advanced-pdes

[modest moth]

Maybe I'm capping, online it says:

[\lvert\lvert f(x)g(x)\rvert\rvert \le \lvert\lvert f(x)\rvert \rvert:\lvert\lvert g(x)\rvert\rvert ]

[untold delta]

Max

[modest moth]

I've reached:
[ \lvert\lvert u(x,t)\rvert\rvert^2_{L^2(\mathbb{R})}\le [ \lvert\lvert e^{-k\xi^4t}\rvert\rvert^2_{L^2(\mathbb{R})}[ \lvert\lvert f(x)\rvert\rvert^2_{L^2(\mathbb{R})}]

But the factor on the front is out by a square.

[untold delta]

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

[lilac barn]

Take the Linfty norm of the exponential out. Get rid of the transform by bounding by L1 norm and then do change of coordinates to get rid of t.

[lilac barn]

This is false

[modest moth]

Is it only true for compact domains ?

[lilac barn]

Not even then because then for f=g, it implies that any L2 function on compact domain is L2n.

[modest moth]

Hmm and is that not true?

I can't see any examples that would counter it

[lilac barn]

X^-1/4 is in L2(0,1) but not in L4(0,1)

[modest moth]

Okay cheers

[modest moth]

Problem is if I use the L1 norm inequality it introduces a factor of $2\pi$ and I'll have a double integral instead of a single integral

, rotate

[modest moth]

Problem is if I use the L1 norm inequality it introduces a factor of $2\pi$ and I'll have a double integral instead of a single integral

[untold delta]

Max

[untold delta]

Max

[modest moth]

NVM I made a mistake

Anyone see what I'm missing?

Each time I try bound it above by L1 of the exponential I get a double integral

[modest moth]

This would complete the proof

🥺

I know it's something super small that I'm missing

[lilac barn]

Are you sure that f is L2 as well?

[modest moth]

f is definitely L2

Were these the steps you intended?

Surely the infinity norm of the function of e would be 1...

[lilac barn]

Not really but I doubt such an inequality could hold. Because the Linfty norm of the exponential thing is necessarily 1 (obtained for xi=0) whcij we can't bound by integral/2kt thing as that will tend to 0 as it goes to infinite

[modest moth]

Bruh, it was so close

What did you mean then?

[lilac barn]

[ \lVert \hat{f} e^{- k \xi ^4 t} \rVert_{L^2} \leq \lVert e^{- k \xi ^4 t} \rVert_{L^\infty} \lVert \hat{f} \rVert_{L^2} \leq 1 \cdot \lVert f \rVert_{L^2} \not\leq \frac{\lVert e^{-x^4} \rVert_{L^1}}{\sqrt[4]{2kt}} \lVert f \rVert_{L^2}
]
as the RHS can be made arbitrarily small as (t \to \infty) so it cannot upper bound (1).

[untold delta]

cocat

[modest moth]

Yea I see

I got an exam tomorrow, hopefully they don't repeat this Q

[modest moth]

How about starting with the infinite norm instead.

[\lVert u\rVert_{L^{\infty}}^2= \frac{1}{\sqrt{2\pi}}\lVert \hat{u}\rVert_{L^1}^2=\frac{1}{\sqrt{2\pi}}\lVert \hat{f}(\xi)e^{-k\xi^4t}}\rVert_{L^1}^2\le \int_{\mathbb{R}}e^{-2kx^4t}dx}\lVert f\rVert_{L^2}^2,]
where the last line used $\frac{1}{\sqrt{2\pi}}\le 1$ and Cauchy schwarz.

[untold delta]

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

[modest moth]

@lilac barn just found out the question is wrong... sorry for wasting your time.

At my university the professors don't update mistakes in exam questions....

[astral vine]

No take the Laplacian on L²(R^n)

It is 0-sectorial, self-adjoint, and all but not invertible.

[frail skiff]

gm

[modest moth]

gm

[solid flint]

Thank you very much!

[rotund jetty]

On a $d$-dimensional compact Riemannian manifold $M$ without boundary, if $u(x) = \sum_{j = 0}^{\infty}\hat{u}(j)e_j(x)$ is the eigenfunction expansion of $u$ in terms of eigenfunctions of $\Delta$, does $\sum_{j = 0}^{\infty}|\hat{u}(j)| < \infty$ imply $u$ is continuous? This is true on the torus $\mathbb{T}^d$, but it makes use of the fact that $|e_j(x)| = 1$.

[untold delta]

L

[crude depot]

Does anyone know how to show that weak solutions exist for the non-stationary Navier-Stokes equations?

[inland sinew]

I am trying to figure out if my solution is valid for the following problem. The problem I am looking at is $cos(y)z_{xx} + cos(x)z_{yy} = -cos(x)-cos(y)$, where I need to find a periodic solution of $z(x,y)$ as opposed to the obvious qudaratic solution that satisfieis it.

I plugged in a two dimensional fourier series in for $z$ and obtained a the coefficients of the solution through harmonic balancing. But I'm worried because I'm double checking the solution through substitution to see how it exactly satisfies the PDE and it does so by producing an alternating series that doesn't converrge. The way things cancel is something like
$cos(x) + cos(y) - cos(x) - cos(y) + cos(2x) + cos(2y) - cos(2x) - cos(2y)...$.
I wanted to see if I could just get an approximation to the solution by truncating it at some index of the fourier sum, but any index I truncate leaves off one term that doesn't cancel.
I'm not really sure how to interpret this.

[untold delta]

pewpew2385

[mint canyon]

@inland sinew z = X(x)Y(y) then divide both sides by cos(x)cos(y)X(x)Y(y) separates the homogeneous problem doesnt it?

then you can just use the greens function to solve the inhomogeneous part

[astral vine]

L² energy estimates

"sur un liquide visqueux emplissant l'espace" Jean Leray 1934 or 1932 I don't remember

[inland sinew]

@mint canyon For the homogenous problem that does work. The greens functions for this equation are constructed from Mathieu's equations, which I was trying to avoid temporarily. You can do it that way though. I took a second look at it and figured it out though. I also noticed s a strange feature of this equation, which is that some of the solutions seems to be invariant under swapping x,y with cos(x),cos(y)

[untold delta]

x17

[mint canyon]

^ This is wrong, my bad

I didnt account that you have to divide both sides by cos(x)cos(y)X(x)Y(y)

so that wont work unfortunately

[inland sinew]

The solution ended up being way simpler than I realized because I overlooked something.

[dapper belfry]

No entiendo ni una pt mrd

XD

[pliant bough]

Hi! Not sure if this is the right channel, this might not seem to fit anywhere, but I've been reading a paper and I'm not quite sure how to make sense of those calculations. I get that its 'symbolical', but differentiating I get rather different relations: $$\frac{\partial}{\partial \alpha} = \frac{-z}{\alpha} \frac{\partial}{\partial z}$$ and $$\frac{\partial^2}{\partial f^2} = \frac{1}{\left(B \varepsilon \alpha \right)^2} \frac{\partial^2}{\partial z^2}$$ Any ideas what I might be doing wrong?

[untold delta]

MaineCoon

[pliant bough]

At A.9a RHS this might have just been a sign error? (minus instead of equal)

I'm definitely missing some terms and I'm just not sure how to understand that notation

(epsilon and alpha don't seem to be dependent on f, nor does f on alpha, varepsilon )

[pliant bough]

Okay maybe it works if it s equal instead of minus in the alpha one

[lilac barn]

That's some cursed notation

[pliant bough]

Ikr

[lilac barn]

You can post this in #odes-and-pdes

[pliant bough]

Hmm nvm, I guess it was just a problem with the equal sign

sorry heh

[pliant bough]

ok its still wrong

[rotund jetty]

Can someone tell me if I'm right: The inclusion $H^s(\mathbb{R}^d) \to C(\mathbb{R}^d)$ regarded in $D'(\mathbb{R}^d)$ is continuous if and only if $s > d/2$. This is because if the inclusion is continuous, then $\delta \in H^{-s}(\mathbb{R}^d)$, so the function $f \in H^s$ with $\hat{f}(\xi) = \langle \xi \rangle^{-s}$ is in $L^2$.

[untold delta]

L

[lilac barn]

That sounds correct but you still need the converse direction

As I am confused what you meant, I am guessing this is your argument:
For f in Hs, |delta(f)| = |f(0)| <= |f||infty <=C|f|Hs so delta defines a linear functional on Hs. This implies delta is in H-s and thus via the H-s norm, we obtain the condition on s.

[lime folio]

Hey y'all - does anyone have a reference for someone with a PhD in pure maths (functional analysis) to learn basics of CFD from a theoretical perspective (e.g. what methods to use when, why they work, convergence rates, ect.)? Bonus points for brevity. Happy to move this question if not the correct channel. Thanks!

[noble bloom]

Hello, I've been learning PDE using the book by Hillen, Leonard, Van Rosen. But I see MIT Open courseware has a PDE section for free. Have any of you taken it? Any feedback? I'm only 2 chapters into my book so I'm not opposed to switching sources, just looking for tips.

Here's the cover of the book I'm using

[rotund jetty]

From a brief inspection, the book you are using seems decent, though it is at a lower level than typical texts like those by G. Folland, L. Evans, and M. Taylor. I would switch to using sources at the level of the three I mentioned if you have the background for it.

[noble bloom]

The only real reason I picked this book was because it has "completely solved problems" LOL.

When I took PDE in uni, they used the Zachmanoglou, Thoe book, and tbh that book made me hate the course LOL.

I'll check out the Evans book again

[gray jasper]

is this the usual characterization of sobolev spaces?

this seems to conflict with evans

this seems to suggest that indeed sobolev spaces instead are larger than the closure of the space of test functions

but then again, perhaps the two characterizations are equivalent when the domain is R^d?

[quaint herald]

bingo. the compact support of test functions forces their closure to consist of things "vanishing at the boundary" in a suitable sense (mapping properties of the trace operator tell you precisely what sense), but R^n has no boundary!

[gray jasper]

ah interesting, so i can just regard W^k,p(R^d) to both be the closure of C_c^\infty(R^d) and the space with weak derivatives up to order k that are all in L^p?

no weird trace stuff

[quaint herald]

Yep. I prefer to take the latter as the definition (functions in L^p are somewhat less clunky objects than equivalence classes of sequences of test functions), but one of the first things you prove after making such a definition is the density of the test functions.

[gray jasper]

cool cool thank you sm

[pulsar forge]

this last remark is a bit odd, cause usually we define W^k,p as a subset of L^p where distributional derivatives to order k are L^p, so in case the domain is R^n, f(x)=1 would not be sobolov in both definitions.

i guess there is a "equivalent" way im not familiar with that doesnt require f itself to be L^p

[quaint herald]

They are talking about the homogeneous Sobolev spaces in that remark. The defining norm only involves k-th (in this case k=1) order derivatives. So the norm is perfectly reasonable and finite for things that are only locally integrable but have L^p first order weak derivatives.

[quaint herald]

(Lower order polys are usually quotiented out in this space, so f(x)=1 is actually just a representative of the same element as f(x)=0. Note that the homogeneous Sobolev norm cannot tell them apart!)

[lilac barn]

This book looks like Bedrossian-Vicol

[gray jasper]

ur right

[pulsar forge]

whats the advantage of this approach for PDEs? shouldn't we care about the behavior of the function itself ?

[slim forum]

hello! I’m studying weak solutions to 2d Euler equations which model inviscid incompressible fluid flow and found this notation—can’t find where it’s defined in the book I’m looking at (vorticity & incompressible flow by bertozzi and majda). If anyone knows what the L^infty set in point (i) is and could help me understand that’d be mega appreciated

[astral vine]

this is equivalent to:

$$\omega\in\mathrm{L}^\infty([0,T],\mathrm{L}^1(\mathbb{R}^2))\cap\mathrm{L}^\infty([0,T],\mathrm{L}^\infty(\mathbb{R}^2))$$

[untold delta]

Functionanatolysis

Functionanatolysis

[astral vine]

is implied.

