal slate]

but also in my question above why can we do this when u doesn't exist apriori

[buoyant pike]

Solving $\mathrm{det}(\Delta v)=f(x,v,\nabla v)$ and then writing $u=v-\eps\varphi$ doesn't require $u$ a priori

[untold delta]

Angetenar

[mortal slate]

oaky

and if everything is C^\infty then the boundary value isn't really important?

[buoyant pike]

I mean the boundary value affects the solution

[mortal slate]

Better question, where should I study this theory? I kinda just want to scrape the basics but I'm using Figallis book right now and he doesn't go in a lot of detail.

or is Figalli the way to go

[buoyant pike]

https://bookstore.ams.org/view?ProductCode=GSM/240

Have you looked at this

[mortal slate]

not yet

thanks

[mortal slate]

So I see this theory for R^n domains, is there Monge Ampere equations for the sphere?

Or is there a general way to convert pdes from the sphere to R^n?

[buoyant pike]

You can always map the sphere to R^2 with a suitable change of coordinates but I do not recommend this

[buoyant pike]

The main thing to keep in mind is that you don't have a boundary

So f will determine the solution

Likely not all f will admit solutions

[mortal slate]

f is C^infty on the sphere

[buoyant pike]

Hmmmm ok

So

For existence, can you do something variational

For uniqueness, you should be able to do something with a maximum principle

[mortal slate]

Regularity?

[mortal slate]

Thanks for the help!

[buoyant pike]

There shouldn’t be any regularity issues

[heady silo]

can someone help me understand the fundamental reason why separation of variables works as a method of solving linear PDEs or point me to a source that explains it, basically I understand why assuming a form of the solution helps finding loads of solutions and then take linear sum of the separated solutions (sum f_i(x)g_i(y)) and you got yourself a whole load of solutions to your linear PDE like I get that but why is it a complete set of solutions?? I dont get it

the Y (spherical harmonics) are solutions to the angular part of the spherical laplace equation in spherical coordinates and the r is the radial part the source states any solution T to this spherical laplace equation can be written as a sum like this but I just dont understand, I know the Y form a complete basis for the continuous functions on the sphere and I guess the radial part do aswell since its polynomials but yeah I just dont get it

I suspect there is a simple elegant answer that has to do with the completeness of the spherical harmonics but im just missing a key piece, like if we have say f_i and g_j as orthonormal basis functions for C([0,1]) say can we generate a basis for C([0,1]x[0,1]) somehow?

something along those lines is what I imagine is the key

[rotund jetty]

if $(f_j){j = 1}^{\infty}$ is an orthonormal basis of $L^2(X)$ and $(g_k){k = 1}^{\infty}$ is an orthonormal basis of $L^2(Y)$, then $(f_j(x)g_k(y))_{j, k = 1}^{\infty})$ is an orthonormal basis of $L^2(X \times Y)$. This fact justifies some but not all of the applications of change of variables.

[untold delta]

L

[prisma pelican]

I don't get why we would search for such a solution?

like, I get evans here just wants to give the quickest way to derive the fundamental solution but like, that scarecely better than just giving the fundamental solution outright if not all the steps make sense 😭

[buoyant pike]

Homogeneous equation -> solution should be homogeneous

[rotund jetty]

the quickest way is fourier transform I feel

you don't need any Ansatz other than qualitative smoothness assumptions

[prisma pelican]

that's in the coming chapter, I feel

I still like this hands-on method of going about it tho

it's instructive, I feel

[prisma pelican]

wait, what's a homogeonous solution?

