advanced-pdes

[lilac barn]

You don't need 35 but 36 tells you the quantities lie in Lq' so by holder the integral is finite for v in W1,q

[untold delta]

tryingmaths

[tropic rock]

dominated convergence i guess..

[opal condor]

the proof here assumes specific conditions upon f

for simplicity, but i'm wondering whether or not the proof works for any f in general

[astral vine]

you can give a sense of this formula for f to be any L1 loc in time with values in S' (tempered distributions)

in particular f in Lp(R_+xR^n)

for any p

And for super fancier fucntion spaces

[opal condor]

does the discussion that leads up to (12_s) also work these f as well?

[lilac barn]

For now you shouldn't concern yourself with full generality but merely learning the method. Pde theory becomes very technical once you start generalizing over basic assumptions

[frank tide]

i see, also in my case the domain is a wedge of a unit circle, how does the regularity change with the angle?

i’m working with Poisson’s eq

u(r,theta)=(1+r^2)r^b sin(b*theta)

where the max angle is pi/b

[quaint herald]

Because of the r^b. Upon being hit by derivatives this function becomes singular at the origin generally.

[opal condor]

ok got it

[frank tide]

so if i had non convex wedge, like b<1, would the solution be in H^2?

[ocean ether]

man FUCK greens functions

the professor wants me to use the greens function representation and this is making me sad

[ocean ether]

is it more complicated to do this by greens functions?

[quaint herald]

nope, because r^(b-2) will blow up faster than r^(-1) which is not in L^2.

[frank tide]

i see that makes sense. since u_{rrr} (third derivative) is only in L^2 for b>2, does that generally mean that u is only in H^3 (or higher) in (0,pi/2)?

since b<2 means the angle is greater than pi/2

its kinda interesting how that plays out then, u in H^3 for first quadrant, u in H^2 for second, u in H^1 for third

[quaint herald]

yep

well your function is changing, it's more about how singular your func is at the origin, which involves b in a very explicit way.

[lilac barn]

Did you actually try to go through the proof of the mean value theorem and see what modifications are needed?

[ocean ether]

the professor said to try using green's functions

[lilac barn]

In that case first start by calculating the Greens function for a ball of radius r

Actually Evans provide the expression in the next few subsections

[ocean ether]

for the unit ball

not for a ball of radius r

i assume its basically the same though

[lilac barn]

Yes, but there will be some changes involved. Try going through his proof and derive it for a ball of radius r.

Actually he gives it for ball of radius r as well

[ocean ether]

is this just turning the point dual to $\frac{rx}{|x|^2}$?

[untold delta]

the breadwave

[ocean ether]

i guess i now show that $\int_{\partial B(0,r)} g(y)\frac{\partial G}{\partial \nu}(z,y) dS$ reduces to mean value of g around the boundary

[untold delta]

the breadwave

[ocean ether]

(z goes to 0)

[ocean ether]

im just gonna assume this is true and the rest is ez

[chilly basin]

Hey guys, do you know if there is any general method that can allow me to investigate the asymptotic behavior of an ODE (at infinity)? Thank you.

[broken hamlet]

Does this sound like a good motivation for studying Schauder estimates? It's just for myself.

We'd like to solve elliptic PDEs by the method of continuity, which require a priori estimates (and so we want to consider strong and not weak solutions). But we'd also like some form of compactness, which is why we need the Holder norm (so we can apply Arzela-Ascoli).

[quaint herald]

You won't get a single "general method", but it's basically the same thing as finding asymptotics of integrals. Tools like stationary phase, method of steepest descent, Watson's method etc are the basics.

For instance these can give you the classical asymptotics for special functions like Bessels in the various regimes.

(Perhaps I am misinterpreting your question, and you are more interested in geometric things like say a smooth flow on a noncompact mfld "at infinity". If so, please clarify.)

[chilly basin]

Sounds like these are what I want, but honestly I don't know much about those, so maybe I will look them up first, thank you!

[inland sinew]

Hi, I was wondering if anyone had any insight into a particular PDE I've been looking at which is of the form $z_{xx}\beta''+z_{yy}\alpha'' = 0$, where $\alpha$ is a function of x and $\beta$ is a function of y, both of which are periodic. I have studied this PDE for a while because of its physical relevance and have been able to derive about 4 types of solutions. However some of these solutions I feel I could have gotten to quicker if I knew more about fourier analysis and I'm wondering if any insight could be gained by doing something like fourier transforming the equation or if there are other integral transforms that could help me study it.

[untold delta]

pewpew2385

[quaint herald]

You might like Olver's book: Asymptotics and special functions. I mostly cobbled together knowledge of this stuff from various sources like lecture notes (I recall a pretty nice set from Cambridge), but this book seems pretty detailed and has a view towards asymptotics of ODE solutions as you desire.

[sonic olive]

Do you have any assumptions on the sign of Beta’, alpha’, beta’’, alpha’’? If you do you can use theory from second order elliptic/ hyperbolic pde depending on these assumptions

[quaint herald]

Found them. These notes are quite compact and cover the basics. Of course the Olver book is more comprehensive.

@quick pagoda what is the meaning of the stare?

[quick pagoda]

As a kind of “I will note this down as a book I was unaware of”

[quaint herald]

https://tenor.com/view/cat-blink-blinkingcat-cute-funny-gif-8859604814399687515

I knew of it but I literally opened it for the first time about 2 mins before making that post :p

just browsed to the ../pure/analysis/asymptotics/ subfolder of my textbook collection, 99.99% of which I never have opened/never will open.

[quick pagoda]

mood, I definitely have a few hoarded texts

Not quite that many to need sub that many subfolders, including one just for asymptotics, yet though

[quaint herald]

20g textbooks folder and 0.7g papers folder needs some amount of organisation haha.

it's a bit suboptimal but whatever, not very important

[quick pagoda]

I’m only at like single digit GB

Clearly gotta step up the hoarding game and grab the books I have on a to-do list

[quaint herald]

Haha my collection had a big head start as it was built off a massive torrent I grabbed as a UG many years ago. High proportion of useless books, but some good ones in there and a nice organisational structure that I built off.

[lilac barn]

Also for special functions there is an online database dlmf.nist.gov

[quaint herald]

It's got cool classical things like the Ramanujan notebooks, classic papers, old tripos exams, essays like mathematicians apology, mathematicians miscellany

[quaint herald]

Yes, of course the database is absolutely the best place to look up specific asymptotics for known special functions, he was asking about learning things like the asymptotic methods though.

stationary phase etc

[lilac barn]

Yes yes just wanted to plug it for people interested

[quaint herald]

yeah that db is super useful, has definitely saved me some elbow grease

[quick pagoda]

Ah a lot of cool stuff but maybe not stuff of particular use or interest. Mine is mostly inflated by books I’ve grabbed as a possible reference to whatever specific topics, or especially padded around stuff I care about

[quaint herald]

yeah of course. everything I have added to this collection since then (which has accrued to be quite a lot by now) has been of this type

But even the other stuff has some uses, I occasionally mine old obscure books for problems when lecturing for example.

[quick pagoda]

Random papers from the 70’s have a lot of random useful results

[inland sinew]

That's a good question. I have really allowed the case where they can have either sign. Initially when I studied the problem I looked at it from the standpoint of elliptic/hyperbolic, but because the coefficients are periodic, the discriminant of the PDE is nearly always mixed and alternates between hyperbolic and elliptic over the entire domain. I sort of abandoned that approach because of the difficulty is trying to glue together these different regions. The most important solution I've found is
$z = \alpha(x) \beta (y) - \int_{x_{0}}^{x}\int_{x'{0}}^{x'} \alpha (s) '' \alpha(s) ds - \int{y_{0}}^{y}\int_{y'_{0}}^{y'} \beta (s) '' \beta(s) ds$ due to the fact that in general, the fact that alpha and beta are periodic will allow for the integral terms to have a constant that is integrated twice, producing a quadratic source term. But I got kind of lucky finding this and needed to figure out how I could have figured this out with a more powerful approach like fourier transforms, harmonic analysis.

[untold delta]

pewpew2385

[cobalt terrace]

How familiar are you with functional analysis?

[inland sinew]

@cobalt terrace My knowledge of it is pretty weak, I'm sure it will be better in 2 semesters though. But anything to guide me in the right direction would be nice. So far I have only gotten any of my results by using fourier series and just doing harmonic balancing to find my solutions. I'd prefer not to do that again to find my answers because its too involved when you have a double Fourier series you need to balance terms for.

[cobalt terrace]

I was reminded of methods that run under the general umbrella of "Floquet theory" or "Bloch theory". This even works if $\alpha$ and $\beta$ depend on both $x$ and $y$ (but with the same periods). The idea is that while the solutions of the equation are not going to be periodic, one can show that every solution is an integral over solutions which are "periodic up to a phase shift" in the sense that there are numbers $k_1, k_2$ depending on the periods of $\alpha$ and $\beta$ such that $z(x, y) = \exp(\mathrm{i}(k_1 x + k_2 y)) u(x, y)$, where $u$ is a periodic function with the same periods as $\alpha$ and $\beta$. But maybe this is not useful...

[untold delta]

wolftoeter

[cobalt terrace]

Did you already consider to seperate the variables, i.e. $z(x, y) = v(x)w(y)$?

[untold delta]

wolftoeter

[inland sinew]

@cobalt terrace I think you are probably right. Someone mentioned this to me before awhile back and I forgot about it. I had read into it a bit but felt it was a bit out of reach for where I’m currently at, but maybe I need to just take the timr to really try to understand it. It will probably help to understand the problem.

I did try separating variables and that is useful because you can get periodic solutions under some conditions. For example if both functions are cosine, you just have to solve two mathieus equations.

