advanced-pdes

[lilac barn]

Wdym?

[pulsar rivet]

i have trouble on how to verify it

.

[lilac barn]

Well check if it satisfies the PDE by calculating it's first partials and plugging them in, and then check if u(x,0)=x

[pulsar rivet]

it does not

[lilac barn]

It should satisfy the PDE, the question would be about the initial condition. Justifying the latter why should be sufficient then.

[pulsar rivet]

w8

oh dy/ds=-x

mb

it is satisfied indeed

[lilac barn]

What about the initial condition? Is that satisfied?

[pulsar rivet]

i think they are

[lilac barn]

What should be u(-1,0) according to initial condition and according to your solution?

[pulsar rivet]

arent they?

they are different

so they are not satisfied?

[lilac barn]

They're not

[pulsar rivet]

what result fo you get from initial condition?

[lilac barn]

The point is that since the curve (x,y)(s) can select the initial condition from either the positive side or the negative side, they either have to be both equal for a solution to exist (for example requiring u(x,0)=|x| or -|x|) or the solution wouldn't exist.

More precisely, as z(s) is constant in s
x₀ = u(x₀ , 0) = z(0) = z(π) = u(x(π), y(π)) = u(x₀ cos π , x₀ sin π) = u(−x₀, 0) = −x₀ which is a contradiction

[river path]

I don't understand this. This is from Galtier's MHD book. b(y) = (bx(y), b0, 0) where bx is a function of y and b0 is a constant. The region is between 2 infinite plates parallel to the (x, z)-plane. "Insulating plates" imposes the boundary condition curl(b) = 0 on the top and bottom plates. Galtier is claiming this implies there's a potential that depends only on y. I'm not sure A. how you can get a potential from just having 0 curl on the boundary and B. why the potential couldnt vary linearly in x.

I suspect it's an error, because curl(b) = 0 immediately implies dbx/dy = 0 on the top and bottom plates (rather than bx = 0)

and getting both dirichlet and neumann conditions on the boundary seems ridiculous, but the resulting equation will be third order so maybe it turns out ok?

I could post more context but I dont think it's relevant

[spare sentinel]

Ok given

$$(\partial_t + (-\Delta)^{1/2}) u \in L^2([0,T]\times\mathbb{R};\mathbb{R}^n)$$

Do i know if $\partial_t u$ and $(-\Delta)^{1/2} u$ exist and are also of the same integrability?

I already seached for energy bounds fitting this form, but couldnt find anything

[untold delta]

KIl

[lilac barn]

Might be missing something but the vanishing of curl of b is equivalent to b being the gradient of a potential. This is just a general fact for vector-fields. Athough I might just be confused with the abuse of notation of mathbf(b) and b.

[river path]

This is just a boundary condition, though

The curl doesnt vanish everywhere

Only on the plates

[lilac barn]

I imagine you can write down the Duhamel's formula for this PDE and then do Fourier transform to obtain some estimates.
Follow the same strategy as for heat equation.

[lilac barn]

So is the claim that b = grad phi not just on the boundary?

[river path]

No, just on the boundary. But is it true that if the curl of a vector field vanishes on a plane, its restriction to the plane comes from a potential? i guess what you do is use stokes on curves in the plane?

But you'd only get that the gradient matches the vector field in directions parallel to the plane i think. but i guess that's enough for me here

i added 3d but then i removed it, idk why i think i need phi to be 3d

hmm

yeah ok i understand

ty

[spare sentinel]

The problem is that the derivatives are together as on operator, idk if one of them even exists on its own, thus duhamel's formulare is not directly applicable the same with Fourier

[river path]

Still, it seems to me like if b is constant and nonzero on the boundary, the curl is just as easily 0 and the potential just depends on x.

[lilac barn]

I meant take the Fourier transform of the equation first and then solve it via Duhamel. This gives you a formula for the Fourier transform of u so you can then do Plancherel and usual tricks to get some estimates

[spare sentinel]

I know, but taking Fourier is the problem on its own, you cant intercahnge the partial_t with eth FT because the partial_t and the Delta are together and cant be seperated, that is the problem

[lilac barn]

You can also do energy bounds by multiplying the equation by (Delta)^1/2u and then play around with integration by parts, Cauchy-Schwartz, Youngs etc.

[lilac barn]

Not sure what you mean by together? Cant you just write par_t u -(Delta)^1/2 u =f and then take the Fourier transform in x?

[spare sentinel]

nop you cannot because they are apllied as on operator

they exist as a distributional dervative together in L^2

?

[untold delta]

cocat

cocat

[spare sentinel]

well no thats not the definition her

here*

[lilac barn]

What's your definition then?

[spare sentinel]

it is as distributional derivative of (\partial_t - Delta thing) no schwartz function what so ever u is only in L^2

of course the test function yeah they work there for sure, but i need infos about u

[lilac barn]

I mean in my stated definition u can be a distribution as well.

Unless your point is that the equality doesn't hold pointwisely merely in the sense of distributions then sure

[spare sentinel]

Like yes that is correct but i want existence of those derivatives in L^2 like stated before the same integrability

derivatives will always exist for distributions no questions asked

[lilac barn]

The strategy for these things is that you first assume your u is Schwartz. You derive some estimates that only hold in L2 (or whatever you like). This tells you what estimates you should expect. Then you can prove in distributional sense by doing the necessary rigor required to make the statement sensible.

Ofc, this doesn't give you the existence of a solution or whatever; just some a-priori estimates.

[lilac barn]

Also this fact holds regardless of you view u distributionally or not. In the case of distributions, you will transfer the derivatives to the test function, split the sum and then transfer the derivatives back to u, to note that this equality does hold.

[astral vine]

Yes Maximal reularity for the Poisson semigroup

More precisely here this is the L² maximal regularity in time

in fact

put

[untold delta]

Functionanatolysis

[astral vine]

Then you can write

[untold delta]

Functionanatolysis

[astral vine]

Up to consider the extension of F to be 0 on the whole real line, i.e. F(t) = F(t) if t>0, F(t)=0 otherwise.

You can build u as

[untold delta]

Functionanatolysis

[astral vine]

Similarly

[untold delta]

Functionanatolysis

Functionanatolysis

[astral vine]

each Fourier symbol is bounded by one in modulus

The Fourier Plancherel theorem yields the boundedness : it is in fact even better than that

[untold delta]

Functionanatolysis

[astral vine]

(in fact there should be an additional term if u is not in H^{1}_0 (R^{1+n}+), but the whole picture is correct)

[spare sentinel]

Isn't it that I have L^2 maximal theory only in this case if I have minimum H^1 initial condition?

And do you have a good source for it, it's my last part of my master Thesis which I just couldn't get on my own🫠

But thank you a lot!

Just for clarity

My u is at most $L^2_{t,x}$

The operation

$(partial_t + Delta)$ are applied as on operator together

The general $L^P$ theory was already applied once for regularity, but at one point I differentiate and get the existence of

$$
\Delta^{s} ( \partial_t u + \Delta^{1/2} u) \in L^2_{t,x}
$$
Now because weak derivatives commute I just swapped the outer Delta inside and asked my self the former question, that's why I have the operator simultaneously, cause I interchanged it with the outer $\Delta$

If I wouldn't have lost that my initial condition is $H^1$, I would be fine from my point of view, because i only hav $L^2$ I dont know

[untold delta]

KIl

[astral vine]

\dot{H}^{1/2} for the initial data

[astral vine]

In fact, start from the whole space case

R^{n+1}

Then use restriction and extension operators to show that both operator necessarily agrees

(d_t + |D_x|) and (d_t) + |D_x| on R^{n+1}

then on R^{n+1}_+ (once you have the R^n+1 case, this is then just Fubini)

The space you are looking for is L^{2}_t(\dot{H}^{2s}_x)

[astral vine]

Concerning this part this is sharp, iff

H^{1} is stronger since it requires one half more derivative

and u_0 to be in L²

For u in L^2

I think there is no direct hope in general

Maximal regularity is an iff type property

Finally a full and definive answer to your question relies on the joint functional calculus of sectorial operators with commuting resolvents.

A.k.a the Dore-venni theorem and its variants

(Which also holds on Banach spaces more general than Hilbert spaces)

[stark thunder]

Is it possible to find a (shauder) basis for the harmonic functions in the space of all smooth functions on R^n?
I essentially want a way to project into the space of harmonic functions
and having an explicit basis sounds like the easiest way to do this

[sonic jacinth]

What sort of topology are you putting on the smooth functions? Harmonic functions (and all their derivatives) are constant or unbounded on R^n, and under most natural topologies, this space is uninteresting.

[stark thunder]

Why are unbounded functions uninteresting. My goal here is to do a hodge decomposition (which is initially proven only on compact manifolds without boundary)

And in particular, I want to see the "harmonic component" of my function

[stark thunder]

Does that help?

[lilac barn]

Would that be unique? Suppose f = g+h where g is the harmonic component. Then lap f = lap h as lap g = 0. But note
f = (g+c) + (h-c) is also a valid decomposition for any first order polynomial/harmonic c and it still satisfies lap f = lap h = lap(h-c).

Now if you somehow quotient out harmonic functions when you take h, then that might work but not sure if there's a decomposition like that

[sonic jacinth]

they aren't uninteresting, my point was that in most natural topologies (sobolev norms, for instance), we only topologize a subspace, which has a 0 or 1 dim subspace of harmonic functions

[sonic jacinth]

also, since you are asking about a schauder basis, you need to decide on a topology to use. Piggybacking off of cocat's answer, you might also want some way to 'force' the decomposition to be unique. In the compact manifold case, we are looking at L^2 of differential forms, which allows us to identify quotients (by closed subspaces) with subspaces of our vector space. In fact, you can do hodge decomposition for L^2 differential forms in R^n, but the subspace of square integrable harmonic k-forms in R^n is zero dimensional for all k

[spare sentinel]

can yoz show me where you have this? I only found it for H^1 intial data, that would tbh solve the whole thing

https://citeseerx.ist.psu.edu/document?repid=rep1&type=pdf&doi=2120ee5e7fb7f75d0cd9db4927190722f07d4611

This is the one i found which i was able to edit to H^1 initial data

[astral vine]

Juste write down the quantities and play with Fourier transform

I don't have any references, except maybe two of my papers. But those treat the very general case

[spare sentinel]

Its alright if it generel

[astral vine]

I mean no L² spaces or anything just abstract Lp maximal regularity for class of Operators.

[astral vine]

First construct the Projections on L²

A Projection on the closure of the range of d

A projection on the closure of the range of d*

I-P-P* is then a projection on the finite dimensional space of harmonic forms

[stark thunder]

How can I get the projection onto the closure of the range of d without first getting rid of the harmonic part

[stark thunder]

Here's my recipie for compact manifolds with no boundary:

From hodge decomp:
w = df + d*g + \xi
Where \xi is harmonic.

First solve:
\Delta u = 0
By hodge thm you know that there's going to be a finite basis for this space corresponding to the cohomology representatives.
You now have a finite basis for harmonic forms and can easily project and then subtract that off w.

So we can assume wlog that we have:
w = df + d*g

To project this we do:
d*w = d*df
d*w = \Delta f (assuming w is a 1-form here)
And this is just a poisson pde which can be solved using standard (Greens function) methods.

Solving this PDE gives me f. And so I can then I've finished my decomposition sice:
d*g = w - df.

What underpins this all is first erasing the harmonic component

otherwise you get something like d*\xi in your pde

NVM I AM STUPID

d*\xi is always zero

as is d\xi

🤦

I'm having trouble doing this in dimension higher than 2

In all dimensions, we get that d*d = \Delta when acting on 0-forms. This lets you solve a poisson pde to do the projection onto im d.

An easy special case is when we're in 2D because dd* in acts like \Delta. This lets you, again, just solve a poisson pde to project to im d*

In 3D I need to solve something like:

curl w = curl(curl w)

I feel like a fundamental step needs to be directly projecting onto the harmonic subspace. But it's really unclear how to do that.

[I can't use the trick I explained above, because I'm wanting to do this on R^n and not something compact]

IMO it's totally reasonable to think that a vector field on R^n could contain an unbounded harmonic component.

[astral vine]

You can build the projector the way I did

However you won't get anything more than a finite dimensional subspace by De Rham Cohomology.

[stark thunder]

I can't do one of the first projections

In particular, the projection associated with the d* image

[astral vine]

Why ?

What do you mean by "You can't" ?

