I think this is close to what I was getting. I need to express separation constants just in terms of C^2 (at least that is my impression). If the 1/r^2 were not there, I could say: $\alpha^2$ + $\gamma^2$ = C^2

I'm guessing you're getting Helmoltz equation from separating time and space from a cylindrical wave equation. If you need to quantize $\lambda$ and $C$, then boundary conditions are needed.

So you suspect that without boundary conditions, I cannot get an expression to show a relationship between separation constants and C^2?

Well, if by a relationship you mean $C = C(\lambda)$, then no. The solution of the system is \begin{align}\Theta(\theta) &= A e^{\sqrt{\lambda} x} + B e^{-\sqrt{\lambda} x} \\ R(r) &= C J_\sqrt{\lambda}(C x) + D Y_\sqrt{\lambda}(C x)\end{align} where $J_\sqrt{\lambda}$ and $Y_\sqrt{\lambda}$ are Bessel and Neumann functions of the first kind, of order $\sqrt{\lambda}$. If no boundary conditions are given, then you have an infinite collection of solutions spanned by both parameters.

Could I divide both sides by r^2 and R to get B^2 by itself?

Do you see a way in which I could get a^2 + b^2 = C^2?

The separation I've shown you is correct. What is $a$ and $b$? As I told you, the relationship between $C$ and the separation constant comes from the boundary conditions. Why don't you edit your question and show us how did you derived the relation $a^2 + b^2 = C^2$? Here is a quick guide on how to typeset equations.

$a^2$ is a constant and $b^2$ is the other constant. Together they add to $C^2$. Imagine if there hadnt have been a 1/r^2 term. We would have had the addition if two constants = $C^2$

I need to state B.C's for just r term. I did that. Then I need to write down any auxillary B.C.'s I would need to solve the problem (not including theta). I think this means I need to show a relationship.

Is there anyway to get this in Helmholtz form?

I don't understand why you have two constants if you've only done one separation of variables. When you say no $\frac{1}{r^2}$ term, you mean no derivatives on $\theta$?

Are you online? Can you chat about this right now?

To give you more understanding of my problem: First I want to derive the ODE for the one dimensional functions f(r) and $g(\theta)$

Then I want to do derive appropriate boundary conditions for the function f(r) - which I have done

Now, I need to State any additional auxiliary conditions necessary to complete the solution, not including Boundary conditions for theta

Meaning I need to come up with some type of equation that relates these constants to each other. The B^2 can get grouped in with other constants, but do you see a way to do this?

I'm here. Can you give details on what you've done?

So I first started with the original equation posted in the forum

I use ansatz (given) as phi(r,theta) = f(r)g(theta)

I plug that in, then divide by ansatz

Then I multiplied by r^2

Now I have separate variables completely

and end up with the equations i've posted, right?

From here, how can I obtain a boundary condition that only depends on separation constants, and are not functions of r or theta?

Correct, I think it is similar (just different notation)

I know that I can say say the the one term only dependent on theta

can be set equal to say, alpha^2

or rather

-alpha^2

But then that would mean the other part of the equation would have to equals negative of that to add up to zero.

But then I just get 0 = 0 and that is not a valid BC

Ah, but what you are doing there is giving boundary conditions to $\theta$, i.e., periodicity of solutions

No, like I said, I am not suppose to give B.C.'s for theta

I am not solving the problem

I just need to set up B.C.'s for f(r) and any others (separation constants)

what's the original problem?

not only the equation, the hole statement

We have an equation in two dimensional polar coordinates

Assuming the solution is separable in r and theta

in the form phi(r,theta) = f(r)g(theta)

a) Derive the ODE for one dimensional functions f(r) and g(theta)

b) Derive appropriate B.C. for f(r)

c) State any additional auxiliary conditions necessary to complete the solution, not including B.C.'s for theta

And for c, I need some relationship between separation constants

It seems to me that the problem is not well stated

What would be your guess as to how to derive a relationship between separation constants?

