advanced-pdes

[wind geode]

v1 v2 v3 span R3 (check det of matrix is nonzero, that will justify this)

x_0 = a_0*v1 + b_0*v2 + c_0*v3

x(t) = a(t)v1 + b(t)v2 + c(t)v3

[north tulip]

by x_0 you mean x at time 0?

[wind geode]

yes

x(t) = a_0 * e^(L_1 * t) v_1 + b_0 * e^(L_2 * t) v_2 + c_0 * e^(L_3 * t) v_3

In general
$\dot{x} = Mx$ has solution $x(t) = x_0 e^{Mt}$

[untold delta]

robert_

[north tulip]

Yes, but where exactly does the initial condition play the role of it being stable or not? Because if you just look at the eigenvalues of M, the initial conditions will not affect it

[wind geode]

Let $v_1 , v_2 , v_3$ be eigen vectors of $M$ with eigenvalues $\lambda_1 , \lambda_2 , \lambda_3$. Write $x_0 = av_1 + bv_2 + c*v_3$
Then $x(t) = a * e^{\lambda_1 * t} * v_1 + b * e^{\lambda_2 * t} * v_2 + c * e^{\lambda_3 * t} * v_3$

[untold delta]

robert_

[north tulip]

That is this in this problem:
\begin{align}
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{align}

[untold delta]

FrankF

[north tulip]

I have already calculated the solution, but it does not help in determining the conditions for z for it to be stable because I don't know what z's would make e^t go to 0 as t goes to infinity

[wind geode]

Is that supposed to be the solution to this problem?

[north tulip]

yes, it is the general solution adapted to the initial condition

For how that solution is obtained, it is described here

and her

[wind geode]

Ah ok thank you, sorry I didn't read this more closely

Yeah for problems like this it's a lot clearer to leave the imaginary unit in the exponent as is

[wind geode]

OK so the answer is going to be anything spanned by [3, 2, 1]. Without assuming the relevant theorems already, you can prove it in a few ways. First check it's true for any vector in the span of [3, 2, 1]. Then check that for any nonzero vector orthogonal to [3, 2, 1], its magnitude (the absolute value) goes to infinity. The "trick" in this part should be that
$e^{a+i} = e^{a} e^{i}$
and $|e^{i}| = 1$

[untold delta]

robert_

[north tulip]

Okay, I'll try to prove it. How did you come up with the answer? I know that it is (3 2 1) is one of the eigenvectors but what is your thought process behind this idea?

[wind geode]

It's a very commonly studied form of ODE, most results proving stability work with it all the time. The hard part in these is always computing the eigenvalues which you did in your answer. The imaginary part of the eigenvalues doesn't matter if there is a nonzero real part. If there is a nonzero real part, then positive means it grows (or in the context of stability analysis it goes away from the steady state) and when you have negative real part in the eigenvalue that means is shrinks (attraction/stability of a steady state).

[north tulip]

So always pick the eigenvector(s) as the span of the initial condition for the solution to be stable whenever the real part of the accompanied eigenvalue(s) are negative?

[wind geode]

yep! with the caveat that if your eigenvalue has zero real part then you need to treat that specially. But so long as all eigenvalues have nonzero real part then you're golden. (Lotka Volterra equations are a famous system where you have zero real part eigenvalues and need to do a bit more work)

[north tulip]

Thanks! Any idea where I can look up the general proof for this? Is this a named theorem by chance?

[wind geode]

Hm unfortunately I'm not sure, I've seen some people give it different names because it's a common computation that gets rediscovered in different applications. Just googling I found this page which you would probably find a similar page in any book on ODEs like Boyce and DePrima for example https://eng.libretexts.org/Bookshelves/Industrial_and_Systems_Engineering/Chemical_Process_Dynamics_and_Controls_(Woolf)/10%3A_Dynamical_Systems_Analysis/10.04%3A_Using_eigenvalues_and_eigenvectors_to_find_stability_and_solve_ODEs

[wind geode]

I used the google search query "linear stability criteria eigenvalues" and found a ton

There's also some references in the relevant wiki https://en.wikipedia.org/wiki/Linear_stability

[north tulip]

Thanks, I'll have a look at it sometime. Furthermore, I noticed that if I just use the absolute value then for the latter two terms, it is of the form e^t*constant. So it does not matter which constant I have because it will always explode

[wind geode]

yep!

[north tulip]

But that contradicts with the span being (3 2 1)

[waxen forum]

@north tulip for future reference, ode questions should go in #odes-and-pdes

[north tulip]

My bad, will do in the future

[tired hollow]

can someone just send me the index notation proof for the identity of the gradient of the scalar product of two vector fields

what we're asked to prove in 11a

i wanna prove LHS => RHS not that both sides boil down to the same thing

[waxen bobcat]

#multivariable-calculus

[zenith saffron]

Does anyone have a favorite way of thinking of log sobolev inequalities?

[buoyant pike]

I simply do not think about them

[bitter yacht]

How is this (bracketed in red) a consequence of uniform convergence?
From uniform convergence I can show that, given any real $\sigma > 0$, for all $(x,t) \in \partial B$ I have
[ (u^{\epsilon_j} - v) (x,t) < (u^{\epsilon_j} - v)(x_0, t_0) + \sigma ]
whenever $j$ is large enough. But how can we from here make the leap that
[ \max_{\partial B} (u-v) < (u^{\epsilon_j} - v)(x_0, t_0) ]
for all sufficiently small $\epsilon_j$?

Context: page 581 in evans pde. u is a viscosity solution to the hamilton-jacobi-bellman equation and the sequence ${ u^{\epsilon_j}}$ converges locally uniformly to $u$.

[untold delta]

hardisc

[lilac barn]

My guess is that we pick (j) large enough so that ( \lVert u^{\varepsilon j} - u \rVert{\infty} < \rho) where ( \rho < \frac{\alpha}{2} ) for ( \alpha = (u - v)(x_0, t_0)- \max_{\partial B} (u - v)). In other words, pick (j) large enough so that the approximations of (\max_{\partial B} (u - v)) and ( (u - v)(x_0, t_0)) are very close, so that the inequality would be preserved.

[untold delta]

cocat

[bitter yacht]

@lilac barn: ah, interesting idea. i did not succeed with recovering
[ \max_{\partial B} (u^{\epsilon_j} - v) < (u^{\epsilon_j} - v) (x_0, t_0) ]
from
[-\rho < u^{\epsilon_j} (x,t) - u(x,t) < \rho ]
Thanks to uniform convergence, here we are free to control $x$.
but i am failing to see how this leads us to the desired result hmm.
I can set $(x,t) = (x_0, t_0)$ and subtract off $v(x_0, t_0)$, hoping to move in the right direction, but not sure where that leads me

[untold delta]

hardisc

[lilac barn]

Suppose a < b and an -> a and bn -> b. Can you show there exists some large N such that for all n> N: an < bn? The idea is the same here.

Point being if the uniform norm is twice less than the difference, then the lower term can upshoot only half the difference and the higher term can lowershoot only half the difference ensuring that the total difference still remains positive.

[bitter yacht]

@lilac barn: yeah I think I could, I'll give it a go. thanks for the input

[blazing ridge]

I have a question. Suppose we are and working in $(0,1)$ and we are trying to work out the norm of an operator in $X’= W^{-m,p}$. Then from functional analysis we know this would be given by $$ |u| := \sup_{|f|{X}=1} \langle u,f\rangle{} $$ where $\langle \cdot, \cdot \rangle_{} $ represents the action of a functional in $X’$ on $X= W^{m,p’}_{0}$. But I read somewhere that this duality action is equivalent to the $L^2$ inner product. Why is that?

[untold delta]

M6LI

[river path]

If you represent your functional in X' as a function, then what you mean by that function is "a functional that integrates things against this function"

@blazing ridge

So it's kinda true by definition that the action of a functional is given by the L^2 inner product. That's how we define what it means for a function to act on other functions as a functional.

Remember that elements of W^(-m, p) are distributions, and we know how distributions which are functions act on other functions: by integration

[astral vine]

Actually people should stop introduce negative Sobolev spaces as the abstract dual of positive ones, because it raises huge issues of definition no one adresses explicitly.

But I get that introduce properly negative Sobolev spaces on domains require a lot of work and so on, but hell we are doing maths not mumbo jumbo physics

And the real reason why it is consistent with L² pairing is a consequence of the construction, because L²(R ^n) is the actual reference space for duality even on domains.

That's why the dual of W^{m,p} on a domain is identifiable with W^{-m,p'}_0

Constructing stuff as abstract dual makes everything more difficult to grasp

[river path]

Yes, it's how negative sobolev spaces are best defined

[astral vine]

I swear to god, one day I will write a damn fucking book on Sobolev spaces, almost exhaustive, with actual good constructions and understandable.

Even if I am not a believer

[bronze gate]

how else do you define it lol

[astral vine]

The right way is the following

[astral vine]

@bronze gate

[untold delta]

Functionanatolysis

Functionanatolysis

Functionanatolysis

Functionanatolysis

Functionanatolysis

Functionanatolysis

Functionanatolysis

[astral vine]

fail on this one

no ' on the RHS

but here is how Sobolev spaces with negative orders should be truly introduced

and how negative Sobolev as duals are in fact a consequence of the construction rather than an (unusable) definition

In general people from applied PDEs and related applied area should stop taking duals of everything, without taking care, because sometimes it led to some illegal manipulations almost one noticed

[bronze gate]

I think ive seen this before, although I definetly didnt realize that it is not a good idea to think of H^-s W^p,-s et al as abstract duals

[astral vine]

Would you like to join my army to evangelize the world about the true definition of Sobolev spaces ?

[bronze gate]

🫡

[blazing ridge]

Thanks guys for the insight. So if I am trying to find the norm of some integrable u in a negative sobolev space (i.e. as an operator), the duality action reduces to the L^{2} pairing because we are considering u to be a distribution when we work in the negative sobolev space. But since this distribution is more regular (integrable) we know from distribution theory that this action is then equivalent to the L^2 inner product.

But what about if I have an arbitrary functional in a negative sobolev space which is not necessarily a function? Then I don't immediately see why the action reduces to $L^2$ one. My guess is that by the 'constructive' definition of negative sobolev spaces (i.e. elements of W^-m,p are just distributional derivatives of L^p functions), you can consider the action of a functional on a function to be equivalent to the action of a distribution on a test function. Then using distribution theory you can shift all the derivatives onto the function you are acting on which reduces you to the case where you have f acting on some g where f is in L^p. Then this action will be the L^2 inner product by my previous comment. Is that sensible?

in other words... if $f \in W^{-m,p'}$ and $f = D^{\alpha}f_{\alpha}$ then for $u \in W^{m, p}{0}$ we have $$\langle f, u \rangle = \langle D^{\alpha}f{\alpha} , u \rangle = (-1)^{|\alpha|}\langle f_{\alpha} , D^{\alpha} u \rangle = (-1)^{|\alpha|} (f_{\alpha}, D^{\alpha}u){L^{2}} = (f, u){L^{2}}$$

[blazing ridge]

This looks useful but I just haven't learnt about fractional laplacians yet so im a bit put off lol

[untold delta]

M6LI

[astral vine]

If you know Fourier transform, and the convolution by the Gaussian to solve the heat equation, that's in fact more than sufficient to make sense of the definition

Proving equivalence of definitions requires more work.

