Mathematics - 2017-07-08 (page 1 of 6)

Q: For all ordinary differential equations, does there exist a corresponding integral equation?

4:13 AM

0

Once or twice I have heard the term thrown around of a corresponding integral equation. Therefore, I ask if the following conjectures are true. For all differential equations $D$ there exists an integral equation $I$, such that $Sol(D) \cap Sol(I) = Sol(D)$. Note that $Sol(x)$ is the functi...

i got an answer here that makes very little sense. Am I missing something or does that answer not address the question?

Well, you did indicate: "I think I can prove it for the trivial case of y′=f(x), but I don't know how to progress with the larger case." The question does address that special case.

That said...ehhh.

yeeeeaaaah

if they said induction for that matter

then I could see it maybe

For linear ODEs, at least, there is one fairly concrete idea.

but even then that rejects nonlinear equations

Yeah.

4:21 AM

yy' = x doesn't work under that method of proof

in fact, feel free to comment regarding that

With linear ODEs, one can recast an nth-order equation as a first-order system of ODEs

And then one has $Y'(x)=F(x)Y(x)$, which at least in principle has an integral representation.

fair enough

For a nonlinear equations, though, that idea fails immediately.

but that doesnt allow nonlinear

For nonlinear things are a lot murkier.

4:22 AM

which is why i asked if there exists integral equations

and then i asked if there was a minimum integral equation solution set with a unique integral equation

i.e. if whether there is one with 10 extra solutions and no others 10 or lower

the question makes sense, right?

i was worried about the logic of the second thingy. Took me an hour to write out.

why is the answer getting upvotes?

that is infuriating

errr... annoying

@Semi any n-th order equation can be made into a first order system

you just dont get the nice linear form

Hmm, yeah.

I just wouldn't count that as terribly useful.

@Typhon I'm not sure whether the logic of the second thing actually holds up in practice, but I can see what you're going for.

What I'd love to see some 'canonical' example of a nonlinear ODE whose solutions can't be represented as integrals.

yeah that would be interesting to see

I think stuff re: the Painleve equation may be an example of that.

do you know of the implied derivative/antiderivative?

4:27 AM

Not sure what you mean by that.

let me find a link

Here's the link to Painleve: en.wikipedia.org/wiki/Painlev%C3%A9_transcendents

Q: Are all solutions to an ordinary differential equation continuous solutions to the corresponding implied differential equation and vice versa?

7

Now I have to heavily emphasize the fact that I have never studied differential algebra or the concept of other types of differentiation (which is what I believe is the concept behind a differential algebra). So, if I am abusing the terminology a little bit, please forgive me. Let us define a di...

differential-equations limits derivatives differential-algebra

3

Q: Are all weak solutions to an ordinary differential equation continuous solutions to the corresponding implied differential equation and vice versa?

Yes,this is very similar to a previous question I asked. That was about normal solutions and not weak solutions. Let us define a differential operator known as implied differentiation. It actually does not have a unique value. Let us denote the implied derivative operator as $I(f)$, where $f$ is...

calculus differential-equations floor-function weak-derivatives piecewise-continuity

(I've actually got a book in my room on Painleve transcendents which I've long given up trying to read.)

Doesn't the usual procedure for a turning a first order IVP into an integral just work always

convert n-th order equation to first order system then integrate

it's not "practical" but it's an integral equation that is induced by the ivp

4:30 AM

This is where my expertise falls short, I'll confess.

im pretty sure it just works

this is how you prove picard lindelof

the step where you convert to the integral equation doesn't make any assumptions on what your system looks like

Hmm.

