# Equilibrium of a composite rod

• Feb 9th 2011, 12:48 PM
dwsmith
Equilibrium of a composite rod
The right end of a rod of length $\displaystyle L_1$ and thermal constants $\displaystyle c_1, \ \rho_1, \ \kappa_1$ is joined to the left end of a rod of length $\displaystyle L_2$ and thermal constants $\displaystyle c_2, \ \rho_2, \ \kappa_2$. The left end of the composite rod is maintained at temperature 0 and the right end at temperature U. Find the equilibrium temperature distribution in the composite rod if the ends are in perfect thermal contact.

I have a lot of haphazardly placed information but not sure what to do with it.

$\displaystyle \text{B.C.}:\begin{cases}u(0,t)=0\\u(L_1+L_2,t)=U\ end{cases}$

$\displaystyle \text{D.E.}: \ u_t=k(x)u_{xx}$

$\displaystyle u(x,t)=\begin{cases}u_1(x,t) \ \ \ 0<x<L_1\\u_2(x,t) \ \ \ L_1<x<L_1+L_2\end{cases}$

$\displaystyle \displaystyle\lim_{\begin{cases}\epsilon\to 0\\{\epsilon >0}\end{cases}}u_1(L_1-\epsilon,t)=u_1(L_1,t)=u_2(L_1,t)=\lim_{\begin{cas es}\epsilon\to 0\\{\epsilon >0}\end{cases}}u_2(L_1+\epsilon,t) \ \ t>0$

$\displaystyle \displaystyle\lim_{\begin{cases}\epsilon\to 0\\{\epsilon >0}\end{cases}}\kappa_1\frac{\partial u_1}{\partial x}(L_1-\epsilon,t)=\kappa_1\frac{\partial u_1}{\partial x}=\kappa_2\frac{\partial u_2}{\partial x}=\lim_{\begin{cases}\epsilon\to 0\\{\epsilon >0}\end{cases}}\kappa_2\frac{\partial u_2}{\partial x}(L_1+\epsilon,t)$

$\displaystyle \displaystyle\text{Flux}: \ \kappa_2\frac{\partial u_2}{\partial x}(L_1,t)=\kappa_1\frac{\partial u_1}{\partial x}(L_1,t)$

What to do?