[viscid parrot]

Would anyone like to talk to me about solutions of PDEs where functional coeffecients may be non-continuous? For example, the heat equation over a rod [-L,L] with heat capacity a1 over [-L,0) and a2 over [0,L]. How do we know solutions exist? From a purely mathematical approach, I would figure solutions would not be continuous since theres a non-continuous function thrown in, but I think from a physical perspective this is totally realistic

For context, Im trying to study this reaction diffusion system $$\begin{align*}
\frac{dN}{dt} &= aN - \beta(x) NV \
\frac{dI}{dt} &= \beta(x) NV - \delta I \
\frac{dV}{dt} &= DV_{xx} + kI - \lambda V.
\end{align*}
where $x \in[-L,L]$, $t\in [0,\infty)$ and
[\beta = \begin{cases}\beta_1, &\quad x\leq 0 \ \beta_2, &\quad x > 0\end{cases}]$$ with all parameters positive

[untold delta]

deadpan2297
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[solid flint]

Hello, Anyone know about any documents about the stability of bounded Analytic semigroups? I read so far that iR^* is in the resolvent of its generator, so if 0 is also in the resolvent the semigroup must be strongly stable. Is there anything to say about other kind of stability?
Thank you in advance.

[astral vine]

not really in general if iR* is not in the resolvent everything could happen

there are several counter examples you can find here and tehre

[solid flint]

Thanks for your reply, in fact I was referring to this theorem

[astral vine]

yeah yeah

I misswrite my thoughts thinking about many things at the same time

[solid flint]

it is ok.
Do you know any farther stability results for analytic C_0 semigroups?

[astral vine]

I only "know" counter examples when assumptions of above theorem are not fullfilled.

[solid flint]

for the case when the pure spectrum isn't in the imaginary line?

[late hound]

i need df(A, b, λ) / dλ for a propagation of uncertainty

i have
g(A, λ, b) = (A - λ * I) * b
g'(A, λ, b) = -b
But i struggle with the divisor a bit

[nocturne wave]

why we can deduce this ? in evan pde page333

[karmic fjord]

what is 29-32 and 36?

[lilac barn]

From 36, use the knowledge that |a| = m+1 so the highest derivatives of f are of power m+1, so the first term in 36 is controlled by m+1 norm of f. For the latter terms, again using that |a| = m+1, the highest order derivative on u is of m+2, so we can again bound the whole expression by C x m+2-norm of u. Now use 32 to further bound the m+2 norm of u and conclude.

[nocturne wave]

And why we use zeta ^2 instead of zeta in the previous theorem of the book?

[lilac barn]

I believe, amongst other things, it helps you in invoking uniform ellipticity

[empty terrace]

If we have $\partial_t u = L u$ with $u|{t=0} = f$ we can write $u|{t=T} = e^{TL}f$ if we have some assumptions on $L$. Does this calculus have a name?

[buoyant pike]

Semigroups

[untold delta]

criver

[empty terrace]

That's a bit too general for me. The above for example is just spectral theory in the finite-dimensional case with a hermitian matrix L.

Ok, I found it, they just call it operator calculus in physics

[astral vine]

This is trace theory

being able to take values of a priori vector valued functions is always about traces

[empty terrace]

Seems way too general compared to what I am looking at

At least from the thing I saw on trace theory

I found a book where they call this type of calculus Feynman operator calculus though

[empty terrace]

Thank you. That was useful actually, I found this book from Engel and Nagel: https://www.math.uni-tuebingen.de/de/forschung/agfa/members/rana/

[late hound]

No help on that?

[buoyant pike]

#multivariable-calculus

[lilac barn]

Although, a bit imprecise, but this is the contents of the Hille-Yosida theorem. In fact, if you look in Evans, he basically proves it via constructing something like e^tA (which is a bit imprecise because he didn't talk about why does this term makes sense or why does integration on a Banach space makes sense). For a rigorous proof, consult Brezis' book.

[empty terrace]

Thank you!

[astral vine]

Actually, you can build it from Riemannian sums so it is not a big deal.

However in this case you cannot use Banach valued DCT

[late hound]

but its a partial differential equation??

I think i got it btw but not sure

Like that its better

[buoyant pike]

You are trying to take partial derivatives

Therefore #multivariable-calculus

[late hound]

Yes but its not "Early university"?

[dim meadow]

Regardless of what context this arose in, we sort questions into channels based off the actual content of the question

The "level of study" labels are approximations

Your question is about taking partial derivatives, hence it belongs in #multivariable-calculus , not in #advanced-pdes

[late hound]

but isn't pde just the short term for partial derivative equation?

and its advanced, so advanced partial derivative equation, or in short adv-pde?

[karmic fjord]

Partial Differential Equation* but your question isnt (directly) related to pdes..

[dim meadow]

If you don't know what a PDE is, you shouldn't be posting in #advanced-pdes .

The "equation" part is pretty important.

[late hound]

"In mathematics, a partial differential equation (PDE) is an equation which computes a function between various partial derivatives of a multivariable function."
Literally my equation, i don't understand your problems, prolly didn't even read my paper ._.

"If you don't know what a PDE is, you shouldn't be posting in adv-pdes ."
And i meant
"but isn't pde just the short term for partial derivative equation?
and its advanced, so advanced partial derivative equation, or in short adv-pde?"
Ironically...

[dim meadow]

"an equation [...] between various partial derivatives"
You are asking for one partial derivative.

In any case, you have been told repeatedly to move to the correct channel; I won't let you continue to lower the quality of this one

It makes this server less usable for everyone around you

The mute expires in 4 hours.

(I am not sure why you'd argue what a PDE is with people who literally dedicate their lives to studying PDEs...)

[late hound]

Ahhhh okay i see
i wont get help on this server bc ppl from multivariable-calculus tell me to ask in advanced bc no one understands it since everyone in the channel is like early university
But when i ask in advanced i get told i should ask in multivariable-calculus and won't get help here since it isn't "directly" related
I just get sent around between "not directly related" and "i should ask in advanced bc my equation is so complex", i feel like asterix and oberlix in the episode where they had to get the passport A38 :))

[lilac barn]

Missed opportunity for applied triangle inequality

[waxen bobcat]

Now, Manju can ask their question here

[waxen bobcat]

#advanced-analysis

[wintry prairie]

proof moderators can't read

[meager dune]

i think this is proof they do read

[hoary pumice]

mod potato

[winged stag]

Hi everyone, I was wondering how I would plot the nullclines/stability of the following system:
I know to set u_t and v_t to 0, but the spatial dispersion/diffusion is throwing me and I don't know what to do with it.

[buoyant pike]

Is D a constant or a derivative

[winged stag]

it's a diffusion coefficient

[buoyant pike]

Ok so a constant

[winged stag]

D, a and b are constants

[buoyant pike]

Well can you solve u_xx+a-bu+u^2/v=0 and Dv_xx+u^2-v=0?

[winged stag]

That's the thing, I'm not certain how to solve it, I'm sure I've encountered the tools needed, but the knowledge escapes me

[tired hollow]

bro channel name

[winged stag]

Yeah, I know that but I'm just looking for a first step. Forgive me if I'm stupid.

[astral vine]

assume you have a solution (u_0,v_0)

consider a perturbation of it (u_e,v_e)=(u_0+e . u , v_0+e . v)

and look at what is the linear limit system as e goes to 0

You can show that this reduce to investigate the spectral behavior of elliptic operators

[winged stag]

I think I get it, I was trying to throw finite difference methods into it - and getting lost in the process. I also have a feeling that to "plot" them I'll need to use MATLAB? or some equivalent coding software.

[untold delta]

Functionanatolysis

Functionanatolysis

[astral vine]

Write as a single linear system and look at the given matrix operator

stability properties will strongly depends on the sign of 2 Gamma_0-b

[unborn jackal]

Not specifically asking for help with a question but more on course choices. I see Calculus of Variations in the chat description so I'm guessing this is a good place to ask

What are potential red flags for someone considering taking this course whose generic Analysis skill is reasonably strong but without strong calculation skills to be aware of ?

This is the course content

FYI this is a masters course and the lecturers' research interest is in Lie Groups

I know they definitely cover Lie Algebras in this course but they're not explicitly listed here

I guess that's because it's so essential it just blends in everywhere

I'm a second year undergrad who gets the option to do some masters courses early
(I'm intending to take courses in Algebraic Curves, Measure Theory and PDE Theory early and am deciding between Hilbert Spaces and Analytical and Geometric Theory of Differential Equations for my last one)

[karmic fjord]

what's agde?

[unborn jackal]

AG DE is

[unborn jackal]

This

[unborn jackal]

Analytical and Geometric Theory of Differential Equations

Which ... includes Calculus of Variations

[primal loom]

ooooo

''''''''''''''''''''''''''''''''''''''

[buoyant pike]

This looks like a fun class

[hard wolf]

This looks interesting + dynamical systems are cool, but I would bet a lot of money you will get much more milage out of the hilbert spaces course

[unborn jackal]

On what grounds ?

[hard wolf]

that elementary functional analysis is useful for far more... one of those core subjects you kind of just.. need

[unborn jackal]

I am taking a Functional Analysis course later

That's more focused on Banach spaces though

[hard wolf]

You will see hilbert spaces more than banach

[unborn jackal]

Out of these I've seen Inner Product Spaces, the identities, Orthogonality, Gram-Schmidt, Orthogonal Complements, Dual Spaces, Riesz Representation and the notion of an adjoint in previous courses.

[hard wolf]

You have seen only a few of however many particularizations of the Riesz representation theorem there are

There are so many

Dual space takes several different meanings in different contexts

But the idea is roughly the same

[hard wolf]

This might look like a linear algebra description, but it is not

Functional analysis/ elementary theory to hilbert and banach spaces absolutely supersets anything you saw in linear algebra

At any rate you have two paths ahead of you

One is pretty cool but you will probably see that and never touch that exact area again unless you pursue it longer term

The other is not really all that sexy but fundamental

There’s no “one is better than the other”

Consider what you like, what else you’ll be taking (have some breadth), etc

[unborn jackal]

Other stuff I'm taking at Masters Level :

||Algebraic Curves
Measure Theory
PDE Theory||

Hence deciding here between

||AG DE (improves my breadth of knowledge as the lecturer introduces Lie Algebras and the course is different from the others I intend to do. I've done almost no ODE's in undergrad so this course would cover a pretty big gap in my knowledge)

Hilbert Spaces (which complements PDEs, Functional Analysis and Advanced Quantum Theory) ||

Later on :

||Number Theory (complements Algebraic Curves a bit, potentially useful for my Masters Project)

Differential and Geometric Analysis (somewhat of an odd one out, taking mainly for breadth and interest)||

And at the end I'm deciding between (choose 2) :

||Discrete Probability (cool material that complements my Graph Theory course)

Functional Analysis (complements PDEs and Hilbert Spaces)

Martingales (complements Measure Theory)

Advanced Quantum Theory||

Whichever of these two (AG DE and Hilbert Spaces) I take, I won't be able to take the other one because of timetabling.

I intend to learn the other course myself anyway, so I guess it partly comes down to which is likely to be easier to self teach.

[unborn jackal]

What I like : AG DE
Breadth : AG DE
Complements to
other courses : Hilbert Spaces
What I think I'll find easier : Hilbert Spaces

For the last point on difficulty I think there's not actually that much in it.

[hard wolf]

Yeah bruh take AG DE

You’ll get your analysis fix out of measure theory

[unborn jackal]

AG DE is also an Analysis course

[unborn jackal]

And PDEs

[wintry prairie]

your courses are really weird, are they shorter than usual courses?

Ive never seen a course that just does discrete probability, or just martingales, or just hilbert spaces

[unborn jackal]

We do them over 11 weeks with 3 contact hours a week (which are split into lectures and problems classes at the lecturers' discretion)

[wintry prairie]

interesting, that doesnt seem short

[unborn jackal]

Lecture notes for our course range from 30-90 pages depending on the lecturers' notes style and the content volume.

The 30 is a dense set of notes with a relatively small amount of very difficult content.