You have closed Hilbert subspaces. Such projection operators always exists

[stark thunder]

Computationally how would I do the projection

For the image of d it's straightforward

Essentially you have to solve a poisson PDE

I think it's going to end up being a weird equation like:

Curl F = Curl Curl Phi

And you're given F and need to find a Phi satisfying this

(in 3d anyway)

[sonic jacinth]

I'm going to repeat this because I feel like my answer got buried: you can do the Hodge decomposition for L^2 k-forms on R^n.
The subspace of harmonic L^2 k-forms on R^n is zero dimensional for all k (by the mean value property).

[sonic jacinth]

If you want to do this not on the space of L^2 k-forms, then things may get more interesting. You also have to decide how to topologize your vector space, and then potentially deal with issues like uniqueness of decomposition.

[stark thunder]

So, if we restrict our attention to only L^2 forms on R^n then the hodge decomposition holds and is unique. "but" the components in the decomposition are also going to be L^2.

I guess the statement says nothing about the uniqueness of the decomposition if we allow for the components to be not in L^2.
For instance, writing an L^2 1-form as a sum w = df + d*g where df and d*g are not in L^2 could be possible (it's not ruled out by the theorem)

I think this is true since for instance: it's true that any nonzero harmonic thing must be not L^2

mm

actually im not sure

[untold delta]

V5_
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[stark thunder]

Their images intersect at zero regardless of integrability

[sonic jacinth]

sorry I'm having trouble typing the * symbol

[stark thunder]

thats fiine lol

[sonic jacinth]

yes

[stark thunder]

okay, i might try to prove that. But there are so many deathtrap theorems here

[sonic jacinth]

it's sort of a statement about chain complexes over a field, for instance, on smooth k-forms, we have the de Rham complex, and each vector space of k-forms splits as a direct sum of the image of d (from k-1 forms), the cokernel of d, and the kernel of d mod the image of d

[stark thunder]

I'm not from pdes, and sometimes random theorems delve deep into like elliptic pde knowledge lol

[sonic jacinth]

this reminds me of when someone I know who does AG told me about having to learn pdes to read the proof of the index theorem

[astral vine]

I did the Hodge decomposition on Rn and Rn+ in a paper

For any LP/Sobolev and Besov spaces

The main point is that the Laplacian is injective on Rn, once you forgot polynomials.

Then the result is a "trivial" (in L² at least) consequence of Fourier Analysis.

[cinder marten]

come read about psidos with me

[cinder marten]

so glad the french are dealing with this for everyone's benefit

[sonic jacinth]

once my term ends

[verbal nebula]

Don't do it, it's a trap

Discord study groups go nowhere

[astral vine]

The so sad truth

[stark thunder]

• what is C_{s,p}?
• What does the "dot" mean over the spaces? From context, I'm guessing it means the closure
• So are you saying that harmonic polynomials are the "only" problematic things in the HHD?

[stark thunder]

Actually, maybe if you DM me the article I might be able to answer these things for myself

[steel umbra]

kind of lost on this

i think i got

$\casedef{v_t-v_{xx}=g'(t),&\bR_+\times(0,\infty)\
v=0,&[0,\infty)\times[0,\infty)}$

[untold delta]

eigentaylor

[steel umbra]

but i don't really understand the odd reflection part

and i don't think this is right

$v(x,t)=\int_0^tg'(s)\int_\bR\Phi(x-y,t-s)\dx y\dx s$

[untold delta]

eigentaylor

[steel umbra]

any help appreciated

[astral vine]

Forget about the dot, this is just fancy stuff here

and if you can can for got about the point you can also remove the considition C_s,p

this is a condition that says "okay if the space is complete/Banach"

For me the closure is the overline

H^{s,p} is the Sobolev spaces of order s over L^p

(think s=k is an integer)

the dot stands for the homogeneous Sobolev spaces.

in the sense of the norms are left invariant by a dilation

somehow

so a lot not necessarily technicalities for you

get over it

s=0, gives you the L^p result

H^{0,p}=L^p

[unborn quiver]

First, you want to check the sign of g’(t), I think it should be -g’(t).
Now, odd reflection of a function f is taking -f(-x).
Once you have a heat equation defined on all of R using odd reflection, you can use the formula for the nonhomogeneous heat equation in Evans.
From here, it should just be some amount of bashing of integrals to solve for v(x,t) in an integral form that looks like what you’re asked to find

[astral vine]

One way to proceed : prove that there is at most one solution

then show that the given formula is indeed a solution.

[unborn quiver]

Oh that’s fun

[verbal nebula]

I think the issue is that the professor said derive

Rather than prove that this is the unique solution

I think following Evans computations is the point of that exercise

Although that is a cute & clever way of going about it

[astral vine]

Yeah I've seen this after I did propose this solution.

[steel umbra]

right so would it just be

$\casedef{v_t-v_{xx}=-g'(t),&\bR\times(0,\infty)\
v=0,&\bR\times[0,\infty)}$?

[untold delta]

eigentaylor

[steel umbra]

and we're saying v(x,t)=-v(-x,t) if x<0?

[unborn quiver]

Yeah, I'd also recommend writing out the equation being solved on R_- explicitly

It will make it easier to write out explicitly because you'll get integrals on R_- and R_+

[unborn quiver]

oh, and shouldn't it be v=0 on R x {0}

[steel umbra]

oh yes! that was a dumb mistake. yeah that's right.

[steel umbra]

yeah that might explain why my integral for v(x,t) ended up just being g(t) by itself. i will try that rn

[unborn quiver]

||Yeah, there's a sign change of the g' on R-||

spoilered bc I just verified it but didn't want to give it away

[steel umbra]

ahhh so basically it should be

$\casedef{v_t-v_{xx}=-\opn{sign}(x)g'(t),&\bR\times(0,\infty)\
v=0,&\bR\times\bdef{t=0}}$

[untold delta]

eigentaylor

[unborn quiver]

Yeah, I saw the x there and at first was very concerned

lmao

[steel umbra]

thank you! i will try it again

[unborn quiver]

Yeah, the hope is that you end up with v(x,t) = -g(t) + blah, where blah is the result you're anticipating

[steel umbra]

[
v(x,t)=
\int_0^t\frac{1}{\sqrt{4\pi(t-s)}}
\int_\bR e^{-\frac{(x-y)^2}{4(t-s)}}(-\opn{sign}(y)g'(s))\dx y\dx s
]
\begin{multline*}
v(x,t)=\int_0^tg'(s)\bigg[
\int_{-\infty}^0\frac{e^{-\frac{(x-y)^2}{4(t-s)}}}{\sqrt{4\pi(t-s)}}(-1)\dx y\
-\int_{0}^\infty\frac{e^{-\frac{(x-y)^2}{4(t-s)}}}{\sqrt{4\pi(t-s)}}(1)\dx y\bigg]\dx s
\end{multline*}
\begin{multline*}
v(x,t)=\int_0^tg'(s)\bigg[
\int_{-\infty}^{\frac{x}{\sqrt{4(t-s)}}}\frac{e^{-u^2}}{\sqrt{\pi}}\dx u
-\int_{\frac{x}{\sqrt{4(t-s)}}}^{\infty}\frac{e^{-u^2}}{\sqrt{\pi}}\dx u\bigg]\dx s\
=\int_0^tg'(s)\bigg[
\int_{-\infty}^{\frac{x}{\sqrt{4(t-s)}}}\frac{e^{-u^2}}{\sqrt{\pi}}\dx u
-\paren{1-\int_{-\infty}^{\frac{x}{\sqrt{4(t-s)}}}\frac{e^{-u^2}}{\sqrt{\pi}}\dx u}\bigg]\dx s\
=\int_0^tg'(s)\bigg[
2\int_{-\infty}^{\frac{x}{\sqrt{4(t-s)}}}\frac{e^{-u^2}}{\sqrt{\pi}}\dx u
-1\bigg]\dx s
\end{multline*}

[untold delta]

eigentaylor

[steel umbra]

am i on the right track or did i mess up somewhere? i tried to leverage that the integral of the exponential integral is 1

i think it's right and that gives me

$\int_0^t g'(s)\erf\paren{\frac{x}{\sqrt{4(t-s)}}}\dx s=g(t)-\int_0^t\frac{e^{-\frac{x^2}{4(t-s)}}}{\sqrt{\pi(t-s)}}g(s)\dx s$

[untold delta]

eigentaylor

[steel umbra]

i think

[unborn quiver]

Yeah, I think it looks good at a glance

If not, it’s fine up to a sign change in the change of variables somewhere

[steel umbra]

but then i need to show that

$-\int_0^t\frac{e^{-\frac{x^2}{4(t-s)}}}{\sqrt{\pi(t-s)}}g(s)\dx s=\frac{x}{\sqrt{4\pi}} \int_{0}^{t} \frac{1}{(t - s)^{3/2}} e^{-\frac{x^2}{4(t-s)}} g(s) \dx s$

[untold delta]

eigentaylor

[steel umbra]

but the RHS looks like the partial derivative of the LHS wrt x...

[unborn quiver]

mhm

[steel umbra]

then are they equal? if so, why is the expression equal to its partial derivative wrt x?

[unborn quiver]

Oh, I misread what you were doing

$F(s)=\int_{-\infty}^{-\frac{x}{\sqrt{4(t-s)}}} \frac{e^{-u^2}}{\sqrt{\pi}}du$
has $F'(s) = -\frac{x}{4\sqrt{\pi} (t-s)^{3/2}} e^{-\frac{x^2}{4(t-s)}}$

[untold delta]

MSC2020 49Qxx

[steel umbra]

agagdhsh right we're integrating with respect to s not x. thank you for continuing to help me through my dumbery

[unborn quiver]

No worries, these computational things can be annoying so I get it

[inland sinew]

This is a silly question I'm asking because it's been a while since I've done problems with laplacians, but if I am looking at the 2d laplacian on a square, with periodic boundary conditions, if the boundary conditions I choose are dirichlet and say the value on every side is the same constant, this implies the solution is just constant over that square domain correct?

[midnight grail]

Yeah, you can use the maximum principle

[gray jasper]

does anyone have any

relatively short expository papers or anything

to get me excited for reading evans

applications based btw

for example how can the theory of weak solutions and the sobolev space stuff be applied

smth historical could work too

financial applications, fluids, other physics, they all work for me

[waxen bobcat]

short read

[gray jasper]

thanks

[waxen bobcat]

oh, and if you're interested in numerical simulations, maybe play around with finite element software, for example https://ngsolve.org/, which is based on the theory of weak solutions and sobolev spaces

[solid flint]

Hi, any references about evolution families that we use to solve evolution PDEs of the form $u_t(x)=A(t)u(x)$ when the operator $A$ is time dependent, please?

[untold delta]

Mikahopff

[astral vine]

if A is an elliptic operator for any time

look at Non-autonomous maximal regularity

e.g.

https://arxiv.org/pdf/1601.08012.pdf

[solid flint]

Thank you very much

[solid flint]

Do you know anything about the other case please?

[astral vine]

About the other cases ?

[solid flint]

when the operator is not necessarily elliptic?

[astral vine]

For hyperbolic systems

there is the theory of friedrichs Symmetrizers

See Guy Metivier's book

Mostly for first order type systems

[solid flint]

Ok ,thank you. it is good to take a look but I am only ineterested in existence of week solutions..etc.
Thank you again.

[lilac barn]

One Parameter semigroups for linear evolution equations Chapter 6.9

[astral vine]

it seems quite good

was not aware of it

thx

[lilac barn]

The book is so good, the authors wrote a baby version after publishing the book because students were getting put off by the breadth of the book xD

[astral vine]

Hahaha

[solid flint]

Thank you very much.

[brittle cradle]

im not sure if this is the right section to ask but, why is the phasor domain of KVL/ KCL equations of a RC, RL, RCL circuit linear. Allowing for an easier approach at solving intergrodifferential equations

[untold delta]

!!

[hoary oak]

Can anyone please refer to easy to understand book for symmetrization & rearrangement inequalities

[velvet pier]

does anyone know a good vector calculus review?

I just wanna speedrun the curl, grad, div, stuff

• green theorem + stokes in 3d + line integrals

for evans

holding on

oh I also have like

0 intution on the leplacian

[prisma pelican]

I see the appendices but there's no proofs for most theorems

[stark thunder]

1. Mean value property of harmonic functions
2. Heat equation to understand it's regulating properties
3. Maybe complex analysis stuff to see it's other connections

[lilac barn]

I would say you don't particularly need any more review on this besides what he covers in the appendix. Even if you haven't covered the proofs, I would say skip them and get started with the book. You can revisit the proof anytime and the more important thing is how to use the statements which Evans will teach.

