@Danu pretty classic joke

slightly related:

@DavidZ do you do a lot of work with PDEs?

Or know of anyone on this site that does?

Too late @usukidoll :(

I'm almost done with it anyway

so no biggie... just I think one more proof and 3 more computations

one is fourier transform heat equation...one is a very nasty computation involving laplace

You could leave your questions here and asked to be pinged

If you get stuck.

Ayo!

Wow you two are on math too

lol

:D

Do you know how to solve the two body problem without using the center of momentum transform?

Nope, sorry

@skullpatrol personally no, I'm not sure if @KyleKanos might

math majors don't do this

lol

@DavidZ Do you? Or am I just saying nonsense?

@skullpatrol why, what did you want to know?

@Anthony the two-body problem? And by "center of momentum transform" you mean the thing where you switch to one radial coordinate and use the reduced mass?

Yeah.

Sorry, in particular I mean something like two masses joined by a spring.

The equations seem utterly useless without defining a radial coordinate, but I feel like I shouldn't need to.

Is this for 1D or 2D/3D motion?

1D.

@DavidZ it was @usukidoll that had some questions

omg laplace equation with Neumann Boundary Conditions could be the hardest in the computation section ugh

@Anthony Ah well in that case, you can write a differential equation using matrices and just solve it directly. At some point you will have to diagonalize the matrix and the radial coordinate will emerge out of that.

@skullpatrol oh ok, well... I dunno, I might be able to say something useful, but mostly PDEs are just a pain in the butt. Personally when I do need to solve one, I do it using Mathematica. cc @usukidoll

so how do I do mathematica? and p.s this is what the problem looks like...in a sec

what the hell?! these B.C's suck that last line should be I.C @DavidZ

@DavidZ I think I understand what you mean, but I have no idea how I would go about it.

it's on a disk, so the B.C should be that $= f(\theta)$ just what is this mess?

This is driving me nuts.

The free oscillation frequency is $\omega_0 \sqrt{1 - (\beta/\omega_0)^2}$, but I don't think that's the same thing as the driven resonance.

@usukidoll don't use Mathematica while you're learning

@DanielSank well the resonance frequency is the oscillation frequency at which the oscillator's amplitude is a maximum

@DavidZ I am not, but do you know what I need to do in this problem? There's so many issues with this thing. I don't know where to begin.]

I thought it was the same as the free oscillation frequency but maybe I remember wrong

@usukidoll well, since you're given boundary conditions in terms of $U$, not $u$, I'd suggest starting by writing the equation in terms of $U(r,\theta)$

I made this a question: physics.stackexchange.com/questions/153197/…

hmm does it have something to do with harmonics.. I know if I am on a disk I should have a $f(theta)$, but since I have Neumann.... the whole thing changes

@DavidZ meaning turn the laplace equation into it's polar coordinate form

@DavidZ: If you calculate the highest amplitude point you get $\omega_0 \sqrt{1 - 2(\beta/\omega_0)^2}$

arg

@Anthony Well, you'd start by writing the equations for the two masses' coordinates, $x_1$ and $x_2$, and then you could write it in matrix notation

You can also do this without using matrices

@usukidoll harmonics don't come into it until near the very end

@DavidZ $U_{rr}+\frac{1}{r}+\frac{1}{r^2}U_{\theta \theta}=0$

I'm utterly lost either way, can I make matrices on here? $\begin{matrix} 1 & 1 & 1\\ \end{matrix}$

a ha so I do need harmonics after all. I was reading a pdf about it

Yes.

@DavidZ Nice ^^

@usukidoll: For what it's worth, understanding how to use matrices here will really help.

@usukidoll depends. If you're asked to find the most general solution (which is a linear combination of individual solutions), then you use harmonics. But at the point you're at now, you don't need them.

next XD

@Anthony seems like you're missing the coordinates $x_1$ and $x_2$

@usukidoll something's wrong with your DE - the units don't match in the $\frac{1}{r}$ term

I'm guessing that should be $\frac{1}{r}U_r$ (or maybe $\frac{2}{r}U_r$, I forget)

aren't the force equations $k(x_2-x_1)=m\ddot{x_1}$ and $k(x_2-x_1) = -m\ddot{x_2}$?

oops forgot a U_r

@Anthony Yep, those are the ones.