[unborn jackal]

Pure tends to be shorter

[hard wolf]

(Agreed but will bet he’s european + we’re american lmfao)

[unborn jackal]

She + UK uni

[blazing ridge]

bath?

these modules seem familiar lol

[unborn jackal]

I can neither confirm nor deny

[waxen bobcat]

#odes-and-pdes

[barren cargo]

should functional differential equations go in here?

because I solved this one today: f'(x) = f(-x)

but i need some guidance with this one: f'(x) = f(2x)+f(x)

[barren cargo]

I suspect f(x) = 0 is the only solution

unlike the first one, where i solved it and got a solution of A(cos(x)+sin(x)) for all values A

[astral vine]

#odes-and-pdes

[nocturne wave]

why partial U appears ?

evan pde page347

[buoyant pike]

What is theorem 1

[buoyant pike]

Ok do you understand why partial U appears here

[nocturne wave]

yes

[buoyant pike]

Ok do you understand how U and V are related

[blazing ridge]

Suppose you have an operator say $F : C^{\infty}(0,1) \to C^{\infty}(0,1)$ and a function $u$ which satisfies some homogeneous PDE $Lu = 0$. Also suppose that $F^{n}u$ also solves $Lu=0$ for each $n \ge 1$. Is there a name given to these sorts of operators, or the study of such function/operator pairs?

[untold delta]

M6LI

[buoyant pike]

Do you have any examples of F aside from the identity

[blazing ridge]

Hmm how about $L = \partial_{t} + \partial_{x}$ and $F = \partial_{x}$

[untold delta]

M6LI

[blazing ridge]

the domain should include time then of course

and F a linear bounded map

[buoyant pike]

I don’t think this pair is very interesting though

You have the solution f_0(x-t) in which case that’s the identity

x derivatives don’t do anything

[blazing ridge]

well what if you had F as something like $F= v\partial_{x}$ where v is some other function

[untold delta]

M6LI

[astral vine]

This phenomemenon is known as some action of invariant "lie" group

For a non trivial PDE

you could think about Non-linear Schrodinger with power non-linearity

and the multplication by a complex number of modulus 1

[blazing ridge]

very cool thanks, will look into this

[dim topaz]

soft question: is it generally more difficult in pde problems to show the existence of a solution, or its uniqueness? I'm writing an essay about pseudo-holomorphic curves where existence is the hard part, and I want to add some context: is this typical or not?

[astral vine]

There is no general principle

it very depends on the structure of the equations

The function space you use to build your function, and the function space you use for uniqueness

[dim topaz]

could you say more? by structure I assume you mean things like is the pde elliptic

[astral vine]

elliptic, hypoellitptic, parabolic

What kind of non-linearity

what kind of boundary conditions

everything plays a role in well-posedness

The chosen function space too

[dim topaz]

I see

thank you

[low turret]

If you wanted to study diffusion phenomena in hyperbolic PDEs with closed timeline curves and to avoid dealing with general relativity, you imposed time-periodic boundary conditions could you do the following? Wick-rotate the time dimension to a space dimension and study diffusion in parabolic PDEs with space-periodic boundary conditions and then Wick-rotate back to hyperbolic PDEs, giving you quasi-quantum diffusions? If so, where would this get you into trouble, as in the Wick-rotation would give you the wrong answer?

[bitter yacht]

I have seen Evans arguing the following several times:

$\phi$ is some expression constituting e.g. a PDE. We obtain $\int_U \phi v dx = 0$, and it holds for every test function $v \in C^\infty_0(U)$. From this he concludes $\phi = 0$.

For instance when he derives the Euler-Lagrange equation (8.1.2, bottom of page 455).

What exactly permits this conclusion? I asked my pde professor this, but right then and there he could not explain it. I suppose since $\int_U \phi v dx = 0$ holds for every test function, the idea is to choose the test function judiciously. But how?

[untold delta]

hardisc

[astral vine]

First, this is just an introductory calculation not very rigorous. For instance u and w are not well defined (i.e. chosen in a fixed and obvious function space like L² or H^{1}_0).

Second, assuming that everything is well defined,

[bitter yacht]

I managed right now to stumble upon 'fundamental lemma of calculus of variations' (e.g. https://www.ucl.ac.uk/~ucahmto/latex_html/chapter2_latex2html/node6.html), might be what I'm looking for

[astral vine]

if such say phi is L^1_loc and has an a.e. pointwise meaning

Then phi is 0 iff its integral against any smooth comapctly supported fucntion is 0

Works with continuous, and C^1, compactly supported, instead of smooth.

[bitter yacht]

Right, and I was wondering if there was an elementary argument to show this.
But judging from the link I found, it seems to require a bit more care

[astral vine]

A two line argument should be : an L^1_loc function is a distrbution

A distribution is 0 iff its action is 0 on all smooth compactly supported functions.

[bitter yacht]

And proving this is the crux of the matter I suppose.
If f were continuous, I suppose contradiction would be apt. We could say f > 0 in some open ball B and select a test function v>0 whose compact support is inside B. Then we would get a positive contribution to the integral, which is absurd. In essence, something along the lines of that?

[sand echo]

i am late but yes this is fine if f is continuous

the way to do it generally is via mollification

[bitter yacht]

interesting, I'll have to give it some further thought and return to it! thanks to both of you

[sand echo]

nwnw i can give u the whole proof if u need, its nicely written up in my notes

[bitter yacht]

yeah sure, if it is fine with you!

[sand echo]

the last line i use another result which is that if g is in L1, then p_epsilon * g -> g in L1

the proof of that is a little long tho but its elementary i think

and yeah the p_epsilon is just the usual standard mollifier

oh and obvs it should be = on the last line not a -> woops

[bitter yacht]

neat, thank you! what does it read where you write "set g = f ? u". Not sure what operation you have written in place of my question mark

[sand echo]

f * indicator of U

[acoustic minnow]

From my prof's PDE lecture notes, if you want details @bitter yacht

The actual proof is on the last picture.
Here, we don't assume that the closure of the open set is compact and instead use a compact covering

why am I using we

[sand echo]

makes mathematicians feel less lonely

[rare oasis]

lol

[karmic fjord]

so true

[bitter yacht]

hey thank you! i'll have a careful look later

[sour condor]

On section 7 of these notes, equation 89 represents conservation of momentum. I get R(X, 0) is a stand in for the mass, I get teh volcity, but what are the nabla monstrosity and g?

[buoyant pike]

nabla monstrosity

That's one way to put it

[sour condor]

Well why is tehre a superscript in there?

[exotic void]

May I say there can't be Carleman estimates for heat equation since propagation speed is infinite?

[buoyant pike]

So that you're taking derivatives with respect to the eulerian x coordinate and not the lagrangian coordinate

Or maybe the other way around

Oh yeah so that you're taking the derivative in the lagrangian coordinate

[sour condor]

I am still a bit confused about that. I am doing this in the context of MPM, what does it mean to take a partial derivative in teh Lagrangian sense? All you have are discrete particles, do you take differences of immediate neighbours?

[leaden ivy]

I'm having a problem here with a sligthly more complex version of the swimming dog problem. The task is given in a slightly different context, but it's completely analogous to the swimming dog problem where the dog swims twice as fast as the stream of the river, and does not have a given starting position. the position could be downstream or upstream and any distance across the river. And the task is to find the dogs distance from it's owner as a function of the angle of its trajectory

And I am very confused

Because in this task the dog could also be swimming directly parallel to the stream, straight towards the origin of the coordinate system, and thus never change the angle while changing the distance

[buoyant pike]

What

[leaden ivy]

You know the swimming dog problem? Otherwise I'll just rephrase it the way the task is given

We have an object A moving in a line at a constant speed. And an object B that is moving twice as fast (also constant speed), and always directly towards object A.

[meager dune]

Oh that one

I remember trying that years ago and getting stuck lol

[leaden ivy]

And from that I am supposed to find a function from the angle between the trajectories of A and B to their distance

How the hell

[buoyant pike]

Ok why is this in advanced pdes

[leaden ivy]

Should I put it in #odes-and-pdes ?

It feels advanced to me, since I'm getting differential equations that I can't solve

[buoyant pike]

You have odes don't you

[leaden ivy]

I had odes last semester, this is actually in a physics course

[buoyant pike]

The differential equations involved in this problems are odes are they not

[leaden ivy]

Well yes

I guess so

[untold delta]

Dwight Shelford

[bronze gate]

#odes-and-pdes

why is it that half the posts here don't belong xD

[rare oasis]

Maybe start by thinking about u_t=u^3 ? Never seen that

[royal nexus]

going to be very hard to solve over time

[bronze gate]

this is not #advanced-pdes content

[cursive sandal]

Guys I am really struggling to solve the system of equations \del_{x_i} f = f for i \in {1, \dots, n}.

It is technically a pde

plz halp

[charred bloom]

Here for each equation you can fix every variable but one to have an ODE and at the end you have an equation on the parameters you exhibited

[buoyant pike]

Have you tried solving this for n=1, 2 and then generalizing

[slender fulcrum]

are there any comparison principles for gradients? Im wondering if something of the following form is true: if $\Omega$ is a smooth domain, $\Delta u \le \Delta v$ and $u|{\partial \Omega} \le v|{\partial\Omega}$, then $|\nabla u|{L^2(\Omega)} \le |\nabla v|{L^2(\Omega)}$

[untold delta]

whzup

[teal harbor]

So I am trying to figure out how to use line integral of a 2nd kind to solve Biot-Savart law as an integral equation. I tired to multiply $d\vec{l}$ with $\vec{r_0}$ but I didnt got the answer that I should get. And I also wasnt able to find any help from google since they all use geometry and I kinda want to avoid that. Biot-Savart law looks like this (for those who dont know):

[untold delta]

MarsalMacola

[teal harbor]

$\vec{\vb{B}}=\frac{\mu_0}{2\pi}\oint_C{\frac{Id\vec{l}\times\vec{r_0}}{r^2}}$

[untold delta]

MarsalMacola

[buoyant pike]

Green’s theorem?

[teal harbor]

oh no my mistake
integral does not have to be closed

let me rewrite it

$\vec{\vb{B}}=\frac{\mu_0}{2\pi}\int_C{\frac{Id\vec{l}\times\vec{r_0}}{r^2}}$

[untold delta]

MarsalMacola

[teal harbor]

what I did before is just use line integral of the 2nd kind and multiply it with r_0 to get x,y and z component of that new field B then use parametarisation of a curve to make one integral out of that
but it does not work

[sour condor]

I was not sure which of the advanced channels this belonged to, i;s related to fluid dynamics so I will ask here.

in their MLS-MPM there is this claim that $D = 1/4 \Delta x I$

That matrix is defined as $w (\Delta x) (\Delta x)^T$ where $w$ is the evaluation of a quadratic B spline.

It's defined on section 5 of this paper:
https://www.math.ucla.edu/~jteran/papers/JSSTS15.pdf

I cannot for th elife of me figure out how they are getting that simplification

[untold delta]

Makogan

[pine oriole]

Nobody sees your problem here except that you are struggling with parametrizing a line integral

[exotic void]

@slender fulcrum I think it should be $\Delta u \geq \Delta v$. In either case you can set $v=3$ and $u=1\pm (x^2+y^2)$ on a disk as a counterexample.

[untold delta]

shiburin

[tardy pike]

Hi, am stuck on this exercice, i need to find the entropy solution to this problem. Could anyone give me indication or point me to a reference/book that could help me ? thx

[buoyant pike]

Leveque’s numerical methods for conservation laws

[teal harbor]

no, look at this. The easiest example is a straight line. So x(t)=y, y(t)=0 and z(t)=t, -a<t<a;
then x component of dl is: dl=x(t)'dt = 1 similar goes for y and z components. r_0 vector is just x, y and z. Vector product of those two vectors will give us x,y and z component of the vector field (also it must be normalized). Then r^2 = (x-x(t))^2 + (y-y(t))^2 + (z-z(t))^2 and all of that in integrated from -a to a
The point is that is wrong because I dont get the same result at the end as I should get it with other methods.
So I just want to know is my setup for this even correct at the first place

[silk eagle]

That's inviscid Burgers' equation. Do you know Method of Characteristics?

