advanced-pdes

[astral vine]

"mdr" or "ptdr"

[dawn lava]

Dying of laughter I assume?

[astral vine]

Yesn but it gives you the function without a proof

[soft summit]

something de rire I'm sure

[astral vine]

precisely

exactly,

[dawn lava]

That's just the second example right?

[astral vine]

"mort de rire", "péter de rire"

[dawn lava]

The one that's something like unbounded in any open set

[astral vine]

Yes

I did the proof

Gosh

it's 5 pages long

[dawn lava]

I've been meaning to

[verbal nebula]

I think the nice thing about Evans is that there a clearly detailed references if you don't follow

[dawn lava]

I no longer mean to

[verbal nebula]

At the end of each chapter

My profs lecture notes were really good for the chapters 5-7 stuff, but there was no homework so none of it stuck

[astral vine]

I lied, 6 pages long

[dawn lava]

My one hang-up with Evans (so far) is that for the heat equation mean value theorem it has this bizarre integral that's Just. Inscrutable.

Like, in the middle of the proof

And it just tells you to sort of accept it as a fact of life

[soft summit]

the one that evaluates to 4?

[dawn lava]

Yeah the one that's a quotient of distances in the heat ball

Distance in time and distance in space divided, iirc

[astral vine]

The page please ? I have the book in my hands right now

[dawn lava]

This bitch

[soft summit]

54

[dawn lava]

Page 54 in my PC version

[soft summit]

in the new one too

[dawn lava]

I've actually got it right besides me too and it's 54 in the new one yeah

I timidly tried to tackle that integral for about fifteen minutes before noping right out

[soft summit]

I looked it up when I went through this
it was much longer than I had expected

[dawn lava]

It's not an especially important part of the proof, mind

[astral vine]

I do a bunch of parabolic equations, even with Laplacian, and I never use this result

[dawn lava]

If you can prove it converges then the gist of the mean value theorem still applies, you just don't know what constant to multiply the integral by

[astral vine]

I even forget it exists

[verbal nebula]

Have you actually worked it out

I kinda just convinced myself it should be true

and moved on w/ my life

[soft summit]

https://math.stackexchange.com/questions/526997/how-to-prove-that-iint-fracy2s2-dy-ds-4?rq=1

[verbal nebula]

that's kind of annoying

I remember doing the log log 1 + 1/|x| problem on Math Stack

I ended up answering my own question

[astral vine]

The only real good part of Evans is between pages 435-451

[river path]

[dawn lava]

Nope

[verbal nebula]

Do I have to get the book out now

[dawn lava]

Made a half-hearted attempt, quickly convinced myself it wasn't worth it

I don't think I could've figured it out

[dawn lava]

Ok I definitely couldn't have figured it out

[half patio]

bdvanced???

[white hazel]

a step beyond advanced

[tired ether]

eddvanced-pdes soon

[clever aurora]

bdvanced stands for?

[plush dust]

ODE -> PDE -> QDE

[arctic carbon]

like advanced pdes but nami is insecure about using the letter a too many times

[granite sun]

Where’d my bdvanced-pdes go

[arctic carbon]

cope

[potent herald]

Yeah sure, what are you confused about in these topics

well whats your interpretation of it

sure thats a specific case

so we are working a bit more generally, in a normed space. a norm can come from an inner product but a normed space doesnt need to have an inner product.

but the idea is correct, i am defining this convergence to be if |f_n(x)-f(x)| approaches 0

well let me think of an example

like you want the norm of f-f_n

which is like, the series with terms after n

so you would usually wanna show that part goes to 0 somehow

right so you can actually try computing this yourself

define f_k to be the partial sum upto k

try computing or bounding f-f_k

lets use k here since they are using n to index

yeah endgoal is to see what happens as n-> infty

but first using the L^2 norm ig whats |f-f_k|

yeah I am assuming thats the norm you want

yeah

fourier series is kind of a statement about L^2 of periodic functions

you can but you dont need to, just for f-f_k is good enough

yeah just plug it in and do the integral

uh thats a bit hard for me to read

but i think you made a mistake there

the basic idea should be that $f-f_k = \sum_{n>k} \dfrac{(-1)^n}{n} \sin(nx)$

[untold delta]

JohnTheCutiePie

[potent herald]

so $(f-f_k)^2 = \sum_{n,m>k} \dfrac{(-1)^{n+m}}{nm} \sin(nx)\sin(mx)$

[untold delta]

JohnTheCutiePie

[potent herald]

and so $\int_{-\pi}^\pi (f-f_k)^2 \dd{x} = \sum_{n,m>k} \dfrac{(-1)^{n+m}}{nm} \int_{-\pi}^\pi \sin(nx)\sin(mx) \dd{x}$

[untold delta]

JohnTheCutiePie

[potent herald]

$= 2\pi \sum_{n>k} \dfrac{1}{n^2}$

[untold delta]

JohnTheCutiePie

[potent herald]

(using orthogonality of the sines in the last line)

and this is known to go to zero

this ended up working out nicely but thats kinda the idea, usually you will have to do more bounding and stuff but you just wanna show it goes to 0

yeah

the x is in the arguement of sin right

but once you integrate it the x dissapears

im defining

$f_k(x) = -2\sum_{n=1}^k \dfrac{(-1)^n}{n}\sin(nx)$

[untold delta]

JohnTheCutiePie

[potent herald]

i probably lost a few constants in the computation but those dont matter

do you mean how i just wrote it as f-f_k? thats just to shorten notation

i do mean f(x)-f_k (x)

why?

oh

right so i just wrote f as the infinite series

hmm maybe this isnt fully logically sound

ok right so i guess we know f is L^2 integrable (when you write it as that sum) and thats really all you need to show that sum converges to x

do it for the infinite series

to confirm it converges

yeah

and fourier series tells you that since it converges, it must coverge to x

(as the coefficients match up etc)

anyhow i think we got a bit off track with this example

like uh main point was

ok this is a computation in a specific example, but this is what it means to converge in norm

(which in this case is the L^2 norm)

also another thing i should mention is that L^2 is a complete space

so you can also use the cauchy criterion to see if a series converges

in this case i saw the computation was easy enough so i didnt go through that

yeah that is to say its something that converges in L^2

also i feel like your class is skipping a lot of details oof

L^2 is probably the most used hilbert space (thats not finite dimensional) but its certainly not the only

also erm do you guys have access to the lebesgue integral?

yeah you cannot really define L^2 without it

you can define a space of square riemann-integrable functions

but its not as nice

for instance its not complete under the L^2 norm topology then

well i guess for the fourier case, you can just show the sequence f_n itself converges, and then since you obtained it as the fourier series of some function f that is what they converge to

idk how true this is but feels true

sure i have seen them used a lot.

but,

i dont know how important they are for your class

also like, this is probably getting too far afield for me, i was kinda planning on explaining theoritically things like hilbert space. I am not that comfortable with like PDE computations and applications so probably not the best person to ask

you could probably search up some uses in physics if you want motivation

i remember using these quite a bit when i did some physics classes

phi_k?

so in general fourier series is like, finding an orthonormal basis for space of square integrable functions

for periodic functions this is sin and cosine

but you can ask about different things too right

same things, sin and cosine can be written interms of those

if you think about it periodic functions are like, functions on the circle right

you might ask about functions on the sphere for example

then L^2 decomposes into these things called spherical harmonics

if you have heard of those

[mystic ginkgo]

the phi_k are the basis functions, as john said

sure

[potent herald]

well you change the e^(ikm) with the appropriate basis

but yeah

yeah cause otherwise |f|^2 doesnt make sense right

L^2 is kind of a specail space in that it not only has a norm

but an inner product

this is why you can do these nice things like orthonormal bases etc

inner product is as you have seen

<f,g> = int_(-pi)^pi f(x) bar(g(x)) dx

give or take a constant

and the norm comes from this, i.e

|f|^2 = <f,f>

.

[mystic ginkgo]

waving my hands wildly, maybe you find it more recognizable when comparing it to the dot product between vectors, which is also an inner products

and it behaves much in the same way

like $u \cdot v = \sum_{n=1}^N v_n^* u_n$

[potent herald]

bar is just when you have complex stuff

[untold delta]

Edd

[mystic ginkgo]

use an asterisk for complex conjugate if you prefer 😛

anyway, if you wave your hands wildly and multiply inside the sum by a \Delta_n and take the limit as \Delta_n goes to 0, this is kinda like an integral

you can probably cook up an argument based on the basis being orthonormal and cauchy schwarz

the d_k are the weights of a linear combination

say you put your phi_k in a vector, call it v, and we have some weights for a linear combination, call them w

then v dot w is a linear combination of the elements in v

it's telling you there is a set of weights w that are optimal in the least squares sense

and these weights are precisely the c_k they defined above

sure, that's what this is telling you

the basis is orthonormal, shouldn't matter

the theorem as stated is independent of what n is

it's telling you it's true for any n

i would suggest to review linear algebra and relate it to that

[tired hollow]

For this, I just set $V=\Lambda(S)$ and solve for $u$ then set $\tau=0$ right?

