advanced-pdes

[buoyant pike]

Please don’t post your question in multiple channels

[misty needle]

I just don't know what the definition of unconditional stability is

In this article

[astral vine]

Again 😮‍💨

[fringe onyx]

We've developed the basic weak existence theory (Lax-Milgram, Fredholm Alternative, spectrum results) for linear elliptic PDES with Dirichlet boundary conditions, but I'm wondering how such results generalize to Neumann conditions

[buoyant pike]

Have you heard of the dirichlet to neumann operator

[lapis creek]

How do the Dirichlet boundary conditions appear in the theorems, if at all? Are the Dirichlet boundary conditions only imposed in some examples?

[buoyant pike]

?

[fringe onyx]

That would be super helpful thanks

[fringe onyx]

Not familiar with that no

[buoyant pike]

Ok so the DtN operator lets you turn dirichlet boundary data into neumann boundary data

[untold delta]

sucksuko

[hearty bolt]

I'm having some trouble deriving this. ($L$ is the Lagrangian.)

I've got up to $\delta S = \sum_i \int_{t_1} ^{t_2} \pdv{L}{q_i} \delta q_i + \pdv{L}{\dot{q}_i} \delta \dot{q}_i + \pdv{L}{\ddot{q}_i} \delta \ddot{q}_i \dd{t}$.

I know that for the first order version, you take $\dv{t} (\pdv{L}{\dot{q}_i} \delta q_i)$ and re-arrange it. So I've tried doing a similar thing, i.e. $\dv{t} (\pdv{L}{\ddot{q}_i} \delta q_i) $ and then differentiate once more wrt $t$, but I'm really not sure if what I'm getting is right.

My working says that $\pdv{L}{\ddot{q}_i} \delta \ddot{q}_i = \dv[2]{t} (\pdv{L}{\ddot{q}_i} \delta q_i ) - \dv[2]{t} (\pdv{L}{\ddot{q}_i ) \delta q_i - 2 \dv{t} (\pdv{L}{\ddot{q}_i} \delta \dot{q}_i ).$

[untold delta]

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

[hearty bolt]

No physiscs package

Damn

[hearty bolt]

So the trouble I'm having is with the $\delta \dot{q}_i$ term right at the end

[untold delta]

Douglas

[hearty bolt]

You can do integration by parts, but this gives

actually, i've got it now

painful question tho

[stark thunder]

All you should be needing to do is apply integration by parts twice

that's why there's two derivatives of time

[desert flower]

can I ask where did you find this? or what book? I might be able to help

[desert flower]

euler lagrangian equation for second order?

[river path]

@hazy mortar clerk told me to tell you to friend his alt "barconstruction" so that he can talk to you about ocaml

hope youre doing well✌️

[buoyant pike]

Lol

[hazy mortar]

LMAO

why did u ping me here lil bro

[river path]

not that disruptive

[buoyant pike]

Does anyone have a reference for Ladyzhenskaya's proof of 2d euler/NS well posedness

[lilac barn]

I believe you're looking for "Solution "in the Large" to the Boundary-Value Problem for the Navier-Stokes Equations in Two Space Variables"

[hexed kernel]

Is there a source where I can read up further on the shallow-water equations?

I'm kinda cramming my brain on two particular sources, which are from Introduction to Climate Modeling by Prof. T. Stocker, and Harvey Segur's Lecture 8 on the shallow-water equations.

[astral vine]

Shallow water waves ?

[hexed kernel]

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

These things.

[astral vine]

https://perso.univ-rennes1.fr/vincent.duchene/files/CoursM2.pdf

[hexed kernel]

I'm mainly wanting to derive them, and the two sources I'm looking at do a good job, but there are some parts where I just want to know why certain things happen in the derivation.

Also, let me take a look!

[astral vine]

Check in the references

[hexed kernel]

There are 3 in the aforementioned document, but I'll take a look at the first two that are mentioned!

[astral vine]

The lecture itself was really good to me when I was a student.

[hexed kernel]

The one by Harvey Segur?

[astral vine]

no the file I sent you

[hexed kernel]

Oh, they're lecture notes. Interesting.

I'll take a look at this, though I do believe that it's probably a bit more complicated than what I'm expected to do.

[buoyant pike]

Which certain things

[steep oyster]

Does anyone know references for eigenvalues of elliptic operators with neumann boundary conditions? Evans does dirichlet only

In particular I want to know conditions for when the operator is pos def

[astral vine]

@steep oyster What do you want to know exactly ?

[steep oyster]

condition for when lambda_1 >=0

my domain has corners if that matters

[astral vine]

Doesn't this follows directly from the construction by sesquilinear forms

[steep oyster]

i'm not sure what that is. are you saying that the condition always holds?

[astral vine]

You want to investigate operators -div(A Nabla .) on bounded Lipschitz domains, A being uniformly elliptic and with boundary condition A Nabla u . n = 0 on the boundary ?

[steep oyster]

yeah i think that's right. specifically, my operator is laplacian + a 0th order term

[astral vine]

Okay the 0 order term is wat can actually would messed everything up

[steep oyster]

the only thing i know on the 0th order term f is int f = 0. in particular, no information on the sign of f

[astral vine]

Like

[untold delta]

Functionanatolysis

[steep oyster]

yeah that's right

[astral vine]

So knowing this has non negative eigen value is the same as knowing that

$$\langle \nabla u, \nabla u\rangle + \langle f u, u\rangle \geqslant 0$$

[untold delta]

Functionanatolysis

[astral vine]

for all u in H^1.

[steep oyster]

agreed this is the weak form of the above

do you know of any non-pointwise conditions on f to gauarntee that?

[astral vine]

only conditions that are not compatible with the mean 0 : f being non negative

(but this implies f =0 if you ask for 0 mean)

[steep oyster]

oh yeah i do want to eventually conclude that f equiv 0.

i wanted to use a spectral argument to go from int f = 0 -> f = 0. but this requires pos def of e.g. the above operator

[astral vine]

The Neumann Laplacian has eignein value 0, and then non negative eigenvalues

[steep oyster]

you mean -Delta u = 0 + neumann conditions? that doesn't invovle f though

[astral vine]

so adding a term that could have negative value as a multiplication operator should shift the spectrum on the left by the essential infimum of f

[astral vine]

I didn't finish my explanation at time

[steep oyster]

sry for interrupting

[astral vine]

My bad

I should have provided the whole explanation in a row

(This is purely moral)

[steep oyster]

so what i'm after is (morally?) equivalent to f^- = 0 in Linf ?

do you have a reference for this spectrum shifting idea

[astral vine]

yeah

There are plenty of references

my favourite being

Cheverry & Raymond A guide through spectral theory.

[steep oyster]

thanks a bunch!

[astral vine]

Exercise 3.15. p.69-70

[rugged leaf]

hi does anyone have An introduction to partial differential equations by Michael Renardy and Robert C. Rogers

a paper im reading quotes theorem 12.44 but im confused about what its stating so i want the original theorem

[hexed kernel]

I didn't get the ping for this, but I think I got it figured out! I went to office hours today with my professor. Thank you!

[valid edge]

I would love some feedback on this attempted solution to the problem in attached screenshot:
For Helmholtz equation:
$$
u''(x) + \alpha^2 u(x) = f(x), x\in [-1,1]
$$
With bcs = $u(\pm 1) = 0$.
We define $L(u) = u''(x) + \alpha^2 u(x)$, and $R = L(u) - f$ and $R_N = L(u_N) - f$.
We seek to find an approximation solution $u_n \in V_n$ where $V_n$ is the span of $\psi_i$, for $\psi_i (\pm 1) = 0$. By Galerkin's method we find $u_N \in V_N$ such that $(R_N,v)=0 \forall v \in V_n$.
Expressing as linear algebra problem:
$$
(u_N'' + \alpha^2 u_N - f, \psi_i) = 0, i \in 0,1,\dots,N
$$
With some manipulation we get:
$$
\Rightarrow \sum_{j=0}^N \hat{u}_j ((\psi_j'',\psi_i) + \alpha^2(\psi_j, \psi_i)) = (f,\psi_i), i \in 0,1,\dots,N
$$
We define $\mathbf{x}$ such that $x_i = \hat{u}j$, $\mathbf{b}$ such that $b_i = (f, \psi_i)$, and $A$ such that $a{ij} = \alpha^2(\psi_j, \psi_i) - (\psi_j',\psi_i')$.
As such, $u_N$ may be found by solving $A\mathbf{x} = \mathbf{b}$ for $\mathbf{x}$.

[untold delta]

madlor

[valid edge]

i hope im in the right channel for this, numerical analysis could perhaps be the correct place too

[buoyant pike]

Numerical analysis

[valid edge]

ok. delete to not clog up channel or can i leave it here?

[hard laurel]

what is this? (Ordinary Differential Equations by Vladimir I. Arnold)

[candid token]

for this problem from evans, doesn't u automatically satisfy the entropy condition since it's decreasing in x? or am i tripping

(ofc i still have to check that it's an integral solution of the PDE)

[grave oyster]

ahh evan's, my favourite past time

[candid token]

it's a pretty nice textbook

its just this one thing that i'm confused over

[candid token]

do we even need to care about entropy? because it's a shock wave not a rarefaction wave right?

[grave oyster]

I have no idea. I only studied the parabolic and the strong/weak maximum principles sections for my MFE

[grave oyster]

@candid token You'll struggle to get help here though. Even though evan's book is very nice, PDEs in mathematics is definitely a minority topic in math departments. I would try out the physics discord channel. They might not know the book, but these PDEs are use a lot in stat mech

[lilac barn]

PDEs in mathematics is definitely a minority topic in math departments
What?

[minor mulch]

lol

They might not know the book, but these PDEs are use a lot in stat mech

i think physics phd students know what evans PDE is lmfao

also why single out stat mech

[grave oyster]

because I'm familiar with it in from my physics undergrad

Could be used in others, but I just know it's true for stat mech

[minor mulch]

could be

like uh

all of the others

general relativity comes to mind as an unbelievably powerful and central aspect of modern physics

[grave oyster]

Generalising statements are pointless. Sure, every field has pde's, but how many use them on a daily basis and require the understanding of fundamental theorems of pde's

[minor mulch]

lol

going to assume thats a rhetorical question

[lilac barn]

There are whole departments of mathematics that are dedicated to studying PDEs, both as PDEs in itself and studying PDEs arriving from different phenomena. Quite a lot of Field medalists have been given their medal for notable contributions in PDEs.

[minor mulch]

im pretty sure PDE is by far the most common thing in math departments worldwide

[grave oyster]

@lilac barn Not my experience at my university

[minor mulch]

go to a better school then

[lilac barn]

Generalising statements are pointless.

[grave oyster]

it's ranked top 3 in Aus

[minor mulch]

Aus

lol

just lol

[grave oyster]

Huh

[minor mulch]

the world is a big place

[grave oyster]

exactly

[minor mulch]

as you will learn

[grave oyster]

unless you're comparing to the resources of a university like oxford/MIT, there's not a whole lot of difference between the universities from the class below.

You would only care about specialisation

[minor mulch]

uhh

idk what that means

[minor mulch]

see here

this is not a “generalizing statement”

you can look at basically any university

[grave oyster]

I mean like, some universities that are very "low ranked" can be decent

[minor mulch]

yours is in a very very small minority

[grave oyster]

in subfields

but bad in other fields. I.e. my local university is in a mining town and money is pumped into the engineering department

[lilac barn]

And you might also not be a good judge of the research interests of the professors at your university. It might be they're tackling some obscure problem but the motivation or the technique could be very PDE.

[grave oyster]

So globally, top ranked 20-50 universities, there's not a whole lot of difference between them.

I don't have enough fingers to count how many of my professors went to princeton for example

[lilac barn]

This is wrong as well but doesn't have much relevance to the current topic. Also, it seems like the discussion is getting tangential so let's move it to #math-discussion if necessary

[buoyant pike]

Just not true

[astral vine]

I actually agree with Ange. @grave oyster Activity about a field can be easily measured by the amount de preprint on Arxiv on a daily basis

Math.AP (Analysis of PDEs) just utterly destroys any other category.

I don't know about the quality of the research itself, but there is so much work uploaded there.

[bronze gate]

and Pde is connected with lots of other areas of math

[unborn gyro]

this seems wrong, combinatorics definitely has way more publications than pde

probability theory also has way more than PDE

[candid token]

my question spawned a flamewar ig lmao

[fringe onyx]

we've discussed how to use the galerkin method to prove existence of solutions to linear evolution equations but I'm having trouble seeing how to apply it to non linear equations. In particular there's two steps in the process bringing complications: first global existence of finite dimensional solutions (the non linear solutions usually make the resulting system of ODEs only admit local solutions) and convergence of the non-linear terms when passing from finite to infinite dimensions

are there general tricks for dealing with those?

[buoyant pike]

No lol

[fringe onyx]

hurb i will cry alone then

[buoyant pike]

If nonlinear equations were easy to deal with then the field of pdes wouldn't exist

[fringe onyx]

yeah fair point haha, i was just hoping for an easy way out of my struggle

[blazing ridge]

Galerkin still can be useful for non-linear equations. Usually to get global solutions you look for additional estimates indep of time and to pass to the limit you'll need some other nice bounds on your solutions so you can pass to the limit in the non-linear terms via some div-curl type lemma

[minor mulch]

i dont think math.PR is at all close to math.AP in terms of activity

you are right about math.CO though

[fringe onyx]

Yeah I ended up figuring it out for my specific case, thanks

[muted void]

v(x)=u(Mx) where M is rotational matrix and u is a harmonic function. we write this to prove that laplace operator is invariant. But my point is since u is harmonic in Rn and any rotation to vector will be inside Rn only so as lap(u) = 0 regardless of what we put in place of x in Rn what is the point in proving it is invariant?

[pine oriole]

is anything even worthwhile if it's not one day used in a PDE /j

[rotund jetty]

Is it possible to have a distribution that is given by integration against a function $f$ but $f \notin L^1_{loc}$?

[untold delta]

L

[minor mulch]

what would that even mean

mollify an indicator of a compact set

[quaint herald]

You can get distributions out of some functions that aren't locally integrable in a couple of ways at least. For simplicity let's assume that there is only one point the function fails to be locally integrable at and we are in dimension 1. The ideas generalise a bit, but not always in a canonical way.

1. By understanding the integral in a a limiting sense (principal value). e.g. p.v. (1/x) is one that shows up often. Integrate your input multiplied by 1/x outside the interval (-r,r) and take r->0.

2. Sometimes when you do the process in 1., the integral need not converge (this is the case with 1/x^2 instead of 1/x for example). Then instead as r->0 you get an asymptotic expansion in r, involving some negative order terms as well. Nevertheless, if you drop you "divergent part" (the negative order in r terms), you can then take the limit and get a well defined distribution. This process is called Hadamard regularisation, and the resulting distributions are finite part (p.f.) distributions.

They behave a bit like one might guess. E.g. p.v.(1/x) has distributional derivative p.f.(-1/x^2).

[gleaming lily]

So hamilton jacobi equation induces the two hamilton equations by method of characteristics

Can you go the other way around?

[quaint herald]

Your question is a bit vague to me, but I think yes. The HJE can come about (amongst other ways) by starting with Hamilton's equations and solving for the symplectomorphism that changes variables to those in which the Hamiltonian is trivial.

The solution to the HJE is then a generating function (in the symplectic geo sense) for this symplectomorphism.

You can pass between all these different mechanics formulations pretty freely. HJE is particularly cool because in some sense it passes closest to quantum mechanics.

