advanced-pdes

[astral vine]

Solve it

Inverse (with Fourier) it to find out an explicit rep. Formula (this is the hard part if you want to do it rigorously)

[tired hollow]

FYI the method you originally wanted hinged on Fredholm Alternative and thats held off until functional analysis (if u wanted to account for any bad behavior). I can’t even think of how to do those proofs from scratch (above person prob can tho)

[fringe charm]

Oh okay thanks! I’ve solved the 1-D wave function with the method I was tackling the 3-D case, but this is more complicated than the 1D one. I was trying to avoid Fourier transforms but it might be easier to tackle it that way so I’ll give it a try. Thanks for the help!

[astral vine]

You need to be aware about stuff like Fourier transform of Riesz potentials...

In the tempered distribution sense

(calculation for the wave eq are mainly obtained in Distribution books)

[fringe charm]

My knowledge on functional analysis is not that rigorous which is probably why I was struggling with this problem, but thank you!

[tired hollow]

Ur good idk shit about func (all this stuff are things I heard from grad and faculty)

[fringe charm]

I’ll be reading abt that thanks!

[fringe charm]

Lol still helpful

[fringe charm]

Btw I just thought of this so not sure if it works but since I have the solution to the 1D equation can I argue that it will be a non trivial solution to the 3D equation and normalize it. Then apply Fredholm’s Alternative theorem and boundary conditions to solve for the 3D green’s function? Like would that make sense

[fickle goblet]

So I clared this up and the solution is apparently:
$$-\varepsilon \Delta u+W'(u)=0$$

[untold delta]

Enoo58

Enoo58

[tired hollow]

Idk how u come up w this. I’ve been trying to find a reason to say no but I can’t find any u technically have initial conditions met with the 1D soln. Lmk if it goes anywhere

[buoyant pike]

[waxen forum]

@craggy pumice dont shitpost here

[little knot]

Can anyone give me a tip on how I can show this? I suppose Hölders inequality is involved, but I don’t have a real approach.

[buoyant pike]

How did you define the W1,p norm

Do you know what it means for norms to be equivalent

[little knot]

No I do not

[little knot]

Give me a sec I need to find the part

[little knot]

This is how me defined the W norm

[buoyant pike]

So two norms $\norm{\cdot}_A$ and $\norm{\cdot}_B$ are equivalent if $c\norm{x}_B\leq\norm{x}_A\leq C\norm{x}_B$ for all $x$ in whatever space you are working in

[untold delta]

梦境倒流

[astral vine]

use equivalent norm in finite dimension on vector (u(x), d1 u(x), d2 u(x), ..., dn u(x)) for a.e. x

then take the Lp norm in x

Oh Ange gave the idea

[little knot]

Thanks for the replies, I'm in a seminar rn but will look into later, looks promising.

[unborn quiver]

$g\in C^2_c(\mathbb{R}^n)$ and I need to show that
[v(x) = \int_0^\infty u(x,t)dt = \int_0^\infty \int_{\mathbb{R}^n} \frac{e^{|x-y|^2/4t}}{(4\pi t)^{n/2}} g(y)dydt]
is well defined. I know that on $(\epsilon,\infty)$, $\int_0^\infty u(x,t) <\infty$, but I'm not sure how much that helps me because the integral appears to blow up as $\epsilon\to 0$. I know assuming that it's well defined, $v(x)$ is a solution to the Poisson equation $\Delta v = g$ over $\mathbb{R}^n$

[untold delta]

kirby

[unborn quiver]

I know that u solves the heat equation, but I simplified it as such to make the line visible on TeXit

[buoyant pike]

How much have you discussed approximations to the identity

Wait nevermind

Do you have initial conditions

[unborn quiver]

For v? No, I just have that u(x,0) = g(x)

Additionally n >= 3, but that didn't really seem to help much

[buoyant pike]

Which integral are you saying blows up

[unborn quiver]

the integral in time

[buoyant pike]

u(x,t)<g(x) by the max principle for the heat equation

Wait that is not true

Lol

Anyways as eps to 0 then the heat kernel becomes an approximation to the identity

[unborn quiver]

Wait yeah, I'm following now, I was overthinking it

[buoyant pike]

So as t to 0 then u approaches g right

[unborn quiver]

Yes, and that's bounded, so as t goes to 0, you get that u is integrable in time

I think

[buoyant pike]

Something along those lines

[unborn quiver]

I spent way too long thinking about that, I got hung up on a bound for u I had that was definitely not helpful

but now my homework in this class is done, so time to start the next one

[astral vine]

Minkowski's Integral Inequality, Sobolev's Embeddings, and Young's Inequality for the Convolution

You should be able to prove it in a row

in order to prove it

prove v lies in Lp, for 1<p<n/2

n being the dimension

(If instead the goal is to prove Sobolev embeddings from it, then do a careful pointwise estimate of the heat kernel depending on region of t,x)

[hearty path]

How do I solve ${\pdv[2]{u}{t}}-{\pdv[2]{u}{x}}-{\pdv[2]{u}{y}}=f(t,x,y)$ if I know $u(0,x,y)={\pdv{u}{t}}(0,x,y)=0$?

[untold delta]

Jayx

[astral vine]

You also need to know the value of u(0,x,y), and what is the open set you consider ?

The whole Rn ?

[hearty path]

The value of u(0,x,y)=0 and the partial with respect to t is also 0

And we're considering t >= 0

But I think I got the answer

[astral vine]

I was thinking about x,y

where does it leave ?

iin a square ? a ball

Dirichlet or Neumann BC ?

[hearty path]

Oh we don't place any restriction on it

[astral vine]

On the whole space R² then

[hearty path]

So you consider $\eta_s$ a family of solutions to the homogeneous wave equation $\pdv[2]{t}-\pdv[2]{x}-\pdv[2]{y}=0$ such that $\eta_s(s,x,y)=0$ and ${\pdv{\eta_s}{t}}(s,x,y)=f(s,x,y)$, then $u(t,x,y)=\int_0^t\eta_s(t,x,y)\dd s$ should be the solution

[untold delta]

Jayx

[hearty path]

Specifically $a_s(t,x,y)=\eta_s(s+t,x,y)$ will be a solution to the 2D wave equation satisfying $a_s(0,x,y)=0$ and ${\pdv{a_s}{t}}(0,x,y)=f(s,x,y)$ so you can use Poisson’s Formula for the 2D wave equation for $a_s$ and find $\eta_s$

[untold delta]

Jayx

[hearty path]

And ${\pdv{u}{t}}=\eta_t(t,x,y)+\int_0^t{\pdv{\eta_s}{t}}(t,x,y)\dd s=\int_0^t{\pdv{\eta_s}{t}}(t,x,y)\dd s$ and ${\pdv[2]{u}{t}}={\pdv{\eta_t}{t}}(t,x,y)+\int_0^t{\pdv[2]{\eta_s}{t}}\dd s=f(t,x,y)+\int_0^t\left({\pdv[2]{x}}+{\pdv[2]{y}}\right)\eta_s\dd s=f(t,x,y)+\Delta u$

[untold delta]

Jayx

[hearty path]

@astral vine what do you think?

[astral vine]

I think this does not fit in advanced PDEs

but fits in #odes-and-pdes

[hearty path]

Oh alright

[viscid sparrow]

Can Galerkin method be used for general elliptic operators? I am having trouble to prove this generalization of Evans problem 4. Evans dealt with $L$ being the Laplacian. The good thing about this is that Laplacian would make the analysis easy by confirming there are positive real eigenvalues..

[untold delta]

Cookieman

[astral vine]

discrete eigen values, with simple eigen vectors should be enough (i.e. compact resolvent, with eigen subspaces of dimension 1).

What is L to be more precise ?

-div (A grad(.)) ?

[viscid sparrow]

It's the general one

[viscid sparrow]

I don't think this one ingeneral have simple eigenvalues, does it?

[astral vine]

I don't think so

[viscid sparrow]

yeah. Which is why I am stuck.. Following the proof of evans for parabolic equations, I'll need to find the coefficient of the series.. Implying I need to solve a system of ODEs.. But I'm not sure if it is solvable without these infos on eigenvalues..

[astral vine]

discrete eigenvalues should be ok

[viscid sparrow]

Is that also a standard ode theory like mentioned in evans?

[buoyant pike]

Have you seen picard lindelof/cauchy lipschitz

[viscid sparrow]

A long time ago..

[buoyant pike]

Well it's the standard existence and uniqueness theory for ODEs

[viscid sparrow]

Ok.. in my case we don’t really have the derivative in the front since it’s not time dependent. So I can’t really use the Picard lindelof theorem

[buoyant pike]

Why not

There is a time derivative on d_m^k

[viscid sparrow]

In my question, we are dealing with elliptic equations. There’s no time derivative in that case

[exotic void]

In this case eq19 read $e^{kl}d^l_m = f^k$ so you just invert that $e^{kl}$ matrix to solve $d^l_m$

[untold delta]

shiburin

[viscid sparrow]

And why is it invertible?

[untold delta]

shiburin

[exotic void]

I was careless. you want to solve Lu = f right?

[viscid sparrow]

This is the part I do not get.. $B[w_l, w_k]$ is positive definite why? The eigenvalues are not necessarily positive, right?

[untold delta]

Cookieman

[exotic void]

Yeah i was making assumptions forget about it

[viscid sparrow]

[exotic void]

Anyway since $Lu =f$ may not be solvable I'll assume $Lu=0$. In this case eq 19 is $\sum^m_{l=1} e^{kl} d^l_m = 0$ and $d^l_m$ is in the kernel of $[e^{kl}]$. We may force $\sum^m_{l=1} (d^l_m)^2=1$ (if the kernel is nontrivial) so $\Vert u_m\Vert_{L^2} = 1$ and the rests are from energy estimates of $\Vert u_m\Vert_{H^1_0}$.

[untold delta]

shiburin

[viscid sparrow]

Turns out the professor has a couple of typos.. The operator L in particular is the second only elliptic operator with only the divergence term..

So I think now the problem is solved..

[junior bloom]

Ay help me interpret some space physically? I've proven that some family of fields is compact in the space $L^p(\mathbb{R},B^{p,\infty}_{-\alpha}(\mathbb{R}^3))$ for some $\alpha<0$, $p\geq1$.

This works for what I need mathematically but I do not know what negative index Besov spaces mean for the space part of spacetime. There's the Paley-Littlewood characterization that I'm staring at but the negative index means I can't convert it into the sum of the functional norm and the Holder-type continuity part that gives me a physical meaning.

(The above means my spacetime field is Lp in time and Besov in space, with negative index.)

[untold delta]

teafortwo

[buoyant pike]

Which index is negative?

You have a negative alpha but alpha is negative

Unless one of those should be flipped

[junior bloom]

Oh alpha is positive, sorry.

I forgot I applied the negative to it

[buoyant pike]

[slender fulcrum]

does anyone here know any concrete applications of geometric measure theory in PDE?

[buoyant pike]

@astral vine

[junior bloom]

See "Young measures" and "k-currents/varifolds". Basically GMT lets us solve manifold dynamics in weak ways the way distribution theory lets us solve PDEs in weak ways.

[astral vine]

Thank you Ange, but tbh my knowledge there is too much elementary.