[slim forum]

firstly tyvm—though I haven’t seen the comma/semicolon notation before. Is that saying that for almost all t in [0,T], \omega(x,t) is integrable over space and essentially bounded over space

[astral vine]

exactly

[untold delta]

Functionanatolysis

[slim forum]

that’s so helpful tysm

[astral vine]

the essential supremum in fact

but you got it

[slim forum]

ofcofc

[mint canyon]

Question: Substitution seems to work generally for reducing systems of linear PDEs to a single scalar PDE, is there a matrix method for this where I can use row operations instead of substitutions. Let me give an example of what I mean:

[mint canyon]

$$ A \partial_x \phi + B \partial_y \theta + C \phi = 0, P \partial_x \theta + Q \partial_y \phi + R \theta = 0 $$ where $$Q \partial_x \partial_y \phi = -P \partial^2_x \phi - R \partial_x \theta$$, gives the first equation as $$A( \frac{1}{Q} ( -P \partial^2_x \phi - R \partial_x \theta )) + B \partial^2_y \theta = -C \partial_y \phi$$, then $$P \partial_x \theta + Q( -\frac{1}{C} (A( \frac{1}{Q} ( -P \partial^2_x \phi - R \partial_x \theta )) + B \partial^2_y \theta)) + R \theta = 0$$, and finally $$-\frac{AP}{C} \partial^2_x \phi = P \partial_x \theta -\frac{1}{C} (-AR \partial_x \theta + B \partial^2_y \theta) + R \theta$$

[untold delta]

x17

[mint canyon]

I guess there would be one more step:

$$ -C \partial_x \phi = A( -\frac{C}{AP} (P \partial_x \theta -\frac{1}{C} (-AR \partial_x \theta + B \partial^2_y \theta) + R \theta)) + B \partial_y \theta $$ is substituted in to the second equation to give $$ P \partial^2_x \theta + \frac{Q}{C} (A( -\frac{C}{AP} (P \partial_x \partial_y \theta -\frac{1}{C} (-AR \partial_x \partial_y \theta + B \partial^3_y \theta) + R \partial_y \theta)) + B \partial^2_y \theta) + R \partial_x \theta = 0 $$

[untold delta]

x17

[mint canyon]

As you can see the reduction is tedious. Is there a faster way to do it?

[quaint herald]

It is not a "different approach", it is just one of MANY different function spaces measuring different quantitative features of functions/distributions. Depending on what is relevant for the problem being studied, different features are important.

[astral vine]

I should add that changing the space change the inner behavior of the problem you are studying.

[astral vine]

Sometimes one needs to measure only the amount of derivatives that shows up in the equation. However the definition here is misleading and partially wrong for homogeneous Sobolev spaces.

Closure of C_c infty with respect to what structure etc... For instance,

[quaint herald]

they probably mean completion rather than closure

[astral vine]

Whatever is the completion of homogeneous W^{1,2}(R²), it CANNOT be identifiable with actual elements of D'(R²)

This is morally due to the critical Sobolev embedding in BMO

There cannot be any continuous embedding from the former to the latter.

It fact it can be proved by contradiction.

[quaint herald]

I don't think they claim otherwise though. they make their actual definition as a completion (which is fine) and then make a handwavy (and I agree misleading) remark about other possible definitions without giving details.

[astral vine]

To get rid of constant by hands there is a huge issue

If you do so

Some how you HAVE to lost completeness of the space

Which seems unbearable

But this is the most fair construction

So "taking the completion" is somehow wrong either

(with respect to the authors' remark)

Here is a MSE post about those issues: https://math.stackexchange.com/questions/4818059/completeness-of-homogenous-triebel-and-sobolev-spaces/4819111#4819111

Middle of Part II] of the selected answer

"But do we really have to lost completeness ?"

[lilac barn]

I don't see the issue here. Per Bahouri and Grafakos, it's clear the issue comes primarily because of the parent space. If you work in the obvious S', you lose completeness whereas in the nebulous S'/P, you retain completeness.
But note that they're working with C^\infty_c so their approach is more in line with Grafakos for which you've completeness in all cases.

[astral vine]

Not according to what said the author in the comment here

[astral vine]

.

End of the screenshot

Choice of not having 1 in the space

is either saying that he used equivalence classes

or he did get rid of polynomial by hands

[lilac barn]

In fact, a very simple example to check is to consider delta *(1_[0,1])^. This will lie inside Bahouris definition but not under their definition (as the element is strictly a tempered distribution and hence not a norm limit of smooth functions)

[astral vine]

if you check with equivalence classes it does

in S'/P

so (C_c^nfty +P)/P identified with C c infty

is dense

[lilac barn]

I am not talking about S'/P. My claim is strictly about the definition they proposed: aka completion of Cc\infty under the homogenous norm

[astral vine]

But you have to choose where you put you compeltion

the only suitable choice is S'/P

otherwise this is not continuously embedded in the dual of some smooth function space

[lilac barn]

I mean I don't have to put them anywhere, I just consider the equivalence class of the Cauchy sequences; but even if I put inside S'/P, then I automatically get completeness as I have completeness inside S'/P

[astral vine]

Yes but the compeltion does not gives elements of D'

so you cannot test them on C_c^\infty function

Do we really want to play with elements that aren't distributions in any sense ?

[lilac barn]

The completion gives you functions in return; which gives you a well-defined distribution.

[astral vine]

That's the actual point

it does not

Which is the most confusing thing

[lilac barn]

Are you claiming that here the completions lie outside of the full space of distributions?

[astral vine]

Exactly

[lilac barn]

Okay what's an example?

[astral vine]

.

H^{-n/2} homogeneous does not contain smooth compactly supported functions

[lilac barn]

I am talking about their definition though. Their definition has nothing to do with Bahouris definition

[astral vine]

What EVER IS THE COMPLETION

By any compeltion

The dual space is not included in D'(R^n)

[lilac barn]

The completion can very well lie inside S'/P (which it should because it doesn't contain any constants or polynomials unlike S')

[lilac barn]

But their n is nonnegative? Your claim might be that their definition falls apart for negative n, but for nonnegative n; I don't see how your argument works.

[astral vine]

n is the dimension

If the pre dual space does not contain C_c^\infty

the dual space cannot be included in D'

[lilac barn]

n isn't the dimension in their definition; it's the number of derivatives. Your example might be proving that H-n/2(Rn) has no sense; which migjt be true but this says nothing about Hn/2(Rn) aka spaces where the regularity is nonnegative.

[astral vine]

when s<n/2

the spaces are complete

no matter what

included in S'

the issue happens for high regularity only

(beyond the critical Sobolev embedding)

[lilac barn]

Your example is for negative regularity spaces; not for positive high regularity.

[astral vine]

No

read again

homogeneous H^{-n/2} does not contain C_c infty this is negative regularity

we agree on that

but then

any completion of Schwartz function w.r.t the homogeneous H^{n/2}-norm can be put in duality with homogeneous H^{-n/2}

Both are Hilbert spaces

Since the pre-dual space does not contain C_c^\infty

homogeneous H^{n/2} (the dual of hmogeneous H^{-n/2})cannot be included in D'(R^n)

[lilac barn]

What could be an example of such an element?

[astral vine]

This just an abstract argument

So I don't have a constructive counter example

[lilac barn]

Wait, if H-n/2 doesn't contain C\infty_c, this doesn't imply Hn/2 doesn't contain C\inftyc. What this implies is that D' is not inside Hn/2 which is true. In particular A subset B implies B* subset A* not B* subset A.
In fact; that's the most likely outcome. We expect C\inftyc to be inside Hn/2 so it implies that we should have H-n/2 is inside D'. Which it does hold.

[astral vine]

I didn't say that

I just claimed H^{n/2} is not included in D'

[lilac barn]

Don't you suggest one can construct a Cc\infty function with infinite H-n/2 norm?

[astral vine]

This is easy

take one with mean value 1

[lilac barn]

I know it's easy; this implies H-n/2 doesn't contain Cc\infty. The implication that derives from this is that D' is not included in Hn/2 not that Hn/2 isn't included in D'.

[astral vine]

Those are Hilbert spaces, but what ever this is not the proof

The actual argument is heavier

Argue by Contradiction

if H^{n/2} is a subset of D'

then one of many version of the closed graph theorem for TVS implies that this inclusion is continuous

this implies C_c^\infty is included in homogeneous H^{-n/2}

contradiction

[astral vine]

this is the tricky part obviously

Only "few"* people are actually aware of those issues in the PDE community

*already a lot actually, but still a lot of people are unaware of it

[bronze gate]

Which version? I don't think that there is one applicable to D' (as it is not even barreled iirc)

[astral vine]

according to Wikipedia it is

D' being barrelled

https://en.wikipedia.org/wiki/Spaces_of_test_functions_and_distributions#Topological_properties

5s search

[inland sinew]

Hi, this is sort of a broad question and I might also ask it in dynamical systems. But I think a class of PDES I'm working on can be rewritten as wave equations in a slightly different dependent variable. Is there any good systematic way of figuring out what this choice would be?

What I mean is I have a PDE L(z(x,y)) = 0, where z is the dependent variable and L is the operator that has a mixed sign on the discriminant(hence a coordiante transform cannot globally turn it into a canonical form of the wave equation). However Im pretty sure there is some function of z, say F(z), which satisfies the 1 d wave equation in x and y.

[rotund jetty]

Are there Sobolev spaces $H^s(\mathbb{R}, L^2(P))$?

[untold delta]

L

[lilac barn]

What's P here?

[astral vine]

Depending on what's P there is no actual answer. But in general one can introduce Banach-valued Sobolev spaces

[rotund jetty]

probability measure

[astral vine]

Yes

You can define what ever vector Sobolev space you want

Moreover, here the Banach space is a Hilbert space so everything goes in an easier way

(you can define Hilbert valued Fourier transform: you still have Fourier -Plancherel etc.)

[rotund jetty]

To define $H^s(\mathbb{R}, H)$, do we need $H$-valued distributions

[untold delta]

L

[astral vine]

Not really

you are in a easy case

But you cna do it that way

[untold delta]

Functionanatolysis

[astral vine]

Then

[untold delta]

Functionanatolysis

[astral vine]

This is overkill tho

[rotund jetty]

Should it be defined in terms of pairings against $\mathcal{S}(\mathbb{R}^n; X)$. Like using the space $(\mathcal{S}(\mathbb{R}^n; X))'$

[untold delta]

L

[astral vine]

This is where you have to be carefull

the problem is about the behavior of the dual space of X

which can be really messy

this why we use continuous linear maps from S or D with values in X

instead of the dual of (S(R,X))'

you need a predual space

which may not happen

in the Hilbert case no one actually cares

Morally (S(R,X))' is S'(R,X')

so your distributions ar X' valued and not X valued

[rotund jetty]

I was thinking of the $L^2(\mathbb{R}^n; H)$ pairing $(u, v) = \int_{\mathbb{R}^d}(u(x), v(x))_H,dx$, and that it should extend to $v \in (\mathcal{S}(\mathbb{R}^n; H))'$.

[untold delta]

L

[astral vine]

For Hilbert spaces everything coincicdes

no troubles

But this is not necessary to build your spaces

[lilac barn]

You don't even need to go at the level of distributions. For nice properties, you consider a rigged Hilbert space (H,V,H*) and require your u to be in, for example: L2([0,T] ; V) and u' in L2([0,T]; H*). For further details, you can look into Evans book

[astral vine]

However this is a bit different in this case

This is the outer space that can "move"

not the innver

The way I would build your Sobolev spaces without distributions theory is the following

Start from the Hilbert Fourier-Plancherel Formula:

[untold delta]

Functionanatolysis

Functionanatolysis

Functionanatolysis

[astral vine]

Reprove using density of Hilbert-valued Schwartz functions that everything coincides: that s actually measure the amounts of derivatives etc.

u in H^{s}(H) <=> u & Du in H^{s-1}(H)

etc.

[rotund jetty]

Here $\mathcal{F}u(\xi) = \int_{\mathbb{R}^d}u(x)e^{-i\xi^Tx},dx$, the integral being a Bochner integral, valid for $u \in L^1(\mathbb{R}^d; H)$?