[gleaming lily]

Basically the professor said that there are three different equivalent ways

Hamilton Jacobi equation is $u_t+H(x,Du)=0$

[untold delta]

Whoever

[gleaming lily]

Hamilton equations are $\dot{x^i}=\partial_{p_i}H$, $\dot{p_i}=-\partial_{x^i}H$

[untold delta]

Whoever

[gleaming lily]

And Lagrangian

Finding ${\bf x}$ minimizing $\int_IL({\bf x},\dot{\bf x})\dd s$

[untold delta]

Whoever

[gleaming lily]

Tbh idk any of the words you said to me 🙃

But the answer was that the solution to the hamilton equations is the path x that minimizes this integral if H = L* which is the Legendre transformation

[quaint herald]

Lol. Well yes these are the three formalisms I was referring to. It sounds like you understand some directions of their equivalence. For a complete picture though you should really learn some symplectic geometry, this is the natural language to speak in.

[quaint herald]

This is passing from Lagrangian to Hamiltonian, not from Hamiltonian to HJE, as you originally asked.

[gleaming lily]

Oh I see

[gleaming lily]

I won't be able to do this in undergrad I'm afraid 🙃

there are other things I will have to learn like

Logic

And algebraic geometry

Actually next semester is geometry time so I might learn some of that

[quaint herald]

But yeah fair, my ug geometry was sorely lacking.

[gleaming lily]

But I have to say this theory is so beautiful to me

[quaint herald]

Yeah I like it too, mostly when the symplectic geometry enters the picture though. Before that it felt kind of ad hoc even though it worked.

[gleaming lily]

Actually I am doing a reading course next semester on some geometry stuff

I forgot the name

But sung jin oh is leading the reading course

[quaint herald]

Cool 🙂

[unborn quiver]

This is a crazy gif

Not Gomez brand

[quaint herald]

It is very gomez brand to 🐶 whoever.

[shrewd epoch]

Hey, I'm trying to work out the action of an operator with Fourier symbol $a(\xi)$ on an oscillatory function but I'm a bit stuck on a part. All is gonna be in $\bR$, it doesn't change much.
Not 100% sure what assumptions are the right ones, but I'm taking $a$ such that for all $n\in\bN$, there is $C_n$ such that $|a^{(n)}(\xi)|\le C_n\sqrt{1+\xi^2}^{1-n}$.
I wanna look at the action of the operator $ia(-i\partial_x)$ on an oscillatory function of the form $u_\varepsilon(x)=b(x)e^{i\frac{S(x)}\varepsilon}$, where, say, $b\in\mathcal S, S\in C^\infty$ and $\varepsilon$ goes to $0$.
Then I got $$a(-i\partial_x) u_\varepsilon(x)=\frac1{2\pi}\int_{\bR} a(\xi)b(y)e^{i\frac{S(y)}\varepsilon} e^{i(x-y)\cdot\xi},\dd y,\dd\xi.$$
Then I can rescale it as $$\frac1{2\pi\varepsilon}\int_\bR a(\frac\xi\varepsilon)b(y)e^{\frac i\varepsilon(S(y)+(x-y)\xi)},\dd y,\dd\xi,$$ and I'm in the right setting to apply the stationary phase approximation, but I'd like to get one more term in the approximation, in order to (ideally, not sure that's the right formula it's just the one I expect to get) reach $$ia(-i\partial_x)u_\varepsilon(x)=ia(\frac{S'(x)}{\varepsilon})u_\varepsilon(x)-a'(\frac{S'(x)}\varepsilon)b'(x)e^{i\frac{S(x)}{\varepsilon}}+o(1).$$Any hint regarding how I can achieve that, or anywhere I can find lower order terms in the stationary phase method?

[untold delta]

upheaval

[unborn quiver]

Could you expand on this btw, I know symplectic geometry is the natural setting for control theory, but I've always been curious

[quaint herald]

At conference dinner, will reply at some point over the next few days.

[unborn quiver]

Thanks! Enjoy dinner

[quaint herald]

Otherwise just read some of Arnold's classical mechanics book for quite a readable exposition of a lot of these ideas. Great book.

Abraham Marsden too, but that is maybe less good for gaining intuition.

[white bay]

So we have initially this perioidic eigenvalue Sturm-Lioiville problem
$$ - u^{''}n (x) + \left( V(x) - \lambda \right) u_n (x) =0,$$
with $u_n (0) = u_n (L)$ and $u' n (0) = u' _ (L).$
Now we plug in $u{\kappa} = q{\kappa} (x) e^{i \kappa x}$ in the eigenvalue problem so basically we replace $u_n$ by $u_{\kappa}$, where $\kappa = \frac{2\pi x}{a}$ and $a=\frac{L}{N}$, we get
$$-e^{i \kappa x} q^{''}{\kappa} - 2e^{i \kappa x} i \kappa q' {\kappa} (x) + e^{i \kappa x}q_{\kappa} (x) * \left( \kappa ^2 + V(x) - \lambda \right)=0,$$
where $q_{\kappa} (0) = q_{\kappa} (a)$ and $q' {\kappa} (0) = q' _ {\kappa} (a)$.
We can write $V(x)$ as a trigonometric polynomial
$$V(x) = \frac{1}{2}V{0} - V{0}\frac{e^{i\frac{2\pi}{a}x} + e^{-i\frac{2\pi}{a}x}}{4},$$
with the fact that $V(x+a) = V(x).$
Now we can solve this with the Frobenius method, but instead with a power series, we use a Fourier series
$$q{\kappa , n} (x) = \frac{1}{\sqrt{a}} \sum_{k \in Z} \hat{q_{\kappa , n , k}} e^{\frac{ i 2 \pi k x}{a}}.$$
Plugging it in the new eigenvalue problem, and applying frobenius method we get
$$\frac{1}{\sqrt{a}} \sum_{- \infty}^{\infty} \hat{q_{\kappa , n , k}} e^{\frac{i 2 \pi k x}{a}} \left( \frac{16 \pi ^2 k^2}{a^2} + \frac{V_0}{2} - \lambda \right) \frac{- V_0}{4} \hat{q_{\kappa , n , k-1}} e^{\frac{i 2 \pi k x}{a}} - \frac{V_0}{4} \hat{q_{\kappa , n , k+1}} e^{\frac{i 2 \pi k x}{a}} = 0.$$
From this we get the following recurrence relation
$$\hat{q_{\kappa , n , k+1}} = \hat{q_{\kappa , n , k}} \left( \frac{64 \pi ^2 k^2 }{a^2 V_0} + 2 - \frac{4 \lambda}{V_0} \right) - \hat{q_{\kappa , n , k-1}}.$$
Now my question is if we set $V(x) = 0 $ then we only have $\hat{q_{\kappa , n , k}}$ the other q's will vanish. How can we solve the eigenvalue problem with $V(x)=0$ then?

[untold delta]

Fractalogist

[white bay]

I need to find the eigenvalues $\lambda _{\kappa, n}$ and their respective eigenfunctions.

[untold delta]

Fractalogist

[stark tree]

Hey for the Navier Stokes equation the left side builds a total differential of the acceleration which is the local acceleration and convective acceleration or the total differential of velocity differentiated with respect to time, however im a little confused on the following

https://cdn.discordapp.com/attachments/423244559682764800/1185330939182719056/182418252830343170.png?ex=658f3876&is=657cc376&hm=5ed0eb4ecc5190ccf28a57c44c471fa64f4fc271c70f00b439165631680c3a4c&

this would just be the way its written down in general

however when you just look at the x-direction a lot of publications show it as the following: https://cdn.discordapp.com/attachments/423244559682764800/1185330239727997009/image.png?ex=658f37cf&is=657cc2cf&hm=410ba0fffe93577ff38cbbfd43567ebe796a3d6efc4599e520d38f9180ac1bc0&

but if the total derivative with respect to time is the following:

then, ignoring the density rho, how is the velocity in x direction u inside the derivative and squared?

[buoyant pike]

So partial(u^2)/partial x=2*u*u_x right

This is what you want but there's an extra factor of 2

[river path]

If you ignore rho and suppose the fluid is incompressible, then d_x u + d_y v + d_z w = 0 (the velocity field is divergence free).

[stark tree]

I know that but thats not what i was originally asking, im still confused why the notation seems so different in different publications

like even Wikipedia seems weird about it

[river path]

The two expressions are equal

[river path]

if you take this into account

in particular, $$\partial_x(u^2) + \partial_y (uv) + \partial_z(uw) = u \partial_x u + v\partial_y u + w \partial_z u = [(\vec{v} \cdot \nabla) \vec{v} ]_1$$

[untold delta]

ryc

[river path]

this isn't quite the formula for the total derivative in the first coordinate

[stark tree]

so that holds when the laplace operator of vec field v = 0 ?

[river path]

you apply a directional derivative to u in the (u, v, w) direction

[river path]

the divergence of v = 0 in this case

[river path]

to see that, we can expand this

using the product rule on the left hand side (i'll even use it on the square), we have

\begin{align*}\partial_x(u^2) + \partial_y (uv) + \partial_z(uw) &= u \partial_x u + u \partial_x u + u \partial_y v + v \partial_y u + u \partial_z w + z \partial_z u \
&= u (\partial_x u + \partial_y v + \partial_z w) + u \partial_x u + v\partial_y u + w \partial_z u \
&= 0 + u \partial_x u + v\partial_y u + w \partial_z u
\end{align*}

[untold delta]

ryc

[stark tree]

omg ok that makes a lot of sense now

went through 3 books on this and no one ever went to the effort to show this and just wrote it down

[river path]

(if you include a density, you no longer have that the velocity field is divergence free. instead you have the conservation of mass equation for the density, which lets you do the same cancellation)

[river path]

i had the exact same experience for ages lol, it's so dumb

[stark tree]

but yeah i guess the math still holds true, so in the end with incompressibility it doesnt matter which notation i use tho right?

of course it doesnt theyre equivelent

[river path]

but it's very very useful to be able to decide whether you want to write $\vec{u} \cdot \nabla \vec{u}$ or $\nabla \cdot (\vec{u} \otimes \vec{u})$ in more advanced pdes, so it's a good thing to remember that divergence freeness kills half the product rule here

[untold delta]

ryc

[stark tree]

im not familiar with the circle and the x

between my velocity fields

[river path]

also happns when you dont just have u twice, but the FIRST vector field is divergence free (not necessarily the second!)

[river path]

it's the outer product (matrix where the ijth entry is u_i * u_j)

[stark tree]

oh

[river path]

so here the stuff in the derivatives is the first row of the outer product: u^2, uv, uw

and the next row would be uv, v^2, wv, and then uw, vw, w^2

and each of those you take the divergence of

[stark tree]

ah right

[river path]

it's a little shorthand

[stark tree]

you mentioned this btw can you expand?

[river path]

you wrote down $\partial_t u + u \partial_x u + u \partial_y v + u \partial_z w$ but i think you meant to write down $\partial_t u + u \partial_x u + v \partial_y u + w \partial_z u$

[untold delta]

ryc

[river path]

the easy way to remember the order is to remember that it looks like $[(u, v, w) \cdot (\partial_x, \partial_y, \partial_z)] u$. it's a directional derivative from multivariable calc

[untold delta]

ryc

[river path]

it's just that the direction is also a function now

and ofc, what i just wrote is exactly $(\vec{u} \cdot \nabla) u$

[untold delta]

ryc

[stark tree]

ok i always get a little confused with the notation since ive also read from literature thats its referred to as pseudovectorial notation

[river path]

ya

it's confusing

like when you apply a differential operator to a vector, you just apply it to each coordinate or whatever

lots of fluids comes down to keeping track of what indices match what haha

[stark tree]

so its v dotproduct grad v right?=

[river path]

yes

and the dot product happens first

then you apply the thing you get (which is a linear combination of derivatives) to each component of v

it says "the x component of v moves in the direction (u,v,w)"
and so do the y component and z component

this makes sense since as the fluid particles move, they conserve momentum (they bring their velocity components along with them in the direction they're moving) until a force or pressure forces it to change

[stark tree]

but if the dot product happens first

wait now im really confused

isnt the dot product that happens first = du/dx + dv/dy + dw/dz

and so if i multiply the scalar with the vector of u again i get u du/dx + u dv/dy + u dw/dz

[river path]

so

really important here

even for single variable functions, the operator $\frac{d}{dx} f$ and the operator $f \frac{d}{dx}$ are not the same

[untold delta]

ryc

[river path]

multiplication of functions and differential operators is not commutative

this sounds ridiculous if you havent seen it before, but this is actually the thing that causes heisenberg's uncertainty principle

[stark tree]

I mean yeah ive been told before that operators are supposed to be treated differently

since they arent really numbers or variables or whatever

[river path]

but in this case, it's important because if we multiply either of these by a function g, we get different answers

[stark tree]

which is also where my confusions from the beginning stems from how the u ended up in the derivative

[river path]

(g on the right)

so in this case, we have (v dot nabla) v

the stuff in the parenthesis first. since v is on the left, we can't swap the order of the derivative and v, so we just leave it next to the derivative. in this case, that gives u d_x + v d_y + w d_z (each component, and then summed)

[stark tree]

oh i see

[river path]

is that ok? after that we can apply to the v outside the parenthesis

now we have (u d_x + v d_y + w d_z) vec(v)

the left hand side is now a "scalar" (its not a vector anymore, after all the dot product outputs scalars)

so we just do scalar multiplication into each component of vec(v)

[stark tree]

yup i see

and now its on the right hand side so it applies to the differential

[river path]

exactly

[stark tree]

ok nice that cleared things up

[river path]

so in the first slot, you get all the differentials applied to u. in the second slot, they all apply to v. etc

[stark tree]

btw while we are on topic do you know perhaps where $\left(\vec{v}\cdot\nabla\right)\vec{v}=\nabla\left(\frac{1}{2}\vec{v}^2\right)-\vec{v}\times\nabla\times\vec{v}$ comes from?

[untold delta]

lennygo

[river path]

hmmm

[stark tree]

or $grad\left(\frac{1}{2}\vec{v}^2\right)-\vec{v}\times rot\vec{v}$

[untold delta]

lennygo

[river path]

anything with these double cross products is a huge mess

[stark tree]

I mean when i went to write it all out it is true

im just wondering how anyone came up with it

It comes from Schlichtings Boundary Layer Theorie if youve heard of it

[river path]

hmm

i dont have a good reason for why this should be true off the top of my head. it's oftentimes useful to find gradients that you can extract from one term or another because they can be stuffed into the p (which doesnt care as long as what you stuff into it is a gradient)

so at least on an algebraic level, it gives you a new nonlinear term v x rot v that you can screw with. which might be nice if you want to say stuff about vorticity (rot v)

[stark tree]

alr but thanks

you cleared up a lot

[river path]

ok

i guess it's the "rotational correction" in some sense

if you're rotating in a circle

rot v points you perpendicular to the plane of the circle

v points you tangent to the circle

so -(v cross rot v) (youve got to check the sign) points you towards the center of the circle

that's just like centripetal force

[stark tree]

huh

[river path]

grad(v^2) just says "go mostly in the direction where the velocity is growing" and -v cross rot v says "but make sure to get nudged towards the direction the veclocity field is turning". which together I guess adds to following the velocity field exactly.

shrug

its a good exercise to try to interpret this stuff as physics so i appreciate being made to do it

[stark tree]

its supposed to represent the convective acceleration btw

[stark tree]

Ok uhm shortly after my previous problem i have another one... has anyone here encountered div and Div?