[opal condor]

this is from evans

how did they take the derivtive if u wrt to t?

is it by chain rule or something?

[pulsar forge]

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

[opal condor]

ok nice thanks

[sinful herald]

Dose some one have any reference for maximal Lp-Lq regularity for stokes operator. In particular I need something like this, $u'(t) + Au(t) = \textrm{div} (f)$ where $f \in L^p(0,T ; L^q(\mathbb{R}^n))$ where $A$ is the stokes operator

[untold delta]

Math&Tea

[modest kettle]

Can someone suggest me a good book on PDEs where I can get the topics like, "Homogeneous Wave Eqns, Vibration of Finite Strings with Fixed length, Nonhomogeneous Wave Eqns," etc. Till now I've been studying the book Linear Partial Differential Equations for Scientists and Engineers by Tyn Myint -U and Lokenath Debnath but the explanations or the language of the book seems horrible imho. As for my background, I am an undergraduate student pursuing a Bachelor's degree in Mathematics. Now, coming back to my issue, it seems the syllabus of the course in our university is directly designed using the book I mentioned. So, if the recommended book that has the same sort of contents as this book, then it'll be very helpful in my case.

[lilac barn]

Look up for papers by Ken Abe. He has some results for Stokes operator

[sinful herald]

Thanks I will take a look

[lilac barn]

Evans should be sufficient

[chrome trout]

Assume $u \in L^{2^*}(\bR^N)$ and $\nabla u \in L^2(\bR^N)$ where two star is the sobolev exponent. I have $\varphi_R$ a mollification of $\varphi$, a smooth function with compact support. How can I estimate $\int_{\bR^N}|u\varphi| | \langle \nabla u,\nabla \varphi_R\rangle|dx$? Also, is this correct $$|\nabla (u\varphi_R)|^2 \leq |\nabla u|^2 \varphi_R^2 + |u\varphi_R| |\langle \nabla u,\nabla \varphi_R\rangle| + u^2|\nabla \varphi_R|^2$$?

[untold delta]

мир

[chrome trout]

Ok, I will try to estimate $$\int_{\bR^N}|\langle \nabla u,\nabla \varphi_R\rangle|^2,$$so that I can Use Holder's inequality. Indeed, using Cauchy-Schwarz $$\int_{\bR^N}|\langle \nabla u,\nabla \varphi_R\rangle|^2 = \int_{supp \nabla \varphi_R}\norm{\nabla u}_2\norm{\nabla \varphi_R}_2 < \infty.$$Does this work?

[untold delta]

мир

[lilac barn]

Here the inner product is just the dot product?

[sinful herald]

where is $\phi$

[chrome trout]

Yes

[chrome trout]

It is varphi

[lilac barn]

[ \int \lvert u \phi \rvert \lvert \langle \nabla u , \nabla \phi R \rangle \rvert \leq \lVert u \phi \rVert{L^2} \lVert \langle \nabla u , \nabla \phi R \rangle \rVert{L^2} \leq \lVert u \rVert_{L^{2^\star}} \lVert \phi \rVert_{L^q} \lVert \nabla \phi R \rVert{L^\infty} \lVert \nabla u \rVert_{L^2}. ]

[untold delta]

cocat

[chrome trout]

ok

thanks a lot

[chrome trout]

one remark: you used Holder to start with but we (at least I) did not know if the inner product of the gradient of u with that of phi_r is in L^2

[lilac barn]

It is as shown in the last inequality

[chrome trout]

then I see that it is because you just use the fact that the gradient is still smooth with compact support and the gradient of L^2

of u is in L^2

yeah indeed

thank you

[gray pebble]

WTH what is this 💀

I'm in the first semester

When I see this, I don't know if I want to keep studying.

😭

[sinful herald]

yes but to apply holder one would need that varphi is in some Lr space? I meant to ask that? was it in some Lr space

[chrome trout]

varphi is smooth with compact support, hence in every L^p

[sinful herald]

ok, I miss understood then, I though you were emphasizing that varphi_R where smooth,

[modest kettle]

https://discord.com/channels/268882317391429632/1293878126002044935

I need some help with this :?)

[chrome trout]

and they are

[chrome trout]

how exactly do you bound the inner product? I am sorry for being picky with the details

We have $$\int |\langle \nabla u,\nabla \varphi_R\rangle|^2.$$Do you use Cauchy-Schwarz? Because it seems like you just did $$\norm{\langle \nabla u,\nabla \varphi_R\rangle}2 \leq \norm{\nabla \varphi_R}\infty^2\int |\nabla u|^2$$

[untold delta]

мир

[chrome trout]

At first I could restrict the domain of integration to the support of $\nabla \varphi_R$. Since it will have compact support I could use Cauchy-Schwarz so that $$\int |\langle \nabla u,\nabla \varphi_R\rangle |^2 \leq \norm{\nabla u}_2\norm{\nabla \varphi_R}_2|supp \nabla \varphi_R|$$

[untold delta]

мир

[lilac barn]

Cauchy-Schwarz for vectors

[lilac barn]

Yes basically that's it

[chrome trout]

But that's not cauchy-schwarz? You would need the 2 norm on \nabla varphi_R?

[lilac barn]

Wdym?

I meant bound the inner-product via Cauchy Schwartz then bound the integral by just taking the Linfty norm out

[chrome trout]

Precisely, like this then

[lilac barn]

You can also just take the Linfty norm out and that's enough

[chrome trout]

But you get $|\langle \nabla u, 1 \rangle|^2$ and this seems different than the 2 norm of nabla u

[untold delta]

мир

[lilac barn]

No, first use Cauchy-Schwarz to get the 2-norms of both and then take the Linfty out.

Here the 2-norm is the sequence norm

The Linfty is the Lp norm

[chrome trout]

I see

[carmine wedge]

Ohhhh it's here

I am given this problem, which I have to solve for its generalized form (i.e. p instead of 2)

I've solved everything except the equation

specifically I think I need an inequality to help, but I don't know how to obtain it

I'm trying to go throw the first variation

But I really don't know how to go from here |Du + eDv|^p to |Du|^p + e |Du|^(p-2) <Du, Dv>

A colleague of mine asked chatgpt

it's just the first order approximation of the function e -> |Du + eDv|^p

[tame bison]

@zealous prairie let’s continue here

i’m 22

[zealous prairie]

ok this is like way out of my league but ok 😂

[tame bison]

so like recap

we are trying to solve the PDE

du/dx + du/dy = 1

PDEs need to be given an initial condition

[zealous prairie]

why do they 'need' it?

or, how do they 'need' it

[tame bison]

cause just like how an ODE determines a solution only up to a constant

a PDE determines a solution only up to some “boundary condition”

[zealous prairie]

but you can still get a family without it? no?

[tame bison]

in this case our boundary comditioj shall be

[tame bison]

yes

but

[zealous prairie]

so it doesnt need it we're just solving for a unique solution?

[tame bison]

it needs to be specified where you are prescribing the boundary condition

in our case let’s take u(0, y) := f(y)

i.e. we prescribe the initial values along the y axis

[tame bison]

And i claim this uniquely determines the solution to the pde

to see this, define for each real number z

the curve c_z: R -> R^2 as i described earlier

[zealous prairie]

u is the vector function we're solving for, and we're choosing x=0 which somehow is an initial condition? not sure i follow

[tame bison]

we are trying to solve for u

god said

[zealous prairie]

and u is a vector function right

[tame bison]

takes in a vector in R^2, spits out a scalar

[zealous prairie]

u takes in a vector in R^2 and spits out a scalar?

[tame bison]

god said, (this is part of the initial problem) that along the y axis

[zealous prairie]

who is god

[tame bison]

yes u: R^2 -> R

The one who gave the problem

[tame bison]

along the y axis

[zealous prairie]

so basically like f(x,y) right?

[tame bison]

yep

[zealous prairie]

just a regular multivariable function

alr

[tame bison]

yep

we trying to solve for that

God said that in this problem

u(0, y) = f(y) for some given function f

i claim that just given this f

i can uniquely determine a solution to the whole PDE

[tame bison]

god gave us only the values on the y axis

now i will extend it to the whole domain and in doing so solve for u

oh no orion is here

[zealous prairie]

💀

[tame bison]

How do u mute seriously i forgot

[zealous prairie]

bro thought he could escape

[tame bison]

It’s distrscting

anyway

c_z := (0, z) + t(1,1)

consider a candidate solution u

[zealous prairie]

if u is our solution, c_z seems like a random vector function?

what is it, and how did you determine it?

[tame bison]

c_z is like a family of limes

lines

that i just pulled out of my ass

I defined it

[zealous prairie]

oh lol

[tame bison]

one line for each z

[zealous prairie]

gotcha

[tame bison]

it starts at (0, z)

yep

so now the magic is

g_z’ = c_z’ (du/dx, du/dy)

= 1

so we can solve for g_z exactly

clearly g_z (t) = f(0, z) + t

[zealous prairie]

so if you plug in t=u
then the derivative of your space curve c_z(u) is 1

[tame bison]

indeed

[zealous prairie]

does that work for any family of space curves you pull out of a hat?

[tame bison]

nope for this pde

i chose those lines specifically

[zealous prairie]

how 😭

[tame bison]

because of the form of the pde

[zealous prairie]

should i just accept this XD

[tame bison]

well the pde says that du/dx + du/dy = constant

so clearly this says that the derivative is constant along straight lines with slope 1

“clearly”

clearly if u already know the method xD

[zealous prairie]

😂

[tame bison]

but yeah that’s th3 gist

now you have g_z (t) = f(0, z) + 1

[zealous prairie]

but if c_z'(u) = 1

[tame bison]

(c_z \circ u)’ = 1

not c_z’ u

[zealous prairie]

d/du (c_z(u)) = 1

?