To derive conditions for f(r) you need, at least the domain

I am given that

0 < r < R

with equal to under <

0 < theta < 2pi

with equal to under <

Ok. Since 0 is in the domain of definition, the radial solutions can't have Neumann functions

As they are discontinous in 0

Should I just set them both equal to separation constants?

and say -alpha^2 = gamma^2?

Not just yet. The angular variable is between 0 and 2pi

Implicitely, you are assuming the solutions must be periodic in theta

right, but I am not suppose to write down B.C.'s for theta

Just one more B.C. as a function of separation constants

well, because the question is not well stated

whats the original domain of the pde?

I just told you

That is the information given

Ok. The thing is that, whit that information, the problem is not well stated

Dont i need to trasnfer this into form of Helmholtz equation?

The pde is the Helmholtz equation.

Let's start from the begining

The assumption that the equation is separable, i.e. Phi(r,theta) = f(r) g(theta), leads to two ordinary differential equations, where a separation constant is used to derive them

what if

the separation constant is a new unkonwn

I Set the whole term dependent on r = constant

then divide by r^2

this gets rid of r^2 in B^2 term

That would be wrong.

ok

The separation constant is determined by the boundary conditions

why?

Because if you don't have boundary conditions, then you have an infinite collection of solutions

so if you were made to create a relationship between separation constants, what would be your guess?

My guess would be that you're expecting periodic solutions in the theta variable

I group r terms

right, but he explicity said not to use that

and to find it based on just seapation constants

then the hole probem is meaningles

you know how we can say that g(theta) term is alpha^2

look at tis

mathworld.wolfram.com/…

they set theta term = -m^2

or m^2

Because they are assuming solution is periodic on theta

ok well lets follow their method

if solution is periodic on theta, then lambda cannot be every value

if I say that it is equal to -m^2

could I not just say the other term equals n^2?

since it all depends on r?

No. You need conditions on the radial solution.

I know those conditions

phi(R,0) = 0

phi (0, theta) < INFINITY

sorry

the first should be

phi(R,theta) = 0

Ok. Now, that makes sence

Sorry for that

do you see how to get a relationship between separation constants? I cannot find a way that i feel like is right

You know that f(r) = A J_m(C r) + B Y_m(C r)

to be clear, is the direction your headed, just a relationship between separation constants and not variables r and theta?

The condition that phi(0,theta) should remain bounded means that no Y_m should appear

yes

Right, I remember him doing somethign like that in class

Meaning B=0

We just have first form of Bessel

and then f(r) = A J_m(C r)

What is C suppose to be?

now, the second condition f(R) = 0 means J_m(C R) = 0

Oh i see, from helmholtz equation

right

Shouldnt it be J_o?

from your boundary condition

no

is J_m

for m integer

But is the C related to the C^2 in the orginal problem?

we are not done yet

we will get there

ok

J_m(C R) = 0 means that C*R must be a zero of J_m

but J_m has infinitely many zeros

ok

from an example he did in class

he had

then G was 0

so f(r) = F*Jo(alpha*r)

then he defined Y to be the zeros of Jo

and said Jo(Y) = 0

sorry, Jo(Yn)

=0

then fn(r) = Fn*Jo(alpha_n,r)

Right. This is the same, only that m is not equal to zero

=

and then Fn*Jo(Yn, r/R)

sorry, my internet is going very slow

thats fine

so whats next?

Bessel functions J_m have iinfinitely many zeros

so whats next?

in the example the teacher gave you, you are working with J_0

so, let alpha{0n} be the n-th zero of J_0

then the condition your teacher gave is C R = alpha_{0n}

or C = alpha_{0n}/R

in your case, you have J_m

i.e suppose alpha_{mn} is the n-th zero of J_m

then C R = alpha_{mn}

or C = alpha_{mn}/R

ok

I am not sure yet how this will get the relationship

hence, for each m, there are infinitely many constants C that fulfill the boundary condition

but I follow what you did

C_{mn} = alpha_{mn}/R

so, solutions are J_m(C_{mn} r)

is that C_mn*r?

or , r

times

that isnt an equation though

do you have chatjax?

does it equal 0?

how do i get it?

math.ucla.edu/~robjohn/math/mathjax.html

never mind, though. Lets do a simpler example

Laplacian phi + C phi = 0, where phi=phi(x,y)

how do i install it?

just drag the link to your bookmarks bar

Well I need to understand my examples

this is the same, just that, instead of bessel functions, you have sins and cos

much more intuitive

I think i am suppose to use bessel

thats the form he used

did you follow all steps until J_m(C R) = 0?