The book author uses div and Div and there is a difference between them but he never goes into detail what the difference is

it just sort of jumps from one equation to the other

[buoyant pike]

Sounds bad

Can you share a picture of such an instance

[stark tree]

Its probably because these books care more about the answers and applications of the theory than the mathematical background

ive got a physical copy but i can quickly write it down

Explicitly for the Navier-Stokes-Equations under a symbolic writing:
$\rho\frac{D\vec{v}}{Dt}=\vec{f}-grad\left(p\right)+Div\left(\tau\right) with \tau=\mu\left(2\dot{\epsilon}-\frac{2}{3}\deltadiv\left(\vec{v}\right)\right)$

[untold delta]

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

[buoyant pike]

Potentially because tau is a tensor

[stark tree]

thats confusing then because matrix dont have divergences no?

or does that only apply to special tensors

[buoyant pike]

https://en.wikipedia.org/wiki/Tensor_derivative_(continuum_mechanics)#Divergence_of_a_tensor_field

[stark tree]

Yeah i was just about to say found the ressources for it, thanks

didnt even know you could do this

[buoyant pike]

But also if you're working with newtonian fluids this doesn't matter that much

[hallow pumice]

terrance tau?

[river path]

In fluids, divergence of a matrix always means "take the divergence of each row vector". The result is a column vector, each entry is the divergence of each row

[stark tree]

You know what im just gonna drop the book for now and switch to the Anderson and see what he has to say, a lot of people recommended that one

I bet Schlichting's book is pretty good but its just not a book that introduces you to the topics you need to know when approaching this stuff damn

[buoyant pike]

Boundary layer theory?

[bitter yacht]

Say we have a uniformly bounded sequence ${u_k}_{k=1}^\infty$ in $W^{1,m}(\Omega)$. Here $m$ is greater than the dimension $N$ of the ambient Euclidean space $\mathbb{R}^N$.

By weak sequential compactness of $L^m(\Omega)$, we have $u_k \rightharpoonup u \in L^m(\Omega)$ as $k \to \infty$.

And for each $i$:th weak partial derivative we have $ \displaystyle \frac {\partial u_k}{\partial x_i} \rightharpoonup \frac {\partial u}{\partial x_i} \in L^m(\Omega) $ as $k \to \infty$.

Say moreover we have a bound $C$ on the gradients, where the bound $C$ does not depend on $k$:
[
\left( \int_\Omega |\nabla u_k|^m dx \right)^{\frac 1m} \leq C < \infty
]

How can I show that this inequality passes over to the limit $u \in W^{1,m}(\Omega)$? In essence, how can I justify letting $k \to \infty$ in the inequality above? Namely I would like to verify
[
\left( \int_\Omega |\nabla u|^m dx \right)^{\frac 1m} \leq C
]

I have only managed to prove for each weak partial derivative that
[
\left( \int_\Omega \left| \frac { \partial u }{ \partial x_i } \right|^m dx \right)^{\frac 1m} \leq C < \infty
]

But how can this be used to obtain the inequality for the whole gradient $|\nabla u|$ and not just its components?

[untold delta]

hardisc

[lilac barn]

Prove that weak convergence, in general, implies Fatou's lemma for the corresponding norm.

As for the Gradient, once you have the m-norm bounded for each component, use lm into l2 inclusion, so you can use "Fubini" to bring the summation outside to get something like a bound NC

[untold delta]

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

[exotic void]

So many typos🤦‍♂️
Anyway I want to know why my proposed variational problem for $u(r, \theta)\in \dot{H}^1_{0, \text{div-free}}(\Omega)$ is wrong

[untold delta]

shiburin

[solemn junco]

are spectral bases like the nearly orthigional hat functions from the fem?

[buoyant pike]

A spectral basis is more like a fourier system

So increasingly oscillatory

[solemn junco]

ah is this related to the reinmann lebeque lemma?

[buoyant pike]

Sort of but not really?

[solemn junco]

alright, im just trying to understand this. my teacher talked highly of this book so im reading it

[buoyant pike]

A spectral basis is a truncation of a fourier series

[solemn junco]

got it, and the sums there look similar it as well

[buoyant pike]

Yes

[solemn junco]

got it, thanks for that. this book is really cool gotta say 🙂

the signials going to zero, and the the inverse problem is amplflying them blew my mind too cool

i appreciate the help man!

[bitter yacht]

thank you for the quick input, and apologies for a delayed response. I can see how by assuming $m$ is strictly larger than 2, we can use Hölder's inequality to show that $L^m$-integrability implies $L^2$-integrability. But I fail to see in detail how "Fubini" comes to our aid.

Of courseby definition we have $|\nabla u|^2 = \sum_{i=1}^N \left| \frac {\partial u}{\partial x_i} \right|^2
$, and this we can certainly integrate, and then interchange summation and integration, as the summation is finite.

How can we appropriately insert $ |\nabla u|^m$ into the picture here? I can certainly bound all the integrals $\int_\Omega \left| \frac {\partial u}{\partial x_i} \right|^2 dx$ upwards in terms of an appropriate constant, which in effect enables me to bound $\int_\Omega |\nabla u|^2 dx$ upwards.

[untold delta]

hardisc

[bitter yacht]

I wish I could say I knew exactly what you had in mind here, but unfortunately no. Fatou's lemma I certainly know, but in the context of norms and weak convergence, I am not quite sure what tool it is and how it potentially leads one right

[lilac barn]

Something like ( \lVert \lVert , \cdot , \rVert_{\ell^2} \rVert_{L^m} \leq C \lVert \lVert , \cdot , \rVert_{\ell^m} \rVert_{L^m} = C \lVert \lVert , \cdot , \rVert_{L^m} \rVert_{\ell^m} =C \left( \sum_{i\leq N} \lVert ,\cdot,\rVert_{L^m} \right)^{1/m})

[untold delta]

cocat

[lilac barn]

So written in gradient language, this means we have (\lVert \nabla u\rVert_{L^m}\leq C \left( \sum_{i\leq N} \lVert \partial_i u\rVert_{L^m}\right)^{1/m})

[untold delta]

cocat

[bitter yacht]

hi again, I have been trying to understand and implement your suggestions over the holidays, but have struggled to do so. For the inclusion, I certainly have that, for any $m$ such that $m > 2$:

\begin{align*}
\int_\Omega \left| \frac { \partial u }{ \partial x_i } \right|^2 dx
& \leq
\left( \int_\Omega \left| \frac { \partial u }{ \partial x_i } \right|^{ 2 \cdot \frac m2 } dx \right)^{ \frac 2m } |\Omega|^{ 1 - \frac 2m }
= \& =
\left( \int_\Omega \left| \frac { \partial u }{ \partial x_i } \right|^m dx \right)^{ \frac 2m } |\Omega|^{ 1 - \frac 2m }
\leq \ & \leq
\left\lVert \frac { \partial u }{ \partial x_i } \right\rVert_{ L^m (\Omega) }^2
|\Omega|^{ 1 - \frac 2m } \leq C^2 \cdot |\Omega|^{ 1 - \frac 2m } .
\end{align*}

For the $\ell^2$-norm of the gradient, I believe we have simply
[
\lVert \nabla u \rVert_{ \ell^2 }^2 = |\nabla u|^2 = \sum_{i=1}^N \left| \frac {\partial u}{\partial x_i} \right|^2
]

This I can certainly integrate term by term and bound upwards, but it seems like that leads us in the wrong direction.

For the $L^m$-norm of the vector $\nabla u$, I believe we have the natural definition:
[
\lVert \nabla u \rVert_{ L^m }^m = \int_\Omega |\nabla u|^m dx
]

But I fail to see now how exactly to implement the suggested "Fubini-maneuver" with the norms, where the $\ell^2$-norm $|\nabla u| = \sqrt{ \sum_{i=1}^N \left| \frac { \partial u }{ \partial x_i } \right|^2 }$ is buried inside the integral and moreover we have a pesky square root.

[untold delta]

hardisc

[bitter yacht]

Moreover, will this really ensure that we obtain $\int_\Omega |\nabla u|^m dx \leq C^m $ from $ \int_\Omega \left| \frac{\partial u}{\partial x_i} \right|^m dx \leq C^m $? It seems like a constant $N$, the dimension of the ambient space, manages to emerge in the sought inequality, but maybe I am just failing to see if and how it vanishes later.

[untold delta]

hardisc

[bitter yacht]

thanks, and happy holidays to everyone

[lilac barn]

I meant change l2 of gradient into lm then you can interchange Lm with lm

[solemn junco]

is b matrix like a matrix of fourier coefs? kinda like a matrix of sin and cos and the highest value at the top is associated with the lowest freq while the bottom v is associated with the highest frequence?

[unborn quiver]

Yes, that seems to be the case at first glance. Fourier coefficients are sometimes viewed just as a complex exponential as working with exponential tends to be cleaner

[lilac barn]

Absolute Fourier coefficients, there is an absolute value on it

[solemn junco]

rgr, yeah i see it. im self taught on these things so im just verifying my intution is correct 🙂

so this reimann lebeque lemma is essentially saying that as you go into the higher frequences (higher order of n in the fourier expansion) the signals (values) become zero right???

sorry this is a book about discreet inverse problems, so i think this the right place to ask, but im sure this is an analysis question im asking

its in relation to this part here

kinda beautiful how bessels/parsvaels inequalities come up

in a way this is similar to saying |bn| <= A/n^4

[lilac barn]

For an integrable function, yes

[solemn junco]

thank you very much 🙂

[tropic bramble]

why is reimann such a common misspelling of riemann

hwo does that even happen

[lilac barn]

Because i before e except before c is an arbitrary rule

[meager dune]

oh i didn't realise that lol cocat

but i think a lot of german names are unfortunately spelled wrong by english speakers lol

[solemn junco]

i didnt mean to disrepsect the legend. im just some dumb redneck 😛

[bitter yacht]

thank you, I think I managed to put it together, but I do have a follow-up question. here is how I went about it.

Hölder's inequality (for series) yields:
\begin{align*}
|\nabla u|^2 = \lVert \nabla u \rVert_{\ell^2}^2 = \sum_{i=1}^N \left| \frac {\partial u}{\partial x_i} \right|^2 & \leq \left( \sum_{i=1}^N \left| \frac {\partial u}{\partial x_i} \right|^m \right)^{\frac 2m} \left( \sum_{i=1}^N 1 \right)^{ 1 - \frac 2m } \leq
\ & \leq
\lVert \nabla u \rVert^2_{\ell^m} \cdot N^{1 - \frac 2m}
\end{align*}

Thus, raising this to $\frac m2$, we have
[
|\nabla u|^m \leq \lVert \nabla u \rVert_{\ell^m}^m \cdot N^{\frac m2 - 1}
]

Integrating yields (here I switch order, but it is trivial, as the sum is finite):
[
\int_\Omega |\nabla u |^m dx \leq N^{\frac m2 - 1} \cdot \int_\Omega \left( \sum_{i=1}^N \left| \frac {\partial u}{\partial x_i} \right|^m \right) dx
= N^{\frac m2 - 1} \cdot \sum_{i=1}^N \left( \int_\Omega \left| \frac {\partial u}{\partial x_i} \right|^m dx \right)
]

Using $\int_\Omega \left| \frac {\partial u}{\partial x_i} \right|^m dx \leq C$, we get
[
\int_\Omega |\nabla u|^m dx \leq N^{\frac m2 - 1} \cdot \sum_{i=1}^N C^m = N^{\frac m2 - 1} \cdot C^m \cdot N = N^{\frac m2} \cdot C^m
]

Raising to $\frac 1m$, we get at last
[
\left( \int_\Omega |\nabla u|^m dx \right)^{\frac 1m} \leq \sqrt N \cdot C
]

So we did not quite get $ \int_\Omega |\nabla u|^m dx \leq C^m $, but rather the factor $\sqrt N$ appeared. But the sought goal was to go from $\int_\Omega |\nabla u_k |^m dx \leq C$ to $\int_\Omega |\nabla u|^m dx \leq C$.

Is this not possible, or am I just failing to see an additional argument?

[untold delta]

hardisc

[lilac barn]

You will get something with N but that's just a determined constant so that can go inside C.

[bitter yacht]

ok interesting, thanks! I should have been clearer in my first question, but the $C$ appearing in both inequalities was intended to represent one and the same constant. So the question was if and how to justify that e.g. $\displaystyle \int_\Omega |\nabla u_k|^m dx \leq 4$ will imply $ \displaystyle \int_\Omega |\nabla u|^m dx \leq 4 $, when $k \to \infty$. but I suppose that is not the case then?

[untold delta]

hardisc

[bitter yacht]

for context I am reading Julio Rossi's paper on tug-of-war games and pde:s, where he argues as follows:

it seems as if the upper bound stays the same, even when passing to the weak limit $u_\infty$, upon letting $p \to \infty$ in the inequality obtained for $\int_\Omega |\nabla u_p|^m$. I do not see why he does not change the upper bound, even if it is not all that important for the subsequent results

[untold delta]

hardisc

[lilac barn]

That's just Fatou's lemma

[bitter yacht]

I thought so too, but we only have weak convergence, not pointwise convergence, right? If we had pointwise convergence
[
\lim_{p\to \infty} |\nabla u_p|^m = |\nabla u_\infty|^m
]
then surely by Fatou
\begin{align*}
\int_\Omega |\nabla u_\infty|^m dx & = \int_\Omega \lim_{p \to \infty} |\nabla u_p|^m dx = \int_\Omega \liminf_{p \to \infty} |\nabla u_p|^m dx
\leq \ & \leq
\liminf_{p\to \infty} \int_\Omega |\nabla u_p|dx
\leq \liminf_{p\to \infty} C= C.
\end{align*}

All weak convergence tells us is for each $i$ that
[
\lim_{p \to \infty} \int_\Omega \frac {\partial u_p}{\partial x_i} \phi dx = \int_\Omega \frac {\partial u_\infty}{\partial x_i} \phi dx
]
for each $\phi$ in the space dual to $L^m (\Omega)$. I think this weak convergence for each partial derivative can be turned into a convergence in $\mathbb{R}^N$ of the gradient vectors, namely
[
\lim_{p \to \infty} \int_\Omega (\nabla u_p) \phi dx = \int_\Omega (\nabla u_\infty) \phi dx \in \mathbb{R}^N
]

Are you saying this can be adapted so that a Fatou lemma emerges and gets us where we want to go?

[untold delta]

hardisc

[lilac barn]

Okay I thought you had pointwise convergence. Regardless, you can use the duality characterisation of norms and as test functions are dense in Lm' and you have convergence against these functions, you get that the norms converges

[mint canyon]

Is there any progress on 2 spatial and 1 time dimensional Inverse Scattering Transform for Solitons?

[bitter yacht]

Hm okay, I am trying to unpack your suggestions here. By "duality characterization of norms", do you mean the following: for any $x$ in a normed vector space $(X, \lVert \cdot \rVert)$ we have (via Hahn-Banach theorem)
[
\lVert x \rVert = \sup \left{ |F(x)| : F \in X', \hspace{5pt} 0 < \lVert F \rVert_{op} \leq 1 \right}
]
Here $ \lVert F \rVert_{op} := \sup { |F(x)| : 0 < \lVert x \rVert \leq 1 } $.

