advanced-pdes

[junior bloom]

actually mostly its either L^2 or its Sobolev or Besov

Young measures in some cases

[astral vine]

Then this book is heavily recommended for you

L², Sobolev, Besov

everythign is treaten in Chapters 1-2

[junior bloom]

I'll go find a copy somewhere

[astral vine]

is available in MyCopy program

[junior bloom]

"somewhere"

[astral vine]

if you want a printed version

and on Springer Link the pdf is free if you can log through your Institution

[junior bloom]

good point

huh

I guess I didn't need to sail the high seas to find a lot of books then

[astral vine]

But some obscure rferences may need it sadly

[junior bloom]

you just changed my world making me realize i can just download the entire springer book catalogue

[astral vine]

Not the whole

Your uni must be a sucriber to some offers or have at least bought the book you want to download

[viscid sparrow]

I'm not seeing why we need all of the argument after 2.1.8... Is it not fine just conclude after 2.1.8?

especially the absolutely continuous property holds for almost every line

[white hazel]

To conclude the proof you need to know that 2.1.8 holds for almost every x

So that’s the point of the stuff after

[astral vine]

Assume you have a like the one above, a PDE you want to solve in a more general form

[untold delta]

Functionanatolysis

[astral vine]

If you can solve both

[untold delta]

Functionanatolysis

Functionanatolysis

[astral vine]

Then U= v+w must be a solution for your first problem

(superposition is generally about linearity of ODEs/PDEs)

[cerulean light]

Alright, thank you.

[cerulean light]

do you mean $u(\cdot,0)= u_0$? in $u(x,0)=5x^2$ t is 0

why is it given that $u(x,0)=5x^2$

[astral vine]

When you to solve evolution PDE you want to know what is the current state of the system you're starring at, described by your PDE

but it cannot start from nothing

You need to know its initial state

here th initial state/data for your problem is u(0,x) = 5x²

and yes your problem is the one with

[untold delta]

Functionanatolysis

Functionanatolysis

[vivid patio]

Hey all, new here so please let me know if I'm posting in the wrong area or anything. Wondering if anyone can walk through finding the general solution of a PDE in the form (partial ^2 u/partial x^2) + (partial u/partial x) + (constant u) = ax +by + c? I believe it's to do with forced oscillations but I am getting twisted up in the lectures and not really following along as well as I'd like. Am I correct in thinking this follows the mx" = -mux' -kx+f(t) format? Rearranged to the above? And if so, how do you then go about finding the general solution? I think I'm confusing myself as all the examples we're given are things like f(t) = 5 cos(3x) or something, but this equation is in the form ax + by + c and has thrown me haha.

[astral vine]

#odes-and-pdes should be a better place

above should also be better in #odes-and-pdes

[vivid patio]

thank you 🙂

[mental geyser]

Quick logic check: if I take the negative of a supersolution of the heat equation that function would be a subsolution of the heat equation correct?

[astral vine]

which one

with what kind of data

notion of super/sursolution seems weird for PDE like Heat Equation

[mental geyser]

Cauchy problem i.e. homogeneous with nonhomogeneous initial condition on the satisfying function

[astral vine]

Then no

since it does not satisfies the initial condition

u0 turns into -u0

but obviously up to flip the sign, then yes

But I don't get what you can do with it since you lose what happen with initial condition

[mental geyser]

I may be wrong about the initial condition. The definition in use for supersolution is $u_t-\Delta u\geq 0$ on a cylinder. Doesn't depend on initial data.

[untold delta]

TheRedLotus

[astral vine]

That's weird, I hope someone knows something about it because to this is unclear if you have to check condition for a fixed initial data

[late moth]

I have the following problem
[\begin{cases}
u_t-\Delta u = 0 &\text{in $ U \times (0,\infty)$} \
u(x,t) = 0 &\text{on $\partial U \times (0,\infty)$} \
u(x,0) = g(x)
\end{cases}]
where $g(x) \ge 0$. I need to show that
[ \frac{d}{dt} \int_U u(x,t) dx \le 0.]
I was able to show in the case $\frac{d}{dt} \int_U u(x,t)^2 dx \le 0$, it's fairly simple because it just employs some integration, but in this case I am left with
[\int_U u_t(x,t)dx = \int_{U} \Delta u(x,t)dx = \int_{\partial U}u\cdot \frac{du}{d\nu}dS(x)]
I am not really sure what I am missing here, unless there's another technique I can't think of offhand

[untold delta]

CirbyMU

[quaint herald]

@late moth

I think your integrand should just be the outwards normal derivative of u in the final integral (no factor "u").

Now, the maximum principle implies that your solution should be non-negative on the interior of U for all t>0, and so this normal derivative is non-positive everywhere, from which the result follows.

(Precise details will depend on assumed regularity of g, U, and what theorems you have at your disposal already. The above argument should serve as a template though.)

(It's also to be expected that you won't get the result by just integrating by parts like you did with the square, because if you flipped the sign of your initial data you would have the opposite monotonicity. The non-negativity of g is important).

[late moth]

Ah yeah, thanks, I hadn't really thought about the maximum principle, but this helped a lot, thanks

[tranquil steppe]

If you're asking whether $u_t-\Delta u \ge 0$ and $v = -u$ implies $v_t-\Delta v \le 0$, then the answer is yes.

[untold delta]

SingularityResearch

[tranquil steppe]

(you haven't specified in what sense these inequalities hold, so I'm assuming it's classical)

[late moth]

I have another question,
I am given a uniqueness proof for a wave-like equation. Going through the motions of starting the proof, I have the following problem
[\begin{cases}u_{tt} - \nabla\cdot(A(x)\nabla u) = 0&\text{ in $\R^n\times (0,\infty)$} \
u(x,0) = 0,\quad u_t(x,0) = 0 &\text{ in $\R^n$}\end{cases}]
where $A(x)$ is an $n\times n$ symmetric matrix with $0\le A \le \Lambda$, i.e. $\Lambda I - A$ and $A$ nonnegative definite where $\Lambda$ I believe is the largest eigenvalue? The notation wasn't explained.

So to show that this is 0 everywhere, my first thought is to use an energy method. This has 2 parts,
\begin{enumerate}
\item Select a backwards light cone with large enough propagation speed (this is as stated in the hint given to me)
\item Determine what the energy looks like for this wave-like equation.
\end{enumerate}

It is also mentioned that the inequality $2\langle A(x)v,w\rangle \le \langle A(x)v,v\rangle + \langle A(x)w,w\rangle$ will be helpful, which I was able to show, but this feels a little mystical right now.

I think that my biggest question is how to handle $A(x)$ in this case because now we have some sort of "weighted" laplacian that makes things weird to work with.

[untold delta]

CirbyMU

[astral vine]

This kind of equation can be treated via Cosine/Sine operator theory (a brief extension of C0 semigroup theory)

This is deeply linked with Kato-square root property for divergence form elliptic operators. the constant Lambda is one of the two ellipticity constant (uniform in x), notice you should have also as you mentioned this is also bounded from below. This provide exactly the following L² estimate

$$\lVert \nabla u\rVert_{\mathrm{L}^2(\mathbb{R}^n)} \simeq_{\lambda,\Lambda} \lVert A^{\tfrac{1}{2}}\nabla u\rVert_{\mathrm{L}^2(\mathbb{R}^n)} $$

[untold delta]

Functionanatolysis

[late moth]

Can you give some reading on that? It seems interesting on first glance, but I’m not sure if it’s immediately helpful to me as it still feels a little mystical.

I’m on my phone so I can’t latex it rn

I don’t really want to expose what uni I go to bc this problem is readily available

regardless, you make a good point

No conditions on f, phi, nor psi

which is why I thought you could take u = v-w, where v and w are other smooth solutions of the wavelike equation
then those all vanish

[quaint herald]

@late moth It looks to me like it will be a pretty similar calculation to how you prove things like finite propagation speed for the standard wave equation, you will just need to choose your energy appropriately. I think something like E(t)=integral of |ut|^2 + <Agrad(u),grad(u)> works.

[late moth]

Okay great, that was the energy I was able to come up with, I am just looking at some of the integration by parts, but that is where that inequality that was mentioned to be helpful will come up

[quaint herald]

Integrals being over a ball of radius c(t0-t) for t in (0,t0)

Then differentiate in t and integrate by parts the second term. The interior integral should totally die from the fact u solves the modified wave eq.

And you should be left with an integral over the boundary you can make non-positive by choosing c high enough. (Something like larger than Lambda will suffice I imagine).

[late moth]

I have a quick question about integraton by parts here (I know it's basic, but I haven't worked with these sort of weighted inner products yet). I'm a little bit confused how to deal with the normal derivative term of Green's formula here. I have
[\int_{B(c(t_0-t))} u_tu_{tt} + \langle A(x)Du, Du_t\rangle = \int_{B(c(t_0-t))}u_t(u_{tt}-\nabla (A(x)\nabla u))dx + \int_{\partial B(c(t_0-t))} \text{some normal derivative term} u_t]
My first thought was that it was $\langle A(x)Du, \nu\rangle$, where $\nu$ is the normal vector, but this doesn't feel correct.

[astral vine]

For above inequality, just write the operator L = -div A grad in the weak sense as a sequilinear form. For above equivalence of norms. ie <Lu,u> = <A grad u , grad u> and use ellipticity of A.

The square root of L, in the unbounded sense given by sesquilinear forms on L², is closed with H1 domain, and generate a Cosine and a Sine functions, see Chapter 3 Subsections 3.14, 3.15, and 3.16, and Chapter 7 of W.Arendt, C.J.K. Batty, M.Hieber, F.Neubrander - Vector-valued Laplace Transforms and Cauchy Problems which will tell you about existence and Uniqueness of solutions for such generalized wave equation.

If you want to learn more about properties second order divergence form elliptic operators via properties of their associated sesquilinear forms and Heat Equation (and not wave because analytic properties is carried over by the parabolic problem, or at least is easier to see from this pov.) you should check El Maati Ouhabaz, Analysis of the Heat Equation on Domains which is a simpler first L²-Lp approch of such problem.

Some work done by people like Pascal Auscher might also be interesting as well.

[vivid obsidian]

What does it mean to prove that we can assume p(x) without loss of generality ?
What is equivalent to prove ?

Q9 says that :
Using a change of variable v(s) = au(bs) for certain constants a and b, show that
that it can be assumed without loss of generality that u0 = 1.

Do I have to prove that [ u is a solution of (7) <=> v is a solution of (7) ] ?

I don't really understand

[tranquil steppe]

Yes, that's one of the things to prove here. You need to choose a, b properly in order for this to be true.

Other than this, the "wlog" thing means the usual in this context:

1. We prove the theorem under additional assumption u0=1.
2. In the general case, we are given arbitrary u. Thus, we choose a, b properly so that v(s) = a u(bs) is also a solution, and v0=1.
3. We apply to v the special case of the theorem we already have. And then infer the claim for u.

[vivid obsidian]

Thanks for your answer but I'm having trouble to understand

I'm not really convinced why I have to prove what I said

My intuition was that I wanted v to verify the equations of the system so that the change of variable is "legit"

[tranquil steppe]

Hm, I don't understand your problem, then.

What do you mean by the change of variable? The act of introducing the auxiliary function v?

[vivid obsidian]

yes

[tranquil steppe]

What do you mean by "legit"? The fact that such v is also a solution?

[vivid obsidian]

exactly

[tranquil steppe]

And you're asking why this is required?

[vivid obsidian]

yep

[tranquil steppe]

If so, the reason is: point 3.

If you want to apply the theorem to some function v, you'd better check that the assumption is satisfied.

[vivid obsidian]

Why showing that v0 = 1 prove that we can assume wlog that u0 = 1

I'm kind of lost I think

[tranquil steppe]

OK, what about the 1,2,3 strategy of proof I wrote?

Do you follow the logic there?

[vivid obsidian]

Ok so first you call "theorem" the system of equations ?

I don't really understand the point 1. because we have to show that we can assume wlog
Why we have to prove with an additional assumption ?

[vivid obsidian]

not really :/
My bad

[tranquil steppe]

So at which point are you stuck?

[vivid obsidian]

first

[tranquil steppe]

Meaning "I don't know how to prove 1" or "I don't know why 1 is needed later"?

[vivid obsidian]

second

[tranquil steppe]

So in point 2 you construct a function v which solves the same equation but satisfies v0. This means that you can apply to v the thing we know from point 1.

And once you know the claim for v, you should be able to translate it into a corresponding claim for u.