[echo garden]

why you want to speedrun? are you in a hurry?

[velvet pier]

yea, I need to catch up to a group of people I'm reading evans with

[velvet pier]

I would do that if I could at the very least understand the theorem statements

I think I'm missing a couple of definitions

[quaint herald]

which definitions?

I think the definitions and notation choices in Evans are pretty self-contained for the most part.

[lilac barn]

Feel free to ask them on the server but as Gomez said, you can find all the definitions at the back of book.

[untold gust]

Is it an online group?

[velvet pier]

like this definition for example,

I can't just look at it and understand what it's saying

which tells me I should prolly read on this stuff somewhere first before going back to evans

[velvet pier]

will do, thankyu!

[velvet pier]

yes

[velvet pier]

the definition right after it too

like, what the heck is \nu?

is it just like, any vector field defined on \partial U?

surely that can't be true

coz otherwise this wouldn't be well defined!

also, what's the difference between dS and dx

I just... don't understand any of it lmfao

[sand echo]

this basically is saying that the boundary is C^k if at each point, if you zoom in far enough, the boundary just looks like the graph of a C^k function

[sand echo]

at each point on the boundary, the unit normal vector is the uni vector which points orthogonally to the boundary in the 'outward' direction, might help to just look up a picture

[prisma pelican]

but why does it exist

and is it well defined?

[sand echo]

because the boundary is C1, its locally just like a graph of a C1 function

and this lets you define it explicitly more less

[sand echo]

probably a good idea to read up on surface integrals though before diving into evans

[prisma pelican]

oki

thankyu!

[quaint herald]

Okay, so you mean that you are having trouble parsing some of the definitions, not that these definitions are not present in Evans?
In the first example you should think in small dimensions first like n=2. This definition means a region like: y > x^2 has smooth boundary, because the function f(x)=x^2 is smooth. It just looks a bit more complicated because you can only hope to represent the boundary of a domain locally as a graph. (E.g. with a disk you would have to use different functions to graph different pieces of the circular boundary, but the point is you can do it "locally" about any point).

For the second, once you understand tangent (hyperplanes) to graphs of functions of several variables, the well-definedness of normal vector fields is a matter of linear algebra. (The orthocomplement of a hyperplane is 1-dimensional) .

[velvet pier]

oh!

okay, that makes sense

[untold gust]

Can I join? I’m re-reading Evans for grad school

[velvet pier]

ohhhh, so that's what evans meant by "relabeling and reorienting the coordinates axes"

you just reorient them so that U \cap ball looks like the graph of some function

[velvet pier]

you should ask @pulsar forge tbh, it's not my reading group

you can also look at his post in #events message

there are certain commitments you have to agree to before joining

[velvet pier]

the answer is yes to the first question btw

but not only that, when I see a theorem or a definition I don't even know if I'm missing something or if I'm just being dum lel

[quaint herald]

In 2 dimensions, relabelling and reorienting just means that either write y as a function of x or x as a function of y to describe the boundary as a graph locally.

hmm good to talk about it with others / try problems etc to figure out exactly what your gaps are if any.

[pulsar forge]

sure i dont mind, more discussions to be had

[velvet pier]

oki

I think I'll read on that stuff seperately rq and see if I don't understand anything

[pulsar forge]

surface measures basically, similar to polar coordinates and how you have a "measure" on the circle using arc-measurment , its just more generalized.

i think folland covers it iirc

[quaint herald]

If your surface is a graph say $f(x)$, then it is parametrised by $x$ and the surface measure is like $\sqrt{1+|\nabla f(x)|^2}, dx$ in these coords

[untold delta]

grobmez

[quaint herald]

if you are reserving x to use as coordinates in the whole space, replace the x in the above formula with like a x' or whatever.

[lilac barn]

If you need more clarifications for definitions like and related to this, check out Folland PDE book. He is much more precise with preliminaries like this.

[untold gust]

Also, it helps to know that one of the reasons we care about the smoothness of the boundary because we frequently use Generalized Stokes Theorem.

[inland sinew]

Does anyone know any good resources for asymptotic methods in PDEs? I am familiar with asymptotic methods for ODE's, but I am less certain how to apply my methods in cases where I cannot reduce my PDE to some ODE's through separation of variables. I want to use it to study mixed type linear PDE's so I can understand how the solution is changing with the introduction of a term that extends the discriminant from being purely elliptic/hyperbolic to a mixed type.

[astral vine]

Not that much actually since one can reach lower the regularity at an absurd level and still have a Stokes theorem. Lipschitz, SKT domains etc.

[astral vine]

The general idea on the whole space is to consider the pseudo-differential operator and to cut it out, in the Fourier side' in various operators that behaves at different regimes. Usually heavily used, if I am not mistaken In problems such as dispersive PDEs as in water-waves systems etc.

This relies generally on Harmonic Analysis methods such as the stationary phases and other related results

[pulsar forge]

simple question, but is this enough to deduce the wave operator is not hypoelliptic? (since it seems to imply u doesn't need to be always smooth)

[lilac barn]

Well this isn't telling you that the solutions cannot be smooth, merely that they're C2. But your conclusion is correct: the wave operator isn't hypoelliptic as it's solutions can be arbitrarily rough.

[native crown]

How normal is trying to find solutions using Sobolev spaces? Just for curiosity

[lilac barn]

Not sure what "using" means. If the question is how usually do Sobolev spaces arise in PDE theory then the answer is all the time. Sobolev spaces give a more flexible characterisation of your original PDE problem

[native crown]

I meant, when ordinary derivatives doesnt work and the "solution" dont satifies the restrictions

Because my thesis is PDE related and Im trying to understand what it motivated Hörmander to consider distribution theory to solve a problem

[lilac barn]

Then check this out https://math.stackexchange.com/questions/3314557/why-should-i-believe-in-weak-solutions-to-pdes

[native crown]

I will check it thanks

[sonic jacinth]

iirc what motivated thinking about them was wanting to have some notion of solutions depending continuously on parameters (such as initial conditions) among other things

[native crown]

Im also taking a course about Sobolev spaces so I will ask my teacher too. He motivated from where this spaces appears and why but Im still a bit confused

[native crown]

Smth similar our teacher told us. I really appreciate that he gave us a math context with dates

[azure blade]

Hi everyone. I'm not sure if this is the best place to ask this question because it's not really PDEs, but I'm using it to perform analysis related to PDEs. Essentially, I'm trying to expand a solution to a PDE using Hermite series (sum_n a_n H_n(x) for H_n the physicist Hermite polynomials). For simplicity, let's say it's a stationary solution for a 1D PDE so there's no need to worry about tensor spaces right now. For the purpose of my analysis, I need a Banach algebra on the sequence space of coefficients. So if w = uv (all three Hermite series), I need |w|_X <= C |u|_X |v|_X (edit: these are norms, but double bars are behaving strangely) for some constant C and X a Banach space. With Fourier series, we usually use l^1 (or a weighted version of l^1) as C = 1, so you don't need to "think about C" in that sense. For Hermite, this isn't so obvious because the product is "weird" (i.e. not a convolution). I'm not necessarily looking for a proof right now, but I've been going at this for a while so I thought I'd see if anyone could provide some intuition here? I'm beginning to think it's not true for l^1, but I don't want to aimlessly try l^2, then l^3, etc. and keep hitting dead ends if I'm looking at it the wrong way. I can provide some work if that helps.

[robust moat]

This is definitely not my area (I deal with Riesz spaces and I avoid norms as much as I can), but I'll soon be meeting with some experts in functional analysis (the people I'll see on Tuesday deal with pre-Riesz spaces, but they like Banach algebras and the people I'll see on the 25th of March are people who work in Banach algebras). If you don't mind waiting (and, probably until the 25th of March), if you remind me then I can pose your question to an expert in Banach algebras and send that response to you. Apologies that I can't be of any help and that it'll take a while until I'm with the appropriate people.

[azure blade]

That would be great! I'm not in a rush to get an answer. I'll also let you know if I find an answer beforehand. I'll be in touch!

[robust moat]

OK, I'll let you know who has what to say!

[prisma pelican]

why does this inequality hold?

I'm confused, shouldn't the integral go to infinity?

\Phi is the fundamental solution of the leplace equation and f is a compactly supported C^2 function

[astral vine]

use Polar Coordinates

then you just have to found 1-dimensional anti-derivative

[candid token]

What is L^p norm on distributions?

like W^(k,p) is the space of functions with kth weak derivatives in L^p

but what does it mean for kth weak derivatives to be in L^p?

[lilac barn]

There isn't one. There is a natural one for distributions that are representable by Lp functions.

[lilac barn]

That's the function representing the kth weak derivative is in Lp

[candid token]

i see ok so it has to genuinely reside in L^p, makes sense

like sin(|x|) is in W^(k,infty) for any k

[astral vine]

An interesting convention is also to say that for distributions that are not identifiable with an Lp function the value is +\infty, as for measurable functions.

Think for instance about Triebel and Besov norms

[brave bay]

Is there a simple relationship between the concepts of the characteristic variety, characteristic hypersurfaces and bicharacteristic curves/strips?

Say we are looking at an m-th order linear PDE on a smooth manifold X of dimension n.

The first is a single 2n-1 dimensional conic submanifold of the cotangent bundle T*X, defined by the vanishing of the principal symbol of P.
The second is a set of of n-1 dimensional hypersurfaces in X, defined by the property that the conormal bundle is contained in the characteristic variety.
The third is a set of 1d curves in the cotangent bundle T^*M, defined as the integral curves of the Hamilton vector field associated to the principal symbol of P.

My understanding:
-The third and first have a clear connection, in that through any point on the char. variety there is a "null" bicharacteristic that stays on the the char. variety. The importance of these curves is results like the propagation of singularities.
-The second is something that can be regarded as an obstruction to solving Cauchy problems. (It essentially means that the m-th normal derivative of a solution to Pu=0 is not determined by the first m-1 normal derivatives). There are further refinements such as the various forms of hyperbolicity that one can ask for that guarantee existence of Cauchy problem solutions, but a non-characteristic initial hypersurface is using imposed to begin with.

However the geometric link between the characteristic hypersurfaces and bicharacteristic curves (or their projections to the base manifold X) is unclear to me. Does anyone know of one beyond their links to the char. variety mentioned above?

[mint canyon]

Does it count as a "relationship" that the bicharacteristic curve is always on the characteristic hypersurface? I think the bicharacteristic is the result of taking the characteristic of the characteristic equation (in the average use case).

[brave bay]

It counts as a relationship, but also one that I was aware of. (See first point under "my understanding"). What I am more interested in is the relationship between the second and third notion. That is, between the notion of a characteristic hypersurface (most relevant for the consideration of Cauchy problems and where to specify initial data) and bicharacteristic curves (most relevant when considering the propagation of regularity for solutions to PDE of real principal type).

Also, I don't really know what you mean in your second sentence.

[mint canyon]

I was referring the the specific case of the second order method of characteristics being solvable by way of Lagrange-Charpit (the fully nonlinear method of characteristics for first order equations)

[brave bay]

Sure well I know how to solve nonlinear first order equations using characteristics, and that in the linear case the characteristics are just the projections of the bicharacteristics in the cotangent bundle onto physical space, and furthermore that the condition of an initial hypersurface being non-characteristic is precisely equivalent to these projected characteristic curves meeting the initial hypersurface transversally. (Which is of course desirable in order to make the method of characteristics work).

All of this is well and good, but all of this is also for first order equations in particular.

The heart of my question is that I am curious if there is an equally nice geometric picture for the relationship between these concepts in the setting of higher order linear equations.

[mint canyon]

The characteristic equation for Second order PDEs is first order

∑ᵢ,ᵣ Aᵢᵣ(∂φ/∂xᵢ)(∂φ/∂xᵣ) = 0

Im not certain if method of characteristics generalizes to higher than second order. I think the principle coefficients in that case dont fit in an NxN matrix

you could use a rank (order) "tensor" I guess

[brave bay]

I don't think it does, but I am not really asking about the method of characteristics in any case. Whether or not one can use the notion of characteristics (in any sense of the term) to actually solve equations, one still cares about:
a) the solution of Cauchy problems (for which the notion of a non-characteristic initial hypersurface is important to solvability)
b) the propagation of singularities (which happens along bicharacteristic curves in the cotangent bundle)

[mint canyon]

Okay, the only context I have heard the word "bicharacteristic" is when you take the characteristic of the characteristic equation.

I dont know enough about geometric PDE theory to probably answer your question, but I do know what all those words mean for general relativity

[brave bay]

in what sense do you mean "take the characteristic of" ?

