Answer to Question #152527 in Differential Equations for Annanya

Question

A rod of length L has its ends A and B maintained at 20 °C and 40 °C respectively
until steady state condition prevail. The temperature at A is suddenly raised to
50 °C while that at B is lowered to 10 °C and maintained thereafter. Find the
subsequent temperature distribution of the rod.

Answer 1

The heat equation:

$\frac{\partial u}{\partial t}=\alpha^2\frac{\partial^2u}{\partial x^2}$

In steady state:

$\frac{\partial^2u}{\partial x^2}=0$

Solving, we get

$u=ax+b$

The initial conditions, in steady–state, are 

$u=20, x=0$

$u=40, x=L$

Then:

$b=20, a=20/L$

The temperature function in steady–state is

$u(x)=20x/L+20$

The boundary conditions in the transient state are 

$u(0,t)=50, u(L,t)=10, u(x,0)=20x/L+20$

We break up the required funciton u (x,t) into two parts:

$u(x,t)=u_s(x)+u_t(x,t)$

$u_s(x)$ is a steady state solution, $u_t(x,t)$ is a transient solution

$u_s(x)=ax+b$

$b=50, a=-40/L$

$u_s(x)=-40x/L+50$

$u_t(x,t)=u(x,t)-u_s(x)$

Putting boundary conditions, we have the boundary conditions relative to the transient solution :

$u_t(0,t)=u(0,t)-u_s(0)=50-50=0$

$u_t(L,t)=u(L,t)-u_s(L)=10-10=0$

$u_t(x,0)=u(x,0)-u_s(x)=20x/L+20+40x/L-50=60x/L-30$

We have:

$u_t(x,t)=(Acos\lambda x+Bsin\lambda x)e^{-\alpha^2\lambda^2t}$

Using boundary conditions, we get:

$A=0, \lambda=\pi n/L$

Then:

$u_t(x,t)=Bsin\frac{\pi nx}{L}e^{-\frac{\alpha^2\pi^2 n^2}{L^2}t}$

The most general solution:

$u_t(x,t)=\displaystyle\sum_{n=1}^{\infin}B_nsin\frac{\pi nx}{L}e^{-\frac{\alpha^2\pi^2 n^2}{L^2}t}$

Using boundary conditions:

$u_t(x,0)=\displaystyle\sum_{n=1}^{\infin}B_nsin\frac{\pi nx}{L}=60x/L-30, 0<x<L$

$B_n=\frac{2}{L}\displaystyle\intop_0^L(60x/L-30)sin\frac{\pi nx}{L}dx=$

$=\frac{2}{L}((60x/L-30)\frac{-cos(\pi nx/L)}{\pi n/L}-\frac{60}{L}\cdot\frac{-sin(\pi nx/L)}{\pi^2n^2/L^2})|^L_0=$

$=\frac{2}{L}(-\frac{30Lcos(\pi n)}{\pi n}-\frac{30L}{\pi n})=-\frac{60}{\pi n}(cos(\pi n)+1)$

$B_n=-\frac{60(1+(-1)^n)}{\pi n}$

$B_n=0$ when $n$ is odd

$B_n=-\frac{120}{\pi n}$ when $n$ is even

$u_t(x,t)=\displaystyle\sum_{n=2,4,6,..}^{\infin}(-\frac{120}{\pi n}sin\frac{\pi nx}{L})e^{-\frac{\alpha^2\pi^2n^2}{L^2}t}$

$u(x,t)=-40x/L+50+\displaystyle\sum_{n=2,4,6,..}^{\infin}(-\frac{120}{\pi n}sin\frac{\pi nx}{L})e^{-\frac{\alpha^2\pi^2n^2}{L^2}t}$