So in that case, since the weak limit $u_\infty$'s each partial derivative lives in $L^m (\Omega)$, we have
[
\left \lVert \frac { \partial u_\infty }{ \partial x_i } \right \rVert_{ L^m (\Omega) } = \sup \left{ \left| \int_\Omega \frac { \partial u_\infty }{ \partial x_i } \phi , dx \right| : 0 < \lVert \phi \rVert_{ L^{m'}(\Omega) } \leq 1 \right}
]

We have
[
\left| \int_\Omega \frac { \partial u_\infty }{ \partial x_i } \phi , dx \right| \leq
\underbrace{\left| \int_\Omega \frac { \partial (u_\infty - u_p) }{ \partial x_i } \phi , dx \right|}{ \to , 0 \text{ as } p \to \infty, \text{ by weak conv.} }
+
\left|
\int\Omega \frac { \partial u_p }{ \partial x_i } \phi , dx
\right|
]

So Hölder's inequality gives us
[
\left| \int_\Omega \frac { \partial u_\infty }{ \partial x_i } \phi , dx \right| \leq
\left|
\int_\Omega \frac { \partial u_p }{ \partial x_i } \phi , dx
\right| \leq \left \lVert \frac { \partial u_p }{ \partial x_i } \right \rVert_{ L^m(\Omega) } \lVert \phi \rVert_{ L^{m'}(\Omega) } \leq C \cdot 1 = C
]

Thus [ \left\lVert \frac { \partial u_\infty }{ \partial x_i } \right\rVert_{ L^m (\Omega) } \leq C ]

This seems to work for each component, if this is what you had in mind that is. But how do we get the same inequality for the whole gradient $ \left( \int_\Omega |\nabla u_\infty|^m dx \right)^{\frac 1m} \leq C $?

[untold delta]

hardisc

[limpid stirrup]

I have great trouble using the geometric method for inhomogenous variable coefficient linear pdes... I've encountered the Lagrange-Charpit equations, are there other good methods?

[atomic pond]

hello, I need to show that the billinear form a is coercieve. It is suggested to show the poincaré inequaility (on the first subspace V for example). Once I showed it, how the coercivity follows? Because we will have |Du|^2_L^2>= C|u|^2_L^2 and the lower bound is not the norm of u in V. How to correct this please?

[spare sentinel]

I've got a question regarding uniqueness for the PDE

$$ \partial_t u + (-\Delta)^{1/2}u = u |d_{1/2} u|^2 $$

where

$$ |d_{1/2} u|^2 = \int_\mathbb{R} \frac{|u(x) - u(y)|^2}{|x-y|^2} dy. $$

Now, I have a weak solution in ( H^1(\mathbb{R};\mathbb{S} ^{n-1})) and in ( C^1([0,T];L^2) ), which is great, but for a robust uniqueness proof, you would typically need ( L^2([0,T];H^1(\mathbb{R})) ).

I already have this in ( L^2([s,T];H^1(\mathbb{R})) ) for all ( s > 0 ).

The problem is that the approaches I have tried all failed because of the pseudolocal behaviour of the fractional Laplacian and the complex combination of Sobolev-Slobodeckij and Bessel spaces, which are both contained in my PDE through ( d_{1/2} ) and the Laplacian. My general question, where I am open to any ideas, is: What methods exist to combine the behaviors of the Sobolev-Slobodeckij and Bessel potential integrals, i.e., the ( d ) and the ( -\Delta ) terms?

What are general approaches to achieving this kind of integrability (in a normal setting, ( H^2 ) would be sought instead)?

Are there other approaches to establishing uniqueness for these kinds of evolution equations?

Thanks in advance for any response; I just need more ideas!

[untold delta]

KIl

[astral vine]

Bessel and Sobolev-Soblodeckij spaces are the same over L²

[spare sentinel]

I know that, its more about the inner part of the seminorms especially locally, like the |d_1/2 u|^2 and the (-Delta)^1/4 u, how are integrals over Omega similar to each other, if they even are

[astral vine]

But you are on the whole line

[spare sentinel]

Yes indeed

But that is the problem

[astral vine]

other wise uniqueness in finite time

[spare sentinel]

Oh sry not the line in time is tge problem, the line in space

[astral vine]

Yes

I didn't mixed up

There is the maximal L^q(L^p) regularity property for the Poisson semigroup

[spare sentinel]

Yes there is, the part which needs local behavior is the integrability of the function over time. The approach if you are looking at the normal harmonic heat flow leads to a localised look on the integrals and then a cover argument, a direct act on the whole space leads to problems,

The problem is that the fractional laplacian isn't a local operator only a pseudo local, which leads to a similar local estimate, but an estimate which is not suitable for covering arguments.

That's why I would like to switch to slobodeckji locally somehow with the fractional gradient as well to get some kind of local behavior, but the different norms and thus different scalar products makes it hard to do that

The maximal property was already used in the duhamels principle for getting a mild and then a weak solution

[astral vine]

but you can also use it to prove uniqueness

https://arxiv.org/pdf/2311.07971.pdf

Theorem 3.6 pages 7-8

[spare sentinel]

Ok wait now I am confused I misread the property you meant, this looks like an idea I can try thank you!

[astral vine]

Play with Sobolev embeddings, Holder and interpolation inequalities

[spare sentinel]

How do you people always have some kind of paper which at least gives one always some ideas, I mean how much paper do you all read😂

[spare sentinel]

Jup thats always the best to do

[astral vine]

I am really use to this topic, so you have been just lucky there

[spare sentinel]

Happy me

[solemn junco]

anyone have any suggestions? ive been on it for like 5 hours and gotten no here

did the previous two

[solemn junco]

do i plug this in for the integral?

for f?

[mint canyon]

What's the most general strategy for Linear PDEs? I suspect it's separation of variables, but what's the systematic method to determine the ansatz in that case?

[tired sinew]

I have trouble with this Green's function. Can someone help me?

[atomic pond]

daniil9274

[ebon shard]

you could try mirroring across the x-axis and the boundary of the circle

[cedar frigate]

"linear PDEs" is a very large class of problems, I don't think there is any "general" technique.

Broadly they break down into elliptic, parabolic, and hyperbolic PDEs, and there are some standard techniques within each family that work fairly broadly.

One example is the method of characteristics for hyperbolic PDEs. It is a way of turning a PDE into an ODE along certain special curves, and those ODE solutions can be combined into a PDE solution.

https://en.m.wikipedia.org/wiki/Method_of_characteristics

PDE is kind of annoying in that there is no single winning technique, there are many, many different tricks, many different ansatz that work in different situations, and many, many PDEs that are not solvable exactly by pen and paper at all.

[mint canyon]

Im familiar with 2nd order method of characteristics, but as you mentioned it doesnt cover the elliptic case. Im also not particularly concerned with boundary conditions since the use case here is for solving the rather arbitrary pdes you encounter when solving the symmetry condition for a nonlinear equation.

[cedar frigate]

There are things you can say about when separation of variables is possible. You can always search for your "separated" solution, the question is whether every solution can be written in terms of the special solutions you find, and you can probably imagine this is a question about whether a set of vectors spans a vector space (the equation is linear after all).

One sufficient (but maybe not necessary?) condition is that your differential operator be self-adjoint. In that case, the spectral theorem guarantees you a basis of eigenfunctions, and separation of variables is possible. There is a short discussion on the separation of variables page on Wikipedia.

[mint canyon]

So for a complex matrix A to be self adjoint, does that mean the real part of the matrix is symmetric and the imaginary part is anti-symmetric?

[mint canyon]

this looks promising

https://ntrs.nasa.gov/api/citations/19700020429/downloads/19700020429.pdf

[cedar frigate]

I don't typically think of operators on a space of functions in terms of a matrix because the space of functions is usually infinite-dimensional.

I would instead look at the inner product definition of self-adjoint. I believe the right inner product to use here is the L^2 inner product.

[mint canyon]

You can, the Heisenberg form of quantum mechanics uses matrices as linear operators instead of the schrodinger equation. Then each basis of the vector space corresponds to some linearly independent function in Hilbert space.

[cedar frigate]

Yeah, but you can get matrices with infinitely many entries

[mint canyon]

in that case it doesnt matter since the higher eigenvalues correspond to energy levels that dont exist in any practical sense. SO you can just cut them off as they dont contribute much to the probability.

[cedar frigate]

Physics gets away with a lot 😋

[mint canyon]

yeah... i think it took mathematicians a few decades to formalize and prove Path Integrals, by that time it was a standard method in physics.

[cedar frigate]

They are still not rigorous except in some special cases believe it or not

We have one of the leading path formulation guys at our school

He worked at Cal Tech and would always run up to Feynman saying "we figured out such and such about the path formulation!". He would say "does it change the physical answers?"...."No..."...."then who cares!"

[mint canyon]

You know, I hadn't considered that functional integration might apply more broadly to the space of linear PDEs than the schrodinger equation, but there does appear to be ongoing work in that area.

https://www.amazon.com/Functional-Integration-Differential-Equations-Mathematics-ebook/dp/B088D73S5L

[lilac barn]

A complex matrix is self-adjoint if it's complex conjugate equals it

[mint canyon]

Yeah, I get that but I was trying to figure out what the conditions for that would be

also its the conjugate transpose

[lilac barn]

You can explicitly write down the condition elementwise if you write down the equality for a general self-adjoint matrix

[mint canyon]

Okay. Did you know of any particularly general approaches to linear PDEs? 2nd order is probably fine, havent run into much higher than that.

I know "completely general" doesnt exist, just like more general than they teach in Uni, and where the eigenfunctions arent in a table somewhere.

,help

[untold delta]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[lilac barn]

A somewhat general approach is method of characteristics.

If you're familiar with Fourier analysis, then that also works in a whole bunch of cases.

There's also semigroup theory and more generally the Duhamel's principle which helps you deal with evolution pdes

[mint canyon]

Fourier analysis relies on complex exponential or sin/cos eigenfunctions doesnt it?

[lilac barn]

Yes

[mint canyon]

if you wanted to use an integral transform on something that separates into Bessel functions you would need a Henkel transform etc.

I mean any linear operators solution "could" be represented by any set of orthogonal functions, but the preference is that they diagonalize the operator, or the known methods kind of break down

[lilac barn]

Yes this is how you can view Fourier transforms as well: an operation that diagonalises the convolution

[mint canyon]

I suppose its also important that these linear PDEs will always be homogeneous, so no need of greens functions or anything

[signal epoch]

https://dsp.stackexchange.com/questions/34842/convolution-of-two-exponential-signals

does anyone know why our integral of convultion limits go from -00 to +00 and it changes to 0 to t

and we multiply our y(t) for u(t) which is the step function

i dont understand why can anyone help?

[inland sinew]

Hi everyone. Im working on a linear PDE right now that I realized has a strange form. It is variable coefficient(they are periodic) and I realized the operator actually decomposes into two parts, where one part is really just a coordinate rotated version of the other one. For example It can be written something like (L1 + L2)z = 0, where if I just had L1, I could change it to look like L2 by the change of coordinate u = x+y, v = x-y. I know how to solve the problem if I just have the single operator L1z=0 or L2z=0, since in essence they are the same type of operator after a coordinate rotation

Is there any literature on this type of problem? Where the differential operators have this relationship?

[mint canyon]

could you show the PDE?

[inland sinew]

(1+cos(x))cos(y)zxx + (1+cos(y))cos(x)zyy + 2sin(x)sin(y)zxy = 0, notice that I can write this as [(cos(y)zxx + cos(x)zyy)] + [cos(x)cos(y)(zxx + zyy) + 2sin(x)sin(y)zxy] = 0. The two parts in brackets are almost the same if you do a rotation and a scaling. So for example, if you take the second part and do a change of variables u = x+y, v = x-y, you'll get the form of the pde in the first bracket with a factor of two. So thats why I was thinking in essence these are kind of the same operator after a rotation and scaling. If you're just trying to deal with cos(y)zxx + cos(x)zyyy for example, you can easily do separation of variables to get a product of mathieu functions for your solution.

The main theme of this problem that is interesting but also extremely annoying is that it feels like there are 4 key directions, not 2. If I just had either operator, the directions would either be (x+y) and (x-y), or x and y. I've noticed this over and over again when I analyze the problem. The main insight I gained is that you can write this problem as ((1+cos(x)cos(y)zx + sin(x)sin(y)zy))x + (sin(x)sin(y)zx + (1+cos(y))cos(x)zy)y = 0, which helped me find its also derived from a lagrangian, and when I have the rewritten in the form I just wrote, it looks like the divergence of a vector field. I figured something in the vector field formulation could help me figure out whats going on with the solution.

[mint canyon]

so is zxx or zxy shorthand for ∂²z/∂x² and ∂²z/∂x∂y?

[inland sinew]

yes

[mint canyon]

The characteristic equation for that should be:
(1+cos(x))cos(y)(∂φ/∂x)² + 2sin(x)sin(y)(∂φ/∂x)(∂φ/∂y) + (1+cos(y))cos(x)(∂φ/∂y)² = 0

which can be solved by lagrange-charpit

(assuming you can figure out the integrals)

(also this only works if the equation is not elliptic, which I didnt check. The signs make it look like it might be elliptic, at least in some regions.)

[inland sinew]

I see. I was looking at that at one point, but these would be characteristics of the energy like function no, not the dependent variable z? That would still be a nice result. The discriminant is also mixed so it would only be valid in certain regions. The domain is thought of as periodic, with each unit cell being mostly elliptic, but the regions between them transitioning from hyperbolic to parabolic.

[mint canyon]

My understanding is hyperbolic, elliptic, parabolic classification is pointwise, so the characteristics might be valid in some places and invalid in others. Keep in mind that unlike first order characterists, second order characteristics are not solutions. They propagate boundary conditions over the region of the solution.

(z is constant along the characteristic)

(I used the variable φ to emphasize that the characteristic is not a solution)

[inland sinew]

I see. I appreciate the help. I am trying to work this out for the easier problem where I am just looking at cos(x)(cos(y))(zxx + zyy) + 2sin(x)sin(y)zxy = 0, because I do know at least one infinite family of solutions to it and then I can try to see how those might emerge from the lagrange charpit equations.

Is there any good interpretation of whats going on when my characteristics transition from regions where they are real in the hyperbolic regions to the complex in the elliptic regions? Because in my case, that will be happening periodically in this medium.

[mint canyon]

I havent finished reading it yet, but if you wanted to continue on your separation of variables direction I found this paper earlier

https://ntrs.nasa.gov/api/citations/19700020429/downloads/19700020429.pdf

Separation of variables seems to be associated with point symmetries, though the symmetry condition is not useful for linear equations since the original equation recurs in the decomposition

[rose creek]

Does anyone know the lowest energy solution (and preferably other solutions too) of the time-independent Schrödinger equation for a logarithmic potential in two dimensions?

[mint canyon]

this −(h²/2m)∇²ψ + ln(x)ψ = Eψ?

[rose creek]

yeah that looks right

I don't care about the constants

so just -∇²ψ + ln(x)ψ = Eψ

[mint canyon]

alternatively there is -∇²ψ + ln(y)ψ = Eψ or ∇²ψ + [ln(x) + ln(y)]ψ = Eψ

[rose creek]

oh

in that case I mean ln(sqrt(x^2 + y^2)) is the potential

[mint canyon]

yeah, sorry there was some ambiguity in the question. Unfortunately standard separation of variables techniques are going to fail and you wont be able to find the eigenfunctions. Supposing there is an eigenfunction we don't know, can we predict the eigenvalues despite that? I need to go look in griffiths for a refresher.