[untold delta]

Sam W

[storm glade]

Does anyone know any reference to inequalities for fractional Laplacians, maybe sth that would look a bit like $|(1 - \Delta)^{\frac{s}{2}} (\psi^3) |{L^2} \lesssim | \psi^2|{L^2} |(1 - \Delta)^{\frac{s}{2}} \psi|_{L^2}$

[untold delta]

Golol

[twilit rover]

@broken sapphire Do you know what an orthogonal projection is? It is elementary to show that the orthogonal projection (if it exists) gives the best approximation.

[buoyant pike]

Why is right angle italicized

[twilit rover]

@broken sapphire I was answering your previous question

This one

[exotic spruce]

Can someone please explain this

How is this integration by parts

He just took the derivative of both functions independently wtfffff

[twilit rover]

He differentiated under the integral in the first line of the "integration by parts" part

[exotic spruce]

So he just took the derivative of the first function with respect to epsilon (I don’t know what that Greek letter is) and then took the derivative with respect to x of f(x)?

[twilit rover]

In the very first line he did $\partial_\xi \int = \int \partial_\xi$

[untold delta]

IlIIllIIIlllIIIIllll

[exotic spruce]

Yeah I get that

But what about the second line

[twilit rover]

the second line is just algebra

[exotic spruce]

Like this is what I’m seeing what he didd

[twilit rover]

no

the first equality is $\partial\xi \int = \int \partial\xi$. The second equality is just algebra, nothing fancy.

[untold delta]

IlIIllIIIlllIIIIllll

[twilit rover]

The third is from integration by parts

[exotic spruce]

I’m just confused because If he is taking the partial with respect to epsilon then why is that 2 alpha coming out

[twilit rover]

expand out the $\frac{d}{dx}e^{-\alpha x^2}$

[untold delta]

IlIIllIIIlllIIIIllll

[exotic spruce]

Yeah

So what I wrote is exactly what he did

He took the derivative of the first function with respect to epsilon and he took the derivative of the second function with resoext to x

Respect

Ugh

WAIT I GET ITTTTT

Omg

Thank you

[twilit rover]

What he's really deriving here are more general identities that hold for "rapidly decaying" functions.

[exotic spruce]

He didn’t take the derivative of that he just wrote it like that so it would cancel out omg

Yeah

So in this case that e ^ - alpha x ^2 is like a function that’s super useful here

Like it just works good

[twilit rover]

namely $F(\partial_{x}f) = i \xi F(f)$ and $F(xf) = i\partial_{\xi}F(f)$.

[untold delta]

IlIIllIIIlllIIIIllll

[twilit rover]

Valid for $f \in \mathcal{S}(\mathbb{R})$

[untold delta]

IlIIllIIIlllIIIIllll

[exotic spruce]

Ok thanks so much!

Wait where did that i epsilon come in for the third line

[twilit rover]

Its $\xi$ not $\varepsilon$

[untold delta]

IlIIllIIIlllIIIIllll

[twilit rover]

It's integration by parts.

What the author is doing is taking the ode $f'(x) = -2\alpha x f(x)$ and taking the Fourier transform of both sides, using the identities established for rapidly decaying functions.

[untold delta]

IlIIllIIIlllIIIIllll

[twilit rover]

To get $F(f') = -2\alpha F(xf)$.

[untold delta]

IlIIllIIIlllIIIIllll

[exotic spruce]

Did I find the fixed points right here

[turbid scaffold]

It looks like the $a_n$ are the Fourier coefficients on the piecewise function on the right side.

What is h again?

[untold delta]

fajitas

[buoyant pike]

#odes-and-pdes

[hidden fractal]

Apologies I'll move it there!

[buoyant pike]

[buoyant pike]

Zworski has a book right

[river path]

zworski's book is very popular

[quaint herald]

definitely don't use monsterbook lol

yes the books you mention are all good, zworski is very good, I learned a lot from Martinez as well

[buoyant pike]

#odes-and-pdes

[solemn ruin]

ok

[clever aurora]

is Maximum principle for wave equation a real thing?

[clever aurora]

\begin{gather*}
\laplacian u = 0 \
u(0, y) = 0
\end{gather*}
on the domain $\Omega = {(x, y) | x < 0}$ then $u\equiv 0$. This is false I can see as $xy, xy+x$ also satisfies the PDE but why is it?

[untold delta]

Ryu

[buoyant pike]

#odes-and-pdes

[harsh veldt]

How do you prescribe an initial condition to the 2D wave equation given that you want a specific displacement on a specific point at a specific time? This sound like an inverse problem which I not know how to do.

by 2D i mean rectangular membrane

[clever aurora]

run the wave eq in reverse

[sudden pumice]

hi guys, anyone know an example of a function that is in L^2(R^n) but not in L^1(R^n)?

[clever aurora]

1/x, modify it a little

[sudden pumice]

thanks!

[solid flint]

Seems to be true after some computations

Can anyone help me figure out what's the best regularity I can get for a function $u \in H^1_0 (\Omega)$ such that $\Delta u \in L^2 (\Omega)$ where $\Omega$ is a bounded subset of $\mathbb{R}^n$ with smooth boundary?

[untold delta]

Mikahopff

[solid flint]

Thank you in advance.

[astral vine]

H3/2

Nothing better

for Lipschitz boundary

which is still "smooth enough" for some people

If by smooth you mean C infty

then H² is exactly what you get

[astral vine]

Finally if by smooth you mean something else, I can not help you anymore

[harsh veldt]

Hi. I'm trying to prescribe an initial condition to the 2D wave equation such that u(1/2,1,10)=1/3. I got to this point but I don't know how to proceed in recovering f(x,y). Any idea?

[harsh veldt]

<@&286206848099549185>

[buoyant pike]

#odes-and-pdes

[harsh veldt]

no one is actually answering me. 🥲

[buoyant pike]

No one is obligated to answer

[solid flint]

Thank you so much, can you mention a reference where I can find the details please?

[astral vine]

Gilbarg Trudinger, or Grisvard's Elliptic Equations in non-smooth domains

For sharp H3/2 regularity in the case of Lipschitz domains, this is a very technical result

#odes-and-pdes

[peak niche]

I did but no answer

Tho

[buoyant pike]

No one is obligated to give you an answer

[astral vine]

Exactly what I was currently writing

to get back on H3/2 regularity for the Dirichlet Laplacian in bounded Lipschitz domains, this is a very hard result, which can be only found in hard papers

mainly published in 90's

the generic Lp case was done by Jerison and Kenig

Not so hard

but needs a lot of knowledge to deal with it

[solid flint]

you mean that's less hard than the L^2 case?

[astral vine]

The involved tehcnics mades it more clear than the first sharp L² result iirc

I have Jersion and Kenig's paper just in my hands right now

[solid flint]

that's surprising at the first glimpse because of the tools that we have in these L^2 related Hilbert spaces

[solid flint]

That's very nice, do you mind sharing it or at least its title

[astral vine]

"The inhomogeneous Dirichlet problem in Lipschitz domains"

[solid flint]

Thank you so much

[astral vine]

but be careful it needs A LOT of knowledge

and being well aware about the content in the references therein

[solid flint]

I may be reading from the books, the paper is feed my curiosity somehow

[river path]

Consider the matrix A = (0 1; 0 0), a 2x2 jordan block. If we had an inner product which made A self adjoint, then this would imply that A has 2 eigenvectors which are orthogonal for this inner product (by the spectral theorem). But it's not true, A has only one linearly independent eigenvector.

I think this is ok

You can probably look at, say, differentiation on some space and get the same thing.

[astral vine]

every self adjoint operator have a spectrum contained in the real line

[solid flint]

Hello, is there any function in $L^2(\mathbb{R})$ that's not vanishing at infinity?

[untold delta]

Mikahopff

[river path]

Here's an option, the indicator of the set $\bigcup_{n=1}^\infty (n - 1/n^2, n)$

[untold delta]

Kanga Gang Drug Mule Ryc

[solid flint]

isn't this zero a.e?

Actually even if we ask that the function is smooth we can find some examples

[river path]

Yep

You can turn what I said into smooth bumps that get thinner and thinner

If you want globally lipschitz, or even uniformly continuous, I think you're going to run into trouble though.