[mint canyon]

what I said about using Lagrange-Charpit earlier

[brave bay]

What you call the characteristic equation is the condition for the level set of a function being a characteristic hypersurface in the sense I mean it

ah okay sure, well then yes that coincides with what I call bicharacteristic (the integral curves of the hamilton vector field of the principle symbol)

So I guess I am not completely satisfied yet, but I do now see that the condition defining characteristic hypersurfaces for an operator P (this condition is a nonlinear first order PDE) has characteristics that coincide with the bicharacteristic curves of the initial operator P in the Hamiltonian sense. That is an interesting (but probably obvious from the right point of view) observation.

Yeah ofc I guess that is the equivalence between Hamiltonian dynamics and the Hamilton-JAcobi formulation

[mint canyon]

when you say Hamilton vector field, do you mean Hamiltonian vector field, or is it something else?

[brave bay]

exactly that. both terminologies are used in diff sources.

[mint canyon]

isnt that supposed to exist on a symplectic manifold instead of a Riemmanian?

[brave bay]

Yes, that is why I said that bicharacteristic curves live in the cotangent bundle, which has a canonical symplectic structure.

what you are calling characteristic curves are the projections of these onto the base space

[mint canyon]

ahhh, okay. I learned a bit about those in Classical mechanics

symplectic form is preserved under canonical transformations and such

[brave bay]

yep

[mint canyon]

did you know second order characteristics can be generalized to the fully nonlinear case? I was pretty surprised by that.

still has to be hyperbolic/parabolic for the characteristics to exist, but feels like it should be mentioned more in the literature

[brave bay]

yes, it is definitely surprising when you first see it!

[mint canyon]

Since you brought up Hamilton-jacobi, there is a particular equation I was attempting to analyze for characteristics

https://en.wikipedia.org/wiki/Hamilton–Jacobi–Einstein_equation#General_equation_(free_curved_space)

[misty needle]

How was this equation derived in the document?

[mint canyon]

in free space, you can drop the sqrt(g)R term

[misty needle]

What should I do if the derivation in your paper cannot be extrapolated

[candid token]

will these also be equal to delta^(alpha+beta) F

[quaint herald]

\partial, not \delta, but yes.

[candid token]

oops yea

[mint canyon]

What is the deal with Method of Constraints? Are you just selecting a constraint equation and hoping there is a Union between the solutions of the original and constraint equations?

[steel umbra]

i'm totally stuck on part b. like i get we want to show the limit of the integral of uxx^2-utt^2 is zero, can we somehow say that ux-ut approaches zero as t goes to infinity?

[sand echo]

its probably easiest to show it using the formula for the solution in terms of g and h

remember that you need to show k(t) = p(t) for all t > T for some finite T. Its not enough for k(t) to approach p(t) or something like that

[steel umbra]

I see. it does have to be in finite time.

[steel umbra]

do you mean d'Alemberts solution?

[sand echo]

yeah exactly

[steel umbra]

so i did the whole computation and when i did ux^2-ut^2 i got a sum of products of like g'(x+-t)h(x-+t), and for large enough t, we can misalign the supports to get it to be zero for all x.
but i think it might be much easier to just write it as u(x,t)=F(x+t)+G(x-t). i think same thing but the computations not as horrible lol

because F and G will also have compact support right?

because we can set the lower bounds of the integral on h to be outside the support of both functions

[sand echo]

yeah so these are not compactly supported

because of the integral of h

the thing you did originally sounds correct

[wind mortar]

Literally my test for PDE today lmao

Just had it

It was straight brutality

Omg the same thing but a little harder lol

[wind mortar]

5 3 Can someone help

[steel umbra]

but the derivatives [g'(x)+-h(x)]/2 will be, right? then i think it does work
ux=F'(x+t)+G'(x-t)
ut=F'(x+t)-G'(x-t)
both have compact support, so ux^2-ut^2=4F'(x+t)G'(x-t)
we can pick t large enough such that we knock them both off their support and F'(x+t)G'(x-t)=0 for all x.
does that pass the vibe check?

[wind mortar]

The product of uxut from negative infinity to infinity goes to zero

[wind mortar]

What exactly is sign(x) defined as?

[rapid vault]

1 if x>0, -1 if x<0

[sand echo]

Yeah I think so

[pulsar rivet]

any help?

[mint canyon]

use separation of variables, U(x,t) = X(x)T(t)

[hasty shard]

potentially a silly question, suppose
\begin{align}
\lvert \int g(x) f(x), dx\rvert^2 = \lvert \int \tilde{g}(x) f(x), dx\rvert^2
\end{align}
for all nice (to a physicist's standards) functions $f$. By taking $f(x) = \delta(x-x_0)$, you can see that $\lvert \tilde{g}(x_0)\rvert = \lvert g(x_0) \rvert $ for all $x_0$, and so $\tilde{g}(x_0) = e^{i \theta} g(x_0)$. From this alone there is no reason to think that $\theta$ is independent of $x$, but I strongly suspect that it is, and that one can see this from taking $f$ to be other functions besides a delta function. \

My question is... is $\theta$ actually independent of $x$? If so, how do you show it, and if not, then why not?

[untold delta]

nay

[blazing ridge]

g is a function from R to R?

[hasty shard]

sorry should have specified, g and f are functions from R -> C

[untold delta]

nay

[hasty shard]

sorry this has errors

corrected: From taking f to be a single delta function $f(x) = \delta(x-x_0)$, we know $\tilde{g}(x_0) = e^{i \theta(x_0)} g(x_0)$. Now we want to show $\theta(x) = \text{constant}$. Take $f(x) = \delta(x-x_a) + e^{i \phi} \delta(x-x_b)$ for some $\phi$ of our choosing. Then we have, upon substitution into the integral and using the result from taking f to be a single delta function:
\begin{align}
\lvert g(x_a) + e^{i\phi}g(x_b)\rvert &= \lvert g(x_a) e^{i\theta(x_a)} + e^{i\phi} e^{i\theta(x_b)} g(x_b)\rvert \
&= \lvert g(x_a) + e^{i\phi} e^{i (\theta(x_b) - \theta(x_a))} g(x_b)\rvert
\end{align}
and then geometrically this can only hold for all $\phi$ if $\theta(x_a) - \theta(x_b) = 0$.

[untold delta]

nay

[inland sinew]

This might be a silly question, but Im looking at a boundar value problem(dirichlet) for the laplacian on a square. Once I have the solution, Im wondering how the boundary changes when I integrate the function with respect to x or y.

Suppose thr boundary condition for the solution at the right edge x = a is f(y). Im wondering how the function at that edge changes when I integrate with respect to x.

I figured there was a straightforward way to see what it becomes because if I integrated with respect to y, its obvious youd just integrate f(y), but when I integrate my solution z with respect to x and evaluate it at x = a, its changing the fourier coefficients if f(y).

[steel umbra]

my question is just about plotting the "exact" solution i'm going to be getting numerically. would we use a very small epsilon and the Cole-Hopf transformation (would that even work?) or is there a simpler way i'm not seeing using characteristics or something?

[sand echo]

you should be able to calculate a formula for the solution using characterstics

[mint canyon]

ln|t| + a = (ln|x| + b)/u = c

[steel umbra]

I broke it up into two Riemann problems and got

$u(x,0.5)=\casedef{
-0.5,&0\leq x<0.25\
\frac{x-1/2}{0.5},&0.25< x<1\
1,&1<x<1.25\
0,&1.25<x<1.5
}$

does that seem reasonable?

[untold delta]

eigentaylor

[steel umbra]

$u(x,0.5)=\casedef{
-\frac12,&0\leq x<0.25\
\frac{x-\frac12}{0.5},&0.25< x<0.75\
1,&0.75<x<1.25\
0,&1.25<x\le\frac32
}$

[untold delta]

eigentaylor

[steel umbra]

or actually I think it'd be this

not sure

[sand echo]

yeah this looks quite reasonable @steel umbra

i can check carefully in a bit

[steel umbra]

that would be much appreciated. I think it should actually be the first one but having someone double check would be great

[sand echo]

okay yep I think i agree with the first one

one sanity check is the bit between the solution changing from u=x-1/2 / 0.5 to u=1 should be continuous

[steel umbra]

yeah definitely

[inland sinew]

Hi. I have a question about the Laplacian on a square again. I am looking at the case where all of the boundary conditions are equal to a quadratic, something simple like x^2. By boundary conditions I mean dirichlet. I'm doing it just by adding 4 of the individual solutions that are zero on 3 sides besides 1. The basis functions for expressing the functions on the profile are something liike sin(pin(x-a)/L), and when I'm trying to find the fourier series for this function that is quadratic using these, and im looking at the partial approximations, it looks like the series is converging extremely slowly and I'm wondering if that is because the basis functions I am using.

[wind mortar]

Can I see it in equations???

[inland sinew]

@wind mortar Let me attach a pdf witht the plot profile and my short calculations attached.

[wind mortar]

Beautiful

[inland sinew]

@wind mortar

lmao nevermind, I got it. I had a typo in my code that was plotting this. It still takes so many terms to get a good approximation though. I needed 200 terms just to get this. It just seems weird to me to approximation functions that look like they are non-zero at the end points of the boundary by eigenfunctions that are zero at the end points of the boundary. I mean in the case of x^2 over the interval [-pi pi], I'm approximating it with these sin((x+pi)n/2) functions that are zero at those end points.

[wind mortar]

Hi I’m back

I don’t open pdfs

Im glad you found it

[flat crow]

I have a question about weak solutions of the wave equation

suppose I have the wave equation $u_{tt} = u_{xx}$ on $\mathbb{R} \times \mathbb{R}^+$ with initial condition $u(x,0) = H(x)$ and $u_t(x,0) = 0$

At the moment, it seems to me that the solutions $u_1(x,t) = H(x-t)$ and $u_2(x,t) = \frac 12 \left( H(x-t) + H(x+t)\right)$ both satisfy the wave equation weakly and the initial conditions ae. What am I missing here?

[untold delta]

jamiecjx

[flat crow]

suppose H were not a heaviside function but something that approximates it smoothly, then only the second solution $u_2$ makes sense.

[untold delta]

jamiecjx

[flat crow]

I think i resolved it, if we are to have a weak solution that takes in to account the initial conditions, then our test function $\varphi$ will somemtimes need to have support containing $t=0$, in which case the integration yields extra terms that involve the initial conditions

[untold delta]

jamiecjx

[inland sinew]

@wind mortar My bad, thanks though. I do have a technical question though about solutions of the laplacian. When solving the laplacian on a square in the problem I mentioned with dirichlet bc, the typical approach Ive seen is adding these 4 solutions that each contain one of the boundary condition.

Does this approach fail to capture any solutions? For example, what about a solution that is bilinear that I also know solves the equation, or constant? Granted I know fourier series can also represent those, but it seems almost like a hastle to write in that manner.

[wind mortar]

@grand garnet add Me, are you doing quantum mechanics ?

[inland sinew]

@wind mortar No, it's sort of hard to explain the problem I'm looking at. It's related to small bending of surfaces. The out of plane component of the deformation for the surface I'm looking at is governed by the laplacian over a square interval [-a a] x [-a a]

[tired hollow]

Is the problem you are concerned with well-posed?

i.e. the solution exists and its unique

If this is the case, the separation of variables method should give us the solution, assuming that the method works.

[unborn quiver]

Yeah, the domain is a square, so with Dirichlet boundary conditions, it'll be fine to just use separation of variables/fourier series to get to a solution.
If this is a numerics problem, you're best to ask in #numerical-analysis

Something like $\Delta u(x,y)=0$ on $[0,1]^2$ with $u(x,y)=g(x,y)$ in $\partial ([0,1]^2)$ is fine

[untold delta]

Kirby

[unborn quiver]

They mention g is a quadratic, so it's smooth

[tired hollow]

That problem is well-posed

Due to the Lax-Milgram theorem

[unborn quiver]

I meant to reply to your first message with that bc I think that's what they're trying to solve, but yes

[tired hollow]

Oh I see

Sorry didnt catch that

[unborn quiver]

No worries, discord likes to drop replies for me for some reason, as in if I click reply, it forgets I clicked reply

When I leave and re-enter the app

[inland sinew]

@tired hollow yes I would say so. I am just looking at the case where each profile on the side is f(x) = x^2 or g(y) = y^2. When I am plotting this however, I am getting these very sharp peaks at the 4 corners of these solutions which I anticipate is due to the fact I'm using eigenfunctions that vanish at the corners where my function takes non-zero values. I figured the appropriate way to handle this is to have my solution be a sum of solutions, one defined on (-1,1)^2, which is my square minus the four corners, and the other solution being piecewise that defines the value of the solution at those 4 corners

[tired hollow]

Your boundary data lack smoothness at the corners

This affects the convergence rate of the series

Back to your original question, this is fine since for wellposedness if i remember correctly the boundary data has to be in H^(1/2)(Γ), where Γ is the boundary.