Is that not what that matrix means?

@Anthony I mean, what you wrote was a matrix equation, but the matrix equation you wrote doesn't match the equation you want.

so... we have Neumann Conditions.. does it use that very long integration type thing

Why not? :/ I don't understanddddd.

@Anthony that equation is equivalent to four individual equations: $k = m$, $-k = 0$, $k = 0$, and $-k = -m$

Oh!

Crap.

Think about how you would write $k(x_2 - x_1)$ in matrix notation. It'd be something like $kM\begin{pmatrix}x_1 \\ x_2\end{pmatrix}$

where $M$ is some 2 by 2 matrix that you should figure out

@usukidoll I don't know what very long integration type thing you're talking about

if harmonics aren't needed, what do I need to do to solve this laplace equation with Neumann Boundary conditions and I'm on a disk

Generally solving a PDE involves converting it to a set of ordinary differential equations. How can you do that?

WHAT THE!

there is no converting in here

uh... wait, how have you solved 2-variable PDEs in the past then?

I am dealing with Linear pdes

@Anthony Yeah, that's it

So then what do I do from there?

I'm guessing you know how to solve the differential equation $k x = m\ddot{x}$?

Yeah.

OK, well... oh, wait, actually that comes a little later.

Well, there are two ways to go from here.

The way I had in mind was that you find the transformation that diagonalizes the matrix on the left. In other words, you find 2-by-2 matrices $S$ and $D$ where $D$ is diagonal and $S^{-1} D S = \begin{pmatrix}-1 & 1 \\ -1 & 1\end{pmatrix}$

for neumann to work, I need to satisfy the compatibility condition'

But that matrix isn't invertible, can I still diagonalize?

oh... hm, yeah, that just means it has a zero eigenvalue.

$\int_{0}^{L} g(x) dx - \int_{0}^{L} f(x) dx + \int_{0}^{M} k(y) dy + \int_{0}^{M} h(y) dy =0$ that's the compatibility condition from my book @DavidZ

Oh wait am I dumb.

It only has one eigenvector.

:(

@usukidoll whoa, I haven't seen that equation in a long time.

I need that @DavidZ or my Neumann condition goes out the window... so do I need this to solve my crazy Laplace problem?

@usukidoll I think you only apply that in the very end, the same point where you worry about harmonics.

Maybe you've learned a different method of solving the equations than I'm familiar with though.

ughhhh so what is the middle in this problem?!!??!?!??!!? @DavidZ

I can't digonalize this then, eh?

@Anthony huh... well, maybe that method doesn't work then.

So why exactly can't you use the difference between the $x$'s as your one coordinate?

I mean I can, and I understand that that works.

I just want to know how to do it another way

Yeah, but why do you expect there to be another way?

I'm not sure, is that a common feature of Differential Equations?

Can't you normally decouple equations?

?

any ideas @DavidZ

@Anthony well, yeah, you can. When the matrix is singular then you have to use more complicated techniques than eigenvalues and eigenvectors.

If you just move the negative sign out of the right side and into the left, then you can diagonalize it.

Maybe singular value decomposition. To be honest I've never really looked into this before; when I'm dealing with singular matrices I can always just add or subtract the equations.

@Anthony hm?

Is that fair?

OH duh, silly me, yes. Actually you have to do that.

I totally missed that your $x$ vector on the left wasn't the same as the one on the right.

You need to write the equation as $kM\vec{x} = m\frac{\mathrm{d}^2}{\mathrm{d}t^2}\vec{x}$

Oh I see.

and what you had before wasn't of that form, because on the left you had $\begin{pmatrix}x_1 \\ x_2\end{pmatrix}$ but on the right you had $\begin{pmatrix}x_1 \\ -x_2\end{pmatrix}$. Sorry, I should have noticed that.

@usukidoll In the very beginning of the PDF you linked me to, it gives you an example of how to solve a 2-variable PDE by separating it into two (1-variable) ordinary differential equations. See what you can get from that.

@Anthony You're still leaving out the $\vec{x}$'s

Like that?

> You're still leaving out the $\vec{x}$'s

Oh you mean you want the actual variable.

How do you make the arrow?

-.- separation of variables? really?

vec

@usukidoll yep. That would be my first approach.

You mean that?