[harsh veldt]

Is the initial conditions imposed on the wave equation over a vibrating circular membrane are both (i.e. initial displacement and velocity) zero, does this mean that the solution is trivial?

[astral vine]

use the energy of the equation

[harsh veldt]

oh to solve for the solution?

how do I get the solution out of the energy?

[astral vine]

Assume you have a solution

ask it to be $H²_{t,loc}L²_{x,loc}$ and $L²_{t,loc}H²_{x,loc}$

[harsh veldt]

okay okay

[untold delta]

Anatole

[astral vine]

the energy of the wave equation is nothing but $$E(t):=\int_{\Omega} |\partial_t u(t,x)|^2+ |\partial_x u(t,x)|^2 \dd x$$

[untold delta]

Anatole

[exotic spruce]

Does anyone know how to solve this stochastic differential equation??

[kind prawn]

So I was reading on the variational form of Poisson's equation from a functional analysis book. Why does the second integral in the functional become a surface integral? It remains a volume integral when I try to do the substitution

This is the original functional

[river path]

It should probably still be over Omega

Unless f is like

A distribution supported on the boundary or something

Yeah to get the right pde at the end it def needs to be all over Omega

[astral vine]

The boubdary have 0 measure with respect to the Lebesgue measure

So I think too, that the integration domain is Omega

[strong maple]

besides, aren't you missing a 1/2 term next to the grad u integral?

[exotic spruce]

I’ll ask again anyone know

[twilit rover]

@exotic spruce What is that supposed to mean?

[exotic spruce]

It’s a stochastic differential equation

The boxed thing

[twilit rover]

looks fancy

[exotic spruce]

Lmao

I don’t think it’s right. Stochastic differentials with trig are almost never solvable

[soft summit]

oh nvm I see why u is constant 🤦‍♂️

[soft summit]

how do you get the last line?

the PDE is

[strong maple]

isnt it just differentiate u-g(...) wrt u?

[astral vine]

Wrt to x

[solid flint]

Hello, are there any video lectures about the Control theory of PDEs please?

[strong maple]

how do you get that 1 then?

[astral vine]

chain rule

just plug the ansatz in your equation

[buoyant pike]

Hello friends

Why is $\int_0^1\frac{1}{r(1+r)^n\log(1+\frac1r)^n}dr$ finite for $n>1$?

[untold delta]

陆景和

[buoyant pike]

The 1+r is not so important because it's bounded above and below I suppose

But the other terms remain

[river path]

#calculus

[buoyant pike]

[verbal nebula]

It's an integral that comes from a sobolev space question

[buoyant pike]

Ryc knows

[astral vine]

change of variable u= 1/r

so it suffices to look at the integrability of u^n/(u(1+u)^n log(1+u)^n) near infty

Since n>=2 this is obvisouly true being aware of bertrand criteria for integrals

[buoyant pike]

I see

That makes sense

Not really but it makes enough sense

[river path]

I remember that

[buoyant pike]

Lexicographic order...

[river path]

isn't this like how the sum of 1/(n (log n)^p) is finite for p > 1

and you prove it by doing the thing where you sub in 2^n and then multiply by 2^n to get 1/n^p

here if you have 1/(u log(u)^n) you do e^x = u and then get 1/x^n and that's it

[solid flint]

Hello everyone, does anyone knows a light introduction to microlocal analysis just to know what is it about and how it is used please?

[astral vine]

Quantum Mechanics books for Mathematicians are quite good

I never finished the second one

but its firts chapters are good and readable

But #book-recommendations was a better place to ask

[solid flint]

@astral vine Thank you so much, I didn't post there because I was looking for some simple explanations maybe. But books recommendation is really good.
Thank you so much.

[astral vine]

Microlocal Analysis is deeply related to Pseudo-Differential operator theory and Quantum Mechanics

The main ideas comes from studying operators given by pseudo differential operators with a paramater h near 0 (the Planck constant in Physics), the simplest one is hLaplacian + potential, but there is way more than this. The better to answer your question is probably @quaint herald .

[solid flint]

@astral vine Thank you so much

[unborn cedar]

@dense hamlet okay so like
Let X be a smooth variety over C and let f be a non-invertible regular function on X.

Theorem: there exists a polynomial b(s) in C[s] and a polynomial P(s) in D_X[s] whose coefficients are differential operators on X, such that P(s)f^{s+1}=b(s)f^s in the D_X-module O_X[s,1/f]f^s.

The polynomials b(s) satisfying the above identity form an ideal in C[s], the monic generator of this ideal is the Bernstein-Sato polynomial of f.

Example: X=A^n, f(z_1,...,z_n)=z^2_1+...+z^2_n. Then we have
∑_i ∂^2_i f^{s+1}=4(s+1)(s+n/2)f^s
so b(s)=(s+1)(s+n/2)

Example: X=A^2, f(x,y)=x^2+y^3. Then we have b(s)=(s+1)(s+5/6)(s+7/6)

etc etc

they are kinda hard to compute in general

I'll explain in a second how you can prove their existence using holonomic D-modules, but first, what can you do with them

Example: if f(z) is a non-negative polynomial in C[z_1,...,z_n] then f^s only makes sense for Re(s)>0. We can analytically continue f^s to a distribution valued function in all s using the functional equation
f^s=(P(s)/b(s))f^{s+1}
and this has poles exactly at the zeros of b(s+n) for integers n≥0

[dense hamlet]

Okay yeah that's what I was wondering about, what s here was lol

[unborn cedar]

Example: Malgrange–Ehrenpreis says that every differential operator with constant coefficients has a Green's function, that is P(∂_1,...,∂_n)u(z)=δ(z) has a distribution valued solution. Here is a two-line effective proof using Bernstein-Sato polynomials

by taking Fourier transforms of both sides this amounts to showing that every polynomial f(z) has a distribution valued inverse g(z); without loss of generality we may assume that f(z) is non-negative, in which case the distribution valued inverse g(z) is the constant term of the Laurent expansion of f(z)^s at s=-1 which can be computed using the Bernstein-Sato functional equation

[dense hamlet]

That's pretty cool huh

[unborn cedar]

yea this absolutely rocks

So here's the proof: the idea is to cook up a holonomic D-module that governs this situation, the holonomic property means the D-module has finite length so chains of submodules stabilize, and this will give you the b(s) we're after

let j:U->X be the canonical inclusion with U=X-Z(f). Let X_s and U_s be the base change of X and U from C to C(s), with corresponding inclusion j_s:U_s->X_s. We have respectively D_X_s=D_X⊗C(s) and D_U_s=D_U⊗C(s)

consider the D_U_s-module M=O_U_s f^s where the action of derivations is given D(gf^s)=D(g)f^s+sg(D(f)/f)f^s (the division of f is fine, as it's invertible on U_s)

Then M is a holonomic D_U_s-module (to see this consider filtration F_fM=O_U_s f^s for k≥0 and F_kM=0 for k<0, then Ch(M)=Ch(O_U)_s=(T*U)_s so this filtration is good hence M is holonomic)

now consider the D_X_s-module N=D_X_s f^s (which is the +-pushforward of M along j_s)

Then N is a holonomic D_X_s-module (to see this consider the maximal holonomic D_X_s-submodule N' of N which is compatible with restriction to open subsets, then N|U_s is a D_U_s-submodule of M and hence N'|U_s=N|U_s. So we have an exact sequence 0->N'->N->Q->0 where Q is supported on Z(f)_s; for the section f^s of N, there exists k≥0 such that f^kf^s is in N', so D_X_s f^kf^s is a D_X_s-submodule of N' hence holonomic, but we have an isomorphism of D_X_s-modules N=D_X_s f^kf^s sending P(s)f^s to P(s+k)f^kf^s hence N is holonomic)

Now consider the chain of D_X_s-submodules ...≤D_X_s f^2f^s≤D_X_s ff^s≤D_X_s f^s. Since D_X_s f^s is holonomic it is of finite length, hence this chain stabilizes. It follows that there exists an integer m≥0 such that f^mf^s is in D_X_s f^{m+1}f^s and hence f^s is in D_X_s ff^s, so there exists Q(s) in D_X_s such that f^s=Q(s)ff^s.