[rose creek]

I already have the solution for a -1/r potential in 2D (just a modified version of the 3D potential) but a real coulombic potential in 2D is ln(r), not -1/r

[mint canyon]

thats a well known problem though, the eigenfunctions in the 3d case are the associated legendre polynomials

(i think)

[rose creek]

legendre polynomials multiplied by e^(-r)

[mint canyon]

if you found the eigenfunctions in some literature, Im not saying you're wrong, Im just saying I dont know them, and the R(r)O(θ) ansatz wont work.

right okay, the product of the two was the spherical harmonics

[rose creek]

why not?

I mean I don't know if the logarithimic potential will have eigenfunctions based on the legendre polynomials, but the solutions should certainly still be decomposable into radial and angular components

[mint canyon]

So we talked about this a while back, but the condition for separability is the differential operator must be self-adjoint, once you know its separable, you still dont know the ansatz though.

(separation of variables does not work on every linear pde)

This paper may also be relevant to you since we are in 2 dimensions

https://ntrs.nasa.gov/api/citations/19700020429/downloads/19700020429.pdf

[steel umbra]

can someone tell me if my understanding is right?

for burger's equation, a smooth solution will only stay smooth if the initial condition is monotonically increasing (so that no shock waves occur. instead it will form rarefaction waves), and discontinuous solutions will stay discontinuous unless they're monotonically increasing (where they will eventually become continuous as rarefaction waves).

[rose creek]

Well technically speaking a discontinuous solution might become continuous and then discontinuous again if it is monotonically increasing in the neighborhood of the initial shock but decreasing somewhere else

[steel umbra]

interesting. so it should instead be that a solution that decreases anywhere will eventually stay discontinuous?

[rose creek]

I don’t know

But if it’s decreasing anywhere I think it should become discontinuous at some point

I don’t know if that can later become continuous again

[mint canyon]

here is the symmetry condition for burgers equation

x̄ = ξ∂/∂x + τ∂/∂t + η∂/∂u + η_x(∂/∂u′) + η_t(∂/∂u̇) + η_xx(∂/∂u″)

η_x = Dη/dx - (∂u/∂x)Dξ/dx - (∂u/∂x)Dτ/dx = ∂η/∂x + ∂u/∂x(∂η/∂u) - (∂u/∂x)[ ∂ξ/∂x + ∂u/∂x(∂ξ/∂u) ] - (∂u/∂x)[ ∂ξ/∂x + ∂u/∂x(∂ξ/∂u) ]

η_t = Dη/dt - (∂u/∂t)Dξ/dt - (∂u/∂t)Dτ/dt = ∂η/∂t + ∂u/∂t(∂η/∂u) - (∂u/∂t)[ ∂ξ/∂t + ∂u/∂t(∂ξ/∂u) ] - (∂u/∂t)[ ∂ξ/∂t + ∂u/∂t(∂ξ/∂u) ]

η_xx = Dη_x/dx - (∂²u/∂x²)Dξ/dx - (∂²u/∂x²)Dτ/dx = ∂/∂x(∂η/∂x + ∂u/∂x(∂η/∂u) - (∂u/∂x)[ ∂ξ/∂x + ∂u/∂x(∂ξ/∂u) ] - (∂u/∂x)[ ∂ξ/∂x + ∂u/∂x(∂ξ/∂u) ]) + (∂u/∂x)∂/∂u(∂η/∂x + ∂u/∂x(∂η/∂u) - (∂u/∂x)[ ∂ξ/∂x + ∂u/∂x(∂ξ/∂u) ] - (∂u/∂x)[ ∂ξ/∂x + ∂u/∂x(∂ξ/∂u) ]) - (∂²u/∂x²)[ ∂ξ/∂x + ∂u/∂x(∂ξ/∂u) ] - (∂²u/∂x²)[ ∂ξ/∂x + ∂u/∂x(∂ξ/∂u) ]

x̄( ∂u/∂t + u∂u/∂x - ν∂²u/∂x²) = 0
-> (∂u/∂x)η + uη_x + η_t - νη_xx = 0

just need to substitute ∂²u/∂x² = ( ∂u/∂t + u∂u/∂x)/ν when ever it appears in the above

its linear in ξ, τ, and η, since those functions only depend on x, t, and u, you can make a decomposition on the derivatives of u into a really large system of usually trivial pdes

[inland sinew]

@mint canyon I looked at charpits method agian yesterday for the more simple case with cos(x)cos(y)(zx^2 + zy^2) + 2sxsyzxzy = 0. The equation reduces a lot because of the trig identitites and because these coefficients are related by derivatives. I feel like I should be able to get an exact integral here but I'm not seeing it

This is what I have.

I left out most previous steps because it's too long. But I feel like I'm overlooking something here.

[mint canyon]

did you try the parameter independent version?

well its easy to derive from the parametrized version

[inland sinew]

Not yet. I sort of have the parameter t in the beginning and then eliminate it through these reductions I guess. I initially thought after rearranging some of these, I could equate the symmetric equations again to integrate

What I mean is after I manipulate the first two and define new variables for u = sinx and v = siny, getting this relatinship dv/du = q/p, Id still be able to say dv/q = du/p = d(p^3)/u = d(q^3)/v, but I think thats not correct.

[mint canyon]

I mean, if its integrable, then integrate it and plug it in to the characteristic equation to see if its satisfied

https://core.ac.uk/download/pdf/51404814.pdf

look at equation 14

[inland sinew]

Okay I will look. I think I saw this paper earlier at one point. Maybe it will be a good read through

[mint canyon]

looking at the differentials I suspect a typo in that figure... my bad

[inland sinew]

I find this PDE strange because in its original form I showed you, I looked at coordinate transforms originally using characteristics to change it into a canonical form, for example eliminating the zuu or zvv coefficients in the new coordinate system. But doing so requires solving PDE

[mint canyon]

one should definitely be dq

[inland sinew]

solving these PDEs that have this quadratic form structure. and then I noticed this entire pde is actually derived from a lagrangian that has the exact same quadratic form structure.

[mint canyon]

you mean its an euler-lagrange equation?

[inland sinew]

yes

[mint canyon]

huh, I wonder if thats a general result or a coincidence

[inland sinew]

So for the equation cosxcosy(zxx + zyy) + 2sinxsinyzxy = 0, the corresponding lagrangian is L = (1/2)(cosxcosy)(zx^2 + zy^2) + sinxsinyzxzy. The class of PDES ive been studying are derived in a certain way so that the coefficients end up being derivatives of a similar functions. So in this PDE, they are second derivatives of a function cosxcosy. The general form of these equations are fyyzxx + fxxzyy - 2fxyzxy = 0, where f is a function you'd know that determines the coefficients. I think thats why it works out this way, but the derivations that leads to that equation doesnt use a lagrangian. The fact that it can also be derived from a lagrangian is just something I noticed the other day which helped a lott

The lagrangian for any of these is based off a quadratic form of the Hessian of the function f(which again we choose), and the "vector" formed by gradient of z(x,y).

[mint canyon]

Okay, you had me intrigued for the wrong reason, I thought you meant the characteristic equation was also an euler-lagrange equation. This is definitely specific to your problem

it would have been neat if there were some kind of variational description of the characteristic

[inland sinew]

Also I was thinking, when one uses characteristics to change coordinates for a second order PDE, they tend to have a standard procedure for eliminating these coefficients. So if you are doing a coordinate change from x,y to u,v, you can try to figure out what u and v have to be to say eliminate the coefficients on certain second derivatives in your new equation. So they have the standard form for parabolic, elliptic, etc. But when they go through this procedure, they usually do something like this.

So for example if the PDE we start with is Azxx + Bzxy + Czyy = 0, then the coefficient on zuu in the new equation would be Aux^2 + Buxuy + Cuy^2, and we would try to set that equation equal to zero and solve to determine u if we wanted. Usually when I see this done, they just divide through by uy or ux, demand that du = uxdx + uydy = 0, and then plug in the relationship for dy/dx and then use the quadratic formula to obtain some equations you need to integral dy/dx for to obtain the characteristics.

Okay my question is, if you get to the point of Aux^2 + Buxuy + Cuy^2 = 0, which you want to satisfy for your coordinate transform, why not just use the lagrange charpit method to figure out some expression, even if its implicit, for u. If I try the standard way of doing this problem where they end up with some step that says dy/dx = -B +- sqrtt(B^2 - 4AC)/2A, its totally impossible to do these integrals in some instances. If I use the lagrange charpit method there to solve that integral, am I gaining anything? The only thing I could think is it might implicitly give me an answer that relates u to x and y but Im not sure.

[mint canyon]

Ive seen the quadratic formula version of the second order method too, but it only works in the 2 variable case, so i just remember the generalized multivariable version

also, if you dont have constant coefficients, do you really want to deal with integration of the quadratic formula?

[inland sinew]

Yeah I agree. I'm just wondering why its so hammered into every second order linear pde approach I see. Like they never use the lagrange charpit method for solving those coordinate transform equations.

[mint canyon]

there may be some condition required for lagrange-charpit to work, but its the best method by far for fully nonlinear first order equations

Symmetry methods are a fallback and guaranteed to exist, but the condition for that can be intractable quite often

[inland sinew]

I haven't found a single example though where lagrange charpit was used on a nonlinear first order equation of my kind though, where its quadratic and has the mixed term zxzy in it. If I found more examples of that, it could help me.

What I mean is any example with azx^2 + bzy^2 + czxzy = something

[mint canyon]

I saw a paper where it was used on Hamilton-Jacobi, thats got quadratic derivatives

Though its actually really easy to solve with separation of variables despite being strongly nonlinear

[inland sinew]

Yeah I need to look more. I do see people use it for hamilton-jacobi which is pretty cool. Do you know any other good resources where I can read up on the lagrange charpit method and examples? I do have that one you sent me going over the rigor of it

[mint canyon]

I had a derivation memorized at one point, since its actually very close to Hamiltons equations of motion, except its in a 2n+1 dimensional "phase space"

(actual phase space is always 2n dimensional)

[inland sinew]

Prob time for me to go but I'll have to look up more resources on this for sure

[mint canyon]

Here's the starting point:

F(qᵢ, u, ∂u/∂qᵢ) = 0 -> F(qᵢ, u, pᵢ) = 0
qᵢ′ = dqᵢ/ds, u′ = ∑∂u/∂qᵢ(dqᵢ/ds), pᵢ′ = ∑∂pᵢ/∂qᵢ(dqᵢ/ds)

Then you choose qᵢ′ = ∂F/∂pᵢ

and substitute into the other equations

u′ = ∑∂u/∂qᵢ(∂F/∂pᵢ)
pᵢ′ = ∑ ∂pᵢ/∂qᵣ(∂F/∂pᵣ)

the pᵢ are an issue since F contains no derivatives of p, but if we look at DF/Dqᵢ expanded with the chain rule we get:

DF/Dqᵢ = ∑ ∂F/∂qᵢ + pᵢ∂F/∂u + ∑∑ ∂pᵢ/∂qᵣ(∂F/∂pᵣ) = 0

rearrange and you see that: ∑∑ ∂pᵢ/∂qᵣ(∂F/∂pᵣ) = - ∑ ∂F/∂qᵢ + pᵢ∂F/∂u

so pᵢ′ = - ∑ ∂F/∂qᵢ + pᵢ∂F/∂u

now get rid of the parameter:

dqᵢ/[ ∂F/∂pᵢ ] = du/[ ∑∂u/∂qᵢ(∂F/∂pᵢ) ] = dpᵢ/[- ∑ ∂F/∂qᵢ + pᵢ∂F/∂u ]

There are your first integrals

[steel umbra]

can we not remove the (4pi vt)^(-1/2) out of the log and neglect it since its constant wrt x?

supposed to derive that in my hw basically and i'm wondering if this step is ok or not (we used epsilon instead of nu)

it seems fine to me, but idk why they wouldn't simplify that further unless maybe something sketchy happens idk

[mint canyon]

This is the differential equation Im most interested in at the moment:

(½gᵢᵣgᵤᵥ - gᵢᵤgᵣᵥ)(δS/δgᵢᵣ)(δS/δgᵤᵥ) = 0
(½gᵢᵣgᵤᵥ - gᵢᵤgᵣᵥ)((∂S/∂gᵢᵣ - ∂/∂xₐ[∂S/∂gᵢᵣ,ₐ])(∂S/∂gᵤᵥ - ∂/∂xₐ[∂S/∂gᵤᵥ,ₐ]) = 0
(½gᵢᵣgᵤᵥ - gᵢᵤgᵣᵥ){(∂S/∂gᵢᵣ)(∂S/∂gᵤᵥ) - (∂S/∂gᵢᵣ)∂²S/(∂xₐ∂gᵤᵥ,ₐ) - ∂²S/(∂xₐ∂gᵢᵣ,ₐ)(∂S/∂gᵤᵥ) + ∂²S/(∂xₐ∂gᵢᵣ,ₐ)∂²S/(∂xₐ∂gᵤᵥ,ₐ)} = 0

[mint canyon]

27 independent variables 🧐

[rose creek]

what's the equation for

[mint canyon]

its a single scalar equation for the Hamilton-Jacobi form of ADM

ADM is the canonical field equations for general relativity, if you want to represent it as a Cauchy problem

well, its the vacuum version, so no matter/sources

[bronze gate]

<@&268886789983436800>

[mint canyon]

did I break some kind of rule?

[pale cave]

It's likely there was spam that was deleted

[mint canyon]

ahh I see

[mint canyon]

I have no idea whats covered in that book, but the book I used was "A First Course in Differential Equations" D Zill

also, there is another channel called odes-and-pdes for more intro questions

[fervent flame]

Suggest me a starter book for pde I tried 2,3 books but I find it hard to learn please help me I liked shepley l Ross book on ode I need book something like this

[mint canyon]

my pdes book for undergrad wasnt too amazing, but it was cheap (dover). let me get the title/author

"Fourier Series, Transforms, and Boundary Value Problems" JR Hanna, JH Rowland

There are better books, but it has the info in there and it doesn't get too bogged down in rigor/formalism

Contrary to the title it also has Bessel functions and Legendre polynomials in it

[fervent flame]

This is my syllabus

[mint canyon]

Yeah sorry, my book focuses entirely on boundary value problems, it has nothing on method of characteristics. Maybe grab a high rated math methods for physicists type book that covers all the pde stuff in the table of contents?

[fervent flame]

I need just a basic starter book

Which cover most of these topics

I am just doing for exam I hat pdes 🫣

[mint canyon]

yeah, thats why I recommended the math methods books, physicists dont care about rigor at all, it will be purely algorithmic discussions in those books

[fervent flame]

Oh! thank you so much

[inland sinew]

@mint canyon I made a little more progress on this general problem. I realized that there is another way you can formulate it even more simply that involves three linear first order pdes of 3 variables. The equivalent formulation is.

ux + fxzx = 0
vy+ fyzy = 0
uy + ux + fxzy + fyzx = 0

Where f is known. If you eliminate u and v from the equation, you get back
fyyzxx + fxxzyy - 2fxyzxy = 0

I like this simpler form better because uts first order and linear but im wondering if its too broad to draw results from. I think I should be able to get something from it, but I need to define new variables of some kind. I mean for example one clear thing is I can get new equations from adding and subtracting these, one being.

(u+v)(x+y) + f(x+y) * z_(x+y) = 0

Which looks very close to some sort of transport equation. In special cases the equations reduce nicely. What do you think? I feel like this set up is actually nicer and easier to draw results from as a starting point for a given f

[mint canyon]

you cant use method of characteristics on systems of equations unfortunately

also, I never know whether im looking at dependent or independent variables when you write your equations like that

[inland sinew]

Thats true. The general approach people use sometimes with systems of first order pdes is to solve for one of the variables. The point here is that that is exhausting and makes the problem harder. I want to keep it in this form with three independent variables and analyze it that way.