[buoyant pike]

[bitter yacht]

I am trying to show the following: Say $\Omega \subset \mathbb{R}^n$ open and connected. If $u \in W^{1,p}(\Omega)$ and all first order weak derivatives of $u$ are zero almost everywhere on $\Omega$, show that $u$ is constant almost everywhere.

Mollifying $u$ gives us the smooth function $u^\epsilon = \eta_\epsilon * u$.
Moreover we have $\partial^\alpha u^\epsilon = \eta_\epsilon * \partial^\alpha u$, where $\alpha$ is a multiindex. The derivative on the left-hand side is classical,
and on the right-hand side it is weak.

Since the weak derivative of $u$ is zero almost everywhere, we get that $\nabla u^\epsilon = 0$ everywhere on $\Omega_\epsilon$ (and the derivatives on $u^\epsilon$ are classic). But I do not know whether
$\Omega_\epsilon$ is connected, because then I could have concluded $u^\epsilon = 0$ on $\Omega_\epsilon$.

My goal somehow is to show that $u$ converges almost everywhere to zero, by first
showing $u^\epsilon = 0$. But I do not see how to progress beyond the obstacle highlighted above.

Certainly the domain $\Omega_\epsilon \to \Omega$ when $\epsilon \to 0$, but
this kind of set-theoretic convergence I am not exactly sure how to define and incorporate

[untold delta]

hardisc

[noble bloom]

Hello, I am trying to find Green’s Function for $\Omega = {x^2 + y^2 < r^2 : y > 0}$

[untold delta]

Hotwind

[meager dune]

For which operator / BCs etc

[buoyant pike]

Lol

So presumably you have some pde

Which pde

[noble bloom]

Well since I’m a noob at this, I’ll just ask this, should i consider this question as polar, then should I then attempt to solve this using electrostatic images

[buoyant pike]

And do you have dirichlet boundary conditions or neumann boundary conditions

[noble bloom]

I’m also gonna be honest, that is the whole question

[meager dune]

Oh okay so Poisson equation with vanishing at infinity probs

[buoyant pike]

Ok what is the context

Do you have point charges

[noble bloom]

Here’s the snippet out of my homework assignment

[meager dune]

Bruh

[noble bloom]

LOL

[buoyant pike]

Are 1 and 2 unrelated

[meager dune]

Nah I mean dw lol this is just physics being physics

ig

[noble bloom]

Yea 1. Is initial value with shock curves, 2. Is easy hyperbolic pde general solution

[buoyant pike]

?????????

potato can deal with this nonsense

[noble bloom]

Believe me, it’s nonsense to me too, but I digress

[empty terrace]

Is there a closed form solution for the Green's function of the 2-dimensional Laplacian with zero Neumann boundary conditions on the boundary of a rectangle? That is, I have something like this $\Delta u(x) = f(x), x \in \Omega$ and $\partial_n u(x) = 0, x \in \partial\Omega$ and I know that $f$ integrates to zero in $\Omega$ (so the pure Neumann problem is consistent).

[untold delta]

criver

[empty terrace]

I have seen only the $G(\vec{x},\vec{y}) = \frac{1}{2\pi}\log|\vec{x}-\vec{y}|$ for the 2D unbounded case.

[buoyant pike]

vec, not ecv

[untold delta]

criver

[empty terrace]

At least for the Dirichlet problem they claim there is no closed form solution here: https://www.sciencedirect.com/science/article/abs/pii/S0955799706000683

[empty terrace]

Would mirroring the problem do the trick actually?

i.e. have this rectangle repeat in a mirrored fashion

since zero Neumann boundaries are reflecting boundary conditions after all

[tranquil steppe]

I guess the problem is nonexistent when Omega is a ball?
If so, you can apply the reasoning to any ball B contained in Omega and conclude that is constant a.e. on each such ball. By connectedness, it implies the same for Omega.

[solid flint]

Hello, is there cases where an m-dissipative operator keeps generating a C_0-semigroup in a Hilbert space even after being perturbed by an unbounded operator?

[solid flint]

I should have said that, but I am not looking for trivial cases.

[solid flint]

@orchid reef I am mainly looking for cases when the perturbation alone doesn't generate a C_0 semigroup.

[astral vine]

The optimal stuff about perturbation theory for operators can be found in Kato.

[woven radish]

The problem asks to find the Natural Boundary Conditions of the following functional and admissible set:

$$J(y)=\int_0^1 (yy'' + xy')dx$$
$$A={y:y\in C^4[0,1], y(0)=1}$$

I then obtained the following NBCs:
$$y(1)=0, y'(1)=0, y(0)=0, y'(0)=0$$

Does this mean there are no extrema in A, since the NBC requires $y(0)=0$, whereas A only contains $y$ s.t. $y(0)=1$? Or did I do something wrong computing my NBCs?

[untold delta]

JacobHofer

[viscid sparrow]

I have shown everything except from the last question.. I am kind of stuck on the last one. Using the hint, I found the relation between $\frac{d}{dt} \int u^2 = \int_{\Omega} u^4 ,dx - 4E(u)$.

[untold delta]

Cookieman

[viscid sparrow]

How is this hint helpful? I do not see how it helps

[viscid sparrow]

Why is $E$ negative?

[untold delta]

Cookieman

[viscid sparrow]

I know the derivative of $E$ is negative, since it is dissipating

[untold delta]

Cookieman

[viscid sparrow]

Oh well, at the initial data

yeah

But again I don't see why that is helpful?

[frigid pilot]

I’m not 100% sure how to solve this problem, but one common technique for using these types of identities to prove blowup is to argue that some quantity you know is positive becomes negative at some later time

for example if you show that the second time derivative of a positive quantity is bounded above by a negative constant, then that quantity is bounded by a quadratic function of time which becomes negative at some point

typically conservation laws can be helpful here to ensure your upper bound is constant in time

[viscid sparrow]

I see. Let me try this idea out.

Thanks!

[quaint herald]

Let $G(t)=\int_\Omega |u|^2, dx$. It suffices to show $G$ cannot be a smooth function on $(0,\infty)$, which we do by showing $G(t)$ is unbounded on $(0,T)$ for some $T<\infty$.

We have $G'(t)=-4E[u]+\int |u|^4\geq A+BG(t)^2$ for positive constants $A,B$ from the energy condition on $u_0$ and Cauchy-Schwarz (I won't bother being explicit with these constants). At this point you can either cite a suitable ODE result to deduce finite-time blowup of $G$ or you can manually prove it via elementary methods as follows.

Since $G'\geq A > 0$, $G$ is clearly increasing and so invertible. Then for large $M$ we have

$$G^{-1}(M)=\int_{0}^{G^{-1}(M)} , dt=\int_{G(0)}^M \frac{du}{G'(G^{-1}(u))}\leq \int_{G(0)}^M \frac{du}{A+Bu^2}.$$

Taking $M\to \infty$ we see that $G$ is unbounded on $(0,T)$ where $T=\int_{G(0)}^\infty \frac{du}{A+Bu^2}$.

[untold delta]

gomez

[viscid sparrow]

Thanks for the help! This is probably trivial but why does it suffice to show G is not a smooth function on (0,infty)?

[quaint herald]

If u is a smooth sol on the domain Omega x (0,inf), (which is what you are proving impossible), then G is smooth on (0,inf).

Upper limit of integration should be inf in defn of T btw. Too late to edit.

[viscid sparrow]

Also I’m not sure about how we arrived at the last statement where G is unbounded.. How does that follow from the bound for G inverse?

[viscid sparrow]

Sorry for the multiple questions..

[quaint herald]

all good

[viscid sparrow]

Also I’m not sure about how we arrived at the last statement where G is unbounded.. Does it follow from the bound for $G^{-1}(M)$?

[untold delta]

Cookieman

[quaint herald]

Sorry, I didn't see you actually asked this additional question lol. Well G assumes the value M at G^{-1}(M). We have shown that for any large M > 0, we have G^{-1}(M) < T, so in other words G assumes any arbitrarily large value somewhere in the fixed interval (0,T).

[frigid pilot]

an alternative way to prove blowup from the riccati ODE inequality G’ > A + BG^2 > BG^2 is to just solve the ODE G’ > BG^2 by integrating both sides, you should get G > 1/(C-Bt) for some C > 1/G(0) (or alternatively 1/G < C-Bt which means the solution can’t exist for all time as G > 0)

[viscid sparrow]

Ohh right that makes so much sense. Thank you! And thanks @frigid pilot for the idea too!

[stable field]

Can someone help me here? I'm trying to solve this particular PDE and I understand how to solve using superposition but I'm having trouble understanding how they find sinh to be a part of the solutions

[buoyant pike]

This should go in #odes-and-pdes I believe

[stable field]

oh ok sorry I didn't realize

[buoyant pike]

No worries

[tired hollow]

seems correct, the NBCs you have mean that there will be no extrema in A, since the NBC requires $y(0)=0$ and A only contains $y$ s.t. $y(0)=1$

[edgy creek]

how can I prove this formula?

[frigid pilot]

is the expression on the right the inner product of e_i with Ae_i ? (e_i being an orthonormal basis?)

[edgy creek]

yep exactly

[river path]

I recommend you write everything out on the right hand side, so that you have 3 summation signs

see if you can swap the order of summation in a useful way

[edgy creek]

I thought about that but it felt like that's too complicated, like there's probably a simpler way

but I'll try that

[river path]

there is, if you know a fact about the trace (which says that tr(AB) = tr(BA) for any matrices A and B)

[edgy creek]

yeah I proved that a few moments ago

[river path]

oh ok cool

then the right hand side is the trace of some operator. what operator is it?

[edgy creek]

AI?

[river path]

what's I?

[edgy creek]

the identity matrix

/operator

[river path]

it's true that the right hand side is the trace of AI (which is just the trace of A) but that's what we're trying to prove so we can't use that

there's a nice operator you can write down in terms of A and (e_i) which directly gives you the thing on the right as its trace

you could either try to find that operator and then use the fact you proved for a one line argument, or you could write out the sums

[edgy creek]

I think we can rewrite
< i | A | i > as
A | i > < i |
because of the cyclic property tr(AB) = tr(BA)

no wait that would result in an operator not a scalar

[frigid pilot]

If A is a matrix written with respect to the orthonormal basis, think about what vector Ae_i is (try it for e_i = [0 … 1 … 0]^T for example). Then think about what scalar you get when you take the dot product of a vector with a basis element

[edgy creek]

Ae_i would be the ith column of our matrix

[frigid pilot]

exactly

[edgy creek]

and dotting them gives you the diagonal elements

[frigid pilot]

yup

[edgy creek]

nice thanks

I'm still thinking about the other proof, the one utilizing tr(AB) = tr(BA)

[frigid pilot]

hmm idk if I really understand that one

[forest cradle]

if you're familiar with bra-ket notation, then you can change basis in trace by using
\begin{align*}
\sum_{j}\langle \psi_j|A|\psi_j\rangle &= \sum_{j,k}\langle \psi_j|A|\varphi_k\rangle\langle\varphi_k|\psi_j\rangle \
&=\sum_{j,k}\langle \varphi_k|\psi_j\rangle\langle\psi_j|A|\varphi_k\rangle\
&=\sum_k\langle \varphi_k|A|\varphi_k\rangle
\end{align*}

[untold delta]

washingbear 🌊🐻

[forest cradle]

where psi_j, psi_k are orthonormal bases

it is kind of similar to the tr(AB)=tr(BA) proof though doesn't use it directly @edgy creek

then just take the standard basis as the varphi_k

[edgy creek]

yeah I'm familiar and I proved that before, but that still didn't give me the A_ii ~ < i | A | i > correspondence

also I just noticed that this is advanced pdes, not advanced analysis

[forest cradle]

what exactly is |i>? the ith standard basis element?

or just any orthonormal basis element?

in the former they are exactly equal, in the latter they are not

[buoyant pike]

Shouldn’t this all be in #linear-algebra

[edgy creek]

how so? shouldn't they be equal anyway because similar matrices have the same trace?