Try reading "Finite Volume Methods for Hyperbolic Problems .... something ..." by Randall LeVeque just a brief, specifically in the early nonlinear chapter that talks about inviscid Burgers'

[pine oriole]

You say you don't get the same results but we don't even see your results

Write in proper latex the whole thing

And this is calculus 3 problem, not advanced PDEs

You're having trouble with parametrizing and plugging things into a formula

[teal harbor]

I dont have it because people are so afraid to do it I do not have exact equation for every single point because its above my class
All I can do is to expand that line to infinity and then I will get formula that is similar but not quite right
this tangens will give me pi and that pi will cancel out with the one from Biot-Savart law and that wasnt supposed to happen

[karmic fjord]

Ask in #multivariable-calculus or something, not adv-pdes

[teal harbor]

I get $\vec{B}=\frac{\mu_0I}{2r^2}$ and I am supposed to get $\vec{B}=\frac{\mu_0I}{2\pi r^2}$

[untold delta]

MarsalMacola

[teal harbor]

this is integral equation lol

[karmic fjord]

this is not a pde...

[teal harbor]

uff read channel description

[karmic fjord]

you're plugging in a formula, not dealing with integral equations..

anyway #multivariable-calculus or #calculus is more suitable

[teal harbor]

this is literally Maxwell equation

[karmic fjord]

yes, but it's not an integral equation

Anyway, you can wait for an answer here lol

[teal harbor]

its literally partial differential equation in integral form 😭

[tired ether]

Still on topic for #multivariable-calculus or #odes-and-pdes

[pine oriole]

It's not. Otherwise calculus 3 is filled with integral equations

Only when you have the same unknown on both sides that it becomes such. This is just calculation

Anyway since you don't want to share the details (probably you plugged something wrong somewhere) we can only tell you to study multivariable calculus

[teal harbor]

I did exact same thing that I wrote on message up above. If its a problem then its my understanding of situation and my model is wrong not math
To be clear I thought all of this to myself since I didnt found anything useful on the internet about solving things in this way. Even in my textbook there isnt ANY example where you find a field in every possible coordinate in the space.
I though that someone did something like this and that someone would tell me where I'm wrong but looks like im always on my own lol
thanks anyway

[pine oriole]

Since you shared no calculations or details, yes you chose to be on your own

[teal harbor]

I literally wrote what I did up above

[teal harbor]

here

[pine oriole]

That's literally nothing

What results did you get and how do they differ

If you want to ask people at least give them complete info

That's basic courtesy. Don't expect others to be telepaths. How would anyone know?

[teal harbor]

you literally have that one integral 1/r that that will be $\arctan(\frac{t-a}{\sqrt{y^2+z^2}})$ from -a to a

[untold delta]

MarsalMacola

[teal harbor]

multiplied with bunch of constants

and resulting field is 0i -zj + yk

[pine oriole]

Write a pdf detailing your problem and your calculations, including the end result

Since you're asking for people to spot a problem in your calculations

[teal harbor]

not calculation, theory behind it

my calculation is right I even did simulation it looks like it should be

my setup is wrong somehow

[pine oriole]

You have to write the whole thing.

[dim meadow]

You've been told repeatedly that this isn't appropriate for #advanced-pdes

Your mute lasts 24 hours

Please listen to people next time

I can't understand people who come in to advanced channels and confidently tell people who spend their lives studying PDEs that they're wrong about what an integral equation is...

[solid flint]

Hello, I don't fully understand this proof.
I didn't get it why he is mentioning $X_1$? if $\lambda -A$ is injective then the inverse exists from $rang (\lambda -A) $ to $X$ and by the closedness of $A$ we can conclude. Can you explain it to me please?

[untold delta]

Mikahopff

[lilac barn]

Existence of the inverse has to do with injectivity on the image. The boundedness of the inverse however has to do with closedness of the range as we can then apply closed graph on c -A and so by inverse bounded (c-A)-1 is bounded.

[solid flint]

I didn't get what do you mean by the first statement ("Existence of the inverse has to do with injectivity on the image"), can you explain it more please?
I agree with the second statement, it is quite clear to me, my problem is that if the inverse exists (i.e. (c-A)^-1), then it is bounded if and only if the range ( range (c-A)) is closed. Then I don't understand why the space where the image of (c-A)^-1 lands matters as far as it is a Banach space and we can apply the closed graph theorem.

[acoustic minnow]

This exact problem was in the hw I handed in today lol.
The idea is that you have a shock wave at x=-1 and a rarefaction wave at x=0.
The both of them "collide" at t*=2. If you consider u(t*,x) as a new initial condition, you see that you have a shock so you have to apply Rankine-Hugoniot in order to find the equation of the shock line

[lilac barn]

By the first part, I just mean that inverse is a set-theoretic object and doesn't have anything to do with the boundedness of the map. So it can exist and we then establish the boundedness in other ways.
I am a bit confused about your question: are you asking why the last implication in the screenshot holds true or why is it needed?

The reason why we needed that the inverse be unbounded in X1 is because we want our approximate eigenvectors to be inside D(A). If the inverse is unbounded but only for elements outside D(A) that's of no use.

[solid flint]

Yes, I am asking why it is needed.

[solid flint]

Thank you, things are clearer now but I am still confused.
The first part of the proof seems to give all what's needed, i.e. the inverse is bounded from the rang of c-A to X1

[lilac barn]

How does the book defines "approximate eigenvalue" or sigma(A)?

[solid flint]

The approximate spectrum is the set of complex numbers c such that c-A is not injective or its rang is not closed

[lilac barn]

I see. So the reason they had to the full detour is that the first implication only says that the range of c - A, as considered as an operator from X1, is not closed. Whereas s(A) are points for which c - A is unbounded from X to X. So just knowing that inverse is bounded on X1 isn't sufficient, because if X1 were dense then the operator extends to the whole space.

Oh wait, you said the range is not closed.

[solid flint]

Here is the thing, since A is closed, then if the rang of c-A is closed, then it is a Banach space, hence from the closed graph theorem (c-A)^-1 is closed in a Banach space then bounded, this is no matter what's the arrival space, X or X1 as both are Banach spaces

Can we join the voice channel?

[lilac barn]

But that's the thing, I think they are changing the domain of c - A in the middle. If c - A : X1-> has closed range is a different thing then saying c - A : X -> has closed range

[lilac barn]

You can dm if you like, but I am not too fond of voice channels.

[solid flint]

Yup, and I am trying to understand that

[sour condor]

Is someone here familiar with Hongkai Zhao's fast sweeping algorithm?
https://graphics.stanford.edu/courses/cs468-03-fall/Papers/zhao_fastsweep1.pdf

I am trying to understand what exaclty it does but I am having problems.

[buoyant pike]

#numerical-analysis

[sour condor]

Ty

[fierce nexus]

i'm reading into perturbation methods, and a lecturer i've been following mentioned the equation $$Lq + Nq = 0$$ where $L$ is a linear operator, and $N$ is not. my question is what $q$ can be, as it was said it is a state variable, but i'm not sure what that is restricted to

[untold delta]

maximo

[fierce nexus]

most examples we've dealt with end up with q being something along the lines of [y,y']^T, or [u_t, u]^T, etc.
is there a right way to pin down what q can and can't be?

[frozen orbit]

I don't know much geometry so kind of struggling with this one. I tried expanding the expression X_aT^abY_b but I don't see how this relates to the time-like condition

I think the first part of the expression contracted with X, Y gives you Xu * Yu, but then I have no idea how to show this is comparable to grad(u)^2. I'm assuming this is why we want both fields to be forward or backward but idk how that translates to Xu * Yu >~ grad(u)^2

[river path]

The time like condition will come up in the second term

Where you end up with -1/2 g^ab X_a X_b del^g u del_g u

Oh ok I see what you mean about showing comparability

But because of negativity of 1/2 g^ab X_a X_b (which ill call -c) this becomes Xu * Yu + c |grad(u)|^2, with c > 0

So both terms are pos def

[rancid tusk]

I'm looking at my notes about using fourier to solve HEQ cauchy problem, I start with
$$u_t - D u_{xx}=0$$
\newline $$u(x,0)=\phi(x)=0$$

[untold delta]

KooKoo

[rancid tusk]

and then, I say that $$\hat{u_t}(\xi,t) - D \hat{u_{xx}}(\xi,t)=0$$

[untold delta]

KooKoo

[rancid tusk]

what does the hat actually mean? I know I did the fourier transform, does $\hat{u_t}(\xi,t)$ mean $\int_{-\infty}^{\infty}e^{i \xi x}u_t dx$?

[untold delta]

KooKoo

[frozen orbit]

I guess im wondering how we get -1/2 g^ab X_a X_b in the second expression, shouldn't it be something with both X and Y?

[river path]

Well, to check positive definiteness you plug in the same vector twice

[frozen orbit]

oh uh

then why does it matter whether Y is timelike?

sorry maybe I dont really understand what it means to be positive definite here

[unborn quiver]

Yes

Working with the fourier transform is nice because it diagonalizes differentiation, or rather, if you view a linear differential operator as a polynomial (in this case $p(x)=x^2$ because you have a $p\left(\frac{d}{dx}\right)u=\frac{d^2}{dx^2} u(x,t)$ term), you see
[\widehat{p(d)u}=p(i\xi)\hat{u} ]
through integration by parts

[untold delta]

Kirbanach-Alaoglu

[unborn quiver]

Reduces the problem of solving u_t - Du_xx=0 substantially

[rancid tusk]

huh

yeah I am giga lost lol

[unborn quiver]

Okay so I just explained the motivation there

[untold delta]

Kirbanach-Alaoglu

[river path]

A matrix Q is called positive definite if X^t Q X > 0 for all nonzero X. Then we call the function f(X, Y) = X^t Q Y a "positive definite quadratic form" (if Q is any other kind of matrix, it's just a quadratic form). What this is saying is that, as a function on the set of all timelike vectors, f(X, Y) = X^t T Y is a positive definite quadratic form, i.e. that T is a positive definite matrix on the vector subspace of timelike vectors.

So Y being timelike here is just us making sense of the fact that as a function of two vectors T is still called "positive definite"

It's kinda like how the dot product of x and y doesn't need to be positive, but the dot product of x and x needs to be positive as long as x is nonzero. We call the dot product a "positive definite function" even though, e.g., x dot (-x) is negative.

[frozen orbit]

okay this makes perfect sense now tysm

[solid flint]

Hello, Can anyone tell me know can we apply the Cauchy's integral theorem to get this formula?

from the resolvent equation I could get the following :
$$R(\lambda, A)= \frac{R(\lambda_0-\lambda) R\big( (\lambda_0-\lambda)^{-1}, R(\lambda_0-\lambda) \big)}{\lambda_0-\lambda}$$ but I don't see how can the Cauchy's integral theorem be applied here.

[untold delta]

Mikahopff

[solid flint]

The formula (1.2) is indeed proved my elementary multiplication. but the other one I don't see how can we get it? Even the authors of the book said that Cauchy's integral theorem is used. Do you see how?

[pine oriole]

Nevermind I misread. You're asking about the second picture

Not the first picture

@solid flint

Still elementary

Implicit assumption in this notation is that stuff commutes. Easily proven

[solid flint]

thank you a very much, it seems that we didn't use Cauchy's integral theorem to get it. it is confusing.

[opal vortex]

Hi, when calculating uxy i keep getting the answer in blue, have i missed anything out here? thanks

[stark thunder]

In the "generalized" Harnack inequality (screenshot from textbook):

Is there a counterexample for when Omega' is not connected?