[tame bison]

d/dt c_z (u(t))

[zealous prairie]

alr

[tame bison]

ye

now c_z \circ u (t) = f(0, z) + u(t, z + t) by definition

[zealous prairie]

huh

[tame bison]

just plug in the definition of c_z

[zealous prairie]

c_z(t) = (0, z) + t(1, 1)
c_z(u) = (0,z) + u(1,1) = (0,z) + (u,u)? im not understanding...

[tame bison]

OH FUCK ME

it needs to be u \circ c_z

directions are hard

xd

[zealous prairie]

uh

[tame bison]

u \circ c_z (t)

[zealous prairie]

u(c_z)?

😭

[tame bison]

yeah

[fluid token]

nty

[zealous prairie]

u(c_z(t))

[tame bison]

yeah

so like g_z (t) := u (c_z(t))

differentiate g_z wrt t, to get g_z ’(t) = 1

[zealous prairie]

hoooold on that was big jump 💀

g is still g_z?

[tame bison]

this one?

yeah

[zealous prairie]

yea

[tame bison]

sorry notation is hard on discord

[zealous prairie]

wait i need to rederive how it was 1

[fluid token]

use latex

[tame bison]

right so

[zealous prairie]

(u(c_z(t)))' = 1
how did we know this?

[tame bison]

g_z ‘ = (du/dx, du/dy) \cdot (1, 1)

just calculus

multi variable calculus

to find the derivative of a composition

[zealous prairie]

g_z is the gradient dot with (1,1)?

[tame bison]

yep

[zealous prairie]

oh i see

[tame bison]

you compose the derivatives of each component

ye

[zealous prairie]

thats how du/dx + du/dy = c

[tame bison]

this was the original defining pde

[zealous prairie]

yeah

[tame bison]

du/dx + du/dy = 1

so if we plug that in

g_z’ = 1 as we desired

and now you can backsolve for u

g_z = f(0, z) + t by our derivation

but also g_z (t) = u (c_z(t)) = u(t, z+t) by definition

so that u(t, z+t) = f(0, z) + t

[zealous prairie]

i feel like im missing something 😭

[tame bison]

oh.l

[zealous prairie]

g_z' or g_z is the gradient?

g_z' i guess?

i dont know why/how we know this

[tame bison]

g_z is the function g_z’ is the gradient

[zealous prairie]

is that how its defined?

g_z is which function again?

g_z is u(c_z(t))

[tame bison]

well we can compute g_z’ = (du/dx, du/dy) cdot (1,1) right

[zealous prairie]

and c_z(t) you pulled out of a hat

[tame bison]

by calculus

Yep

[zealous prairie]

uh

[tame bison]

this is just du/dx + du/dy

which is 1 by our defining pde

[zealous prairie]

we can compute (du/dx, du/dy) . (1,1), but how do we know thats g_z'?

[tame bison]

g_z is

[zealous prairie]

wait

[tame bison]

u \circ c_z

[zealous prairie]

oh a vector c_z(t) is inputted into a function of two variables (or a function of a vector) u(x,y)

g_z = our solution u(x,y)

[tame bison]

so g_z’ = u’ \cdot c_z’

yeah

[zealous prairie]

and since its a solution we know that du/dx + du/dy = 1

or whatever

= c

[tame bison]

g_z is not ur solution

[zealous prairie]

oh

[tame bison]

g_z is u (c_z)

[zealous prairie]

ohh

[tame bison]

or as i like to write it more unambiguously , u \circ c_z

[zealous prairie]

its our random vector inputted into our solution

[tame bison]

Indeed!

along the curve which we think of as parametrised by time t

differentiate this to find that u changes along the curve by 1

per unit time

[zealous prairie]

right...

[tame bison]

and then you can extend the god given values of u on the y axis

To any location in R^2

That’s the method of characteristics in a nutshell

We end up getting u(t, z+t) = f(0, z) + t

[zealous prairie]

g_z = u(c_z)
so somehow, a random vector inputted into our solution is the gradient vector.
do we know this BECAUSE its a solution to the differential equation?

[tame bison]

no,

we defined g to be that

and then differentiated it and plugged in our pde to find that g_z’ = 1

[tame bison]

and then used that to backsolve for u

now if you want a solution in the form u(x, y) = something

change variables

t -> x, z -> y - t

[zealous prairie]

we defined g_z to be the gradient? or we defined g_z to be u(c_z)?

[tame bison]

latter

[zealous prairie]

ok so since we differentiated u(c_z)

and got uh

(1,1)?

wait no we got 1

[tame bison]

we get (du/dx, du/dy) \cdot (1,1)

which is equal to 1 by our pde

[zealous prairie]

u(x,y) is always 1

so u(c_z) is 1 as well??

is that the argument

[tame bison]

u(x, y) is not known to us yet

we are trying to solve for u(x, y) for general x, y

that’s what it means to solve a pde

[zealous prairie]

sorry i meant

idk what i meant wait what ._.

[tame bison]

ahaha

[zealous prairie]

💀

[tame bison]

okay take it easy ahah

it’s a lot to digest over a discord comvo

convo

[zealous prairie]

fr 💀

[tame bison]

but i think u can read evans ch1 now and have a decent time

[zealous prairie]

alr

[tame bison]

Thanks for listening to me rant

[zealous prairie]

thanks for tryna teach me 💀

poor orion

[azure frigate]

these results all follow from convex analysis? I haven't done any convex analysis so im not sure

uniform convexity for the continuity for all but at most countable x and strictly convex would imply uniqueness?

i can post context if needed

[sonic olive]

Non decreasing is what they used for continuity up to countable set

This is true for all monotone functions on R

@azure frigate

[azure frigate]

ohh im stupid thanks

[untold delta]

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

[arctic whale]

I am trying to do this exercise.

[untold delta]

ru0xffian

ru0xffian

[arctic whale]

This of course gives that there is some x_0 in Gamma and a sequence x_n in P so that x_n converges to x_0 but I can not move from there to show the existance of some ball around x_0 on which u >=0

[lilac barn]

Expand out nabla u dot nabla u

[arctic whale]

Could you elaborate more?

[lilac barn]

Evans PDE C2 theorem 3

[untold delta]

Rotta

[quaint herald]

This is a great problem. One approach is to enlarge your domain slightly near a point p on the boundary portion where we have vanishing. Consider the extension of u by zero on this enlarged domain \Omega'. Using the fact that u is harmonic you can show the extension v is harmonic (some work is required here, but not much). Then you are done by analyticity in the interior.

I don't know of any simpler way, in particular it is not clear to me what cocat's intended solution was. If you have trouble showing v is harmonic, I can give additional hints.

[arctic whale]

just done this with the equivalence of weak harmonic with harmonic

[quaint herald]

fantastic

[arctic whale]

Didn't get what cocat was hinting to either

[quaint herald]

sounded like the standard energy estimate way of showing that dirichlet and neumann data vanishing everywhere implies u=0.

but ofc we don't have that.

[arctic whale]

My initial idea was a bit complicated I guess and I could not finish it anyways.

[quaint herald]

yeah unclear to me how to make that kind of idea work.

[noble gate]

How would you guys approach this

I dont know what a transport equation is

[rotund jetty]

This is $h(x, y) = \nabla h(x, y) \cdot (a, b)$. Fix $y$. Along curve $z(t) = h((0, y) + t(a, b))$ we have $z'(t) = z(t)$, so $h((0, y) + t(a, b)) = e^{t}h(0, y)$.

[untold delta]

L

[noble gate]

I see

[spiral geode]

does 1d anisotropy makes any sense, since anistoropy usually indicates the difference of behavior accross different axes

[carmine wedge]

I'm trying to derive the Poincare Wirtinger inequality for the case H1_0 under the interval [-1,1]

I'm not sure if it's possible to go through Euler Lagrange

by this theorem, we know that the functional with lagrangian f(x,u,u') = u'^2 - pi^2u^2 has a minimizer call it uB for u bar = \overline u

Would it be possible to prove that uB is in H^2_0 and so through the E-L equations obtain the class of minimizers and the inequality?

[void flame]

I know for part b I should consider phi squared, but is the fact that the characteristics line “collide” prove that the solutions don’t exist in a classical sense? Or do I need to argue more

if anyone could help I rlly appreciate it!

[arctic whale]

Let $G(x,y)$ be Green's function on $\Omega$. Let $f$ be a bounded integrable function on $\Omega$. I was trying to prove that $\int_\Omega G(x,y)f(y) dy \to 0$ as $x \to \partial \Omega$. I could solve it using generalized DCT and weak Lp convergence. But I was thinking about something else. If $f$ were continuous, then we know that the integral satisfies the Dirichlit problem $\Delta u = f, u = 0$ on $\partial \Omega$. So the result holds trivially when $f$ is cont. Is there a way to carry some approximation argument and prove it for integrable $f$?

[untold delta]

ru0xffian

[lilac barn]

Mollify f then study the difference of G mollified f and G f

[lilac barn]

Indeed: more precisely, there will be two characteristics propagating different values. To get a picture of this, check Evans Page 140.

[void flame]

Oki will do, thank u!

[lilac barn]

I haven't checked it for your case but most likely if there's an issue it would be like this.

[void flame]

oki oki

[untold delta]

22

[carmine wedge]

Does anyone know why the author does this step?

this is the setup and, the thing I'm confused about in a proof relating this is:

I'm confused about, why can't we a priori differentiate g wrt epsilon?

[sand echo]

differentiating g wrt epislon is no problem, since epsilon -> g(x,epsilon) is C^1. But the point is that passing the limit epislon ->0 into the integral requires justification. So the point is you use DCT, and to get the upper bound on the difference quotient for g you naturally use the mean value theorem

[carmine wedge]

sure, but why use the MVT?

