Answer to Question #137069 in Differential Equations for Markiv

Question

A bar 100cm of along with insulted sites has its ends kept at zero degree Celsius and under degree Celsius until studies trip reveals two ends Har then suddenly insulted and kept so find the temperature distribution

Answer 1

Let length of the rod denoted by $l$ and temperature at other end is 100.

Heat equation is given by $\frac{\partial u}{\partial t} = \alpha^2 \frac{\partial^2 u}{\partial x^2}$ (1)

When t=0, heat flow was independent of time.

Equation will be

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

Solution of the equation, $u = ax+b$

since u=0 at x = 0 and u=100 at $x=l$

hence equation will be $u = \frac{100}{l} x$

Boundary conditions will be

(i) $u(0,l) = 0$ $\forall t \geq 0$

(ii)$u(l,t) = 0$ $\forall t \geq 0$

(iii) $u(x,0) = \frac{100x}{l}$ $\forall 0\leq x \leq l$

Then solution of equation (1) will be of form,

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

using boundary conditions, we get

$A = 0, \lambda = \frac{n\pi}{l}$

Then equation (2) will be

$u(x,l) = Bsin(\frac{n\pi x}{l})e^{-\frac{n^2\pi^2\alpha^2 }{l^2}t}$

Most general solution is

$u(x,l) = \Sigma_{n=1}^{\infty} B_nsin(\frac{n\pi x}{l})e^{-\frac{n^2\pi^2\alpha^2 }{l^2}t}$

Applying last boundary condition in the above equation,

$u(x,0) = \Sigma_{n=1}^{\infty} B_nsin(\frac{n\pi x}{l})$

$\frac{100x}{l} = \Sigma_{n=1}^{\infty} B_nsin(\frac{n\pi x}{l})$

$B_n = \frac{2}{l} \int_0^l \frac{100x}{l}sin(\frac{n\pi x}{l})dx = \frac{200(-1)^{n+1}}{n\pi}$

Hence general solution is

$u(x,t) = \Sigma_{n=1}^{\infty} \frac{200(-1)^{n+1}}{n\pi}sin(\frac{n\pi x}{l})e^{-\frac{n^2\pi^2\alpha^2 }{l^2}t}$

Putting $l = 100 cm = 1m$

$u(x,t) = \Sigma_{n=1}^{\infty} \frac{200(-1)^{n+1}}{n\pi}sin({n\pi x})e^{-{n^2\pi^2\alpha^2 }t}$

which is the required solution.