[untold delta]

L

[astral vine]

yes

[rotund jetty]

I have a process $Z(x) = \sum_{j = 0}^{\infty}(1 + \lambda_j)^{-s/2}\xi_j f_j(x)$, where $f_0, f_1, \dots$ are eigenfunctions of $-\Delta$ on a $d$-dimensional compact Riemannian manifold $M$ and $\xi_j$ are i.i.d. $N(0, 1)$, and I showed it was $C^k(M, L^2(P))$ when $s - d/2 > k$ by basically reproving Sobolev embeddings, but it seems that we only have $Z \in H^{s - d/2 - \varepsilon}(M; L^2(P))$, which is quite a bit weaker. Is there a way to prove $C^k$ using the Sobolev spaces $H^t(M; L^2(P))$?

[untold delta]

L

[quick pagoda]

Is there a good reference for these Hilbert or Banach valued distributions/Sobolev shenanigans?

[lilac barn]

Pick pretty much any graduate parabolic PDE Book, it should contain it. You can look into Evans, Eberhard etc.

[quick pagoda]

Evans as in the giant PDE book?

(Or is it some other one he also happened to write)

[lilac barn]

Yes the giant PDE Book

[quick pagoda]

Right right, I’ve just been had before by an author name referring to a very different book than the “big one” they’re known for

[astral vine]

Those notes by Herbert Amann for abstract vector valued Distribution theory

Then its book about Parabolic equations

Which does not contains the notes.

[pulsar forge]

I'd ask in #book-recommendations but i think this channel is more appropriate for this specific request. I need a good book for a introduction to viscosity solution theory.

[patent fjord]

the inequality is easy to see, but why it implies that Ker(I-K) = {0}?

i tried to use the inequality to verify that I-K != 0 so that would then imply that \phi = 0 hence Ker(I-K) = {0}

[proper knot]

the inequality implies that $\varphi$ must be zero, else you get a contradiction (namely $y \leqslant y/2$)

[untold delta]

Rudolf

[patent fjord]

ah yeah clearly the inequality cant hold for all x and the maximum is achieved hence which leads to contradiction namely $y \leq y/2$

[untold delta]

Rottawarlock/Rottamage

[patent fjord]

thanks

[quick pagoda]

Thanks

[patent fjord]

any idea how to solve this monster ? 😄

[rotund jetty]

Maybe rewrite it in complex exponential form

for (my) convenience

[patent fjord]

hmm

[astral vine]

By contradiction assumme it has at least one eigen value and one eigen vector

write down the equation

take the scalar product your basis

you will have a contradiction with the fact that the set of eigenvalues is non-empty

[patent fjord]

basis of L^2(0,2pi) ?

[astral vine]

yes

the one given by the sines and cosines

[patent fjord]

ye e^(int)

[astral vine]

if you prefer

but regarding the Kernel

[patent fjord]

gets messy with lots of sines and cosines

[astral vine]

the sine and cosine basis seems more convenient

[unborn gyro]

notice that for convinence you can simplfy using trignometric identities

[untold delta]

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

Rottawarlock/Rottamage

[lilac barn]

It's my night brain speaking so I haven't worked it out but it should work.
First recognize that the summand is -(k+1)^-2 sin((ks+s)+kt).
Now you have int K(s,t) f(t) dt = cf.
Taking Fourier transform, you should have something like
1/((k+1)^2) int e^{ikt} delta_{n - (k+1)} - e^{-ikt}delta_{n+(k+1)} f(t) = c fhat(n).
Thus we have for each n:
(1/n^2 int e^{i(n-1)t} - 1/(-n)^2 e^{i(n+1)t} ) f = c fhat(n).
Thus c fhat(n) = 1/n^2 (fhat(n-1) - fhat(n+1)) along with fhat(0)=0.
This should kinda give you a sequence of coefficients that shouldn't hold for an L2 integrable function unless f is 0. Like we notice that fhat(1)= -1/c fhat(2) and so on.

Might be too convoluted

[patent fjord]

interesting approach

[rotund jetty]

We have $$K sin(kt) = \frac{1}{(k + 1)^2}\cos((k + 1)t)$$ and $$K \cos(kt) = -\frac{1}{(k+1)^2}\sin((k+1)t)$$

[untold delta]

L

[patent fjord]

yeah then the kernel reduce to K/(k+1)^2, but i dont see the steps yet

[patent fjord]

what is K ?

[rotund jetty]

$K : L^2(M) \to ?$ is the operator $Kf(x) = \int K(x,y)f(y),dy$

[untold delta]

L

[patent fjord]

ahhh i see

[patent fjord]

i just end up with an integral $\int_{0}^{2\pi}\sum_{k=0}^{\infty}\frac{1}{(k+1)^2}(cos((k+1)s)sin(kt)-sin((k+1)s)cos(kt))(\frac{a_0}{\sqrt(2\pi)}+\sum_{n = 1}^{\infty}(a_ncos(nt)+b_nisin(nt)))dt$ but dont really see where to go here

where $u(t) =(\frac{a_0}{\sqrt(2\pi)}+\sum_{n = 1}^{\infty}(a_ncos(nt)+b_nisin(nt)))$

[untold delta]

Rottawarlock/Rottamage

Rottawarlock/Rottamage

[rotund jetty]

essentially you swap integration and summation

[rotund jetty]

In the end you will need to know where $K$ sends each basis element $1, \cos(kt), \sin(kt)$.

[untold delta]

L

[patent fjord]

yeah it basically reduces to a question what is an integral of $\int_{0}^{2\pi}sin(tk)cos(nt)dt$ and $\int_{0}^{2\pi}sin(nk)cos(kt)dt$

[untold delta]

Rottawarlock/Rottamage

[patent fjord]

lol i over thought the integral

[patent fjord]

as a result i get $-a_0\sqrt{2\pi}sin(s)$ which is odd something is missing

[untold delta]

Rottawarlock/Rottamage

[patent fjord]

this is odd shouldnt the operator map 1, cos and sin to zero? atleast i get zero

[rotund jetty]

no. Fix $j \in {1, 2, \dots}$. Then by using the orthogonality relations, $$(K \cos(jt))(s) = \int_{0}^{2\pi}K(s, t)\cos(jt),dt = -\frac{1}{(k+1)^2}\sin((j + 1)s)\pi$$

[untold delta]

L

[patent fjord]

but integral from 0 to 2pi of $cos(jt)(sin(tk)-cos(tk))$ with respect to t is zero even if j = k

[untold delta]

Rottawarlock/Rottamage

[rotund jetty]

No, $\int_{0}^{2\pi}\sin(kt)\cos(jt),dt = 0$ for all $j, k \geq 1$.

[untold delta]

L

[rotund jetty]

you can easily see it is 0 over -pi to pi since the integrand is odd

[patent fjord]

but $\int_0^{2\pi}cos(jt)cos(kt)dt$ is also zero for all $j,k \geq 1$

[untold delta]

Rottawarlock/Rottamage

[rotund jetty]

$\int_0^{2\pi}cos(jt)cos(kt)dt = \pi \delta_{jk}$ for $j, k \geq 1$.

[untold delta]

L

[patent fjord]

oh man i really need to check my computations

[rotund jetty]

It's easy to do these using complex exponentials

Namely $(e^{ijt}, e^{ikt}){L^2([0, 2\pi])} = 2\pi \delta{jk}$, $j, k \in \mathbb{Z}$

[patent fjord]

i used Product-to-sum identities

[rotund jetty]

sure that's basically the same

[patent fjord]

1/2(cos(jt-tk) + cos(jt+tk))

[untold delta]

L

[patent fjord]

yeah i made a silly mistake in my calculations..................... namely cos(0) = 0 (which isnt not true) xDDDDDDDDDDDDDDDDDDDD

ok i see the result now .D

😄

thanks @rotund jetty

[rotund jetty]

If $M$ is a compact smooth Riemannian manifold of dimension $d$ and $s > d/2$, what is the closure of the span of ${\delta_{x} : x \in M}$ in $H^{-s}(M)$?

[untold delta]

L

[rotund jetty]

I think it's all of $H^{-s}(M)$?

[untold delta]

L

[astral vine]

I think you are right

Because those spaces are Hilbert spaces

so assume you have an H^{s} function u that vanishes on all the dirac masses

then u(x) = 0 for all x by definition

thus u =0

the orthogonal of dirac masses is the null space and then the dirac masses are dense in H^{-s}

[rotund jetty]

Given a Hilbert schmidt self-adjoint operator $T : H \to H$, and an orthonormal set $\psi_1, \dots, \psi_n$, is there a way to bound $\sum_{j = 1}^{n}|(T\psi_j, \psi_j)|^p$ by something like $\sum_{k}|\lambda_k|^p$, where ${\lambda_k}$ are the eigenvalues of $T$?

[untold delta]

L

[lilac barn]

If you can do this when p =2 then you can do for any p>=1 as on finite dimensions all norms are equivalent

[rotund jetty]

I want to do it for $p < 2$ because I am sending $n \to \infty$.

[untold delta]

L

[rotund jetty]

that equivalence of norms comes with a constant that is too large for me when $p = 2$.

[untold delta]

L

[rotund jetty]

I got $\sum_{m = 1}^{\infty}|(T\psi_m, \psi_m)|^p \leq \sum_{j}|\lambda_j|^p$

[untold delta]

L

[proper knot]

is the space separable?

[vestal cypress]

Hi!
I am interested in Finite Element Exterior Calculus (FEEC), which is a finite element theory that it formulated in terms of differential forms and exterior calculus.
I'm programming a FEEC library for solving PDEs in rust, that supports arbitrary spatial dimensions.
If you know anybody about one of these topics, then come talk to me about it :)

[rotund jetty]

yeah separable Hilbert space $H$.

[untold delta]

L

[proper knot]

ok then I got the same

[patent fjord]

do there exists cacciopoli inequality for a case when $\Delta(u) - Vu = 0$ where $V$ is some bounded function

[untold delta]

Rottawarlock/Rottamage

[patent fjord]

i tried to multiply the equation by $\theta^2u$ st $\theta = 1$ on $B(x,r)$ and $supp\theta$ is in $B(x,2r)$ and integrate by parts over the set $\supp \theta$ i obtain $\int_{B(x,r)}|\nabla u|^2dy \leq M\int_{B(x,2r)}|u|^2dy - \int_{supp \theta}u<\nabla (\theta)^2, \nabla u>dy $

[untold delta]

Rottawarlock/Rottamage

[patent fjord]

i want to obtain $\int_{B(x,r)}|\nabla u|^2dy \leq 2M\int_{B(x,2r)}|u|^2dy + \frac{c}{r^2}\int_{B(x,2r)}|u|^2dy$

[untold delta]

Rottawarlock/Rottamage

[patent fjord]

on the another hand we could just integrate over the set $B(x,r)$ and integrate by parts and obtain $\int_{B(x,r)}|\nabla u|^2dy \leq 2M\int_{B(x,2r)}|u|^2dy$ and then add the term $\frac{c}{r^2}\int_{B(x,2r)}|u|^2dy$ since its positive and we obtain the result, but this feels odd

[untold delta]

Rottawarlock/Rottamage

[patent fjord]

https://www.math.cmu.edu/cna/LectureNotesFiles/Keniglecture1.pdf its here 2nd page

[sand echo]

sorry this is all rough working im in a slight hurry @patent fjord

writing it in the weak formulation and then select the test function as they suggest, the key point is you can control the derivative of 'eta' by 1/r, and then the other key step is to apply youngs inequaltiy carefuly so you can rearrange the '|Du|^2' term onto the left hand side

hopefully its okay

[patent fjord]

ah i see, the weak formulation + "cut off" function, wasnt sure about the test function

[patent fjord]

i think u forgot 1/2 from the exponent when u apply the cs

since we dont know $<\nabla \theta, \nabla u> \geq 1$

[untold delta]

Rottawarlock/Rottamage

[sand echo]

so im applying it like $ab = (ta)(b/t) \geq 1/2 [ a^2 t^2 + b^2/ t^2] $

idk how latex works

not sure what you mean sorry

[patent fjord]

when you use the cs 1st time to obtain $<\nabla \theta, \nabla u>\leq|\nabla \theta| |\nabla u|$

[sand echo]

right, so whats the problem sorry

[untold delta]

Rottawarlock/Rottamage

[patent fjord]

there should be $|\nabla \theta|^{1/2}|\nabla u|^{1/2}$

[untold delta]

Rottawarlock/Rottamage

[sand echo]

when i write |v| it means sqrt(v1^2 +.. + vn^2)

| .| is eucldiean norm

[patent fjord]

ahhh i see

[quick pagoda]

Any good reference for the DG-ish stuff in DEs?