Now clear denominators and write Q(s)=P(s)/b(s) with P(s) in D_X[s] and b(s) in C(s). Then b(s)f^s=P(s)ff^s and we are done.

one can play around with D-modules more and show Kashiwara's theorem that the roots of Bernstein-Sato polynomials are all negative rational numbers

One last example application: the analytic continuation of all Archimedean zeta functions

Let f:R^n->R be continuous with f(x)>0 for all x in R^n and let φ:R^n->C be continuous and compactly supported. Consider the Archimedean zeta function

Z_φ(s)=∫_R^n f(x)^sφ(x)dx

basic complex analysis shows that Z_φ(s) is analytic on the right half plane. The claim is that this admits an analytic continuation to a meromorphic function on C whose poles are of the form a-n where a is a root of the Bernstein-Sato polynomial of f and m≥0; in particular the poles are all negative rational numbers

take H_m={Re(s)>-m}. We're already analytic on H_0, the Bernstein-Sato functional equation shows that you can extend this to a meromorphic function on all H_m by induction on m≥0, and the functional equation + Kashiwara's theorem gives you the required pole structure

[river path]

😵‍💫

[buoyant pike]

[astral vine]

nice maths

[soft summit]

just to clarify
in method of characterstics is it possible that for some x in the interior of U no characteristics lines from the boundary reach the point?

[buoyant pike]

Yes

You can see with this burgers in 1d with correctly chosen initial data

You get something called a rarefaction wave

[soft summit]

I see ty

[soft summit]

what does he mean exactly with this regularizing effect

[astral vine]

Because the heat equation have fully smooth solutions on (0,+oo)x R^n

[soft summit]

ah ok I was overthinking

ty

[astral vine]

Kruskov trick yeay

[late moth]

if f’(x) is piecewise smooth and the fourier series is continuous, then you have term by term differentiability iirc

for the fourier series to be continuous, you need f to be continuous on [-L,L] and \hat{f}(-L)=\hat{f}(L) where L is the period

[late moth]

yes, and iirc can be approximated(?) by C^1 functions on (-L,L) [check the definition, i don’t remember]

lipschitz > uniform

there are uniformly continuous functions that are not lipschitz. Take \sqrt{|x|}

ah, it’s not guaranteed to have existence, even take a holder continuous x^(1/2)

wait this is existence not uniqueness

give me a sec

iirc the proof of picard’s theorem relies on the contraction mapping theorem

don’t really get it from the problem or you don’t get how it gets you local/global existence

ohh i see

yes, it’s lipschitz continuous on any finite interval, so on a finite interval (which you can make however big), you have existence

yes, if what you’re saying is that for all $x_0\in \R^n$
[x(t)=x_0+\int_0^t f(s,x(s))ds]
is unique for all $t\in \R$

[untold delta]

kirby

[late moth]

i messed up the quantifiers

but that’s beside the point i’m on my phone

yes iirc Any uniformly continuous function on R is lipschitz continuous on a finite interval [you should probably prove this to verify]

rectangle is an interval in R^2

the ode is nonautonomous (i.e. it is of form f(t,x) instead of f(x)) all that matters is its spatial component is lipschitz wrt x for picards

okay so on the interval [a,b], you have lipschitz, so existence and uniqueness is achieved by picard’s theorem

of course show that to be the case

local lipschitz means you have global uniqueness, but not necessarily global solutions (existence)
i’m not entirely sure what you’re meaning by global, like just with a domain T = [0,infty)? you will have that.
now, local existence will allow you to kind of extend the reach you have, but there are some conditions for this that i forget

all of this reminding me to go over this part of my odes course again in my study for my final next week bc i am not doing a good job explaining this oops

yes
to clear my unfinished ideas:
if |f(t,x)|<g(x) for some g such that int_x0^infty dx/g(x)=infty, you have global existence

here are my lecture slides regarding locally lipschitz functions

i don’t think so?

the requirement is for your f to be lipschitz for picard’s

yea, anyway i should probably sleep before i become even more wrong/nonsensical lmao

lipschitz, uniform continuity gives you locally lipschitz if the derivative is bounded in an interval (i forgot that part)

[buoyant pike]

#odes-and-pdes

[harsh veldt]

small question: is Lp space dense in the Sobolev space?

[twilit rover]

$L^p(\mathbb{R}^n)$ is not contained in $H^{1, p}(\mathbb{R}^n)$

[untold delta]

IlIIllIIIlllIIIIllll

[harsh veldt]

Ohhh haha thanks! lmao. I was having a problem writing the proof that C^infty with compact support is dense in H^1,p with compact support so I was thinking of using Lp spaces lmao

[buoyant pike]

Convolution

[harsh veldt]

oh so do I use mollifers then?

[solid flint]

Anyone can enlighten me about the ill-posedness of the wave equation in L^p spaces for P different than 2, please?

[soft summit]

is this R the propagation speed of hamilton-jacobi

or maybe has something to do with it
because this cone definitely looks like the one used with wave equation in energy methods

[solid flint]

The Lebesgue spaces with exponent different than 2, or any other spaces that you know about

I would love to know about whatever results you know about

but based on what I have heard, this is established on bounded domains at least

[carmine blaze]

Find the solution to the initial value problem for the Schrodinger equation $u_t = iu_{xx}$, $x \in \mathbb{R}$, $u(x,0) = e^{-x^2}.$

[untold delta]

Dpao

[carmine blaze]

this showed up in my complex analysis pset and I have no idea how to do it

does it require any outside knoweldge? I found this https://www.physics.drexel.edu/~vogeley/Phys326/fourier.pdf

[solid flint]

Fourier Transform seems to work here,

even semigroups theory if you are familiar with it

[carmine blaze]

can you get that just from $u_t = iu_{xx}$ and $u(x,0) = e^{-x^2}$ or do you need other stuff

I tried something like this which maybe gives you coefficents for the wave function but I dont know anything about mechanics and its entirely outside the scope of the class
\begin{align*}
f(\xi)&=\frac{1}{2\pi}\int_\mathbb{R} u(x,0)e^{-i\xi x}dx\
&=\frac{1}{2\pi}\int_\mathbb{R} e^{-x^2}e^{-i\xi x}dx\
&=\frac{1}{2\pi}\lim_{h\to\infty}\int_{-h}^{h}e^{-x (x +i \xi)}dx\
&=\frac{1}{2\pi}\lim_{h\to\infty}\Bigr[\frac{1}{2} \sqrt{\pi } e^{-\frac{\xi ^2}{4}} \left(\text{erf} \left(x+\frac{i \xi }{2}\right)\right)\Bigr]_{-h}^{h} \text{ error func. }\pm 1\text{ resp. at }\pm\infty\
&=\frac{1}{4\sqrt{\pi}}[ e^{-\frac{\xi ^2}{4}}(1-(-1))]\
&=\frac{e^{-\frac{\xi ^2}{4}}}{2\sqrt{\pi}}
\end{align}

[untold delta]

Dpao

[brisk granite]

Can anyone pls tell me if this is possible? If not why? If yes how?

Please feel free to ping me if any of you has an answer

The question is: is it possible to integrate like that?

[carmine blaze]

$\int u du=?$

[untold delta]

Dpao

[brisk granite]

That's a del x

$\int x δx=?$

[untold delta]

ブラボー

[brisk granite]

That's my question

Is it possible to integrate?

If not why? If yes how?

[buoyant pike]

Meaningless

Also inappropriate channel

[brisk granite]

Can you please tell me how is it meaningless? If you don't mind of course.

[astral vine]

the notation means absolutely nothing

this just a bunch of symbol

[brisk granite]

That notation stands for partial derivative if I'm not mistaken

[astral vine]

No.

[brisk granite]

Then pls tell me what's the symbol for partial derivative?

[astral vine]

$\partial$

[untold delta]

Anatole

[brisk granite]

Ok thanks

So my question meant to be

$\int x \partial x$

[untold delta]

ブラボー

[brisk granite]

Is that possible?

[astral vine]

Same comment

the notation means absolutely nothing
this just a bunch of symbol

[brisk granite]

Thanks

[astral vine]

Did I say something weird ?

[buoyant pike]

We don't integrate with a dx because d is a derivative???

It's purely notational

[brisk granite]

Wait what?

I'm not going into the notations I'm finding the existence of a mathematical equation that is absurd but it should exist. But it doesn't.

If we can partially differentiate, we should be able to partially integrate as well that's what I'm proposing. It's the why and how that's troubling me. Maybe my basics are weak. I'm definitely wrong but i could be right. Idk. But it's just a theoretical question.

[buoyant pike]

#multivariable-calculus

[brisk granite]

Ok I'm new here thanks

[river path]

"partial integration" is just denoted with a d whatever variable

denoting partial derivatives differently is a stupid historical holdover

and was always a mistake

[brisk granite]

But why if you could give me an explanation

[untold delta]

Kanga Gang Drug Mule Ryc

[brisk granite]

No they don't

[river path]

you just only use the first one when there are other variables

for no reason

well, they both mean nothing at all

[astral vine]

there is no real explanation

[river path]

both of them are just a bunch of symbols that appear as part of two notations

[astral vine]

because integral over many variable works roughly as if you fixed all the other variable, and perform a one variable integration

that's what says the Fubini Theorem

[brisk granite]

No they don't. It may look the same but they have different purposes.