Since the problem is well-posed, the solution exists and its unique, so your method should lead to the one only solution.

[inland sinew]

@tired hollow Okay I figured. Would using different eigenfunctions like cos help my convergence? Because theyd be nonzero at the corners or would this make no difference. I just figured it would be more natural to choose for these boundary conditions.

[inland sinew]

Does anyone know any good literature on linear second order pdes with periodic coefficients? (in such a way that the discriminant is mixed and alternating between hyperbolic and elliptic periodically in the domain).

[astral vine]

There is only two/three references about PDEs with periodic coefficients and this is mostly for elliptic/Parabolic and Schrodinger equation, like PDEs

Search for Bloch-Floquet Theory

Kuchment is a first but hard reference

but prolly the msot important one

[unborn quiver]

If I have something written like
$\mathrm{tr}(D^2F(\grad w))D^2w)=0$ (in the case of strictly convex functionals $F$, with enough regularity on $w$). We can "ignore dependence of $D^2F$ on $\grad w$ and write $A(x) = D^2F(x)$ uniformly elliptic." I am curious of the motivation of when we can remove dependence like this when studying solutions

[untold delta]

Kirby

[unborn quiver]

I'm currently reading Vasseur's notes on De Giorgi's method for Hilbert's 19th problem

[mint canyon]

has anyone messed with semigroup methods for linear operators?

[astral vine]

I'm involved with the topic

what are your questions

[mint canyon]

@astral vine The following link contains an integral formula that "exponentiates" the infinitesimal generator to the representation of the semigroup, but its kind of confusing what they mean by Inverse

https://en.wikipedia.org/wiki/Analytic_semigroup#Characterization

Im assuming the operator is taken to be a vector field in the coordinate basis of the partial derivatives.... though Im not sure what that means if you have 2nd derivatives

[mint canyon]

I guess it would just be another vector, but Im not certain what it would be in terms of the coordinate basis

[unborn quiver]

Idk why this got a thumbs up, it was a question about some formal calculations in these notes on page 4 https://web.ma.utexas.edu/users/vasseur/documents/preprints/DGPekin.pdf

I understand the method, I'm just curious on when we can ignore dependence like this

[astral vine]

Ok so if I am correct : what is confusing you is definition of

[untold delta]

Functionanatolysis

[mint canyon]

That's correct

[astral vine]

Okay so the definition of such an operator is given if you can solve

[untold delta]

Functionanatolysis

[astral vine]

provided f is known and u is the known

if you have existence and uniqueness of such solution

then you set

[mint canyon]

I'm also not certain I understand what the identity vector is

[astral vine]

This is not a vector

This the identity map

so for the Laplacian

you want to solve the (Helmholtz) equation

[untold delta]

Functionanatolysis

[astral vine]

I'm back on this part

such an unique solution allows to define

$$ (\lambda \mathrm{I} - A)^{-1}f := u $$

[untold delta]

Functionanatolysis

[astral vine]

because if you apply the operator

$$(\lambda \mathrm{I}-A) [(\lambda \mathrm{I}-A)^{-1}f]= (\lambda \mathrm{I}-A) [u] = \lambda u-Au = f$$

[untold delta]

Functionanatolysis

[mint canyon]

Okay, I get that. The inverse in this case seems like it would be some kind of "integral operator", so Im guessing that cant be treated as a vector in the original formula?

[astral vine]

What do you mean by vector

u and f are the vectors here

everything else is about linear operators in such setting

[mint canyon]

∂/∂x is the basis vector in the x direction, I was assuming the operator A was taken to be a vector field in that way

[astral vine]

You are mixing geometry stuff and differential operators formalism in functional analysis

Both are denoted the same way

[mint canyon]

well, infinitesimal generators are vector fields in Lie Groups

[astral vine]

but are distinct objects

[astral vine]

Can be representend by*

I get that point but you are mixing up the different type of stuff

there is an identification made there

those ∂/∂x (from geometry) and ∂/∂x (from FA) are NOT exactly the same objects

But okay

what ever that was jsut confusing for me

[mint canyon]

well, the idea is if you have some parametric curve along coordinate curves, with parameter x, the derivative gives you the tangent vector which is the basis vector. Though it does have to ACT on something to be meaningful (a curve)

[astral vine]

I know geometry a little bit

[mint canyon]

I hear you though, functional analysis is different, I'll keep that in mind

[astral vine]

This is the same semigroup theory behind those evolution equations

When you want to say that a "vector field X" is the generator of a semigroup, you have to forgot (for a short time) the geometric POV somehow

[mint canyon]

So Im feeling like if you can invert $$ (\lambda \mathrm{I} - A)$$

[untold delta]

x17

[mint canyon]

and you have a standard evolution equation, dont you already have a solution to the X separable part of the equation?

[astral vine]

what do you mean by the "X separable part" ?

[mint canyon]

$$ \partial_t u = Au $$ is separable isnt it?

[untold delta]

x17

[astral vine]

Not really

[mint canyon]

does A have t dependence?

[astral vine]

No but I don't really get what you mean by separable

However

Yes

having the good information on L*I-A

gives you the semigroup

then the unique solution to your linear Abstract PDE

provided you have some prescribed initial data

[mint canyon]

$$ \partial_t u = \lambda, Au = \lambda $$

[untold delta]

x17

[astral vine]

No you cannot do that

You forgot a u and the two right handsides, first.

And second, this assume that the time derivative or A at least do have eigen vectors and eigenvalues

(without the u's as you wrote, it the equation is not false but won't lead anywhere)

[mint canyon]

the u on the right comes after the initial separation and can vary a bit with the choice of ansatz

[astral vine]

This won't work even for the basic equations

[mint canyon]

you might need to divide by some extra function to get separation for instance

[astral vine]

Maybe in your super specific case

But otherwise

for general A

it doesnot really makes sense

Think about A a matrix acting in dimension 2.

say a rotation matrix

(since you deal with Lie Groups this should be familiar to you)

[mint canyon]

Okay, so because u isnt known to be separable in general, Au might have t dependence, and u_t might have x dependence, even though A has no t derivatives or coefficient functions in t

[astral vine]

Yes

If this is what you meant with separable

[mint canyon]

general separation of variables can get pretty complicated, you have R-separation and the general ansatz $$ \Sigma_i^N X_i(x)T_i(t) $$

[untold delta]

x17

[astral vine]

This may not happen even for the easiest linear equations

This very specific to A when it has spectrum made only of eigenvalues, whose eigenvectors forms a Basis of your space

[mint canyon]

Yeah true, even with some really exotic ansatz, most linear equations arent separable, which is why Im looking into semigroup

I got distracted, Im sorry

[astral vine]

No worries :)

It is easy to get lost

[mint canyon]

What does it mean to integrate over the inverse operator, you said I should think of it like a matrix?

[astral vine]

if A is a matrix that's not a trouble right ?

if you are on a Banach space

the equality

[mint canyon]

is it correct to call $$(\lambda I - A)^{-1}$$ the resolvent in this casee?

[untold delta]

x17

[astral vine]

Yes

and applied to a fixed element of your Banahc space

this lead to an holomorphic function whose "poles" are the spectrum (eigenvalues or not)

[mint canyon]

so have do you translate the resolvent from its operator to the associated matrix?

[astral vine]

Don't get what you mean

[mint canyon]

is it like in Quantum where operate the Hamiltonian on an eigenfunction and then left multiply by its adjoint eigenfunction, and the 2 indices give the matrix element?

[astral vine]

Note really I was talking about matrices because semigroup theory contains exponential of amtrices

everything still hold true for matrices

instead of general linear operators

[mint canyon]

i did a bit of matrix exponentiation with generators of GL(n)

again back in Lie group stuff

So when a start with a linear PDE, A is a differential operator. The resolvent of a will be... something else maybe a Volterra operator??? but it wont be a matrix, how do I get the resolvent into a form where I can use the formula?

[astral vine]

In general there are no explicit form for the resolvent

You can just show that the operator and its resolvent exists and that's all

There are even operators that CANNOT have integral representation of solutions

[mint canyon]

So what is the procedure for using the formula and generating a solution (in the cases where that is possible)?

[astral vine]

I don't know

just the semigroup and the resolvent exist

and in general additional regularity properties from the operator A are sufficient for what people wants to achieve

[mint canyon]

bummer I was hoping I could use this to solve elliptic equations with arbitrary coefficient functions

[astral vine]

The theory, if the coefficients are not too rough, is already known

[mint canyon]

is it the transformation to standard form using the characteristics?

[astral vine]

Absolutely not

[mint canyon]

when I use Lie's transformation group method (differential invariants), I get a bunch of arbitrary linear pdes that need to be solved, most of the time they are easy, but some times not. So im trying to learn methods for as many linear PDEs as possible

wait... isnt the resolvent just Greens function of the operator $$(\lambda I - A)u = 0$$

[untold delta]

x17

[astral vine]

instead of 0

[mint canyon]

delta function, right. my bad

[astral vine]

you should put the dirac mass

But It may not exist green functions in general

it does for elliptic oeprators with smooth coefficients

[mint canyon]

There is a slow converging method of fourier expanding the Green's function

[astral vine]

this only works in the periodic setting, or on the whole space when you have smooth coefficients

again

There is still somesort of Fourier approximation methods if you can show that the spectrum of the operator is purely made of eigenvalues whose eigenvectors are Basis of L²

Which again may not happen

[mint canyon]

suppose you could get a closed form (not expanded) greens function, then you can directly compute the semigroup representation?

[astral vine]

This is true is you have a closed form for the resolvent problem yes

This is still an integral formula

but

somewhat explicit

but in general to compute an explicit formula for resolvent problem may be a hard task

[mint canyon]

okay, thanks alot. i think I understand the procedure and limitations of the method now.

[astral vine]

No worries

I hope this was not too convoluted or anything

[mint canyon]

Im a physicist, so i tend to lean more toward algorithmic thinking, sorry if that was creating disconnects, im mostly looking at methods as tools.

[spare condor]

I have a sequence of functions $u^n$ and strong convergence $u^n \rightarrow u$ in $L^2((0,T) \times \Omega)$ on $\Omega = [0,1]^2$. I also have a weak convergence $f(u^n) \rightharpoonup g$ in $L^2((0,T) \times \Omega)$. What criteria does function $f$ need to satisfy so that I have a weak convergence $f(u^n) \rightharpoonup f(u)$ in some $L^2((0,T) \times \Omega)$ and $f(u) = g$ a.e.?

[untold delta]

Xilexio

[lilac barn]

This is equivalent to f being a closed mapping from this space endowed with strong topology to this space endowed with weak topology

Now you can look up for references dealing with your hypotheses to see whether they imply closeness of the operator. A fairly general condition is f is a continuous/linear-bounded operator between the strong topologies

[spare condor]

Thank you for your response. I have C^1 f that scales quadratically. The problem is that while in this case I was able to prove a strong convergence because |f(u^n) - f(u)|_L1 is bounded by C * |u^n - u|_L2, I do not know a theorem or method to get weak convergence. And I have just L2 convergence of u, not pointwise, so I don't see why it couldn't explode to infinity in some points as long as it integrates. Can you advise on what am I missing?

I'm having trouble finding a specific theorem about weak convergence of mapping with strongly converging arguments. Would you be able to name a theorem I could look up? Continuous mapping theorem works only for continuous u.

[lilac barn]

This tells you that f(u^n) converges strongly to f(u) which implies weak convergence

[spare condor]

L1 convergence does but do I absolutely need strong convergence of f(u^n) to get weak one but in better space? Note that now I'm searching for a more general principle because I can imagine some bounded f(u^n) that converge to something weakly but not converging strongly even in L1.

Specifically, if u^n converges strongly to u in L2 and f is continuous and nonlinear and f(u^n) is bounded in space L1 (so converges to g), is g = f(u) a.e.?

And is proof of boundedness of f(u^n) required or does strong convergence of u somehow circumvent that?

[lilac barn]

In what space is f continuous?