Afraid that’s not something anyone where I’m at really seems to work on to ask

[patent fjord]

at the page 2 i see the resulting limes when $\varepsilon \rightarrow 0$, but dont why we actually need the dagger inequality. i tried to apply DCT to switch order of integration and limes, but the dagger seems bit odd in that case.

[untold delta]

Rottawarlock/Rottamage

[patent fjord]

but somehow we need to justify switching order of integration and limes

[void flame]

In Strauss PDE Chapter 9 when trying to derive the 3D wave equation solution by method of spherical means, the author wrote that we can define v this way to use D'Alembert's formula. However I don't recognize this as being D'Alembert's formula, since it should only contain one integral, but this formula for v contains two integrals and the second integral I don't even recognize. Can someone please explain to me what's happening here?

[patent fjord]

my question is how do they apply the dagger at the page 2?

[void flame]

I have a question about some statements in Strauss, I've been rereading chapters 2 and 3 and I saw statements claiming the uniqueness of the solution to the diffusion equation on the whole line without proof or state the uniqueness in a theorem(he only proved the uniqueness in a rectangle in the book). He also uses this uniqueness in later proofs, which doesn't feel very rigorous. Here he assumed the uniqueness of Diffusion with a source, and I'm not sure if it's a rigorous statement or not, and it's bugging me a little bit....

And I saw this which seems to suggest that the solution is not necessarily unique, which is confusing: https://mathoverflow.net/questions/125166/wild-solutions-of-the-heat-equation-how-to-graph-them

[robust moat]

There's a free paper I haven't read so I have no idea if this will help you, but the title is convincing.

[void flame]

Wha

Thank you but... that looks a little more than what I was looking for

[robust moat]

Ahhhh, sorry.

I need to have a sleep. I'll look at it a bit later and see if I can provide any assistance.

[void flame]

Oki, thank u so much!

[robust moat]

You're welcome, but hold thanks. You never know how stupid I can be on any given day, 🤣.

[somber veldt]

Hello I am stuck on question 2 from this sheet. I’ve added my working but not sure where I’m going wrong. Please help me understand.

[quaint herald]

I will have a look at Strauss in a second, but I suspect that it is just a little bit of handwaving because the book is not overly theoretical.
The important thing to remember about heat equation uniqueness is that the "wild solutions" (i.e. nontrivial solutions from 0 initial data) all grow very rapidly in the spatial directions. So the uniqueness theorems are usually all the form "there exists a unique solution in the class of functions F" where F only needs to be restrictive enough to rule out things that grow very fast.

[void flame]

Oh, I see! I didn't really see any theorems stating the uniqueness so I agree it is a little handwavy. I'll search for more detailed results related to this. It's just a little confusing in the book because the author in fact uses this uniqueness to prove other things, without proving or stating the uniqueness of a certain problem properly.

[kind wren]

Hi. In Sobolev Spaces, for a rectangle $K = (a,b) \times (c,d)$, we know $H^2(K)$ is continuously embedded in $C^0(\overline K)$. We then have that there is $M > 0$ such that $||f||{C^0(\overline K)} \leq M ||f||{H^2(K)}$ for all $f \in H^2(K)$. My question is: are there good estimates for $M$ in terms of $a,b,c,d$?

[untold delta]

phao

[rotund jetty]

I found the result for the torus: https://www.ma.imperial.ac.uk/~jdg/JMP_BartGib_2011.pdf

dunno if the proof can be adapted for the unit square

[kind wren]

THank you

[serene mica]

Im trying to get from the first equation here to (2.2), and im struggling to do that. Clearly its enough to show that 2.2 with BAu instead of the commutator is equal to 0. I tried flipping to the Fourier side, but I ended up with a convolution I wasnt sure how to handle. And if I just keep it on the spatial side and apply integration by parts as well as the assumption u is divergence free, it also doesnt seem to help. So if anyone has any hints or suggestions for how to do this, Id very much appreciate it.

[untold delta]

Functionanatolysis

Functionanatolysis

[serene mica]

But why exactly? @astral vine

[lilac barn]

Recall a . (u . grad) a = 1/2 u . grad |a|^2. Then use integration by parts.

[buoyant lantern]

woww a lie bracket 😍

[astral vine]

Just a commutator, what's rely more on Lie theory is "u dot nabla"-stuff which stands for the Lie derivative following the direction given by u

[buoyant lantern]

oh wow

i thought the lie derivaitve was denotes as L_u(v)

[astral vine]

In this specific case both coincides

[serene mica]

Thanks!

[buoyant lantern]

thanks

not sure if this is pdes or diff geo but for the alembert wave equation, u_tt = c^2 div(grad(u)), how does one account for waves on a sphere

or does the equation apply there too

[brave bay]

It applies on any Riemannian manifold. div and grad have general definitions in terms of the metric tensor.

[mystic lynx]

not sure whether this or adv probability is the right channel, but can anyone explain to me bourgain's GG* argument?

[blazing ridge]

If you have a conservation equation $\partial_{t}\rho + \partial_{x}(\rho u ) = 0$ and you know that $\partial_{x}u < 0$ then heuristically this should mean that for any point x the function $\rho(t,x)$ is an increasing function of time. Does anyone know somewhere I can find a proof of this?

[untold delta]

M6LI

[blazing ridge]

I guess the idea behind the proof would be to assume that it does not hold and then show that mass conservation would be violated somehow.

[blazing ridge]

nvm got it

[inland sinew]

Are there any theorems about the existence of solutions for linear, second order PDE's with variable coefficients that are periodic? I am looking at problems with mediums that have periodic qualities(like stiffness), and these types of equations show up a lot.

[astral vine]

Periodic coefficients, but not necessarily periodic solutions, say ?

Or Periodic coefficient with periodic solutions ?

The first kind goes back to tools such as the Bloch Transform

Which allow to diagonalize somehow the operator to fall in the range of the second case

Which is easier to deal with.

See e.g. Bloch Transform.

See for instance Peter Kuchment's book.

[novel veldt]

What’s a good source on weak solutions in sobelev spaces

[hidden coral]

Evans is the standard reference I believe

[novel veldt]

Tysm

[inland sinew]

Thank you for the reply. I have slowly been trying to read up on the Bloch Transform, but I only have knowledge of basic fourier series/transform so its taking me a bit of time. I asked because I was looking at a PDE of the form. f(y)zxx + f(x)zyy = -Af(x) - Bf(y).

A and B are constants, and f is a periodic function that can be written as a cosine fourier series. I am seeking solutions for z that are periodic. Because the obvious solution to the problem is just to choose a quadratic function which can immediately satisfy the problem with the correct constants.

I found a solution to this general case based on just plugging in a 2d fourier series for z, and spending way too much time figuring out the coefficients, but there is something very special going on with the coefficients that made me realize there is probably something going on I don't understand and I have a hard time believing its not related to integral transforms.

[turbid scaffold]

Hey can someone tell me if I'm computing the commutator of this pair of operators correctly? I am not sure how to compose differential operators like LB because some terms of B have functions as coefficients

[buoyant lantern]

3rd partial derivatives

[turbid scaffold]

Are you talking to me? Can you tell if I computed LB or BL wrong?

[compact stag]

Does anyone have a good reference for optimal control (not the basic Linear Quadratic Regulator, Harmonic Oscillator basic stuff) without going through the whole physics stuff
The goal is to eventually get to stochastic optimal control

[buoyant lantern]

not sure

maybe try oing [B,L]..?

[rotund jetty]

what is $\frac{d}{dt}L$?

[untold delta]

L

[rotund jetty]

is it composition?

if so, your $\frac{d}{dt}L$ is missing a term from the product rule. Look at $\frac{d}{dt}Lf$ for smooth f

[untold delta]

L

[kind olive]

Is u a function of the variable x, or of two variables t and x ?

[kind olive]

I did the computation, and got a plus on the orange and blue terms of BL

but otherwise same

(if you look at the signs, BL and LB are both with only pluses since L and B can both be factored by the - sign)

[wind mortar]

where can I find this? is it in a book? so U am guessing this is the s power of the expected value of the del operator?

[astral vine]

wdym ?

[untold delta]

Zanarcane

[lilac barn]

This book is Intro to navier stokes by Vlad Vicol and the angle bracket here is the Japanese bracket. In particular, this operator is defined by the Fourier symbol (1+|x|^2)^s/2

[low turret]

That's his book with Jacob Bedrossian, right?

[shadow topaz]

I have a system of differential equations : u'+v=f(t) , v'-u=g(t)
(u(0)=0 , v(0)=pi/2)

So I let Y=(u , v) , and get to solve Y'=((0 1) (-1 0)Y +(f(t) g(t) ) (Matrix and Vector)

For the homogeneous part I get Yh(t)=(acos(t)+bsin(t) , -asin(t)+bcos(t) )

To this point , I wanna use the "constant variation" method, but I can't seem to figure out how to do so.

In 1D , the method is to let Yp(t)=z(t)Yh(t) , to plug in this Yp in the original equation, then solve for z'(t) and get back to Yp. Any Idea/hints ?

[astral vine]

#odes-and-pdes

[vapid apex]

@knotty river I forgot to mention yesterday: There is so far only one computer algebra system (CAS) for Lie point symmetries (the method in those books) I know of, which is a module in Maple

https://www.maplesoft.com/support/help/maple/view.aspx?path=dsolve%2FLie See the navigation hierarchy bar here

Sadly Maple is not available for a lot of people. It would be extremely nice to have a FOSS CAS for this, say in Python, SAGE, or Julia

Might bring some popularity to the method too

[knotty river]

Will check it out

In the way of CAS generally I’ve only used Mathematica so far

And some matlab

[knotty river]

Makes sense yeah

[vapid apex]

Well, I think more people have Mathematica/MATLAB access than have Maple

[knotty river]

yeah def

[reef tartan]

Has anyone seen works on FEM stability analysis that takes the influence of a curved mesh into account?

[misty moss]

How do u get the bottom one from the above mesh

[hidden coral]

this should be in #numerical-analysis, but the idea is that $\Delta u(x_i, y_i) \approx -4u(x_i, y_j) + \cdots$, so if you replace $u$ with $\Delta u$ and expand usng the approximation again you get the bottom mesh

[untold delta]

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

[misty moss]

Oh sorry I wasn't sure of where to post it

[devout arch]

Hi,
I am trying to solve the Linear Stability Theory of a fluid flow given an input of the base flow. I have done the simulations and extracted the flow. A breif overview of the LST is:

1. Linearise the navier-stoke equations
2. decompose the variables into the base flow and fluctuating component. Apply the wave decomposition to the fluctuating component and substitute it into the linearised navier-stoke equations
3. The resulting system can be simplified into a second order ODE: [AD^2 + BD + C]*q =0; where A, B, C are 5x5 matrices and q is a vector of the variables in the navier stoke equations: q = [u,v,p,T,w].transpose --> 5x1 matrix
4. The boundary conditions of the ODE are: u(0) = v(0) = p(0)-1 = T(0) =w(0) =0 and u(inf) = v(inf) = p(inf) = T(inf) = w(inf) = 0

to solve this system requires an eigenvalue problem (EVP) being created: Lq = lambdaq; where L is the operator [AD^2 + B*D + C]; q is the same vector but represents the eigenvectors of the solution and lambda is the eigenvalues

I am stuck on how to solve the EVP, given that there is an ODE inside it? I have applied a Fourier transform to the L operator but an stuck on how to apply the boundary conditions and solve the EVP. a tricky part is that the matrices B and C are functions of the wavenumber (which are the output), so i have to define an initial gues for them and solve the solution then apply an interative technique to get the final answer. Can someone help with this?