[buoyant pike]

Do you know what it means for a function to be homogeneous

[prisma pelican]

oh

I see, hm, well, the above isn't exactly homogeonous but I can see it

I think the idea is that we leave alpha and beta unspecified so that we can easily change them later to simplify the problem

[heady silo]

this shit is complicated bruh

another question I didnt even think of... we dont usually have continuity of the differential operator so it might not preserve infinite sums even

bruhh

[rotund jetty]

yeah, the standard way to make it rigorous is view the sums as converging in some Sobolev space. This works flawlessly for the Laplacian

[heady silo]

alright I guess that makes sense since the coefficient are the thing that shrink to zero as you add more and more terms so with some bounds on the derivative you should be good I guess

or something like that

I guess you need to check it for the specific basis you are using it doesnt work in full generality

[buoyant pike]

They don't change partway through the problem

There are a few ways to reason about the correct alpha and beta, but one of the easier ways is dimensional analysis

[lilac barn]

Not the quickest if you include developing Fourier theory as part of developing the answer

[prisma pelican]

hm, that doesn't sounds accurate, evans seems to casually fixing those constants as it suits his fancy

[prisma pelican]

wait, how was this computation done?

ig I can justify it heuristically by like, the definition of a derivative but is there some rule I'm unaware of?

[buoyant pike]

Chain rule + FTC?

[prisma pelican]

I'm sorry, I don't see how this is the chain rule 😭

where is the composition?

[buoyant pike]

Oh sorry not chain rule

Product rule

[prisma pelican]

oh, well, even then it's not clear to me, 😭

like, I see a product but \Phi isn't even differentiated in the end result

[buoyant pike]

It is, but then you use the fundamental theorem of calculus

[prisma pelican]

ah

oh

I genuinely still don't get it, like, if I understand you correctly you're saying the second term is
[\int_0^t\int_{\bR^n} \Phi_t(y, s)f(x-y, t-s)\dd y\dd s]
but I don't get how we can use the ftc since the inscribed functions depends on $t$

[untold delta]

DarQ!

[prisma pelican]

like, as far as I can tell, if we can apply the ftc here we can apply it to the other integral term

I think, even if you defined
[F(t)=\int_0^t f(t-s)\dd s]
heurestically,
[F_t(t) = \int_0^t f_t(t-s)\dd s+ f(t)]
no?

[untold delta]

DarQ!

[prisma pelican]

I think I can straight up just prove this, I dunno why I'm too hung up on this

[buoyant pike]

Oh wait my bad

When you differentiate u, you pick up two pieces

The first piece is from the derivative of the integrand

The second piece comes from the cancellation of the s integral and the t derivative

[prisma pelican]

that's what I thought, but you use the above, right?

[prisma pelican]

this here

I just never came across this before and evans never proves it and I'm wondering if you could derive it with just simpler rules

[buoyant pike]

It would be +f(0)

https://en.wikipedia.org/wiki/Leibniz_integral_rule

[prisma pelican]

oh right

yes

[prisma pelican]

thankyu!

[steady flicker]

guys. is there a standard textbook (preferably a relatively popular one) covers foundimental solutions of helmholtz equation? deviation and all that. physics books also welcome

[noble gate]

I think Evan’s does

[steady flicker]

Lawrence Evans? I have the book but I don't see the derivation. it only mentioned Helmholtz in two places. It's more theoretical

[buoyant pike]

https://math.stackexchange.com/questions/2235615/fundamental-solution-of-the-helmholtz-equation

What do you want that this MSE post does not cover

[steady flicker]

ok this is good. at least there is some reference

thank you

[white veldt]

i think evans does something related to the fourier transform pair mentioned in the mse in a chapter (at least during my pde class i think i stumbled on this when going throughn evans)

[buoyant pike]

<@&268886789983436800>

[scenic niche]

When you're working with functionnal spaces such as $L^1 \cap L^r (\Omega)$, which norm is usual ? People don't define if they use $\left| . \right|{L^1(\Omega)}$ or $\left| . \right|{L^r(\Omega)}$. Is there something usual or it always depends of the context ? The notation used is $\left| . \right|_{L^1 \cap L^r (\Omega)}$

[untold delta]

quentin

[quaint herald]

The sum of the two norms is a common choice, as convergence in this norm is equivalent then to convergence in both of the parent spaces.