Since g is C^1 wrt to epsilon, and pointwise in x, all of the arguments are C^1 functions in epsilon

doesn't the chain rule already imply

$g_\epsilon(x,0) = f_u(x.u,\nabla u) \varphi + \langle f_\xi(x,u,\nabla u);\varphi\rangle$

[untold delta]

Trivial Lemma

[carmine wedge]

Both of those functions are L^1 by Cauchy-Schwarz + (H3) + Young's inequality

So there's absolutely no need to do all this magic, innit?

or do we need it to prove the first line to the second?

[sand echo]

exactly this, the DCT is needed to pass the limit into the integral

and do apply DCT, your 'dominating function' has to dominate the converging sequence, not just the limit

the converging sequence here are the difference quotients g(x,ep) - g(x,0) / ep, which you need MVT to estimate

[carmine wedge]

can't I estimate them using the first order taylor series expansion at eps = 0? With an o(e)

[sand echo]

MVT is basically a special case of taylor series in that sense

its the same thing

[carmine wedge]

so like g(x,e) = g(x,0) + [the chain rule]e + o(e)

since o(e) is in O(1) and Omega is bounded, it would follow right?

[carmine wedge]

just feels more convoluted to me to use MVT*

Anyway, thank you in any case ^^

[sand echo]

MVT is precisely taylors theorem with the remainder form, in the case 'k=0' where k is the degree of your polynomial

so its just identical

but from a moral standpoint i would prefer to call it MVT haha

[void flame]

could anyone give me a hint on this? I wrote it out using the definition but just struggling to find the right bounds..

[quaint herald]

Hint: duality.

(and this is the hint for most things you prove about distributions).

[void flame]

sorry i dont follow

[quaint herald]

Prove a related statement about how the FT acts on a convergent sequence of Schwartz functions, and leverage this to get the tempered distribution result for free.

[void flame]

sorry but I'm struggling to understand how duality comes into play here

;_;

[rotund jetty]

how do you check whether $\hat{g_n} \to \hat{g}$ in the sense of tempered distributions?

[untold delta]

L

[astral vine]

<Fu,v>:= <u,Fv>

where F is the Fourier transform

u your tempered distribution

v a Schwartz function

[opal condor]

reviewing for exam, and im not sure how to do part 2 of this problem, help wouls be appreciated thanks

[void flame]

Ok I think I got it, thank you!

[void flame]

Could anyone help me with this? The incompressible transport equation being this:

[strong maple]

I think you need to differentiate the integral, conmute the derivative with the integral, do chain-rule and replace u_t with -a . grad(u), then integrate by parts to use div a = 0

[void flame]

yes thats what i tried, but i couldnt get it to work, couldnt get diva

it works if we're integrating u but here we're integrating u^2

[strong maple]

Do you get div(2ua)?

[void flame]

yes

[strong maple]

Use product rule: you should get an integral that has an opposite sign than a previous step

[void flame]

sorry wdym, use product rule on div(2ua)?

[strong maple]

Yes, div(2ua) = 2 grad(u). a + 2 u div(a)

[void flame]

but then grad(u) is nonzero

oh I just realized that I could write grad(u^2)= 2u grad(u). Is that what you mean?

[strong maple]

I mean, you can. I was saying that, at some point, your integral reads -int( 2ua.grad(u) ), but after integration by parts and product rule it reads int( 2ua.grad(u) ). That suffices to conclude the integral is zero

(I was implying we did the same steps, my bad)

[void flame]

sorry, I dont get why that suffices to conclude the integral is zero. I thought we only know that u vanishes outside a compact set and that div(a) = 0. Could you elaborate?

[strong maple]

A number is both nonnegative and nonpositive so it must be zero A number is equal to its opposite, thus zero. Whoops

[tired hollow]

Does anyone have a maths server about PDEs or maths in general that has an active VC?

It might not be the best channel so excuse me

Just said maybe some PDEs lovers here might recommend a server

[lilac barn]

Yes precisely: agrad u times u = a dot grad|u|^2/2 then IBP and Divergence free

[void flame]

oki, that’s what I did! Thank you ❤️

[opal condor]

is d'alembert's solution unique for wave equations?

1d wave equation

[ocean ether]

so for part a, i think i need to show that $u_xu_{xt}=-u_{xx}u_t$?

[untold delta]

the breadwave

[ocean ether]

which idk how to do

[lilac barn]

Show that derivative of k + p wrt to t is 0

[ocean ether]

yes i know

i just need to show that if we differentiate p wrt t the inner integral reduces to -laplacian times u_t

[azure frigate]

integration by parts

[ocean ether]

A.

ok

[ocean ether]

ok yeah this is trivial with compact support

[verbal nebula]

Any good way to get into dispersive PDEs or Potential Theory?

[ocean ether]

having some trouble with part b

i think i got to that $k(t)-p(t)=\frac{1}{2}\int_{-\infty}^\infty h(x)^2-g'(x)^2 dx$?

[untold delta]

the breadwave

[ocean ether]

idk if i differentiated d'lambert's formula correctly tbh

[azure frigate]

you didnt

[ocean ether]

yeah

i thought so

i got like

$\frac{1}{8}\int_{-\infty}^{\infty}(g'(x+t)-g'(x-t)+h(x+t)+h(x-t))^2-(g'(x+t)+g'(x-t)+h(x+t)-h(x-t))^2 dx$

[untold delta]

the breadwave

[ocean ether]

but now t doesn't matter because x covers the real line so the value of the integral doesn't change if we kill t

so what goes on with that @azure frigate

[azure frigate]

regardless what is important is the compact support

[ocean ether]

yes i am aware that compact support is important

i think i killed t wrong then

[azure frigate]

i just dont think its necessary but you can do it i think

[ocean ether]

im just confused because its like

x covers the real line so even though t may be huge, when x is huge h(x-t) and g'(x-t) may not be 0

wait a minute...

oh nvm this problem is trivial

if we simplify the integral we get $\frac{1}{2}\int_{-\infty}^\infty -h(x+t)g'(x-t)+h(x-t)g'(x+t)-g'(x-t)g'(x+t)+4h(x-t)h(x+t) dx$

[untold delta]

the breadwave

[ocean ether]

and now since we have 2 functions of compact support h(x+t)g'(x-t), when t is large 1 of them is gauranteed to be 0 as if h(x+t) is not 0 then x is close to -t, so -x-t is huge and therefore g'(x-t) is 0

[ocean ether]

nvm i got i tnak you

(does my reasoning look right)

[astral vine]

Dispersive PDE : Ponce Linares' book or Tao's book

For Potential theory, it may have different meanings and then underlying goals so, be more clear please

[buoyant pike]

Hi anatole

[astral vine]

Oh The Return of the King

How are you buddy ?

it's been a while

[buoyant pike]

I'm good how are you

[astral vine]

Fine, a bit busy tho. What are doin now, are you still doing research ?

[buoyant pike]

Yep phd year 4

[manic tinsel]

Hello. I have a question about the book Partial Differential Equations in Action (Third Edition) by Sandro Salsa. In section 4.6.2, it says that u_L is unstable and u_R is asymptotically stable. Also, somehow this leads to the phase diagram on the right being the correct phase diagram for the equation when q''<0. Could someone explain to me why they have that stability/instability and why this leads to the subsequent phase diagram? I think it might be something simple that I am just having trouble finding information about.

[verbal nebula]

Are you still doing fluids

[buoyant pike]

Geophysical fluids, yes

[upbeat spade]

im trying to learn microlocal analysis and feeling very dumb. how do you show that the typical polynomial function with multiindex degree m is in the symbol class S^m?

this is from thibault lefevre's book

[verbal nebula]

Oh dang that's computational stuff

[upbeat spade]

i figure once i see the most basic example it should be obvious but everywhere online assumes that it is obvious

[buoyant pike]

Ok so what does it mean to be in symbol class S^m

[upbeat spade]

figured it out yesterday, i think the key was the compactness in the x variable

what not doing analysis for a year does to a man

[round zinc]

Hi guys. Could someone tell me what the notation $(a \otimes b):(c \otimes d) $ means?

$a,b,c,d$ are vectors. I think the $:$ might denote the frobenius inner product, but I only know how it works for matrices. Is it maybe supposed to mean $\sum_{i} a_i c_i + b_i d_i$? I have never seen this notation

[untold delta]

mcdnldsmngr

[buoyant pike]

I would interpret that to be the matrix double dot product

a otimes b and c otimes d are both matrices

a otimes b = a dot b^T

[round zinc]

Is that common notation? Because i think we use the abstract tensor product

[buoyant pike]

Common notation for the matrix double dot product?

[round zinc]

This

[buoyant pike]

To be clear, $A:B=\mathrm{tr}(AB^T)$

[round zinc]

Ye

[untold delta]

Angetenar

[buoyant pike]

otimes for the vector outer product is quite common

[round zinc]

Hmm

I hate this course 😆

[buoyant pike]

Lol well good luck

[round zinc]

ty

[spiral geode]

is the diffusivity coefficient just a weight function in general case of diffusion models ?

no matter how i look at it, it is just a weight function, and the word is just charged to indicate diffusion instead of weight which is just very broad to indicate any behavior

[spiral geode]

[spiral geode]

heat equation

[astral vine]

in many cases it can be interpreted as such.

[spiral geode]

I see, ty for insight

[hard moth]

Does wave equation count as a pde?

[buoyant pike]

The wave equation is indeed a pde

[hard moth]

Then I choose the wave equation

[white sequoia]

Are there notable uses of the logarithm of a dirac delta? Does that work well - what notable properties does that distribution have?

[buoyant pike]

How did you come across the logarithm of the dirac delta....

[white sequoia]

Thinking of the behaviour of score functions, alla log(p(x)), for sharp concentrations

[buoyant pike]

How do you propose to interpret log(dirac delta)?