[edgy creek]

it should, I thought about advanced-analysis cuz bra-ket notation but yeah now that I think about it linear-algebra would probably be better

[forest cradle]

I thought it was trace class operators

[unborn quiver]

I'm doing a question on the Dirac operator, and for the life of me, I can't figure out the calculations for these parts of the problem (second picture) because nothing seems to cancel out. The definitions are in the first picture.

I presume $\partial_t \rho^{\pm}$ is just $2|u^{\pm}|u_t^{\pm}$
For $\operatorname{div}(J^{\pm})$, I have reduced it down to
[\sum_{j=1}^3 (u^{\pm})^H_j\gamma_j^H\gamma_0 u + (u^{\pm})^H\gamma_j^H\gamma_0 u_j]
I know these calcs are elementary, but it's like weird to work through all of these

[untold delta]

a KuwubY (Kirby)

[unborn quiver]

nvm got it

[viscid sparrow]

I am tackling this problem

Question 6 (the contunuation of previous problem)

I was able to prove all of these other results:

However, I am completely stuck on 6..

Like how do I even get started?

How to get started on finding the supersolution...? Taking the infimum of a family of supersolutions somehow?

[rotund jetty]

https://en.wikipedia.org/wiki/Wave_equation what is the equation for this pulse here

the pulse on the string

[buoyant pike]

https://commons.wikimedia.org/wiki/File:Wave_equation_1D_fixed_endpoints.gif

The details are described here

[rotund jetty]

wow

[wooden sundial]

Epic wave!

[rotund jetty]

I don't think it solves the wave equation

[frigid pilot]

the code you see is a numerical solution

[rotund jetty]

@orchid reef Yeah I am trying to find a solution that starts as a bump at $x_0 \in (0, 1)$, goes to $1$, bounces back, and arrives at $0$. But I don't know why I should expect some solution like this to exist, or how to find it.

[untold delta]

IlIIllIIIlllIIIIllllIIIIIlllll

[rotund jetty]

@orchid reef I know we can write any solution as a Fourier series, using basis functions $\sin(n\pi x)$, but how do I create a pulse like that

[untold delta]

IlIIllIIIlllIIIIllllIIIIIlllll

[buoyant pike]

#odes-and-pdes

[rotund jetty]

Does anyone know an English reference for theorem 2.3.1

[frigid pilot]

you could check if some version of that result is here https://web.stanford.edu/~jluk/NWnotes.pdf

[rotund jetty]

I'm confused how the trace is even defined here

$u$ is only $L^2$, so the trace isn't naturaly defined for it

[untold delta]

IlIIllIIIlllIIIIllllIIIIIlllll

[rotund jetty]

apprently there is an estimate that applies specificly to solutions of the wave quation

[rotund jetty]

it's an interesting "hidden regularity"

[rotund jetty]

How is the formula for $\lambda_n$ correct?

[untold delta]

IlIIllIIIlllIIIIllllIIIIIlllll

[rotund jetty]

I think it should be $\lambda_n = \sqrt{-n^2\pi^2 - \alpha}$

[untold delta]

IlIIllIIIlllIIIIllllIIIIIlllll

[rotund jetty]

@orchid reef https://cel.hal.science/cel-00392196/document

@orchid reef I guess by heat equation you mean to rewrite the wave equation in the form $\Phi' = A\Phi$?

[untold delta]

IlIIllIIIlllIIIIllllIIIIIlllll

[potent ice]

.

[buoyant pike]

?

[slender delta]

nonlinear pdes?

[buoyant pike]

Those are linear

[slender delta]

oh

i want to learn more about diffeq now

[kind wren]

Hi. I think this might be the appropriated place to ask about Sobolev spaces.. Right? I'm looking for results of the kind... if $u \in H^1((0,1)^2)$, then for almost all $y \in (0,1)$, $u(\cdot, y) \in H^1(0,1)$. Are there results like that? What should I be looking for?

[untold delta]

phao

[kind wren]

This seems like it'd be something on traces.

Btw... idk if that is indeed true.

[astral vine]

This is not true

When you fix one variable in a Lp-based Sobolev on a (n-dim.)cube, a (n-dim.)Torus, or the (n-dim.)whole space then you lose exactly 1/p derivative

so u(.,y) belongs to H^{1/2}((0,1))

for almost all y in (0,1)

This follows from the standard H^s trace theorem on the whole space R^n with a translation argument. To get back on cubes, use and extension-restriction argument

[astral vine]

This first step yields an embedding

[untold delta]

Functionanatolysis

[astral vine]

Whenever s>1/2

The extension-restriction argument makes it clear I guess

[kind wren]

Ok, but this isn't exactly what I'm looking for, I think.

[astral vine]

It gives you this

n=2, s=1

[kind wren]

Ah ok.

Isn't there some kind of special case/exception for when n=2?

[astral vine]

I don't think so

What do you really want to prove ?

[kind wren]

I'm trying to understand somethings in a paper.

[astral vine]

What's the paper ?

[kind wren]

Wait a moment...

Supraconvergence and Supercloseness of a Scheme for Elliptic Equations on Nonuniform Grids
by Ferreira andGrigorieff

https://www.researchgate.net/publication/243047159_Supraconvergence_and_Supercloseness_of_a_Scheme_for_Elliptic_Equations_on_Nonuniform_Grids

[astral vine]

which part ?

[kind wren]

First paragraph of (pdf) page 12

it's assumed u is in H^2(Omega) (you can assume Omga is an xy axis-aligned rectangle in R²)

and in the end of the first paragraph of that page, it says

$(x_j, x_{j+1}) \times I_\ell \subset \Omega$

[untold delta]

phao

[astral vine]

I see

[kind wren]

I mean... it could be wrong, of course.

I don't know enough about Sobolev spaces to tell, tbh.

[astral vine]

Maybe the argument is just a Fubini argument ?

[kind wren]

Ok. It's something to be tried. I haven't tried to justify it like that.

[astral vine]

like u is H² of [a,b]x[c,d], then in particular u, d_x u and d_x² u are in L²([a,b]x[c,d]).

but this implies by Fubini Tonelli that u(.,y), d_x u(.,y) and d_x² u(.,y) are in L²([a,b]) for almost every y

other wise ther full H² norm on the rectangle would blow up

(by contradiction)

[kind wren]

Right... The issue would be the "integration by parts" equality

[astral vine]

Maybe a tensor product argument for the test fonction

[kind wren]

I'm sorry. Idk what that is.

[astral vine]

write integration by parts on product of a function of x and a fucntion of y

for which each one is a test function

while the x one stays arbitrary

replace the y-one by a mollifying sequence ?

[kind wren]

I'm sorry. I don't follow.

[astral vine]

I'm probably not clear

[kind wren]

@astral vine do you know any books that talk about these things?

[astral vine]

Not really I just tried to give a proof on the fly

or at least few elements

maybe @buoyant pike or @river path , if they are available, could help ?

[kind wren]

I see. THank you.

I'll check out some traces theorems and try also the Fubini/Tonelli argument

About the tensor product argument you've mentioned. Do you know of any place in which it appears so that I can look at an example of it being used?

(A book if possible)
Any one you'd know? @astral vine

Sometimes, we only see these things in class or something I guess.. hehe

[astral vine]

I have no book in mind sorry

[kind wren]

Ah I see. No problem.

[kind wren]

@astral vine Hi

There is this Theorem on Evans' Measure Theory book.

Could (i) be it?

[astral vine]

yes probably this would fit

[nocturne wave]

what is perquisite knowledge of studying advance PDE?

[kind wren]

There are many... Measure theory, functional analysis, good grasp of n-dim euclidean space analysis

it also depends on what kind of PDE I guess.

maybe ask a professor at your university

I think asking a professor there is a good idea because a career possibility for you, studying PDE, would be to follow something a local PDE researcher is doing. You'd go under his/her advisorship and get going into his/her thread within the whole PDE subject.

[kind wren]

Hi... sort of random question...

the whole sobolev spaces thing is used to study (that I know, of course) linear elliptical pde's

Is it also used to study non-linear evolution (e.g. difusion, reaction, advec, wave, etc) equations as well?

[buoyant pike]

Yes

[quaint herald]

absolutely, and even in the linear world sobolev spaces are used everywhere, not just the elliptic setting.

[tired hollow]

They are particularly very important in hyperbolic PDEs since they usually dont have maximum principles

[frigid pilot]

yes, for some nonlinear evolution equations there are criteria which more or less tell you that if a certain sobolev norm is a priori uniformly bounded in time, the solution exists for all time (and if it blows up, the solution does not exist globally). this is often how global existence/blowup results for dispersive equations like wave and schrödinger are proved

[astral vine]

There are similar criterion for NSE equations : the Serrin-Ladyzhenskaya criteria

If mixed some mixed Lp(Lq), for one (or equivalently all) couple (p,q) satisfying some scaling invariant conditions, is uniformly bounded then there exists a global solution to NSE

There are similar criterions fro Sobolev/Besov spaces

[tired hollow]

Weak solution

[astral vine]

ah yes my bad

Thank you

[tired hollow]

im only in high school

but this is a insane class

im usin heat equation to model chemotaxis flux

with crank nicholson solution

and i dont get anything lmao

[frigid pilot]

there could be a lot of places for mistakes - make sure you implemented the boundary conditions correctly, make sure you’re correctly coding the linear system and solving it, etc

i forget if there’s a stability condition for crank nicholson (there probably is) but also make sure your spatial discretization and timestep obey the stability condition

[tired hollow]

Crank Nicolson is an implicit scheme, it is unconditionally stable

But how we implement boundary conditions can sometimes destroy the stability, which is expected since von neumann analysis did not take into account boundary conditions treatment

Unfortunately implementing boundary conditions is not simple, in fact this was my PhD

Even before talking about numerical implementations, we have to be aware that not all boundary conditions lead to a well-posed problem

By well-posed, I mean well-posed in the Hadamard sense

[slender delta]

where can i read a proof of method of characteristics working?

[fresh anchor]

Any good resources for solving PDEs given a set of boundary conditions through finite difference numerical analysis?