Kind of like that, but it's still not of the form $S^{-1} D S$ on the left

You need to make it so that the first matrix is the inverse of the last

Is it not...?

@DavidZ what occurs after separation of variables?

@usukidoll then you solve each of the individual one-variable equations and construct the solution to the PDE as a sum of products of the individual solutions

It's all in the PDF file

but how the heck do I find $\alpha$ @DavidZ

@Anthony nope, $\begin{pmatrix}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{pmatrix}$ is not the inverse of $\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}$

@usukidoll first you have to solve the PDE for arbitrary $\alpha$, and then at the end you see which values of $\alpha$ are compatible with the boundary conditions

solve the pde for alpha? huh

No, I said "solve the PDE for arbitrary $\alpha$." Just ignore the fact that you don't know $\alpha$ for now.

I don't see why... $\frac{1}{2}+\frac{1}{2} = 1$, $\frac{1}{2}-\frac{1}{2} = 0$....

@Anthony oh, whoops, I miscalculated, sorry. Comes from trying to do four things at once I guess.

?

You're right, it is.

I suppose. :P I appreciate it.

Anyway so your equation is of the form $kS^{-1}DS\vec{x} = m\ddot{\vec{x}}$

Yeah.

then you multiply by $S$ to get $kDS\vec{x} = mS\ddot{\vec{x}}$

and then if you calculate $S\vec{x} = \begin{pmatrix}r_1 \\ r_2\end{pmatrix}$ (i.e. find $r_1$ and $r_2$ in terms of $x_1$ and $x_2$)

then you have uncoupled differential equations for $r_1$ and $r_2$

Haha. I hate it. :(

Try it ;-)

I mean, you can multiply both sides by $S$ surely?

Yeah. I meant I hate it because I think it's going to give the same thing.

But now I'm confused again, but okay multiply. Then you multiply $DS$.

Well yeah. If it didn't give the same thing it wouldn't solve the problem!

lol

@Anthony no, you leave $D$ and $S$ separate

Oh, alright.

It's $S\vec{x}$ you want to find (and call that $\vec{r}$)

It's just $\begin{pmatrix} x_1 + x_2 \\ x_1 -x_2 \end{pmatrix}$

Well, $S$ was the other matrix, wasn't it? $\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}$

But yeah, that is the idea

Haha. Sorry, yeah.

So now you have $kD\vec{r} = m\ddot{\vec{r}}$

where $D$ is diagonal

and thus the equations are uncoupled

I see, however, we used a change of variables.

Is this sometimes necessary for differential equations?

@Anthony Yes... this is what I meant when I said the switch of coordinates pops out of the math.

I know sometimes you can solve for one in one equation, then plug it into the other one.

In general, methods of solving ODE's are just... Scrambled?

I'm not sure if you can do that here.

Agh.

So frustrating.

I wonder why not...

@Anthony yeah, there's no overriding logic to it. There's an assortment of methods and you have to hope one works for your specific case.

I guess what you could do is take the derivative of one of the equations twice, and then substitute into the other

maybe

because you'd then have one equation involving $x_1$, $x_2$, and $\ddot{x}_1$, and the other involving $\ddot{x}_1$, $\ddot{x}_2$, and $d^4 x_2$

(hope it's clear what I mean by $d^4$)

it seems like it would turn into a mess of differentiating things to try to get some variables to match up

Haha, yeah.

I wonder if that would work though.

I get worried when I start taking so many derivatives, I feel like I'm applying math incorrectly.

and even if you do get it to work, you'd have a fourth-order differential equation at the end, which means you'd need three additional boundary conditions

Oh, I see.

I guess you do need 3 BCs anyway, so that's not such a big deal

@Anthony well, as long as you don't actually make mistakes doing the derivatives, it's technically valid, it just may not be helpful

Substitution is better for when you have a system like $y' = f(x)$ and $x' = g(y)$

i.e. there's only one variable on the right side of each equation, or at least on the right side of one of them

Mmhm....

Anyway, for completeness I might as well mention the other method of solving, which goes all the way back to when you had $kM\vec{x} = m\ddot{\vec{x}}$

Wait you can't do derivatives.

hm?