[white sequoia]

I'm basically asking if there's topics where it popped up, places where the various formal manipulations (whether there \delta is interpreted as extreme statistical location concentration (and where the log then ties to information functions) or not) have been performed and clarified, in the distribution sense

[ocean ether]

so what exactly is duhamel's principle

[buoyant pike]

Duhamel's principle is something that you learn once to solve the non-homogeneous heat/wave equations and forget thereafter

[patent fjord]

$\partial_v u$ means conormal derivative of $u$.

Let $u \in H^1(\Omega)$ be a solution for a problem $Lu = F$ in $\Omega$ and $u = g$ on $\partial\Omega$, where $L$ is an elliptic operator.

Then i want to show that $|\partial_v u|{H^{-1/2}(\partial\Omega)} \leq C|u|{H^1(\Omega)}$, but i was able to show $$|\partial_v u|{H^{-1/2}(\partial\Omega)} \leq C(|u|{H^1(\Omega)} + |F-cu|{L^2(\Omega)}),$$ where $|c|{L^\infty} \leq M$. Is there a way to get rid of the $ |F-cu|{L^2(\Omega)}$ somehow? i was thinking something like $ |F-cu|{L^2(\Omega)} \leq |u|_{H^1(\Omega)}$, but i dont see that currently

[untold delta]

Rotta

[void flame]

I have a question about this paragraph about the deconvolution being ill-posed, they wrote that it is basically because of deconvolution, but where does deconvolution appear? Is it because we're taking the inverse fourier transform of phi over u?

[rotund jetty]

It's an ODE thing. See pages 73-74 of https://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/analmv.pdf. You can extend it to PDEs by rewriting the PDE as an ODE with values in a function space.

[verbal nebula]

I think I'm required to say Navier-Stokes or Euler

[low turret]

Can't you say EFE with fluid matter sources?

[waxen bobcat]

What's EFE?

[low turret]

Einstein Field Equations

[waxen bobcat]

Ah, alright

[verbal nebula]

Seeing how none of that ever shows up in the research that I'm doing, no

[fringe portal]

Hey guys, sorry for jumping in out of nowhere, but I am dealing with the equation in the pic attached for p integer and p > 1. It resembles a thin-film equation but its not quite there. Has someone worked with something along this lines?

[buoyant pike]

Well what do you want to do with the equation?

[fringe portal]

So as a little backstory, it stems from a physical model of a roughening solid surface.

Honestly, classification as a start so I can search works up on it would be great. Ideally, understanding of solution behvaiour (e.g. does it blow up in finite time) so I can implement a numerical scheme for it.

I've already played around with it's linearization and a family-viscek-type ansatz, though the resulting ODE in the second case wasn't really useful

[astral vine]

This is a consequence of the ontoness of the trace operator from H¹(Ω) to H½(dΩ). It has nothing to do with the elliptic operator L. Actually this is needed to define properly L as an operator.

[astral vine]

Or to define its Neumann BVP counterpart.

[buoyant pike]

If we take the approach of only caring about the highest spatial derivative, we can simplify [A\frac{\partial^4u}{\partial x^4}\Rightarrow Apu^{p-1}\frac{\partial^4u}{\partial x^4}] and then you just look at [\frac{\partial u}{\partial t}=Cu^{p-1}\frac{\partial^4u}{\partial x^4}]

[untold delta]

Angetenar

[buoyant pike]

For which you can try searching for quasilinear biharmonic equation theory

[spiral geode]

i believe they did define what they meant by it, on the last paragraph, it is basicly the way g is defined taking u(x,t1) as input

[void flame]

oki, ty!

I have a question about this: So I derived the solution using Fourier transform wrt to variable x, but it also asks me to "be precise about the regularitary conditions", and I dont know how to get that.

I assume it makes sense to require g to be L1, (so the convolution with the heat kernel is L1 too) but what about the source term?

[spiral geode]

I hope someone provide some insight, i don't know really the answer, but i am assuming it wll boils down on the whatever regularity you used during the process to this point, so i would revise all what i have done and what kind of regularity was needed to make certain equalities or inequalities valid

[buoyant lantern]

poissons eq

vibrating drumhead is a fun one

ricci flow

[patent fjord]

i think the trace theorem doesnt work for the negative s

[astral vine]

Think about it that way :

if you take the trace of u you end up in H½(dΩ), assume Ω is the half-space, then the then the normal derivative, you lose one derivative. we are perfectly fine

What CANNOT happen for the trace is when you have u in H^{s}(Ω),, with s<1/2. Without any further assumption, you cannot take the trace of u in H^{s-½}(dΩ), this is wrong in general.

[quaint herald]

How are you taking the normal derivative $\textit{after}$ taking the trace? The trace lies in $H^{1/2}(\partial \Omega)$, there is no normal direction to differentiate in.

I don't think an estimate like $$|\partial_\nu u|{H^{-1/2}(\partial \Omega)}\lesssim |u|{H^1(\Omega)} $$
can hold for $u\in\mathcal{C}_c^\infty(\Omega)$ without additional assumptions (*).

Density of $\mathcal{C}c^\infty(\mathrm{int}(\Omega))\subset H_0^1(\Omega)$ would imply that $\partial\nu u$ should vanish for $u\in \mathcal{C}_c^\infty(\Omega)$ vanishing on the boundary. But this is not true of functions like $u(x_1,x')=x_1\cdot \chi(x)$, where $\chi$ is a smooth cutoff equal to $1$ in a neighbourhood of the origin.

(*) Solving an elliptic eqn like in the original question is the kind of assumption that can save you.

[untold delta]

grobmez

[astral vine]

This was a moral idea. Of course you cannot since you are fixed within the realm of tangential variable.

[trim skiff]

Somebody here can share me a pythagoras poster?

A poster about pythagoras and stuff

Really wanna learn math

[astral vine]

You don't need to solve the PDE just to be in the closure of some domain given by a non-negative sesquilinear form, if I am correct.

Defined over H¹.

Okay one may say "it solves a PDE" in some sense.

[patent fjord]

Let $\Omega' \subset \Omega$ be a lipshitz domain, then if u solves a PDE $Lu = h \text{ in } \Omega$ and $Lu = 0 \text{ in } \Omega\backslash\Omega'$ does that imply that the conormal derivative of u is zero on $\partial \Omega'$?

[astral vine]

Here is a proof.

For the Laplacian

This is H½ of the boundary in the two last lines.

[quaint herald]

Huh? Maybe I am just sleep deprived (sleepless night last night), but how are you bounding H1 norm of v above by H1/2 norm of it's trace in second last displayed eqn? I agree the trace is surjective and so open, but it's certainly not injective on H1 so not bounded below.

I still think H1 is not enough, for the reason in my last post.

[astral vine]

You only need surjectivity to get the reverse estimate

You have to think the otherway

for each prescribed boundary value in H^{1/2} I can find a v such that it is the trace, and since the functional cancels out H^{1}_0 this is fully well defined

This is the same as proving divergence free L2 functions admits a partial vanishing trace in H^{-1/2}

[quaint herald]

How so? Just abstractly, an estimate ||u|| =< C||Tu|| cannot hold for noninjective T...

[astral vine]

It is more like

I choose the extension operator from the H^{1/2} of the boundary to H^{1}

if you prefer

Because i have shown my functional only depends on the boundary value of v.

So let g in H^{1/2}, and say instead I apply my functional to E(g)

Just the open mapping theorem does not say anything about bounded linear right inverse

But it says you can always chose some preimage uniformly in some ball

So either you use the open mapping theorem, either you use the extension operator from the boundary to omega

[quaint herald]

Okay, so you are NOT claiming that $$|v|{H^1(\Omega)}\leq C|v|{\partial \Omega}|_{H^{1/2}(\partial \Omega)} $$ for arbitrary $v\in H^1 $ then? This is what your secondlast line appeared to be claiming, which is what I was objecting to.

[astral vine]

No absolutely not

[untold delta]

grobmez

[astral vine]

You are right pointing this

My bad on abuse of notations

[quaint herald]

Yeah its not at all clear the way it was written, but then what is your final conclusion? That there is a well-defined neumann trace operator from the range of any extension op H^(1/2)(boundary) -> H^1(domain)?

[astral vine]

H^{1}(O) -> H^{-1/2}(boundary)

u is fixed

I want to make sense of neumann trace of u

[quaint herald]

So you are claiming existence of a bounded neumann trace map H^1(O) -> H^(-1/2)(boundary) ?

If so, I think we are still in disagreement lol.

[astral vine]

K_u is a functional on H^{1/2} so I represent it by an element of H^{-1/2}

That's what I proved

this element must be the Neumann trace

Or seemed to prove

according to you

[quaint herald]

well like we just discussed, your estimate in the penultimate line holds for v in the range of a fixed choice of extension operator, not for general v in H^1

[astral vine]

We don't care, the estimate is valid for all v

And I did show in the middle it only depends on the boundary value of v

The extension operator was to make it simple

But the open mapping theorem make it fully arbitrary

and intrinsic

[quaint herald]

it can't be, like I said that would imply EVERY element of H^1 has "trace" zero.

[astral vine]

where ?

[astral vine]

.

?

wait wait wait wait wait.

[quaint herald]

Yes. That initial post of mine is pretty much the shortest summary of why (unless I am still misinterpreting your claim) we are in disagreement.

But I am also a zombie right now, so I apologise if I am totally missing your point.

[astral vine]

Probably me, then

Probably I did something illegitimate

But that's not the open mapping part, since this is the standard argument.

Oh yeah

wait no

[quaint herald]

My sleepy take is that I think it comes down to the fact that when you formalise your penultimate line (e.g. using an extension operator), you are only getting the |v| =< |Tv| bound on the range subspace of your (noncanonical) choice of extension op.