[slender delta]

bump

[cobalt drift]

Method of characteristic in how many dimensions

[cobalt drift]

I can give u my lecturer's notes if u need- https://www.damtp.cam.ac.uk/user/examples/B8Ld.pdf

chapter 9 is characteristics

[slender delta]

sorry i meant in general but this is good thanks

i dont know any textbooks that approach pdes rigorously

or ig not handbook books

[high mulch]

What are handbook books

[bronze gate]

there are lots of rigorous books (eg. evans)

[slender delta]

books that usually go through methods of solving pdes without being super rigorous but give you intuitions for how the solutions are derived

[high mulch]

Would you say those are different than survey books (a bunch of tools’ and techniques’ applications without needing exact proofs)

[slender delta]

ig not really but the purpose of handbooks afaik is just good introduction to thr solutions/references

or maybe they are i didnt really make the term

[forest cradle]

besides evans there's also Brezis, Gilbarg Trudinger, Taylor, Fritz John that are all fairly common

[nocturne wave]

why (ii) imply (i)?

[simple bridge]

Suppose the maximum occurs on the interior. Then u is constant and in particular the limit as you go to the boundary will also be that constant since u is continuous on the closure of U.

Otherwise, the maximum occurs on the boundary [it needs to occur somewhere again because u is C(\bar(U))]

[lilac barn]

The above is the essential step however there are few more considerations to be made before applied in i since ii requires U to be connected which we aren't given. So to adapt, start by taking the connected components Ui of U. Since u is harmonic and thus continuous on compact Ubar, we have that the maximum is achieved on Ubar. If it on the boundary then we are done. So suppose it in U and hence in Ui for some i. Then by ii, restriction of u is constant on Ui, implying that the maximum occurs on boundary of Ui. Since for a locally connected space, the boundary of connected components are contained in the boundary of the set, the maximum occurs on the boundary of original set.

[nocturne wave]

why the corrector function phi^x disappear?

[river path]

Smoothness means that after epsilon is taken to 0, that term should disappear

So they're just dropping it early

This happens cause the ball gets smaller and smaller

Meanwhile phi^x w is a bounded function in some B(x, r). So the integral is bounded by area of ball * bound on phi^x w

That doesn't happen for Phi since Phi probably has a singularity at z = x

[river path]

Oh, that makes sense

Thanks

[astral vine]

For a linear operator, eigen vectors belongs to its infinity iterated domain

i.e.

what is C^2_infty ?

Regularity of elliptic operators is really hard question

on R^n, there is no eigen value for the Laplacian

the spectrum of the Laplacian is purely continuous

Describing exact domain of an elleiptic operator is hard in general

see the Girsvard's book for instance

for really smooth domains, people uses pseudodifferential operators

they localise the operators via smooth partition of unity and charts

Then pullback everything on R^n

show that the operator preserves regualrity and so on

For rough domains

everything falls in the scope of Harmonic Analysis

If your operator is an actual Laplacian, then your eigen functions are fully smooth in the interior

But you can't say more

Usually yes via sneiberg arguments on the interpolation scale

I have similar problems to characterize some boundedness of operators in some unbounded Lipschitz domains

special ones

that are a pure Lipschitz graph

You cannot do that much yes

What's your actual problem ?

Maybe I have some references that could help you ?

lol this looks like one of my friends problem

he wants to the same things with polylaplacian instead

So you want to look at

$$ \lambda u-\Delta_\mathcal{N} u + (-\Delta_\mathcal{N})^{\frac{s}{2}}u = f $$

?

[untold delta]

Functionanatolysis

regular boy

[astral vine]

this sounds like hell

issues with proper deifnition with this kind of stuff

I was exactly thinking about the first one

yes the regularity of the full power will dominate

Ok so I get thisq

If you use like the Fractional Laplacian on rn with vanishing conditions outside the domain

you want to know

when will its fractional power will dominate the actual regularity ?

is that it ?

the one you prefer

probably yes

Did you review a wide part of Abatangelo's work ?

I'm far from being a specialist of fractional (whole space) Laplacian restricted on domains

I'm more a functional calculus person

For actual regularity of Neumann on lipschitz case

and other rough domains

check out Mitrea family's work

like Fabes Mendez Mitrea

Grisvard's book

I absolutely don't know

In some sense yes

I'm focused on bent half spaces

with possibly irregular boudnaries

in the scope of applying it to fluid dynamics

Haha, I'm stuck on the problem for a year and a half now

But thank you very much

Good luck

[plain yoke]

in the last minute of 3b1b video on raising e to the power of a matrix he talks about raising e to the power of the derivative operator, but doesnt explain it. Can someone guide me where to learn about this? or explain it to me

no idea what youre talking about

go easy on me new to the stuff

[untold delta]

regular boy

[plain yoke]

I dont think ill understand, i am asking for an introduction on raising exponentials to the power of derivatives

maybe its in my textbook further down the line, ill look

[untold delta]

regular boy

[plain yoke]

and the proof/intuition for this?

so e^D is a shifting map?

[untold delta]

regular boy

[plain yoke]

Ah

got it

thank you

[astral vine]

the intuition if f is smooth

is to write roughly the exponential

then you will end up with the Taylor series of t mapsto f(t+s) near t=0

$$e^{t\partial_t}f(s) = \sum_{k=0}^{+\infty} \frac{t^k}{k!} (\partial_t)^k f = \sum_{k=0}^{+\infty}\frac{(t-0)^k}{k!}f^{(k)}(0+s) = f(t+s) $$

this is purelly moral

but this gives good insights in my opinion

[untold delta]

Functionanatolysis

[nocturne wave]

how to compute that intergal? E(1)=E(0,0:1)

[nocturne wave]

fundamental solution of heat equation

[nocturne wave]

what is (2.) doing ? can someone draw me a picture?

[rocky mountain]

what is U_T?

anyway, this seems to be an assertion and proof of the fact that any two points in a connected open subset U of R^n can be joined by a path obtained by concatenating finitely many line segments, with each contained in U.

[nocturne wave]

what is the idea of (29) last inequality?

[lilac barn]

Are you asking how did they get 29 from the previous?

[nocturne wave]

yes

[lilac barn]

Just replace the given substitution in the previous inequality.

[nocturne wave]

i cannot see how

[lilac barn]

Replace (T + e) with 1/4(a + gamma)

[nocturne wave]

but why<=sup g

[lilac barn]

What's Gamma?

[lilac barn]

On page 441 of Evans, he shows how to solve this system using semigroup theory and the idea is that we consider the semigroup S_t: L2(U)->L2(U) generated by -L so that we can solve via v(t) = S_tg.
This solves the first and third boundary conditions, why does it solve the second boundary condition?

oh nvm, he accounted for second one by setting the solution space to be H1_0(U)

[astral vine]

He did not "set the solution to be in H^1_0"

the operator L is built so that has its domain included in H^1_0, and the range of the semigroup lies in the domain of L for all t>0, therfore in H^1_0

the boundary condition is a part of the operator

which is really not the same, although we end up with a similar conclusion

[lilac barn]

Sorry, I should have been precise. I meant that he considered A := -L and reduced its domain to be inside H^1_0(U). This ensures that the semigroup generated S(t) lands inside H^1_0(U) provided the input is in D(A). So if g satisfies the required conditions, we have that S_tg satisfies the required boundary conditions, first and third by being a A-generated semigroup and second via its image being inside H^1_0(U).

[astral vine]

neither : "I meant that he considered A := -L and reduced its domain to be inside H^1_0(U)."

even if I understand what you meant again

[lilac barn]

That's what he does in the next line.

[astral vine]

give me the page ?

[lilac barn]

442

Yeah, my guess is that g is taken to be inside D(A) = H1_0(U) cap H2(U)

[astral vine]

In fact the true meaning of L is in the weak sense, by sesquilinear forms,

if you put domain H²(U) on L without H^1_0's intersection

-L does not longer generate a semigroup

We truly should write L_D instead of just L

since L is subordinated with a Dirichlet bc

You may also think about L_N, the same operator with adapted Neumann BC

[untold delta]

Functionanatolysis

[astral vine]

But for general set of BCs say J, it is not clear (and false under a lot of circumstances) that -L_J generates a semigroup

This is kind of stuff in Evans is part of the reasons why I don't like that much this book

[lilac barn]

I hope Haim Brezis writes a PDE bible soon

[astral vine]

For L² or Lp semigroups for most part of Elliptic operators

Ouhabaz's book

Analysis of the Heat Equation on domains

a must have

[frosty viper]

hey guys, im stuck with the application of laplace equations and harmonic functions. Is anyone avaliable to help?

its for part c and d of this question, been stuck on it for almost an hour now lol

would I use seperation of variables and the fact that Z(0) = Z(c) = 0

And that Y(0) = Y(c) = 0

?

[frosty viper]

???

[plain yoke]

how to solve this

i wrote it in matrix form

[bronze gate]

#odes-and-pdes

[exotic void]

Can we have C^{1,a} regularity of solution Lu=0 where ellptic operator L in divergence form with C^0 coeffs? (For context I only find C^a of u in Chap 8 of Gilberg Trudinger)

[exotic void]

Let's say the domain is just a ball centered at the origin and the boundary data is C^0 (btw I just want C^{1,a} interior estimate of the solution)

[exotic void]

Is there any ref or (counter)example?

[astral vine]

this false for uniformly elliptic Lipschitz Matrix

lol I had way too much complicated things in mind

[fickle goblet]

How can i proof that if morreys inequality holds so:
$$|u| _ {C^{0,\gamma}(\mathbb{R}^d)}\leq C| |\nabla u| |{L^p}$$
then:
$$p > d, \gamma = 1- \frac{d}{p}$$
I have as a hint that i should look at scaled functions $u\lambda(x):=u(\lambda x)$

[untold delta]

Enoo58

[fickle goblet]

so i did that but how does that tell me anything about the coefficients

I am new to hölder and sobolev spaces and didnt really get it i think

Oh i think i got it

[astral vine]

the dilation argument will give you the good gamma

Now to prove the estimate choose u smooth

and check than you can write

[untold delta]

Functionanatolysis

[astral vine]

then bound the whole stuff

[harsh veldt]

Hi. I want to rewrite a non-conservative form of a PDE into a conservative form. How do I know when I reach the conservative form?

[tired hollow]

If the PDE is derived from physical laws then it should be straightforward to write it in a conservative form by looking at the derivation

[harsh veldt]

Gotcha! I read in a thread that usually, if not always, PDE in conservative form would look like u + div (f(u)) = F. Is this correct? My problem is in my case F has a biharmonic operator so I’m not sure. I can include it in the divergence on the left side but idk if it’s necessary

[untold delta]

P(∂)E= 𝛿 (legal laplacian)

[buoyant pike]

u_t presumably

[native moat]

\begin{equation}
u_{t}-\frac{1}{t}u_{xx}=2\cos(x),0<x<\frac{\pi}{2},t>1
\end{equation}
\begin{equation}
u(x,1)=cos(3x),u_{x}(0,t)=u(\frac{\pi}{2},t)=0
\end{equation}

[untold delta]

alexix21

[native moat]

What method can be used to solve this problem? I couldn't solve it using the method of separation of variables.

[blazing wasp]

When you take a diffeomorphism of some system x(t) we just apply phi(x(t)) right?

and the other system they refer to is d phi(x)/dt

which is M^(-1)f(x) based on the givens

but somehow M^(-1)f(x) = g(x) is 0 for x=0

idek

[bright idol]

Use the finite difference method

\begin{equation}
u_{n+1} - \frac{2}{h^2}u_n + u_{n-1} = 2\cos(x_n)
\end{equation}
where $h$ is the step size and $x_n$ is the corresponding point in space

[untold delta]

Nats

[native moat]

ty @bright idol

[fickle cipher]

I know this is a PDE channel, but can I ask an ODE here? I just want to know how to prove or give a counterexample of that given a homogeneous linear ODE, a_n(x)y^(n)+…+a_0(x)y=0, a_n doesn’t equal constant 0, dimension of linear space of solutions is n.