[scenic niche]

Oh yeah thanks you that's simple and that make sense

[void flame]

I'm a little confused when reading this, in Evans they are proving that the solution given by duhammel's principle solves this cauchy problem with a source. But why do we need to prove this? I thought Duhamel's principle is rigorous as long as we assume some regularity conditions (which we assume in this theorem). So why are we proving it from scratch again?

[astral vine]

You can claim it as a rule of thumb : that if you have such representation formula then the double convolution adding time dependency should solve the PDE you desire. And you would be morally correct.

However, If you are rigorous, you should always verify that the given formula actually solves your problem

Having a formula for solving gives you necessary conditions saying that it may be possible to solve the problem. Now prove it actually solves it.

for problems such as simple PDEs as the Heat Equation, no worries this is very simple if you are in standard function spaces 'smooth function with enough decay, say)

but for more complicated "possible solution formula" for more complicated PDEs, especially on the whole or the half-space, if you no longer assume compact support of your datas, even if you assume smoothness, then it may holds that the formula you did guess does not longer solve your problem anymore. Or it may but not the way you intended it to.

This kind of trouble actually almost never happens for standard PDEs studied in the standard cases: Parabolic in finite times or Elliptic PDEs on bounded smooth domains, say for square integrable functions as datas, or on Lp spaces with 1<p<+oo.

[serene mica]

This is more a vector calc question than a pde question, but I figure since it comes from a pde paper, it fits here. Anyway, can anyone see why |B_{\xi}| = 1? This essentially is equivalent to |A_{\xi}| = 1, but I cant see any reason why that is true, at least from just the given info

[serene mica]

Ive come to the conclusion the given info on A isnt enough to force its modulus to be 1, pretty easy to construct counterexamples. So I think this is just a typo

[buoyant pike]

Does kpz universality tell us anything about fluids

[heavy latch]

anyone here has studied leray’s article “on the motion of a viscous fluid filling space”?

im having trouble understanding what the T_{ij} oseen’s fundamental solutions are

[astral vine]

Okay, when you have an evolution equation say

[untold delta]

Functionanatolysis

Functionanatolysis

[astral vine]

The Oseen Kenerl is when you choose

[untold delta]

Functionanatolysis

Functionanatolysis

[astral vine]

I messed up with silent variables

After the bracket, replace tau by z, and say t in the last terms by tau.

[buoyant pike]

To elaborate on this thought

There is the kinetic theory of gases

Which comes from brownian motion

And can be used to derive navier stokes in the hydrodynamic limit

And tasep is another interacting particle system

And in the hydrodynamic limit, 1d tasep becomes burgers equation

[untold delta]

<Andrew>

[exotic lava]

this fails even for H=C

uh

still fails

if you take f_n to be like

split (0,1) into intervals of length 1/2^n

and alternate between -1 and 1

then this converges weakly to 0

but no subsequence can converge pointwise to 0 anywhere

which is the same as pointwise weak convergence for H=C

[solid flint]

Hello, In a Banach space $X$, suppose we have two strongly continuous semigroups of bounded operators $(S(t){t \geq 0})$ and $(A(t){t \geq 0})$ with generators $A$ and $B$ respectively. Moreover, suppose that the two semigroups commute. My question is what is the generator ( and its domain) of the composition of the two semigroups? we know that in $D(A) \cap D(B)$ the generator acts as the sum of the two operators $A+B$. However, its domain can but bigger. Anyone knows any results in this direction please?

[untold delta]

Mikahopff

[junior bloom]

The current belief is no

KPZ noise is an order less singular than thermal noise in fluids.

Also there's some sophisticated coupling-scaling argument, let me see if I can find it

[junior bloom]

There are issues with the hydrodynamic scaling limit. You can see a sort of updated version of the Quastel-Yau stuff in

https://arxiv.org/abs/2311.02223

[minor mulch]

also i thought the hydrodynamic limit is ignoring the fluctuation behavior

which is what KPZ universality is about

like i thought the philosophy is that fluid dynamics <-> large deviations for interacting particle systems

[junior bloom]

that's macroscopic fluctuation theory in general yeah. see bertini et al.