Thanks

[frank tide]

i’m not familiar with what your doing at all but i’m curious, do any predictor-corrector type methods exist for that last part you mentioned

[devout arch]

Im not sure what you mean. currently my initial guess for them is 0, then work from there

I initially didnt have an iterative method, but realised that was wrong.

there is a normalisation condition that needs to be met, which helps with the iterative process

for convergence

ps if I dont reply nudge me with a dm

[frank tide]

in numerical methods for odes there’s predictor-corrector type methods where usually the predictor is an explicit method and the corrector is an implicit method so i was curious if one existed for something like this

so like for eulers method you “predict” with
$$y_{k+1}=y_k+hf(t_k,y_k)$$
(forward euler) and then correct with
$$y_{k+1}=y_k+hf(t_{k+1},y_{k+1})$$
(backward euler)

[untold delta]

ansh

[frank tide]

no idea if something like that exists for what you’re working on

i suppose it’s similar to what you have but instead of the initial guess being arbitrarily 0 you could maybe do better if there’s some sort of explicit formula for it

[devout arch]

I was reading this (https://github.com/ElsevierSoftwareX/SOFTX-D-22-00069/blob/main/src/PSE.py) open-source code which doing something similar, but solving the exact stability equations. To do so it first solves the LST which an input to the PSE solver. In the solve function (line 1305), it has this comment:
# Description
# Call the relevant function to build, apply BC, compute
# NLT terms and solve ths system. The marching procedure
# performed using backward differentiation formula (BDF)
# The Normalization condition is enforced locally, at every
# station.

[gritty jackal]

I know very little about pdes and came to the following problem from generating function land, hoping y'all can lend some insight. I have a function $F(x, y, z)$ of three variables and I know that:

\begin{enumerate}
\item $F$ is symmetric in the inputs,
\item $F(x, y, 1) = xy$,
\item $\frac{\partial F}{\partial z} (x, y, 1) = \frac{xy}{x + y - xy}$.
\end{enumerate}

I would like to at minimum conjecture some possible shapes of some $F$s which satisfy these conditions; I don't expect them to be enough to pin it down. Anyone have any ideas?

Bonus points if, for each $m$ and $n$, $\frac{\partial F}{\partial y^m\partial z^n} (e^x, 1, 1) = \sum_i b^{(m,n)}_ie^{\lambda^{(m,n)}_i x}$ (where all of the $b$s, the $\lambda$s, and the range of the sum depend on $m$ and $n$), although getting some candidates which satisfy the first three would already be really very helpful. Thanks in advance :).

[untold delta]

every place I ever loved

[last cloud]

do you think we can create a large language model only trained on analytically solving differential equations

[bronze gate]

#chill

[rotund jetty]

Does mathematica count?

It's not LLM though

[chrome trout]

How can I compute the weak formulation for $\textup{div}(A \nabla u) = f$ where $u$ is the variable function and our operator then reads $L := \textup{div}(A \nabla \cdot)$. Here, $A$ is a smooth matrix of suitable dimension

[untold delta]

мир

[waxen bobcat]

Multiply with a test function, integrate over the domain, and use integration by parts

[chrome trout]

yes

I know that steps

the author gets this

And I have $\textup{div}(A \nabla u)$ which is different from $(\nabla A , \nabla u) + A \Delta u$ since $A$ is a matrix and not a function

[untold delta]

мир

[chrome trout]

ot is it the same?

I could also try to just use the divergence theorem on $\int \textup{div}(A \nabla u)\phi$ but this would be $\int_{\partial \Omega} (A \nabla u) \phi$??

[untold delta]

мир

[waxen bobcat]

\[\divg(A \grads{u} \phi) = \divg(A \grads{u}) \phi + \angles{A \grads{u}, \grads{\phi}}\]

[chrome trout]

Why did you put v inside?

is your v the test function?

[waxen bobcat]

Yeah

[untold delta]

Zanarcane

[chrome trout]

I see..

thank you

[chrome trout]

I think I forgot the answer but can we find an explicit solution, provided all the necessary conditions on the functions, for the Poisson equation with reaction term? $-\Delta u + \lambda u = f$

[untold delta]

мир

[chrome trout]

I mean, this looks like an inhomogeneous Sturm-Liouville problem so it might be possible using Green's function

Yeah.. the solution is $u(x) = \int G(x,\xi)f(\xi)d\xi$.

[untold delta]

мир

[chrome trout]

It's a matter of finding G associated with L = -Delta + lambda

[lilac barn]

Evans page 320 example

[astral vine]

the explicit Kernel is actually super ugly and goes back to special function afair

Hankel functions

(not sure, I have to check)

However the Fourier side is nice

another nice form is

[untold delta]

Functionanatolysis

[tribal breach]

Hello, basics of pde and which level do you study them and list of books along with their authors name...

[astral vine]

Read Evans' book. Then near the end of move on Brezis' book.

[hidden coral]

is there anything that Brezis' book covers that Evans doesn't cover?

[quaint herald]

I'd regard Brezis more as a functional analysis text with a view towards PDE than a PDE text. Both cover plenty the other does not, and the emphases are quite different. I wouldn't say one naturally follows the other sequentially.

I haven't thoroughly looked at Evans for a while so I cannot recall exactly how much FA is covered, but much of it is probably relegated to the appendices without proof, e.g. spectral theory.

[frank tide]

are semi group methods a common approach for solving pdes?

noticed that evans has a section on it but haven’t seen it pop up in many books

[mint canyon]

I looked into it and the method does not seem terribly practical, there were multiple variants of it though, so maybe one of the other ones is more generally applicable.

the one I looked at required pdes of the form:

$$ \partial_t u = D(x) u $$

[untold delta]

x17

[mint canyon]

where D(x) is some linear differential operator in x

[lilac barn]

Brezis does everything in a more theoretical fashion whereas Evans does it in a more practical fashion.
This ends up meaning that Brezis becomes easier to read (both notationally and theoretically) whereas Evans is hard to read. On the other side, PDE in Brezis starts at chapter 7 whereas in Evans you get to it on the getgo.
The right order should be to read Brezis then Evans unless you're studying for a PDE course in which case Evans.

[hidden coral]

I'm not sure what is meant by practical fashion, because Evans also uses a lot of functional analytic tools in part 2 of the book

[lilac barn]

They're one of the first immediate methods you would do on a linear pde if your PDE admits a nice semigroup operator.
Then any further analysis of the linear PDE can become just the analysis of the semigroup operator. In case of nonlinear (which is where the most magic happens) is a different story.
I would recommend to check out the one-parameter semigroup book.

[frank tide]

i see thanks! i’m not too familiar with them but are they related to group theory at all in that they use methods from it to solve pdes or is it called semigroup theory simply because you utilize a semigroup operator

[lilac barn]

By a practical fashion, I mean you're doing FA for the sake of PDE instead of doing FA for it's own sake like Brezis. This ends up meaning that you dont "waste" time but also ends up convoluting the picture a little (for example, some proofs become more convoluted like the Lax-Milgram proofs in Evans or even the handling of multiple inner-product on subspaces in Galerkin part)

[frank tide]

also are there any topics which are a full on intersection of gt and pdes?

or abstract algebra more generally

[lilac barn]

Not to mention, doing FA for it's own sake implies you end up becoming a lot more comfortable with evaluation and norm notation which can make reading proofs much easier than keeping track of integrals.

[lilac barn]

It's semigroup theory only precisely because the operator is a semigroup when regarded as a function of time. I don't think you particularly need techniques from group theory. That said, I won't be surprised if someone did end up using techniques from gt to solve some niche problem in PDE.

The nice thing of having a semigroup is that you can define a "derivative" of that operator and from there, it's a one-line away from solving an evolution equation.

[chrome trout]

Imagine I have domain $\Omega \subset \bR^N$ with boundary $\Gamma$. Suppose I have a diffeomorphism $\Phi(\cdot,\cdot) : [0,T) \times \overline{\Omega} \mapsto \bR^N$. Then $\Omega_t = \Phi(t,\Omega)$ and $\Gamma_t = \Phi(t,\Gamma)$. In particular, if I have a Dirichlet problem on $\Omega_t$ then $u_t = u(\Phi(t,x)) = 0$ for each $x \in \Gamma$. I want to differentiate this to $t$ but am I getting $u'\frac{\partial \Phi}{\partial t} = 0$, whereas the book gets $u' + \nabla u \frac{\partial \Phi}{\partial t} = 0$ on $\partial \Omega$.

[untold delta]

мир

[lilac barn]

Since Phi is a homeo in Rn so the right form is u(t,Phi). Then taking the partial with respect to t should give you the right equation

[chrome trout]

trueee

thanks

[lilac barn]

Yw!

[chrome trout]

the notation u_t(phi) let me in mistake

thinking it was just a function of phi

[ocean ether]

what happened from 11 to 12

how did evans end up with this inequality

[hidden coral]

this looks like holder inequality

[ocean ether]

oh it is

thx

[ocean ether]

my guess is that C is a typo then?

because $\norm{D^2f}_{L^\infty(\mathbb{R}^n)}}$ should be $C$

[untold delta]

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

[ocean ether]

as f is twice cts differentiable and therefore it's bounded by some C

[chrome trout]

The supremum is over R^n whereas the integral is over the ball. Might be it?

[ocean ether]

yeah

[chrome trout]

So yeah, if the supremum is outside of the ball and is bigger then C > 1?

[lilac barn]

He is taking the supremum out but laplacian is the sum of same partials whereas D2f is the l2 norm of the second partials. You need a Holder inequality to convert the l1 norm into l2 norm which brings out the C.

[tribal breach]

Thanks...

the basics are the math tables. then move to algebra. when you solve these equations, taught in eight class, here, in your mind to get that initial point we have to graph it. this gets the approximate answer. then using the usual methods we can get the exact answer.

this the pathways to them. Math. tables. algebra. calculus. ode and pde. simple plot them. and geometry is a great application of them. even architecture. a computer program helps. geometry. point, line, triangle, square, rectangle, curves and planes and polygons. whole maths right here in five minutes. do tell more. bye...

physics, chemistry and even biology. With maths...

for example to solve a quadratic equation, we have to get root of it. root means that a point which fits in the solution. like y=f(x). a basic solution is to keep x at zero then find that answer by substitution of zero in fx. 0,x gets it. this a generic solution. if the equations if plotted does not intersect with any axis. then we have to use b2-4ac methods. again more solutions to it. applications are when we have to find an answer when any two given lines intersect. or in linear algebra we have to plot multiples equations or lines and get the optimal answer. by finding the intersection of system of lines. simplex method is another example of these. and a number of solutions. for each situation. bye...

Hi there! Let me introduce my self here. I am a BCS or Bachelor of computer science graduate. from University of Poona, maharashtra. India. and you are welcome.

[mint canyon]

This is not the channel to talk about elementary ODEs and PDEs, we have another channel #odes-and-pdes

[ocean ether]

ok so the C in the middle part of the inequality is a typo

(i asked my professor)

it's just a regular application of holder

[hidden coral]

what cocat is saying is that $\Delta f = \sum \partial_i^2 f$, whereas $\norm{D^2 f}{\infty} = \sup \abs{D^2f} = \sup \qty(\sum{i, j} (\partial_i\partial_j f)^2)^{1/2}$, and the latter does not necessarily bound the former.

[untold delta]

Alphyte

[hidden coral]

now by equivalence of norms there does exist some C such that the L^1-norm of D^2f is bounded by C * L^2-norm of D^2f

so I believe the C is not a typo here

[ocean ether]

based on the errata uhh

https://math.berkeley.edu/~evans/errata.PDE.second.ed.1.pdf

no

i think i had to bring up that $f$ is in $C^2$?

[untold delta]

the breadwave

[ocean ether]

oh wait

i think i see what you mean?

it's 100% an error though

[hidden coral]

the C² thing?

or the constant

[ocean ether]

the constant

[hidden coral]

wdym

[ocean ether]

the constant in the middle part of the inequality is a typo

[hidden coral]

you need the C to bound $\abs{\Delta f} \le C\norm{D^2 f}$ though

[untold delta]

Alphyte

[ocean ether]

why are we even in the L^2 norm

[hidden coral]

cuz that's the convention

[ocean ether]

H^2?

[hidden coral]

nvm ignore me

just woke up

[quaint herald]

It is not an important choice at this stage. Any choice of p other than 2 would also give you an equivalent norm.
This leads to another point, the leading constants in these inequalities are not important and not something you should get hung up on. The important thing is that there IS such a C giving you a bound of this form.

So sometimes C will get written whether or not it needs to be there.

[lilac barn]

Although I'll say that if it's confusing you then you should try to figure it out because the techniques involved are usually elementary and it'll help when the same technique is used elsewhere.