I added it but when I go to bookmarks and click it

it doenst do anything

No? how do you see $J_m(C r)$?

oh it works!

;)

can you copy paste what you had earlier, it be clearer to me

Ok. Lets go to the top.

( I really appreciate this)

You have $\Theta'' + m^2 \Theta = 0$

Where $m \in \mathbb{R}$

no

sorry

$m \in \mathbb{Z}$

ok

I would certainly agree with you there

the function of theta seems easier than the function of r

Then $\Theta$ is not a function of $\theta$ only, but also of $m$. This is usually denoted as $\Theta_m(\theta) = A \sin(m \theta) + B \cos(m \theta)$

the m is just the constnat?

Yes, but not any real constant

it has to be a natural number

any

this is to ensure periodicity of solutions, i.e. $\phi(r, \theta + 2 \pi) = \phi(r,\theta)$

my teacher (For that part) had:

so, you have a family of solutions

$g(z)=C1sin(\lambda*z)+C2*cos(\lambda*z)$

then

$g_m(z) = A_n*cos(\lambda_m*z)$

ok, so whats next?

The reason behind that is that, you have a second order pde on the disk, right?

you need a function that is, at least, of class $C^1$, right?

yes

then if you assume separation of variables for your solution, it must have the same value for $\theta=0$ and $\theta=2 \pi$

right

and so its derivative

this translates to $\phi(r,0) = \phi(r,2\pi)$ and $\phi_\theta(r,0) = \phi_\theta(0,2 \pi)$

ok

that makes sense

just periodicity

wiat

how could $\phi$ at r,0 be the same as at 0,2pi?

Exactly, you dont want your solution to jump when you do go around the origin.

Ahhh, that's the question

ok

so

we assume $\phi(r,\theta) = f(r) g(\theta)$

right

then $$\frac{r^2 f''(r) + r f'(r)}{f(r)} + \frac{g''(\theta)}{g(\theta)} + r^2 C^2 = 0$$

Then $$\frac{r^2 f''(r) + r f'(r)}{f(r)} + r^2 C^2 = - \frac{g''(\theta)}{g(\theta)}$$

i follow well so far

and the left side is a function of $r$ while the right side is of $\theta$. The only way this equality is fulfilled for all $(r,\theta)$ is that both sides are equal to a constant

i.e. $$\frac{r^2 f''(r) + r f'(r)}{f(r)} + r^2 C^2 = \mu = -\frac{g''(\theta)}{g(\theta)}$$ where $\mu$ is a separation constant

so \begin{align} r^2 f''(r) + r f'(r) + (C^2 r^2 - \mu)f(r) &= 0 \\ g''(\theta) + \mu g(\theta) &= 0 \end{align}

so far so good?

looks good

:)

Good!

Now, to this point, we don't know anything about $\mu$

what we know is that $\phi(r,0) = \phi(r,2 \pi)$ and $\phi_\theta(r,0) = \phi_\theta(r,2 \pi)$

meaning $\phi$ must be differentiable inside the disk of radius $R$

Translating this into our anzats, we have $g(0) = g(2 \pi)$ and $g'(0) = g'(2\pi)$

$g$ must be periodic

of period

$2 \pi$

ok

Given that $g'' + \mu g = 0$ implies $g(\theta) = A \sin(\sqrt{\mu} \theta) + B \cos(\sqrt{\mu} \theta)$

$\mu$ cannot be any constant

right?

we need period $2 \pi$

right

hold on one second

wouldnt it be easier to call constant $\mu^2$

that way we dont have radical?

yes, but then you would be assuming that is positive

and you don't know that just yet!

