Separating Partial Differential Eq

Question

I have a PDE:

$$ \frac{\partial^2\phi(r,\theta)}{\partial r^2} + \frac{1}{r}\frac{\partial\phi(r,\theta)}{\partial r} + \frac{1}{r^2}\frac{\partial^2\phi(r,\theta)}{\partial\theta^2} + C^2\phi(r,\theta)=0 $$

I need to separate the PDE (just functions of r,theta) and show the relationship between the separation constants and $C^2$. I need to use solution of $\phi(r,\theta)$ = $f(r)g(\theta)$. When I do that, then divide by solution, I do not see how I can separate $g$ from $1/r^2$ without having the $C^2$ change to $C^2r^2$.

Any ideas?

I have no idea what $+C^2(r,\theta)$ means. Would you please write out the original PDE so that it can be clearer to the rest of us? — Cameron Williams, Nov 25 '13 at 03:24
I would but I am not sure how to use mathjax. 1/f(r)d^2f(r)/dr^2 + 1/f(r)1/rdf(r)/dr + 1/g(theta)1/r^2d^2g(theta)/dtheta^2 + C^2 = 0.
This is after I applied solution then divided by solution — Jackson Hart, Nov 25 '13 at 03:25
original is d^2phi(r,theta)/dr^2 + 1/rdphi(r,theta)/dr + 1/r^2d^2phi(r,theta)/dtheta^2 + C^2*phi(r,theta) — Jackson Hart, Nov 25 '13 at 03:26
To write fractions, the notation is \frac{numerator}{denominator} and to write in Mathjax, put your LaTeX code in dollar signs. — Cameron Williams, Nov 25 '13 at 03:27
Good! Now to do Greek letters, you do \phi, \theta, etc. Or if you want the upper case Greek letters, \Phi, \Theta, etc. — Cameron Williams, Nov 25 '13 at 03:30
And as a stylistic choice, do not include asterisks in LaTeX code (except in very specific instances). Asterisks are generally frowned upon. — Cameron Williams, Nov 25 '13 at 03:32
I just voted to delete other post. I completely wrecked the mathjax — Jackson Hart, Nov 25 '13 at 03:37
@CameronWilliams I need to express the separation constants in terms of C^2. Normally this is easy. However, here, the 1/r^2 must be multiplied meaning C^2 gets group in and I do not know how to make a relationship between C^2 and separation constnats — Jackson Hart, Nov 25 '13 at 03:43
I see your point. This is a pretty tricky one to me. I'm not sure how to do it immediately. Perhaps you'll need to do some sort of Bernoulli trick for the radial part of the SoV. — Cameron Williams, Nov 25 '13 at 03:52

Pragabhava · Accepted Answer · 2013-11-25T20:43:32.670

1

Well, if you suppose $\phi(r,\theta) = R(r) \Theta(\theta)$, then

$$ R'' \Theta + \frac{1}{r} R'\Theta + \frac{1}{r^2} R \Theta'' + C^2 R \Theta = 0 $$

Multiplying by $r^2$, you have

$$ r^2 R'' \Theta + r R'\Theta + R \Theta'' + r^2 C^2 R \Theta = 0 $$

Dividing by $R \Theta$

$$ \frac{1}{R}\left(r^2 R'' + r R'\right) + \frac{\Theta''}{\Theta} + r^2 C^2 = 0 $$

Then $$ \frac{1}{R}\left(r^2 R'' + r R'\right) + r^2 C^2 = - \frac{\Theta''}{\Theta} $$

And given that the left hand side depends only on $r$ and the right hand only on $\theta$, you have

$$ \frac{1}{R}\left(r^2 R'' + r R'\right) + r^2 C^2 = - \frac{\Theta''}{\Theta} = \lambda $$

where $\lambda$ is a constant. Then \begin{align} r^2 R'' + r R' + (C^2 r^2 - \lambda) R &= 0 \\ \Theta'' + \lambda \Theta &= 0 \end{align}

The first one is a scaled version of Bessel equation, while the second one is the harmonic oscillator.

EDIT

Here is the transcript of the chat I with the OP, where a full answer is given.

edited Nov 25 '13 at 20:43

answered Nov 25 '13 at 03:54

Pragabhava

4,903

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 – Jackson Hart Nov 25 '13 at 03:57
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. – Pragabhava Nov 25 '13 at 04:01
So you suspect that without boundary conditions, I cannot get an expression to show a relationship between separation constants and C^2? – Jackson Hart Nov 25 '13 at 04:05
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. – Pragabhava Nov 25 '13 at 04:09
Could I divide both sides by r^2 and R to get B^2 by itself? – Jackson Hart Nov 25 '13 at 05:31
Do you see a way in which I could get a^2 + b^2 = C^2? – Jackson Hart Nov 25 '13 at 05:41
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. – Pragabhava Nov 25 '13 at 06:04
$a^2$ 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 = $C^2$ – Jackson Hart Nov 25 '13 at 06:28
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. – Jackson Hart Nov 25 '13 at 06:42
Is there anyway to get this in Helmholtz form? – Jackson Hart Nov 25 '13 at 07:04
let us continue this discussion in chat – Pragabhava Nov 25 '13 at 15:58

Separating Partial Differential Eq

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