At least that i recall

Practical example, then. The Painleve I equation is given by $y''=6y^2+x$

This can be turned into the first-order system $v'=6y^2+x$ && $y'=v$. How does one proceed from there? @EricSilva

@Semiclassical the implied derivative is just name I have for a multivariate operator that essentially treats step functions as constants whilst preserving the general properties of differentiation. My conjecture is that the continuous solutions to the corresponding implied differential equation are the solutions to the corresponding integral equation as I have defined it.

see the links for the full definition

i am too lazy to write it out here

:p

4:34 AM

That's a bit much for me to chew on at this time of night, tbh

@Semiclassical fair enough. I am intending to one day try my hand at constructing something similar but for periodic functions. All my attempts have so far failed.

likely due to there being no simple characteristic for periodicity

So i'd write this as $(x_{1}, x_{2})' = F(x_{1}, x_{2}, x) = (x_{2}, 6x_{1}^{2} + x)$

(suppressing the $x_{i}(x)$)

For reference, these notes on Picard-Lindelof for first-order systems seem pretty decent: wwwf.imperial.ac.uk/~jswlamb/DynamIC/M2AA1-2010/…

then you just do the normal thing

if it makes you feel better swap $x$ with $t$, it makes more sense

Sure. That's what it was originally, heh, I just changed it for cosmetic reasons

4:40 AM

@Semiclassical if you look at all functions of the form $f(g(x)^2)$, did you know that includes all periodic functions?

same is true for any even power

but if you look at any odd power... you get all functions.

@EricSilva Judging from the notes, though, it seems like Picard-Lindelof rests on an iterative procedure. I'm not convinced that would correspond to an integral equation in the desired sense.

neat, huh?

Hmm. I'll have to think about that.

$f(\sqrt{(x \mod 1)^2}) = f(|x \mod 1|) = f(x \mod 1)$

the last term is all periodic functions

@Semi Picard lindelof iterates because it lets you find a solution of the integral equation usign a fixed point theorem

the derivation of the integral equation as equivalent to the ODE happens before you start iterating

4:43 AM

True.

if the ode is sufficiently ugly the iterating procedure just might not give you a solution

So one would end up with, if I'm following right:

glad I sparked a discussion

(switching from x to t to avoid confusion)

the only thing is that our system isnt autonomous (but that doesn't matter because you can always turn a non-autonomous system into an autonomous one)

4:46 AM

@Semiclassical so I am right in saying that answer was weird, right?

$(x_1,x_2)'=(x_2,6x_1^2+t)\implies (x_1,x_2)(t)=\int_0^t (x_2(t'),6x_1(t')^2+t')\,dt'$

It certainly wasn't clear in its meaning.

the answer?

Yeah.

or my question?

I mean, it may be enough to talk about the first-order case.

4:47 AM

ya sniped me

But one needs to explain why that's enough.

@Semiclassical true but then we need the inductive step

@Semi in fact because of our discussion there is only the first order case

@Typhon you do not need induction

uuh

yeah you do

no you don't

4:47 AM

yes you do

you can reduce any n-th order ode to a first order system

Eh, induction really isn't needed in the reformulation I wrote above.

you need to show that if it is true for Nth order then it is true for (N+1)th order

Not really.

no you don't

you can just show it's true for any n directly

4:48 AM

im referring to the second statement

@EricSilva proving it for the first order serves as a base case

Again, not really.

no dude im telling you, it doesnt use induction

not talking about the existence of an integral equation

Then I'm not sure what you're talking about.

im talking about the uniqueness of an integral equation with the least number of superfluous solutions.

4:49 AM

im unclear what you're talking about then

oh that

the second statement in my question...

I'm unclear on if that makes sense

@EricSilva in what sense?

I'm not sure there would be any superfluous solutions when you do Picard-Lindelof, at least if the boundary conditions are specified as "y and its first n-1 derivatives at 0"

Once you start talking other boundary conditions, by contrast, I have no idea.

the number of superfluous solutions produced by an integral equation is either infinite, uncountably infinite, or finite.

4:51 AM

well you might not get unique solutions

if your ODE isn't regular enough

Ugh, true.

i don't think there would be a superfluous solution. banach fixed point theorem guarantees uniqueness of fixed point, doesn't it?

mayve i'm wrong

@Balarka he made no restrictions on how regular the ODE is

gotcha

i am arguing that there is a unique integral equation such that the number of extraneous solutions is finite and minimal

4:51 AM

you're definitely right if you have like local lipschitz at least

Las Vegas Raiders

hi chat

right

@EricSilva not unique solutions. I'm talking about a unique integral equation I such that $|Sol(I)|$ is minimized

im very unclear on how we can even get multiple integral equatoins

@Typhon I know i wasn't talking to you

errr

4:53 AM

I'll admit, this is where my interest starts to flag. Give me a specific ODE and I'm happy.

im assuming one can do that?

