Answer to Question #146208 in Differential Equations for Abhishek mr

Question

A bar 40cm long with insulated sides had its ends A and B maintained at 0 degree and 100 degree Until steady state condition prevails Then the temperature at A is suddenly raised to 20 degree and at the same time that of B is lowered to 60 degree. Find the temperature distribution in the bar at time t

Answer 1

The one dimensional heat flow equation is given by:

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

In steady-state:

$\frac{\partial u}{\partial t}=0$

So:

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

Solving:

u=ax+b

Initial conditions in steady-state:

u=0 at x=0

u=100 at x=40

Then:

u=2.5x

Hence the boundary conditions are:

u(0,t)=20

u(40,t)=60

u(x,0)=2.5x

The solution of one dimensional heat flow equation is given by:

u(x,t)=(A*cos($\lambda$x)+B*sin($\lambda$x))*exp[$-\alpha^2*\lambda^2*t$ ]

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

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

To find u_s(x):

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

u_s(x) = ax+b

u_s(0)=20

u_s(40)=60

Then:

u_s=x+20

To find u_t(x,t):

u_t( x,t) = u(x,t) –u_s(x)

u_t(0,t) = u(0,t)–u_s(0) = 20–20 = 0

u_t(40,t) = u(40,t)–u_s(40) = 60–60 = 0

u_t(x,0) = u(x,0) –u_s(x) = 2.5x-x-20=1.5x-20

u_t(x,t)=(A*cos($\lambda$x)+B*sin($\lambda$x))*exp[$-\alpha^2*\lambda^2*t$ ]

A=0; $\lambda$ =(n$\pi$)/40

u_t(x,t)=B*sin((n$\pi$x)/40)*exp[$-\alpha^2*((n\pi)/40)^2*t$ ]

Most general solution:

u_t(x,t)=$\displaystyle\sum_{n=1}^\infty$ B_n*sin((n$\pi$x)/40)*exp[$-\alpha^2*((n\pi)/40)^2*t$ ]

Using condition:

B_n = 2/40*$\int_{0}^{40}$ (1.5x-20)*sin(n$\pi$x/40)dx =

u=u_s+u_t