[spare condor]

Let me write it down better.
I have $f \in C^1(\mathbb{R})$. I have strong convergence $u^n \rightarrow u$ in $L^2(\Omega)$. I have bounded $f(u^n)$ in $L^2(\Omega)$, which gives me existence of $g$ such that we have weak convergence $f(u^n) \rightharpoonup g$ in $L^2(\Omega)$. I also know that $f(u) \in L^1(\Omega)$. I wanted to ask how to prove that $f(u) = g$ a.e. or equivalently that we have weak convergence $f(u^n) \rightharpoonup f(u)$ in $L^2(\Omega)$ or possibly in other space. Using quadrating scaling of $f$ (extra property), I was able to prove that I have strong convergence $f(u^n) \rightarrow f(u)$ in $L^1(\Omega)$, but I'd like to learn if there is a way to prove any weak convergence of $f(u^n)$ to $f(u)$ without relying on additional properties of $f$ and instead relying just on previous info.

[untold delta]

Xilexio

[shrewd epoch]

Hi,
I'm trying to study the discrete spectrum of a 1D Schrödinger operator on an interval, either by finding it explicitly or by finding asymptotics as the size of the interval goes to infinity. Basically the operator I'm looking at is $-\frac{d}{dx^2}+V$ where the potential $V$ is given by some nice symmetric "single-well" $L^2$ function like for instance $\tanh^2(x)-1$. Operator is acting on $H_2\cap H_1^0(-\frac1\varepsilon,\frac1\varepsilon)$ (acting on functions on that interval with Dirichlet boundary conditions).

[untold delta]

upheaval

[shrewd epoch]

I know that the discrete spectrum of those operators will "converge" in some sense to the spectrum of the same operator on $H^2(\mathbb R)$, and I can get a basic description of this spectrum as the potential is in $L^2$ (I know that its point spectrum lies in $\mathbb R^-$ and 0 is the only possible accumulation point)

[untold delta]

upheaval

[shrewd epoch]

but I'm kinda stuck as to how to get estimates on the eigenvalues

(I don't need anything really sharp, just asymptotics of the spectral gap would be enough)

[last cloud]

apply semiclassical analysis techniques. For a potential $V(x) = \tanh^2(x) - 1$ and considering the operator on a finite interval with Dirichlet boundary conditions, the asymptotic behavior of the eigenvalues $\lambda_n$ as the interval size tends to infinity can be summarized by the formula:
[
\lambda_n \sim -\left(\frac{n\pi}{\log(\varepsilon^{-1})}\right)^2
]
for large $n$, where $\varepsilon$ characterizes the size of the interval.

[untold delta]

aafd

[hoary oak]

If f is a compactly supported smooth function on R^n then show that its symmetric decreasing rearrangement f* is also a compactly supported smooth function.

[blazing ridge]

anyone have a good reference for derivation of compressible navier stokes?

or i will settle for incompressible

[mint canyon]

compressibility comes from the equation of state and equation of continuity which relate pressure to temp and density and which relate change in density change to flux of momentum, in the incompressible case you just set the density to a constant parameter

[sick maple]

What is the difference between these two definitions of the fractional sobolev space H^s?

f in H^s iff (1+|n|^s)|f^(n)| in L^2

f in H^s iff f in L^2 and (1+|n|^{2s}) f^(n) in L^2

[unborn quiver]

Where are you getting these definitions? The second definition doesn’t have order s, is it missing a square root?

[sick maple]

https://arxiv.org/pdf/1104.4345.pdf

First is from evans pdes

[tight dome]

Yeah you forgot a square root in the second one

Since when you take the L^2 norm the first term is squared

[sick maple]

What

But see the defn i sent

[tight dome]

You need a square root around the factor (1 + |n|^2s)

Since you’re taking an L^2 norm

[sick maple]

Then why doesnt that definition say that

[unborn quiver]

What

[tight dome]

Because that definition is phrased in terms of an integral

[sick maple]

Oh wait

So the two defns are equivalent? Or

[tight dome]

If you add the square root, yes

[sick maple]

Doesnt the sqrt go on the integral though

[tight dome]

Write out the integral which defines the L^2 norm of (1 + |n|^2s)f^(n) and compare it to the one in the paper

[unborn quiver]

You said that (1+|n|^2s)f^(n) is in L2, which is not what this says

We’re saying that they’re equivalent if the (1+|n|^2s) term has a square root so that it’s order s

[sick maple]

How would that make it order s

You cant just apply the sqrt to the 2s

[tight dome]

It’s effectively order s

Both are order 2s when squared

You have inequalities like C_1(1 + |x|^s)^2 <= (1 + |x|^2s) <= C_2(1 + |x|^s)^2 which is all you need to show the norms are equivalent

[unborn quiver]

Yeah

[sick maple]

Oh

So theyre equivalent norms

[unborn quiver]

The order of that scalar is all that’s important here. In some places, you’ll also see $(1+|\xi|^{2})^{s/2}$ (denoted as $\langle \xi\rangle^s$) instead

[untold delta]

Kirby

[unborn quiver]

@sick maple this is a good exercise to verify that they’re equivalent

[sick maple]

$(1+|n|^s)\hat u(n)\in L^2 \Leftrightarrow \int (1+|n|^s)^2|\hat u(n)|^2 dn<\infty$

[untold delta]

kevinhardy2

[sick maple]

and youre saying that

ok

yeah

youre saying that the right side of that is basically $\int(1+|n|^{2s})|\hat u(n)|^2 dn$

?

[untold delta]

kevinhardy2

[blazing ridge]

Sure but I’m looking for details

[mint canyon]

what specific details?

also, the momentum transport equation is just F = ma when you volume integrate both sides

I have a book I read about the derivation in if thats what you want, let me find the amazon link

https://www.amazon.com/Viscous-Flow-Theory-I-Laminar/dp/B000E5WOLO

[unborn quiver]

Yes, for the reasons frank mentioned. I’d recommend you work through why the norms are equivalent

[untold delta]

Rudolf

[proper knot]

nvm I figured it out

[blazing ridge]

so i've seen how to derive the continuity equation from mass conservation but didnt get how balance of momentum leads to the momentum equation

the book im reading just says its basically the same argument but i dont see how

[mint canyon]

the (mass) density * the material derivative of the velocity is basically the total time derivative of the momentum density, you replace the dx/dt, dy/dt, dz/dt terms with the velocity components though

on the other side you have the divergence of the stress tensor, which is just the Force density

[sick maple]

for a similar reason |n|^{2s}|\hat u(n)|^2 is also equivalent right

[pulsar forge]

how does the red part follow from compactness here?

[sand echo]

its not; it follows from the previous bit of the proof

[quaint herald]

I am out for lunch atm but the estimate should just be from interior approximation, (i.e. the previous part of the proof). the compactness gives you that there are only finitely many of these needed.

[pulsar forge]

ah, i see it now

[mellow totem]

If $f\in W^{1,p}(\Omega)\cap C(\overline{\Omega})$ then can we say that $\mu(\operatorname{supp}(f)-{f\ne 0})=0$? Assume $\Omega\subset \mathbb{R}$.

[untold delta]

contrapositive

[bright idol]

in one dimension, the boundary of an open set has lebesgue measure zero

for $f\in W^{1,p}(\Omega)\cap C(\overline{\Omega})$ with $\Omega\subset\mathbb{R}$

[untold delta]

Nats

[bright idol]

so yeah

we can indeed say that

[tight dome]

I think this is false, I think there are open sets whose boundary is a fat cantor set

Yeah, take the complement of a fat cantor set

[bright idol]

yeah, it's not always the case that it has zero lebesgue measure, that's true

[balmy solstice]

check this out -

I tried doing this and got stuck halfway

[tight dome]

For open subsets $V \subset\subset U$, the standard elliptic estimate for an order $m$ elliptic operator $L$ looks like
$$
|u|{H^k(V)} \leq \mathrm{const}(|Lu|{H^{k - m}(U)} + |u|_{L^2(U)}).
$$
Is there any way to get such an estimate to hold with $V = U$? Possibly after shrinking the domains?

[untold delta]

Frank

[shell jackal]

I don't have time to elaborate or confirm but I think that if u vanishes on the boundary of U then you can take a bunch of functions u_k that are given by taking u and making it go to 0 on smaller and smaller collars about the boundary of U which limit to u

[tight dome]

Ah yes, I think in that case it works

But I think when u doesn’t vanish on the boundary you really can’t get anything :’(

[pulsar rivet]

i have to show that

i know its with fourier series but i am stuck, any help

[mint canyon]

$$ \int_0^1 dx x(2-x) sin( \frac{2m + 1}{2} \pi x) = B_n \int_0^1 dx sin( \frac{2n + 1}{2} \pi x) sin( \frac{2m + 1}{2} \pi x) $$

[untold delta]

x17

[mint canyon]

it'll be zero unless n=m , so just do the right integral for that case to get the normalization constant and divide both sides by it

[pulsar forge]

can i get a hint for showing this ? i tried to think of functions that "blow" up on the boundary like ln(|x|), such that there derivative involves some form of dirac delta to avoid them being in W^1,p, but im not able to see the conclusion, how would i show the trace cant exist, because im basically trying to show any extension of some L^p function wont be L^p on its boundary but i dont see where the issue will arrise.

maybe it has to do with its derivative being badly behaved ? but then again i dont see how the derivative behaving badly implies the function itself is not integrable on boundary after some extension (at least not explicitly)

[pulsar forge]

actually i have a idea, what if in R i take ln(|x|+1/n) which is integrable on (0,1), then the L^1 norm on the boundary is ln(1/n)+ln(1+1/n) while its L^1 norm is finite, so passing to the limit the operator cant be bounded

and i guess in case of W^1,1, what makes sure these things dont happen is the norm of the derivative making the boundary norm "well behave"

if im right

[quaint herald]

I think you are overcomplicating things.

Take some continuous compactly supported function f in the half-plane (x,y:x > 0) that is nonzero somewhere on the boundary {x=0}. Then consider the one-parameter family of functions g(x,y;t) := f(tx,y) for t large. What are the restrictions to the boundary? How are the L^p norms of the restriction and g itself related?

Then understand why this example captures the general situation.

(Whilst doing this you should also think about the derivatives of g to understand why it does not violate the actual trace theorem.)

[lilac barn]

A simple method would be to consider mollifications of the boundary which would have constant Lp(boundary) but vanishing Lp(U)

Even simply, linearly interpolating between 0 and 1 near the boundary of U will also be sufficient

[quaint herald]

the method I gave above needs no computation with explicit functions at all, you win simply from the jacobian. although it is similar in spirit to the linearly interpolated example.

[lilac barn]

I wanted to give the mollification example as it works in quite a lot of different situations but then realized the interpolating example would be good for motivation

[pulsar forge]

so if i got this right. denote the half plane by $\Omega$ we have that $$||g||{L^{p}(\Omega)}^{p} = \int{\Omega} |f(tx,y)|^{p}dxdy$$ which goes to 0 as $t \to \infty$ (by for example LDCT) but on the boundary we have $||g||{L^{p}(\partial\Omega)}^{p}= \int{\partial \Omega} |f(0,y)|^{p}dy$ which we can make positive by our choice of $f$, violating continuity of trace if it exists, but the derivative saves us cause then we have $D_{t}g(x,y)=xf(tx,y)$ which also vanishes on the boundary

[untold delta]

James Banach

[pulsar forge]

i should say ||g(t)|| to be precise

[quaint herald]

First part yes. Second part careful, the relevant derivative is in x, not t.

[pulsar forge]

ah, yeah i see the mistake

[quaint herald]

Also one should not be considering the derivative of g at the boundary. The point is that our example does not have derivatives going to zero in norm in the interior, although the Lp norms of the function themselves do.

[pulsar forge]

so what happens to ||g(t)|| in W^1,p? i cant see why ||grad f||, or even just \D_{x}g would not go to zero, its just integral of tDf(tx,y) but Df(tx,y) still goes to zero as t gets very large

[lilac barn]

Write down the integral, then do a change of coordinates from tx = u so that a t will pop-out in the integral and that will help you

[pulsar forge]

oh ofcourse 🤦 missed the easy detail

thanks both

[kind wren]

Hi. I'm trying to look for materials on the following system of two PDEs.
Consider $\Omega = (0,1) \times (0,1)$, and $T > 0$.
[
\begin{cases}
\partial_t u(t, x) = -u(t,x)v(t,x), & 0 < t \leq T, x \in \Omega;\
\partial_t v(t,x) = \Delta v(t,x) - u(t,x)v(t,x), & 0 < t \leq T, x \in
\Omega.
\end{cases}
]
Boundary conditions: $u(t,x) = v(t,x) = 0$ for $0 < t \leq T$ and
all $x \in \partial \Omega$.