(I don't know what that claim is, by the way, but this is the least important)

[vivid obsidian]

Why point 1 is even needed ? I thought we have to prove that we can assume wlog and not that we have another assumption

[tranquil steppe]

You seemed to be confused by the use of "wlog" here, am I right?

[vivid obsidian]

I think so

[tranquil steppe]

The whole point of my explanation was to give a plain strategy of proof, avoiding the use of the phrase "wlog".

That's why I asked you to stick to 1, 2, 3. Once you understand the proof written this way, you'll know what "wlog" was supposed to mean.

[vivid obsidian]

How do you understand what does "wlog" mean

Does it mean that the system with u0 = 1 is a "general" system ?

[tranquil steppe]

In this context, "wlog A" means
"Let's add A as an additional assumption. The general case can be reduced to the special case where A is satisfied."

[vivid obsidian]

Okay I agree

I understand now why you told me your strategy of proof

[vivid obsidian]

What does reduce mean in our context ?

[tranquil steppe]

It means exactly what was done in points 2 and 3.

That is, you can somehow use the special case to prove the general case.

[vivid obsidian]

Ok. And to do that, we use v as an intermediate function that is going to help us to say that the general case can be reduced to the special case

I think I get it

[tranquil steppe]

Yes, exactly.

[vivid obsidian]

What does the equivalence I stated in the first place proves ?
You told me it's one thing to prove

[tranquil steppe]

Actually, you don't need an equivalence. You just need one implication: if u solves the equation, the so does v. That's part of point 2.

[vivid obsidian]

Ok that's what I thought

[vivid obsidian]

Why do I need to show that v0 = 1 ? Because if v is a solution v0 must be equal to u0 right ?

[tranquil steppe]

Oh, instead of "the equation" I should say "the first and the third part of (7)".

Or in other words, v satisfies (7), but with the second condition replaced with v(0) = v0 = 1 (its own initial condition).

[vivid obsidian]

Ok so this is the whole point of the question

Ok I think I get it now. I will try to write it rigorously tomorrow.
Thanks a lot for your help @tranquil steppe

[west reef]

anyone here study epidemiological models, like SIR , SEIR?

i have some doubts

[lilac barn]

Why does Evans construct F_e at the end? Don't we already know that the W^{1,q} norm is bounded of u?

[tranquil steppe]

I don't think I understand the question. The definition of F_e has nothing to do with the W^{1,q} norm, but rather with pointwise bounds on u and its derivative.

[lilac barn]

Okay, so I am just confused about why is he introducing F_e. Do you know a reason as to why he is introducing? I can post the following page of the proof if you want.

[tranquil steppe]

Yes, that would be helpful.

[lilac barn]

Here is the entire thing. We also have that U is bounded with smooth boundary.

[tranquil steppe]

You can see that Evans refers to (19) (which is the definition of $F_\varepsilon$) when justifying (24): $\int_{G_\varepsilon} L(Du,u_k,x) \to \int_{G_\varepsilon} L(Du,u,x)$ as $k \to \infty$. Do you know how to show this convergence?

[untold delta]

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

[lilac barn]

Here is an answer from MSE.

[tranquil steppe]

Right. So here it's important that we only consider the derivatives of L on a compact set, which is ensured by the bound on du.

[lilac barn]

But at the same time, we need a similar bound on Du_k (since it is the lower bound on the integral). I am guessing that bound comes from sup_k |u_k|_{W^1,q}, but that bound would also apply for Du

[tranquil steppe]

Do you need a bound on Du_k, though? This quantity doesn't even appear in (24).
And to stress it once more - you need pointwise bounds here, Sobolev bounds are insufficient.

[lilac barn]

Oh okay, thank you very much for clarifying. I think I get now: D2L is bounded in the third variable since that it is over U. It is bounded in the second variable since {uk} is a uniformly convergent sequence and so uk, and u are both bounded.
It is bounded in the first variable since Du is bounded in Fe
Just one last thing. Do we really need the pointwise bound for u in Fe since {uk} being uniformly convergent should give us a pointwise bound on u as well?

[tranquil steppe]

I don't see a direct way to avoid using the pointwise bound for u.

[lilac barn]

For each e> 0, we have that u_k uniformly converge to u on Ee. In particular, we have that {u_k} is uniformly bounded on Ee. This implies that u is pointwisely bounded on Ee. So if we don't put the pointwise bound on u inside Fe, we still have the pointwise bound on u.
So my point is that it seems that the definition of Fe can be weakened to {x in U: |Du x| ≤ 1/e} since uniform convergence on Ee furnishes the pointwise bound on u.

[vivid obsidian]

Hello, I have a question for this ode again
What does happen when u_0 = 0 ?

If someone else can help me too I'd be a pleasure

I'm studying that case because I take a = 1/u_0 so I need u_0 =/= 0

[tranquil steppe]

I don't follow. If u(x)=1/x and u_k(x)=1/x+1/k, then still u_k tends to u uniformly, but u is not bounded.

[tranquil steppe]

That's a good question! I would believe that in this case u(r)=0 is the only solution, but I don't see a direct argument for uniqueness for this particular ODE.

[tranquil steppe]

How does it relate to the equation with -|u|^(p-1) u on the right hand side?

[vivid obsidian]

The Q10 asks us to solve it and we assume p = 0 so we have -|u|^{-1} * u and naively, depending on the sign of u, I said it's equal to 1 or -1

[vivid obsidian]

Hello, how do you show that if p = 0, the right hand side is + or - 1 ?
I want to show that for all r, u(r) =/= 0 so it allows me to say that u > 0 or u < 0

because I guess u is continuous ?

[tranquil steppe]

If p=0, the term u/|u| is ill-defined at points where u=0, so the question is also ill-posed. Unless you're given some specific meaning of the equation (e.g. viscosity sense).

[vivid obsidian]

I guess it is just not defined for r s.t u(r) = 0

They didn't initialise u

[vivid obsidian]

I didn't tell you but my project is a "simple modelisation of a star"

and the function u , here, was "initialised" when we did a change of variable at a previous questions which was
p = c * u^p where p is the density function from B(R) -> R+ where B(R) is our star

[forest cradle]

If anyone is familiar with the torsion problem (https://arxiv.org/pdf/1702.01258.pdf), can you say a bit about why the problem is significant and why people are interested in the product G(\Omega) = first ev * maximum of torsion function?

Or physically, what the torsion function represents?

[untold delta]

squirtlespoof

[astral vine]

This kind of solution exists, only if you can split the (space) domain into a cartesian product, and the whole domain must be bounded (compact) so being basically a square or a rectangle.

if the domain is not a square or a rectangle but is still a bounded (compact) one

the elementary solutions are of the form B(x,y)c(t), where you cannot split B into a product

[untold delta]

squirtlespoof

squirtlespoof

[brave spindle]

Does anyone have any examples of PDEs where traditional numerical methods fail?

^that are somewhat physically motivated / no delta functions

[forest cradle]

I would guess things with highly oscillatory terms could be problematic

if stationary phase doesn't work

[junior bloom]

spectral decomposition on irregular bounded domains with atypical boundary conditions, literally impossible or somehow possible in some systems?

sorry bit of a general question

im just searching far and wide for tools to avoid having to do any gmt

[astral vine]

It depends on the boundary conditions, and how much irregular is the domain

[junior bloom]

As irregular as you can make it, let's say the BC is something reflective

[astral vine]

Several results for second order (and order 2m) elliptic operators with Lopatinski-Shapiro boundary conditions still satisfies nice spectral decomposition on bounded Lipschitz domains

but nothing explicitly "computable"

[junior bloom]

aaaaaaaaaaaaaaaaaaa

rip

OK, if the best thing I can find is an obscure subarea of PDEs this approach won't work, never mind but thanks for the help

[astral vine]

You cannot compute it for some Laplacians with neumann or Dirichlet Bc on general Lipschitz domains

what did you expect

[junior bloom]

not much

well

lots of things have nice spectral expansions that make everything really easy in an infinite domain

tricks that can make pdes into systems of odes, which is a good trick when there are stochastic terms because nobody wants to do rough paths theory

[astral vine]

This only works on the half space

and some results can be transported via perturbation techniques on some rough bended half-spaces (as perturbations)

[astral vine]

'Nobody wants to do rough paths theory'

[junior bloom]

its the probability equivalent of gmt, dont do it!

[astral vine]

GMT ?

oh Geometric Measure Theory

[astral vine]

Yes

[slate prism]

Got an exercise that I'm stuck on.

Let $\lambda,k > 1$ and consider the PDE

$$\left{
\begin{aligned}
&u_t = u_{xx} - \lambda u - u^k &&\quad x \in (0,\pi) &&\quad t > 0 \
&u(0,t) = u(\pi,t) = 0 &&\quad&&\quad t >0 \
&u(x,0) = \varphi(x) &&\quad x \in (0,\pi)
\end{aligned} \right.$$

where $u := u(x,t)$. I would like to show this equation has a positive and unique equilibrium solution $u_+(x)$ which satisfies

$$\begin{aligned}
&u_+(x) > 0 \text{ on } (0,\pi) \
&u_+'(0) > 0 >u_+'(\pi)
\end{aligned}$$

Clearly, to be an equilibrium solution, $u_+$ must then satisfy the ODE IVP given by
$$
\left{
\begin{aligned}
&0 = u_+''(x) - \lambda u_+(x) - u_+^k(x) &&\quad x \in (0,\pi) \
&u_+(0) = u_+(\pi) = 0
\end{aligned}
\right.
$$

However, I'm not sure what to do from here. Solving this is certainly not feasible. Supposing the nonexistence of such a $u_+$, then one of the conditions must be violated.

For instance, if $u_+$ is not strictly positive, it must be negative somewhere (say $x_0 \in (0,\pi)$), and at that point we'd have

$$u_+''(x_0) = \lambda u_+(x_0) + u_+^k(x_0) < 0$$

(though I imagine we have to constrain $k$ further, e.g. to integers, so that a negative to a power is well-defined). That would mean we are concave-down at $x_0$, but I'm not exactly sure what that would get us.

I'm even less sure about how to get anything meaningful from the conditions on $u_+'$ being violated.

Some reading suggests that maybe the maximum principle would be useful and the way to go, but I'm not entirely sure how it could be properly used here.

Thanks for any ideas you might have.

[untold delta]

Eevee Trainer

[west reef]

has anyone here does SIER models
[5:49 PM]
SIR models
[5:49 PM]
i need some help understanding the time series code

[slate prism]

addendum: should be +λu (or +λ(u_+)) throughout, if it makes any difference for anyone

[untold delta]

homo-genius

[river path]

Value of fourier transform at 0 is one way to see it

[astral vine]

With no further assumption on u ?

[river path]

Yeah this definitely depends on something

Hmm

[astral vine]

what ?

This make no sense to talk about (- Delta)^s u , since there is no canonical Laplacian on bounded domains with boundary

you have to chose if your Laplacian is either, Neumann, Dirichlet, or more generally Lopatinski-Shapiro like BC

If you are on Rn

like for a function in H^{2s,p} of Rn, is it true that

[untold delta]

Functionanatolysis

[astral vine]

Then yes since the space

[untold delta]

Functionanatolysis

[astral vine]

is dense in any H^{s,p}, s in R, 1<p<+\infty

[river path]

Yes

[river path]

I meant to look at things which were nice at 0 and then use density when I said this. I should have been clearer

[astral vine]

But for general functions it can fail

The map for a function look at its integral, is general not continuous since there is no continuous embedding from Hs,p to L1

but the spirit is, if it must have a value then it should be 0

On domain the question is way more complicated

Because as I said there is no canonical Laplacian

you need to chose the kind of Laplacian you want to look at its fractional powers

Again there is no canonical definition

It has

people that say its 'The Laplacian to the power s' in their paper are kinda lying

The definition of "The Fractional Laplacian " through Gagliardo weighted finite differences is kinda a lie

the definition is ture for Rn

not for bounded domains

neither for domains in general

You can show that this is equivalent to some singular Fourier Multiplier

No

You cannot defined it like this

and Omega being what you want completely breaks the symmetry

On domain you have to chose a different way to define the fractional Laplacian

Chose your favorite Laplacian with your Favorite BCs, then use what ever, Philipps Functional Calculus, Holomorphic Functional Calculus, L² Functional Calculus (via spectral measure) to construct its powers

There is nothing to ensure it

give me the exact reference of the paper

(it seems to me you want to reproduce a proof on Rn, to turn it in a proof on domains, in general this kind of approach completely fails, since domains involve boundary, and you need to control what happens on it. The later on is in genral not possible, that's why you need to get into boundayr value problem)

[astral vine]

In Ros-Oton & Serra, they investigate a special kind of problem, prescribing some 0 boundary value (=0 on the complement of Omega), and in deed it is not a Laplacian on Omega, but the Laplcian on Rn, with some additional conditions.