Once you see/figure it out a couple of times, the Cs will become second nature and you won't get bothered.

[ocean ether]

i think i should read the appendix before i move on

[quaint herald]

At least what the notations mean, certainly.

[untold delta]

cocat

cocat

[ocean ether]

yeah i thought D^2f was still in L^1

[lilac barn]

A good reason to keep the (\ell^2) notation is because the integration by parts formula is nicer:
\begin{align*}
\langle D^2 f, D^2 f \rangle &= \int \lvert D^2 f \rvert^2 = \int \sum_{i,j = 1} (\partial_{ij} f)^2 = \sum_{i,j = 1}^n \int \partial_{ij} f \partial_{ij} f \
&= \sum_{i,j = 1}^n \int \partial_{ii} f \partial_{jj} f = \langle \Delta f, \Delta f \rangle .
\end{align*}
If we used the (\ell^1) notation then
\begin{align*}
\langle D^2 f , D^2 f \rangle &= \int \left(\sum_{i,j = 1}^n \lvert \partial_{ij} f \rvert \right)^2 = \int \sum_{i,j,k,l = 1}^n \lvert \partial_{ij } f \rvert \lvert \partial_{kl} f \rvert
\end{align*}
and you then don't have an easy conversion between the two.

[ocean ether]

so what gomez said

[untold delta]

cocat

[ocean ether]

it all works out regardless of what l^p norm im in?

[lilac barn]

The inequalities will work out due to norm equivalence. My point is about why do we fixate on choosing p=2 than some other p-value.

[ocean ether]

ah ok

thanks all

[chrome trout]

what are examples of linear and nonlinear effects in this sense?

[lilac barn]

Linear effects are when small initial data gives you small effect. But the soap film oscillations gives you large amplitude even though the thickness can be quite small or quite large showing that the amplitude depends nonlinearly on the thickness

[ocean ether]

do we even need smoothness to prove this?

[quaint herald]

I assume you are still reading Evans, and that the definition in this exercise of subharmonic is being C^2 with non-negative Laplacian (there are more general defs).

Indeed you should be able to complete this exercise for just C^2 \phi.

[tired hollow]

Could someone who specializes in the pure and applied PDEs to tell me about the beauty of the pure and applied PDEs?

[quaint herald]

Pure and applied PDEs are beautiful.

[tired hollow]

yeah? why?

in your knowledge

[astral vine]

Tbh Applied is more like a bet you may lose

Like there are a lot of wonderful things. But hell a lot of ugly things too.

[hidden coral]

the world is described with PDEs

[tired hollow]

could you explain more?

[tired hollow]

I'd like to have much details as possible especially from people who are into PDEs

sorry if that's too much to ask

your sentence is so deep and pretty interesting tbh, it made me so curious to know why you have said that @hidden coral

[quaint herald]

I agree, I just didn't feel like properly engaging with such a broad/vague question without context.

[mint canyon]

@tired hollow Any system you can think of can be decomposed into a set of attributes represented by functions/fields. The natural behavior of these fields is to change. They change with time, they change from external forcing, they change with position, and they change in response to changes in other attributes of the system. A PDE is a mathematical description of how fields or systems of fields change with respect to each other and their independent variables. It describes how some initial conditions and boundary conditions will evolve over the life of the system, and there are as many solutions as there are initial and boundary condition configurations. (So infinitely many)

[lilac barn]

This belongs in #math-discussion

[tired hollow]

This is pretty good explanation, I highly appreciate it. you made things clear and easy to understand. thanks

[tired hollow]

okay, my bad

[ocean ether]

wtf are green's functions

[hidden coral]

fundamental solutions but with boundary conditions

[ocean ether]

ok better question to ask

why exactly do we call fundamental solutions "fundamental"

why is $\Phi(x):=\frac{-1}{2\pi}\text{log}|x|$ of particular interest

[untold delta]

the breadwave

[hidden coral]

if G is the fundamental solution to some linear PDO L, then the PDE Lu = f is solved by u = f * G, where * is convolution

the easiest way to see this is like electrostatics

coulomb's law tells us the electric field induced by a point charge (i.e. dirac delta fn)

[ocean ether]

so are all linear PDEs easy to solve because of convolutions?

[hidden coral]

then, to find the electric field induced by some distribution of charge you essentially integrate coulomb's law across the distribution

[hidden coral]

no

[ocean ether]

why not?

[hidden coral]

for one, finding a fundamental solution is pretty difficult for most PDEs that even admit one

[ocean ether]

ah

ok

we just know poisson super well because of laplace then?

[hidden coral]

well what is the difference between poisson and laplace

[ocean ether]

because of the fundamental solution

is what i mean

[hidden coral]

yeah

[ocean ether]

wait this is just like the fundamental solution in cookbook ODEs

[hidden coral]

not sure what the fundamental solution in cookbook ODEs is

if it's "impulse response" then yea

[ocean ether]

something something 5y''+2y'+y=0 something something use it to solve 5y''+2y'+y=e^x

[hidden coral]

that's homogenous solution

I think

[ocean ether]

so is the concept similar

use homogenous solution to find nonhomogenous solution

[hidden coral]

not really

you are using the solution for the dirac delta and linearity to find a nonhomogenous solution

[stone rampart]

Hi

[mint canyon]

has anyone used methods associated with exterior differential systems of PDEs? How general are they for linear equations and what are their limitations?

[astral vine]

Lol

[astral vine]

By exterior differential system you mean PDEs on manifold which obeys certain additional structure ?

If this is the case, I can answer: even doing the easy cases properly in other spaces than L² can be an absolute nightmare to write down carefully. The same goes even on smooth domains or the whole space.

[mint canyon]

@astral vine Im still very early on in learning about it, but the decomposition of the PDE into a system of differential forms is called a differential ideal and solutions are the integral manifolds of the differential ideal?

[orchid wraith]

I'm no expert in this field, but I think I can speak a bit to why these algebraic considerations are desirable. The overall objective is to gain topological insight into manifolds associated to PDE's.

Consider for example an oriented, Riemannian manifold $M$.

[untold delta]

TennisSteve42

[orchid wraith]

At each level of the exterior algebra, the space of $k$-forms decomposes by $\Omega^k(M) = \text{im}d \oplus \text{im}d^* \oplus \ker \Delta$ where $\Delta =dd^*+d^*d$ is the Hodge Laplacian.

[untold delta]

TennisSteve42

[orchid wraith]

Observe that the differential ideal associated to the equation describing harmonic forms ($\Delta \omega = 0$) is precisely the set $I = {\Delta \omega , | , \omega \in \Omega(M)}$.

[untold delta]

TennisSteve42

[orchid wraith]

Hence, at each level of the exterior algebra, the decomposition $\Omega^k(M) = I_k \oplus \ker \Delta$ takes place, where $I_k$ denotes the set of $k$-forms in $I$.

[untold delta]

TennisSteve42

[orchid wraith]

Now here's the kicker (connection to topology): the harmonic $k$-forms correspond to de Rham cohomology classes. Hence, $\ker \Delta \cong H_{dr}^k(\Omega)$ for each $k$.

[untold delta]

TennisSteve42

[orchid wraith]

Therefore, $\Omega^k(M) \cong I_k \oplus H_{dr}^k(M)$.

[untold delta]

TennisSteve42

[orchid wraith]

Thus, this differential ideal $I$ is somehow capturing all of the topological information that the de Rham cohomology groups capture about $M$, because it is complementary to these groups in a sense.

[untold delta]

TennisSteve42

[orchid wraith]

More generally, it now seems reasonable to expect that the differential ideals that you build from PDE's would have something to say about the topology of the corresponding integrable manifold that you mention.

And most importantly, the fact that this is an algebraic structure allows one to study solution sets of PDE's up to diffeomorphism. That is, invariantly.

Also, I forgot to explicitly mention above that $I \cong \text{im}d \oplus \text{im}d^*$.

[untold delta]

TennisSteve42

[orchid wraith]

Hope this helps! Also, if anyone more experienced can add to this example or connect it more explicitly to the PDE construction which op mentioned that would be cool 🙂

[astral vine]

Useful for Fluid Dynamics and Electro-magnetism

Like the Leray projection, fondamental operator for the study of incompressible fluids

is canonically the same has the Hodge projector on N(d*).

However, the approach given from geometry people only is not that much useful in this case since the decomposition holds only in the Hilbertian sense

and for non-linear PDEs one needs to use Lp- based Sobolev spaces

On the whole space R^n, there is no trouble but the decomposition fails to be topological on Lp for p=1 and p=infty.

But there is a weird issue on the whole space R^n for the decomposition u =dv+d*w, for u in Lp(R^n), 1<p<+\infty

v and w CANNOT be in Lp as well, but dv and d*w are.

However, for smooth manifolds/domains with smooth boundary

everything is fine

the problem is then what happens if we lower the regularity of the boundary, and does it happens for higher order Sobolev spaces.

[orchid wraith]

Also, I know that sometimes it is desirable to study solutions of PDE's up to a group action. If the space has compact quotient under the action, one can study the quotient instead.

[astral vine]

This is another thing I am not that much aware of

[orchid wraith]

This is an approach developed by Atiyah to build cohomology groups for non-compact manifolds (dependent upon a group action) using L^2 harmonic forms.

But the machinery here is weird, because one needs von Neumann algebras and von Neumann dimension to define the size of these new cohomology groups. This leads to non-integral betti numbers (even irrational in some cases).

[astral vine]

"non-integral betti numbers" bro wtf

[orchid wraith]

Yeah, it's kind of messed up, and it doesn't generalize to other PDE's other than the equation $\Delta \omega = 0$ to my knowledge. So it's more a topological tool than anything else.

[untold delta]

TennisSteve42

[orchid wraith]

Actually, I think he develops something for elliptic operators in general, but I don't know much about that.

[mint canyon]

@orchid wraith I've read and appreciate your responses, I unfortunately know nothing about topology beyond the standard "donut and ball are topologically different" example

well, I guess I did a bit with homotopy transformations too, but that was all purely applied math stuff, there is a PDE approximation technique called HAM.

based loosely on Gromov's H-Principle

I know my geometry a bit better but it mostly comes from GR background, and Im familiar with forms and Hodge duals etc

[mint canyon]

Okay, I roughly get the idea of a chain complex for a Homology, and the 2nd exterior derivative always gives 0 is where your composition of sequential groups maps to 0 in the deRham cohomology etc. Is it a cohomology instead of a homology just because forms lie in the cotangent space of the manifold?

[orchid wraith]

Ah, no worries at all. Here's a quick crash course on de Rham cohomology. I have a bit of a backwards understanding myself, because I know very little about ordinary homology.

Recall the fundamental theorem for line integrals in calc III, which says that if you integrate a conservative vector field F$ around a closed loop, you get zero.

[untold delta]

TennisSteve42

[orchid wraith]

This is a sort of 3d fundamental theorem of calculus, because as a consequence of this, the integral of $F$ over any path is just the difference of its values at the endpoints.

[untold delta]

TennisSteve42

[orchid wraith]

When your space has a hole in it, the fundamental theorem of calculus fails.

Consider $R^2$ without the origin.

[untold delta]

TennisSteve42

[orchid wraith]

Then the integral of $-y/(x^2+y^2)dx + x/(x^2+y^2)dy$ around the unit circle is not vanishing. And yet on $R^2 \backslash {(0,0)}$ this differential form is smooth.

[untold delta]

TennisSteve42

[orchid wraith]

You can think of de Rham cohomology groups as just the vector space consisting of differential forms like this, whose line integrals along specific paths are nonvanishing.

In this sense, the de Rham cohomology measures the extent to which the fundamental theorm of calculus fails on your manifold.

So, on $R^2$, the de rham cohomology is trivial, because the fundamental theorem of calculus holds.

[untold delta]

TennisSteve42

[orchid wraith]

On $R^2 \backslash {(0,0)}, the de Rham cohomology is 1-dimensional, because of the above example, and the fundamental theorem of calculus fails.

In general, if you poke $k$-holes in the plane, the de Rham cohomology will be $k$-dimensional, because there will be a form like the above for each hole.