[fickle cipher]

Nvm found reference

[tired hollow]

f(u) can contain derivatives of u

for example, the diffusion equation can be written as u_t + div(-grad(u)) = 0

[harsh veldt]

Yes! I was able to figure it out. The funny thing is mine actually even have a curl! I got rid of the biharmonic though. using other derivation so I don’t have to deal with higher order derivatives

[empty terrace]

I remember reading in some book a formulation with a determinant based on boundary conditions of a PDE, where if the determinant was non-zero, the PDE had a unique solution (i.e. the number of boundary conditions were sufficiently many). Does anyone know what book I am talking about?

[brave spindle]

Does anyone have source for bounding the solutions of PDEs? Particularly interested in Poisson's

[buoyant pike]

Have you seen the maximum principle

[astral vine]

in which sense ?

L-infty bound ?

L-p bound ?

L² bound ?

H^s bound ?

[brave spindle]

Forgot to specify but L-infty

[astral vine]

then yes maximum principle

Ange is right

[brave spindle]

thanks I've seen those before but not too useful for what I'm looking at

essentially have a class of quantum algorithms to solve PDEs, having a Linf bound for the solution even if very loose would be useful

e,g more stuff like this

https://math.stackexchange.com/questions/4121200/upper-bound-of-u-l-infty-of-solution-to-poisson-bvp

interested in any PDEs really

[chrome trout]

I am newbie in PDE, can anyone provide a friendly introduction to the subject? I want to study the laplace operator
I've read a little bit of Evans but it was too fast

[buoyant pike]

Which bit of Evans did you read

Chapter 2?

[chrome trout]

Not much, I started at section 2.2 and barely finished
I think I will give it another try tho

[buoyant pike]

I do think that Evans has a style which can make it hard to follow, but Evans ch. 2 is fairly standard

[chrome trout]

Got it thanks! I will push through!

[frozen orbit]

If f is an a.e. function in a Sobolev space W^k,p, is |f| also in W^k,p?

[buoyant pike]

What does it mean to be in a sobolev space

[granite sun]

When solving Delta(u)=1 for tempered distributions, you typically Fourier transform to get r^2 u-hat = delta. In order to solve this, it seems we are restricted to the R^n, n>=3 case so that 1/r^2 is integrable near the origin. How then can we find solutions for the n=1 case?

[buoyant pike]

(don't you normally solve Delta(u)=delta for tempered distributions)

Anyways

What is the laplacian on R

[granite sun]

Yep, sorry.

[granite sun]

D^2 of course

[buoyant pike]

How do you solve d^2u/dx^2=delta

[granite sun]

Ah I see - something something |x|. Though, I’m still a bit shaky on handling distributions, are there any nuances to be wary of?

[buoyant pike]

At least in this case, that's all there is to it

You can check that this works out distributionally with test functions but shrug

[granite sun]

I’m assuming this is with the understand of also checking for homogeneous solutions?

[buoyant pike]

What do you mean?

[granite sun]

Well |x| is obviously a particular solution. But the general solution will likely need to include distributions that satisfy the laplace equation, no?

[buoyant pike]

Oh sure things in the kernel of the 1-d laplacian

[granite sun]

And, to make sure, everything in the kernel is an (integrate against) polynomial?

[buoyant pike]

There is a straightforward classification of the kernel

[frozen orbit]

function and it’s first k derivatives are in L^p

i think it should be true because derivative of |f| is sgn(f)•f’ which is L^infty • L^p but my hw suggests the proof is more involved

[bronze gate]

depending on how you define derivative (eg. convergence in Lp of a certain function) it could change

[frozen orbit]

distributional derivative, say

[bronze gate]

literally just integration by parts

[frozen orbit]

yeah

[frozen orbit]

are you saying derivative of |f| in L^p follows from integration by parts?

[astral vine]

This is true for Sobolev spaces including dirichlet boundary conditions up to regularity index 1+1/p (i.e. when p=2, for elements of H^{s}_0 0<s<3/2 (s=0 is obvious))

beyond this treshold, everything fails

a very easy proof is for elements of H^{1}_0

H^{1} should work too I don't remember clearly to be honest

hmm

the main idea is to approximate |f| by sqrt(1/n+|f|²)

then show that the distributional derivative converge

weakly

[quaint herald]

This is a good question. After taking the Fourier transform you should get instead that $|\xi|^2\hat{u}=1$ (I will be a bit lazy about constants here).

As you know, for $n\geq 3$ we can simply take $\hat{u}=|\xi|^{-2}$ as this is a locally integrable function, hence a distribution and in fact a tempered distribution which we can inverse FT to obtain $u$.

For $n=1,2$ this function is no longer locally integrable and so it is not quite as simple as setting $\hat{u}=|\xi|^{-2}$...but one CAN use a regularisation process on this function to obtain a distribution (read about Hadamard finite parts integrals, or pseudofunctions). It is similar to/related to the Cauchy principal value.

If you do this carefully, you obtain a tempered distribution that is equal to $|\xi|^{-2}$ away from the origin and inverse Fourier transforming you get the standard fundamental solution in these dimensions too. (Of course n=1 is easier, just antidifferentiate twice).

[untold delta]

gomez

[frozen orbit]

i see, thank you

[mint canyon]

There is a topic I wanted to read up on, but I cant for the life of me figure out what it's named in the literature, could I get some help?

[astral vine]

Tell us, describe it.

[mint canyon]

I posted in the adv-analysis channel since no one seemed to be here, want me to copy what i typed there?

This was the start of the thread: "So in Calculus of Variations you have the functional derivative operator, and Ive seen equations that use those operators to define an unknown functional, what are those types of equations called in the literature?"

Here is a much simpler linear example:

A(δ/δf(x)) S[f(x)] + BS[f(x)] = 0

[astral vine]

It looks like an Euler-Lagrange equation in a specific form right ?

[mint canyon]

The euler-lagrange equation is used to evaluate a functional derivative, its sort of the definition o the operator, and you use it when you want to find the variation of a KNOWN functional, it would definitely play a role in solving them... I think?

In the variation you want an unknown f(x) that extremizes a known S[f], but in these equations you are looking for S[f]

[hallow pumice]

why isnt it possible to solve the schrodinger equation for a general potential V(x, t)?

[astral vine]

what do you mean by solve ?

[astral vine]

oh okay my bad, I am not aware about that

[arctic whale]

Let $k(x,y) : L^2[0,1]^2 \to R $ defined as $x(1-y)$ on $0 \leq x \leq y \leq 1$ and $y(1-x)$ on $0 \leq y \leq x \leq 1$. For the integral equation $e^{-x^2}\int_0^1 k(x,y)f(y)dy + f(x) = \sin(\pi x)$. I've shown that a unique solution exists. Is there a way to find this solution explicitly ?

[untold delta]

ru0xffian

[buoyant pike]

Oh my

[safe crater]

same reason you can't "solve" all pdes. the form of the function is of course going to determine the solution, but some functions don't have analytic integrals or analytic diffyq solutions.

[safe crater]

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

That wiki page has lame examples

less lame examples are integro differential equations, with the "integro" coming in from a functional that is an integral

actually...none of those use functional derivatives...my bad

https://www.researchgate.net/post/How_to_solve_functional_differential_equations_that_involve_functional_derivatives

gives https://www.researchgate.net/deref/http%3A%2F%2Fquantum.phys.unm.edu%2F523-14%2Fch15.pdf as an answer

that has a "functional differential equation" that this time actual involves functional derivatives (expressed in bra-ket notation)

[deleted link to textbook as i don't have information pertaining to the legality of this upload]

on "functional differential equations"

that considers diffyqs that look like $\dot x = F x$ where F is an operator on functions, in analogy to the ODE version, but this time F is a functional

this is "INTRODUCTION TO THE THEORY OF
FUNCTIONAL
DIFFERENTIAL
EQUATIONS
METHODS AND APPLICATIONS"

by "N. V. AZBELEV
V. P. MAKSIMOV
L. F. RAKHMATULLINA"

[untold delta]

Xela

[empty terrace]

I am studying interpolation using polyharmonic splines. That is, given a set of points $\mathcal{X}={\vec{x}_1,\ldots,\vec{x}_N}\subset \mathcal{M}$ I solve the weak formulation of:
$(-\Delta)^{m} u(\vec{x}) = 0$ for $\vec{x}\in\mathcal{M}\setminus \mathcal{X}$, and $u(\vec{x}i) = y_i$, and $\partial{\vec{n}}(-\Delta)^lu(\vec{x}) = 0$ for $\vec{x}\in\partial\mathcal{M}$ and $0\leq l\leq m-1.

[untold delta]

criver

[empty terrace]

For m>d/2 the space H^m(M) is a reproducing kernel hilbert space (https://arxiv.org/pdf/1905.10913.pdf) and thus the interpolation problem should be well-posed (potentially with the addition of some extra constraints)

I made a mistake above, the boundary constraints should have read $\partial_{\vec{n}} (-\Delta)^lu(\vec{x}) = 0$ for $\vec{x}\in\partial\mathcal{M}$ and $0\leq l \leq m-1$.

[untold delta]

criver

[empty terrace]

the addition of these extra constraints ought to make the reproducing kernel positive definite instead of only positive semi-definite

So I decided to study the weak formulation for $d=2$ and $m=2$ in order to better understand what is going on

[untold delta]

criver

[empty terrace]

$\int_{\mathcal{M}\setminus\mathcal{X}}\Delta v \Delta u = \int_{\Gamma} \partial_{\vec{n}} v\Delta u - \int_{\Gamma} v \partial_{\vec{n}}\Delta u + \int_{\mathcal{M}\setminus\mathcal{X}} v \Delta^2 u$

[untold delta]

criver

[empty terrace]

Here $\Gamma = \mathcal{X} \cup \partial \mathcal{M}$. For the part of the boundary $\partial\mathcal{M}$ the terms vanish due to $\partial_{\vec{n}}v = 0$ and $\partial_{\vec{n}}\Delta u = 0$. What troubles me is the $\mathcal{X}$ part of the boundary and the term $\int_{\mathcal{X}}\partial_{\vec{n}}v \Delta u$.

[untold delta]

criver

[empty terrace]

I would generally not want to prescribe additional derivative information at the interpolation points, so d_n v does not have to be zero there. As long as Delta u is not singular at X, I assume that in 1D I could argue that since v is differentiable, then then integral evaluates to zero because I have the left derivative being equal to the right derivative. In 2D and higher dimensions the set X has zero measure, so I am guessing the integral should evaluate to zero as long as the terms d_n and \Delta u are non-singular?

What bothers me is that usually in the literature for H^2 one prescribes both a d_n u and u constraint on the boundary. However, here I have omitted the derivatives constraint at the interpolation points.