ok

is the periodicity that ensures you that $\mu$ is positive

and the period $2 \pi$ means that $\sqrt{\mu} = m$, where $m$ is an integer

so $\mu = m^2$

ok

I understand

(the difference of sign form wolfram's site is because I assumed $\mu$ as separation constant. If I'd assumed $-\mu$, it would be the same)

so, $\mu = \mu_m = -m^2$

periodicity implies that not every real constant is usefull, only squared integers

why is it negative?

my mistake

$\mu_m = m^2$

sorry about that

thats fine

then $g(\theta) = g_m(\theta) = A_m \sin(m \theta) + B_m \cos(m \theta)$

and $$r^2 f''(r) + r f'(r) + (C^2 r^2 - m^2) f(r) = 0$$

This is a scaled version of Bessel's differential equation of order $m$

and it has as solutions $$f(r) = c_1 J_m(C r) + c_2 Y_m(C r)$$

The condition that $\phi$ remains bounded on $r=0$, implies that $c_2 = 0$, so $$f(r) = c_1 J_m(C r)$$

are you following so far?

one question

just to be clear

earlier

$\phi_\theta$ is derivative?

yes

ok

then i follow

great

where do we go from here?

now, the next condition, $\phi(R,\theta) = 0$ implies that $f(R) = 0$

or $J_m(C R) = 0$

meaning $$C R = \text{zero of } J_m$$

just as $\mu$ couldn't be any constant, $C$ can't be either

$$C = \frac{\text{zero of } J_m}{R}$$

ok i follow

in our notation $$C = C_m = \frac{\text{zero of } J_m}{R}$$

i.e., for each $m$, there is a $C$, or $C= C(m) = C_m$.

right

as it happens, $J_m$ has not just one zero, but infinitely many zeros

If we denote them by $\alpha_{nm}$

then $J_m(\alpha_{mn}) = 0$

So

for each $m$

there is not only one $C_m$, but infinitely many

$$C_{mn} = \frac{\alpha_{mn}}{R}$$

that fulfill the boundary condition

but where can I get a relationship just with constants?

That's your relatioship

the $\alpha$'s are constants

but what happened to the mu's?

they are fixed now

Ok, I guess the main concern I have is this:

My professor wants an auxillary equation in which we do need to apply Boundary Conditions to thetea

$$C(\mu) = \frac{\alpha_{\sqrt{\mu} n}}{R}$$

I feel like we used that extensively

Now I see what your teacher was asking

What do you mean?

We used the B.C. for theta

he said not too

yes

is as if we should started the other way around

im not sure that i follow

suppose we don't know the bc's for $g$

right

we know $$r^2 f''(r) + r f'(r) + (C^2 r^2 - \mu)f(r) = 0$$

right

with $|f(0)| < M$ and $f(R) = 0$

then $$f(r) = c_1 J_\sqrt{\mu} (C r)$$

ok?

ok

so, $f(R) = 0$ means $$C = \frac{\text{zero of } J_\sqrt{\mu}}{R}$$

that is the relationship you are looking for

If we called the zeros $Y$

then it would be

C = $Y_(sqrt(\mu))/R$

right?

I'm guessing you are trying to write $$C = \frac{Y_{\sqrt{\mu}}}{R}$$

yes

and you are saying that would be the auxiliary equation with just separation constants?

this is done by typing C = \frac{Y_{\sqrt{\mu}}}{R}

exactly

the thing is that you have infinitely many zeros

not just one

has is the infinitely many zeros expresed here

shouldnt there be an m ?

so, the auxiliary equation would be $$C = \frac{Y_{\sqrt{\mu} n }}{R}, \qquad \text{where } Y_{\sqrt{\mu} n} \text{ is a zero of } J_\sqrt{\mu}$$

for each $\mu$, there are infinitely many $C$'s

ok

and we are done?

yes

ok

i think i understood the process

i appreciate your hlep

a lot

No problem. I'll edit my answer for further reference :)

sounds good!

I accepted it

;)

See you around then. A good book for reference is Churchill and Brown's Fourier Series and Boundary Value Probelms