Give me an arbitrarily pathological one, though, and I find it hard to sustain my interest.

yeah no this is the kind of thing where i dont really feel it's easy to start approaching because finding ways to transform ODE is usually very ad hoc

@Semiclassical i like differential equations that involve floor. I like solving them with special methods. But now that I have advanced more I should justify those methods.

This is literally the foundation of building up to the proof of that justification

XD

@Typhon To be fair, problems like that are important in application

4:55 AM

@Semiclassical differential equations involving the floor function?

e.g. differential equations with time-dependent coefficients.

O.O

i can trivialize them with certain means

including ones that might have discontinuous time dependence

eliminate the need to pull out the laplace transform

and get the same solutions it gives

yeah and it usually doesn't help to do the thing i said: re making them autonomous by increasing dimension, because you're just shifting the difficulty

4:56 AM

of course... probably a transition to an integral equation going on there

In math there's a principle of conservation of difficulty.

^^^

so one way to deal with floor is to increase the order?

lol

This more has to do with the number of derivatives, I think.

Higher order ODE = first-order system with more components.

fixed

huh

interesting

4:58 AM

@Semi we're all confused over here and I have a professor who has told me "Yeah from a mathematical perspective ODE are basically trivial"

the implied derivative thingy's solutions are trivial

rip

The idea, as alluded to before, is to write $y=x_1,\,y'=x_2=x_1',\,y''=x_3=x_2',\,\cdots$.

since in that 'space', piecewise constants and constants are essentially one and the same

i don't like DE's

evil little things

4:58 AM

@BalarkaSen well we do

D:

@Balarka don't u like foliations

ye but i don't think of them in terms of DE's

@BalarkaSen if you don't feed the differential equations between 11 pm and midnight, you'll be fine. Put them in water and they multiply and give you lots of fun challenges.

idt there's really a "DE" perspective though

I find asymptotic questions in DE's to be the most interesting.

5:00 AM

it's such a gigantic field

Is there integer $n$ such that the equation $\log_b(n+b) \in \Bbb N$ has three non-trivial integral solutions for $b$?

well, let's put it this way. i have not seen a super-inspiring corollary of Frobenius integrability

theoretically, it's cute. but

frobenius isn't a super interesting DE theorem tho

if it were interesting it would be harder

@Semiclassical imagine your hardest problem you could have to solve involving differential equations and floor. Now imagine if you could start off by just substituting all the piecewise constants with constants and proceed like normal. That is essentially how it works. Then you just find the continuous solutions. That's the trickier part. Involves a lot of series and partial sums of series and lots of no closed forms.

or would you rather use the laplace transform... oh wait the function increases faster than e^nx!

Here's the DE I have in mind as far as asymptotics go

5:05 AM

i think equations involving $e^{\lfloor x \rfloor^n}$ are good candidates to mention

since they break the laplace transform

and are still solvable methinks

@Eric hah fair

let me try

@Balarka also have heard from a professor: "If a theorem about PDE was discovered before world war 1 then it wasn't really a theorem about PDE"

lol

$$-i\dfrac{d}{dt}\binom{x_1}{x_2} = \begin{pmatrix} t & \alpha \\ \alpha & -t\end{pmatrix}\binom{x_1}{x_2}$$

I think that's the one I'm remembering.

5:06 AM

let's try to solve $y'' + 2e^{\lfloor x \rfloor^2}y' + 4e^{2\lfloor x \rfloor^2}y = 0$

or rather i will

note, if you're favorite thingy can transform it, i'll just nest more e^___'s

The idea with the one I'm talking about right now is that, if $\alpha=0$, then this problem decouples in the obvious way.

@Semi does the $-i\frac{d}{dt}$ thing show up a lot in physics

auxiliary equation: $r^2 + 2e^{\lfloor x \rfloor^2}r + 4e^{2\lfloor x \rfloor^2} = 0$

@EricSilva yuuup.