[steep oyster]

i'm getting a bit confused on weak solutions and a.e. stuff. for starters, i'm not sure if a weak solution in a sobolev space is a legitimate function, or an eq. class of functions.

a more serious problem (that's closely related) is the following. suppose we want to solve a 2nd order elliptic eqn Lu=f. at each stage of "improving regularity", it seems that although we exchange one solution "u" for another "v" with better regularity but Lv=f is only true a.e. so for example, if my coefficients and f are all smooth, they we end up with a smooth solution v, but one that only solves the PDE a.e.

i guess my question is what i wrote above correct; are we actually exchanging "u" for another "v" which differs a.e. when we improve regularity? if so, why is it acceptable to call "v" a solution to the original PDE? my conception of a "solution" means solving for each point, not a.e.

[buoyant pike]

Yes so elements of sobolev spaces (and others, like L^p), are equivalence classes of functions that are equal almost everywhere

[steep oyster]

ok so you don't solve Lu=f pointwise, you solve it a.e. if you use sobolev methods?

[buoyant pike]

Yes you could say that

[steep oyster]

ok thank you. i guess that feels very odd to me.

[buoyant pike]

Yes so you want to leave the physical interpretation behind when you start thinking about this

[mint canyon]

@safe crater Yeah, I was confused because on wikipedia "functional equations" and "functional differential equations" referred to equations with ffunctions of modified arguments like f(x) = kf(x+h) - g(x) or something.

[empty terrace]

Is there a straightforward way to trade Navier boundary conditions $\Delta u = f$ with directional boundary conditions $\partial^2_{\vec{n}} u = g$?

[untold delta]

criver

[empty terrace]

In 2D, the only thing I can think of is to rewrite $\Delta u = \partial^2_x u + \partial^2_y u = \partial^2_{\vec{n}} u + \partial^2_{\vec{n}^{\perp}} u = f$

[untold delta]

criver

[empty terrace]

However I am still left with a tangential term

[mint canyon]

is there a good paper at introducing methods from the Homotopy Principle to solving PDEs from more of an Applied perspective?

[buoyant pike]

What do you mean by a more applied perspective

Like numerically?

https://arxiv.org/abs/1111.2700 discusses the homotopy principle a bit with regards to fluids

I do not think this gets used numerically though

[mint canyon]

Well, I sort of mean without going super rigorous on proofs and stuff, Im definitely interested in the analytic solutions

Also, thanks for the link. I'll check it out

[astral vine]

Analytic solutions in which sense ?

[mint canyon]

i was taught solutions to differential equations were either analytic, quasi-analytic (perturbative), or numerical.

Analytic being the only one where you can plug in the solution and it balances both sides o the equation.

[hard wolf]

sorry if I'm interrupting -- a bit rusty and not especially knowledgeable from the get-go about explicitly solving PDEs. pointer in the right direction/ something to read for working on part (i)?

[mint canyon]

Oh Analytic also means an infinitely diferentiable function too, I guess that could cause confusion?

[hard wolf]

I am assuming I can mess with the fundamental solution to the poisson problem, but the addition of k(x) in there is... disconcerting

[mint canyon]

@hard wolf I know they used partial deriatives, but that looks like an ODE

[hard wolf]

it's the poisson problem in 2d

one time and one space

[mint canyon]

u' = a/k -> u = { ax/2 + b on [-1,0], ax + d otherwise -> a(-1)/2 + b = 0 -> b = a/2 -> a(-1) + d = 3 -> d = 3 + a, I wonder if you are supposed to assume the constant of integration is the same in both piece wise regions?

[hard wolf]

alright am done procrastinating answering light questions for my TA job

[hard wolf]

where is this u' = a/k ansatz(?) coming from?

going to try throwing IBP at this...

[mint canyon]

the derivative of something = 0 means that its a constant. Then ust divide k on both sides

[hard wolf]

mmmmmm

this just feels like an incorrect hack to make but I can't offer an immediate reason as to why

[mint canyon]

well you can integrate both sides if you like, the integral of 0 is 0 + a constant.

[hard wolf]

your detail about it being constant is trivial but the claim is then that the partial_x u * a discontinuous function is constant

im like intuitively willing to accept that u ends up being convex in x, sure, though

[hard wolf]

I am not sure how to rectify that with your method

then working back to justifying it as the weak solution seems like it would end up being handwavy asf

[mint canyon]

I treat k like a normal function till I do the final integration, then I split the domain o the integration. I you like I guess you could restrict manipulation to (-1,0] and [0,1) and then ust do the exact same thing in each case?

if you do that I guess you have a1 and a2 as constants o the irst integration, Im not sure there is enough info to show they are the same (or if they aren't to evaluate them)

you only have 2 boundary values, so you can only afford to have 2 constants of integration

[mint canyon]

Another thought, maybe you could write k using heaviside step functions? Then you could use the product rule and your k' term could become dirac delta functions

[hard wolf]

maybe but nothing like that/ trace operators were ever introduced

so the "heavy machinery" solutions are maybe out the window

in light of the hint, it seems like I am to solve it over the subdomains and join it together at 0 assuming continuity as mentioned

[lime chasm]

I am stuck after applying the second boundary condition for Y(0) i find 0 = cos(2pix)

any idea what should i do?

[astral vine]

@lime chasm #odes-and-pdes

[lime chasm]

oh i apologize

[astral vine]

No problem :)

[lime chasm]

no one is answering either way haha

[astral vine]

this kind of question have been answered several times

like hundred time a year in #odes-and-pdes

[lime chasm]

can u quickly help me identify the problem then? it seems the boundary condition does not work

[lime chasm]

can someone help me ?

[hard wolf]

BTW this one was really not so bad

I was just not convinced of the domain splitting and overthought it

split into (-1, 0) and (0, 1), solve those restrictions, get piecewise linear function, rectify the properties and ensure it solves the PDE

[frozen orbit]

are all spectral methods considered non local analysis?

[steep oyster]

i have a few questions on schauder estimates i'm crossposting here: https://math.stackexchange.com/questions/4639363/two-questions-on-schauder-interior-estimate-theorem-6-2-from-gilbarg-trudinger

[buoyant pike]

Yes I suppose

[tired hollow]

I'm in a senior level pde class, would questions about this be better suited for this or #odes-and-pdes

[astral vine]

it depends more on the question itself than the class level

so ask here

we will redirect you if it doesnot fit

[tired hollow]

My professor just wrote $\frac{\partial u}{\partial t} = \frac{\partial u}{\partial t} \frac{dz}{dt}$ and I'm so confused because $\frac{\partial u}{\partial t} \neq 1$

[untold delta]

Iced Sugar

[tired hollow]

U is a function of x and t

[buoyant pike]

We're going to need more context

[tired hollow]

Yeah I figured that might be the answer

He just wrote that and moved on and it super confused me

[acoustic minnow]

Also, what's z?

[buoyant pike]

It sounds like you should ask your professor

[mint canyon]

z(t) = t + c according to what he wrote, or I guess z(x,t) = t + f(x)

[buoyant pike]

So dz/dt=1

[mint canyon]

cant be anything else I could think of

[buoyant pike]

Oh are you just guessing

I personally suspect that it should be partial u partial z

[mint canyon]

oh, you mean the prof made a mistake while he was writing the notes? That happens

we should have asked what the class was, might have given insight what equations they were working with

[untold delta]

hardisc

[buoyant pike]

If you want an answer you should tell us what the problem is

[bitter yacht]

Yeah I realized I made a mistake and deleted my question.
Could not delete this last TeXit bot-message though

[bitter yacht]

I am having a go at problem 17 in chapter 8 in evans.
I managed to follow his hint, all the way until he concludes $Du/u = D\hat u/\hat u$ almost everywhere.

If I integrate [|Dw|^2 \leq \eta \left( s \left| \frac {Du}u \right|^2 + (1-s) \left| \frac{D \hat u}{\hat u} \right|^2 \right) = \frac 12 |Du|^2 + \frac 12 |D \hat u|^2 ]

I get
[ I(u) = I(\hat u) = \int_U \eta \left( s \left| \frac {Du}u \right|^2 + (1-s) \left| \frac{D \hat u}{\hat u} \right|^2 \right) dx = I(u) = I(\hat u) ]

To obtain this I used that $u, \hat u$ are both minimizers and therefore $I(u) = I(\hat u) \leq I(w)$.

Somehow I suppose I should force the integral to be zero, but I have not been successful in attempting this

[untold delta]

hardisc

[bitter yacht]

Giving it some further thought, I suppose we do get the equality
[ \int_U \eta \left| s \frac {Du}u + (1-s) \frac {D \hat u}{\hat u} \right|^2 = \int_U \eta \left( s \left| \frac {Du}u \right|^2 + (1-s) \left| \frac {D \hat u}{\hat u} \right|^2 \right) ]

If this implies the integrands must be equal almost anywhere (which I have not been able to justify as of yet), then we get
[ \left| s \frac {Du}u + (1-s) \frac {D \hat u}{\hat u} \right|^2 = s \left| \frac {Du}u \right|^2 + (1-s) \left| \frac {D \hat u}{\hat u} \right|^2 ]

But $s$ is neither 0 nor 1, since that would contradict positivity of the minimizers.
So the equality we obtained should be problematic, since I think the map $x \mapsto |x|^2 $ is strictly convex. To save the equality we obtained, maybe this forces $Du/u$ and $D\hat u/\hat u$ to equal one another almost everywhere? Does this work?

[untold delta]

hardisc

[blissful quiver]

Hi, i have a question about approximation of sobolev functions by smooth functions. In Evans Theorem 2, he says that, with respect to the Sobolev Norm, $C^{\infty}(U) \cap W^{k,p}(U)$ is dense in $W^{k,p}(U)$. Now in Theorem 3(Approximation up to boundary by smooth functions) he says that $C^(\overline{U})$ is dense in $W^{k,p}(U)$. Now my questions are: Can't we also say already that $C^{\infty}(U)$ is dense in $W^{k,p}(U)$ in Theorem 2 and also say that $C^{\infty}(\overline{U})\cap W^{k,p}(U)$ is dense in $W^{k,p}(U)$ in Theorem 3, since any convergent sequence with respect to the sobolev norm has to consist of sobolev functions from a certain index on?

[untold delta]

chrisply

[lilac barn]

The reason we drop the Wᵏᵖ(U) in the next theorem is because being C(U̅) implies continuous functions attained their maximum on the bounded sets which automatically imply that they are integrable because U is bounded. So no need to incorporate integrability. In the previous case, we had to put Wᵏᵖ(U) because functions may not be bounded anymore so no reason to expect integrability. So we instead, reduce the space to just both smooth−integrable functions.

[blissful quiver]

ok, so we could also intersect with Sobolev FUnctions in Theorem 3 and it would still be fine, but we can't omit the intersection in THeorem 2?

[lilac barn]

Yeah, you can intersect in theorem 3 if you like. The very fact that we are taking the Wᵏᵖ norm in theorem 3 implies that our u_m lie inside Wᵏᵖ.

[quaint herald]

yes, the function and it's derivatives can blow up at the boundary whilst still lying in C^inf(U)

so that intersection gives you an integrability condition on your derivatives.

[blissful quiver]

ok yeah makes sense

however

Closure of a set is defined as all limit points with respect to a norm, in this case the sobolev norm. So if we would just say that the closure of $C^{\infty}(U)$ with respect to the Sobolev Norm is $W^{k,p}(U)$ that would be fine, right? Since any convergent sequence with respect to that norm must be comprised of mostly Sobolev Functions (that is all but finitely many) due to triangle inequality?