Similar to how with a state space form, its easier to draw conclusions frok that formulation instead of solving for one variable and looking at its higher order ode

My bad. u,v, and z are the dependent variables of x and y

[mint canyon]

is fx f*x or ∂f/∂x?

i dont remember coefficients of x in the original pde, so i dont think they would arise from simple coord transforms

unless its like x = arcsinx' or something

[inland sinew]

Yeah when Im writing fx or zx I mean the partials

[cedar frigate]

This is a classic written by masters that is more explicit than Arnold

https://www.amazon.com/Differential-Equations-Dynamical-Systems-Introduction/dp/0123820103?ref=d6k_applink_bb_dls&dplnkId=132136d1-d6b9-4118-97ea-b2aaed4e8640

Though Arnold is awesome imo

Definitely gotta do some leg work to read it

[river path]

Holy based

I recommended the same book to them in another channel

hirsch smale devaney!!!!!!!

[cedar frigate]

Good taste lol

[magic nymph]

hi , how one can get the second inequality here $u \in W^{2,p}(B_{4} )$; $B_{4}={x\in R^{n},|x|<4 } $ , $\overline{\nabla u}=\int_{B_{4}}\frac{\nabla u}{|B_{4}|}$ and there is no assumption for n and $p>2$ , i tried to use Sobolev inequalities but i failed , any hint? okey i think that i solved that : i will apply mean value property for \nabla^{2} v since it is harmonic because v is

[untold delta]

amani

[pure field]

I know this isn't really the place to ask so I apologise however I asked this a while ago in book recommendations and got no response. Does anyone know of any PDE books (preferably more applied) that cover coupled pdes or systems of pdes.

[harsh veldt]

the one I read is Numerical Method for Conservation laws by LeVeque if you are interested in hyperbolic systems. do you have an idea of the applications you’re interested in?

[stark wraith]

is this substitution correct?

[pure field]

Ah, I am currently just doing a methods course so little to no theory at the moment. I think I found a section in some small dover book. The reason why I was interested is because I have a more theoretical course next year and thought since I have the time and I am interested, might aswell have an early look. Thanks though, I will keep that book in mind 🙂

[pure field]

https://discord.com/channels/268882317391429632/908110794569449553

[stark wraith]

iam not getting any reply there

[mint canyon]

this is the pde channel, we dont talk about odes here

[wicked tendon]

I need some help with this question

I introduced function h(x) and h'(x) that satisfy the needed conditions

then y + ah(x) and y' + ah'(x) are 1 parameter (a) functions at which K is stationary

But I never used the PDE anywhere in my answer

[inland sinew]

I dont think you need that EL pde to prove that right?

I think that is just a statement about the condition for stationarity.

[wicked tendon]

The way I see it, I need to prove: EL pde holds => K is stationary w.r.t small variations

[inland sinew]

Ah you're right actually. I misread it

[inland sinew]

Are there any theorems on any analysis done on how the solution to a lagrangian, say L = 0, relates to the solution of its corresponding euler lagrange equation?

What I mean is L = 0 might be a pde/ode in its own right that one could solve, and the correspinding EL equation is a higher order equation thats typically harder to solve. Like when could I get away with saying the solution to L = 0 satisfies the corresponding euler lagrange equation

[steel umbra]

gonna be real, i have no idea what to do here. im reading the proof of the mean value formula in Evans (where the problem comes from) and i do not see at all how to get to this. are we supposed to use something like
u(0) = convolution of the fundamental solution with f = the average integral over a ball centered at 0
and then somehow manipulate it to get there? idk

[unborn quiver]

In Evans, there should be something defined like
[\varphi(s) = \frac{1}{n\alpha(n)s^{n-1}}\int_{\partial B(0,s)}u(y)dS(y)] in the proof of the mean value formula. Look at $\int_{\epsilon}^r \varphi'(s)ds = \varphi(r)-\varphi(\epsilon)$. I think the motivation of this problem is to arrive to $u(0)$ independently from the fundamental solution

[untold delta]

カービィ

[unborn quiver]

epsilon here is set because phi(s) is clearly not defined at 0 so you need to be careful

I think the involved part here is calculation, and there should be enough information that pops out of this that you might be fine

[tired hollow]

What are the textbooks for beginning graduate students or advanced undergraduates who want to look into PDEs?

[lilac barn]

Evans' PDE

[tired hollow]

After finishing, I would be PDE master?

[lilac barn]

In front of highschool students? Sure

[tired hollow]

What comes after Evans?

[lilac barn]

Quite a lot of things, the book itself will inform you on that

[river path]

If you want to learn about the important stuff that forms the backdrop for modern PDEs, I really recommend Distribution Theory and Fourier Analysis by Strichartz. Evans doesn't really get into that stuff, Evans is more traditional. Evans + Strichartz is a great foundation.

The most important parts of evans are probably chapters 2, 3, 5, 6, and 8? If you're looking for a subset to focus on

[civic turret]

This is a Millenium Problem solution, right? https://arxiv.org/abs/2401.17147

[minor mulch]

what do you think

[civic turret]

i think Dr. Xiangsheng Xu will be a millionaire if the proof is valid

[minor mulch]

not necessarily, the tax rate may be quite high

anyway it is very likely wrong

there are so many red flags

[civic turret]

F

[minor mulch]

and i say this as someone with zero knowledge of navier stokes or fluids in general

so i imagine it will be retracted soon

[civic turret]

like the single-author?

[minor mulch]

but who knows

no that is fine

but the big problem is

the claimed result is the opposite of what most of the community expects

also it is very short

[civic turret]

i thought the community was expecting the existence of global smooth solutions 😶

[minor mulch]

i dont think so

why, where did you hear that?

[civic turret]

for the zero viscosity limit, Euler, probably no, but i thought the extra laplacian term was expected to induce smoothness, as it does for the scalar case

[minor mulch]

idk

there is this though https://arxiv.org/abs/1402.0290

[fringe onyx]

so i know that generally solution operator for linear dispersive PDEs doesn't have bounded L^1 operator norm, but are there cases where S(t)u_0 is not even in L^1 for u_0 in L^1?

[steel umbra]

yeah so in class we sorta did that defining it as the average integral which is exactly that. but we showed that its derivative is actually zero for a harmonic function. so wont phi(r)-phi(ep)=0 for all epsilon>0?

[lilac barn]

You should be able to get this via the solution operator for the linear Schrodinger. Your S_t looks like e^i|x|^2 with something and then taking L1 norm of S_tu_0 an infinite upper bound via Young's inequality.

[minor mulch]

your question was not about a harmonic function?

[steel umbra]

oh my god you're right asudhsdsa

[tired hollow]

Hi guys, I’m extremely sorry if I’m asking dumb question but let me ask.
How to start understanding characteristic curves of first order linear PDE? Like any good recommendations like articles/blogs/links?

[lilac barn]

Check this out

[tired hollow]

Oh okay thanks

[untold delta]

unstable-like

[mint canyon]

@tired hollow The implicit form of a solution surface is u - u(x,y) = 0, from vector calc we know the normal vector field to that surface is the gradient (∂u/∂u)k - (∂u/∂x)i - (∂u/∂y)j = 0. The vector field Ai + Bj + Ck is a tangent vector on the surface if its dot product with the gradient is C(∂u/∂u) - A(∂u/∂x) - B(∂u/∂y) = 0. A tangent vector field on the surface will have integral curves on the surface specified by the parametric equations du/dt = C, dx/dt = A, dy/dt = B.

[tired hollow]

Uh oh, let me read this and understand what you’re saying

[mint canyon]

tangent vectors and normal vectors would always be orthogonal right?

[tired hollow]

Yes, true

[mint canyon]

quasilinear first order pdes ARE the condition for orthogonality then

[tired hollow]

Uh so implicit function I understand

[mint canyon]

an integral curve is just the name of the associated parametric curve to some tangent vectors of that curve.

[tired hollow]

Okay I understood why u said implicit function = 0 can be represented as surface

But I didn’t get why u said the expression u mentioned is normal of surface?

[mint canyon]

you should have learned that in vector calc

[tired hollow]

At least I know partial derivative, directional derivative and freshet derivative. But I’m surprised about your expression on normal of a surface, I’ll check it out in vector calculus, no issues.

[mint canyon]

https://en.wikipedia.org/wiki/Normal_(geometry)#Calculating_a_surface_normal

[fringe onyx]

well in principle yes this is what would motivate the existence of such a solution but I'm not sure that it actually exists

we could for example just have that the L^1 norm of some solution grows very fast

[mint canyon]

Has anyone here read about Order Completion methods?

[hollow flume]

would a first course in fourier series and such go in here or better suited for the other chat?

odes and pdes in early university

[verbal nebula]

If it's at the undergraduat elevel, most likely the odes & pdes channel

Or the advanced analysis channel if you're brave/if it's appropriate

[hollow flume]

im never brave in math but thank you!

not anymore anyway

[steel umbra]

i need to compute the wave breaking time for a general convex flux function $f(x)$
$$u_t+\paren{f(u)}_x=0,\quad u(x,0)=u_0(x)$$

[untold delta]

eigentaylor

[steel umbra]

but i'm having a hard time understanding the process

my understanding is to look at the characteristics, but i never really learned how to find them lol

would a general characteristic curve be something like
$$x=x_0+f'(u_0(x_0))t$$

[untold delta]

eigentaylor

[steel umbra]

if so then i think i get
$$t=-\frac{1}{f''(u_0(x_0))u_0'(x_0)}$$

and then i pick the minimum of that. which i think is consistent with the result for burger's equation but it seems a bit... sus lol

[untold delta]

eigentaylor

[steel umbra]

the pure mathematician in me wants to be like "what about when its not defined or if f' isn't continuous etc."

but this is for a numerical PDEs class so maybe i should just be like "yep definitely works no problems to consider here"

and IF i'm on the right track, should it be
$$T_b=\min\paren{-\frac{1}{f''(u_0(x_0))u_0'(x_0)}}$$
or $$T_b=-\frac{1}{\min\paren{f''(u_0(x_0))u_0'(x_0)}}$$

[untold delta]

eigentaylor

[tired hollow]

Uh, latex typing here is interesting though. Some features are not working…

[magic nymph]

hi i have some questions about Vlasov-poisson equation , can someone help me?

[tired hollow]

Not sure, u can post the qn here I guess

[tired hollow]

Yeah it makes sense, generally we assume that the solution exists $u(x,t)$ and upon solving ODE $dx/dt = f’(u)$, we expect that the characteristic curves $x(t)$ exists. So the existence of characteristic curve depends on f(u) function. And doing the whole process, I guess u correctly stated the characteristics equation (straight line). But the function f(u) dictates the slope, so it can cause rarefaction regions / shocks all that I guess.

[untold delta]

Grand Duke

[mint canyon]

@eigentaylor that family contains equations fully nonlinear in ∂ₓu, you can only use method of characteristics on quasilinear equations. My understanding is Lagrange-Charpit (which would work in that case) is also a characteristic method, but I cant say if the characteristics you get from it have all the same properties as the quasilinear case.

[modest crescent]

You can also try appealing to the closed graph theorem.

[weak birch]

Hi i'm currently working through M. Taylor P.D.E II, specifically on Pseudodifferential Operators. I'm trying to understand the lemma

My current try looks the following:

However im having a hard time understanding the requirement that $\delta<1$. Couldnt i just bound the $<\xi>^{m+\delta |\alpha|}$ term by some polynomial term of higher order?

[untold delta]

hrbibbi

[lilac barn]

I think the 1 isn't important; just that you need some specific bound on delta so that you have a fixed seminorm to bound (20). But I might be wrong

[weak birch]

but then the lemma itself doesn't really make sense to me, since $p(x,\xi)\in S_{\rho,\delta}^m$, and thus delta is already bounded

[untold delta]

hrbibbi

[modest crescent]

The derivation is a bit hard to follow because it's seems bit inaccurate after line 17 -- for example, from (18) to (19) he seems to be trying to use the symbol bounds $p \in S^m_{\rho, \delta}$ without bringing the absolute values inside the (oscillating) integral. I would focus on (17).

[untold delta]

Casey

[modest crescent]

$D\alpha_x (x^\beta pv)$ is a family of Schwartz functions in $x$, parametrized by $\xi$. So you want to see how the Schwartz seminorms depend on $\xi$.

[untold delta]

Casey

[modest crescent]

How do you normally show that the Fourier transform of a Schwartz function is Schwartz? If $f(x)$ is Schwartz, then $\xi^\alpha \hat{f} = (-1)^\alpha \int[ (i \partial_x)^\alpha f (x)\ e^{-ix \xi} , dx$ which you can bound using a Schwartz seminorm of $f$ by a constant times $|f |_{\alpha, n+1}$ where $n$ is the dimension.

[untold delta]

Casey

[modest crescent]

Now suppose $f$ actually depends on a parameter $\xi$. If $|f(\cdot, \xi)|\alpha \langle \xi \rangle^{\delta |\alpha|}$ and $\delta < 1$, then you can move the powers of $\xi$ on the right side to the left to obtain $ |\xi|^{(1-\delta) |\alpha|} |\hat{f}| \le M{\alpha}$ for all $\alpha$.

[untold delta]

Casey

Now suppose $f$ actually depends on a parameter $\xi$. If \|f(\cdot, \xi)\|_\alpha \langle \xi \rangle^{\delta |\alpha|}$ and $\delta < 1$, then you can move the powers of $\xi$ on the right side to the left to obtain $  |\xi|^{(1-\delta) |\alpha|} |\hat{f}| \le M_{\alpha}$ for all $\alpha$.
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[modest crescent]

Sorry for the LateX errors I'm new to discord -- but hopefully you get the idea

[weak birch]

thank you very much for the help. I was kind of suspecting that i had made some kind of error in my derivations, ill try to look from (17)

[quaint herald]

Just consider the estimate when $\beta=0 $.

You have $$|\xi^\alpha p_v(\xi)|\leq \int |D_x^\alpha (pv)|, dx .$$

This produces terms of the form
$$\int |D_x^{\gamma}p(x,\xi)||D_x^{\alpha-\gamma}v(x)| , dx $$

The symbolic estimates give the bound
$$|D_x^{\gamma}p(x,\xi)|\leq C_\gamma \langle \xi\rangle ^{m+\delta |\gamma|} $$
and the second factor is simply Schwartz, so certainly this integral in $x$ is finite, with the estimate
$$ |\xi^\alpha p_v(\xi)| \leq C_\gamma \langle \xi\rangle ^{m+\delta |\alpha|}$$
where I have redefined my constant.
We have this for every $\alpha$ so this implies
$$|p_v(\xi)|\leq C_\gamma \langle \xi\rangle ^{m+(\delta-1) |\alpha|}$$

To get that this is finite (no matter what $m$ is), we simply take this last estimate with $|\alpha|$ sufficiently large. This does not work if $\delta=1$.

[untold delta]

grobmez

[quaint herald]

It is. ^

(replace "finite" with "bounded by <\xi>^(-N) for arbitrary N" in the last para.)

[weak birch]

okay and in the case where $\beta\neq 0$ we just get the term
$$\int |D_x^{\gamma}p(x,\xi)||D_x^{\alpha-\gamma}(x^\beta v(x))| , dx $$
where the second term again is Schwartz and the argument holds again

[untold delta]

hrbibbi

[quaint herald]

yep

[weak birch]

great thank you

[quaint herald]

no problem

[tired hollow]

Guys, can anyone can pls throw some light on Monge cones?
My prof brought that up in PDE class, any helpful video/article or even an explanation would be very helpful!!