[astral vine]

The reason I used open mapping instead

basic Functional analysis

for Advanced PDEs

for undergrad PDEs vector calculus

[quaint herald]

Yeah, I think whatever you are doing there is what I disagree with (or do not understand), but we'll see how it looks to fresher eyes tomorrow.

[astral vine]

Sure

Of course especially within Stochastic PDEs

[zenith hound]

are there any applications relating to isotopes

[buoyant pike]

Do you mean chemistry isotopes

Or another isotope

[arctic whale]

How would someone come up with such function v? I see that the first term is somewhat natural but the rest seems magical.

[quaint herald]

Okay, so I now know what you meant in the open mapping theorem part. Indeed that argument is a standard one, and is fine. This means I agree with you that your \gamma_u defines an element of H^(-1/2) for any u in H^1.

I am still not certain about why there is a discrepancy between your construction and my proof that no bounded Neumann trace map H^1 -> H^(-1/2)(boundary) can exist, but here is a very basic question: How exactly are you defining the Laplacian as a bdd map H^1 -> (H^1)*? Simply taking weak/distributional derivatives should instead land you in H^-1 = (H_0^1)*, so it looks like there is a duality pairing being used instead.

My current guess is that with the definition needed to make your construction work, it ceases to coincide with the classical trace for C^inf functions (apart from say if u is harmonic).

[astral vine]

Actually I did think about it for the whole day

and we ARE both correct, but still you more than me

Because the Laplacian as a bounded operator from H^{1} to (H^{1})* is actually the Neumann Laplacian which is a fucking isomorphism.

The unbounded L² realization of the Laplacian given by the associated sesquilinear form (or the friedrich extension) is the Neumann Laplacian

So I just built the 0 map.

lol

my proof is not wrong just pointless

because the normal derivative was actually already 0, before I even know it

So we don't contradict each other, at least not directly with my useless proof. The claim of of the boundedness of the trace operator is wrong tho

and your argument is fully correct.

[quaint herald]

Haha right exactly, it was a subtle point. Okay, glad we are on the same page now and we both learned some things :).

Of course exactly your argument is how one proves Neumann traces exist in certain settings, e.g. for solutions to the Dirichlet problem.

[astral vine]

But I did retry the argument, Du in L² and Lu in L², L being a nice elliptic operator in divergence form possibly with lower order terms, with the same argument give some "twisted" Neumann trace in H^{-1/2}.

[astral vine]

yeah.

[quaint herald]

Yeah that kind of statement I believe.

Glad it was resolved, this would have tormented me if it was still unclear after today.

[astral vine]

This point is actually necessary to build the Robin Laplacian properly

[astral vine]

Me too

[arctic whale]

one more magically cooked comparison function.

[modest kettle]

While studying about PDEs I found that we can calculate the general solution of a PDE from it's complete solution.

The method goes like this:

If say, a PDE $F(x,y,z,p,q)=0$ where $p=z_x,q=z_y$ is given, and $f(x,y,z,a,b)=0$ is a complete solution of $F,$ then assume $b$ to be an unknown function of $a$ say, $b=\phi(a).$

So, the complete solution $f$ becomes, $f(x,y,z,a,\phi(a))=0.... (1)$

Now, differentiating the above equation wrt $a$ we get, $f_a(x,y,z,a,\phi(a))+f_b(x,y,z,a,\phi(a))\phi'(a)=0.... (2)$

In principle we can solve $(1)$ and $(2)$ to get $a=a(x,y,z).$

Substituting this in $(1)$ we have,
$f(x,y,z,a(x,y,z),\phi(a(x,y,z))=0.$ This is precisely the general solution of the given PDE as, $b$ is an arbitrary function of $a.$

---

Now, I wanted to implement this method using an example. So, suppose we have a PDE $z=pq$ having a complete solution $z=(x+a)(y+b).$

Now, proceeding in line of the above method, let $b=\phi(a).$

So, the complete solution becomes $z=(x+a)(y+\phi(a)).$

Differentiating the above equation wrt $a,$ we have,

$$0=(x+a)\phi'(a)+(y+\phi(a))$$

So, we have got $z=(x+a)(y+\phi(a))$ and $x\phi'(a)+a\phi'(a)+y+\phi(a)=0$ as the two equations.

But I don't know how to solve for $a$ such that $a=a(x,y,z)$ from the last two equations.

This makes me doubt about the validity of the method I described earlier for $(2)$ always has an unknown term (or function) $\phi(a)$ due to which finding $a=a(x,y,z)$ does not seem to be feasible at all.

Can someone please help me with this?

[untold delta]

Franklin244

[ocean ether]

where did evans pull this

[pulsar forge]

whats your question? are you asking how the strong maximum principle implies this?

[ocean ether]

yea

[pulsar forge]

suppose that u is not positive everywhere in U and apply the principle to -u

[ocean ether]

wait a minute yeah this is trivial

[pale dragon]

Hey guys. I am bit stuck at this exercise from Dorina Mitrea's book. Intrinsically, I guess u hat should be naught or multiple of a dirac function. But clearly it's not true. Could you tell me what's wrong?

[astral vine]

Not a multiple but a sum of derivatives of Dirac masses.

because fourier transform exchange multplications by powers of x into derivatives

[lilac barn]

First reduce to the case where the support is at 0 using the property of translation and transform. Then note that not only uhat has support 0 but also so does xi uhat. What can you conclude from this: aka what's the general distribution that you can construct which takes advantage of it)

[fast orchid]

would someone please mind helping a gal out? This has been driving me crazy: #help-29 message

[fast orchid]

I expanded on the question a bit as a post: https://discord.com/channels/268882317391429632/1304123540047532102
Would appreciate some help if you can

[wintry delta]

How to approach this?

[buoyant pike]

Do you know what lambda is

[wintry delta]

average propagation speed?

$\lambda = (f(u_r) - f(u_l))/(u_r - u_l)$

[untold delta]

Lesage

[buoyant pike]

Ok what manipulations have you tried to apply to the inequality

[wintry delta]

I stuck in case where k is between u_l and u_r

[buoyant pike]

Ok, what have you worked out in that case

[wintry delta]

$\lambda (u_r + u_l - 2k) >= f(u_r) + f(u_l) - 2f(k)$

[untold delta]

Lesage

[wintry delta]

So I don't know how to proof this inequality

[buoyant pike]

Do you have any conditions on f

(most authors restrict f to be convex don't they)

[peak ridge]

Hey, let r denote the distance from the origin and suppose G is a scalar function. Can anyone help me to compute $|\nabla (G(r) x_i /r)|^2$?

[untold delta]

Buffet

[peak ridge]

The gradient is supposed to be the euclidean, so I believe we are seeing $r = |x|$

[untold delta]

Buffet

[peak ridge]

So, can anyone help me computing $|\nabla ( G(|x|)x_i/|x|)|^2$?

[untold delta]

Buffet

[waxen bobcat]

What's there to compute?

[peak ridge]

A starting point would be the gradient of $G(|x|)x_i/|x|$ but I am getting because I am getting Kroenecker deltas..

[untold delta]

Buffet

[peak ridge]

there might be something easier

[buoyant pike]

Do you know what the gradient is in spherical coordinates

[final sand]

is this channel appropriate for asking questions of ode?

I cannot find one dedicated to ode, and my question is not quite the early uni level

[buoyant pike]

#advanced-analysis for advanced odes

[final sand]

thank you!

[wintry delta]

If I remember correctly it can be either strictly convex or strictly concave, but not both (convex on one part, concave on the other)

[buoyant pike]

Ok so if f is strictly concave, negate it so that it's strictly convex

[buoyant pike]

Then apply convexity to this inequality

[wintry delta]

I don't understand what it means? Convex means that second derivative is greater than 0, right? How to apply it here?

[buoyant pike]

What other properties do convex functions have?

[wintry delta]

$f(λx1​+(1−λ)x2​)<λf(x1​)+(1−λ)f(x2​)$

$f(\lambdax1​+(1−\lambda)x2​)<\lambdaf(x1​)+(1−\lambda)f(x2​)$

[buoyant pike]

You'll want to fix your latex but yes, you will want to use this

[wintry delta]

I guess I can redefine k as $\alpha * u_r + (1 - \alpha) * u_l$?

[buoyant pike]

Yes

[untold delta]

Lesage

[calm falcon]

Could someone give me some ideas on this

the method of characteristics tells me that there exists a unique solution in the neighborhood of the initial curve

I have that the solution is of the form u = u_0(s)*e^{-2t} and I know s is implicitly a function of x and t

not sure what other conditions need to be imposed

[buoyant venture]

11

[velvet grail]

bro your dept?

you are from iitg?

[buoyant pike]

#odes-and-pdes

[opal condor]

this is from second page of evans

second editions evans pg 193

what does the notation |.,.| mean in this case?

[shy narwhal]

this is what my copy says

[opal condor]

what |x,y|=1

means

it's used right before 21

[shy narwhal]

those are just absolute value bars

it’s a typo

[opal condor]

oh wait i see it now

[opal condor]

for the same example

i thought that they let v(x,t)=u(x, it)

so that they can plug in it in place of t

but wouldn't that solve $iu_t-\Delta u$

[untold delta]

lmao

[opal condor]

instead of $iu_t + \Delta u$

[untold delta]

lmao

[opal condor]

oh wait nvm i should be v(x,t)=u(x, t/i)

[void flame]

Uh could anyone help me with this (a)? I'm just so gigantically stuck, I know it is supposed to be zero by using the laplacian and the fourier transform property but I just cannot do the direct computation and see how it is equal to zero.