[lilac barn]

If you replace U with Ubar, then yes that is precisely what 3 is saying

[quaint herald]

IT is not a good idea to say that in Thm2 because C^inf(U) is not a subspace of the Sobolev space.

But yes you could state these in terms of density, so C^inf(clos(U)) is dense in the Sobolev space in 3 and C^inf(U) intersect the Sobolev space is dense in the Sobolev space in 2.

[quaint herald]

Closure is defined for a subset of a topological (in this case normed) space, so it does not make sense to talk about the closure of something that is not a subset.

[blissful quiver]

yeah makes sense

thank you both, that clears up my confusion

[blissful quiver]

is there a space we could take, that has both smooth functions and Sobolev Functions as subsapces, for which the Sobolev Norm would still be a norm, so that the closure of the Smooth functions are exactly the sobolev functions, or can such a space not exist?

[quaint herald]

some smooth functions simply don't lie in the sobolev space, so no

[blissful quiver]

ok yeah, that was kind of a dumb question xd

but i think i got it now

thank you

[north tulip]

Can someone help me to understand this proof. I do not see how continuity is applied here

[astral vine]

by contradiction assume t^* is such that

|u(t^*)-x_0 | < R

continuity arguments leads to a contradiction since this would give you that the undelrying interval is closed and open then connected

continuity preserve connected sets

[north tulip]

Can you elaborate on the underlying interval is closed and open then connected? It is puzzling for me where that comes from

[untold delta]

Functionanatolysis

[astral vine]

continuous right ?

if you if fix t

but the preimage of [0,R] should a closed open set

(with respect to induced topology)

therefore the whole preimage should be closed subset of (t_0,t)

so either of the form (t_0,s] (for some t_0<s<t)or to be the whole (t_0,t)

However t^* is defined as the sup of all t such that (t_0,t) is in the preimage of [0,R] through aabove map

can you connect the dots from there ?

[nocturne wave]

why there is only finite term?

[quaint herald]

Because the terms in the sum are supported in W_i so in shrinking neighbourhoods of the boundary. The compact set clos(V) has a positive distance to the boundary.

[mint canyon]

What are the challenges in writing down a symmetry condition for Backlund transforms in the same way you can for Point-wise transforms?

[nocturne wave]

Do you mean that V is covered by finite many W_i?

[plain yoke]

y"-x^2y=0

Does anyone have an idea how to solve?

[buoyant pike]

#odes-and-pdes

[mint canyon]

at first I read that as y" - x^(2y) = 0

[arctic whale]

Just checking is this true by just considering that bilinear form $B(u, v) = \int \nabla u \cdot \nabla v + \int c u v$. We can just show it's coercive $B(u, u) \geq ||\nabla u| |^2 + \lambda | |u| |^2 \geq 2 \lambda | |u| |^2$ and we have that $\lambda$ is just the principal eigenvalue of principal elliptic operator so it's positive. So we just conclude by Lax-Milgram that there is a weak solution for each $f$ ? Is this correct ?

[untold delta]

ru0xffian

[slender fulcrum]

in your coercivity estimate it should say $$B(u, u) \ge |\nabla u|{L^2(\Omega)}^2 - \lambda_1(\Omega)|u|{L^2(\Omega)}^2.$$ then you can use the minimality property of $\lambda_1(\Omega)$ to show coercivity (in the $H^1(\Omega)$-norm!)

[untold delta]

whzup

[arctic whale]

yeah i figured that. Thanks a lot !

[topaz falcon]

hi! currently reading up on variational elliptic pdes and just need some help understanding thus

what does the elements of the set A look like? also don't understand what is meant by a matrix vector field sorry

[astral vine]

This just a matrix whose coefficients vary

Each coefficient is a function on Omega with value in R or C

A is not a set

It is a matrix given by n² coefficients

[versed shoal]

@buoyant pike pls enlighten

i feel like I should split the matrix and use the Trotter product formula (because we used it for H = -Laplacian + V) but honestly I don't even know how to take powers of that matrix, can you just treat the entries like you would scalars...?

also I don't think there should be a -I in the (2, 2) entry oops

[astral vine]

yes no -I

for the exponential

go on the fourier side

[versed shoal]

I should Fourier it before looking at the matrix right?

will see what that gives

[astral vine]

Fourier the matrix or Fourier the equation

this is the same

[versed shoal]

ah ok

nice

[versed shoal]

So you would split out the I in the (2, 1) position and use Trotter on the two matrices?

(haven't done the computation yet tho)

[astral vine]

Just do it roughly, everything commutes, do the comuputations on the Schwartz class

and look separately at odd and even powers

this will give you Sine and Cosine of (I-Delta)^{1/2}

[versed shoal]

what does (I - Delta)^{1/2} mean here? (makes sense as an abuse of notation)

just the thing that comes out of the functional calculus? will go through it and see what comes up

[astral vine]

(1+| xi |²)^1/2 on the Fourier side

which coincides with holomorphic functional calculus, and the Borel functional calculus, and the Philip Functional calculus, etc.

[buoyant pike]

Ah yes the Japanese angle bracket functions

[astral vine]

Yes

[buoyant pike]

Yes

[astral vine]

Or also called the Bessel potential of first order

[versed shoal]

had a lapse of concentration and was thinking about the un-Fourier stuff again, so you're just looking at the nth power of [{0, -1}, {-|x|^2 - 1, 0}] then doing the infinite series, then un-Fouriering, seems easy enough

[astral vine]

yes

[versed shoal]

got it lol

idk why i was confusing myself

[astral vine]

When you are on the whole space with constant coefficients you should think first about Fourier

(even in a more instantaneous way when you are on L²)

The method in general still works in the Lp setting,, but the meaning of Fourier symbols is a bit tough to get unless you had a good Harmonic Analysis lecture before

[stark thunder]

Hey! I just made this video on gradient flows/energy minimisation, let me know what you think! 😄 https://www.youtube.com/watch?v=tXcvkGqCcR4

[buoyant pike]

#optimization may also be interested

[stark thunder]

awesome thanks:)

[wind geode]

Nice video!

[untold delta]

realbluelion17_the_great
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[deft saddle]

I was wondering if there are any y like this$: x^{2} = \sum_{n=1}^{\infty} \frac{\partial^ny}{\partial x^n}$

[untold delta]

realbluelion17_the_great

[astral vine]

#calculus

[river path]

the best way to handle something like this is to use "neumann series", which is where you treat a sum of operator powers like a geometric series.

so we would try to see what we get if we look at $x^2 = \left( \sum_{n=1}^\infty \left(\frac{d}{dx} \right)^n \right) y$

[untold delta]

ryc

[river path]

the series "should" sum to "d/dx / (1 - d/dx)" which ofc is nonsense, but we interpret the division as an inverse. Then we're looking at x^2 = (1 - d/dx)^(-1) dy/dx. multiplying 1 - d/dx to both sides, you get x^2 - 2x = dy/dx. now the answer we get is just x^3/3 - x^2 + C.

did these formal manipulations (which are completely unjustified) give us a sensible answer? they did - the series turns out to be 0 after n = 3 and we're just looking at (x^2 - 2x) + (2x - 2) + (2) + 0 + 0 + ...

which is x^2.

now you might ask, shouldn't this have more than one degree of freedom since has derivatives of every order? well, maybe, but i think if you take any solution and apply (1 - d/dx) to both sides of the equation, you'll find that it has to satisfy the x^2 - 2x = dy/dx equation we came up with.

unless there are nonconverging sums (in which case i would not call that a solution)

[astral vine]

There is a very important underlying problem, about giving the meaning to all of this

[river path]

yes definitely

[astral vine]

you need a very rigid ambient structure

Like everything lying in an appropriate Banach space of continuous functions

[river path]

i'm taking it at face value, and what an analyst does in practice is throw neumann series at it and then try to figure out based on that what the "correct" meaning to give to the problem is

i didn't really need to do that since i ended up just finding out that the sum terminates if you want to interpret the problem vaguely classically

[astral vine]

but in this case x^2 doesnot seem tobelong to any of such reasonable spaces, unless we are on an interval, and we add appropriate boundary conditions

but in this case we don't solve it on the whole line

[river path]

weighted L^p space

[astral vine]

Yes but that's not a natural structure people, asking this kind of question, may have in mind

they are like "well let us solve this diff eq with smooth solutions, in the classical sense"

[river path]

yeah i agree. i just think if you're going to ask a question like this there's a good chance you don't know to look for structure.

[astral vine]

Exactly

[river path]

i only answered it to be able to bring up neumann series, cause i think thats important

[astral vine]

You are right but the additional underlying concepts are very important to make sense of it

Otherwise this no longer actual maths just bunch of weird notations

and few computations

[river path]

yeah i agree

[lilac barn]

To add onto this point, for the Neumann series to work, we would need the sum to be able to converge in the underlying Banach space, which is equivalent to asserting that the operator must have norm strictly smaller than 1. This isn't true for derivatives in general so we would need a nice enough space to work with where we can guarantee this condition.

[quaint herald]

You can still carry out manipulations "formally" to get plausible solutions, and then verify they are indeed solutions, which is all the first guy wanted. Here if you differentiate under the sum, subtracting this from your first equation gives x^2-2x=y', giving y=x^3/3-x^2+C at the very least as a solution in the most classical sense.

[astral vine]

But you need additional framework to show uniqueness (up to a constant)

Checking that you make sense of the equality owing you have this specific solution, is okay

but a priori the expression does not have an actual sense, which seems to be required to prove uniqueness

[quaint herald]

Yes I made no claim about uniqueness, I was just answering the literal question that was originally asked.

[solemn gyro]

I am studying the trace theorem from Evans , and i get confused by the very first part of the proof, may i ask, how to show such smooth compact supported function exist?

[wind geode]

Urysohn's lemma

It should be in there, check the index

The example he has in mind is this https://en.wikipedia.org/wiki/Bump_function though I never heard it called "the bump function" we just called it "the standard mollifier"

[solemn gyro]

I dont know how to extend Urysohn lemma to smooth case , can you give me some help?

I know the standard mollifier but i cant see which scalar function should be multiplied with it in order to make $\xi \equiv 1$ on $\hat{B}$

[untold delta]

mikeqwertyuiop

[buoyant pike]

Can you see how you would construct such a function for a closed interval of R

[wind geode]

Suppose you use a standard mollifier with radius $\delta / 2$ and you have a hat function $\mathbb{1}B'$ where $B \subset B'$ and every point $x$ with distance less than $\delta / 2$ from $B$ is also in $B'$. This will guarantee that $\phi{\delta / 2} \star \mathbb{1}_B'$ will be 1 in $B$.

[untold delta]

robert_

[wind geode]

i.e. the idea is mollifying a hat function who's set is the set you want plus some buffer around it so the mollifier keeps the convolution result being 1 within B

[solemn gyro]

is this the hat function that you are mentioning?

[wind geode]

no I mean 1_B(x) = {1 if x in B, 0 otherwise

I think it has another name too, like "the set function" I think? I couldn't find it.

[solemn gyro]

Indicator function?