Because any substitution you make is going to put in something involving higher order derivatives, won't it?

yeah, but it reduces the number of variables

That's kind of a general rule, you can trade derivatives for auxiliary variables

well, maybe not a general rule, but in the cases where you can actually solve for the right variables, when you can do a substitution, you trade a derivative for a variable

When your original equation is first-order in two variables, you can convert it to a second-order equation in one variable which might be easier to solve

oh yeah I see.

Anyway, on that other solution: you know that the solution to $kMx = m\ddot{x}$, if $M$ were just a number and $x$ were an unknown function (not a vector), would be $x = A\sin(\sqrt{kM/m}t + \phi)$ or something

Yeah.

so you can actually just write the solution to the matrix equation as $A\sin(\sqrt{kM/m}t + \phi)$, where the sine of a matrix is defined by its Taylor series expansion

Here $\phi$ would be a 2-by-2 matrix

Or something like that anyway. I might have missed a factor or something.

I see.

Thanks @DavidZ.

No problem

Anyway I should get dinner :-P

Later pal

(going back) @usukidoll honestly I think you might need to find someone who can help you in person for this. It would take a very long time and a lot of effort to work through it here in chat.

@skullpatrol see ya :-)

thanks for helping out

wait @DavidZ is going?

Perhaps @usukidoll you can post your work on the physics main site.

nah I need to go study some japanese before I sleep

See ya later my friend

@skullpatrol @usukidoll that probably would not be on topic for the main site. This isn't a homework help site. You'd have to ask about a very specific physics issue, but it doesn't seem like that's what you're asking.

Your question wouldn't be specific, and it wouldn't be about physics either. Chat is the right place for that sort of thing. Though in this case, as I said, it would take a long time to work through the details.

Actually @usukidoll what you could do is check the math homework forum at physicsforums.com. They don't have the length restrictions that our chat does, so maybe you can put all the relevant details in your post over there and get some better help.

@DanielSank: I sent you an email? Did you notice it? I am saying from the beginning that the problem is difficult and I don't exactly have hopes that someone can answer, except if serious research is done on the issue. Just, please be kind and tell me if my email came to you. It may be that the problem indicates that the QM treatment is unfit, and it should be replaced by a QFT treatment. But, this is just a thought.

Interesting that as I post this, the only answer to the actual question asked is in a comment

@ACuriousMind: since you've mentioned Emilio being the only score-40 member a couple days ago, I've moved up 3 from 26 to 29 (2 from 2 new badges, 1 from moving from 11k to 12k rep). Soon I too will be at 40 (in like 4 months)

@KyleKanos Someone's motivated :D

I am

I mean, it's probably entirely meaningless to get that 40

Nah, it's not meaningless. Just don't let it blow-up on you.

:D

Oh Cagney

The Classic.

@skullpatrol You were the one who asked DavidZ about solving PDEs last night, yes?

No, I was asking him to help usukidoll @KyleKanos

Ah, got it

She is very impatient.

A lot of people are impatient

True.

There are no royal roads.

hi guys

any nice posts lately?

:: shrugs ::

LOL I just saw this youtube link Danu had shared hahaha

wtf!!

When do you guys get your hats?

@Phonon Define nice post

7.5 hours?

Do we get hats?

We should get hats

Winter bash gives you one for free.

You mean real hats?

Define real on the internet :-)

Real hat = one that was sent to my home so that I can wear it on my physical head

Then the answer is no.

Okay. So you mean Winter Bash hats

stackexchange.com/promos/14/winter-bash-2013

yes

Well it should start tomorrow then

14/12/15 through 15/01/04, I think

guys how to explain to someone that you cant create torque with a force parallel to the axis of rotation, i cant put it into proper words i guess.

winterbash2014.stackexchange.com

The count down on that thing is pretty neat looking

@Gowtham tell them to try to open a door by pushing towards the hinges :-)

i gave them an analogy like holding a book on one of its corner and letgo. But i just dont know how to answer his comments, how to say that the weight of a door produces zero torque about the vertical axis. oh well time to sleep :/

sigh

Stupid little question: How can I do one of these 'chat quote' things

where I quote an earlier message from chat, with some kind of special layout to show that it's an earlier post

@Danu Just post the permalink to that message

It turns them automatically into that "quote", then

...ah, but only if you post only that, maybe?

yeah

thanks

Right, it does that only if the link in the only thing in your post