The fact that fractional powers of Elliptic Operators are in general non local is the main interest in the Theory

i.e. does not preserve support

[junior bloom]

I'm trying to recall a type of argument using the Banach-Alaoglu to establish the existence of a weak solution to a PDE.

Is anyone with a better familiarity with PDEs able to direct me to an example of this?

Should be basic, class-level stuff.

I remember this from a class.

[forest cradle]

I think I found it in my PDE notes for Dirichlet solution in a Sobolev space

maybe more generally existence of solutions to 2nd order elliptic equations

It looks like you take a minimizing sequence v_n for the associated minimization problem E[v], use Banach-Alaoglu to conclude there is a weak limit v0, and then you show liminf E[v_n] \ge E[v0] so that v0 is a minimizer and therefore a weak solution

this is in $H^1_0(U)$

[untold delta]

washingbear

[lilac barn]

I don't think Banach Alaoglu gives you a weak limit. It gives you a weak*-limit. What you might be referencing is that for reflexive spaces, in particular Hilbert spaces like H^1_0(U), a norm-bounded sequence has a weakly converging subsequence. This is known as Eberlein–Šmulian theorem.

A comment from MO

https://mathoverflow.net/questions/357889/usefulness-of-the-sigmal-infty-l1-topology-in-the-context-of-differential

[forest cradle]

maybe I have called it the wrong name, I mean the theorem that if you have a bounded sequence in W^{1,p}(U), then there is a weakly conv subsequence

[lilac barn]

Yes, that is Eberlein–Šmulian theorem. In full generality, the theorem is: Suppose X is a Banach space.
Then X is reflexive iff every bounded sequence in X admits a weakly convergent subsequence.
This applies for W^{1,p}(U) for 1< p < infty since in those cases, the space is reflexive.

[tranquil steppe]

@lilac barn Can you point me to a reference for that statement? All I could find online was the equivalence of various notions of compactness.

[tranquil steppe]

In general, Banach-Alaoglu gives you weak-* compactness. If your space is reflexive (and W^{1,p} is), then it's the same as weak compactness. And if it's separable (and W^{1,p} is), then the weak topology on bounded sets is metrizable, and compactness translates into the possibility of taking convergent subsequences. This is what you're usually interested in.

[lilac barn]

Page 70 of Brezis' books gives out several references which covers the theorem.

The forward implication is spelled out, after building some theory of separable spaces. The backward implication is left for references.

[tranquil steppe]

OK, so I see that for reflexive spaces one can drop the separability assumption (since it's possible to restrict to a separable subspace anyway), that's nice.

Interestingly, Brezis only uses the name "Eberlein–Šmulian" for the inverse theorem (Thm. 3.19), and doesn't give a name for the theorem we're discussing (Thm. 3.18).

And the proof given in the book is the one I described above, so one way or another Banach-Alaoglu is the source of compactness here. (OK, if you dig deeper, it's Tichonov)

[bronze magnet]

I do not understand how the last part of equation (6) is obtained. Any help?

[quaint herald]

each of the norm terms in the middle expression is bounded above by the H^1_0 norms of v (respectively, u), just from the definition of the H^1_0 norm.

[bronze magnet]

Oh, it can be deduced from $|u|{H^1}^2 = |u'|{L^2}^2 + |u|_{L^2}^2$ as a definition of $H^1_0$ norm?

[untold delta]

fnechz

[quaint herald]

yes

[junior bloom]

So given the Sobolev space $H^k$ we "naturally" define $H^{-k}$ to be its dual. If we have $\mathcal{H}^\alpha={u\in H^\alpha:\sum_{j\in\mathbb{Z}}\lambda_j^\alpha|\langle u,e_j\rangle|^2<\infty}$ for a spectral basis ${e_j}{j\in\mathbb{Z}}$ in what way is it natural to to have $\mathcal{H}^{-\alpha}={\omega\in H^{-\alpha}:\sum{j\in\mathbb{Z}}\lambda_j^{-\alpha}|\langle\omega,\varepsilon_j\rangle|^2<\infty}$? There's a name for this right?

[untold delta]

teafortwo

[astral vine]

Depends on what is Lambda j

[junior bloom]

The eigenvalues

of some compact Hermitian operator

[astral vine]

In this case this is the closure of the range of the alpha-th power of tour Hermitian operator

(if it is injective with dense range oc)

[junior bloom]

Oh that's a good observation, I hadn't thought about that

[astral vine]

You can also think of it as the domain of the -alpha-t power of your operator

[astral vine]

Sadly not in my knowledge that's why I started to write some "lecture notes"-pdf about it

Triebel does it in weird, but also tedious generality

Hitchikers' guide do not talk about Bessel Potential space for the case p=/= 2

The Fractional spaces treated in it are only Besov spaces

That's why I don't like it

Yes

Other book ta deal with it are Functional Analysis books

like ones on Interpolation theory

Functional Calculus etc.

But they generally stop with the Rn case only

But here it is

Last part

deals with all kind of fractional spaces except Triebel-Lizorkin

you have no idea

...

This is the more important case

for BVP

p =/= 2

But Generally there is no simple accessible references about it, like general trace theorem with exact regularity

[waxen bobcat]

Oh, I was wondering why it's hard to find references about the trace theorem

[astral vine]

Because proofs are highly non trivial and done in full generality, with geometric measure theory etc.

Jonsson & Wallin proved it

But hell, it's unreadable

Only Zhonghai Ding's proof is readable

an still require complex interpolation and is made in the case p=2

[waxen bobcat]

Thanks! I found the one by Zhonghai Ding

[astral vine]

Notice the proof cannot be adapted straight forward for case p =/=2, due to the use Fourier Plancherel Equalities in the last part of the proof

[waxen bobcat]

And Jonsson & Wallin proved it for 1 <= p < infty? Also for higher derivatives? Which paper / book are you referring to exactly?

[astral vine]

Jonsson and Wallin proved it for 1<p< infty for restriction to (d-1)-set for subset of Rn in the Alfhor-Davies sense.

The book is called 'Function spaces on subset of Rn'

for p less than one in the case of Besov/Triebel-Lizorkin space is only made on Rn with trace in R^{n-1}

[waxen bobcat]

Ah, this one, sadly, I don't have digital access to it

[astral vine]

Above result for p less than 1, is made in Sawano's book on Besov spaces, a proof for homogeneous function space case is done in some paper written by B Jawerth

[cold basin]

Hello! I'm not sure whether I should be here or in #advanced-probability , but here goes:

I am wondering whether it's possible to show that a random walk on some lattice will converge to Brownian motion, and whether one can find the corresponding PDE?

[tranquil steppe]

Sure, one can approximate the Brownian motion by random walks. What's your question, exactly?

I'm not sure what you mean by "the correspoding PDE", but the elliptic operator associated with the Brownian motion is (half) the Laplace operator.

[cold basin]

Say we've got the triangular lattice like this, can we make the grid smaller so that we approximate the brownian motion with the random walk? And can we then get a solution to the probability distribution p(x,t) for the particle in the continuous case? The larger triangle is the boundary

[astral vine]

@orchid reef There is the book of Runst & Sieckel's book "Sobolev Spaces of Fractional order, Nemytskij Operators [...]"

[tranquil steppe]

Oh, I didn't expect a triangular lattice. I'm pretty sure this converges to the Wiener process too, and you can probably use the central limit theorem to that end, although I don't have a full justification in my mind.

"And can we then get a solution to the probability distribution p(x,t) for the particle in the continuous case?" A probability distribution of what? The exit point? In that case, what do you mean by "getting a solution"?

[cold basin]

well, ideally I'd like to get the distribution of the first exit time

the bolded sides of the triangle are an absorbing boundary

[tranquil steppe]

That seems like an interesting problem. I don't have an idea how to obtain a closed form for this distribution (if it's even possible).

[junior bloom]

yes, donskers theorem should reproduce

with all regular lattices

you can actually do it with a random lattice increasing with depth

[cold basin]

what do you mean by this?

[junior bloom]

n-dimensional uniform random variables in rd , as n goes to infty

the graph laplacian on this random set converges to the laplacian

both generate the random walk

[cold basin]

do you know if there's a way to find the solution for p(x, t) here once we have the PDE? I'm kind of out of my element here, but I came across the Feynmann-Kac formula. Do you know what else could be useful? Or could you point me in any particular direction?

I'm also looking to find the distribution of the first exit time of the brownian motion particle from the triangle. Initially I looked into the harmonic measure, but it doesn't quite give me what I want (I think)

like I mentioned above, bolded sides of large triangle are absorbing/have dirichlet boundary conditions of p(x, t) = 0

[junior bloom]

oh thats an interesting and tough problem. the c0 cusps on polygons tend to break a lot of bm theorems

[cold basin]

oh I see ...

how so?

[junior bloom]

idk lol, its just kind of singular. intuitively imagine that lots of bm properties come from reflection and bm likes to get lost in cusps, e.g. bm can enter a fractal boundary and never leave in extreme cases

there are a lot of higher dimensional first exit theorems but usually for radially symmetric things, my suggestion is to look at the fokker planck evolutions at the borders

alternatively monte carlo it

[cold basin]

Hmmm. I see. Essentially I've got the answer after some pretty unwieldy counting/combinatorics for the discrete case (which is what the figure above represents), but I wanted to see what happens if we try to go to brownian motion in the same space. It does seem pretty hard... which is why I was stuck lol

So would this be a problem in the same way if we literally just had like a box in R^n? Because of the cusps?

Like not even the fact that I've got this triangle lattice or anything

[junior bloom]

the box is easy because first exit is just the min of first exit in each dimension

which is symmetric

anyways no, i think this wouldnt be an issue either

becayse its not a true cusp

im just intuitively describing why c0 is not nice for bm boundaries

[cold basin]

but the triangle is symmetric too, and the next dimension would look like a tetrahedon and so on. They were chosen specifically for their symmetry (because it made the counting way easier)

[junior bloom]

yeah i had that thought, the infinite limit of a triangular graph with the graph edge metric would turn the triangle boundary into a sphere or something

it depends on the metric of your graph

if you pick a euclidean metric between nodes then it should remain a triangle bpundary

but if the distance between nodes decreases in a certain way 🤷‍♂️

[cold basin]

so where exactly is this important?

[junior bloom]

its important because its implicit in the graph laplacian

think about it, if the graph boundary is all equidistant from some center point then in the limit the metric has to make it a sphere

which you may be able to use to find first exit time from a sphere

which is very well known

[cold basin]

hmmm. How does this usually work if you're dealing with Z^d (like the cube grid) and you take the limit? Is it not the euclidean distance?

[junior bloom]

good question

lol

[cold basin]

thank you for your patience 😂

[junior bloom]

my naive sense is yeah, the cube becomes a sphere for l1 metric

in a zd walk

in limit an l2 sphere

[cold basin]

I'll have to look into this...

[solid flint]

Hi, Is there a Lyapunov theory for polynomial stability of PDEs/ODEs?

[astral vine]

Not in my knowledge, but similar results are sometimes obtained for reaction-diffusion equation by Lp-Lq time decay estimates on semigroups generated by (negative) elliptic operator

Like Lp-Lq estimates of the Heat semigroup

In general a semigroup have exponential decay iff 0 is in the resolvent set of the operator

When you want to perform bootstrap techniques for mild solutions of some nonlinear PDEs, both approach can be used cutting the spectrum of the operator

M.A.Johnson, P.Noble, L.M.Rodrigues, K.Zumbrun - Nonlocalized modulation of periodic reaction diffusion waves: nonlinear stability

But this is done for a very particular kind of problem here (since we want to recover some orbital stability)

[solid flint]

Thank you @astral vine

[dusky glen]

hello! anyone familiar with zubov's method and zubov's equation?

[errant gust]

I already know that log|x| is harmonic - does anyone know how to proceed?

[lilac barn]

Rewrite ∂n as ∂r and dS = r dθ for |x| = r and see how far can you proceed

Once you have simplified it enough, the identity [ \frac{1}{2\pi} \int_{\partial B_r(0)} u d \theta =\left( \int_{\partial B_r(0)} \frac{\partial u}{\partial r} dS \right) \log r + u(0)]
should provide the rest.