It's basically saying "oh hey I'm doing Schrodinger"

sooo

5:08 AM

i saw it in an analysis class where we proved some theorem

that was actually just a physics theorem

$r = 2e^{\lfloor x \rfloor^2}$

but i didnt understand that perspective

actually, I should probably not have the minus sign. it doesn't matter, of course.

therefore $y(x) = C_1(x)e^{2e^{\lfloor x \rfloor^2}} + C_2(x)xe^{2e^{\lfloor x \rfloor^2}}$

now to find piecewise constant C's such that f is continuous

help

heh heh

good luck man

5:12 AM

actually it aint that hard

just need the integral of floor squared

XD

wait no

How to find the maximum velocity from this graph?

i did this all wrong

therefore $y(x) = C_1(x)e^{2xe^{\lfloor x \rfloor^2}} + C_2(x)xe^{2xe^{\lfloor x \rfloor^2}}$

What's the definition of velocity?

5:14 AM

i completely forgot how to do auxiliary equations.

@Semiclassical Displacement/time (rate of change of displacement with time)

right. (change in s) / (change in t). or at least that's how one does average velocity; one takes a limit for instantaneous velocity.

@Semiclassical yes

@Semiclassical integral of $e^{\lfloor x \rfloor^2}$ would be helpful

XD

at the point of inflection

5:15 AM

@Semiclassical How do I find that using the graph?

but if i pick two points on the graph, then the change in s is just the vertical rise, and the change in t is just the horizontal run.

no result found - wolfram alpha

fuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

here it's where the tangent lines go from being below to above the graph

so the average velocity defined by those two points is just the slope of the line between them.

hrm

gotta rethink this

5:16 AM

locally i should say

Homer Simpson Swearing

►

@Semiclassical I know that. How do I find the maximum velocity?

well, where is the graph the steepest?

You've only got a drawing, so you'll have to settle for a visual estimate of where it's steepest.

@Semiclassical Between 10 to 14 s?

something like that. I'd say maybe 11-12, but absent a specific function it's impossible to be precise.

5:19 AM

id say it's between 12 - 14 no

i guess 12ish makes sense

yeah, you're right---I forgot that the boxes have width 2 s not 1.

yeah

12-14 is what I intended.

This is also the inflection point of the graph. But I don't really have the patience to explain that, tbh

i can never remember like all the terminology

i remember learning that the inflection points are where the derivative changed sign

rather than just its critical points

which doesnt make sense to me

i have inflection point as concave up -> concave down or vice versa.

(could also have something like x^4 of course, but w/e)

5:23 AM

oh i always used convex concave

i like up/down just because of how visually immediate it is.

i think that's concave up?

unsure

1. mentally associate "velocity" with "slope"
2. ????
3. profit

I forget tbh.

x^{2} is convex

5:25 AM

So that'd be concave up.

@Semiclassical I ain't getting the right answer if I consider that time interval

the reason i use convex concave is because everything that has ever mattered in my life has been convex

shrug @abcd

Shadow

Can anyone help me understand the question?

Ten points in the plane are designated. You have ten circular coins (of the same
radius). Show that you can position the coins in the plane (without stacking
them) so that all ten points are covered.

Why can't I put all the coins on the ten designated points?

@shadow without stacking them

suppose all ten points are really close together.

5:26 AM

@LeakyNun ?

@Abcd can you show us the whole question?

if you try to put the coins on top of the designated points in that case, they'll overlap each other and would require stacking

@LeakyNun Ok

Shadow

Oh I see

you can still cover them in that case, though, just by using one coin to cover all of them at once and then put the rest wherever

Shadow

5:28 AM

@Semiclassical Thanks

but that's a specific case.

maybe u can apply a covering theorem from gmt /s

Probably you shouldn't start with 10 coins. Maybe just do 3 coins to start with.

@EricSilva err, that would make $x = 0$ not an inflection point of $x^3$?

5:29 AM

@LeakyNun ^

@Abcd and what is the answer?