[minor mulch]

instead of large deviations i mean macroscopic behavior

ok

so yeah they are just not really related i think

[junior bloom]

the hydrodynamic limit is problematic for physical reasons

[minor mulch]

the macroscopic behavior of e.g. tasep is just not universal for 1d growing interfaces

[junior bloom]

but I don't know if this is published anywhere

mathematicians dont talk to physicists

[minor mulch]

why

[junior bloom]

the quick version is that you have to renormalize effective viscosity as you take the noise to 0 in the large deviations way.

it's absolutely not clear that this doesn't ruin things. mathematicians generally aren't aware that effective viscosity changes as you rescale.

so their hydrodynamic limit is very naive

[minor mulch]

i see

[minor mulch]

i think its sort of a coincidence that the KPZ theory applies to tasep in the first place

like tasep happens to biject to corner growth model/exponential last passage percolation

maybe there is a conceptual explanation coming from algebra or something

and perhaps a physical one too

but ive never heard one

from https://dept.math.lsa.umich.edu/~baik/Publication_files/ICM.pdf

h is the height function of corner growth model

[junior bloom]

I believe there has to be a deep reason why KPZ and TASEP appear to coincide.

Besides the obvious asymptotic one

[minor mulch]

why

i mean like yes, there must be, but i think it should be the same deep reason for why the KPZ theory describes many other growth models

dont see why theres anything special about tasep

[junior bloom]

I'm trying to gather my thoughts on this formally tbh. Hairer's regularity structures seem to be general but it feels like they actually have to describe some specific type of coupling in the regime where you aren't just lifting lower-level objects up into the KPZ world. That's my very loose feeling.

The reason I feel this way is

If you do Zwanzig-Mori style derivations for Boltzmann systems in an effective field theory way

[minor mulch]

what do regularity structures have to do with this?

[junior bloom]

a lot of them just skip the KPZ level and become unamenable to that general family of approaches

They are a framework that make sense of the KPZ formally and in a physically realistic way renormalizes it.

[minor mulch]

that is about the KPZ equation

i am talking about the broad phenomenon of KPZ universality

[junior bloom]

Yeah I know.

I'm just saying it feels like that kind of technique slices out a level of dynamics

anyways I can't formalize this, it's just a feeling

[minor mulch]

the 1:2:3 scaling doesnt preserve the KPZ eqn

[junior bloom]

idk what that is

[minor mulch]

the thing which “KPZ universality“ is about

[minor mulch]

like this

[junior bloom]

ok sure

[minor mulch]

you can think of KPZ equation at long times

with the same rescaling as in that screenshot

[minor mulch]

maybe i misunderstood but doesnt this just mean those approaches are studying different things?

like i dont see why the failure of those methods to probe the fluctuations necessarily means there is a deep reason why the fluctuations are what they are

[junior bloom]

it's more like counting levels of criticality in noise

well i don't know what the count is

and the criticality is in a loose sense endowed by hitting he noise term with derivatives

and how many you add seems to control what kind of structures you need to understand it.

i mean yes i agree it's not a very deep reason

it just seems structural

because derivatives are basically local interactions

[minor mulch]

criticality in the PDE sense? like when you rescale the nonlinearity does X?

[junior bloom]

yeah

[minor mulch]

hmm

[junior bloom]

to wit there's some sophisticated reason constructive field theorists like to tell me that fluids have to be an effective theory but kpz scaling is true up to infinity because that's where the dynamics come from when appropratitely renormalized.

that has to do with short and long range couplings

i dont fully get it lol

but anyways this coupling appears in equations in terms of the bad derivatives

because when you coarse grain miscroscopic descriptions thats the terms they become

[minor mulch]

idk about fluids but yeah the real KPZ universality theory is about what happens after long times

i think less is understood about intermediate times

at short times it is just gaussian

[minor mulch]

where does tasep come into this though

[junior bloom]

no clue lol. it fits in kpz universality and the general particle interactions seem to have similar correlational structure/interacting kernel.