[wind geode]

Yes that's a name for it too

[solemn gyro]

or characteristic function

[wind geode]

yes that one too lol

[solemn gyro]

Thanks 😄 ,I see how this works now

[wind geode]

😄

[rancid vine]

Hi i am stuck with this problem 4.10.2 I need to show that the EL equation( for second order) becomes an identity iff the integrand is of the form as shown in the ss

So far what I have done is set the coefficient of 4th order derivative of y to 0

Giving me

g(x,y,y')y''+h(x,y,y')

Then I put this back in the EL equation and found the coefficient of second derivative of y

And that turned out to be this monstrous pde

How do I show that g and h are of the form as given in the question?

[buoyant pike]

Does anyone know of a reference that derives the convolution theorem for spherical harmonics

[tired hollow]

Is the region of influence and domain of influence in wave equation PDE problems the same thing?
Or are they different

[buoyant pike]

Those are the same

[tired hollow]

ok thanks

[buoyant pike]

Or they should be at least

[ashen niche]

My PDE professor burst into today's lecture by proclaiming that he cried when he reviewed the midterm submissions. Tears of joy no doubt.

[tired hollow]

if i see $\alpha^{+}$ what does that mean?

[untold delta]

KooKoo

[buoyant pike]

Ummm

Context?

[somber blaze]

Hi! I have this bilinear form:
$a(u,u)= \int u'^2 + bu'u + cu^2 \ dx \ b,c > 0$
where $u \in \mathbb{H}^1_0$.
How can I show if this PDE is coercive or not? I've looked at examples where both $bu'u$ is gone & $cu^2$ but I seem to not be able to handle nor find a solution for this one. Here's the definition for coerciveness in the doc I'm reading.
The problem is that I can't seem to find a way to define the lower bound when taking into account all these arbitrary constants to show that some alpha exists which satisfy the inequality.
(A very similar question on mathSE, however $bu'u$ is gone in this one
https://math.stackexchange.com/questions/2727116/show-that-the-bilinear-form-is-coercive)

[untold delta]

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

[astral vine]

b is a cosntant

so buu' integrate into b u(1)²/2 - b u(0)²/2 = 0-0=0

because u is assumed to be in H^1_0

then everything works as in the mathstackexchange post you linked

b has no relevant contribution in the energy formulation

if b was a bounded (even smooth) measurable (or just in some Lebesgue space) function the answer would be a bit harder, using sharp Sobolev embeddings does the job

[somber blaze]

God... that did the trick. Thank you very much. I've been trying to figure this one out for too long now

[untold delta]

cocat

[mellow totem]

I am new at this. I just want to ask , to understand Fundamental solution in pde do i need distribution theory?

I haven't yet study distribution theory.

[astral vine]

Depends on what you mean by "understand"

Like intuitive meaning, and construction of some fundamental solutions, no you don't need it.

But if you want to prove things in careful way, and have actual rigorous results, then yes this is necessary.

(to learn Distribution theory)

[mellow totem]

Oh ok.

[frozen orbit]

Let $H\subset R^n$ be a half space, $V$ a bounded function, and $u:R^n\to R$ a locally square-integrable function that solves the equation $-\Delta u+Vu=0$ in $H$ with Dirichlet boundary conditions ($u\equiv 0$ on $\partial H$). I am wondering whether it is possible to extend $u$ to a solution of this equation on all of $\mathbb{R}^n$

[buoyant pike]

u=0 on all of R^n

[frozen orbit]

oh just extend by zero? I'm not sure how to show that the first derivatives match at the boundary though

[buoyant pike]

u=0 is a solution in H for your problem

[frozen orbit]

sure

I haven't seen a uniqueness result for $L^2_{\text{loc}}$ functions

[untold delta]

Lakshay

Lakshay

[frozen orbit]

okay so this won't work in general -- consider the Half space $H={x\geq y}\subset R^2$, $V=0$, and $u(x)=x^2-y^2$. Then extension by zero is not $C^2$ and isn't harmonic on the boundary ${x=y}$

[untold delta]

Lakshay

[wise hare]

In solving heat equation on a finite interval say $(0,L)$ with Dirichlet Boundary conditions why does one consider odd periodic extension and not even periodic extension?

[untold delta]

K零ꓘ

[wind geode]

I just spent like 20 minutes looking through Walter Strauss book to find an answer. He demonstrates it several times with a guess and check method but never explains underneath why it would be the odd periodic extension and not the even one. (Basically posting to save anyone thinking the same the time)

[buoyant pike]

Separation of variables in #odes-and-pdes

[astral vine]

First you can prove uniqueness by energy methods, without requiring any extension procedure

then since you have at most one solution

it suffices to build a solution

And this procedure gives you one, so the solution

[north tulip]

I've managed to do a and b, but im not sure how to do c. I guess I need to show that u(t) is not equal to 0 as t goes to $t^+(x_0, 0)$. However, I don't see how I can start with that

[untold delta]

FrankF

[north tulip]

nvm, I think I got the answer which is:
Since, $\abs{u(0)} \leq 1$, by (b), we conclude that for all $t \in J$ such that $t \geq 0$, it holds that $\abs{u(t)} \leq 1$. This means that $u(t)$ can never blow-up as $t \leftarrow \infty$. We know that at the boundary of the maximum existence interval it is either the case that $u$ blows up or that $u$ reaches the boundary of $I$. Therefore, $u$ must reach the boundary of $I$, which is $\infty$ in this case.

[untold delta]

FrankF

[untold delta]

tywungu

[hallow pumice]

isn't that a PDE book for like engineers and physicists? (I.e not meant to be rigorous)?

[wind geode]

It’s a comprehensive treatment of just about everything you can say about PDEs in 1-3 dimensions while only using calculus.

[wind geode]

It was the undergrad PDE text for math majors when I was at NYU.

[hallow pumice]

Wouldn't math majors use something like evans?

[buoyant pike]

So Strauss is a book to learn how to solve the 4 pdes that you can actually solve explicitly

Evans is a book to learn the theory of pdes more generally, requires a lot more background, and is not so concerned with solutions

[bronze gate]

(disclaimer: I hate Evans): the first part of Evans is undergrad, the rest is grad pdes

[north tulip]

I am not sure how to continue for this question because I have that I need to choose z such that $e^t((2z_3-z_2)\begin{bmatrix} \frac{\cos(t) - \sin(t)}{2}\cos(t) \cos(t) \end{bmatrix} + (2z_1-5z_2+4z_3)\begin{bmatrix} \frac{\sin(t)+\cos(t)}{2}\ \sin(t)\ \sin(t)\end{bmatrix})$ goes to 0 as t goes to infinity. Below is what I have tried so far for this question:
\newcommand{\C}{\ensuremath{\mathbb{C}} }
The characteristic polynomial is $$-\lambda(\lambda^2-4)+7(2\lambda+4)+8(-4-2\lambda)=-\lambda^3+2\lambda-4.$$
Equating it to 0 gives
\begin{align*}
\lambda^3-2\lambda+4 &= 0\
\lambda = -2 \lor \lambda = 1 \mp i
\end{align*}
The eigenvectors are:
\begin{itemize}
\item $\lambda = -2 \implies \begin{bmatrix} 3\2\1\end{bmatrix}$.
\item $\lambda = 1 + i \implies \begin{bmatrix} \frac{1+i}{2}\1\1\end{bmatrix}$.
\item $\lambda = 1 - i \implies \begin{bmatrix} \frac{1-i}{2}\1\1\end{bmatrix}$.
\end{itemize}
The general solution is
\begin{equation}
u(t) = C_1e^{-2t}\begin{bmatrix} 3\2\1\end{bmatrix} + C_2e^{(1 + i)t}\begin{bmatrix} \frac{1+i}{2}\1\1\end{bmatrix} + C_3e^{(1 - i)t}\begin{bmatrix} \frac{1-i}{2}\1\1\end{bmatrix}
\end{equation}
with $C_1, C_2, C_3 \in \C$. Note that
\begin{align*}
e^{(1 + i)t}\begin{bmatrix} \frac{1+i}{2}\1\1\end{bmatrix} &= e^t(\cos(t)+i\sin(t))(\begin{bmatrix} \frac{1}{2}\1\1\end{bmatrix} + \begin{bmatrix} \frac{i}{2}\0\0\end{bmatrix})\
&= e^t(\begin{bmatrix} \frac{\cos(t) - \sin{t}}{2}\\cos(t) \\cos(t) \end{bmatrix} + i\begin{bmatrix} \frac{\sin(t)+\cos(t)}{2}\ \sin(t)\ \sin(t)\end{bmatrix})\
\end{align*}
The real representation of the general solution is
\begin{equation}
u(t) = C_1e^{-2t}\begin{bmatrix} 3\2\1\end{bmatrix} + e^t(D_1\begin{bmatrix} \frac{\cos(t) - \sin(t)}{2}\\cos(t) \\cos(t) \end{bmatrix} + D_2\begin{bmatrix} \frac{\sin(t)+\cos(t)}{2}\ \sin(t)\ \sin(t)\end{bmatrix})
\end{equation}
with $C_1, D_1, D_2 \in \C$.

\newcommand{\R}{\ensuremath{\mathbb{R}} }
Filling in $t = 0$ gives
\begin{equation}
\begin{cases}
3C_1+\frac{1}{2}D_1+\frac{1}{2}D_2&=z_1\
2C_1+D_1&=z_2\
C_1+D_1&=z_3\
\end{cases}
\end{equation}
Solving it gives
\begin{align*}
D_1 &= 2z_3-z_2\
C_1 &= z_2-z_3\
D_2 &= 2z_1-5z_2+4z_3
\end{align*}
Since, $z \in \R^3$, we know that with $C_1, D_1, D_2 \in \R$. Plugging it in the general solution gives
\begin{equation}
u(t) = (z_2-z_3)e^{-2t}\begin{bmatrix} 3\2\1\end{bmatrix} + e^t((2z_3-z_2)\begin{bmatrix} \frac{\cos(t) - \sin(t)}{2}\\cos(t) \\cos(t) \end{bmatrix} + (2z_1-5z_2+4z_3)\begin{bmatrix} \frac{\sin(t)+\cos(t)}{2}\ \sin(t)\ \sin(t)\end{bmatrix})
\end{equation}
When $t \rightarrow \infty$, $e^{-2t} \rightarrow 0$, so the first term of $u(t)$ vanishes.

[untold delta]

FrankF

FrankF

[wind geode]

those with FA would use evans, usually grad school/masters level at least

[buoyant pike]

This is an ODE

#odes-and-pdes

[wind geode]

All z in the span of the eigenvectors of the matrix that have negative eigenvalue

[north tulip]

Did you conclude that from the last u(t)?

i.e. \begin{equation}
u(t) = e^t((2z_3-z_2)\begin{bmatrix} \frac{\cos(t) - \sin(t)}{2}\cos(t) \cos(t) \end{bmatrix} + (2z_1-5z_2+4z_3)\begin{bmatrix} \frac{\sin(t)+\cos(t)}{2}\ \sin(t)\ \sin(t)\end{bmatrix})
\end{equation}

[untold delta]

FrankF

[wind geode]

it's just a general theorem about ODEs of that form (1st order linear homogeneous)

[north tulip]

And if I want to prove that for this particular case how would I do so?

[wind geode]

but to prove it:

let v1, v2, v3 be the evecs of the matrix with evals L1, L2, L3