[river path]

[astral vine]

Bro

[river path]

alright

in differential geometry, the first one describes a vector field while the second describes a differential form

[astral vine]

Did you ever learned, measure theory, or a real definition of the Riemann integral ?

[astral vine]

Bro don't

please

save his soul

[river path]

in calculus/analysis, they mean the exact same thing

except for the fact that the ambient space of one is one dimensional, and of the other isnt

[brisk granite]

I'm absolutely fine.

[kindred aurora]

When you're integrating something you're using a measure, depending on how you're working, and how the mathematical objects you're using, you have to use this measure if you're using measure theory.
Writing $d\lambda(x)$ is just a notation saying you're integrating with respect to a certain mesure.

[untold delta]

Anarchy

[kindred aurora]

That's a measure theory explanation

[river path]

anyway this is just notation and is completely nonmathematical, it's like arguing with someone about semantics

[brisk granite]

It is not semantics I'm talking about like what....

Have you done method of separation of variables?

[astral vine]

We did plenty of things

What's your point

[brisk granite]

My point is it's not semantics, partial derivative and ordinary derivative mean differently although they are the "same"

[astral vine]

but the d in integration IS NOT FUCKING RELATED to derivatives

[kindred aurora]

What you're not understanding is that integrating partially doesn't have any meaning

when you integrate, you integrate a function of variables, the $d$ only relates to which variable or thing you're integrating for

[untold delta]

Anarchy

[brisk granite]

Ok.

[dim topaz]

wdym? if you want we can continue this in #calculus or somewhere more relevant

i feel like this discussion is outside the scope of a pdes channel

[brisk granite]

Which channel do I go to?

[tired ether]

#multivariable-calculus

[dim meadow]

#multivariable-calculus

[brisk granite]

Ok

[astral vine]

Sorry I losed my patience

[harsh veldt]

How can I use mollifiers to prove this one? I'm not very familiar on how to do this one 😦

[white hazel]

doesn't look like you need mollifiers

i would just take derivatives and just estimate the L^p norms

[buoyant pike]

No mollifies, direct computation of W^1,p norm

[harsh veldt]

Oh like bound the function with some other familiar functions?

[buoyant pike]

Just compute the Lp norm of u and u’

[harsh veldt]

Okay thank you so much! Will try this one. So the L^inf case is just noting that it's not bounded?

[buoyant pike]

And show that it’s finite

Yes

[harsh veldt]

Okay thank you!

regarding this, I'm not sure how I can estimate the integral of |log(log(2/|x|))|^p

it has a singularity at 0 right?

[white hazel]

try a taylor polynomial or something

[buoyant pike]

Poincare inequality

[white hazel]

cause this is log(a + log(x)) - log(a) with a = log(2)

[harsh veldt]

Ohhh okay so I only need to compute for the p norm of the derivative?

[harsh veldt]

Ohhh yeah this also makes sense thanks I'll also try this

[buoyant pike]

Yes

[harsh veldt]

thanks!

[harsh veldt]

does my weak formulation look correct?

[harsh veldt]

Hi I tried doing this and got this

Not sure how to say it is finite

[buoyant pike]

https://math.stackexchange.com/questions/3827326/application-of-the-bertrand-integral

[harsh veldt]

Ohhh thanks! Will read this

[white hazel]

Seems like overkill

I would just switch to polar coordinates

[buoyant pike]

I had a similar homework problem and it boiled down to this so I would not say that it's overkill

Even after you switch to radial coordinates you still have some r*log(r)^n terms in the denominator

[white hazel]

A u sub should work since this is an easy case right?

Since p=n

[harsh veldt]

yeah I tried this too and it was difficult lmao

[harsh veldt]

like u=log(2/|x|)?

[white hazel]

U = log(r)

Something like that

Then it should end up being roughly the integral of 1/u^n on (1,infty)

And n>=2 so it converges nicely

[harsh veldt]

oh I will try this! That might be interesting. Thanks!

[astral vine]

the gradient is false, it is not even a vector

[white hazel]

It’s almost correct at least

Should be x/|x| instead of |x|/x (second thing doesn’t make sense)

[harsh veldt]

Thanks for pointing out. Let me fix it

[astral vine]

I have the result if you wanna check it

[harsh veldt]

Yes please!

[harsh veldt]

@astral vine I submitted my work but I'm really curious what the "correct" way of doing this problem is so whenever you have the time 👌🏻

[astral vine]

$$\nabla u(x) = \frac{x}{\frac{|x|^2}{2} \log\left(\frac{|x|}{2}\right)}$$

[untold delta]

Anatole

[astral vine]

so $$|\nabla u(x)| = \left|\frac{1}{\frac{|x|}{2} \log\left(\frac{|x|}{2}\right)}\right| = \frac{1}{\frac{|x|}{2} \left|\log\left(\frac{|x|}{2}\right)\right|}$$

[untold delta]

Anatole

[astral vine]

then using polar coordinates you will obtain a one variable integral about the radius times a constant which nothing but the area of the unit sphere of R^n

Make a change of variable 1/r

and use the Bertrand Criteria for integrals

[buoyant pike]

Yes I said bertrand integrals yesterday

[astral vine]

Oh sorry I didn't see it

my bad Ange

[buoyant pike]

No worries

More like

Vindication

Because young_smasher said it wasn't necessary

[astral vine]

I do not know how to justify finiteness without it to be honest

[harsh veldt]

Thank you!!!

[junior bloom]

I remember running into a characterization of when and why the Green's function for an elliptic operator is singular. Anyone know what I'm talking about? It's on the tip of my tongue so to speak but I can't remember what I'm thinking of.

Or rather, that it's always a singular kernel in order to attain some well-behavedness in solution.

[astral vine]

Iirc this is mentioned several times in Mitrea family's work

But I can't tell you where to find a explicit reference about it

[junior bloom]

[astral vine]

I think this deeply related to Pseudodifferential Calculus

[junior bloom]

I can't read anything Mitrea

Taylor-tier writer

lmao

[astral vine]

Me neither, but I need those for my PhD thesis

sadly

[astral vine]

They are collaborator

so

[junior bloom]

Unsurprising

[astral vine]

Absolutely hahaha

Iirc the main way to deals with singular Kernel of elliptic operators with not so bad coefficients (Smooth is the best obviously), is to get the problem on the whole space, then introduce the Schwartz Kernel which will be singular thanks to Hormander and Calderon-Zygmund theory

Then you get back on open set considering only boundary (layer) operator

I am not sure about it

but

This is well known but I don't know any good synthetic reference about it

Maybe @quaint herald can help us please ?

[junior bloom]

that description does sound familiar

if you find a reference or gomez pops by id appreciate a ping, thanks for stretching your brain for me

[quaint herald]

@junior bloom @astral vine
Here is a rough explanation, not sure if it answers your question entirely but hopefully it helps.

Suppose P is an elliptic differential operator, say of positive integer order k, then the pseudodifferential calculus yields a "parametrix" of order -k, that is an operator Q such that PQ-I and QP-I are smoothing, a.k.a residual a.k.a pseudodifferential operators of every negative integer order. An equivalent characterisation is having smooth Schwartz kernel. If you haven't worked with these things before, think of them as "approximate inverses" (they are incredibly useful and their existence is one of the most immediate and powerful benefits of entering the world of pseudodifferential operators).

If G is a Green's function for P, then convolution by G is a linear op L so that PL=I, hence P(Q-L)=(I+R)-I=R is residual, where Q is a parametrix for P. Applying Q from the left to this equation tells you that in fact Q-L must be residual, i.e. L differs from Q by an operator with smooth kernel. Since pseudodifferential operators (in particular Q) have singularities only on the diagonal x=y, this tells you that the same must be true of L. Moreover, it must have SOME singularities, as otherwise it would be smoothing which would force I=PL to be smoothing as well.

Oh I suppose I should explain the name smoothing, it means they gain arbitrarily many orders of regularity. So they map distributions to smooth functions (without being too specific about the class of distributions because this and the pseudo calculus itself depends on the geometric setup a little).

[astral vine]

Thank you very much for this explanation, moreover TeaForTwo and I asked for a synthetic reference, do you know one that you particularly love ?

[quaint herald]

not really for this specific question, but to learn about pseudodifferential operators in general: grigis-sjostrand is great, treves, shubin also.

not to mention hormanders 4 volumes, but they get harder to read as you go further along

if you had to only open one, then grigis-sjostrand is probably a good choice, but they have all taught me things.

[astral vine]

Thank you again. I'll probably take look on it later, since this result is always used without any mention.

[junior bloom]

Not familiar enough with PDOs to grokk this explanation but I know what I need to read up on.