[untold delta]

TennisSteve42

[orchid wraith]

This is only the 1st de Rham cohomology, but it generalizes. Observe that above I just used 1-forms, which can be integrated over paths. In general, the same idea holds for higher de Rham cohomology groups, you're just integrating $k$-forms, and searching for higher dimensional holes in your space. For example, if you consider a ball with a small ball at the center chopped out, any path can still be shrunk to a point in this "3d annulus". But a 2-form, integrated over the "shell" surrounding the hollow center, can detect this 3d hole.

[untold delta]

TennisSteve42

[orchid wraith]

To sum up, the basic idea is that you're measuring how badly the fundamental theorem of calculus fails on your manifold to get an idea of how many holes your manifold has.

I think there is some connection to group cohomology which arises from passing from a group to its classifying space. I don't know enough about it to say though.

[orchid wraith]

Ah, sorry I didn't account for this in my exposition. The definition of the $k$-th de Rham cohomology group is take the closed $k$-forms ($d\omega = 0$) modulo the exact $k$-forms ($\omega$ is $d$ of something). In the calc III example I gave above, the form in question is closed but not exact, and thus constitutes a nontrivial element of the first de Rham cohomology.

[untold delta]

TennisSteve42

[mint canyon]

This is slight off-topic, but if you consider a manifold with no holes and homotopy transform it the limit it pinches off into having a hole, then take a manifold with 1 hole and homotopy transform it the limit the hole disappears (basically approach the same "inbetween" manifold from the 2 different topologies) what does that do to the deRham cohomology? Is there some kind of critical point between simply connected and 1 hole?

[orchid wraith]

Smooth homotopies $f$ should induce linear maps $H_{dr}^k(f(M)) \rightarrow H_{dr}^k(M)$ via pullbacks (note the direction of the arrows). So I suppose one could introduce a hole, because then the de Rham cohomology could become larger. However, if two manifolds are homotopy equivalent, then their de Rham cohomologies should be isomorphic.

[untold delta]

TennisSteve42

[orchid wraith]

I keep saying "should" because I've never studied the connection to homotopy.

[mint canyon]

Im using a very layman understanding of the word, that its basically just a smooth transformation that preserves the topology of the manifold (donut into a coffee mug etc.)

[orchid wraith]

If it's invertible (a smooth diffeomorphism), then it will induce an isomorphism of the de Rham cohomologies.

If it's not invertible, and just a smooth map $f:M \rightarrow N$, then there will be linear maps $H_{dr}^k(N) \rightarrow H_{dr}^k(M)$ given by pulling back forms under the map. Indeed, $d$ commutes with taking pullbacks, so this makes sense. Then you could try the rank theorem to get some formula involving the dimensions of the groups. Idk though

[untold delta]

TennisSteve42

[mint canyon]

@astral vine was saying if you get certain properties of the manifold for a differential ideal it creates a lot of difficulties in finding the integral manifolds? Assuming smooth coefficient functions, how general of a method is it for Linear PDEs?

[astral vine]

?

[mint canyon]

It looks like its not 1 particular method, but rather a bunch of special cases where integral manifolds can be found. Also, not every integral manifold is a solution, and has to satisfy additional constraints.

[mint canyon]

This comment*

[astral vine]

I do believe if everything is smooth enough, just localize everything go back on R^n, use Hormander's machinery for Pseudo-differential operators should do the trick, then revert back on your manifold.

[shell jackal]

the de Rham cohomology is homotopy invariant so it wouldnt change if such a homotopy hypothetically existed: actually the de Rham cohomology is a way to see that such a homotopy can never exist (since the procedure you are describing would in fact change the cohomology)

[mint canyon]

well the process I desrcibed was 2 different homotopies up to the limit of the same state, I understand the topology of the manifold cant actually change under the rules of a homotopy

[mint canyon]

the cohomology being homotopy invariant makes sense, thats why there would be no "transition", because it wouldnt change at all.

[lapis summit]

anyone here knows of a good book on control of PDEs?

[royal niche]

Depends on what you are looking into particular.

Curtain and Zwart's book on Infinite Dimensional Linear Systems is a classic text and was used as a reference for Doom's short course on PDE control.
(There seems to be a newer version as well by the same authors, not sure about the differences yet - might be worth checking out).

Old One - https://link.springer.com/book/10.1007/978-1-4612-4224-6
New One - https://link.springer.com/book/10.1007/978-1-0716-0590-5

Be wary though, the resource does expect some prior background in Algebra and Fun.Anal.

The other common reference tends to be this one Bensoussan, Prato, Delfour and Mitter but the prereqs. are more or less similar (It just has a more control-oriented viewpoint with a small review of finite dimensional/ODEs before going to the semigroup theory that's developed in Curtain and Zwart).

https://link.springer.com/book/10.1007/978-0-8176-4581-6

Anything more specific than above leads to research monographs and review papers on the specific subject.

[lapis summit]

Thanks, I did both in my masters

Fun anal and algebra

This is good, ty

[royal niche]

You are welcome!

[tropic rock]

Do someone know good youtube videos on proper course on calculus of variation? rigorous one at grad level

[sand echo]

I dont know of any, but I'm studying the book by rindler which is quite a comphrensive introduction https://link.springer.com/book/10.1007/978-3-319-77637-8

[tropic rock]

Thank you .. seems good..

Seems it's beginner friendly and pure maths oriented ..but I have to find out

Are you following some syllabus or reading on your own??

[sand echo]

reading on my own

[azure frigate]

i dont understand (b)

[lilac barn]

What don't you understand. Did you try checking the hint?

[azure frigate]

i dont understand how to show it applies for all T>0 and how that even relates to the fact that u(s,x)=0 for all s in R

so understand nothing :)

[lilac barn]

Once you show the hint for T-t, you will be able to change the 0 into a t via some translation substitution on the integral and the T will then function as a s.

[azure frigate]

what about t=T is that not a problem for the inequality?

[lilac barn]

It shouldn't be as you're only concerned in the limit.

[pulsar forge]

#book-recommendations message linking here, basically asking if taos pde book is worth reading

[lilac barn]

It's very worth reading, not just for the dispersive part but also the "high-level" coverage of elementary topics in the first two chapters. For dispersive, it's pretty much a staple.

[pulsar forge]

ah, neat i better get started with it then

[azure frigate]

sry for late reply was looking at smth else. so is the argument that after applying (a) you can translate under the integrand of U and it doesnt change the integral because its over R^3, so the inequality still holds? and then by setting t=T-s you get |u(s,x)|=|v_T(t,x)|=< 1/4pi(T-t) U(R^3;T) where the RHS goes to 0 as T-> inf so u(s,x)=0.

[lilac barn]

Well first the integral is on BT-t but after translating so that the second argument of U is t, you can upper bound the domain by R3

[azure frigate]

yes thank so much

[shell jackal]

Im reading about the local regularity of the CR equations on an a.c. manifold and there is a claim that is not making sense to me: so the proper set up is, let S be a riemann surface, V an a.c. manifold with J the a.c. structure, and the CR equations can be written as (i + J(f)) del bar f = (i - J(f)) del f, which is equivalent to del bar f = q(f) del f, where q = (i + J)^-1 (i - J). this source is claiming that by replacing f by x |-> b^-1 f(a x) and q by v |-> q(b v) you can take the norm of q in the C^r topology to be arbitrarily small

but i do not see what choice of b is supposed to shrinkg the norm like this? clearly this replaces q with (i + bJ)^-1(i - bJ), but then if you take b small this just gets close to being 1, so that cannot be the right answer. instead it seems like you want to choose b such that i and bJ get close to each other everywhere on the unit ball (since we are summing the L^infty norms of the derivatives plus a holder piece). but for arbitrary J i see no reason why this should exist?

perhaps you can choose a local form of J for which it does

(if its not clear, here im using i to be the matrix i * id)

[solid flint]

Hello, Anyone knows about books, monographs, notes about non autonomous evolution PDEs please? especially the case of non constant domains

Thank you in advance.

[opal condor]

what's the difference between this page and the odes and pdes page?

[warm quail]

"advanced"

[astral vine]

#odes-and-pdes is more like about undergraduate type questions

it concerns basic solving of ODEs, PDEs, "by hands"

and multivariable Calculus.

Here #advanced-pdes is more about fine properties with a more "global" point of view, which involves other subfields of maths such as Functional Analysis, Differential Geometry, Harmonic Analysis etc. Those topics are usually not reached before the end of undergraduate studies/beginning of graduate studies.

The treatment is more theoretical : we usually do not look for, and don't care about, explicit formulas. We check out the sharp regularity of solutions, existence and uniqueness in broader context. (Changing the set/space where you are looking for solutions implicitly changes the equation and its behavior even if it is formally "the same equation")

Somehow it deeply interact with #advanced-analysis since it uses a lot of tools concerned in this channel.

Am I somehow clear ?

[lilac barn]

For future reference, click on the top on the title of the channel and it'll show you a description.

[verbal nebula]

This makes me the big sad. I love explicit formulas

[quick pagoda]

Any good reference for free boundary things and boundary regularity tech?

[azure frigate]

I had a course about introduction to PDEs we just did first order PDE: linear, quasi linear, general (hamilton-jacobi). In each exploring the existence and uniqueness of non characteristic cauchy problems. Did the same for wave equations and heat equations. After this course if I want to continue self study and get really into the theory of PDEs what book(s) would you guys recommend me.

or even prereqs books i might need

[lilac barn]

Evans PDE book

[azure frigate]

thanks. recommend all problem?

[lilac barn]

Sure why not

[azure frigate]

ok i start thank

[rotund jetty]

"Introduction to Partial Differential Equations" by G. Folland, "Partial Differential Equations I" by M. Taylor.

[ocean ether]

this is so fucked up

[opal condor]

im literally studying the same syllabus as you

wanna try to understand the book together lmao

[ocean ether]

wait are you in my course

whats the first letter of the university you go to

[opal condor]

probably not since my syllabus has different format from the pic above

[ocean ether]

A.

nvm

[hidden coral]

yeah no distribution theory

[kind wren]

Hi. Say I have a diffusion 1D equation, $u_t = D u_{xx}$ for $D > 0$,
over the interval $(0,1)$. I'm thinking a no-flux condition on one end and a null Dirichlet condition on the other end. I'm wondering, if I go ahead and defined the mass of the system as $m(t) = \int_0^u(t,x)dx$ at the time $t$, is it too difficult to deduce a field $F$ such that $m'(t) = F( m(t))$? And have $F$ explicitly given in term of $D$ somehow?

[untold delta]

phao

[kind wren]

If I set $u_x(t,0) = 0$ and $u(t,1) = 0$, it's easy to show that $m'(t) = Du_x(t,1)$ assuming enough regularity of $u$. That's as far as I could get.

[untold delta]

phao

[kind wren]

Essentially, I'm wondering if there a way to deduce an ODE for the mass. The idea being that if I only want to know the mass, I'd bypass the PDE, hopefully.

[kind wren]

It should be integram from 0 to 1 in here.

[lilac barn]

What properties you fix on F?

[kind wren]

What do you mean? I'd like it to be C¹ if possible.

[upbeat spade]

grisvard has an account of weak variational solutions to the dirichlet and neumann problems belonging to H^2 (for f in L2). is there an analogous account for the robin problem somewhere?

[opal condor]

This is problem 3 chapt2 evans, I don't want a solution but can any tell me wether this works for any function f or only the ones that are differentiable

[rare oasis]

I would first try to prove for continuous f, doesn't look like diff is needed here

it might be true for even weaker funcions like L2 or something

[steady flicker]

wow i just amazed to see that in measure theory book brownian motion is related to PDEs. man math is so vast and amazing. I'm nobody

[lilac barn]

Brownian motion is not only related; it very much is a (solution to) (stochastic) differential equation: it can be used to build a solution for the heat equation quite naturally as the molecular behavior of particles under heat diffusion is precisely the Brownian motion.

[steady flicker]

that actually makes a lot of sense physically.
but it blow my mind that something stochastic process can be the solution to deterministic PDE

[lilac barn]

Stochastic thing can be related to deterministic stuff if some operation on the stochastic thing makes it deterministic.
For example, averaging the stochastic behavior of molecular particles gives you a (relatively more) deterministic quantity of temperature.

[steady flicker]

yeah I know from statistical mechanics. macroscopic property determined by microscopic behaviors. But to see it actually in math is cool. thanks

[rare oasis]

does anyone have text recomendation on the topic of pdes on graphs?