@LeakyNun Answer of b is 25 cm/s

@Balarka ?

what are you referring to

you said "i remember learning that the inflection points are where the derivative changed sign"

you can click on the little grey arrow at the front of the message to go to the message i am replying to btw

oh sorry i meant where the second derivative changed sign

not first

oh that's a cool feature

5:32 AM

oh i see

i want them to just be the crit points of the derivative tho

@Abcd if you look at t from 11 to 12 and get 30cm/s I would say it's pretty dem close

It's also helpful to draw a line of slope 25cm/s just to get a sense of how hard it is to distinguish various slopes precisely.

Te amo, te amo

he said to me

(25 cm/s = 100 cm/4 s = 1 m / 4 s)

5:35 AM

somebody tell me what he said

don't it mean I love you?

@Twink yes

it is Spanish for "I love you"

he said te amo then he put his hand around my waist

and also Latin

Out of context song lyrics remain confusing.

For reference: azlyrics.com/lyrics/rihanna/teamo.html

yes but I changed it

5:38 AM

Te amo can just mean that you like someone depending where you're from

Doesn't make it any less of a non sequitor.

think it means I love you

te quiero means i love you

@EricSilva that's for a friend

@EricSilva I would not call $x = 0$ an inflection point of $x^4$.

you're just saying you just want the second derivative to be 0, right?

5:40 AM

@Leaky depends where you are

lots of parts of LA use te quiero for a significant other

@EricSilva but me gustas tu is universal?

(it's also a title for a k-pop)

uhh in some parts of latin america that's very awkward phrasing so no

@Balarka i would tho

eh

i just want to think of that saddle point picture

i think that's fair

I told him, listen we can dance, but you gotta watch your hands

5:42 AM

but the thing is that i dont like this terminology at all

cause i dont think it's ok to use in higher dim

but the thing is that i dont like this terminology at all

because I understand that we all need love and I'm not afraid to feel the love but I don't feel that way

@EricSilva alright

then he said Te Amo

Does out-of-context song-lyrics-quoting count as noise?

like im not a native spanish speaker but ive been to most of latin america so i speak only about those basically

5:45 AM

@LeakyNun And I'm gone gone gone / Now I'm older than movies
Let me dance away / Now I'm wiser than dreams

etc

Rihanna - Te Amo

►

think it means "I love you"

@LeakyNun Actually, I should have said "IT GOES IT GOES IT GOES IT GOES GUILLOTINE YAH"

i feel like a third of the chat is probably just noise

5

@LeakyNun How to solve this^?

@Abcd no idea

5:58 AM

ok

Anyone else?

Under those conditions, what are the particle's radius vectors as a function of time elapsed $t$?

@Semiclassical There are two particles. (I can't understand your question too)

particles', then.

What's the position of each particle as a function of time?

@Semiclassical r1(t) and r2(t)

And these equal...?

You know the initial positions and the constant velocities. That's enough to say what r1(t),r2(t) are.

6:07 AM

@Semiclassical $r_1(t) = v_1*t$

If that's true, then $r_1(0)=v_1\cdot 0=0$.

So that'd require $r_1(t)$ start at 0 at time t=0.

oh

@Semiclassical $v_2 = (r(t) - r_2)/t$

Okay. So $\vec{r}_2(t)=\vec{v}_2t+\vec{r}_2$.

@Semiclassical yes

How about $r_1$?

6:11 AM

@Semiclassical $r_1(t) = v_1t + r_1$

(with vector arrows above the letters)

Sure, as vectors

Now, if they collide, they must be at the same position at a given time.

So what equation does that give?

Liad

6:44 AM

@Semiclassical hi

i need to find this integral where $\gamma$ is a curve around the ellipse $(\dfrac{x}{a}) \ ^ 2 + \dfrac{y}{b}) \ ^ 2 =1 $ . i used Green's theorem and got $0$. but when i open this integral by definition i get $2\pi ab$. so, how could it be ?

@PVAL-inactive hi

Liad

7:26 AM

Also when I want to calculate the volume of the region above the cone and inside the ball $ x \ ^ 2 + y \ ^ 2 + (z-1) \ ^ 2 \lt 1 $ , we have $\phi \in [0,\pi /4] \theta \in [0,2\pi] , r \in [0 , 2cos(\phi) ] $ and the volume turns out to be $3\pi$. this make no sense because the volume of the ball is $4 \pi /3$. someone can help ?