[tired hollow]

I guess I’m getting some understanding from this notes:

https://www.math.iitb.ac.in/~siva/ma515152022/3MA515Notes.pdf

We fix a point (x0, y0, z0) in R3 of a possible integral surface (integral surface = solution surface for nonlinear pde: f(x,y,u(x,y),ux,uy)=0)

(Don’t ask me how is that unknown z0 is fixed, I’m assuming that we assume it exists and fix it somewhere and let the other variables play 🤣) then we have
f(x0,y0,z0,ux,uy) = 0

My confusion is that we assume some solution z0 exists, can’t we assume that in same way it’s partial derivatives also exists? Like p0 & q0. Why do we have to assume that z0 is fixed and there we look at all possible tangent planes satisfying the eqn f(x0,y0,z0, ux,uy)=0?

(Maybe I’m not convinced that the solution to f exists in R5 plane not R3 )

[bitter yacht]

I am trying to understand the following proof, which is necessary for e.g. proving weak derivatives are unique:
https://math.stackexchange.com/questions/3212146/two-proofs-of-the-fundamental-theorem-of-calculus-of-variations-one-correct-o

I understand every step, save for the statement that $u(x) \omega_\epsilon (x) $ converges to $u(x) w(x) $ as $\epsilon \to 0$, for almost every $x$. From Lebesgue's differentiation theorem it follows that the mollified $\omega_\epsilon$ converges to $\omega$ almost everywhere. But how is this passed on to the product $u \omega_\epsilon$? I understand it should suffice to look at $x$ that live in the support of $\omega_\epsilon$, but how do we control the value of $u(x)$? If $u(x)$ were bounded, we could simply just consider
[
| u(x) \omega_\epsilon(x) - u(x) \omega(x)| = |u(x)| |\omega_\epsilon(x) - \omega (x)|
]
$|u(x)|$ we could then bound, and the difference $|\omega_\epsilon(x) - \omega (x)|$ can be sent to zero due to almost everywhere convergence of $\omega_\epsilon \to \omega$.

[untold delta]

hardisc

[lilac barn]

It seems like they prove for you that |uw(x)| ≤ |u(x)| so u is a majorant and hence DCT is applicable

[bitter yacht]

yes the aim is eventually to deploy DCT. but the hypothesis remaining for DCT is pointwise convergence a.e., namely, $u \omega_\epsilon \to u \omega $ as $\epsilon \to 0$, so that we can pass to this limit in the integral $ \int u \omega_\epsilon $. And that pointwise convergence I do not see how to confirm

[untold delta]

hardisc

[lilac barn]

Ah you don't understand how to get pointwise convergence?

If you get that we(x) converges to w(x) pointwisely, then for each fixed x, we(x) is just simply a sequence indexed by e converging to w(x). Now for a fixed x, u(x) is just some number so just recall the effect multiplication by a number have on a converging sequence

[bitter yacht]

ah crap, was it really that simple?

we have that $N = { x : \omega_\epsilon(x) \not \to \omega(x) }$ has measure zero.

fix an arbitrary $x \not \in N$. then $u(x)$ is just some finite number. Then
[
|u(x)| | \omega_\epsilon (x) - \omega (x)| \leq |u(x)| \cdot \epsilon \to 0, \text{ as } \epsilon \to 0
]

Thus $u \omega_\epsilon \to u \omega$ fails only on a set of measure zero, so the convergence is almost everywhere.

like that? last question remaining is then what if $u(x)$ is not finite at a certain $x$? e.g. if $u(x) = + \infty$?

[untold delta]

hardisc

[lilac barn]

You only care about almost sure convergence, so remove the zero-measure set of u = infinte (prove for integrable u, such sets have zero measure)

[bitter yacht]

oh right. if the set of u = infinite had positive measure then we could integrate |u| over that set, getting an infinitely large integral, and that would contradict the integral of u over the entire space being finite. thus that set must have zero measure

huge thanks to you @lilac barn cocat, appreciate it so much

[lilac barn]

Yw!

[unborn gyro]

I was recently attempting a problem from Evans' PDE text and I solved it, but I'm not sure my answer is correct since it looks a bit fishy. I was hoping someone could verify it and correct me if I'm wrong.

The problem goes as follows:

Let U = ${|x_1| < 1,|x_2| < 1}$ denote the open unit square in $R^2$ centered at the origin. Define the function $u:U \mapsto R$ as follows:

$$ u(x_1,x_2) = \begin{cases}
1-x_1 \text{ if} & |x_1| \leq |x_2|, x_1\geq 0, \
1+x_1 \text{ if} & |x_1| \leq |x_2|, x_1\leq 0, \
1-x_2 \text{ if} & |x_2| \leq |x_1|, x_2\geq 0, \
1+x_2 \text{ if} & |x_2| \leq |x_1|, x_2\leq 0 \
\end{cases}
$$

For which $1 \leq p \leq \infty$ does u lie in $W^{1,p}(U)?$

<br/><br/>

Solution:

My answer is that u lies in the Sobolev space for all $1\leq p\leq\infty.$

<br/>

Sketch of the argument:

First note that u is bounded on a set of finite measure so it lies in $L^{p}(U)$ for each $1 \leq p \leq \infty.$

Now it suffices to show that the weak first partials $u_{x_1}$ and $u_{x_2}$ also lie in $L^{p}(U)$ for each $1 \leq p \leq \infty.$ In particular first we show that these weak derivatives are actually functions and then we show that they are bounded.

To this extent we let $\phi$ be a test function on $U$ and consider the quantity:

$$\int_U{u \phi_{x_i}}$$

In order to simplify this quantity we divide the open unit square into 4 open regions $R_1,R_2,R_3, \text{ and } R_4$.

$$R_1 = {|x_1| < |x_2|, x_1 > 0}$$
$$R_2 = {|x_1| < |x_2|, x_1 < 0}$$
$$R_3 = {|x_1| > |x_2|, x_2 > 0}$$
$$R_4 = {|x_1| > |x_2|, x_2 < 0}$$

we orient each of these counter-clockwise and use Greens Theorem to simplify our integral as follows:

$$\int_{R_j} u \phi_{x_i} = \int_{\partial R_j} u \phi \nu_{i} dS - \int_{R_j} u_{x_i} \phi$$

Here {nu} denotes the outward pointing normal.

[untold delta]

taco

[unborn gyro]

We note that $\phi$ vanishes on the outer boundary of the unit square being smooth and compactly supported in u (it's a test function).

If we picture each of the regions to be triangles then u agrees on the shared boundary of any two regions (the boundaries are precisely $x_1 = x_2$ and $x_1 = -x_2$).

However note that for two regions sharing a boundary their corresponding normals point in precisely opposite directions (due to being oriented in opposite directions).

So we have that:

$$\sum_{j=1}^4 \int_{\partial R_j} u \phi \nu_{i} dS = 0$$

We are left with the expression:

$$ \int_U{u \phi_{x_i}} = \sum_{j=1}^4 (\int_{\partial R_j} u \phi \nu_{i} dS - \int_{R_j} u_{x_i} \phi) = - \sum_{j=1}^4 \int_{R_j} u_{x_i} \phi) $$

However, we see that u is differentiable on each of the $R_j$'s and we may plug in $u_{x_i}$ for each of the regions for $i=1,2$ to find the weak partial derivatives of u i.e.,

$$u_{x_1} = 1_{R_1} - 1_{R_2}$$
$$u_{x_2} = 1_{R_3} - 1_{R_4}$$

However, both these (weak) partial derivates are bounded on sets of finite measure. So we see that they are in $L^{p}(U)$ for each $1 \leq p \leq \infty.$

We can finally conclude that u lies in $W^{1,p}(U)$ for each $1 \leq p \leq \infty. : \square$

To reiterate, please let me know if this is correct or not, if not please let me know what the error is.

[untold delta]

taco

[unborn gyro]

I've made a corresponding stack exchange post if this is not readable: https://math.stackexchange.com/q/4861856/1289223

[frozen orbit]

I'm in a numerical PDEs class and we have a section on numerical method for solving the advection equation $$\partial_tu = -a\partial_xu$$. I'm a little confused on why we'd need special methods to solve this, becaue if the initial data is given by $u_0(x)$ then the exact solution is just $u_0(x-at)$.

[untold delta]

Lakshay

[astral vine]

One way to think about it (but actual numerical/PDE analysts would answer better than me), it is because in interesting cases a=a(u) depends on the solution u itself, so that you need first to have sufficient knowledge on what happens in case were the coefficients are "frozen", to see what actually behaves well and to guess what could behave worse when trying to reach a nonlinear theory.

[inland sinew]

I'm going to post this here as well because I didn't see an ODE channel. Anyway, I'm wondering how to handle an equation of the form f(x)y" + y = 0, where f(x) is a periodic function that crosses zero. For example I'm looking at equations of the form y" + cos(x)/(1+sinx) y = 0. I can't get any solver to continue outside of the interval (0,2pi). Some of the solutions look nice inside of it, but outside of that, it basically brakes. I haven't seen a singularity like this covered in perturbation theory literature.

[untold delta]

Removed the studying! role from you.

[neon urchin]

I'd like to solve a non-linear, coupled Klein-Gordon equation. Can someone lay out the techniques/approaches I need to study to achieve that?

I've found https://eqworld.ipmnet.ru/en/solutions/npde/npde2107.pdf In my particular case the non-linear function is a polynomial and I wonder if I can apply boundary conditions such that the function vanishes at infinity. Actually, I'd rather work with 3-dim space, too. And technically I have a system of 2 coupled equation, but I'm willing to simplify.

Any advice what I could read to do this task as I don't fully understand how to approach it?

For the pdf link I also wonder: Is the implicit traveling wave solution useful to do this task? And for the functional separable solution: why is xi quadratic in the variables x,t now? And why is the ODE for w(xi) of a different form and includes xi now?

[mint canyon]

@neon urchin have you tried a transformation group method to find a similarity variable?

[neon urchin]

I'm afraid I have only a very basic understand of PDEs and I'd need a book to learn that. Anything suitable for someone at physics undergrad level? Ideally to make me able to find a solution to a non-linear wave equation in a small timeframe.
If I assume my equation is w_tt = w_xx + w_yy + w_zz + b w^3, what would your suggestion mean? Is it related to point (3) in https://eqworld.ipmnet.ru/en/solutions/npde/npde2101.pdf ? I'm not sure what the F() means. Is it standard knowledge, free to chose, hard to express or does he just not write it out?
Is there a simple recipe to extend solutions from just w_xx as in the PDF to w_xx+w_yy+w_zz?

[mint canyon]

I read that paper and it doesnt really show any sort of method to solve the equation, it just kind of shows a solution

[neon urchin]

I'd be satisfied if I get a solution for something like a traveling wave 🙂 Should there be a single answer? Or are multiple forms possible?

[mint canyon]

There are many different methods for nonlinear pdes, the one Im most familiar with is transformation group methods (by Lie)

[neon urchin]

Lie sounds good. I always wondered how they are related. what's a modern introduction to learn that?

[mint canyon]

"symmetry methods for differential equations: a beginners guide" -P Hydon

do you want a brief overview?

[neon urchin]

If there is a way for have a short overview, I'd be very interested. It helps mapping out where the journey is going.
In particular I'm interested if it can help me find some kind of discrete spectrum for solutions which vanish at infinity in some way (like boundary conditions).

My main equation is just the wave equation with the d'Alembert operator. It can be a coupled system of equations and non-linear though

I've found papers about similar equation, but I don't understand what is going on there. Is it hard to write down a solution? Are there open questions?

[mint canyon]

So a Lie group can be thought of as a set of closed parametric curves called orbits

imagine these curves map points on a solution to a DE to points on other solutions uniquely

Then we say the differential equation is a differential invariant of the Lie group

that is, the group maps solutions to other solutions by way of some continuous parameter

The tangent vectors of the orbits are called "infinitesimal generators of the group"

[neon urchin]

is there a way that boundary conditions are included and then my solutions split into unconnected parts?

[mint canyon]

Im getting to that

[neon urchin]

ok, I'm reading 🙂

[mint canyon]

The condition for a differential invariant is that the infinitesimal generator acting on some expression is 0

(a vector is also an operator)

If there are derivatives in the expression, the infinitesimal generator must be "prolonged" (extended to a higher dimensional space, the space of derivatives)

You act on the differential equation and boundary conditions with the infinitesimal generator to figure out what the components of the vector field should be.

This condition is always linear

The similarity variable is then the solution to the invariant surface condition (which uses the infinitesimal generator). If you make a coordinate transform using the similarity variable, the number of independent variables is reduced by 1.

So a PDE with N independent variables and M order can be reduced to an algebraic equation by N+M symmetries (N similarity reductions and M order reductions)

[neon urchin]

sounds like an attractive method. for now I can only memorize and write down what you wrote and keep coming back to it while reading the book.

[mint canyon]

The infinitesimal generator (not prolonged) looks like this: χ = ξ(x,t,w)(∂/∂x) + τ(x,t,w)(∂/∂t) + η(x,t,w)(∂/∂w)

note a partial derivative here is the same as a coordinate basis vector

[neon urchin]

thanks! I'll try to imagine what this means and read the book. Hope it enables me to find some kind of discrete spectrum of solutions with boundary conditions

[mint canyon]

for your case the Lie group exists in the space of point transformations in x, t, and w, and maps every solution (point by point) to every other solution

[neon urchin]

could I have some conserved quantities which would take discrete values under some conditions?

[mint canyon]

X(x,t,w;ε), T(x,t,w;ε), W(x,t,w;ε) would be the orbits

[neon urchin]

like some integral over the whole space which for some conditions takes only discrete values

[mint canyon]

if you go to the space of contact transformations, you have Noethers Theorem....

[neon urchin]

I think I will get something conserved. I mainly wonder where to find the discreteness

for some boundary conditions

[mint canyon]

contact transformations are the generalization of canonical transformations in Hamiltonian mechanics

[neon urchin]

I'll get the book to understand these terms. Any additional book recommendation for beginners just in case it doesn't match my style?

[mint canyon]

they are transformations in the space of points and first derivatives of the dependent variables

yeah, I didnt want to burden you too much with contact transformations, but that is where conservation laws arise

[neon urchin]

I think I know what my conservation law will be. is there are chance that this conserved quantity takes discrete values if I impose additional constraints on my solution?

[mint canyon]

when I think of something as discrete or continuous, I usually think of a set. what set are you referring to here? the set of conserved quantities?

like {energy, momnetum, angular momentum} ?

[neon urchin]

some kind of real number which will take discrete values. for example the function value at zero. or the integral over the function over the whole space. or the conserved quantity. something that will need to be from a set of real numbers [E1,E2,...]. yes, like energy. like discrete energy levels

all this only if I impose some boundary conditions

or other kind of conditions on the solution

[mint canyon]

ahhh so you meant the set of possible eigenvalues for some observable

[neon urchin]

not sure if I can already relate this to observables. but naively speaking a real number in the solution which will be like discrete energy levels

like sine-waves in a box which can have only some wave lengths

[mint canyon]

the observable there corresponds to the Hamiltonian operator

(time-independent anyway)

[neon urchin]

maybe I need to get started with the book to understand better what I need

[mint canyon]

sorry, I shouldnt have used observable and operator interchangably. You're right that they are distinct

Klein-Gordan is for the relativistic quantum mechanics of a scalar field right?