[buoyant pike]

Ok, how do you start with computing the fourier transform

[void flame]

like this

I feel like I'm doing something not quite right, but Idk what it is

[buoyant pike]

I think that you are missing something in the first integral

[void flame]

what

I included the phi at the end

in case thats what you mean

oh I missed the integration limits

there should be four

[buoyant pike]

Oh nevermind

I thought you were doing something else

[void flame]

do you have any suggestions

[buoyant pike]

Shouldn't you be taking a fourier transform and not integrating against a test function

[void flame]

i thought the equation means if we look at it as a distribution it is equal to zero

thats why I integrated against a test function

am i wrong

because i am certainly stuck lol

[buoyant pike]

No I don't think you should do that

[void flame]

oh

[buoyant pike]

Take the fourier transform of something

Maybe eq. 1

[void flame]

take the fourier transform of the fourier transform of u?

[buoyant pike]

Well

Fourier transform of abs(xi)^2*u hat(xi)

[dusty swan]

i believe $\hat{u}$ is notation for the fourier transform of $u$, so you have to take the fourier transform of just $u$

[untold delta]

rain

[dusty swan]

no need to take fourier transform of eq 1

[untold delta]

cocat

cocat

[void flame]

Yeah I think we are not allowed to do that here

The simpler way I mean

[lilac barn]

That’s why I provided the longer way first. The simpler way is for your own education

[void flame]

I acc got the exact same proof of you did but I was agonizing over it for hours…. Tysm though

I was agonizing over how I could multiply the two Dirac deltas while in reality I shouldn’t multiply them out in the first place, I should just separate them and treat it as two different transforms

I didn’t write it the way you wrote it, I wrote out the integral. That’s why I thought it would be the multiplication of two Dirac deltas

Well, the second derivative I mean

[lilac barn]

I hope the presentation I provided is clear

[void flame]

Yeah it is much cleaner and I see why now

Thank u so much ❤️

[lilac barn]

Yw!

[dusk juniper]

Does the pde in this channel stand for partial differential equations

[lilac barn]

No it stands for potentially dangerous equations

[dusk juniper]

Nice

[void flame]

I got a question that asks me to prove that the solution of this eqn is unique by energy method, so I supposed there are two solution u1, u2 and w=u1-u2, and then I did the calculations hoping to conclude something. But this is what I got and I cant really determine the sign. Did I do something wrong here?

[lilac barn]

You wanna get an energy inequality and Gronwall and use the fact that the difference is 0 when time is 0

[void flame]

oh I didnt know gronwall, I'll look it up, thanks!

[void flame]

Oh, actually I think we dont need to do that. For this, we denote E(t)= the integral of w^2 at t, and then the equation I got says E'(t)<= a constant times E(t). But E(0)=0 and E is nonnegative. Hence E must be 0 for all t>0

[quaint herald]

How is your "hence" following without Gronwall?

[void flame]

I thought its like a calculus argument

[quaint herald]

Well "Gronwall" is a calculus argument.

I just ask because this is a point at which I have seen a few erroneous arguments from students before. Afaik Gronwall (or equivalent) is the main way to draw such a conclusion.

[void flame]

E'(t) <=0 at t=0 but then E(0)=0 so it must decrease, but then E>=0 so it must stay zero. If we suppose there exists a point where E>0 then it must be E'>0 for some t but then that leads to a contradiction since we can "push" the t back to t=0 to reach a contradiction.

I know its not rigorous and I should write it out rigorously, but this is a 5-minute quiz question so I thought an answer like this should've been enough

[void flame]

Does this sound ok

[quaint herald]

The "pushing the t back to t=0" sounds fishy.
Here is a related question to build intuition. Suppose we instead had:
E(t) >= 0 for all t
E'(t) =< 2E(t)^(1/2)
E(0)=0.

Would these hypotheses also imply E is identically zero in [0,inf) ?

[void flame]

I think so yeah

[quaint herald]

E(t)=t^2

[buoyant pike]

Picard lindelof moment

[quaint herald]

Yep, you can think of this as related to the fact that local uniqueness of ODE sols breaks down when we are no longer Lipschitz.

[void flame]

wha

im confused

[quaint herald]

It's a very common and extremely natural trap to fall into, otherwise I wouldn't have said anything in the first place.

[shy narwhal]

E'(z) = 0 tells you that a linear approximation at z will look flat
it doesn’t say you can’t approximate your function by a quadratic

[quaint herald]

It is probably worthwhile to pretend you haven't seen my counterexample, and try to write down a proof of your original claim with full details/generality.

Likely you will understand the issue whilst doing so.

[void flame]

Sorry, do you mean I can just use this inequality? And I think I can take a to be zero, so it is bounded by zero for all t. Does that sound right?
I was not aware of this inequality, and I now see that I was being sloppy. But I dont have the time to review the details rn so I will have to go back to it later.

[quaint herald]

Yep, this inequality is how one correctly draws the conclusion that E is identically zero. My message was just about you clearing up your personal confusion about your original sketch involving "pushing t back to zero" not working, and leading to an incorrect conclusion in the modified situation I gave.

[void flame]

Does this inequality not work if u is replaced by u^2?

[quaint herald]

Idk exactly what you mean by this, but in the original energy question, if you had an estimate like E'(t) =< E(t)^2 you would still be okay. An example argument is as follows.

For any s at which E(s) vanishes, you get E(t) =< 1 for t near s by continuity, and so E'(t) =< E(t)^2 =< E(t) and the usual Gronwall then gives you vanishing in this small interval about s. If E were not identically zero you could carry this out at s = inf {t : E(t) nonzero} to get a contradiction.

@void flame

[void flame]

Ohhhhh

yeah my calculus skills are rusty, or I'm just too stressed to think properly lol

Thank you so much❤️

[quaint herald]

no worries, you'll be fine :). like I said, this is an extremely common trap.

hope I didn't add to your stress. for short quizzes such technical points probably won't be super relevant.

[void flame]

no your explanation was really good, I think the lecturer realized this question was too involved (nobody did it) and changed it to a homework question instead. So the details are much appreciated!

[lilac barn]

There is a way to make this sort of an intuition rigorous and it's called the bootstrap/continuity argument and it's pretty much equivalent to Gronwall

You can look into Tao's book on dispersive PDEs for more info on it

[void flame]

Oh, oki will make a note of it

[bold violet]

Is this an okay channel to ask perturbation theory questions? I'm confused about this problem in Bender and Orszag:

This is the statement of the problem, which is less than helpful:

[buoyant pike]

Ask in #modeling I will answer there

[arctic whale]

This is from Gilbarg-Trudinger. Is there any other reference for this result? I would like to see a different proof or maybe another exposition of the same proof.

[lilac barn]

Check Evans

[arctic whale]

Evens refers to G-T for a proof

[lilac barn]

Well he does give a proof in a simpler setting.

[arctic whale]

Yeah i wanted to see if there is another exposition of the proof in G-T. There are some points that are confusing me. I will just drop them here later.

[opal condor]

for this question in chapter 5 of evans

can i assume the sets here are open

[lilac barn]

Yes

[patent fjord]

any idea from which book this defintion is or can someone give me a similiar result in any book? its from here https://math.stackexchange.com/questions/612047/sobolev-spaces-on-boundaries

[tired hollow]

can anyone help me solve this?

i can show my work- my professor has an interesting way of doing it.

[buoyant pike]

I think this is more appropriate for #odes-and-pdes

[undone sable]

Assuming $\partial \Omega$ is a smooth manifold, you can just use the definition of Sobolev spaces on compact manifolds $M$ given in https://link.springer.com/book/10.1007/978-3-031-33859-5. For smaller smoothness $k, \alpha$ of the boundary, the same definition should work; you just have to be careful about the invariance of the Sobolev space under coordinate transforms of the given smoothness.

[untold delta]

L

[patent fjord]

Big ty

[river pumice]

Seeing these evans questions gives me mega trauma

[buoyant pike]

[pulsar forge]

<@&268886789983436800> ??

[buoyant pike]

[shy narwhal]

did someone post navier-stokes

was it that traumatizing

[buoyant pike]

[shy narwhal]

f

[buoyant pike]

[storm narwhal]

Does someone have an exact definition of what a breather solution is?

[pale dragon]

Does anyone know how to get all smooth solution of a constant coefficient PDE? Characteristic method applied to first order case. I don't know how to deal with higher order. I believe there is a general solution.

[sinful herald]

Can someone help me understand the last line, I do not understand when it says "is the domain of the part..."

[storm narwhal]

Does it not just mean the restriction of A^k in D_A?

[sinful herald]

ohh, I was confused with the wording they have used, so it says, D_A (\theta + k,p)

is domain of restriction of A^k in D_A?

[outer ruin]

Do you reckon this could be true

[outer ruin]

$\frac{\partial^2}{\partial \phi^2}$ is frechét derivative

[untold delta]

greenface98

[buoyant pike]

Have you checked that it is true for a few simple examples

[outer ruin]

Seems to have some issues

[buoyant pike]

Well then

It's also worth thinking about what you can let k depend on

[opal condor]

this is from evans appendix

ive never learnt lesbesgue'es diffeerentiaiotn theorem before but can someone tell me why its almost everywhere converges

and at which x is this not true?

[sand echo]

Functions in L^1 are only defined up to null sets. For example, the LHS of (i) is independent of the representative for f but the RHS is not

[opal condor]

so you're saying that f(x0) doesn't exist for x0 in a null set?

[rotund jetty]

If you change $f$ on a set of measure $0$, $\frac{1}{Vol(B(x_0, r)}\int_{B(x_0, r)}f,dx$ does not change, whereas $f(x_0)$ can be modified arbitrarily.

[untold delta]

L

[rotund jetty]

that's how you know that (i) can at best hold for a.e. x_0

[opal condor]

ok thanks i think i see it now

[opal condor]

i have a question related to this question of chapter 5 of 11

but if $u$ were differentiable everywhere but the $Du=0$ almost everywhere then is $u$ still a constant?