[minor mulch]

sure

[junior bloom]

one could view everything in physics being built from two point correlations, and kpz universality applying to a slate of similar interacting kernels that give said correlation structures.

[minor mulch]

but my point is that the fact that it lies in the KPZ class doesnt necessarily mean its like deeply connected

[junior bloom]

is quite a coincidence

[minor mulch]

like many objects lie in the gaussian universality class

[junior bloom]

ok sure but that does have a deep reason

donsker embedding is deep

[minor mulch]

despite having little to do with each other

right but that isnt a deep statement about the individual models

[junior bloom]

it says that the noise isn't singular enough to escape donsker

in the models

[minor mulch]

its a deep statement about the universal mechanisms driving fluctuations

[junior bloom]

right, yeah that's what i mean by deep

who cares abotu the rest

[minor mulch]

btw do you know how the KPZ eqn was first derived

[junior bloom]

dont remember, but i should 😰

[minor mulch]

its so annoyingly simple

[junior bloom]

defending within 2 months and i might be asked this

[minor mulch]

what is your thesis on?

[minor mulch]

actually is it true that other interacting particle systems are supposed to be in the KPZ class? ive never looked into this

like obviously various flavors of asep should be

but beyond that idk

[junior bloom]

don't want to doxx myself but basically finding the right RG flows to rigorously show a class of spontaneous symmetry breakings in condensed matter macrodynamics from mesoscale descriptions of micro observables.

im a mathematical physicist in the math dept

[minor mulch]

and kpz is relevant?

[junior bloom]

only as a tangential example

[minor mulch]

interesting

id be curious to read your thesis

[junior bloom]

using some similar stuff to regularity structures, not quite as strong.

sorry, i dont want to doxx myself.

[minor mulch]

yeah np

[quick pagoda]

What's a good reference or two for KPZ, the KPZ universality, Kolmogorov universality, and related things

[junior bloom]

A lot of people will hate this suggestion but Friz and Hairer

they discuss KPZ at the very end after spending like 250 pages developing the theory to talk about it

but I just learned this stuff from reading papers, that's the best way if your goal is to understand KPZ first

[quick pagoda]

Which one is this, one of Hairer's arXiv docs?

[buoyant pike]

I mean isn't kolmogorov turbulence theory about long time behavior as well

[junior bloom]

no, sorry, their course on rough paths

[quick pagoda]

Gotcha

I have a lot of Hairer docs downloaded but not that one it seems

[minor mulch]

KPZ universality is pretty different from the solution of the KPZ equation via e.g. regularity structures

there is https://arxiv.org/abs/1512.07845 but by and large, studies of kpz universality are mostly using lattice models

there are some surveys

https://arxiv.org/abs/1606.06602 and https://arxiv.org/abs/2110.11287 are good and pretty different in scope

[junior bloom]

Yeah KPZ universality is a lattice theory, rough paths for KPZ is an RG theory

but they probably correspond in some way

[minor mulch]

well like hopf-cole solution of KPZ is well understood

and we can prove that various discrete things converge to it

and otoh hairer proved his thing coincides with the hopf-cole soln

not sure if anyone has tried to work out a direct correspondence between the regularity structures and some discrete data though

[junior bloom]

yes but the point is that the Hopf-Cole pops out of lattice theory, but the reason why you want to work with the continuous version is the continuous version actually pops out as an effective field theory

so they are two different physical systems that approach the same formal description and have the same solution class

[minor mulch]

not sure what the distinction is actually

[junior bloom]

the lattice theory posits a cutoff that is like, molecular or whatever

[quick pagoda]

Added to the list, thanks

[junior bloom]

the continuum version says we can write certain kinds of field interactions whose coarse-grained description is effective as a continuum theory only, not a lattice theory

[quick pagoda]

Noted

[junior bloom]

think of it as kind of like

you can't build a lattice minkowski theory

well you can but at the naive level you cant

so GR will always have to be a continuum theory

except thats not the case but at the naive level it is

its like that

[quick pagoda]