[quaint herald]

how about simply take

\Omega to be the union of two disjoint bounded open sets

u to be 0 on one of them but nontrivial on the other

and \Omega'=\Omega

[stark thunder]

trueee

I'm a dumbass I didn't think of zero lmao

I guess now I need to think of a harmonic function which is zero in some open set and nonzero elsewhere

Because I can't really see how you'd have an open set of zero be in a harmonic function

Like, if u: Omega -> R
u != 0
and G open subset of Omega where
f: G -> R
f = 0

[quaint herald]

just define it piecewise. if the two open components of Omega are A,B, let f=0 on A and f=your favourite nonzero harmonic function on B (e.g. a linear polynomial if we are working on R^n).

The piecewise definition is fine because it is still smooth everywhere in its domain and satisfies Laplace's equation.

(for any B you will have to choose your linear function suitably to get your desired non-negativity on B, e.g. if B was supported in the upper half plane of R^2, then f(x,y)=y would be good.)

[stark thunder]

You've created components for Omega, we need to create Omega' they are different.

For example: if we are working in Omega=Rn with u harmonic on Rn. We then need to find A, B subsets of Rn where u(A) = 0 and u(B) != 0. But I don't think that's what you've found exactly.

For example: Your piecewise function on Omega will have a kink and since all harmonic functions are smooth, this doesn't work

Or am I fumbling

Ohh

\Omega'=\Omega

in what you said at the start

lol

okay

[quaint herald]

In my example, yes.

"yes" to your "lol okay", to clarify :p

[stark thunder]

okay makes sense

@quaint herald Will there be examples where Omega' is contained in Omega (without touching the boundary or something)

Basically considering "non-boring" cases

Because tbh, that's more what I was trying to figure out originally

Oh wait

everything still works

I just take Omega' slightly contained in Omega

lol

[quaint herald]

yeah

[rocky ridge]

P

[buoyant pike]

?

[blazing ridge]

So i've seen some people write things like $f \in L^{2}(0,T; X) + L^{2}(0,T; Y)$; what does the + mean? is it the same as intersection?

I also saw somewhere $f \in C(0,T; X- Y)$

[untold delta]

M6LI

M6LI

[blazing ridge]

more specifically it was f \in C(0,T; X- \sigma(X,Y)) so my guess its to do with the weak topology

[astral vine]

f belongs to the sum space X+Y off there exists a couple (a,b) of XxY such that f=a+b

The norm on the space X+Y is the infimum of the quantity IIaII_X +IIbII_Y over all such decomposition (a,b) of f

[blazing ridge]

Aha I thought it was something different because i couldn’t find any pair associated with f but actually I realise now there is

Thank you

[nocturne wave]

why possion kenrel for the ball is zero on x_n=0?

[buoyant pike]

x_n=0?

Are you confusing the poisson kernel for the ball and the poisson kernel for the upper half space?

[blazing ridge]

simple question... does periodic boundary conditions always mean that the functions AND their derivatives have to be equal at the endpoints?

i guess its a matter of definition. ive read a lot of papers where people take $\Omega = \mathbb{T}$ the torus and this is apparently equivalent to studying the problem in a domain like $[-L,L]$ and imposing 'periodic boundary conditions'. But i've never seen anyone mentiion precisely what that means.

[untold delta]

M6LI

[buoyant pike]

Well it depends on what order PDE you are solving

1st order = function values agree

2nd order = function values and first derivatives agree

And so on

[blazing ridge]

why would it depend on the order of the pde and not the regularity of the solutions involved? If we have a second order pde and work with solutions in say H^{3} for example why would we not require all three derivatives to match at the endpoints?

er two i should say since 3rd may not be defined pointwise

[buoyant pike]

Well you need enough boundary conditions for the pde to be well posed right

And you can't have too many or else you don't necessarily have solutions

[rancid tusk]

Why aren't there pde calculators? I'm thinking and there's derivative calculators, integral calculators, ode calculators

What about pdes makes it so you can't create software that does them for you

[unborn quiver]

This is a very big question. So even a lot of ODEs don't have closed form solutions, and there's still an entire field devoted to behavior of ODEs in higher dimensions.

The problem for PDEs is that the solutions largely don't come in closed form, and if they do, there is case by case exploiting of structure of the PDE (Fourier, separation of variables, etc.) I don't know the inner workings of a software like WolframAlpha, but even then, it's not going to be practical.

Maybe you're talking about numerics? Even then, there are plenty of reasons to ask if your numerical solution is even correct because of error picked up by roundoff error, or even if your PDE solver is stable enough to give you a solution (in even first order 1d nonlinear pde, you can have shocks that result in blowup)

I'm sure others can give more precise arguments, but this is generally what I see

[blazing ridge]

Ah right yes because otherwise it could be underdetermined or overdetermined , thx

[mint canyon]

I have a method of characteristics question for second order PDEs. If I find a characteristic of the PDE and it intersects with a boundary surface, suppose the region of interest is hyperbolic and the boundary is hyperbolic, but between those 2 areas, the characteristic curve passes through an elliptic region. Does this invalidate the solution in the region of interest?

[untold delta]

mjachi

[mint canyon]

Has anyone seen this paper? https://math.uit.no/seminar/Preprints/04-06-BKVL.pdf

Looks like it came out in 2004~ish, Im wondering if it stood up to scrutiny

Lagrange-Charpit for systems of PDEs

[untold delta]

shiburin

[exotic void]

nvm I am dumb

[buoyant pike]

You’re welcome

[solemn viper]

Can someone help me visualise a PDE? Also visualise how the lagrange-charpit equations really work. (in the context of solving quasilinear PDEs) I already know they use characteristic curves along which the PDE reduces to an ODE but I wanna know how exactly the auxiliary equations achieve this.

My prof told me the solution f(u,v) where u and v are functions of x,y and z (z is also a function of x,y) gives a curve which on varying gives a surface which is the solution to the PDE

My objectives are to visualise a function of the form f(x,y,z) where z is a function of (x,y) -> visualise a PDE -> visualise the lagrange-charpit aux equations and possibly -> visualise lagrange-charpit equations for non linear PDEs

[buoyant pike]

For those not familiar, Lagrange charpit is just the method of chatacteristics

[high mulch]

This might belong in #odes-and-pdes

What do u mean by aux equations

The method of characteristics tries to reduce the differential equation by choosing (possibly lower dimensional) patches of the solution surface along which the differential equation is (in practice) algebraically simplifying

If you introduce one new variable which is implicit then you are looking at curves that go along your solution surface to the DE. If you introduce two new implicit variables then you may now be looking at surfaces that go along your solution surface to the DE; I think of these surfaces as being built up of new curves which are dictated by the introduced implicit variables u and v. After applying initial conditions and doing a routine check on existence uniqueness and cts dependence on initial data, this is where the method of characteristics ends.
Looking for specific solutions in the set of all possible solutions with this method would go into other fields which may care about finding the solution in the set of possible solution subsurfaces that is minimized with respect to some kind of curvature

[solemn viper]

This is the equation for quasilinear cases, I'd like to understand how this equation is reducing the PDE to an ODE along some curve

And this is for non linear PDEs

[solemn viper]

That's really interesting, can you give me a more fundamental explanation as to how introducing a new implicit variable can be thought as a curve along the solution surface?

I also lack the visualization of a PDE, I can do it for simple cases like such as the 1D heat equation but in higher dimensions I dont know how to visualise it.

Once I get that down, I think I'll be able to understand how the solution surface works and how its the result of varying the implicit variables u and v

[high mulch]

z=f(x,y) is a surface. z=f(x(t),y(t)) is a curve in R^3. z=f(x(u,v),y(u,v)) MIGHT be a curve in R^3; it also has a chance of being a (SMOOTH!) surface in R^3 because you have two new variables, which is the same number of linearly independent variables you originally had (space filling curves are NOT smooth!)

[rancid tusk]

When my professor writes $L_2(0,L)$, is this a set of functions? I know L2 is a metric

[untold delta]

Avina

[rare oasis]

He likely means the square integrable functions defined in the [0,L] interval

It is technically a set of equivalent classes where two functions are identified if they differ only in a null measure set

[river path]

What sydd said. There's the space L^2(0, L), which is a vector space that has a norm on it called the L^2 norm: ||f||_2. And the L^2 metric is given by d(f, g) = ||f - g||_2.

It's a space of functions: specifically, the space of functions for which ||f||_2 is finite. (So, functions whose square is integrable)

[rare oasis]

Given a subset S in RN, what conditions on S make the heat equation well-posed for any initial data?

[astral vine]

what do you mean by any initial data ?

Heat equation with what kind of boundary condition ?

[blazing ridge]

On this topic.... suppose we have the linear heat equation with variable diffusion coefficient $\partial_{t}u - a(x,t)\partial_{x}^{2}u = f$ on $\mathbb{R} \times [0,T]$ with $a(x,t), f(x,t)$ smooth and initial data $u_{0}(x)$ smooth as well. Is there any reference where I can find out about the existence of solutions to this problem? I know that in the bounded domain Evans proves existence of smooth solutions via a galerkin scheme but I can't find anything on $\mathbb{R}$

[untold delta]

M6LI

[exotic void]

A possible argument would be (though you may need some control at infinity, but don't quote me on this): solve this equation on bounded domain and do some a priori estimates. Now take a subsequent limit as this bounded doamin approaches infinity to obtain a solution. Ladyzenkaya had written a monograph called "Linear and Quasilinear Equations of Parabolic Type" which has all sort of general results

[rare oasis]

Let me rephrase it, take a two dimensional compact smooth surface for now, so I wouldnt have to worry about boundary conditions, just an initial function u(x,y,0)=f(x,y), right?

Also, just thinking about the usual linear problem with constant diffusion coefficient

[astral vine]

what does it mean to you "any initial data" ?

[rare oasis]

My question is: is this problem well-posed for f in H^s for every real number s?

[astral vine]

okay

Yes.

The heat flow preserves at least H^{s} regularity

even more it preserves H^{s,p} regularity

This should also depends on your deifnition of the Laplacian on your compact surface this won't give you the exact same heat equation but yes

(there are various different kind of Laplacians)

[rare oasis]

The way Im thinking is the Rn Laplacian restricted to the surface

Oh

[astral vine]

This is ill posed then

You need boundary conditions

[rare oasis]

Like what?

[astral vine]

But the intrinsic Laplacian on say S² (the surface of the unit ball of R³) (which is somehow the euclidean Laplacian + a correction term) doesnot encounter such troubles

[astral vine]

Dirichlet, Neumann, Robin, Lopatinskii-Shapiro

etc.

A Laplacian depends, as an operator that generate a flow, strongly on the open set/manifold you consider and the related boundary conditions

You need to have a "suitable operator"

on compact manifolds (without boundary) "the boundary condition" is encrypted in the manifold itself and the chosen Laplacian

Like on the Torus T^n, periodic boundary conditions, etc.

[rare oasis]

Yeah I never seen this intrinsic way of doing EDP on surfaces, its probably very pratical

[rare oasis]

This correction term depends on the geometry of the surface?

[astral vine]

exactly

Pruss (Rip), Simonett and Wilke did a paper about Navier-Stokes on surfaces

in particular S²

[rare oasis]

What I was wondering is if I can lose this property for some weird domain geometry

Like what if my surface has a corner or something? Any hope?

[astral vine]

Yes obviously this could become false

this is already the case for euclidean Lipschitz domains

Again it depends on the Laplacian you chose

for Dirichlet this should holds true for H^{s}, -3/2<s<3/2, with some weird stuff at s=+/-1/2

[rare oasis]

But is there some canonical choice? If that correction term you mentioned is unique for instance

[astral vine]

The correction term for comapct surface of R^3 is unique in general

[rare oasis]

Oh nice, that was what I was thinking

[astral vine]

For the Hodge Laplacian on k-forms, this could become false too

-1/2<s<1/2

only

etc.

[rare oasis]

Is this all like in Evans book? Or should I look at some geometry book to find?