[untold delta]

lmao

[opal condor]

been trying to prove this for a long time but can't

[rotund jetty]

wlog $U$ is a ball. I think using an approximate identity might work.

[untold delta]

L

[verbal nebula]

Use a mollifier

[opal condor]

yeh tht works i think

[astral vine]

You can redefine A^k as an unbounded operator on D_A(theta,p) as a new ground space instead of X, then the domain of A^k defined on D_A(theta,p) is D_A(theta+k,p).

x in D_A(theta,p) (included in X) such that A^k belongs to D_A(theta,p).

More generally when it is well defined: For Y a Banach space which is a subspace that imbeds continuously in X, A an unbounded operator on X. The part of A in Y is the ""restriction-co-restriction"" of A to Y with domain { y in D(A)nY s.t. Ay belongs to Y}.

A condition to show it is well defined, is stability of Y by the resolvent of A

which trivially holds for operators that generates a bounded analytic semigroup, and Y to be the associated real interpoaltion space.

[outer ruin]

Without loss of generality assume one may assume every Banach space of interest is a Hilbert space

Also ‘normed ‘ is assumed for any Banach space

[astral vine]

actually I did precise normed because in my work I may use normed but non complete spaces, and consider the part of A in such kind of non-complete real interpolation spaces.

But you are right

[arctic whale]

I am trying this problem and want to check my idea. We pick and increasing sequence of annuli $A_i \subset A_{i+1} \subset \cdots$ Without loss of generality we can assume that u is bounded below and that $\inf u$ on the exterior domain is 0. Now we notice $\sup_{x,y \in A_i}|u(x) - u(y)| \leq \sup_{A_i}u - \inf_{A_{i}} u \leq (k_i-1) \inf_{A_i} u$. The inf over $A_i$ goes to zero and so we would be done if we can bound the sequence of constants $k_i$

[untold delta]

ru0xffian

[arctic whale]

I have a vague idea of how to bound the constants but want to make sure that so far I am not doing something stupid.

[outer ruin]

Say I had this functional on L_t on hilbert space H

Where epsilon is distributed according to isotopic Gaussian on R^d

Acting on maps phi from R^d to R^d

And say I had a C_2 curve of functions F_t

How might I show this

Where that operator is the frèchet laplacian

Should my strategy be to express F_t w.r.t so ON basis for my Hilbert space and say that laplacian is the sum over second the directional derivatives over all those basis functions?

And that maybe hope I can show that objective is diffetentibale in time without blowing up?

Frechet derivatives remain very exotic to me

It’s kinda hard to find any info on functional valued PDE’s

[buoyant pike]

The sum of differentials things is not necessarily differentials

Is L_t your original expectation?

[outer ruin]

Each value t produces a loss functional L_t which is that expectation over epsilon when plugging in that value of t

And I wanna show something about the curve F_t induced by the minimization of each of those functional. Take F_t to be C_2 smooth ( in time) curve satisfying F_t = armin L_t

It’s very hard to find theory on functional valued pde’s

Are they to nasty to even warrant consideration?

[buoyant pike]

What have you looked for?

There should be lots of books on theory for calculus of variations

[outer ruin]

Yeah that’s def where to look, ive just been looking at ‘functional pde’s’

Or variational pde’s

But overwhelmingly it’s people using variational methods to solve pde’s of finite spatial dimensions not people studying pde’s on functionals

[sinful herald]

I see, Now it is much more clear. Thanks a lot

[outer ruin]

Why didn’t any of you nerds tell me that the laplacian of a quadratic functional over a Hilbert space is infinite

I’ve been trying to say something about it this whole time and didn’t realize one of the terms in my equation wasn’t even finite

[untold delta]

bard

[ocean ether]

not sure where i should start on problem 4

problem 3 is just an ez pz cole-hopf transformation

[lilac barn]

What progress did you make with energy estimates

[astral vine]

For many reasons very deep, such kind of estimate cannot hold in general.

C will depend necessarily on \varepsilon and T.

[astral vine]

Moreover, you also need to control the time regularity of your boundary value.

This is an L^q-maximal regularity estimate for the initial Boundayr value problem.

Assuming $\varepsilon=1$ for simplicity, one requires $$\partial_x h(\cdot,\cdot,0)\in \dot{\mathrm{H}}^{1/4}t(\mathbb{R}+,\mathrm{L}^2_x(\mathbb{R}_+))\cap {\mathrm{L}}^{2}t(\mathbb{R}+,\dot{\mathrm{H}}^{{1}/{2}}x(\mathbb{R}+))$$

This is necessary and sufficient.

[untold delta]

Functionanatolysis

[astral vine]

If you want further insights on why such spaces, feel free.

[ocean ether]

where do i start

[unborn quiver]

First, recall the divergence term in the E-L equations, and, as per the hint, think about ways that you can get (close to) the left hand side of the equation as the divergence of something

Further hint: ||In 1D, ((e^x)u’(x))’= e^x u’(x) + e^x u”(x) = (e^x)(u’(x)+u”(x))||

(The further hint is not too much of a spoiler if you are still unsure)

[ocean ether]

aight cool

alli need to show is $e^{-\varphi}DuDv=ve^{-\varphi}(D\varphi Du-\Delta u)$ and gg?

[untold delta]

the breadwave

[ocean ether]

is there a general rule of thumb for which transformation i should try using whenever i see a PDE

[unborn quiver]

It’s not only that (bc you’ll need to get it in the right form), but the leading e^(-phi) term is right

[unborn quiver]

It’s not extensive, but usually the answer really is some type of exponential for linear problems. For parabolic problems, for example, something like e^(b•x+at) is usually enough to simplify the problem to a simpler form

[sinful herald]

In above $\mathcal{L}_D$ is the Dirichlet Problem $\Delta u = f$ with boundary data $g$. I was reading some paper where they mentioned this. As far, I can only show similar result in $\mathbb{R}^n$ with zero boundary data using Calderon Zygmund theory but am not able to prove it for bounded domain and boundary data, dose someone know about any reference for this.

[untold delta]

Math&Tea

[astral vine]

This is heavily non-trivial. This is not only Calderon-Zygmund theory, but also boundary layer theory, and use perturbation of invertible operator by a compact operator (or more generally by operator with Fredholm index 0)

The two main references for this are Jerison & Koenig(1995) and Fabes Mendez & Mitrea (1999). Where this is done in the more general framework of bounded Lipschitz domains for 2-e<p<2+e. The C¹ case with the full range of Lebesgue exponents 1<p<+∞ as a special case.

[ocean ether]

my gleeby deeby ass does not know complex analysis

[lilac barn]

This is the standard Laplace and inverse Laplace transform

[ocean ether]

i need to show that the inverse is indeed correct

[rain phoenix]

as in derive the formula?

[ocean ether]

no

just show that the integral does indeed become u(t)

[ebon shard]

Let $\overline{\mathbb{D}}$ denote the closed unit disc in $\mathbb{C}$ and $\varphi \in C^\infty(\overline{\mathbb{D}})$. Is there a simple way to show that a solution $\psi \in C^\infty(\overline{\mathbb{D}})$ exists for
$$ \frac{\partial \psi}{\partial \overline{z}} = \varphi, \quad \textrm{where} \quad \frac{\partial}{\partial \bar{z}} \coloneqq \frac{1}{2} \left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right)? $$

[untold delta]

Yilian

[ebon shard]

i don't really need to know the exact form of the solution, just the existence (i.e. i want to show that the operator $\frac{\partial}{\partial \overline{z}}$ is surjective as a map from $C^\infty(\overline{\mathbb{D}}) \to C^\infty(\overline{\mathbb{D}})$.

[untold delta]

Yilian

[ocean ether]

d

not sure where to start

[astral vine]

implicit derivation formula

[rain phoenix]

well, its stated on the left that your taking an inverse transform of a transform, effectively recovering the original function

or if u(t) is a known function then you can prove it pretty easily

[patent fjord]

$\partial_v w$ is conormal derivative of $w$.
Suppose that $\Omega$ is a Lipschitz domain and let $\Omega'$ be a compact subset of a $\Omega$.
Let $w:\Omega\rightarrow \mathbb{R}$ be a solution for a problem \begin{equation*}
L^w = \begin{cases}
h &\text{in } \Omega'\
0 &\text{in } \Omega\backslash \overline{\Omega'}
\end{cases},
\end{equation} $w = 0$ on $\partial\Omega$, where $L$ is an elliptic operator and $h \in H^1(\Omega')$ is a solution to a problem $Lh = 0$ in $\Omega'$.

Do we have an estimate $|\partial_v w|{H^{-1/2}(\partial\Omega')} \leq C|w|{H^1(\Omega)}$?

By the Green's identity we have
$\Phi(w,Eg) = (Eg, L^*w){L^2(\Omega)} - (g , \partial_v w){L^2(\partial\Omega)}$, where $E:H^{1/2}(\partial\Omega) \rightarrow H^1(\Omega)$ is an bounded extension operator. It follows that
$|(g , \partial_v w){L^2(\partial\Omega)}| \leq C|g|{H^{1/2}(\partial\Omega)}|h|{L^2(\Omega')} + C|g|{H^{1/2}(\partial\Omega)}|w|{H^1(\Omega)}$, dividing by $|g|{H^{1/2}(\partial\Omega)}$ by together taking the supremum yields us $|\partial_v w|{H^{-1/2}(\partial\Omega)}\leq C(|h|{L^2(\Omega')} + |w|{H^1(\Omega)})$. I think using the assumption that $L^*w=h$ might give us an estimate $|h|{L^2(\Omega')} \leq C|w|_{H^1(\Omega)}$, which then would imply the the inequality.