[quaint herald]

@junior bloom The simple idea they are based on is that you can write constant coefficient differential operators of order k on R^n in the form: fourier transform -> multiply by polynomial -> inverse fourier transform. Replacing the polynomial in this process with a more general smooth function in the Fourier transform variable is one example of a pseudo. There are certain estimates that such a smooth function must satisfy in order for this class of operators to be well-behaved, but roughly speaking, if they are bounded by xi^k, then you get a pseudodifferential operator of order k.

And then composition of such "Fourier multiplier" pseudodifferential operators gets intertwined to multiplication of these smooth functions, which is of course the familiar way of using the Fourier transform to solve simple PDE.

The smooth function a(xi) in all of the above is called the "symbol" of the pseudodifferential op, and more generally PDOs take smooth functions a(x,xi) to a class of operators, and allow you to pass between operator composition and multiplication of symbols.

[junior bloom]

Yeah, I think that's the example I ran into in one of my PDEs courses, and what I was trying to recall. I should become more fluent in the topic so I can better relate the structure of PDOs to the singular kernels popping out of elliptic operators, though it's pretty clear in the case of the heat equation that this Fourier multiplier approach applies directly so it's a good first step.

[astral vine]

Is someone here well aware about behavior of dual of Sobolev Spaces on (smooth) open sets ?

[quaint herald]

@astral vine maybe. what specifically did you want to know?

[astral vine]

It said in some places in Literature that, for non negative s, $$(H^{s,p}_0(\Omega))'=H^{-s,p'}(\Omega)$$, where $H^{-s,p'}(\Omega)$ is the restriction space of $H^{-s,p'}(\mathbb{R}^d)$, and $H^{s,p}_0(\Omega)=\overline{C_c^\infty(\Omega)}$ the closure in $H^{s,p}(\mathbb{R}^d)$ (equivalently in $H^{s,p}(\Omega)$). I get a first embedding, I don't get the second one.

The first easy embedding is $$ H^{-s,p'}(\Omega)\hookrightarrow (H^{s,p}_0(\Omega))'$$, which follows from the definition, and the coincidence of duality Bracket

[untold delta]

Anatole

Anatole

[astral vine]

But I don't get the other one, I don't get how to see an element of $ (H^{s,p}_0(\Omega))'$ as the restriction of an element in $H^{-s,p'}(\mathbb{R}^d)$ (i.e. $H^{-s,p'}(\Omega)$)

[untold delta]

Anatole

[astral vine]

(Here to me Sobolev spaces are Bessel Potential spaces)

[astral vine]

What I did for the first embedding if it can give idea or hints for the reverse one.

[astral vine]

Oh, maybe I did it

Wait

Maybe I didn't

[patent vigil]

Hello,
I am trying to solve the following ODE:
x''(t) + f(g(t)) x(t) = 0
And I know the solutions for
x''(t) + f(t) x(t) = 0

Is there a way I can find the solutions of the first ODE using the second? I think I can't in general, they seem to be completely different equations, but for g(t) = t, the solution are the same, obviously. Or if g(t) = -t, then φ(-t) is a solution for the first if φ(t) is a solution for the second. I notice if F(r,s) = A(r)B(s), where A is a solution to the first ODE and B is a solution to the second, then F_rr(t,g(t)) = F_ss(t,g(t)). Idk if this helps. Any ideas/hints are welcome ( I am a newbie in differential equations)
Thanks

feel free to ping me

[buoyant pike]

#odes-and-pdes

[astral vine]

I did stuff and it comes to prove the following $$\sup\limits_{\substack{||u||{H^{s,p}(\mathbb{R}^d)}\leqslant 1\ \mathrm{supp}, u \subset \overline{\Omega}}} ,\langle {\phi}, u \rangle = \inf\limits{\substack{\tilde{\phi}\in H^{-s,p'}(\mathbb{R^d})\ \tilde{\phi}{|\Omega}=\phi}},\sup\limits{||u||_{H^{s,p}(\mathbb{R^d})}\leqslant 1},\langle \tilde{\phi}, u \rangle $$

[untold delta]

Anatole

[astral vine]

seems pretty intuitive

but I need a proof

[crisp heart]

[scarlet fog]

is this the right place to ask a question about a pde with dirichlet boundary conditions?

i have not studied PDEs, this came up in a book on fourier analysis

[junior bloom]

i think you're looking for #category-theory

[astral vine]

Ask, I will try to answer

[sinful jetty]

Does anyone know how to start a?

I may be misunderstanding the problem

But it seems like I’m taking the Wave equation bounded from a (0,L) to an equation that solution is on the interval of all real numbers

Am I scaling the function?

[astral vine]

#odes-and-pdes

[scarlet fog]

thanks! i solved it eventually, it was about the steady-state heat equation

[astral vine]

Steady state is kind of silly name, talk about Poisson problem, this is more convenient

[tired hollow]

anyone with solid background in PDE (involving functional analysis) available to tute? dm please :)

[astral vine]

Depends on the topic

[astral vine]

Ask your question with the whole context here, I will give it a try

[tired hollow]

reviewing material Sobolev/Holder Spaces, weak solutions, Dirichlet Principle (energy minimization), and calculus of variations

[astral vine]

The first two topics are okay to me, but the last ones, I know have few informations but I'm far from being a specialist

[turbid scaffold]

Why not put pdes and dynamical systems in the same thread?

[onyx heart]

because broadly they arent the same thing

[buoyant pike]

Because I do pdes but not dynamics, and slim does dynamics, but not pdes

[river path]

Not for long. Scott is going to break him

[buoyant pike]

Anyways

Morally they are quite distinct

[junior bloom]

Dynamics is Kantian whereas PDEs is postmodern.

[forest cradle]

you can do a bunch of discrete stuff in the dynamical systems channel but probably pde people don't like that here

[astral vine]

Answer, I changed of strategy, now it's done. Hence equality like $H^{-1}(\Omega)=(H^{1}_0(\Omega))'$ is no longer an "abstract" definition to me but a concrete identification between two fully well defined space, it makes things more clear to me.

[untold delta]

Anatole

[vagrant stone]

Is there a simplified reference on how to solve advection-diffusion-reaction equation using finite/control volume approach without vector calculus? I've been searching for whole days but I didn't find any. (Never learned mass transfer before)
This is the equation

[kindred aurora]

Hey, I have a code to solve the advection/diffusion equation in C++ and CUDA, and a pdf about it from a class, are you interested ?

[vagrant stone]

Does it include finite/control volume approach?

[kindred aurora]

Yes

It's in French tho

[vagrant stone]

Sure

[kindred aurora]

there's the method detailed after my screenshot

I think you could understand the math

[vagrant stone]

Yup I can try to look at the equation as long as it does not have too much vector calculus

[kindred aurora]

idk why the PDF has trouble going through

[vagrant stone]

May I have the screenshot on equations?

[kindred aurora]

oh

you don't accept DMs from people that aren't your friends

that's why

[vagrant stone]

Wait lemme accept you as friend

[astral vine]

#odes-and-pdes

Or #numerical-analysis

[astral vine]

And finally «Langage C++ et Calcul scientifique» is a really good book, the online version is free. It's in French and a paper back version is available for only 16€ iirc.

[kindred aurora]

https://membres-ljk.imag.fr/Pierre.Saramito/cours-cxx.pdf

[astral vine]

Exactly

[kindred aurora]

https://membres-ljk.imag.fr/Pierre.Saramito/books/
This page is also interesting

Anyways, his webpage is actually a goldmine of fluid mechanics classes/resources

[kindred aurora]

And to add onto what I said, here is the discretization method for NS equations in a video: https://membres-ljk.imag.fr/Pierre.Saramito/video_fnn/fnn-2013-ch1b.mp4
Since NS equations are very similar to advection/diffusion equations, you can find the explanation easily

[astral vine]

I am quite curious , is there Numerical methods for more exotic function spaces ?

Because generally evry one works in L², H1

[kindred aurora]

exotic function spaces, wdym?

oh

[astral vine]

Is there some attempt for space like

Besov spaces

Or even Lp-based Sobolev spaces

Triebel-Lizorkin etc...

Maybe be I should go in #numerical-analysis

[kindred aurora]

yeah I think so

[tired hollow]

dm? :)

[buoyant pike]

What

[tired hollow]

!