[untold delta]

cocat

[lilac barn]

Actually, u(0) should be replaced with β in the above identity, which will go to u(0) as t → 0

[dusky glen]

I am reading section 34 (Zubov's method) in the book named stability of motion written by Wolfgang Hahn.

The proof of theorem 34.1 seems only proves one side: $A$ is a subset of domain of attraction. But the other side, i.e. the domain of a attraction is a subset of $A$ is not proved.

Could someone help to whether I misunderstood something?

Thank you so much!!

[untold delta]

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

[dusky glen]

🥺 anyone could help....

[errant gust]

it seems that this identity is kind of what we're trying to prove, and we can't use it

do you know how to surround this problem?

[tranquil steppe]

For the first term - log |x| is exactly log t. Putting that in front, you're left with an integral of size t (assuming the derivatives of phi are bounded). The product goes to zero.
For the second term - it's even easier. If you evaluate the normal derivative of log |x|, you're left with 2*pi times the average of phi (over the circle). Since phi is continuous (I guess), that average tends to phi(0).

[errant gust]

Thank you so much! I think I was able to figure this out

[sullen flax]

Ok can anyone in here explain how exactly pur shock functions interplay with navier stokes? NS basically says mass and momentum are conserved and shocks just include energy and then give a wierd solution right so it shouldn’t violate NS but i can’t see how continuous flow or wave functions can support a shock, even though they most be possible to model simply since shocks are a stable and reproducible flow feature not magic yknow- interested both intuitively and numerically here as im working up to numerically modeling some interactions

[lilac barn]

Why do we have that u is a minimiser of I over A? In the previous sections, the crucial point was that I is weakly semicontinuous where I[w] = \int L(Dw,w,x) and L is smooth, bounded below and is convex in the first variable.
In our case, L(p,z,x) = 1/2 |p|^2 - fz and so it is not bounded below. Why does weak semicontinuity for I still follow?

(Page 468 of Evans)

[blissful dagger]

I am studying some Calculus of Variations...does anyone know how to set up this problem so that it can be solved with Euler-Lagrange,
\[\text{maximize} \; \; \int_{0}^{25} -\rho(t) \log(\rho(t)) \; dt \]
subject to,
\[\int_{0}^{25} \rho(t) \; dt = 1 \]
\[\int_{0}^{25} t\rho(t) \; dt = 15 \]
I am trying to find $\rho(t)$, a probability density function. I figured,
\[J = \int_{0}^{25} (-\rho(t) \log(\rho(t)) + \lambda \rho(t) + \theta t\rho(t))  \;dt  \]
but I am not sure that this will yield a solution.

[untold delta]

Skid

[blissful dagger]

Note, this yields p(t)= e^(lambda + t*theta - 1)

Which may be the solution I just can't solve for the multipliers without a numerical solver so I am unsure if it is right

[tranquil steppe]

Maybe L(p,z,x) is not bounded from below, but I[w] is, by Poincare's inequality. The term \in fw is controlled by the term \int |Dw|^2. I guess this should be enough for your purposes.

[lilac barn]

I don't see how would that work. The proof for the weak semi-continuity uses L is bounded below to obtain this first inequality. If it was merely the case that I[w] is lower bounded then we can't restrict to a small subset.

(unless you are suggesting that there exist a different proof which assumes merely that I is lower bounded in place of L)

[brave bay]

When you pass to an strongly L^2 convergent minimizing subsequence won't that negative term in the functional converge to the integral of -fw automatically (by say Holder's)?

Doesn't that mean you just need to prove your weak semicontinuity for the positive part of the functional? (and here the general convex/bounded below criterion is applicable).

[tranquil steppe]

You're right, this part suggests that the stronger assumption L >= -beta is actually needed.

In your case, the functional has a non-negative part (1/2 |Dw|^2) and a continous linear part (f*w), so you can probably get away with using semicontinuity of the first part only.

[mental geyser]

Does anyone have recommendations on obtaining Green's function for a heat equation with Robin boundary conditions? Have done it previously for both Cauchy-Dirichlet and Cauchy-Neumann.

[astral vine]

In general Green Functions are not really computable, you generally obtain those via abstract theorems

[mental geyser]

Yep. We've been using methods of images and Duhamel's principle to obtain them from fundamental solutions or their extensions.

I just haven't encountered a Robin condition until now, and can't find much in the literature.

[astral vine]

You may infer it using what happens on the half space using the fact the Robin laplacian is given by the sesquilinear form

<Du, Dv> + <alpha . u, v>

where the last bracket is expressed on the boundary

[narrow ledge]

so is evans the goto for a beginner?

[quaint herald]

It is a pretty good first book imo, yes.

[versed shoal]

i have this question but I don't understand the solution

specifically these lines

idk why those screenshots are so blurry, lmk if they're illegible

but I kind of understand what they're doing because $$E''(t) = 2 \int_\Omega ((\partial_t u)^2 + u \partial_{tt} u)$$ and $\Delta u = \partial_t u$ so they want to transform one half of the integral to the first term and the other half to the other

[untold delta]

George!

[versed shoal]

I'm just not sure how we're getting from $(\Delta u)^2$ to $u \Delta^2 u$ (not sure what this means - double laplacian?) to $u \partial_{tt} u$

[untold delta]

George!

[versed shoal]

oh I guess you get $\Delta^2 u = \partial_{tt} u$ just by differentiating $\Delta u = \partial_t u$ wrt $t$ but I'm still not sure how to get from $(\Delta u)^2$ to $u \Delta \Delta u$

[untold delta]

George!

[versed shoal]

oh its just green's identities twice (??)

[solid flint]

Hello, any good references about constructing Green's functions? especially for elliptic PDEs with variable coefficients in bounded domains with two different kinds of boundary conditions, please?

[bitter hollow]

What exactly do you mean by constructing a Green's function? Do you essentially mean, how do you create a direct solver for computing the solution to a variable coefficient elliptic system?

[astral vine]

@quaint herald If I have a (Pseudo-)Differential operator, in some appropriate symbol class, say P(x,D) : Is the domain on H^{s,p} (i.e. the set f in H^{s,p} such that P(x,D)f is in H^{s,p}) always the closure of Schwartz functions under the graph norm ?

If this is true, do you have any references please ?

[solid flint]

I mean the techniques used to find an explicit solution to certain pdes

yes

[bitter hollow]

By explicit solution, do you mean analytical? If so then the answer is that there are essentially none. In some simple cases you can express the solution in series expansion.

If you mean numerical, then this is a very active field of study, and I was briefly involved in such research myself in the early days of my career.

[solid flint]

yes, I mean analytical when it is possible

[bitter hollow]

I think series solutions for nice domains and fairly simple coefficients might be the best you can get, constructing Green's functions is HIGHLY nontrivial.

To the extent that an elliptic system is sort of like an arbitrary positive definite linear system on a Sobolev space, then constructing a Greens function is like determining an analytical formula for the inverse of such operators in general. Of course there is some nicer structure than this but I'm just trying to say that it's a hard problem in full generality (variable coefficient, mixed BC, arbitrary domain)

That being said, there are a lot of results that study the behavior of Greens functions

Intuitively speaking, we have full understanding of what greens functions for free space Poisson look like. Then a variable coefficient problem can be viewed as some perturbation of that, so estimates can be derived for the nature of the singularity, growth rates, etc

[astral vine]

A way to do so is to extend the problem to whole Rn and use Pseudo Differential operator Theory, then correct what happens on the boundary via some integral operator called boundary layers

This is a very technical high end approach

This was used to derive the analycity of the Stokes Dirichlet Semigroup on domains from what happens on Rn, and gosh this is so heavy and technical

even for a "simple laplacian" with no boundary conditions

[quaint herald]

honestly not sure off the top of my head.

[astral vine]

Even in the case of differential operators with constant coefficients (polynomials in xi) ?

[quaint herald]

I mean if it's true for constant coeff DEs then it probably won't be that hard to prove, but I have not seen such a result.

[astral vine]

But we agree that it seems kinda natural, right ?

[quaint herald]

By general principles, I think its true for self-adjoint P (so this would include the Weyl quantisations of real-valued symbols) when p=2, but in microlocal analysis you almost entirely work with L^2 based spaces, so it is quite rare that I think at all about mapping properties on L^p based ones.

Kiiinda, but I wouldn't like bet money on it.

I'd try to prove it for constant coeff DE first before growing more confident. (And I personally won't do this right now lol.)

[astral vine]

For fully general pseudo maybe some ill examples , but yeah cconstant coeff first. Hmmm

Maybe for general pseudo we need almost polynomial growth in x (which is implicitly states in main class of symbols iirc) othewise we couldn't even define symbols on S

I will give it a try

[solid flint]

Thank you @bitter hollow

[solid flint]

Thank you, can you provide any links please?

[solid flint]

Thank you

[astral vine]

This is true for constant coefficients with even a su Space of S of Schwartz functions s.t. their Fourier transform is identically equal to 0 in the vicinity of 0

I just checked it

[astral vine]

on Lp (p>1)

[narrow ledge]

Should I learn Functional Analysis before PDEs?

[astral vine]

This is mandatory to learn at least Hilbert/(reflexives) Banach spaces theory, Closed Graph Theorem, some unbounded operators.

In Hilbert spaces Theory it includes Lax-Milgram stuff

deeper/other kind of Functional Analysis will depend the way you want to investigate PDEs

[narrow ledge]

Cool, guess I'll get rudin's book. Although many don't like him, those who do like rudin style say Functional Analysis is even better than his other books

[astral vine]

Rudin's Functional Analysis is great but probably not for PDE purposes

It contains at least distribution theory, and some very general Functional Analysis on Banach spaces

but it is already too much oriented for C*-Algebras etc.

which is far from basic Functional Analysis for PDE considerations

[narrow ledge]

(1) So you're saying Rudin goes further than what's needed for PDE?
I ask because I know the professor who gives Functional Analysis in my university uses Rudin.
Before that course we will already have covered Banach and Hilbert spaces in "Mathematical Analysis II" (Which is two units in calculus theorems, two chapters on banach and hilbert spaces; and Kolmogorov's functional analysis book is in the bibliography).
(2) Maybe this "Mathematical Analysis II" course would already be enough to start getting into PDEs?

Average Rudin Enjoyer

[astral vine]

Rudin's book are very good, and I personally like the Functional Analysis' book, but the treatment is not adapted in my opinion for a PDE point of view : the way results are stated or proved is in a too much different spirit.

[junior bloom]

Taylor's first PDEs book, Evans's PDEs book, and Brezis's functional analysis and PDEs books are my usual references

One day when I really need to learn PDOs I will learn Taylor volume 2

[astral vine]

To not look like a Fanatic, I expected someone else to recommend Brezis, everything goes according to the plan.

[quaint herald]

To look like a fanatic, you should read Hormander I-IV :).

[astral vine]

oh no

[quaint herald]

On a more serious note, I think most analysis people would get something of value out of at least the first volume.

The other 3 are amazing, but more references than a place to learn the heavy machinery.

And for a first course in PDE I wholeheartedly agree with like Evans/Taylor1/Brezis

[astral vine]

In my opinion, difficulties for Hörmander's books lies in the fact that (almost) every single result is sharp in both the statement it self, and the proof

So if a proof have to be complicated but needs to be "at the beginning" then it is

[quaint herald]

yeah he aims for like maximal generality, he is famous for this stylistically.

it's beautiful for what it is

[astral vine]

I agree

But it makes it really difficult to read it and to learn simply form it to me

[quaint herald]

Yep, like I said, better as a reference.

[hushed wasp]

How do I show a function is borel measurable? First I show the function is in the borel set? and that the preimage of the set is in the sigma-algebra generated by the borel set?

[river path]

It's similar to showing a function is continuous in topology. You want to show that the preimage of any borel set from the codomain is also a borel set in the domain. Well, since open sets (or closed sets) generate the Borel sigma algebra, you can just look at preimages of each open set (or of a generating collection of open sets, like open balls in a separable metric space) and see that those preimages are measurable.

You dont need to show that those preimages are open. The function is not necessarily continuous.

Another way of proving a function is measurable is to write it as a limit of functions that you already know are measurable (like step functions on measurable steps). Pointwise limits of measurable functions are measurable, so this works.

[slender fulcrum]

I need to bound $||\phi|^{10/3}{L^{\frac{10}{3}}} - |\psi|^{10/3}{L^{\frac{10}{3}}}|$ for two $\phi,\psi\in H^1(\mathbb{R}^3)$ by $C|\phi - \psi|_{H^1}$. I probably have to apply Sobolev embedding somewhere but I dont see how to get there with these nasty exponents. does anyone have some clue how to do this?

