# Circular plate temperature distribution.

1. May 6, 2015

### davidbenari

1. The problem statement, all variables and given/known data
The temperature distribution of a circ plate with radius $a$, $f(r,\theta,t)$, follows the diffusion equation $\nabla^2f=\frac{1}{\alpha^2}\frac{\partial f}{\partial t}$, with a temperature of zero along the border. The initial temperature is $f(r,\theta,0)=100rcos\theta$. Find the temperature distribution.

2. Relevant equations

3. The attempt at a solution
I've found the typical solution that is $\sum_{m=1}^{\infty} \sum_{n=0}^{\infty} J_n (\frac{k_{mn} r}{a})(A_{mn}cosn\theta+B_{mn}sinn\theta)exp(\frac{-\alpha^2}{a^2} k_{mn} t)$ Where $k_{mn}$ is the mth zero of the nth bessel function.

Now I've found this at the initial time. $\sum_{m=1}^{\infty} \sum_{n=0}^{\infty} J_n (\frac{k_{mn} r}{a})(A_{mn}cosn\theta+B_{mn}sinn\theta)=f(r,\theta,0)=100rcos\theta$ To use an approach similar to finding a Fourier series. But I'm stuck here. I haven't a clue what to do here. Integrals are way to tedious here, and I don't see any way I'm getting a simple result.

Any ideas?

thanks

Last edited: May 6, 2015
2. May 6, 2015

### davidbenari

UPDATE:

I've obtained that the only term that survives is the Am1 series and have calculated that each Am1 is

$A_{m1}=\frac{200}{a^2 J_2 (km1)} \int_{0}^{a} J1 (\frac{km1}{a}r)*r^2dr$

Last edited: May 6, 2015
3. May 6, 2015

### davidbenari

Im just proud of my result so I'm gonna post it, hoping that someone will tell me if its wrong right or too complicated and can be simplified...

The integral $\int_{0}^{a} J1(\frac{Km1}{a}r)*r^2dr$ can be solved using the recursion relation $\frac{d}{dx}[x^2J2(x)]=x^2J1(x)$. Wit this you obtain, after some manipulation:

$r^2 J1(\frac{Km1}{a}r)=\frac{a}{Km1}\frac{d}{dr}[r^2J2(\frac{Km1}{a}r)]$

If this is substituted in the integral you'll get that the integral is equivalent to $\frac{a}{Km1}\int_{0}^{a}\frac{d}{dr}[r^2 J2(\frac{Km2 r}{a})]dr = \frac{a^3}{Km1}J2(km1)$

And therefore substituting this into the previous post one gets that

$Am1=\frac{200a}{Km1}$

This result looks so simple and familiar that I'm worrying that I could've done things simpler...

Any ideas?

4. May 6, 2015

### davidbenari

The final distribution $f(r,\theta,t)=\sum_{m=1}^{\infty}\frac{200a}{Km1}J_1(\frac{Km1}{a}r)cos\theta e^{\alpha^2 \frac{Km1}{a^2}t}$

5. May 6, 2015

### Zondrina

I haven't written all the calculations out, but looking over your work it all seems okay.

You mentioned the wrong Bessel identity, but you used the correct one when actually calculating the final result. I think you meant:

$$\frac{1}{\alpha} \frac{d}{dr} [r^{\nu} J_{\nu}(\alpha r)] = r^{\nu} J_{\nu - 1}(\alpha r)$$

Where $\alpha = \frac{k_{m1}}{a}$.

6. May 7, 2015

### davidbenari

my professor says in one of his notes that f=0 at the borders then this means your solution doesn't depend on $\theta$. Isn't this completely wrong?

7. May 7, 2015

### HallsofIvy

Staff Emeritus
If the function value is a constant at the circular boundary (not just 0) then the solution is independent of $\theta$.
In your original post you say
If there is "a temperature of 0 along the border" as a boundary condition, the initial temperature cannot be $100 r cos(\theta)$.

8. May 7, 2015

### davidbenari

I know this seems weird but it is what is accepted in my book (Boas mathematical physics) and in my class.

9. May 7, 2015

### davidbenari

10. May 7, 2015

### davidbenari

There you clearly have the same problem with discontinuity

11. May 7, 2015

### davidbenari

1. Find the steady-state temperature distribution in a metal plate 10 cm square if one side is held at 100◦ and the other three sides at 0◦. Find the temperature at the center of the plate.

This problem would also necessitate using discontinuity. The inferior side can have 100º then, when you hit the vertical wall, you'll get 0º suddenly.