[astral vine]

I do not know an explicit book /reference about it it

it cames through reading papers here an there

And this is obviously NOT in evans

[rare oasis]

Lol ok

At least my question is non trivial I guess

[astral vine]

Heavily non-trivial

[rare oasis]

Ill look it up

[astral vine]

For what happens for Dirichlet Laplacian on lipschitz domains, which aren't surfaces

Jerison and Kenig 95, Ian Wood's paper (don't remember the year), Fabes Mendez Mitrea (1999 afair)

For the Hodge Laplacian on non smooth riemannian manifolds (with poorly regular metrics), Mitrea Mitrea Taylor 2001

But that's a lot of Harmonic and functional analysis with weird notations

[blazing ridge]

i tried to do this (study the eqn on [-L,L]) but my a-priori estimates seem to blow up as L \to \infty, because the RHS has the H^2 norm of a which may become arbitrarily large as L \to \infty. I have the LZ book so I will try to see if i can find the result there

I feel like studying the limit L \to \infty should work though..

i tested the equation with u to get the a-priori estimate

[blazing ridge]

I should've mentioned that for the problem on [-L,L] I am working with u = 0 on the boundary, and a is bounded uniformly on whole real line,. But actually i think the a-priori estimate does not blow up as L \to \infty. And I think what you can do is test the equation with higher derivatives of u and show by induction that the derivates of u will have norms which remain bounded as well.

what I am now wondering is how one would rigorously argue the existence of solutions on the whole real line R from here. What I would like to say is that if you take the limit as L to infty in these estimates then u remains in these spaces. but is that really enough to show the existence of solutions to the problem on the real line?

[exotic void]

The a priori estimate I was thinking is Shauder estimate (and for C^0 control on solution you use comparison principle etc), so if it happens that your data is C^1 bdd (and a(x,t) is uniformly elliptic) then I think it's fine. (but maybe that's an uninteresting case)

[blazing ridge]

my data meets these requirements i was assuming smooth data

but for less regular data i was thinking that using the density of smooth functions would work, but that is tangential to my question. I will have a look at Shauder estimates, thanks

[astral vine]

@blazing ridge In general non-autonomous paraoblic equations are very difficult to solve (in the sense of proving GWP, stability and stuff) and you need several assumptions.

on a(t,x)

The most used trick is to consider some regulairty in t, and x as well as some mixed integrability

then to consider the operator L_t := -a(x,t)D²_x, t being fixed here

so that for any t_0, L{t_0} generates a semigroup

see eg.

https://arxiv.org/pdf/2208.02527.pdf

(this is for divergence form systems, but the idea is similar)

[rare oasis]

One of these days my prof said that one of the nice things about the KdV equation is that it has “infinite conserved quantities” and I also heard it is “completely integrable”. Even though Im unsure about the meaning of these concepts, are they in any way equivalent? Or does one imply the other at least?

Also, what is the simplest possible example of a conserved quantity?

Hopefully Im not butching any of these terms…

[junior bloom]

look up the virasoro algebra for kdv

idk what completely integrable means wrt just integrable

i dont do integrable systems

yet

allah willing ill learn something about it

[exotic void]

You probably have figured this out already, but what I meant is you can pick subseqences thanks to Arzela-Ascoli and do a diagonal argument (and this is why a uniform estimate and assumption like C^{0,1} bounded are needed, but it may be too restrictive)

[blazing ridge]

that's interesting because i always thought it was the opposite, especially since it is a linear equation. And when I look at some sources like Ladyzhenskayas book, it seems that very general results for linear/quasilinear parabolic equations are known. Though I didn't look at the assumptions very closely so it may be that all these results have strong conditions on the coefficients.

[blazing ridge]

yeah i have to think about it a bit more but i guess im just a bit confused on how you can be sure that the limit satisfies the equation on R with a vanishing far field condition. so suppose you have a solution u_{L} to the system on [-L,L] for any L > 0 (with u_{L} = 0 on boundary) and somehow you find that you can bound the norms of u (lets say H^2 for now) independently of L. then you can apply AA to find that there exists a subsequence converging strongly to a limit u. but can you argue from this alone that the limit satisfies the eqn on R? i feel like there is some detail missing

I hope im making sense i might be mixing ideas up. I think the confusing part is that each member of this sequence is defined on a different domain

[astral vine]

"of course", L² in time-space well posedness is in general "easier" to prove than any other setting.

(actually not that much, but sesquilinear form technics requires not that much additional assumptions)

[exotic void]

Each u_L restricted to a smaller set (say [-L+1, L-1]) still satisfies the equation in the interior (even you impose different boundary conditions on some [-L, L], but let me emphasize the boundary condition at infinity sneaks in when you do the uniform estimate) and each higher derivative of u_L converges to derivative of u so it satisfies the equation as well. By the way, if I recall correctly the "method of frozen coefficients" mentioned by Functionanatolysis is related to method of continuity (comes in handy sometimes).

[rancid tusk]

my professor wrote that if $u_t-Du_{xx}=0$, then $(u_t,u) - D(u_{xx},u)$ also must equal 0. I'm confused what $(u_t,u)$ actually denotes, though

[untold delta]

Avina

[rancid tusk]

I know what u_t and u are but what does the pair of them together mean?

[wintry prairie]

inner product?

[buoyant pike]

Yes inner product

Some people use (-,-) instead of <-,->

[rancid tusk]

oh, inner product

ok

[verbal nebula]

Usually the angle brackets are use (by some) for finite dimensional vector spaces while the round parantheses are used for infinite dim'l inner prodcuts, or L^2

[junior bloom]

Hmm maybe this is obvious but I'm not seeing it. Does anyone know if I can get integration by parts formulae for pseudodifferential operators, finite difference operators (still under continuum integrals and not sums), and operators that look like $T_kf(x):=\sup_{|\xi|\leq1}\lVert\xi\rVert^{-k}\lVert f(x+\xi)-f(x)\rVert$? I'm just messing around with concepts of weak solutions where the usual derivatives don't get me what I want.

[untold delta]

teafortwo

[blazing ridge]

right that makes sense, my previous question was about smooth solutions of course

[blazing ridge]

cool I think I have a better idea about how to complete it now

Thanks guys

[astral vine]

Smooth solutions are constrained by the L² setting in general, and relies on operator theory and classical ODEs regularity theory

[gleaming lily]

So we introduce the green's function to get a general form for the poisson's equation on an arbitrary domain, but how easy is it to solve for the green's function?

[buoyant pike]

Not easy

There are some tricks you can do to get a green's function for some nice domains (balls, half spaces) but in general finding the green's function is as hard as solving the associated pde

[gleaming lily]

Ah ok

But at least only the domain is arbitrary, we don't need to deal with an arbitrary function anymore right?

The arbitrary boundary condition

Which is why this is better

[buoyant pike]

By the conservation of difficulty, it is still hard

[astral vine]

In general a way to obtain integral representation, is to use the whole space Green function and a correction boundary term (called boundary layer) to obtain your solution.

The correction term on the boundary with a full closed form is the hard part to get.

[gleaming lily]

I feel like there are probably better bounds for C_k than this

Because we are bounding k/(k-1) <= 2

Which in the limit is very bad

Is there a better version of this?

[hollow venture]

i suspect you could just chase down evans proof to get a sharp yet ugly bound

i think

equality would probably hold for some radially symmetric harmonic function

[buoyant pike]

This screenshot is from Evans

[rocky mountain]

what’s so bad about k/(k-1) <= 2?

[sand trout]

its not sharp obviously

analysts want to cut stuff

[rocky mountain]

it literally is sharp, take k=2.

[sand trout]

but its not sharp for other k

thats the paradox of analysis, the sharpest inequality is junk <= junk

[rocky mountain]

it’s not sharp asymptotically, but surely the content here is that there’s an O(1) bound?

[sand trout]

same junk on either side

yet they insist on bounding things by other things

[rocky mountain]

certainly k/(k-1) > 1, so even if you try to sharpen, say by establishing a bound 2 > c >= k/(k-1), in that argument you’ll still pick up a factor of the form c^{n+k-1}.

anyway this seems inconsequential given the 1/alpha(n) term…

[hollow venture]

yea unfortunately i recognize it

[buoyant pike]

[astral vine]

Be more precise pls, may be I will confess my love to you

[hollow venture]

tbf ive only worked through evans ch 2 lol

im too lazy to work through the rest but

i have some pde problems im interested in so

someday

somedayyuy

[astral vine]

I am mostly concerned with the "unfortunately" part

[hollow venture]

unfortunately i have read evans chapter 2

unfortunately because i used to do algebra

and im making a massive 180

[buoyant pike]

[modern stream]

Is there a way to describe solutions to the set of PDEs df/dx_i = g(x1, x2, …, xn)xi for all i in Nn. Like find some solution of f(x1, x2, …, xn)

[hollow venture]

my jus woke up brain be like

[df = g \sum_ix_i dx_i=gd(x^2)]
[0=dg\wedge d(x^2)]
which is usually nonzero so you cant find a solution for most g

[untold delta]

ari 亲

[hollow venture]

ima head back to sleep brb

[hollow venture]

ok now that i have a new working brain cell

it also means that

[g(x_1,x_2,\dots,x_n)=g(x^2)]
[df=\frac12 gd(x^2)]
we now know that $f$ is a function of $x^2$ too so we can write

[\frac{df}{d(x^2)}=\frac12g(x^2)]
[f(x)=\frac12\int g(x^2)d(x^2)=\frac12 G(x^2)]

smt like this where G is the some function with G'=g

[untold delta]

ari 亲

[gleaming lily]

Why do we think of a solution to the heat equation to be physically correct if it satisfies the growth condition |u(x,t)| <= Ae^{-a|x|^2}?

[quaint herald]

Well for physics reasons (conservation of energy), we expect solutions to the heat equation that actually model the way heat flows through a solid to obey a conservation law, in particular the L^2 norms over any timeslice should be the same.
You can show formally this is the case by using the heat equation and integrating by parts, but to rigorously justify this you need to impose some sort of growth conditions at infinity. It turns out this isn't just pedantry, because if you DON'T impose such conditions, there ARE other solutions that are quite wild.
For example if you start with u and its time-derivative being 0 at t=0, there is an obvious trivial solution u=0, and conservation of L^2 norm suggests this should be the actual physical behaviour. However, there are other solutions to the heat eqn with these ICs that grow rapidly at |x|=infty if you do not impose some condition ruling this out (look up Tychonoff solution).

You can get away with much milder growth conditions than the one you mention btw, the wild solutions really are quite wild. But anyway, the point is you need to work in a class of functions when your growth at infinity is constrained if you want uniqueness, because of the example mentioned above.

[gleaming lily]

Oh ok

But does the growth condition we put has any physical significance?

It just that it happens to make everything work

[quaint herald]

well that particular one you put is the actual decay rate of the fundamental solution to the heat equation, so it is not a completely randomly chosen bound. you couldn't demand any more decay of a nonzero solution.

but yeah even much milder restrictions will single out the one "physical" solution given any particular ICs, so don't worry too much about the actual bound here.

[gleaming lily]

Ah ok that makes sense thx

Is there a way to rigorously justify this using distributions?

[quaint herald]

yep, as usual with distributions you can do so by pairing with test functions and "integrating by parts"

[gleaming lily]

(Also I don't really know distributions yet, my friend only explained to me intuitively what they are)

Where can I read about it?