[untold delta]

whzup

[tranquil steppe]

They way you wrote it, it cannot be true. Both sides have different scalings, so if you multiply both phi and psi by the same (large) constant factor, the inequality fails.

[astral vine]

This argument would be true if he was looking at homogeneous estimates

by the way he have to remove the 10/3 power exponents to make it easy to prove

[astral vine]

remove the 10/3 power

[slender fulcrum]

ah yes excuse me i forgot to add that $|\psi|{H^1}, |\phi|{H^1} \leq R$ for some fixed $R > 0$

[untold delta]

whzup

[astral vine]

Then there is almost nothing to prove in this case

[slender fulcrum]

im doing b) in this exercise and i know to bound all the other terms in $\mathcal{E}^V(\phi)$ by the difference of the $H^1$ norms but not the 10/3 part

[untold delta]

whzup

[whole badge]

say I have a 1st order elliptic operator restricting to $D:H^1 \rightarrow L^2$. Can I conclude somehow that the orthogonal complement of its image is contained in $H^1$? I can provide more context if necessary

[untold delta]

Brad

[astral vine]

It is not true for general operators

consider the gradient

its dual operator will be -div

the orthogonal complement will be divergence free vector fields with 0 tangential conditions

which is not included in H1

[whole badge]

wow ok, thanks. How about Dirac operators? I'm reading the proof that the Dirac operator is Fredholm on page 210/211 of Salamon's Seiberg-Witten invariant book, somehow they use this fact saying "by elliptic regularity" but it doesnt seem to follow

[astral vine]

D= d* +d ?

[whole badge]

most other resources I can find use a different approach by considering its closure but it seems quite a bit more complicated

its the one that acts on sections of the spinor bundle

on a compact spinc manifold

[astral vine]

You could check Mitrea's work

I guess its done

but you need to be aware about a lot of technnology

In general the domain of Dirac may not be H1 unless everything is fully smooth

[whole badge]

alright, I'll check it out. I was trying to get through lawson and michaelson's section on it today, but it is very dense and uses distributions which I havent learnt yet. And yeh, in my case the Dirac operator is defined on smooth sections first, before they talk about this restriction

[astral vine]

You will always need distribution theory

Yes

provided s>0

[whole badge]

yep I figure its time to learn some... I'm pretty new to pdes, they've only come up once I started looking at this topic

[astral vine]

Prove for H^1 thne use complex interpolation to carry over the result for 0<s<1

[astral vine]

Dirac operators are very difficult to handle

There is already a lot of trouble already on Rn

the whole space

where you have access to Fourier Analysis etc.

[whole badge]

what about them makes them difficult would you say? they seem to come up naturally from the differential geometric point of view

[astral vine]

A lot of things : but in general they full meaning only in the L² sense

carry over an appropriate Lp meaning, with nice behavior like bisectoriality or what ever relies on difficult Harmonic Analysis questions

This is interesting for many reason, but one is to Generalize properly operators that comes from Electromagnetism and Fluid Dynamics to carry over a nice generalization

for instance magnetic field B in Electromagnetism is a 2 form, such that d°B=0

in dimension 3 one would say curl (B) = 0

but this is no longer the case in dimension n

[whole badge]

nice, that seems cool. thankfully L^2 seems good enough for my purposes, but its neat that it can be generalised that way

[astral vine]

The L² setting have already big troubles

if you are on bounded Lipschitz domains

you can build functions such that div(u) and curl(u) in L² but \nabla u not in L²

So you don't have access to H1

[whole badge]

is that an easy counterexample? im on smooth compact manifolds which are surely nice enough right?

[astral vine]

It's a paper of martin Costabel I don't remember the name

but since it is Costabel

probably not a simple counterexample

[whole badge]

lol alright

[astral vine]

If you are on smooth opensets of smooth compact manifold

then everything is okay iirc

according to Pseudo-Differential operator Theory

[whole badge]

ahh ok I need to start reading about pseudo-differential operators next week

before trying to understand the atiyah singer index theorem

[astral vine]

for me smooth says C infty

not C² or Ck metrics

[whole badge]

yep

well thank you very much for your help, I need to sleep now though

[astral vine]

Haha, I didn't help that much

Good night

[junior bloom]

I studied functional analysis with one of Hormander's advisees.

Fun guy

Anyways what I really need right now is a big list of theorems guaranteeing the convergence of Schwarz distributions

But being a relatively esoteric subject this amounts to me having to go through all my references and work through all notions of convergence

[astral vine]

Chapters 1 & 2 of Bahouri Chemin Danchin contains some results about convergence in nice subspace of S'

like sufficient conditions s.t. the limit won't contain any Fourier transform of Dirac masses and other possible weird measures

[junior bloom]

Thanks for the reference.

My problem is I have a bunch of weak solutions to an (S)PDE changing the parameters and I'm trying to suss out the property of the thing in the limit. For the physical system I'm studying to be a good field theory in the limit it should be a solution for another PDE.

My problem is I have some weaker than weak energy estimates (they're probabilistic, in fact), but even restricting to the set where they're just reasonable energy estimates I'm trying to wonk out some ways to characterize the limit's behavior.

I'll check out BCD!

[untold delta]

Haris

[hushed wasp]

@quaint herald

[crude hill]

Can someone help me with the second part? I get that u(x,y)= f(x+y)+g(x+2y/3) and that for the second part it will be f(2x)+g(5x/3)=cos(x) but i dont know how to find f or g??

[meager dune]

Yeah so note that we can rewrite that as f(x) + g(5x/6) = cos(x/2)

So if we pick g to whatever we want, f is determined by this

[crude hill]

so g(5x/6) would be cos(5x/6) for example anf f(x)= cos(x/2)-cos(5x/6)?

[meager dune]

I meant more generally pick any g and then let f be defined by f(x) = cos(x) - g(5x/3)

There isn't a unique solution since they've only specified it on one line i believe

[crude hill]

yes okay thank you so much

[meager dune]

Np

[stark thunder]

does anyone here know enrico valdinoci

[cloud glacier]

Hello,

I'm looking for help to solve this PDE with P1 - FEM

Here is the domain

I know how to get the variational form, but i dont know what to do to with it. In fact that's the first time i encounter such domain condition

I usually solve such PDEs with FEM over domain that don't have "holes", and kind of simpler boundary conditions

[pale grail]

@cloud glacier Well, we can start by multiplying the PDE by a test function v and integrate. Use integration by parts on the laplacian of u.

You should end up with a bilinear form a(u,v) = F(v).

Then you might try to apply the Lax-Milgram lemma to guarantee a unique weak solution.

Probably for you, when you do the integration by parts, you will have boundary integral terms, and you just need to use the boundary conditions provided to get that bilinear form a(u,v).

[cloud glacier]

Thanks for you answer. Well in fact i end up with this equation, and i got to find a way to compute (numerically, with code) each of these integrals

I know how to compute the first two, they are Mass and Stifness matrices that i can get by assembling elementary matrices

But i'm kinda lost for the others, should i get them by quadrature techniques ?

[pale grail]

@cloud glacier _I see, you are jumping into the approximation.

Just to take a step back. The variational problem consists of finding $u \in H^1(\Omega)$ such that $a(u,v) = F(v) \quad \text{ for all } v \in H^1(\Omega),$ where $a(u,v) = \int_\Omega \nabla u \cdot \nabla v + \int_{\Omega_2} u v + \int_{\partial \Omega_0} uv$ and $F(v) = \int_{\Omega_2} v + \int_{\partial \Omega_0} v$.

[untold delta]

Xillicis

[pale grail]

Then you can define your finite element space (P1 continuous) as something like $$V_h := { v_h \in C^0(\Omega) : : : v_h|_T \in \mathbb{P}_1, : \forall T \in \mathcal{T}_h }$$
where $\mathcal{T}_h$ is the elements of the mesh. This space has as a basis the $\varphi_j$ tent functions as you have written.

[untold delta]

Xillicis

[pale grail]

Once you go through the discretization, the boundary integrals can be computed exactly. But they are linear on each edge, so a single point quadrature rule will give you the result.

[dense badge]

help

[hushed wasp]

Does this make sense in order to be borel measurable?

[pliant osprey]

My question is not about pdes directly but as hyperbolic polynomials arose from pdes my question might fit best here. A polynomial f is called hyperbolic w.r.t. $e\in\mathbb{R}^n$ if $f(e)\neq 0$ and $t\mapsto f(x+et)$ is real rooted for every $x\in\mathbb{R}^n$. One can say something about the coefficients of $f$ but I am not sure what it was exactly and I struggle to find a reference. One of the following should be true: a) One can assume the coeffients of f to be real. b) The coefficients of f have the same phase (and thus can be assumed real positive after division by a normalization factor).

Does anybody know whether a) or b) is true and possibly has a reference for this?

[untold delta]

Stephan

[narrow ledge]

damn Evans is such a boring looking book, Brezis looks so much more interesting

[astral vine]

Truth

[tame igloo]

Im working on numerical methods for PDEs and specifically with the heat equation and i was asked if it would be better to use the ADI Method or the Crank Nicolson method. If these are both implicit methods how am i able to determine which method is more efficient to use ?

[lament agate]

The ADI method is in some sense the canonical best choice for the pure heat equation as it's strictly cheaper than Crank-Nicolson for the same problem

Specifically, the ADI gives tridiagonal matrices whereas CN doesn't

Though CN matrices are banded also, albeit with a larger bandwidth (I can't remember the bounds OTTOMH)

@tame igloo Is your problem in 1D or 2D?

[tame igloo]

@lament agate it is in 2D, that makes sense if it is cheaper i was just thinking about how they are both implicit so i could not decide

[lament agate]

Yeah in 2D ADI will be cheaper

🙂

Also there's more to a numerical method than just whether it's implicit/explicit

Stability and convergence is also something to read about

[tame igloo]

got it thanks so much!

[verbal nebula]

Evans is actually pretty good

[astral vine]

Translate the domain in the whole space

yes

[late moth]

I'm trying to work through some problems right now, and I'm a little bit lost on this uniqueness problem.

Suppose $U\subset \R^n$ is a bounded domain with smooth boundary. Show that the initial boundary value problem
\begin{align*}
u_t &= \Delta u + \int_U u^2dx &\text{in $U$ for $t > 0$} \
u&= 0 &\text{on $\partial U$ for $t > 0$,} \
u(x,0) &= g(x) &\text{for $x\in U$},
\end{align*}
may have at most one solution $u\in C^2(\overline{U}\times [0,T])$, for any $T > 0$.

My first thought was to do the usual method of multiplying $u$ and integrating over $U$ with
[\int uu_t dx= \int u\Delta u dx + \int u(x) \int u(y)^2 dy dx ]
then applying the divergence theorem as necessary, but the second term (if I've written it correctly) is confusing as I'm not entirely sure how to tackle it.

[untold delta]

kirby

[astral vine]

Hint :

Look at the derivative of

[untold delta]

Functionanatolysis

[astral vine]

(for g=0)

then check what happens for g non zero considering two solutions

u and v, and then w=u-v

the next step is to think about Gronwall Lemma type inequalities

then integrate it

to obtain

[untold delta]

Functionanatolysis

[astral vine]

(it is almost finish you need two shorts extra steps)

(I don't know if there is a more elementary simple proof, I don't think so)

[brave spindle]

does anyone have a strict proof in the sense of distributions for the heat kernel converging to the delta function?

[tranquil steppe]

Sure. Just look up chapter 2.3 of Evans' book on PDEs (or probably any other PDE book).

[astral vine]

prove it for Schwartz functions, then the result will hold by duality

[untold delta]

kirby

[astral vine]

@late moth no

from there

use the fact that |a²-b²| = |a-b||a+b|

then use

[untold delta]

Functionanatolysis

[astral vine]

it gives then

[brave spindle]

@tranquil steppe that book doesn't have a strict proof and thanks @astral vine

[untold delta]

Functionanatolysis

[astral vine]

then apply Gronwall's inequality in its integral form

not derivative

Then automatically

[untold delta]

Functionanatolysis

[astral vine]

If

Then

but here K=0, t_0 = 0

[late moth]

Right, this makes sense, then of course, w must be 0. This helped a lot, thanks so much!