Im kinda lost with IBP using green identities involving vector field and scalar field

u - solution of PDE
$-\Delta u + \nabla\cdot (\vec b u) = f$ in $U$

$u = g$ in $\partial U$

trying to convert to weak form, to make use of the given $||\vec b||_{L^n} < C$ to use Lax-Milgram theorem to prove uniqueness of smooth solution

multiplying by arbitrary $v \in W_0^{1,q}(U) $

$\int_{U} [-\Delta u + \nabla \cdot (\vec b u)]vdx=<f,v>$

first term is $\int \nabla u \cdot \nabla v dx=\int Du \cdot Dv dx$ since
$\int_{U} [-\Delta u v]dx=-\int_{\partial U} \frac{\partial u}{\partial n}\cdot v dS +\int_U \nabla u \cdot \nabla v dx$ boundary integral cancels since v = 0 on $\partial U$

now dealing with low order term

$\int_U\nabla \cdot (\vec b u)vdx=\int_U[u(\nabla \cdot \vec b)v+\vec b \cdot (\nabla u)v]dx$

Trying to apply Green's identity to make up weak formulation but messing up with vector and scalar fields. :(

Any help would be appreciated.

[untold delta]

DvaNapasa

[buoyant pike]

Is b a constant

[tired hollow]

arbitrary vector field with norm Ln bounded

R^n

sry $||\vec b||_{L^n} < C$

[untold delta]

DvaNapasa

[buoyant pike]

Ok so it's a constant vector?

[tired hollow]

its a vector function

with Ln norm bounded

[strong maple]

i think you can use integration by parts (div thm w.r.t. (v * (bu)) ), so you get (bu).grad v

[tired hollow]

wait up you mean integrate nabla dot (bu) dirrectly ?

and this is not needed?

[strong maple]

yep

[tired hollow]

i dont really feel how because

[strong maple]

div( (bu) v) = div(bu) v + (bu) grad v, integrate and use divergence theorem

[strong maple]

Btw this formula doesn't have a div term

[tired hollow]

this is the only IBP i know, and divergenge thm too

you wanna say one can write $\int_U \nabla \cdot (\vec b u) v dx = \int_{\partial U} \vec b u v dS$ ?

[untold delta]

DvaNapasa

[tired hollow]

@strong maple

[strong maple]

@tired hollow Integrate this and use div thm on the left-hand side

[tired hollow]

isnt the multiplication by v applied after nabla dot bu (whic h is grad bu)

maybe im wrong

cuz want to IBP this

[tired hollow]

:D seems i cant

[strong maple]

bu is a vector field

Because b is a vector

[tired hollow]

yes :)

[strong maple]

Therefore, you can take the divergence of bu

[tired hollow]

right but it makes hard when multiplied by v at the end

:(

[strong maple]

Wdym

I wrote down an identity

You only need to use div thm

[tired hollow]

lemme write by hand and send photo ok?

$\int_U [\nabla \cdot (bu)]vdx$

[untold delta]

DvaNapasa

[tired hollow]

this is the order right?

[strong maple]

Yes

[tired hollow]

$\int_U [\nabla \cdot (bu)]vdx = \int \nabla \cdot [(bu) v]dx - \int (bu) \nabla vdx$

[untold delta]

DvaNapasa

[tired hollow]

?

[strong maple]

Yes

[tired hollow]

now applying div thm for first summand

on RHS?

[strong maple]

Yes

[tired hollow]

$\int_U [\nabla \cdot (bu)]vdx = \int_{dU} \vec \nu \cdot [(bu) v]dS - \int_U (bu) \cdot \nabla vdx$

right? :D

[strong maple]

Yes

[tired hollow]

WHOOO HOOO

[strong maple]

And v has null trace

(Btw highlight that is bu dot grad v)

[untold delta]

DvaNapasa

[tired hollow]

?

[strong maple]

[brave bay]

@strong maple how are you justifying the integration by parts? wasn't their b only assumed to lie in L^n?

[tired hollow]

b is n dimensional vector field whose norm is bounded in L^n

[buoyant pike]

Integration by parts is defined for weak derivatives isn’t it

[tired hollow]

ill perhaps apply Lax Milgram theorem that states that if bilinear mapping B[u,v] is bounded and coercive then exists and unique is the solution

[tired hollow]

just need weak formulation of the PDE

then can make use of bilinear mapping

[strong maple]

(L^inf)^n?

[tired hollow]

L^n?

[buoyant pike]

Oh b is smooth

Lol

C infinity

[brave bay]

lol

[tired hollow]

D: is this bad?

[buoyant pike]

That and b in Ln are very different

[brave bay]

kinda important/useful info to include

[tired hollow]

i want to make use of this theorerm

but i only need to show uniqueness, without existence, maybe applying this theorem is wrong

for part b cant prove that mapping B is bounded (iit feels like)

[buoyant pike]

#odes-and-pdes

[tired hollow]

this statement is strong :|

[buoyant pike]

Someone deleted their message

[strong maple]

is this an exam?

[astral vine]

I'be seen your DM, and gosh there is very much. But bro' I am not at your disposal. I have my own work to do and many other things.

[tired hollow]

thank you :)

its ok

[astral vine]

Assume there is two solutions, take the difference and wrote the associated variationnal problem for the variationnal solution

[tired hollow]

thanks a lot! =)

[astral vine]

To solve it You will need Sharp Sobolev embeddings and Holder Inequality

Then it implicitly assume that n, the dimension, is greater than 2

[astral vine]

I will do (a) for you arguments should work similarly for (b)

[untold delta]

Anatole

Anatole

[strong maple]

i like and hate the fact that you have a formula for the constant lol

[astral vine]

The formula is Kind of "simple" in the L² case

[strong maple]

ig, but isnt it sufficient poincare ineq and set c0 small enough?

[astral vine]

No

How could you get a control on the Ln norm of b with only use of Poincaré inequality ?

[strong maple]

wdym? we want control on the bilinear form in function of the norm of b

(and the norm of u)

[astral vine]

Yes but it is asked for Ln norm of b

espicially Ln norm of b

how do you make it appear

[strong maple]

the domain is bounded, so holder

[astral vine]

That's what I did

If you want to avoid Sobolev embeddings just write exponents

[strong maple]

i didnt noticed that you didnt replaced grad u with u

[astral vine]

you will see you are going to deal with some troubles

[strong maple]

i mean, you need to bound below by the norm of u in the right space

thats why i said poincare

[astral vine]

Yes

I use indifferentfly $H^1$ norm or homogeneous $H^1_0$ norm, on $H^1_0$ since as you said, Poincaré inequality gives equivalence

[untold delta]

Anatole

[astral vine]

But if you use it

you must ask for a smaller value of c_0

[strong maple]

yeah, ik

[astral vine]

So you disallow few solutions

[strong maple]

but for the sake of simplicity ig

[astral vine]

there is not simplicity since you know the fact, right ?

Oh fuck

I just realized

to me $d=n$

[untold delta]

Anatole

[astral vine]

in above solution

[tired hollow]

It seems that this course is giving many mental issues not only to me

[verbal nebula]

Grad PDEs tends to wreck people

[astral vine]

It was the converse, I was like "Oh my god it's so cool and so fun"

my last grad year was the best year of my life

[eager bison]

Hello! can anyone recommend a book or any material that explains sturm-liouville problems intuitively?

[astral vine]

what do you mean ?

[eager bison]

most of the pdfs I've seen were too abstract

[astral vine]

I don't know a systematic treatment of Sturm-Liouville problems in "non-abstract" way

( "non-abstract" in the sense that it does not require Functional Analysis, Spectral Theory in the case of Hilbert spaces)

In fact Sturm Liouville operators are generally unbounded self-adjoint operators, with respect to specific scalar product on a weighted L²-space, in particular when things are nice you can show that it admits a spectrum only made of isolated values that are in fact eigen values, and the eigen vectors asssociated with those eigen values are in fact an orthonormal basis with respect to the previous introduced scalar product.

to make a systematic treatment of Sturm Liouville operators, and not just few specific cases, you need to understand each term in my previous post

[eager bison]

These definitions are known to me, I was looking for geometric interpretations of eigenvalues and eigenfunctions but I think I figured it out. Thank you!

[astral vine]

geometric interpretation ?

[eager bison]

yes, let's say you get a solution sin(pi n x/L) and where pi n/L is an eigenvalue (n = 1,2,3..) and when you enter those n values the eigenfunction satisfies boundary conditions in [0,L] interval

When I graphed these solutions I saw how the sine curve was satisfying boundary conditions

but this is a simple case, there are more complicated boundary conditions and I was trying to see example solutions to those

[astral vine]

Try to investigate the following Sturm Liouville operator on [-1,1], there is no such "geometric interpretation", as you said. Consider $$L= -\frac{\dd}{\dd x}\left[(1-x^2)\frac{\dd}{\dd x}\right]$$

[untold delta]

Anatole

[astral vine]

No boundary condition because this is invariant rotation Laplacian on the unit sphere

How do you find eignen values in this case

then the eigen vectors ?

etc.