[quaint herald]

working right now and meeting shortly so won't elaborate, but you should learn distribution theory properly at some point, it is not really covered in Evans beyond his discussion of stuff like weak derivatives

[gleaming lily]

Cause it seems like distributions aren't part of evans

[quaint herald]

yep they aren't. a self contained and quite nice book is friedlander-joshi: intro to the theory of distributions

also volume 1 of Hormanders 4 volume series is excellent, and much gentler than the subsequent books

[gleaming lily]

👍 ok thanks

[gleaming lily]

Assuming $u,\tilde{u}$ are both solutions to heat equation on a domain $U$ with same value at $U\times\brc{T}$, $\partial U\times[0,T]$, then we want to show that $u=\tilde{u}$ for all time between $[0,T]$. Evans defines the energy function $e(t)=\int_Uw(x,t)^2\dd x$ where $w(x,t)=u(x,t)-\tilde{u}(x,t)$, then it claims that $\ddot{e}(t)=-4\int_UDw\cdot Dw_t=4\int_U\Delta ww_t$, I don't see how the second equality holds

[untold delta]

Whoever

[gleaming lily]

So Green's Theorem gives that $\int_UDv\cdot Du=-\int_Uv\Delta u+\int_{\partial U}v\pdv{u}{\nu}$ where $\nu$ is the outward normal, but where did the $\int_{\partial U}$ term here go?

[untold delta]

Whoever

[unborn quiver]

oops I wasn't in the right channel

[gleaming lily]

Wait how I don’t see

You should have a term like $\int_{\partial U}w_t\pdv{w}{\nu}$ right?

[untold delta]

Whoever

[gleaming lily]

@unborn quiver

[unborn quiver]

yeah

if w=0 on partial U...

what about w_t

[gleaming lily]

Oh fck

I see

I forgot the integral is over the x variable not t

[unborn quiver]

I think even if it were over t, it would still be 0 bc w=0 on partial U x [0,T]?

I don't want to look at book bc I am in bed

[gleaming lily]

I mean just because the function is 0 doesn't mean the derivative is zero right?

That was my reasoning

[gleaming lily]

We are looking at a specific t so no

Anyways thanks

[unborn quiver]

Oh yes I see

but yeah

the derivative is zero because it should still be C1 on the boundary

I forget the regularity defined here but it's constant so no worries

[hollow venture]

(rudin func anal chapter 6 is a cute fast intro too)

[oak badger]

Folland chapter 9 as well

[hollow venture]

actually, im trying to get a sharp bound on this (as an attempt to sleep) and you could probably get a good one via the poisson formula

[buoyant pike]

???

"as an attempt to sleep"

[hollow venture]

looks promising

but i shall sleep now i cant be bothered to differentiate 1/|x-y|^n wrt xi

and to maximize it

idk my rough sketch looks like equality can never be achieved for n>2 but you can always get arbitrary close and compute the constant

ill compute tomorrow

[gleaming lily]

So true bestie

I agree I realized that it kinda doesn't matter

[hollow venture]

im lazy to check if i did it correctly but like the idea is you can holder the last part to ||...||^p||f||^q and take a limit as p->infty to get a sharp bound

this also gives you an f and i suspect equality can hold

and you can probably extend this to arbitrary derivatives by induction since f is smooth

i may work it out later or something

lol

[steep oyster]

For a 2nd order linear elliptic pde with non dirichlet boundary conditions (really I care about mixed) is it still true there’s a fredholm alternative or are there obstructions? And does anyone have a reference? Evans only does the Dirichlet case.

[forest cradle]

For 2nd order elliptic pde check Gilbarg Trudinger

[mellow totem]

Hello, everyone I am new at PDE. I want to approach PDE. How should I start? Can anyone share their journey experience please. I know functional analysis, measure theory, and little bit of Fourier analysis. But never studied PDE in a well manner. Please suggest me some book to start with.

[hollow venture]

evans

we can die in evans tgt

[astral vine]

Hail Brezis

[bronze gate]

🙄

Brezis >>> Evans fr

[astral vine]

For a pure Fourier Analysis point of view with some (but not that much) functional analysis :

Bahouri, Chemin & Danchin's book

Mostly deal with Navier-Stokes, Euler, Non-linear Schrödinger and non-linear wave equations on R^n

But with Fourier techniques

[astral vine]

Even if I don't like Evans, to be honest it offers a wider review of most usual and common techniques, for many different classes of PDEs

HOWEVER

It is really lacky on the functional setting (on purpose). For instance modifying some assumptions, or a little bit the structure of your PDE systems, will modify so much the underlying structure of your problem, that you would need these informations and technical points that are not raised in Evans.

Brezis deals with fewer different classes of PDEs, but the methods and the point of view are general enough to be adapted in case where the assumptions are stressed.

[verbal nebula]

That's why you go read the books recommended at the end of each section if you're interested in that particular section

I think Evans is a very good intro to PDEs

[blazing ridge]

very general question, but is there any particular significance behind 'backward problems'? Where we work on [0,T] and we specify final data u(T, .) rather than initial data u(0, .)

in one chapter of some text i am reading the authors start considering a transport equation as a backward problem and i have no idea why

[astral vine]

for most of (linear) transport equations it does not matter to be forward or backward

but it has to be shown I guess

[buoyant pike]

Backwards heat equation is ill posed

[blazing ridge]

right it shouldnt matter for the transport equation

okay nvm think i figured out why it makes another condition they impose on the coefficient a little simpler

[buoyant pike]

Practically speaking, many problems are backwards problems where we can observe some real thing and ask how it got to that state

[blazing ridge]

yes that does make backward problems sound quite natural actually

[buoyant pike]

Indeed

One of my advisors has worked on backwards evolution of the earth moon system

[steep oyster]

Checked here they only mention mixed conditions once in relation to maximum principle

[astral vine]

What kind of properties are you actually expecting to investigate?

[steep oyster]

Omega bdd domain with boundary divided into D and N imposing dirichlet and Neumann conditions. When does the following bvp have weak solutions?
-Lu=0, L is linear 2nd order
-u=f on D , du/dn=g on N

I’m not sure to what extent this matters but in my specific case D has measure 0, and g=0 (for now at least)

[astral vine]

Okay.

[nimble jolt]

my textbook says, when the dimension is 1, functions in $H^1_0$ ($W^{1,2}$ satisfying vanishing boundary condition) are Hölder with exponent ½, and thus they are continuous.

[untold delta]

Mattuwu

[nimble jolt]

hmmm, that sounds magical, $H^1_0$ are equivalence classes of functions ignoring differences on measure zero sets...if every function in $H^1_0$ is continuous, it almost feels like $H^1_0$ is the same as $C^1_0$

[untold delta]

Mattuwu

[buoyant pike]

Have you heard of sobolev embeddings before

[bronze gate]

As ange mentioned, google sobolev embeddings :)

[nimble jolt]

thanks~

[gleaming lily]

Is the trace operator on sobolev spaces unique?

Ok I think it’s yes because of global approximation by smooth functions

If the domain is bounded and the boundary is C^1

[astral vine]

Yes it is even for Lipschitz domains

the idea is to build distribution theory on the boundary

and to invoke uniqueness for the trace as a distribution

(There is some like technical issues for defining distributions on the boundary, but nothing unreachable)

For the specific case of a connected bounded C^1 domain with compact boundary the distribution space is the topological dual of C^1(Boundary)

for Lipschitz domains the dual of Lip(Boundary)

• an other way : the universal approximation property in Sobolev spaces.

Say you have a Lipschitz domain,

[untold delta]

Functionanatolysis

[astral vine]

This is a bit more stronger than the usual density result for Sobolev spaces

now since s_j>1/p_j (the result above does not need it, I just need it for existence of trace estimates on a dense subset)

You have a trace in L^{p_j} of the boundary (in fact in W^{s_j-1/p_j,p_j}=B^{s_j-1/p_j}_{p_j,p_j} of the boundary)

extracting subsequences shows that both must be equal almost everywhere.

[solid flint]

Hello, How can I solve the following PDE explicitly please?
$u_{tt}-u_{xx}+\alpha u_{txx}=0$ where $\alpha$ is a positive constant.

[untold delta]

Mikahopff

[astral vine]

As for the wave equation

Put it into a single evolution equation of order one on time

Introducing the right matrix of operators.

Check that the matrix of operators as a single operator generates a semigroup

Compute the semigroup explicitly.

[solid flint]

can you elaborate on how to compute the semigroup explicity please?

[astral vine]

Do it on the Fourier side

Diagonalize the matrix to compute the exponential

Revert

[solid flint]

Thank you a lot!

[nimble jolt]

With the Lebesgue measure and a bounded set, how does $L^p$ embedding work? For example let $Ω = (0, 1)$, which one of $L^p(Ω)$ and $L^q(Ω)$ includes which, for $1≤𝑝≤𝑞≤∞$?

[untold delta]

Mattuwu

[buoyant pike]

What have you tried?

[nimble jolt]

I have just tried stackexchange and it works, thanks!
https://math.stackexchange.com/questions/66029/lp-and-lq-space-inclusion

[buoyant pike]

….

[astral vine]

[cobalt geyser]

[astral vine]

I'm dying for real

[river path]

As great as this is, the intuition is that on a finite measure set, the thing that can go wrong is singularities growing too fast. If you take a higher power, singularities grow faster. So the L^p spaces get smaller as p grows for Omega finite measure.
Easiest check for this is that bounded function (L^infty) are integrable if the measure is finite. So L^infty is a subset of L^1 (and all the others are between)

Meanwhile, on a discrete measure space the opposite happens. The thing that can go wrong is that your function/sequence doesn't decay fast enough. Taking higher powers makes decay faster, so it's easier to be in higher l^p's (and they're bigger)

Again you can use that a summable sequence is bounded to remember the l^1 l^infty inclusion

[solid flint]

Hey, what type is this equation please?

$u_{tt}-\Delta u+\Delta u_t=0$
How do we characterize evolutionary PDEs please?

[untold delta]

Mikahopff

[buoyant pike]

What do you mean by "type"

Parabolic/hyperbolic/elliptic?

Those only apply to 2nd order pdes and this one is 3rd order

[hollow sorrel]

Does anyone know if the spectrum of the laplacian on a compact region is discrete?
We can check them for a square and a circle and i think they do.

[buoyant pike]

What have you tried

[astral vine]

Stackexchange and it works!

Sorry. (not that much actually)

[buoyant pike]

💀

[astral vine]

To be a bit more serious: before asking "what have you tried?" maybe we should also ask "what kind of fact do you know ?"

Which is a bit messy

But depending on the amount of technology we have at our disposal , this will deeply change how we should provide the answer/insights, right ?

[hollow sorrel]

I don't remember my real analysis so i don't have a big arsenal. I was mostly just curious, particularly if you need any big machinery.
If you restrict the operator domain to functions with a zero Neumann boundary condition or a zero Dirichlet condition you can at least probably prove that it's self adjoint for some Hilbert space.
RieFrech theorem also can be used here.
If the operator is continuous then that should be nice too.
I'm expecting the answer to be semi trivial, but if not then it'll be a problem for another day

[hollow sorrel]

Have you tried chatgpt? It's really good. :^)

: )

[buoyant pike]

Bad

[astral vine]

What is your knowledge about L² Sobolev spaces and sesquilinear forms on Hilbert spaces ?

[hollow sorrel]

I know all those words.

Should be okayish

[buoyant pike]

Do you know them together though

[astral vine]

Compact Sobolev embeddings and Rellich-Kondrachov too ?

[hollow sorrel]

Also I coul work in C but i was concerned with R.
Compact Sobolev embeddings is shakey.
The second no.

If it requires looking up much I'll do it tomorrow. No worries. I still love you

[astral vine]

Operator theory only (completely) works for complex valued setting, then you can go back usually to the real valued case

The overall strategy is the following:

• Write your (negative) Laplacian as the unique closed Friedrich extension over L² of the symmetric sesquilinear form a(u,v):=<Du,Dv> over D(a)=H^{1} or H^{1}_0.
• Prove that D(a) embbeds compactly in L² (this is where you used rellich Kondrachov, assertin' that H^{1} embbeds in L² with compactness).
• As a direct consequence the resolvent of your Laplacian is compact.
• By the Fredholm Alternative, operators with compact resolvent has only pure discrete spectrum made of eigenvalues.