[astral vine]

You are welcome

[vital leaf]

Does anyone know any resources that try to answer this older SE question? https://math.stackexchange.com/questions/450367/is-there-any-theorem-that-tells-us-how-many-ics-or-bcs-are-needed-for-getting-th Like the OP I've tried looking for a general discussion on this but I I can't find any (or am not connecting the dots; does Cauchy-Kovalevskaya indirectly answer this? i dont know)

[tranquil steppe]

Do you have anything specific in mind? I would trust Willie Wong's comment that very few general rules are known for PDEs.

[vital leaf]

Well yeah, a general theorem doesnt seem to exist so it would be nice if instead there was a discussion about how one would determine the number of IC's/BC's for different classes of pde's. But maybe this is asking too much since it seems in practice people look at a specific pde and then try to prove that it is well posed for a given number of IC's/BC's...

[slender fulcrum]

im no expert but maybe something like algebraic analysis might be helpful

[twilit rover]

What's a good PDE book for a beginner to PDEs?

Folland or Evans?

I read some of each and it seems that Folland uses distributions and Fourier transforms from the start, which illuminate some things.

[late moth]

Evans is a good beginning graduate level text that's also readable, but I've heard good word about Strauss as a more introductory text

I don't have much experience with Folland

[astral vine]

Brezis + a complementary book involving Distributions like Folland

This more for the general Theory of function spaces used in PDEs and basic Functional Analysis techniques to deal with non elliptic problems, or some general evolution problems involving linear parts (quasi and semilinear évolution PDEs from Semigroup Theory, like Heat, Wave or Schrodinger equations)

Evans is, to me, more about cooking recipes more than general Theory, which is nice to have some insights about what happens, but very bad, imo (not so humble tbh) to go on/to adapt to more general settings

[tranquil steppe]

I suppose it can differ depending on which way you want to go next. I would strongly recommend the Evans book, as it gives a light overview of many classical and modern methods. The fact that the book doesn't try to develop "the general theory" is actually a good thing, from my point of view. It lets you see the main idea without caring about technical details.

In my personal experience (which is mostly with harmonic maps and their flows, as well as their p-harmonic counterparts) the general theory is not helpful anyway - most of the proofs in the literature use methods described in Evans book, but adapted to the specific setting.

And if you're a beginner, your future toolbox will probably be a very small fraction of what Evans and Brezis give in their books. Which fraction, it depends on what area of PDEs you choose. So it makes sense to avoid technical things at first (theorems with greatest generality, optimal boundary assumptions etc.).

[astral vine]

I strongly agree about the point "depends on what's next"

I'm more into sharp Regularity Theory via Interpolation Theory on Lipschitz domains, Functional calculus stuff for appropriate operator and some nonlocal linear elliptic Parabolic problems (Stokes/Navier Stokes equations in geometric euclidean settings for instance)

Optimal (Maximal for me) Regularity is very important for general PDEs

That's my point promoting sharp things, general thm/settings, that goes beyond the L² or H¹ case

[bitter hollow]

The strength of Evans' book lies in the latter half which covers some techniques for nonlinear equations, not in the functional-analytic theory where linear PDE can nearly be abstracted away as linear operators

[uneven wadi]

I highly recommend Salsa's book "Partial Differential Equations in Action: From Modelling to Theory" as it gives intuition about what is going on. The first part of the book is essentially an introductory course to PDEs, it deals with the basics (heat equation, Laplace's equation, linear conservation laws, wave equation...). The second part of the book starts with some functional analysis and later on it introduces distributions, Sobolev spaces, and variational and weak formulations

[solemn brook]

Hello, I am working on a diffusion-reaction equation and I am stuck on a system of differential equations (see picture). I don't think I can use linear algebra with the eigenvalues/eigenvectors due to the laplace operator (I am not used to work with that 🙂 ). I have also non dimensionalized the system to see if I could obtain something easier to solve but I am still stuck. If you someone have any ideas/hints on how to proceed that would be nice ! Have a nice day 🙂

[bronze gate]

(sorry that Im being like this) Could you actually attach the picture (i dont want to install malware (not that I don't trust you))

[solemn brook]

Yeah of course:)

[twilit rover]

@astral vine Brezis doesn't seem to even cover PDE until chapter 7. Does this mean PDE is a branch of functional analysis?

[river path]

PDEs is a branch of functional analysis, yes

[lilac barn]

Per Evans, no.

[astral vine]

In my opinion, there is few schools to think about PDEs

Mathematical Physics,
Applied PDEs,
Applied Functional Analysis PoV.

Evans thinks only the 2nd one is the really valable way using sometimes Functional Analysis as tools and only tools.

This point of view is not very consistent for several reasons especially : For almost 15-20 years a substantial part of papiers in various PDEs subfield contains very deep exploration of some pure Functional Analysis and pure Geometric (Differential Géo or Geometric Measures Theory) tools which translate them into a sub part of those fields. People belonging only to the two first schools are less numerous than they were 25 years ago, because new appraoches require new deep tools.

Methods and schools are important

Because they determine somehow a philosophy to solve PDEs

This is important for transversal purposes

Like proving a results in some general way, allows people to use it in other fields like pure geometry

Or arithmetic

You could think about Optimal Transportation for instance

[bitter hollow]

Craig Evans' point is just that a huge amount of PDE theory is fairly orthogonal to linear and functional analytic viewpoint, and even though that viewpoint can grant valuable insight, it is never the "whole" picture. He has done a lot of work on fully nonlinear problems for which the functional analytic viewpoint is not as dominant. He would certainly agree with the notion that there are lots of deep connections to geometry. He wrote a book on geometric measure theory as well.

Also, I would dispute the claim that the applied PDE field is smaller now than before. There's been quite a massive explosion of activity in that area since the 2000s

[river path]

Meh

I basically think of all that more hands-on stuff as fringe functional analysis, but if that's not how these people perceive it then whatever

The flavor may be different but... is it really THAT different? I've never felt that it is.

[astral vine]

People being at the interface of 2 out 3 of any above areas are more massive there is less people doing only one type

That was what I claimed

[agile flame]

I think most people could perceived it really different, though..

Yes, Functional Analysis give a really nice insight about what happens with PDEs..

Yet, that doesn't mean every person could look at it and think fluently that it is Functional Analysis..

That's what I'm perceived when I met most of the faculty in my institute that works in PDEs though (which actually around 99%, only a miniscule that really dive deep in Analysis of PDEs)..........which mostly used numerics and felt that Functional Analysis is too abstract for them.

They know they work in certain settings of Sobolev Space, etc., yet they just felt to mention it once in the beginning and never talk about it again, just go on with how the result fit in with the physical system (most of my faculty work with engineers, though, so it kinda make sense to me)..

[astral vine]

Maybe Numerics could be understood as a part itself, to be fair I didn't think about it when I was writing above messages

[agile flame]

And complementing my message above, I think it worth to note my master study curriculum quite different, both in positive and negative way, because Fourier Analysis (which can be said as one of most important tools in PDEs) kinda reserved only for those who works in Mathematical Analysis group..

My PDEs course list as course of the Applied Math group and it kinda left behind (I mean, they used Strauss books for Advanced Partial Differential Equations courses, and Fourier Analysis just been touched when discussing Sturm-Liouville, so, yeah, I felt overwhelm when everyone around the world using Evans for their master study textbooks PDEs)..

[bitter hollow]

I believe it would be accurate to say that PDE is intimately tied with many, if not most, topics in real and complex analysis. Functional analysis does connect a lot of those topics too, of course. But I think it would be strange to regard, say, the areas of statistical mechanics and probability that surround nonlinear stochastic differential equations to be merely "fringe functional analysis". At some point if you take a sufficiently expansive definition of functional analysis, then perhaps, but such a definition is probably just the same as saying "analysis".

[river path]

Hmm

I suppose so

SPDE is a great example of something that significantly moves away from that

[bitter hollow]

Even if we're only looking at traditional PDE, then it's certainly true that differential things are precisely those things that have local linearity and functional analysis is basically about linear things, but that would also apply to anything about differential geometry. And at that point once again it's clear that functional analysis is pervasive, but the idea that it's all fringe functional analysis would be a very expansive definition of that field!

I myself usually spend a lot of time thinking about series expansions and such for PDE, so connections to linear spaces of functions is fairly clear, but even then, since I'm trying to construct specific expansions or functions, it starts to lose its functional analytic flavor

[tired axle]

i read this and thought "hm ange would get p mad at this take" and only just realized he left ages ago

[astral vine]

We deeply miss him

[river path]

Hahaha

Yes

I was thinking that too when I posted it

"This is an egregious opinion to have, but i vaguely believe it and it's a perfect ange troll so i have to post it"

[lilac barn]

Why was it necessary to recall that "\bar{u} has compact support" to use theorem 1? Theorem 1 doesn't make any assumptions on boundedness of U so I don't see why did Evans state that?

[tranquil steppe]

But you see that the statement of Theorem 1 only gives you local convergence, not global?

Since u bar has compact support, you can choose V to be e.g. a very large ball containing U, and apply Theorem 1 on the open set V.

Or to put it more simply, just use the convolution u^eps and see what happens.

[lilac barn]

Aah, right right, thanks!

[light parrot]

Hello, I am looking for coordinate-free techniques for solving linear tensor differential equations (in other words, Linear Ricci Calculus Differential Equations). Can anyone point me in a direction?

If that is confusing, another way of saying this is that I am simply looking for coordinate-free methods for solving linear partial differential equations with non-constant coefficients.

[agile flame]

I don't think I have specific books for this, I apologize..

However, you could try to seek from Exterior Differential Systems, which can be said as coordinate-free PDEs things.. Hopefully it is suitable with what you looking for..

[lilac barn]

How does one prove that u in W¹,∞(U) admits a continuous representative? For p < ∞, Evans does by considering the limit of smooth functions approximating the extension in the Holder space but he skips the case for p = ∞, and we can't use the same method to obtain a continuous representative since we no longer have approximation by smooth functions.

[tranquil steppe]

If the lack of smooth approximations in W¹,∞ bothers you, you could always lower your p. Just say that your u lies in W¹,2n (at least locally) and conclude it's continuous.

[lilac barn]

I mean idk, I am trying to stick with the material Evans has already introduced by that point, and there isn't any mention of lowering p's just yet. So do you have any idea what Evans is trying to do? (chapter 5 Morreys Theorem)

[tranquil steppe]

Wait, but doesn't it solve your problem right away?

[light parrot]

Thank you. I guess it isn’t too surprising that that’s what they might be called.

[agile flame]

Most books of PDEs that I know have geometric taste in their work mostly geared toward Theoretical Physicists that works with GR, so it kinda difficult to find something with coordinate-free approach for mathematical purposes only (mostly the physics prerequisites makes the reader feel uncomfortable)

Another Geometric PDEs that I known lurking around is in the field of Yang-Mills Theory, Gauge Theory, which quite hard, in my opinion, for first time read..😅

[lilac barn]

It might but I would like to stick to the resources Evans has introduced upto that point. He hasn't introduced any talk of lowering p's so I think there's a more elementary way to go about it.

[tranquil steppe]

Your use of words 'resources' and 'elementary' suggests that my point didn't come across. I simply said that a bounded function is locally in any L^p, which lets us apply the other results already available to you.

But I understand the need to know what was in the author's head (at least in the case of Evans, who is a great author in my opinion).

[lilac barn]

Ahh right, I was confused by the choice of 2n in "W¹,2n" so I supposed you were using some specific result. On further reflection, the choice of 2n seems appropriate.
In that case, would the following argument work: "Since u is in W¹,∞(U) so it is in W¹,2n(U) and so there exists a continuous representative u*. Moreover, for any , n< p < ∞, we have |u*|ₐ ≤ C |u|ₚ where | |ₐ is the Holder norm with a = 1−n/p and |u|ₚ is the Sobolev norm. "Letting p go to ∞", we obtain the result for p = ∞. This argument does seem easy and direct as Evans point out, but I am unsure if we can take the limit of Holder norms (Sobolev should follow from limit of Lp-norms)

[astral vine]

2 things this proof works only After localisation techniques

Hence produce estimates with constant C depending onp blowing up w.r.t p

But proved the continuity on each compact open set

Thus obtained estimates can not be straigthforward adapted to Rn since decaying Embeddings

[lilac barn]

Yes, C depends on n, p , U. I wasn't being precise, sorry.

[astral vine]

(but, again,, it still shows continuity which is a local property)

[tranquil steppe]

I thought your (cocat's) problem is just the continuity of functions in W^{1,infty}. But you're aiming at showing they're locally Lipschitz?

