# Check: Temperatue Distribution On a Long Hollow Rod

1. Nov 9, 2008

### Schmoozer

1. The problem statement, all variables and given/known data

Consider a hollow rod whose initial normalized temperature is unity. At time zero
its normalized outside temperature is suddenly changed to zero. The inside surface
is perfectly insulated. The normalized equation and the BCS and IC are given as:

urr+$$\frac{1}{r}$$ur=ut
ur(a,t)=0
u(1,t)=0
u(r,0)=1

Find the function for expressing the temperature distribution.

2. Relevant equations

None

3. The attempt at a solution

$$\frac{\partial u}{\partial t}=\alpha[\frac{\partial^{2}u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r}]$$

u(r,t)=R(r)G(t)

$$\frac{1}{\alpha G}\frac{dG}{dt}=[\frac{1}{R}\frac{d^{2}R}{dr^{2}}+\frac{1}{Rr}\frac{dR}{dr}]=-\lambda^{2}$$

$$\frac{d^{2}R}{dr^{2}}+\frac{1}{R}\frac{dR}{dr}+\lambda^{2}R=0$$

$$R(r)=A J_0(\lambda r)+B Y_0(\lambda r)$$

$$J_0(\lambda_n a)=0$$ because $$A J_0(\lambda_n a)\neq0$$

$$u(r,t)=\sum_{n=1}^{\infty}a_n J_0(\lambda_n r)e^{-\lambda_n^{2} \alpha t}$$

$$u(r,0)=1=\sum_{n=1}^{\infty}a_n J_0(\lambda_n r)$$

$$A_n=\frac{\int_0^{a}r J_0 (\lambda_n r)dr}{\int_0^{a}r J_0^{2} (\lambda_n r)dr}=\frac{2}{a \lambda_n J_1(\lambda_n ,a)}$$

$$u(r,t)=\frac{2}{a}\sum_{n=1}^{\infty}\frac{1}{\lambda_n}\frac{J_0(\lambda_n r)}{J_1(\lambda_n a)}e^{-\lambda_n^{2} \alpha t}$$

Looks like an expression for temperature distribution to me. What do you guys think?

Last edited by a moderator: Apr 23, 2017 at 9:51 PM