[void flame]

Could I get some help with this please? I'm asked to solve this transport IVP in three dimension. I'm aware that I'm supposed to solve it using method of characteristics but it just doesn't make any sense to me because the u is inside the divergence. I read this https://math.stackexchange.com/questions/4710174/solve-transport-equation-with-variable-velocity-vx-t that makes some sense but when I try to apply this method to my problem I fail because u(t, x) = (0, x2, x3). How do I get the anti-derivative of this u??

[lilac barn]

You are given the explicit formula for u. So just plug it in?

The divergence will expand into p(div u) + u dot grad p. Div u you can explicitly calculate and u dot grad p you can write down

[void flame]

Now I’m stuck on finding the U

I’m confused about why it turned out like this

[lilac barn]

You are given the U. It is u(x,y,z) =(0,y,z)

[void flame]

Ok thank you, sorry I realized I forgot that rho is a scalar function and that's why I got stuck.

I have a new question: how does one solve this by characteristics? This is what I tried and I could only get what x+z is but how do I continue to seperate x ? I'm confused..

what I tried

[lilac barn]

Note that dx/ds= y+z and dz/ds= x-y and dy/ds=y. So
d2x/ds2=dy/ds+dz/ds= y+x-y. This gives you ddx=x which you can explicitly solve.

[proud elm]

Hey I need help finding an appropriate stability condition for the 2 dimensional wave equation because Im gonna vary c during simulations

[void flame]

Does anyone know what I did wrong here? I got the answer and verified it and it was off by a factor of x so it doesnt cancel correctly.
I dont understand what is wrong with my process, I've been staring at this for hours. Would appreciate any help ;_;

[lilac barn]

You cant treat y as a constant and then integrate (imagine y was like sine). The right way to handle the coupled system is to differentiate y twice and make the equation for y purely in terms of y.

[opal condor]

this is a problem from evans

i was wondering if the inequality written here is incorect

isn't it suppose to be

u(x/|x|^2)|x|^(2-n) instead of u(x/|x|)|x|^(2-n)

[lilac barn]

Yeah that looks correct.

[autumn stratus]

You guys have any resources I could look to for the method of characteristics?

[pulsar forge]

Lawrence evans is nice for basics

[pliant bough]

Does anyone have a good source for proof of kolmogorov equations for multidimensional SDEs? I'm looking at the SABR model and I have no clue how to derive the equations for this, given the correlation between the processes.

[autumn stratus]

Thanks bro, I've been grinding through a few PDE books and feel way less dumb already

[pine oriole]

has anyone ever heard of a situation where an elliptic differential operator has symmetric second-order components and skew-symmetric first-order components

On the Fourier sides it becomes a symmetric matrix but not Hermitian.

is there a name for this kind of "symmetry" and do we know anything about its spectrum / eigenvalues? do variational methods even work for them?

somehow general Navier-Stokes yields this operator

[autumn stratus]

let's say I have an eikonal equation like u_x^2+u_y^2=u^2 with u(cos(x),sin(x))=1, how do I deal with this type of condition

[astral vine]

Actually this is some kind of weighted Hodge Dirac squared system, similar to the Lame system but with additional anisotropy

Shouldn't be too hard to cook a sesquilinear form from which one can inherit the operator.

And then recover the standard spectral properties. Like sectorial etc.

If, instead, by spectral properties you meant to diagonalize the system and to study the behavior of eigen values near "critical Fourier modes", notice that this is usually not that much interesting for heat-like equations, as it seems to be one here if I am not wrong. (it can be meaningful for few selected "concrete" applications, but that's not "that much common")

[pine oriole]

thanks. Do you have some references for what you are talking about?

it is indeed a heat-like equation because it is a linearization of general Navier-Stokes
I am working on a modification of what is shown in picture (by adding gravity) and I need to know whether there is an eigenvalue with positive real part (which will imply instability)

[astral vine]

For linearization of compressible Navier Stokes this does not happen. The spectrum will lie in a Parabola that hits 0. Which does not imply instability or stability.

Adding gravity may change things of course, but it depends on where the gravity term will show up.

[chrome trout]

The sobolev injections for $H_0^1$ are much different than those for $H^1$?

[untold delta]

мир

[pine oriole]

it's not just compressible navier-stokes
it's compressible navier-stokes-fourier with heat conductivity

and the gravity indeed plays a crucial role in astrophysics. with integration by parts you get a slightly positive term in the end

[pine oriole]

all the same though, do you have a reference for this thing you were talking about?

[rare oasis]

Im looking for a class of maps (change of coordinates) that preserve the elliptic operator $\partial_{xx}+\gamma^2 \partial_{yy}$, just like the laplacian is preserved through conformal maps. Is this something that exists?

[untold delta]

Sydd

[astral vine]

no just take the closure of Ccinfty in the targeted space

[lilac barn]

Most likely, you would wanna mess with the CR equations to construct a different class of functions that can preserve this. My best guess is that you would wanna put a parameter in one or the equations for CR

In fact, more simply, you can do a scaling in y to recover the laplacian case and then scale back to see the required condition on conformal maps.

[pine oriole]

Perhaps you mean something like this?
https://www.sciencedirect.com/science/article/pii/S0022039618301852

so the operator is not "symmetric" but it is "maximal accretive" ?

[pine oriole]

it seems then that I want to know about the "principal" eigenvalue, or at least the eigenvalue with maximal / minimal real part. is there a general principle to get that

[lilac barn]

I am not sure how useful this will be here but a systematic way of doing this is resolvent estimates (These are also the usual recipe in establishing hydrodynamic stability results in Navier-Stokes).

That, on its own, is mostly a vague buzzword but searching up along with words surrounding your research title should give you a good direction.

[astral vine]

sesquilinear forms allow resolvent estimates "almost" for free.

which was the reason I suggested this

[lilac barn]

I don't think there is an easy way to construct a sesquilinear form in this case. Much easier would be good old energy methods to establish a resolvent estimate.

[astral vine]

The symmetries are actually simple for the initial system, however in presence of gravity

maybe eveything will become messy

[pine oriole]

do you have an example for using resolvent estimates to get principal eigenvalues or stability

[lilac barn]

"Resolvent estimates for Plane Couette Flow" by Brez E Silva. You should check out the references in the introduction as they go into a much intro dive in to the method.

[pine oriole]

thanks will do

oh wait maybe wrong Silva paper

[lilac barn]

You can also check "Resolvent Estimates and Quantication of Nonlinear Stability"

[pine oriole]

oh....
I am guessing you are referring to this
it basically establishes where eigenvalues do not exist (stability)
on the other hand, what I want is the existence of a principal eigenvalue that leads to instability

[lilac barn]

Ahh okay, yeah these methods work nicely when you wanna prove stability.

[pine oriole]

For principal eigenvalue I don't think variational method still works since operator is nonsymmetric.
And Max min representation may have trouble with vector valued stuff.

In my situation however, second order components are symmetric while first order components are antisymmetric. This might be graded antisymmetry, or something very close to symmetry. Perhaps a change of variables would yield true symmetry?

[astral vine]

But this mostly works when the domain is bounded (implying having discrete spectrum/eigenvalues), for PDEs on unbouded domains such as the whole space, this kind of technique is usually not really efficient.

[pine oriole]

Oh great. I'm working on a compact manifold with boundary

It's T^2 x [0,d]
as a model for astrophysics

I really need to look into methods people use to derive principal eigenvalues for vector-valued stuff. Is there a place that lists them

[solid dagger]

What’s a road map for pdes research

I don’t have much course space left in my undergrad due to double majoring in cs, but will be doing a masters (and will self study from books)

[wooden sundial]

Hi can anyone help me interpret this integral?

u is a solution to the wave equation on a hyperbolic manifold of dimension m

[wooden sundial]

Apparently if f is part of the initial condition data for u, then

[radiant flower]

im looking for a substitution for y(x) so I can seperate the the x from the 2xy, I've tried y = e^s(x) it almost works but theres an ugly e^s(x) term still remaining which I would like to remove

[lilac barn]

#odes-and-pdes

[ocean ether]

based on how evans solved the transport equation my guess is to reduce the question down to an ODE via $z(s)$ where $z'(s)=Du(x+sb,t+s)\cdot b+u_t(x+sb,t+s)+cu(x+sb,t+s)=0$

[untold delta]

the breadwave

[ocean ether]

but then if i integrate it i get $u(x+sb,t+s)+\int cu(x+sb,t+s)=0$

[untold delta]

the breadwave

[ocean ether]

which is NASTY

[rotund jetty]

Write down the ODE that $z$ satisfies

[untold delta]

L

[lilac barn]

First try to guess what should be the general solution when b=0. Use this knowledge to reduce down the PDE into the transport PDE solved by Evans

[ocean ether]

i guess we basically have $Du(x,t)+cu(x,t)$ so uhh

[untold delta]

the breadwave

[ocean ether]

$u(x+sb,t+s)e^{cs}$

[untold delta]

the breadwave

[mellow cloud]

Wait Anatole why are you emeritus

T_T

Did u quit?

[astral vine]

Because I am no longer an active guy on this server as I used to be.

"very active" role.

Etc.

[opal condor]

in this problem in evans do we assume that g(t) is differentiable?

or do we have to?

[chrome trout]

Hello

In trying to prove the Gagliardo-Niremberg inequality for $\bR^N$: that is, given $p < q < p^$ with $p < N$, hence $p^ = Np/(N-p)$ I get using interpolation, $$|u|_q \leq |u|p^\theta |u|{p^*}^{1-\theta}.$$Now I can use a Sobolev injection so that $$|u|_p \lesssim |u|_p^\theta (|u|_p^p+|\nabla u|_p^p)^{1/p(1-\theta)}.$$How can I continue?

[untold delta]

мир

[chrome trout]

Notice that I cannot bound the p norm since Rn is not bounded (cant use Poincaré)

[lilac barn]

He usually writes the assumptions at the start of the question section. If not given then just assume integrability

[lilac barn]

Check if you can bring the powers inside

[chrome trout]

wdym? I think the result I need is the Sobolev inequality $| \nabla u|_{p^*} \leq |u|_p$

[untold delta]

мир

[lilac barn]

Something like this cannot hold I think. To see this, take a mollification of a bump function, the integral will remain small but the slopes are getting larger and larger

[astral vine]

This kind of inequality does not hold ine general unless you are super localized on the Fourier side.

[ocean ether]

wha

[chrome trout]

oh

this is the one that I want to use?

[lilac barn]

Look at the direction of the sign

[ocean ether]

...oh my god i was doing problems from the 1st edition instead of the 2nd edition...

[chrome trout]

true

thanks

[ocean ether]

i still require ideas on how to tackle this

i know $u(0)$ basically reduces to $-\int_{\partial U}g(y)\frac{\partial G}{\partial\nu}(0,y)dS+\int_Uf(y)G(0,y)dy$

[untold delta]

the breadwave

[sonic olive]

This problem doesn’t require you to use the Greens function

Let phi(s) be the average value of u on the boundary of the ball of radius s centered at 0. Calculate phi’(s) like in the text. Then use the fundamental theorem of calculus to get a formula for phi(r) - phi(s) and send s to 0

[tired hollow]

Without context when people say boundary conditions do they mean dirichlet

[frank tide]

what is $H^2$ in terms of polar coordinates?
In cartesian you can have
$$H^2={u\in L^2:|: D^{\alpha}u\in L^2, \quad |\alpha|\leq 2}$$
but if i wanted to check if $u(r,\theta)\in H^2(\Omega)$ for some polar region $\Omega$, how do I check this?

[untold delta]

ansh

[rotund jetty]

if the region $\Omega$ is a positive distance from $0$, then in terms of polar coordinates, $H^2$ has the same formula, where $D^\alpha$ should be interpreted as with respect to the polar coordinates.

[untold delta]

L

[quaint herald]

You'd also want boundedness of the domain. Otherwise in general you will need to replace Dx and Dy with Dr and (1/r)Dtheta.

[mellow cloud]

Oh look grumpy catvman

[tropic rock]

why 35 and 36 is required to conclude 38 is valid for any $v \in W_{0}^{1, q}(U)$. Couldn't understand why checking bound of DL helps

[untold delta]

tryingmaths