[astral vine]

Using you argument yeah

Which is maybe a more understandable proof with appropriate insights about what happens

[lilac barn]

I am trying to prove this. For p = infinite case, Evans states that the proof is easy to do directly. So first I tried to prove the existence of conitnuous representative which follows directly from the fact that u is in W^1,2n. Now I am trying to show the inequality and was wondering whether directly taking the limit of p works.

[tranquil steppe]

As for the constant C, one has to be careful, but this is actually not a problem. Evans does state does C depends on p, but this is only because it blows up when p -> n (for understandable reasons). However, if you inspect the proof, you will find that C stays bounded when p -> infty.

[astral vine]

Prove it on Rn with homogeneous estimates then localize

Here gamma =1

[lilac barn]

If the constant C isn't a problem then wouldn't just taking the limit of p work? (assuming we have convergence of holder norms with coefficient 1-n/p as p goes to infinite)

[astral vine]

Should work

[tranquil steppe]

There's a number of ways to prove the p=infty case, it's hard to say which one Evans had in mind.
One easy way is to use convolutions. If u is in W^{1,infty}, then so are its convolutions u_eps (with the norm bounded by that of u), but they're smooth at the same time, so they're Lipschitz (with Lipschitz constant at most || du ||_infty). It remains to show that their limit is indeed u.

[lilac barn]

Evans does this exact thing later on in page 294 so I doubt that is what is intended. But I agree its a reasonable way to argue.

[astral vine]

Prove first continuous representative with your first argument then prove separately the estimate

Thus everything should be okay

Checking convolution limites for appropriate mollifiers acting on appropriate families of L infty functions (or also action of regularizing semi-groups on Linfty functions) can be sometime very tedious

[tranquil steppe]

That indeed sounds discouraging, but it doesn't apply to this case. Our u is (in particular) in L^1, so the convolutions converge to u in L^1, and in consequence (on a subsequence) also a.e. Since they converge to a Lipschitz function at the same time, the problem is solved.

[astral vine]

This dies not prove the LInfty convergence

Just Lp convergence for all finite p

[tranquil steppe]

You're right, it doesn't (and it shouldn't, since continuous functions are not dense in L^infty). But it's irrelevant - we only want the limit function to be Lipschitz.

[astral vine]

So thé estimate have again tonproved separately

[tranquil steppe]

No, not really.

As I mentioned earlier (assuming e.g. compact support to get rid of technicalities), Lip(u_eps) = ||d u_eps||_infty <= ||d u||_infty, so u_eps are all Lipschitz with the constant ||d u||_infty, in consequence their limit u is also Lispchitz with this constant.

Unless you had some other estimate in mind?

[astral vine]

The Lipschitz estimate

If d u eps converge to d u in L infty norm then du should be continuous hence u must be C1 which is false

There exist Lipschitz functions that are not C¹

[tranquil steppe]

Yes, that's what I wrote two posts earlier.

[astral vine]

then I don't get why this should be true ?

oh okay maybe I got it

My bad

[untold delta]

kirby 🌈

[slender fulcrum]

i think you could use the energy to estimate the $L^2$-norm of $u_t$ by the $L^2$-norm of $f$ and then use cauchy schwarz to estimate the scalarproduct

[untold delta]

whzup

[late moth]

Ah yes, this is kinda what I was thinking about before I fell asleep. I’ll see if I can make any further progress using those ideas. Thanks!

[lilac barn]

For this problem, would it work if I define the function fₑ := ηₑ * characteristic of ∂U and consider fₑ|_U where ηₑ is the mollifier? This seems like it should work because it converges to characteristic of ∂U in Lp which has non−zero norm on the boundary and 0 on U

[astral vine]

By contradiction assume the operator is bounded

Construct a bounded sequence of smooth functions in Lp such that the trace blows up

[lilac barn]

Isn't my example doing the same=opposite thing? The trace is identically 1 but the bound on the sequence goes to 0 and so we have a contradiction

[astral vine]

Ah okay I didnt get you explanation

So yes

[quaint herald]

I don't understand your construction. The characteristic of the boundary of U is 0 a.e. in R^n. Doesn't this make your convolutions identically zero?

If you flatten out the boundary it is pretty easy to construct a counterexample using scaled bump functions in each coordinate.

[lilac barn]

My idea was that they are identically 0 in Lp(U) but not identically 0 in Lp(∂U). Is that not the case?
Moreover, how would I construct a bump function for an arbitrary U?

[quaint herald]

I have to go in a sec, but the indicator of the boundary is zero a.e., so your convolution is the integral of an a.e. zero function...

You can instead convolve with the indicator of the complement of U for example.

and you construct a bump function by using a diffeomorphism to straighten the boundary, it suffices to work in the upper half of R^n near the origin

with the first coord as a boundary defining function.

will be back in like an hour and a half if you still have questions, or if you clarify something in the meantime.

[lilac barn]

Does it not matter that if we consider the indicator in Lp(∂U) then it's not identically 0 a.e. since its support is ∂U which should have non-zero measure in Lp(∂U)?

[quaint herald]

convlution here is a map from pairs of funcs on R^n to functions on R^n, it is defined by integration wrt to Leb measure on R^n. Did you mean somethimg else by convolution?

[lilac barn]

Ah no, I see your point. I'll study about your bump function example because I haven't encountered bump functions yet. Thanks a lot!

[quaint herald]

@lilac barn how did you go? also note as I mentioned that you could instead consider convolutions with the indicator of the complement of U.

nice to get used to "flattening things out" though, as this is often how you deal with boundaries in general.

[lilac barn]

Oh right, I see, the convolutions with the complement of U works out nicely. But would I need to restrict the convolutions to the set U? and would that break smoothness? I believe, I would have to restrict but I am not sure about the smoothness of the restrictions

[quaint herald]

smoothness is irrelevant here, you just need to work with u continuous up to the boundary.

[lilac barn]

Oh right, in that case, the continuity of the restricted convolutions. Would the restrictions remain continuous?

[quaint herald]

And for any finite epsilon your convolution is actually going to be smooth over all of R^n, not that it really matters.

For small epsilon it will just look like a smoothed out version of the indicator of the complement of U, that is what convolution by an approximate identity/mollifier does.

In the limit it will be 1 on the interior of the complement of U, 0 in U, 1/2 on the boundary.

Rough rule of thumb to remember about convolutions is that they will be as smooth as the smoother of the two factors.

(Compare this with pointwise products, which are generally only as smooth as the rougher of the two factors).

[lilac barn]

Perfect, thanks a lot. I was worried also about whether the restriction to closure of U is going to be continuous as well, which I guess is a nonsensical worry since restrictions of continuous functions to subsets is continuous

[quaint herald]

yep nothing to worry about there

[untold delta]

Gewisser Fler

Gewisser Fler

Gewisser Fler

[tired hollow]

Where t is the tangential vector

[astral vine]

@tired hollow First you have to be careful about the sense you give to the normal derivative of f on the boundary which is not clear for f in H1 unless you do it in very weak sense, or if f lies in the L² domain of the weak neumann laplacian

Now for your question

Your deifnition of the 2 dimensional curl is also not the usual one

it is more like the orthgonal gradient

sometimes denoted by

[untold delta]

Functionanatolysis

[tired hollow]

Yeah that would make more sense, but idk we defined it like that 😄

So for the derivatives, we always consider them in the weak sense

[astral vine]

I am not talking about derivatives

[tired hollow]

I dont know what a weak neumann laplacian is sorry

[astral vine]

but the meaning of the BC for partial_n f

[tired hollow]

Ah so you mean that $\Delta f$ lies in $L^2$ here?

[untold delta]

Gewisser Fler

[astral vine]

yes

[tired hollow]

Ok now i see the problem

[astral vine]

This could be a good condition to define it in a weak sense

[tired hollow]

Ok, lets maybe just assume it is in L^2

[tired hollow]

Yeah this argument doesnt work, i was thinking about tangential derivative

[astral vine]

Assuming the normal derivative vanishes at the boundary

there is a vector identity that says that the trace of vector fields tangential part vanishes iff it have a whole 0 boundary condition on it

hence gradient of f must that 0 boundary conditions

which is not necessarily true right ?

[tired hollow]

Wait one sec: Im still thinking about the normal derivative. Couldnt we just say $\frac{\partial f}{\partial n} = \nabla f \cdot n$? That should be well-defined

[untold delta]

Gewisser Fler

[astral vine]

this is not continuous with respect to the H1 norm

hence not that much well defined

[tired hollow]

Oh ok

[astral vine]

'because this is a boundary term'

But that'"s not my main trouble

I believe your claim is not true

but I'm very tired now, and I'm not able to find a counter example on the unit ball

[tired hollow]

Is cool, u helped me a lot already

[astral vine]

I did nothing

[tired hollow]

I need to think more deeply about things before starting to prove stuff

[astral vine]

the identity is

$$u = \langle u\cdot n\rangle n - n \times (n \times u)$$

[untold delta]

Functionanatolysis

[astral vine]

applying it to

[untold delta]

Functionanatolysis

Functionanatolysis

[astral vine]

thus the term you want to see to vanish

vanish iff

[untold delta]

Functionanatolysis

[tired hollow]

What exactly is the cross-product here? Since we are only in 2d

[astral vine]

I forgot I had to adapt

lemme few minutes

okay the identity should be like

[untold delta]

Functionanatolysis

[astral vine]

where

[untold delta]

Functionanatolysis

[astral vine]

I miss a n

before the equal sign

[lilac barn]

I am not quite sure how is this the actual negation. What if instead of this, we had that C depends on some another quantity as well? They didn't really show that C must depend only on n,p,U and nothing else whatsoever.
For them to claim this to be the negation, they have to first show that C can only depend on n,p,U, and then they can fix n,p,U and claim this is the negation.

[quaint herald]

I don't think I understand your objection. If they successfully establish a contradiction, they have shown that for any fixed n,p,U, some C must exist. Because you have fixed n,p,U, these are what your C depends on. (In particular, C does not depend on u).

[lilac barn]

My point is that the actual negation of the statement is
"Either what they have said (the strict inequality) OR C depends on more things besides n,p,U". They haven't disproven the second point.

[quaint herald]

Uh yes they have. The only "other thing" in the estimate is u, and if C cannot be chosen uniformly in u, this is the same as the assertion that an sequence of u_k exists with the inequality (2) satisfied.

[lilac barn]

I mean there can be infinitely many things doe? How do they know that it doesn't depend on say the regularity of the boundary in the following way: If the boundary is C^k for k even then C = 2 and C= 1 for k odd. In this way, you don't get the same strict inequality because this implies that k cannot be larger than 2

[quaint herald]

"the regularity of the boundary" is information that is contained in knowing U.

The only quantities/objects that appear at all in (1) are u, U, n, p. (Excluding the constant C ofc.)

[lilac barn]

We also have (u) which assumes a measure, so can it not be the case that C depends on the measure we pick? For example, instead of dx we had 2dx, they have't shown that C doesn't depend on this information.

[quaint herald]

Well all of this is working in R^n, where unless otherwise stated you are working with the standard Euclidean measure (and induced measure on submanifolds etc).

[lilac barn]

Hmm, I guess then the only free variables in this whole statement are these ones, and all the others are considered to be fixed.

[quaint herald]

yep

[lilac barn]

Perfect then! Thanks a lot for the explanation!

[quaint herald]

theres a degree of common sense involved in interpreting statements like "only depends on blah", because certain things are always fixed beforehand.

sometimes it is just bad writing, but usually the authors intentions should be pretty clear, especially when you look at the proofs and see what they actually prove.

[astral vine]

Chad Gomez

[valid edge]

what does it mean to "set up an integral which is to be stationary" ?

this is in the context of calculus of variations

[winter dock]

means its variational derivative is 0 probably

[valid edge]

right

i interpreted it as finding the function F(x,Y(x),Y'(x)) where it makes it so dF/dY = 0

and dF/dY' does not depend on x

im having a hard time making that happen for a particular integral

Also made this attempt

[quaint herald]

I imagine the question is just to solve the Euler-Lagrange equations. Your mistake is in your differentiation of $\frac{\partial F}{\partial y'}$. Note that $y$ (and so $y'$) are functions of $x$, and so your total derivative $\frac{d}{dx}\left(\frac{\partial F}{\partial y'} \right)$ is NOT the same thing as $\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial y'}\right)$, the latter being what you wrote down.

[untold delta]

gomez

[valid edge]

Is this line the problem yea?

[quaint herald]

yep

[valid edge]

right

i did end up using
$$
\frac{y'x}{\sqrt{1-y'^2}} = k
$$
instead

[untold delta]

madlor

[valid edge]

where k is arbitrary constant

solved for y

[quaint herald]

yep, that's what you should be doing.