Initial conditions: $u(0, x) = u_0(x), v(0, x) = v_0(x)$ for all
$x \in \Omega$. They ($u_0, v_0$) are all non-negative throughout $\Omega$.

The idea is that this is a reaction diffusion equation.

I was looking to understand the kinds of solutions for this equation. It seems intuitive to me that they'll all be bounded. Intuitively, it seems to me that: $0 \leq u \leq u_0$ everywhere, $0 \leq v$ everywhere. It also seems intuitive to me that there is a global upper bound for $v$.

Do you guys know where I can find stuff about these kinds of equations?

[untold delta]

phao

[kind wren]

I mean websites... books... a chapter on a book, etc.

The laplacian operator $\Delta$ is spacial, meaning: $\Delta v = \partial^2_x v + \partial^2_y v$ in here.

[untold delta]

phao

[steel umbra]

this is literally how my professor gave this homework problem and i feel like an idiot because i can't parse it

this was my best guess but what the heck are v and V here? functions from Rn to R or vector valued functions?

is v grad v a row vector applied a matrix resulting in a row vector?

[mint canyon]

in GR thats a directional covariant derivative of a vector along itself. If it equals 0, then the integral curve of v would be a geodesic.

$$u^\mu \grad_\mu u^\nu = 0$$

[untold delta]

x17

[mint canyon]

if you assume euclidean geometry (and carteesian coords), the connection goes to zero, and you just have partial derivatives. I would assume thats your case?

$$ (u^1 \partial_1 + u^2 \partial_2+ u^3 \partial_3) u^\nu$$

[untold delta]

x17

[lilac barn]

The usual notation is to write ( v \cdot \nabla v) and is to be interpreted as the matrix (\nabla v) acting on the vector ( v). More precisely, ( (v \cdot \nabla v){i} = \sum_j v^j \partial_j v^i). An important consequence of this that gets routinely used is that if (v) vanishes on the boundary then
[ \langle v \cdot \nabla v, v \rangle{L^2} = \langle v , \tfrac{1}{2} \nabla |v|^2 \rangle_{L^2} = - \tfrac{1}{2} \langle \nabla \cdot v, |v|^2 \rangle_{L^2} = 0
]

[untold delta]

cocat

[lilac barn]

V is some generic function and v is in terms of V. You would wanna plug in the form of v to obtain an equation for V.

[steel umbra]

i ended up with

$Dp=\frac{V,DV}{(1-t)^{2a+b}}-
a\frac{V}{(1-t)^{a+1}}
-
\frac{bx,D V}{(1-t)^{a+b+1}}$

[untold delta]

eigentaylor

[steel umbra]

but i don't know what i'm supposed to know about p to determine a and b

i always thought "man it's so extra to have to always specify what every single function is a map between" but now i get it. this is so annoying

is p like a function of only x?

if i just say "we need the 1-t exponents to match", then that just gives me b=0 and a=1. which could be the answer but it seems... naive? what do i know though

[steel umbra]

this is what my classmate got though. ||but their PDE work has been a bit suspect in the past >.>||

[lilac barn]

For p, take the divergence of the first equation and use the divergence free condition of v

[steel umbra]

you mean of v D v - D p - v_t? what the heck is the div(vDv) 0_0

[lilac barn]

Actually, you may not have to solve for p. You just wanna balance out a and b. So use this equation and write one for div v in terms of V and try to balance out a and b using those

[steel umbra]

so you mean like div(v)=div(V(x/(1-t)^b))/(1-t)^a=0 and work that out?

i'm multivar calc stupid ngl. i never learned advanced multivar calc, which is probably why im so bad at PDEs

[lilac barn]

Yes, this combined with the one previous should give you two equations in a and b that u can work out provided u did the math correctly.

[steel umbra]

i don't see how calculating the divergence is going to help though. it seems like i'll just get something like div(V)/(1-t)^(some power)=0 which just tells me div(V) is also zero.

[lilac barn]

So from here, it tells you that your powers of (1-t) should balance out and so 2a+ b = a+1 implying a+b = 1.

[mint canyon]

@steel umbra Do you know the continuity equation? it reduces to div(V) = 0 if the density is constant.

[steel umbra]

not at all

[steel umbra]

okay so that was on the right track

[mint canyon]

$$ \partial_t \rho + \grad \cdot (\rho \vec{v}) = 0 $$

i hate latex

[steel umbra]

wait doesn't 2a+b=a+1 also have to be equal to a+b+1 or why does that last term not matter?

[untold delta]

x17

[lilac barn]

The last term doesn't matter because it comes with x as well. So ideally, you would multiply by (1-t)^a+1 the whole equation so that the first two (1-t) disappear and for the last one you have x/(1-t)^b which is precisely the same argument as contained in V. So then you would set y = x/(1-t)^b hoping that there isn't any x or t left in the equation

[steel umbra]

i'm not following why having an x makes it irrelevant. and are we assuming Dp=0 so that we can multiply everything by (1-t)^whatever?

[mint canyon]

the time change in density is and the flux of the density through some Gaussian surface is 0. So the total quantity is of whatever rho is, is conserved.

[lilac barn]

Because you will do the change of variable y = x/(1-t)^b and since your V only depend on y so your hope is that the resulting equation only depends on y and any other contributions (like of t) vanish out (due to cancellation of powers)

We aren't assuming Dp= 0. There is usually a similar self-similar form given for p but they haven't provided it to you. Ideally you would assume that p = P(x/(1-t)^b) 1/(1-t)^c and get a similar condition for c.

[steel umbra]

oh so basically because we have (x/(1-t)^b)*Dv/(1-t)^(a+1) like the other term

[lilac barn]

Idk what you mean

[steel umbra]

$Dp=\frac{V,DV}{(1-t)^{2a+b}}-
a\frac{V}{(1-t)^{a+1}}
-
\paren{\frac{x}{(1-t)^b}}\frac{b,D V}{(1-t)^{a+1}}$

[untold delta]

eigentaylor

[steel umbra]

so if 2a+b=a+1 we can cancel the denominators besides having a y

[lilac barn]

Yes so the part in the bracket becomes y and the remaining non-y terms shouldn't be present so it becomes an PDE in terms of V which you can solve much easier

[steel umbra]

i'm still just confused what to do with Dp if we aren't given that form for it

[lilac barn]

Don't worry about it as they aren't asking you to find what p is. Just give the condition for the powers.

[steel umbra]

well i get that a+b=1, but it seems like i'm supposed to narrow it down to specific values

but idk where to go from

$(1-t)^{a+1}Dp=V,DV-aV-ybDV$

[untold delta]

eigentaylor

[lilac barn]

You don't have to go anywhere as it would be for solving V and p. They're not asking you to solve V and p. That is highly non trivial

For now, you just need to suggest that a+b = 1 and that's it. This tells you that there is a full family of self-similar solution depending on some parameter. There are a bit more constraints but u won't be able to derive it without difficult analysis.

[steel umbra]

alright fair enough

[lilac barn]

For curiosity, check "On self similar solutions to the incompressible Euler equations" section 2

[steel umbra]

i do have another question. i was doing separation of variables on another problem and got

$\paren{\frac{X'}{X}}^2\paren{\frac{X''}{X}}
+2\paren{\frac{X'}{X}}^2\paren{\frac{Y'}{Y}}^2
+\paren{\frac{Y'}{Y}}^2\paren{\frac{Y''}{Y}}
=0$

but idk what to do with it since i can't actually separate them

[untold delta]

eigentaylor

[steel umbra]

i did suppose exponentials X=e^(ax) and Y=e^(by) and i got (a^2+b^2)^2=0 so that doesn't really help

unless i'm allowed to use u=e^(x+iy) but i dont think so lol

i just need to find a nontrivial solution

original was $u_x^2u_{xx}+2u_xu_yu_{xy}+u_y^2u_{yy}=0$ in $\bR^2$

[untold delta]

eigentaylor

[mint canyon]

@steel umbra Try the separation ansatz u = X(x) + Y(y), I think it'll make the cross term vanish

[steel umbra]

oh yeah that'll delete the u_xy. so then basically it's
(X')^2 X''=-(Y')^2 Y''=C

nice thanks

or i guess more simply, just a single variable function u=X(x) would work if we just need one solution (since it's symmetric in x and y)

[mint canyon]

ODEs should be solvable with reduction of order

[steel umbra]

or wait... that just leads to a trivial solution nvm. i think you need both

i think they're just separable in X' (or Y')

[mint canyon]

X" = C/(X')^2 -> V' = C/V^2 -> dV(V^2) = Cdx

V = 3(Cx + A)^(1/3)

[steel umbra]

yeah pick A to be 0 for simplicity (we just need one solution)

i got x^(4/3)-y^(4/3) so imma just quickly check that's a solution

[gray jasper]

any idea on how we got this estimate?

[untold delta]

sean

[gray jasper]

in this form of gronwall

i don't rlly understand how they got this estimate, or how the conditions for gronwall are satisfied

[lilac barn]

Note that each component of the Jacobian can be bounded by Linfty norm of nabla u, using Gronwall. Since this holds for each component, we can control the full Jacobian via the same bound, atleast up to a constant

[lilac barn]

They're using the differential form of Gronwall here, look up the statement on Wikipedia, or differentiate the assumption.

Btw the book is pretty good, stick with it even though it might get overwhelming with the huge emphasis on Fourier

[gray jasper]

ohh i see i get it now

[gray jasper]

ah alright thanks for the advice

i have a month to read the first chapter for a research project and its alright so far

[gray jasper]

wait just to make sure, the L infinity norm is this right

im not used to working w vector fields compared to scalar fields so

is this equivalent to the operator norm?

[hoary pumice]

No

Operator norm is the largest singular value of A^TA I think this is something else

[waxen bobcat]

Yes, if you choose the maximum norm in the input and output space

[candid token]

if $U$ is bounded and $\partial U$ is $C^1$, is $f=1$ necessarily in $L^1(\partial U)$?

[untold delta]

minitarrasque

[candid token]

(equivalently, does the boundary have finite arclength?)

[candid token]

are functions which are equal weakly necessarily also equal a.e.?

(on sufficiently nice domains, if necessary)

oh wait of course yes

since int (f-g) phi = 0 for all phi in C_c^infty

so f=g

[turbid scaffold]

Has anyone heard of a linear hamiltonian system?

[modest crescent]

If $U$ is bounded then isn't its boundary compact?

[untold delta]

Casey

[chrome trout]

I don't think so, the function needs to be of bounded variation

consider a circle with a lot of oscilations

like a lot

the perimeter goes to infinity

but it bounds a domain

[quaint herald]

casey didn't say anything about finite length, merely that the boundary is compact. this is true (it is closed and bounded in R^n).

and in the original question, C^1 regularity of the boundary was assumed, which rules out the behaviour you mention.

[candid token]

it would be compact in R^n but not necessarily as a R^{n-1} manifold i think

[lyric cairn]

Guys i have some questions with pde
u_t = u_xx + 1
can i ansatz that u = X(x)T(t) and find some solutions for u_t - u_xx = 0
and then find the fouriers series that is equal to 1 and them compare them?

[mint canyon]

thats an inhomogeneous equation

you need the greens function

[lyric cairn]

thanks

[mint canyon]

Though... you could also use method of characteristics if you want. The characteristic equation would be:

$$ A_{\mu \nu } \frac{\partial \phi}{\partial x^\mu} \frac{\partial \phi}{\partial x^\nu} = 0 $$ where $$ A_{xx} = 1 $$ and all other components are 0

[untold delta]

x17

[mint canyon]

so basically $$ \phi(x, t) = c $$

[untold delta]

x17

[lyric cairn]

green function is for x \in [0,1] right?
i am looking for the (0,1)

[mint canyon]

you need boundary conditions to evaluate the greens function if thats what you are asking

you need them to propagate values along the characteristics too

[lyric cairn]

so method of characteristic is my best friend here?

[waxen bobcat]

more information on the problem is your best friend

[mint canyon]

I just mean you need boundary conditions, and defining boundary conditions on an open boundary is... not something Ive done, except in the asymptotic case

[lyric cairn]

u(0,t)=u(1,t)=0 and u(x,0)= some fucntion

[mint canyon]

right, but 0 and 1 are not part of the solution

[lyric cairn]

i was thinking as a diffusion equation,
with 1 constant

[mint canyon]

I guess you would use right and left limits for the conditions in that case?