Generally speaking Sturm-Liouville operators have no such "geometric interpretation", since the eigen vectors here won't be sin/cosine functions

[verbal nebula]

I digged PDEs when we were doing weak solutions, regularity, and variational stuff. When we got into semi group dynamics it was pure pain

Now I'm trying to learn more about Monge-Ampere

[astral vine]

I'm checking everything, except variationnal problems.

[astral vine]

Those kind of PDE are difficult to deal with, because the non-linear part and the linear part are of order 2 right ?

Orlicz-Lebesgue /Sobolev spaces

[verbal nebula]

I have no idea why they're difficult

It just came up in my research with my prof

And he just said "This is a 34 page argument to understand why, but we ultimately get what we need"

[verbal nebula]

So I'm reading this Elliptic Partial Differential Equations: Second Edition (Courant Lecture Notes) by Han and Lin

Then fully non-linear elliptic equations by Xabre and Caffarrelli

[junior bloom]

Han and Lin good, need to read it too

[verbal nebula]

It's pretty short

[junior bloom]

It's PDEs, it's never short

1 page of PDEs is like 100 pages of normal math

[verbal nebula]

LMAO

Han and Lin seems to retread a lot of the same basic ground for the first two chapters

but I guess it doesn't hurt to go back through harmonic stuff again

[strong nimbus]

I'm trying to find the stationary condition for this functional

The Euler-Lagrange equation gives me $1=Cy\sqrt{1+(y')^2}$

[untold delta]

ImHackingXD

[strong nimbus]

Did I make a mistake midway? I don't see a way to solve for y here

[late moth]

what was your E-L before simplifying? You should
get something like
[-\frac{\sqrt{1+(y’)^2}}{y^2}-\frac{d}{dx}\left[\frac{y’}{y\sqrt{1+(y’)^2}}\right]=0]

[untold delta]

kirby

[late moth]

@strong nimbus

[strong nimbus]

Since my function doesn't depend explicitly of x, I used the formula
$$
f-y^{\prime} \frac{\partial f}{\partial y^{\prime}}=C
$$

[untold delta]

ImHackingXD

[strong nimbus]

where $f=\frac{\sqrt{1+\left(y^{\prime}\right)^{2}}}{y}$

[untold delta]

ImHackingXD

[late moth]

oh indeed

yeah your result is correct

I don’t see the issue

you can get it in terms of y’ then you have a separable ODE

[strong nimbus]

Separable how? Probably being blind right now, sorry

[late moth]

y’ = \pm \sqrt{1-(Cy)^2}/(Cy) @strong nimbus

[strong nimbus]

Oh, so separation of variables between y' and y? Didn't think I could do that! Thank you very much for your help

[late moth]

mhm, you just have to solve for y’
recall that y’=dy/dx

The point of E-L is to create a minimizer that creates some sort of diffeqs that are hopefully solveable

if they aren’t, you still get a lot of information about y

[junior bloom]

If I have $\int_{\Omega}\nabla\cdot\mathbf{v}(\mathbf{x}),\mathrm{d}\mu(\mathbf{x})=0$ for some compact $\Omega\subset\mathbb{R}^3$, what are some fairly minimal conditions I can add on to any part of this system to get $\mathbf{v}:\mathbb{R}^3\to\mathbb{R}^3$ to vanish $\mu$-a.e.? I'm brainstorming because I feel like I'm missing something trivial, maybe even calculus-level.

[untold delta]

teafortwo

[astral vine]

Is Omega smooth ?

[astral vine]

is mu a Borel measure ? or absolutely continuous wrt the Lesbgesgue measure ?

because if mu is the Lesbesgue measure, there is plenty of function with 0 divergence on Omega, that are not 0.

in particular there is many more functions such that the integral on Omega of its divergence is 0.

so you cannot deduce anything without very strong additional assumptions

[junior bloom]

@astral vine I'm "allowing" it to be anything. Loosely speaking I'm trying to model something where I know that the integral of divergence vanishing strongly predicts the flow itself vanishing, but it's not obvious at all what additional structure I should be look for that makes this so.

I'd say \mu should be absolutely continuous w.r.t. the Lebesgue. Omega is not necessarily smooth.

So I'm just shopping around for some conditions to brainstorm.

Omega is compact and strongly connected, but \partial\Omega is only C^0

in guarantee

[astral vine]

Without smoothness assumpiton , say C² boundary, you cannot expect to control the gradient with the curl and the divergence

[junior bloom]

Hm, since I'm fine with purely local control, I think we can assume \Omega is a smooth ball.

In this case.

[astral vine]

But the interesting fact is the boundary

right ?

I don't know if C^0 boundary is sufficient to guarantee a Stokes a theorem

[junior bloom]

It's hard for me to reason out if I particularly care about the boundary in this case, and it's true that even for local control I need some kind of global dependence on the "true boundary" data.

C^0 doesn't guarantee Stokes. OK, let's assume smoothness, what other things can I add on for now?

[astral vine]

To be sure, say that the boundary is at least C^{2,alpha} alpha in [0,1]

Now

mu is absolutely continuous with respect to the Lebesgue measure

do you have any explicit acess to the Radon Nikodym derivative of mu w.r.t. the lebesgue measure ?

does it have additional properties ?

[junior bloom]

Haha we were thinking about the same thing. I dont' formally know I've got it but I believe I can.

Since this system is physical and unlikely to have any singularities.

[astral vine]

Ask v to be in H^1_0

[junior bloom]

We can say it behaves like a density.

[astral vine]

this became equivalent to say that $v$ lies in the orthogonal set of $\nabla f$ in the L² sense

[untold delta]

Anatole

[astral vine]

or weak sense

or whatever

f being the distribution

A distribution annihilate smooth compactly supported function iff it is 0

Hencef must be constant

so mu must be a renormalized lebesgue measure

I said shit

v is a selected vector field not any vector field

wait

but you want that being in the annihilator implies to be 0, right ?

[junior bloom]

Yes, I'm fine with it only being weak, but I'm still puzzling out whether I can be happy with v Sobolev.

[astral vine]

it does not change my previous statments

The problem is ill posed to me

because we write, for $v\in C_c^\infty(\Omega,\mathbb{R}^d)$, $$\int_\Omega \div v\dd \mu = -\langle \nabla f, v\rangle = 0$$, and we want $$v\in N(\nabla f) + [\text{conditions ...}] \iff v=0$$

[untold delta]

Anatole

[astral vine]

N(T) being the nullspace of operator T

if $\nabla f$ is in L², then asking to be in $N(\nabla f)^\perp$ should be sufficient

[untold delta]

Anatole

[junior bloom]

That's a hard ask to give physical meaning.

[astral vine]

We don't do that here

But orthogonality with respect to a gradient should have a deep physical interpretation right

?

[junior bloom]

But anyways thanks for helping me think through this. I'm actually concerned about the weak formulation because I'm afraid the derivatives are singular at the boundaries, but passing it through naively I should still think about what it means for v to the perp to the null of grad f.

Well the issue is I'm already sitting on an assumption that the measure is a density, I think this is the case but it's actually something that should pop out of a Girsanov transform applied to importance sampling.

So it's unclear what the gradient should correspond to.

[astral vine]

L²(\Omega,\mathbb{R}^d) = { \nabla f}^\perp \oplus \mathrm{Span}(\nabla f)

[brisk granite]

Is this an appropriate channel to ask about hermite function?

[astral vine]

Depends on what properties of Hermite functions

which I recall are not Hermite polynomials

[brisk granite]

So can I ask about it? I have a troubling problem

[astral vine]

ask

[brisk granite]

I do not understand how the first step after eqn 1 is happening.

[astral vine]

Putting 0 in the whole equation

then expend the product

terms by terms at the top and the bottom of the fraction

[brisk granite]

Then how do I get n! Here?

In the denominator

I've done I've still not understood

[astral vine]

can you give me the book reference please ?

[brisk granite]

Reference meaning, the name of the book?

[astral vine]

yes

[brisk granite]

It's HK Dass & Dr. Rama Verma

Mathematical Physics

[astral vine]

there is an error

there is many errors

[brisk granite]

Okay what is that error(s)?

[astral vine]

${H}{2n}(x) = \sum{k=0}^{n}(-1)^k \dfrac{(2n)!}{k!(2n-2k)!}(2x)^{2n-2k}$

[untold delta]

Anatole

[astral vine]

This is the real formula

[brisk granite]

Sure?

[astral vine]

Sure

[brisk granite]

Okay

[astral vine]

some k's where replace by few x

[brisk granite]

But what happens if I put x=0? Won't the term become 0?

[astral vine]

no

[brisk granite]

The 2x part

[astral vine]

what happends to the term k=n

you get 0^0