[hollow sorrel]

👍

[hollow sorrel]

That looks tidy, thank you

[astral vine]

Step 2 is the one where it requires compact/bounded domain. otherwise the compact embbeding never holds.

[nimble jolt]

is the following intuition true?
compactly supported smooth functions are exactly the smooth functions that vanishes close to the boundary
in other words
a smooth function is compactly supported if and only if it vanishes close to the boundary

I want to try to write it down rigorously and prove it.

Here is my attempt at rigorizing it:
$Ω ⊂ ℝ, Ω$ open. $f$ is a real-valued function. $f ∈ C^∞(Ω)$

$$\mathrm{supp} f := {x ∈ Ω | f(x) ≠ 0}$$

Claim: $\mathrm{supp} f$ is compact if and only if
$$∀y ∈ \partial Ω. ∃ ε > 0. ∀z∈ B_ε(y) ∩ Ω. f(z) = 0$$

[untold delta]

Mattuwu

[buoyant pike]

This is not true, consider the standard bump function e^(-1/(1-x^2)) on (-1,1)

[solid flint]

Yes

[solid flint]

Actually, thiq existes for any order

[buoyant pike]

Ok well you tell me then

[nimble jolt]

you are right, but I gave a wrong definition of supp too, I was looking at the wiki page for "compactly supported", and I think it involves taking the closure in Ω, (instead of in ℝ), and thus the bump function doesn't have a compact support in this case

[quaint herald]

Yes, you are right. $f(x):=\exp(1/(x^2-1))\notin \mathcal{C}_c^\infty((-1,1))$.

Your original claim is true, modulo the incorrectly stated definition of support. The proof is almost trivial. If $\Omega$ is your domain, assumed open, then $d(x,\partial\Omega)$ is a continuous and positive function of $x\in \Omega$. By compactness, it attains some minimum value (say $\epsilon$) on the support of any $g\in\mathcal{C}_c(\Omega)$ and so the support is contained in ${x:d(x,\partial \Omega) \geq \epsilon}$ for some $\epsilon$.

[untold delta]

gomez

[quaint herald]

(the converse is false though because \Omega could be unbounded and supp(f) could have positive distance from the boundary but still be unbounded itself. you would want to assume \Omega is open AND bounded if you wanted that implication to hold).

[quaint herald]

There are various generalisations. It is pretty straightforward for elliptic, and there are a few different known choices for hyperbolic, but I don't even know of useful generalisations to parabolic. If you have particular definitions in mind you should definitely state them, there is no exhaustive classification of PDE, even linear ones.

[gleaming lily]

Evans claims that heat equation is unique on bounded domains, so if we have a solution on the entire R^n, can’t we just say it agrees with a particular solution on each B(0, r) x [0,infty) to conclude uniqueness?

Because there are nonzero solutions on R^n with 0 initial condition

[quaint herald]

I don't quite follow your question, can you explain clearer what you are asking? Like when you are talking about the heat equation on a bounded domain you are also probably imposing a BC.

[unborn quiver]

There’s a fairly famous counterexample for failing uniqueness if you don’t have a sufficient growth condition (Me^(Ax^2))

[quaint herald]

I am pretty sure he knows this based on his last sentence.

[gleaming lily]

But on a bounded domain you can always get a growth condition by choosing large enough constant?

If your solution was already continuous on the entire R^n

[quaint herald]

The growth condition is a statement about the heat equation on R^n, not on bounded domains.

It's still quite unclear to me what you are asking lol.

[gleaming lily]

I see, but my point is that since we have uniqueness on bounded domains, two solutions on R^n will agree on each ball so aren’t they the same

[quaint herald]

We have uniqueness on bounded domains when we impose some kind of boundary condition. Where are your boundary conditions coming from when you try to apply this result to get conclusions about the R^n heat equation?

[gleaming lily]

Let’s say the boundary condition is always 0

[quaint herald]

Huh? I still don't see how you are getting anything in R^n. Go through it really slowly for me. You claim that you can use uniqueness (for bounded domains with zero boundary trace) to deduce the (known to be false) result of uniqueness (for R^n). How does this argument go?

[gleaming lily]

So let’s say we have two solutions on R^n with 0 boundary, then the two solutions are also solutions on B(0, r) for any r, since this domain is bounded, the two solutions agree on B(0, r)

[quaint herald]

what is "a solution on R^n with 0 boundary"?

[gleaming lily]

So at time 0, the function is 0

[quaint herald]

okay, I would call that 0 initial conditions, but fine

and indeed these two solutions agree on B(0,r), but there is no contradiction here. the uniqueness result for bounded domains is for the heat equation with an imposed boundary condition

namely the value of your solution u on the boundary of B(0,r) is specified for all time

not just the IC

[gleaming lily]

Oh

Ah I see

[quaint herald]

nor should we expect uniqueness without such a BC. imagine a metal bar with some initial heat distribution. we should expect very different behaviour in its evolution depending on what kind of external heat we apply to the ends of the bar.

[gleaming lily]

Right ok that makes sense

[quaint herald]

👍

[gleaming lily]

Ok I see how the term boundary condition was very misleading

[quaint herald]

yeah i mean it is a boundary condition in spacetime, but its only one of the boundary faces

and to avoid ambiguity "initial condition" is probably better terminology

[gleaming lily]

Right ok

[buoyant pike]

Whoever learns what a boundary condition is

[unborn quiver]

Yeah I see that now, I couldn’t parse the question either so I just went for what results do exist for solutions on Rn with 0 IC

[gleaming lily]

🐶

(For myself of course)

[nimble jolt]

this seems true though:
Let 𝑓 be a smooth ℝ → ℝ function.
Let (closed) supp 𝑓 be bounded (⇒ compact),

then: 𝑓 vanishes before infinity, in the sense that ∃ 𝑎∈ℝ. ∀𝑥. 𝑥 > 𝑎 ⇒ 𝑓 (𝑥) = 0 and ∃ 𝑏∈ℝ. ∀𝑥. 𝑥 < 𝑏 ⇒ 𝑓 (𝑥) = 0

[astral vine]

Just for informational purpose : the actual (sharp) trace theorem for boundary values on a domain is built from the theory of initial value problem for Banach-valued differential equation.

this is not the most well known construction

for pedagogic purpose usually people prefer to teach other ones.

Just a fun fact about " "initial datas" and "boundary values" are in fact very deeply related"

this is not just some informal thoughts

[quaint herald]

I know, and never said anything about informality or lack of relation. This was just a matter of nomenclature, because Whoever was getting confused about the statement of the basic existence/uniqueness results for the heat equation on bounded domains.

[quaint herald]

yep 🙂

[astral vine]

I didn't want to be pedantic or anything, it was juste to give some complements for anyone reading this discussion. We have already discussed about it a while ago iirc, I know you knew :)

[simple void]

cant figure this one out

[buoyant pike]

#odes-and-pdes for separation of variables

[simple void]

ok

[blazing ridge]

suppose $u \in BV((0,T) \times \mathbb{R}) \cap Lip([0,T]; L^{1}_{loc}(\mathbb|{R}))$ and $v \in Lip ([0,T] \times \mathbb{R})$

Is there an obvious way to make sense of the equation $\partial_{t} (u\partial_{x}v) + \partial_{x}(u\partial_{x}v) =0$ apart from in the distributional/weak formulation sense?

iim pretty sure the answer is NO but i would like to be sure. Lip functions are W^1, \inf in 1D but not necessarily W^2,p

[untold delta]

M6LI

[steep oyster]

Is the containment of W^{1,p} into Lp still compact on a bounded domain of R^n if p=n? I know rellich and morreys give p<n and p>n respectively.

[astral vine]

no restriction on p

I don't know the proof you looked at

but it weren't has any restriction on p for me

[steep oyster]

i am looking at evans

[astral vine]

The issue for compactness of embeddings of Sobolev spaces is when you consider the critical embedding W^{s,p} (ot H^{s,p}) in L^{q} with 1/q=1/p-s/n

but otherwise the embedding is always compact from W^{s,p} to L^{r} with 1<= r <q.

in particular r=p

[steep oyster]

ah ok i think i misread the statement

thanks

[astral vine]

you where talking about Theorem 1 p 288 ?

[steep oyster]

thats right and also the remark after the proof

[astral vine]

using fractional sobolev spaces and sharp Sobolev embeddings make it easier to obtain the result somehow, but too much technology

[steep oyster]

i'm not sure what those are unfortunately

[astral vine]

Not really important here continue your journey through Evans.

[untold delta]

cocat

[lilac barn]

As an exercise, extend this to W1,n(U) is compactly embedded in Lp(U) for all 1 ≤ p finite. @steep oyster

[dusk mist]

<@&268886789983436800>

[blissful quiver]

hi i got stuck in an argument. I want to proof that the Dirichelt Energy (E(u) = \frac{1}{2} \int_U \abs{\nabla u}^2 , dx ) is weakly lower semicontinous. I konw that norms in general are weakly lower semicontinious and we have a norm on (H_0^1 ) via ( (u,v) = \int_U \nabla u \cdot \nabla v ) then we have [ 2 E(u) = \norm{u}^2 \leq (\liminf \norm{u_k})^2 ] and this is not really what i want. I would want to have the square inside the liminf, but i don't see how that would be possible. Also i want weakly lower semicontiuity for functions in (H^1) and not just in (H_0^1). How do i prove this, or does anyonre know a reference?

[untold delta]

dr.marionette

[blissful quiver]

and i know of some convexity results, but i don't want to use them

can this approach be repaired so that it works?

i am pretty sure that i have seen this exact argument once in some calculus of variations lecture notes but i can't find them right now

ok i found it again, it's page 18 here https://www4.math.uni-hamburg.de/home/schmidt/lectures/CalcVar.pdf is this just wrong?

[blissful quiver]

ok, i think it's just $ (\liminf_{n \to \infty} a_n)^2 \leq \liminf_{n \to \infty} a_n^2 $. I think this will become an equality because of boundedness of the sequence

[lilac barn]

Look up the section of Calculus of Variation in Evans. He gives a sufficient condition for the Lagrangian so that the energy functional is weakly lsc

[buoyant pike]

Ok so

We have this statement:

So this is saying that the only harmonic functions on a sphere are constants

Because spheres are compact manifolds

However

I have these functions

They are non-constant and I have determined that they are harmonic on the sphere???

[cinder marten]

So this is defined for 0 <= theta < 2pi and 0 <= lambda < 2pi

you need to find another harmonic function on the other hemisphere and show that they agree on the overlaps

[buoyant pike]

That is the whole sphere?

[cinder marten]

nope, we have slices cut out

[buoyant pike]

The sphere is theta from 0 to pi and lambda from 0 to 2pi

Which slices?

[cinder marten]

you can't have globally defined theta and lambda coordinates on the sphere, so theta = 2pi and lambda = 2pi I believe

otherwise it wouldn't be a sphere, it would just be the plane

[buoyant pike]

Sure, there is a discontinuity in lambda along a meridian but because the only function of lambda is cos(Rlambda) this is the same whether lambda=0 or 2pi

Theta goes from 0 to pi, 0 is the north pole and pi is the south pole and there are no singularities there because sin(theta) is 0 there

[cinder marten]

my point more or less is that you need to use a different set of translated coordinates to solve for harmonic functions on another patch, and this will forcibly introduce a discontinuity

[buoyant pike]

?

There is a spherical laplacian that doesn't use local coordinates

[cinder marten]

yes, but when you introduce this form you are implicitly choosing a coordinate patch on the sphere

this is what it looks like in those particular coordinates

those theta and phi do not cover the entire sphere

[buoyant pike]

???

Do you know how spherical coordinates work

[cinder marten]

yes. if we had global coordinates on the sphere it would not be a sphere