[lyric cairn]

was thinking you can use fourier to find a interval from 0 to 1

[mint canyon]

independent of any method, you still need to specify the boundary conditions. I guess like I said, you could use 1-way limits

[lyric cairn]

thanks for the help. i am just realizing now im bad at maths.

[mint canyon]

(0,1) is an open interval, 0 and 1 are excluded from it

so saying u(0,t) or u(1,t) does not make sense, they arent in the the domain

[lyric cairn]

that makes sense

lets assume that it is closed

[mint canyon]

maybe $$ lim_{h \rightarrow 0-} u(h,t) = 0 $$

[untold delta]

x17

[lyric cairn]

isnt the equation saying the partial u/ partial x (which is temperature gradient) is equal to the uxx + 1

[mint canyon]

okay, if you can assume its closed then your BC are fine

[lyric cairn]

can i use for example uxx + uyy = delta u for divergence u or something

im going way off topic here my bad

[mint canyon]

the greens function would come from:

$$ (\frac{\partial }{\partial t} - \frac{\partial^2}{\partial x^2} ) G = \delta(x, x', t, t') $$

[untold delta]

x17

[mint canyon]

I'll leave it as a research task to figure out how to solve that, there are plenty of online resources, and explaining it on discord would be messy

basically you split the solution in parts and make them agree where the delta spike is

[lyric cairn]

thanks i will look into it more

[acoustic minnow]

Is there a way to solve an equation of the form $$a(t,x) \partial_t u + \partial_x(a^2 u) = 0,?$$

[untold delta]

afqt

[acoustic minnow]

If a only depended on t or x, I guess it'd just be a case of method of characteristics

[waxen bobcat]

Does it not?

[brave bay]

Looks like characteristics to me. A particularly simple version too, because this is just a linear transport equation.

[mint canyon]

you need to write it in a different form like: $$ a\partial_t u + \partial_x(a^2)u + a^2 \partial_x u = 0$$

[untold delta]

x17

[acoustic minnow]

Yeah, I figured it out. We get something like $\frac{du}{u} = - 2 \partial_x a$ along the characteristic

[untold delta]

afqt

[acoustic minnow]

Thanks

[acoustic minnow]

Yeah I was kinda obsessing with keeping a^2 u together instead of taking the \partial_x a term as a source

[bitter yacht]

Basic question about Sobolev spaces:

I am reading about the fundamental solutions to the $p$-Laplace, which are constituted by $ |x|^{ \frac {p-n}{p-1} } $ where $n$ is the dimension of the ambient $\mathbb{R}^n$. I've read that it does not belong to $W^{1,p} (\Omega)$ when $\Omega$ contains the origin. In order for the fundamental solution to be defined at the origin we need $p > n$. By asking for $n \geq 2$, the fundamental solution gets a cusp at the origin.

How can one prove that it is not weakly differentiable at the origin? I tried my hand at a wee simpler example to try to grasp functions endowed with a cusp, like the one-dimensional $f(x) = \sqrt{|x|}$ for $x \in (-1,1)$, which certainly has a cusp at the origin. The absolute value $|x|$ is weakly differentiable at the origin surely, but it seems like $|x|^\lambda$ for $0 < \lambda < 1$ is not.

Anyway, weak differentiability of the fundamental solution would mean there is $f' \in L^p (-1,1)$ such that for each $\phi \in C^\infty_0 (-1,1)$ we have
[
\int_{(-1,1)} \sqrt{|x|} \phi ' (x) dx = - \int_{(-1,1)} f'(x) \phi (x) dx
]
I suppose one can assume that it is, and try to derive a contradiction by appropriately choosing a sequence of test functions and subsequently performing an appropriate limiting procedure. But I have not been successful with this

[untold delta]

hardisc.

[pulsar forge]

one possible way forward is to try to find its distributional derivative and show its not L^1_loc (typically by a dirac delta showing somewhere)

[brave bay]

Isn't $f'(x)=\frac{\mathrm{sgn}(x)}{2\sqrt{|x|}}\in L^1(-1,1)$ a perfectly good weak-derivative for your function $f(x)=\sqrt{|x|}$?

[untold delta]

4453

[brave bay]

It should be negative exponents when things start to go bad.

(at least when one demands weak derivatives in L1, and is working in 1d)

When things are genuinely bad, a simple way you can show the distributional derivative is not in L^p_loc for whatever p you like is just to calculate the actual derivative as a smooth function away from 0, and test the integrability of the p-th power of this. For radial things like in your question, polar coordinates will be very fast.

[bleak pond]

Does anyone know where I can find information on convolution of a given function with a Hankel function of first kind ?

[bitter yacht]

ah very nice, seems so! i guess it was simpler than i thought, thank you!

[bitter yacht]

yes seems so, but in that case the function is not even defined at the origin, so it does not even possess a strong derivative, whence the question is moot, no?

[brave bay]

What is the relevance of the existence of strong derivatives to your original question? A function does not need to be defined everywhere (we typically are working in L^p spaces if not distributionally, so pointwise evaluation does not even make sense) or have strong derivatives to talk about weak derivatives.

In fact even the sqrt function discussed above doesn't have a strong derivative at 0.

[bitter yacht]

sure, but sqrt(|x|) is defined at $x = 0$, whereas $ |x|^\lambda$ for negative $\lambda$ is not defined at $x = 0$.
that was what i meant

[untold delta]

hardisc.

[brave bay]

yeah but we are talking about taking weak derivatives of some L^p_loc function, also living in some L^p_loc space. such things are not sensitive to individual points. E.g. the function f(x)=1/sqrt(|x|) is a perfectly good definition of a function in L^1(-1,1). Set its value at 0 to be whatever you like, this does not change it as an L^1 representative.

[bitter yacht]

right my bad, the origin has lebesgue measure zero, and thus invisible in integration

[brave bay]

and it still makes sense to ask if it has a weak derivative also in L^1, and indeed it doesn't

roughly because the obvious candidate will blow up like 1/|x|^(3/2) about 0, and thus fail to be integrable.

[bitter yacht]

ah, right! i think it is much clearer now, thank you very much

[brave bay]

no problem

[pine oriole]

assume u: Rⁿ → Rᵐ
often in PDEs we do a change of coordinates (diffeomorphism) in the domain to simplify the equations. For instance instead of u(x) we look at u(Φ(x))

but how about in the codomain, like looking at Φ(u(x)) instead

this seems to be very rarely done for PDEs because it often messes up the equation form. But it is done all the time in ODEs. Nevertheless, in which PDEs have you seen people "change the codomain" ? The ϕ I have in mind is the diffeomorphism between the open ball and Rⁿ.

[lilac barn]

Well, Fourier transform is a major example of this. And it appears a lot in PDEs, especially for dispersive and Navier-Stokes

But yes, not a diffeomorphism

[pine oriole]

yeah it seems like a rare operation in PDEs even though ODEs use it a lot

[candid token]

in this comment on the proof Evans has of the Lax-Milgram theorem, why is this true? Wouldn't we need to show that H is complete under the induced norm?

[pine oriole]

because you can prove the norms are comparable up to a constant

[lilac barn]

Not really. We don't argue via an induced norm. We just define a mapping A : B[u,v] = (Au,v) where A is well-defined due to boundedness of B and thus Riesz.
Moreover, A has closed dense range and injective due to coercivity of B.
As such, A is bijective so you're done.

[pine oriole]

Where is symmetry used here

Also "closed dense range"? That's just... full?

I am guessing you used symmetry to turn injectivity into dense range.

And then the linear map is bounded below so the range is closed?

If that's the argument, that works too, though I doubt that's what Evans intended. Since he mentioned Riesz with symmetry specifically and your argument didn't need symmetry for Riesz.

[candid token]

this is the argument he gave for proving the lax-milgram theorem, but this corollary afterwards i think is not supposed to use that route, i think it is trying to argue "call H' = H but with the new inner product ((u,v)), then by RRT since (f,.) is a bounded linear operator, there exists u such that ((u,v))=(f,v) and we're done"

but this only works if H' is actually a Hilbert space

i.e. it needs to be complete

but I think Delerik's argument works for this

[pine oriole]

Yep equivalent norms give the same topology. This is a fact so basic that in diff geo people only bother to define norms on tensor fields up to equivalent norms.

[vast tapir]

hey

[candid token]

hi?

[vast tapir]

sorry to bother you but can you maybe help me with a question that i wrote it in "real-complex-analysis", if you have a bit time im going to freak out

[vast tapir]

oh i litearlly just did it now thx anyway

[quaint herald]

Without symmetry, why should B be an inner product? This is needed for Riesz.

[pine oriole]

are you asking about my method or cocat
mine just uses symmetry -> inner product + riesz
cocat is bilinear form -> linear operator A via Riesz

[lilac barn]

The coercivity automatically implied closed range and injectivity. For density, pick v such that (Au,v) =0 for all u in H. Then v is 0 again by coercivity, (using the characterisation that a subspace is dense if any functional vanishing over it vanishes everywhere)

[quaint herald]

I think they just forgot to state they were using symmetry at that point. It doesn't make sense to invoke Riesz without having this.

[lilac barn]

Yes the point is that if you had symmetry then it's a trivial consequence of Riesz but since you don't have symmetry, it's remarkable that we still get the same resukt out.

[pine oriole]

I think the person was asking about why the so-called "trivial consequence of Riesz" holds

(not about lax-milgram without symmetry)

[lilac barn]

Ohh, I see. I thought they needed help in understanding the italicized statement. Then yes, equivalence of norms provided symmetry.

[hoary oak]

This is from the Lieb & loss Analysis book. They deduced $\chi_D = R_{\theta}\chi_D$ a.e. for all $\theta$.
From this they are saying that $D=F^$ , I didn't understand how they are getting $D=F^$. Can anyone please help

[untold delta]

souravghosh0465
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hoary oak]

@lilac barn

[lilac barn]

#real-complex-analysis

[inland sinew]

I have sort of an open ended question here so bare with me.

L1(z(x,y)) = 0. Which I know exact solutions for(an infinite, orthogonal family), and Where L1 is a second order differential operator with variable coefficients.

Now say I am looking at the following pde.

L1(z) + L2(z) = 0, where L1 is the same linear operator as before, but L2 is a new operator defined in the following way.

Say I take my original equation L1(z) = 0 and coordinate transform it by the substitution x -> x+y and y->x-y. The resulting operator you get in this new equation is exactly L2.

So in essence, the equation (L1 + L2)z = 0 is sort of like L1 plus its rotated version by 45 degrees(with a small scaling factor).

An example of this time of equation would be zxx - zxy + zyy = 0, with L1z = zxx - zyy .

Is there any more general procedure to sort of tackle this problem (L1 + L2)z = 0 when I exactly know how to solve the easy case? Or even say something about possible solutions

[quick pagoda]

@pine oriole any good reference on what’s going on in “parabolic” situations?

[solid flint]

Hello,
Let $L$ be a sectorial operator, with a bounded inverse. then $D(A^\alpha)=[X, D(A)]\alpha$, where $X$ is a Hilbert space and $[X, D(A)]\alpha$ is the complex interpolation space between $X$ and $D(A)$
Anyone knows where I can find more details and the proof please?

[untold delta]

Mikahopff

[lilac barn]

You can find this in One Parameter Semigroups for Linear Evolution Equations Chapter II.5

[pine oriole]

that's just the heat equation
any PDE book will teach you first elliptic (Laplacian) then parabolic (heat equation)

heat semigroup is born for such

[quick pagoda]

I see

[solid flint]

Hello,
Are injective sectorial operators invertible?
I know they have dense range but I am not sure if this can be used to derive the boundedness of the inverse ??

[modest moth]

I have the PDE that becomes:

[\hat{u}(\xi,t)=\hat{f}(\xi)e^{-k\xi^4t},]
after taking Fourier transforms. I want to prove the inequality:

[ \lvert\lvert u(x,t)\rvert\rvert^2_{L^2(\mathbb{R})}\le \frac{\int_{\mathbb{R}}e^{-x^4}dx}{\sqrt[4]{2kt}}\lvert\lvert f(x)\rvert\rvert_{L^2(\mathbb{R)}}^2]

So I start with the LHS, I use plancherels theorem (ie. L2 is preserved under Fourier).

My next step is writing the norms in the integral definition. The problem is then splitting the integral into a product:
[\left(\int_\mathbb{R}\lvert \hat{f}(\xi)e^{-k\xi^4 t}\rvert^2d\xi\right)^2,]

then using inverse Fourier transforms to get to the result. Cauchy schwarz was my first idea but I get too high power inside the integral and too small of a power outside.

[untold delta]

Max