[neon urchin]

Of a complex scalar, yes

but non-linear, not the standard one

[mint canyon]

yeah, they have a nonlinear schrodinger equation too

[neon urchin]

right, I saw papers about it. I need specifically $\Box w=f(w)$.

[untold delta]

Gerenuk

[mint canyon]

well, if transformation groups end up being too big an undertaking, there is also generalized separation of variables, and method of constraints if your conservation law is valid. good luck

[neon urchin]

is that also mentioned in the book you suggested, or do you have another recommendation for this?

[mint canyon]

I have yet to find great resources on them, and Im still learning them unfortunately. They are narrower in scope, but just wanted to provide other options

[neon urchin]

ok, I'll take a note of these options

[weak birch]

Im trying to understand proposition 2.1 from M. Taylor P.D.E II as shown in picture. However it is the first time I have encountered the notion of a distribution being $C^\infty$ does anyone have a resource or an explanation on what this means?

[untold delta]

hrbibbi

[lilac barn]

Recall that function induce a natural distribution. The claim here is that there exists a smooth function which induces a distribution that equals this distribution

[weak birch]

Ah okay makes sense. I must have overcomplicated it for myself…

[bitter yacht]

When proving uniqueness of minimizers, one technique is to exploit strict convexity of the integrand of a functional. E.g. $F(u) = \int |u|^p$.

I have tried understanding how to prove $\mathbb{R}^n \ni x \mapsto |x|^p$ is strictly convex for $p > 1$. One can show $ |(1-t) x + t y|^p \leq (1-t) |x|^p + t |y|^p $ for arbitrary $x,y \in \mathbb{R}^n$ and $t \in (0,1)$ e.g. via Hölder inequality. Here is an argument for the one-dimensional case: https://math.stackexchange.com/a/2200275. This is convexity, and for strict convexity further analysis is required.

For strict convexity one has to check that equality forces the vectors $x$ and $y$ to be equal. The one-dimensional case is treated in the link above, and there one reaches $|x| = |y|$, where one can account for the signs of $x$ and $y$ being equal, it follows $x = y$.

The same argument leads one to $|x| = |y|$ in higher dimensions. But how would we obtain $x = y$ in higher dimensions from $|x| = |y|$? This is the gap I have not been able to bridge.

[untold delta]

hardisc

[frozen orbit]

I see, thank you!

[minor mulch]

i think sign is not the right way to think about the 1d case

in R^d, if x and y were not equal

then the convex combination would lie in the interior of the ball of radius |x|=|y|

the same argument works in 1d, but the boundary of the ball is two points so it is misleading

[bitter yacht]

thanks for the response, I managed to solve it. we wind up having equality in the triangle inequality, which forces x = C*y for some non-negative constant C, and in tandem with |x| = |y|, we obtain x = y (assuming implicitly x and y are both non-zero).

[dense scaffold]

I’m trying to derive the pde of the heat equation but cant find a resource that describes it step by step. Anyone who knows one or can help?

[verbal nebula]

Boyce & Di Prima's book does this

[dense scaffold]

Is there a soft copy?

[wild patrol]

Do you want the physical derivation?

Like, as in the physics behind it?

[dense scaffold]

The physics and calculus behind it

[dense scaffold]

Yeah

[wild patrol]

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

Just scroll down to Fourier's Law midway down.

[modest crescent]

A typical example is when $K$ represents the identity map -- $K(x,y) = \delta(x-y)$ is smooth (identically 0) away from the diagonal $x=y$.

[untold delta]

Casey

[steel umbra]

can anyone provide some insight as to where this comes from?

[midnight grail]

If you extremize that functional, with Euler Lagrange, you obtain the Poisson equation

[minor mulch]

https://en.wikipedia.org/wiki/Obstacle_problem#Motivating_problems

[lilac barn]

Multiply the equation by w and integration by parts. You will get this functional. The admissible set is taken to account for the boundary

[steel umbra]

"what is the intuition behind this"

[untold delta]

lobby scammer

[last glen]

this is from fritz john p. 65, but is there an actual name for this $$f \in C_{M,r}(y)$$ definition?

[untold delta]

wciscnet

[bitter yacht]

I am studying the p-laplace and in such contexts the following identity and the next two inequalities it implies are ubiquitous (see attached picture). I managed to verify the first identity. The middlemost inequality number one I also managed to obtain. But I have not at all been able to read off inequality number two. Judging from the author's presentation, it should perhaps be obvious, but I cannot see it.

The middlemost inequality number 1, I have shown as follows:
Suppose $|a| \leq |b|$ without loss of generality. Then $|a|^{p-2} \leq |b|^{p-2}$ and $|a|^2 \leq |b|^2$, so by the second term in the identity, inequality number one follows readily.

For inequality number two, my attempt only reached this far:
Supposing now strictly $|a| < |b|$ (we will soon divide by $|b-a|>0$), we then have $|a|^{p-2} < |b|^{p-2}$, and so
\begin{align*}
2^{-1} \left( |b|^{p-2} + |a|^{p-2} \right) |b-a|^2 > 2^{-1} \cdot \left(2 |a|^{p-2} \right) |b-a|^2 = \|a|^{p-2} |b-a|^{2-p} |b-a|^p = \left( \frac {|a|}{|b-a|} \right)^{p-2} |b-a|^p
\end{align*}
Unfortunately I do not see how to control $\frac {|a|}{|b-a|}$. If it is larger than $\frac 12$ then we are done, but I am not sure that holds for arbitrary $a$ and $b$ such that $|a| < |b|$. Perhaps my downward estimations have not been sufficiently sharp?

[untold delta]

hardisc.

[bitter yacht]

hm, interesting, how exactly? do we not need p -2 > 1 for convexity? e.g. $x \mapsto |x|^{\frac 12}$ is not convex I believe

[untold delta]

hardisc.

[lilac barn]

There should be a (\mathbf{2^{-1}} 2^{2 - p} \lvert b - a \rvert^p) at the end in which case, the argument follows from ( p \geq 0 : )
[ \lvert a -b \rvert^p \leq \lvert 2 \max_{}, { \lvert a \rvert , \lvert b \rvert} \rvert^p = 2^p \max_{}, { \lvert a \rvert^p , \lvert b \rvert^p} \leq 2^p ( \lvert a \rvert^p + \lvert b \rvert^p).
]
To see a counterexample, set ( b = 3a) and (p = 2). It should give you a contradiction.

[untold delta]

cocat

[bitter yacht]

ah so it simply boiled down to $ |a + b| \leq 2 \max{ |a| , |b| }$, neat! big thanks.
however, I tried your example in the original inequalities number one and two (in my picture),
did not seem to produce a contradiction. or perhaps I have misunderstood?

just $p=2$ alone yields
[
2^{-1} (1 + 1) |b-a|^2 \geq 2^0 |b-a|^2
]
which certainly holds irrespective of $a$ and $b$.

[untold delta]

hardisc.

[lilac barn]

Right, my bad. Try a = 0 and b =1, 0<p<2; this should give a contradiction

[stray forum]

What happened in the statement of Marcinkiewicz interpolation theorem if I replace the condition of simple function class by schwartz class function? Then still the marcinkewicz interpolation is true?

[stray forum]

I want to know this because while reading duoandixeta's book they have shown hilbert transform is weak (1,1) and strong(2,2) for schwartz class function. From there how they can concluded by interpolation it is strong (p,p) ?

[stray forum]

@lilac barn can you help me in this case.

[bitter yacht]

oh right, your hint a=0 helped with finding a counterexample. thank you so much

[astral vine]

Interpolation does not depend on the choice of subspace as long as it is a dense one.

[unborn quiver]

Don't ask the same question in multiple channels, you've been repeatedly told this in the past. I'm not really sure what you mean by simple function class here

[astral vine]

And contained in both endpoint spaces.

[stray forum]

I don't remember such things. Sorry I somehow did not access the group adv pde after posting. And in future if it happens again I don't care about reposting.

[stray forum]

No marcinkewicz interpolation proof I have seen using only simple functions.

[unborn quiver]

That shouldn't matter though, density should get you the same result

[stray forum]

See this statement from the book of grafakos.

I don't see how this statement is valid for any dense class? Can you please provide me the general statement, I did not find it in any book at all. @astral vine @unborn quiver please.

[stray forum]

Actually this is the main reason if the marcinkewicz statement is true for any dense class then, why did showing it for schwartz class , javier again proved it strong (p,p) nature for any L_p function from scratch? I did not get that. That's why I am caring about the general statement.

[unborn quiver]

The statement I know is a subspace that's closed under truncation. I.e.
[f_t(x) = \begin{cases}
f(x) &|f|\le t\ 0 &|f|>t\end{cases}]
and $E$ (subset of measurable functions) is closed under truncation if $f\in E$ implies that $f_t\in E$ for all $t\in[0,\infty)$

[untold delta]

MSC2020 49Qxx

[stray forum]

What is it?

How is it connected to the interpolation theorem at all?

[unborn quiver]

You asked about what you can do instead of simple functions

[lilac barn]

He proves the relevant Marcinkiewicz interpolation theorem at the start of the book

[stray forum]

But that is not for schwartz class.

[lilac barn]

He proves for Lp iirc

[stray forum]

How can he use marcinkewicz in the second page? In the first page they wrote marcinkewicz interpolation theorem?

[stray forum]

Please see this. @lilac barn

[lilac barn]

What's your question here? Wdym by how can he use

[unborn quiver]

Schwartz class functions are Lp for every p in [1,infty]

[stray forum]

In the proof of statement of theorem 3.2 he showed (2,2) and weak (1,1) for schwartz class after that how he concluded it for scwartz class for p€(1,2) by interpolation ?

Just tell me this thing rigorously.

[stray forum]

That I understood.

[lilac barn]

You consider an operator that is defined on L1+L2 mapping to the space of measurable functions.
Further impose that the restrictions to L1 (aka L1+0) is (1,1) and restriction to L2 (0+L2) is (2,2) bounded. Then Lp subset L1+L2 gets mapped to Lp

[stray forum]

The problem is why it should be weak (1,1) for any L^1 function? I am asking this since where he proved it until that portion he only defined hilbert transformation for schwartz class function.

For L^2 it is fine by extension as a operator.

[stray forum]

So your (1,1) does not make sense for L^1 function except the schwartz class. Please read it precisely.

[lilac barn]

It doesn't but then you use density

[stray forum]

Then tell me how you can use density to this equation,
$\lvert Hf(x)>\lambda\rvert<= c|f|_1/\lambda$. How can you use density to prove this equation hold for $L^1$ function.

[untold delta]

Vishnu das

[stray forum]

From wealk boundedness you could not extend the definition from schwartz class to $L^1$ function.

[untold delta]

Vishnu das

[stray forum]

The biggest problem is you still do not know how hilbert transform is defined for $L^1$ function?

[stray forum]

From wealk boundedness you could not extend the definition from schwartz class to $L^1$ function.

[untold delta]

Vishnu das

[lilac barn]

You pick a Schwarz limit of a L1 function. The sequence is L1 Cauchy so their Hilbert transforms are Cauchy in measure, via the L1 weak bound. This implies there's an a.e limit. You define that to be the Hilbert transform of L1 function

Now prove that this transform is well-defined and the L1 limit still satisfies the (1,1) bound (it's simply triangle inequality)

[late lichen]

this isn't 'advanced' relatively, but this was the closest channel i could find that mentioned pdes

i am trying to analytically derive a solution for poisson equation based on very simple ICs and BCs and 2nd and 3rd BCs

and then model it with FDA and FEA

and compare answers

is this the room for that stuff?

thank you

[bitter hollow]

I think the closest thing would be the numerical analysis channel: https://discord.com/channels/268882317391429632/576514592725794816.

[late lichen]

thank you for your response. for the analytical derivation would this still be the best channel for that? Just not the FDA and FEA stuff?

[bitter hollow]

I think so, yes.

[late lichen]

thanks

[dusky gyro]

Not sure this is "advanced" PDEs for this #advanced-pdes chatroom but I'm trying to get intuition on the Poisson equation (ideally non physical) I made a question and just added a bounty

https://math.stackexchange.com/q/4822486/452937

[minor mulch]

do you know any probability

[dusky gyro]

not really

[minor mulch]

hmm

[dusky gyro]

I mean a bit but probably not much more than a basic undergrad course

[minor mulch]

do you know what brownian motion is

[dusky gyro]

yes

[minor mulch]

ok

consider for simplicity dim = 1

and consider a solution u to poisson eqn u'' = f on (0,1) with u(0)=u(1)=0

[lilac barn]

For good motivation in PDE perspective, you can look into Evan’s or Folland’s PDE book. For a statistical interpretation, Complex Made simple by David Ullrich provides a very nice motivation that isn’t very probability-theoretic, but suited for those without any stochastic background.

[minor mulch]

actually lets take a step back

first we can try to understand the green function

of the laplacian

[dusky gyro]

-1/2|x-x0| I believe since dim is 1

[minor mulch]

let's define a "brownian bridge" to be a brownian motion B(t) started at B(0)=0, and conditioned to satisfy B(1)=0

(this is not obviously well-defined but just imagine a brownian motion joining the points (0,0) and (1,0))

[untold delta]

memorylessfunctor

memorylessfunctor

[dusky gyro]

I understand what this means but not sure why this is true

[minor mulch]

also if you view this identity as the definition of B, you can extend this to any dimension by just using the corresponding green function. in dim=2 the analogue of B looks like this (called a gaussian free field)

[minor mulch]

its just a computation

there is some reason why it should be true though

there are some other facts about harmonic functions

relating to brownian motion

like you can rephrase the mean value property as u(x) = E[u(B_T) | B_0=x] where B_T is brownian motion stopped when it first hits the boundary of the domain

[minor mulch]

not a particularly easy computation if you do not know how to define brownian bridge, though

iirc E[B(s)B(t)] = min(s,t)(1-max(s,t))

something like that

[untold delta]

memorylessfunctor

[minor mulch]

so with that formula you can just compute

[blazing ridge]

maybe the wrong place, but does anyone know much about 'solving' PDEs using neural networks? Has it seen much success recently?

[mint canyon]

neural networks are fundamentally approximations, whatever you get as a result will likely be equivalent to a numerical method, though I guess you could maybe train a really sophisticated neural network a known analytic method?

[pulsar rivet]

yo any idea for b)?

[lilac barn]

Try writing the usual characteristics

[pulsar rivet]

i did

and i have x(s)=x0sins and y(s)=x0coss

[lilac barn]

And what information can you deduce for the solution u(x(s),y(s))=z(s)?

[pulsar rivet]

that z(s) is a constant

[lilac barn]

Good so u(s) = z(s) = z_0. Now what do you know about z_0 and can you then write the resulting form in terms of x and y?

[pulsar rivet]

z=g(x0)=g(sqrt(x^2+y^2))?

[lilac barn]

What's g here?

[pulsar rivet]

u(x,0)= lets say g(x)

so u(x0,0)=g(x0)

[lilac barn]

But you also know that u(x,0)= x at the boundary

So what should g(x0) be?

[pulsar rivet]

sqrt(x^2+y^2)?

[lilac barn]

How's g defined? What's g(x)?

[pulsar rivet]

g(x)=u(x,0)

[lilac barn]

And what's u(x,0)?

[pulsar rivet]

x

u(x,0)=x

[lilac barn]

Indeed so g(z) = z. Now plug this here

[pulsar rivet]

so sqrt(x^2+y^2)

[lilac barn]

Now verify if it does indeed solve the equation with the right initial condition

[pulsar rivet]

thats where i am a little fuzzy