Metric geometry
I'm in

[minor mulch]

ok i mean i get this

but im not seeing why this distinction is actually important for the phenomenology

in the kpz case

[junior bloom]

i cant really give a good account of this. constructive field theorists on the physics side tell me it's meaningful but after hearing their explanation twice ive still failed to understand it properly. the distinction has something to do with the nature of field couplings.

lol

[minor mulch]

like why cant i say that the kpz equation is the hopf-cole solution

[junior bloom]

it's on my docket to properly understand

[minor mulch]

(which btw can be formulated entirely in the continuum)

[junior bloom]

yeah, because the lattice theory shrinks slower than the noise in some sense

as for kpz, idk, it applies to things that aren't literally surfactant physics as it purports to describe, so physicists care about it

as for why i cant just call its description the hopf cole

it's because the RG technique used for it is generally useful

across a broad class of similar problems

[minor mulch]

what physical situation do you picture of when you think about the kpz eqn?

i usually never think about continuum models of anything

[junior bloom]

i sometimes picture an interface between two fields without a meaningful wavelength cutoff.

[minor mulch]

oh

[junior bloom]

an example of where this might be the case is in two-fluid plasmas.

though i dont believe this is in KPZ universality

[minor mulch]

so not like matter sticking to a surface or something

[junior bloom]

yeah no

[minor mulch]

ok i can imagine this would demand a continuum object

and hopf cole is indeed maybe not helpful

since it is basically just exploiting the connection to directed polymers

[minor mulch]

is there somewhere i can read about this perspective?

[minor mulch]

and also this but ig you probably havent seen a good reference either

[junior bloom]

honestly, no idea. I'll ask my advisor, he's a constructive field theorist who knows this stuff cold.

[buoyant pike]

kpz is two phase hele-shaw?

[minor mulch]

idts

[junior bloom]

old mathematical physicsts are extremely smart. I only very recently learned a thing he doesn't know, namely Hodge theory

i took that as a personal victory

he's not that impressed though. insofar as hodge theory is useful its trivial to derive in the easy cases.

lol

the general discussion given by voisin is

i haven't seen it used in proper generality, or any real appearance of sheaf cohomology in physics

[junior bloom]

btw, this isn't in the kpz regime but the reason why this perspective is generally useful is, is that many molecular reactions can only be solved in a very weak sense, so all descriptions of the real system are coarse grained because they're measured at the mesoscopic scale. because the descriptions are coarse grained, it may well not be meaningful to talk about them in terms of cutoff theories. the effective theories are continuums even for molecules.

this is can reference

Polonyi and Schwenk "Renormalization Group and Effective Field Theory Approaches to Many-Body Systems" kind of discusses this

hmm

well thats good enough for now, anything more would require some lit diving.

[minor mulch]

i dont see why it might be wrong to use cutoff theories

like if the cutoff is smaller than whatever precision we can measure

i guess the point is that we want a theory that remains useful as our experimental abilities improve?

[junior bloom]

you can't generally recover a coarse grained solution with a cutoff. there's also an issue where the nuances of geometry of the cutoff can sometimes really matter. this is the case in hydrodynamic derivations from a more robust boltzmann theory, rather than in simplified quastel-yau like lattice gases.

[minor mulch]

oh rly

[junior bloom]

for an example of this look at uhh

Español, Pep, Anero, Jesús G., Zúñiga, Ignacio (2009) Microscopic derivation of discrete hydrodynamics. I think

there's also this Large-distance and long-time properties of a randomly stirred fluid by Forster et al. just to showcasse some difficulties in microscopic derivations that are fully general

warning, neither of these are rigorous math lol

physics math

it's an open problem to make them rigorous

the first one is doable i think. the second one is way out of reach.

[minor mulch]

ig fluids are messed up

[junior bloom]

plasmas are worse, did you know nobody has actually, even just formally, derived Hall MHD from microscopic observables in a physically realistic manner

[minor mulch]

natural world bad