[valid edge]

neat

thank you

[quaint herald]

👍

[solid flint]

Hello, Do you know any concrete examples of functions that are equal in $L^2(\Omega)$ but they differ in $H^1(\Omega)$ ?

[untold delta]

Mikahopff

[astral vine]

this does not exists

because they are equal in distributional sense

then so are their derivatives

L² equality implies equality in D' , thus if one of both function lies in H1, the second one too, and they are equal a.e.

[solid flint]

Thank you very much

additional question, are their normal derivatives equal if it happen that we have one of them in some sense?

[astral vine]

necessarily yes

[solid flint]

Thank you!

[frail vale]

does anyone know how to show that time-minimizing curves are straight lines in the theta-t space?

we know the brachistochrone curve is just a cycloid, but is there a way of showing that the curve in theta t plane is a straight line, where theta is the angle between tangential velocity and the vertical

[pseudo sonnet]

Hello! Suppose that A is a second-order differential operator on $L_2(\mathbb R^n)$ with domain $D(A)=C_c^{\infty}(\mathbb R^n)$, such that A is symmetric (i.e. formally self-adjoint) and positive (i.e. $\langle Au,u\rangle\geq 0$ for $u\in D(A)$), so A is not necessarily elliptic. Is it true that A is essentially self-adjoint (i.e. its closure is self-adjoint)?

I know that A admits self-adjoint extensions (e.g. Friedrichs extension), but my question is specifically about the closure of A, in particular whether Friedrichs extension coincides with the closure of A. I also know that there are examples of positive symmetric differential operators on some subsets of $\mathbb R^n$ that are not essentially self-adjoint but in those examples that I know the problem arises because of poor choice of boundary conditions, and in the case of $\mathbb R^n$ there is no need for boundary conditions.

I would be grateful for any relevant answers or helpful references.

[untold delta]

Hewman

[astral vine]

For properties of some general second order Elliptic operators and to be more specific their functional Analytic properties you should check out Pascal Auscher's work, but I don't know if there is such general discussion

Generally there is assumptions made on the coefficients (smoothness/growth), the (non-)degeneracy

But generally for this class of operators you look at /you ask to have,

[untold delta]

Functionanatolysis

[astral vine]

so non degeneracy implies that the L² form domain by completion will be necessary H1

[pseudo sonnet]

Thank you for your input! I'll check out the reference you provided. I do require coefficients to be smooth and to have less than quadratic growth, but if I understand correctly what you mean by non-degeneracy I'm actually interested exactly in degenerate case, when $\langle Au,u\rangle$ could be 0 but $\nabla u$ is not zero. For example, $A=-\frac{\partial^2}{\partial x^2}$ on $C_c^{\infty}(\mathbb R^2_{xy})$.

[untold delta]

Hewman

[pseudo sonnet]

Do you by chance know any references where degenerate case could be considered?

[astral vine]

I do not know references on Rn

but on domains there are several treatments

[pseudo sonnet]

I think that still could be helpful

[astral vine]

Analysis of heat equations on domains.(LMS-31) - ‎Ouhabaz

[pseudo sonnet]

Thank you again for your help!

[lilac barn]

How does he go from saying that u = 0 is the only weak solution of (26) to saying that u = 0 is the only solution of (27)? Shouldn't he still write that u = 0 is the only weak solution of (27)?
Page 324 of Evans.

Are we employing some sort of "compact operators send weak limits to strong limits in reflexive spaces" idea here?

[tranquil steppe]

As far as I see, (27) is not a PDE, so it wouldn't make sense to talk about weak solutions, right?

[lilac barn]

I mean, saying that u is a weak solution of an equation is basically saying that we can test it against some functions via integration by parts, so we can say the same here? Regardless, if it doesn't make sense to talk about weak solutions, why is he justified in saying that u = 0 is a (strong) solution to (28) when the previous equalities only hold weakly.

[tranquil steppe]

Wait, so is it a PDE or it isn't?

[lilac barn]

26 is a PDE, 27 is just a linear equation

[tranquil steppe]

So why would you test (27) with some function?

If it's just an equality between two things in some Banach space?

[lilac barn]

I mean K does contain L in it which is basically the second-order partial differential operator in divergence form. So yes, even (27) is a PDE, if by PDE we mean an equation containing partial differentials.

My point is that I don't know what exactly he mean by u = 0 is the only solution to (27). The only reasonable interpretation I can think of is that he means u = 0 is the only weak solution to (27) but Idk if that's what he means.

If he does mean u = 0 is the only weak solution to (27) then idk if its consistent to say that x is an eigenvalue of T if x is a weak solution of Tx = cx (where T is a pde operator)

[tranquil steppe]

Calling (27) a PDE is very confusing, it's just an equality understood in the usual sense (and without any quantifiers).

If you're asking about the equivalence between (26) and (27), it's probably explained before that.

[lilac barn]

My point was that you can test (27) with functions so there is an interpretation for u being a weak solution to (27)

[tranquil steppe]

I expect L_gamma is introduced as an operator that provides the only weak solution.

[tranquil steppe]

What is this intepretation?

[lilac barn]

Let me rewrite my question again a little more clearly: Before (28) he writes "Now if u = 0 is the only solution of (27)"
Does he mean strong solution or weak solution? If strong solution, how does he go from saying "u = 0 is the only weak solution to (26)" to saying that u = 0 is the only strong solution to (27) just by rearranging the equation.

[tranquil steppe]

Once again, there's no notion of a strong or weak solution of (27). It's just an equality that doesn't require clarification.

I guess what he means is what he writes: "u = 0 is the only weak solution to (26)" is equivalent to "u = 0 is the only solution to (27)".

Why it's true, that's another question, probably addresed in the previous part.

[lilac barn]

This is the entirety of the result and he doesn't discuss the equivalence anywhere else. He takes the equivalence between these statements as obvious or something, I guess from the fact they are just algebraically equivalent when treated as simple equalities. For context, Lᵧ = L + γ and K = γ Lᵧ⁻¹

[tranquil steppe]

That's not the whole thing. If you look it up, L_gamma^-1 and K appear on page 321 and there you can find the explanation.

[lilac barn]

Hmm, yeah I do know the relationship, I have stated it above "For context, Lᵧ = L + γ and K = γ Lᵧ⁻¹"

Or is it something else you are referring to?

[quaint herald]

He is (I assume) talking about (14) on p321, in which the notation L_gamma^-1 is introduced. It is essentially shorthand for "the unique weak solution to blah".

[lilac barn]

Ahhh, I see, so the weakness of the solution is basically coded inside the definition of Lᵧ and so we preserve the initial weakness of the solution u! Thanks a lot.

[quaint herald]

Well, in the definition of L_gamma^-1, not the definition of L_gamma. Don't think of L_gamma^-1 as an inverse to L_gamma, at this point in Evans I don't think it has been proven that L_gamma is invertible as a map between any two particular function spaces.

All that has been proven (via Lax Milgram etc) is that a weak solution exists to L_gamma u = f, for suitable gamma.

So it is just convenient to use this inverse notation to refer to this weak solution (at the cost of sometimes confusing students!).

[lilac barn]

This makes sense. So immediately after this, he proves a "boundedness of inverse" theorem. I am guessing all that means is that "Lᵧ⁻¹ is bounded" and not that Lᵧ⁻¹ is an inverse of Lᵧ and the former is bounded (which would also resolve my confusion about why we aren't just using the inverse bounded theorem from Functional analysis and just proving that Lᵧ is bijective and bounded)

[quaint herald]

Yep, exactly.

[quaint herald]

Note that at this point the spaces don't even quite make sense for an invertibility statement. For f in L^2 you get a weak solution u in H^1 with zero trace. But L is a second order operator, so it is not even obvious that Lu is in L^2, let alone that it is actually equal to f.

And that's where the "elliptic regularity" results coming up are relevant. They show that the weak solutions are actually H^2 and are actually solutions in the strong sense that Lu=f.

[lilac barn]

Makes perfect sense! Thanks a lot!!!

[quaint herald]

No worries :).

[regal cypress]

Hello, I'm currently have quite some trouble understanding the proof of a theorem (more specifically the calculations sadly...), If someone could maybe guide me through it I would be very grateful.
I'm following the book "Hyperbolic systems of conservation laws" by RAVIART AND GODLEWSKI from the collection MATHEMATIQUES & APPLICATIONS.
I'm trying to produce a simpler proof for the Kruzkov Theorem which can be found on page 77 (the proof is for d dimensions but in this case I want a simple case where d=1).
The theorem is a proof of the uniqueness of entropy solutions of systems of conservation laws. Thank you in advance!

[astral vine]

I'm not sure but Bresson's book may contain a proof of the 1-d case

[regal cypress]

would you be willing to help just with the last step of the proof? I'm pretty sure it's not pde related and just calculations but i can't figure it out

[astral vine]

It depends on which part : it is kinda late right now to check out this kind of heavy arguments/calculations

[regal cypress]

last step

how do I use the triangle inequality? I was hoping it would be just a clever way of summing or subtracting that I didn't notice

[astral vine]

\begin{align}
\int_{|x|\leqslant R+M(t-\delta)} {|u(x,\delta) - v(x,\delta)|} \dd x \leqslant &\int_{|x|\leqslant R+M(t-\delta)} {|u(x,\delta) - u_0(x)|} \dd x\ + &\int_{|x|\leqslant R+M(t-\delta)} {|v(x,\delta) - v_0(x)|} \dd x\ + &\int_{|x|\leqslant R+M(t-\delta)} {|u_0(x) - v_0(x)|} \dd x
\end{align}

[untold delta]

Functionanatolysis

[astral vine]

letting delta goes to 0

(1) & (2) disappear

so only (3) remains

and use the first inequality given from your above screenshot

[regal cypress]

but is the modulo of the sum of what's inside the other 3 modulos the same as the original one?

cuz isnt the inequality |a+b+c|<=|a| + |b| + |c|

(in our case)

[astral vine]

it is

u(x,d) - u_0(x) + u_0(x)- v_0(x) + v_0(x) - v(x,d)

everything reduce to u(x,d)-v(x,d)

[regal cypress]

oh that's right... I tried summing them, guess I just messed up. Thanks a lot!!

[astral vine]

No problem you are welcome :)

[regal cypress]

wait

why is the last v_0 positive?

[astral vine]

?

a-b = a - a_0 + a_0 -b

=a - a_0 + a_0 - b_0 + b_0 -b

-b_0 + b_0 =0

right ?

with modulus trying to make appear b_0-b is exactly the same as trying to make appear b-b_0

do we agree ?

[regal cypress]

yes

[astral vine]

Do you see, then why everything is fine ?

[regal cypress]

yes I see now |a-b|=|b-a|

my bad

[verbal nebula]

I'm working through elliptic PDE stuff right now, and I've programmed into my head that Lv stands for Louis-Vuitton. I always say "You can't spell love without LV baby"

So now I just have expensive purses crowding my mind

[regal cypress]

Hi guys, I'm following the proof of this theorem (again...) and I'm really confused with what it means when it says (4.13) follows at least formally (right at the end of the 2nd picture). Could someone please explain pls?

[bronze gate]

formally usually means you ignore certain issues (eg. ignore whether or not a given something converges) and do the operation as if you could

see: https://math.stackexchange.com/questions/152034/usage-of-the-word-formally

[regal cypress]

I see, and what issues are we ignoring at the moment? Is it obvious from the statement of the problem?

[bronze gate]

(i am not an expert in this field so discount my answer if it seems wrong): i think the implicit assumption is that you can integrate the inequality (4.2)

so saying "(4.13) follows at least formally" means that if you ignore issues about integrability it follows

[regal cypress]

so we're assuming that u and v are sort of well-behaved and it this case we can integrate

[bronze gate]

basically

[regal cypress]

Alright, thx a lot!

[verbal nebula]

Follow up to this, I'm reading Han and Lin right now and my God there are so many typos. I've been cross checking the maximum principle sections with Evans

But there's just so much either left out, or just plain "wrong"

[quaint herald]

Exercises! 😺

[verbal nebula]

I do think that it does force me to learn it better

Because I'm like "wait a minute" then I spend an hour digging around trying to think of examples or why their reasoning is flawed

[quaint herald]

exactly

[verbal nebula]

My current PDE Path is Han and Lin ---> Fully Non-Linear elliptic Equations (Xavier and Cafferelli) ----> Monge Ampere by Figalli