Answer to Question #152365 in Differential Equations for Kumar

Question

A uniform bar of length which heat flows is insulated at its sides . The ends kept at zero temperature . If the initial temperature at the interior points of the bar is given by k sin^ 3 ( pi x l ). find the temperature distribution in the bar after time ' t'

Answer 1

Let the equation of conduction of heat be,

$\dfrac{du}{dt}=\alpha^2\dfrac{d^2u}{dx^2}$

The boundary conditions are                             

$(i)u (0,t) = 0, t ≥ 0\\ (ii) u (l,t) = 0,t>0, \\ (iii) u (x,0) = f (x)=ksin^3(\pi l), 0 < x < ℓ$

Taking fourier sine transform of above equation ,

$\int_0^{\infty}\dfrac{du}{dt}sinwxdx=\alpha^2\int_0^{\infty}\dfrac{d^2u}{dx^2}sinwxdx$

$\Rightarrow \dfrac{d}{dt}\int_0^{\infty}usinwxdx=\alpha^2[-\omega^2u_s]\\\Rightarrow \dfrac{du_s}{dt}=\alpha^2(-\omega^2u_s)\\\Rightarrow \dfrac{du_s}{u_s}=-\alpha^2\omega^2dt$

Integrating both the sides and we get,

$logu_s=-\alpha^2\omega^2t+logc\\\Rightarrow \dfrac{u_s}{c}=e^{-\alpha^2\omega^2t}\\\Rightarrow u_s=ce^{-\alpha^2\omega^2t}~~~~~~~~~-(1)$

Given initial temprature is-

$u(x,0)=sin^3\pi l$

Taking finite fourier sine transform-

$u_s=\int_0^lsin^3\pi lsin(\dfrac{n\pi x}{l})dx$

$\Rightarrow u_s=sin^3\pi l|cos(\dfrac{n\pi x}{l})|_0^l\dfrac{l}{n\pi}$

$\Rightarrow u_s=sin^3\pi l(cosn\pi-cos0)\dfrac{l}{n\pi}$

$\Rightarrow u_s=sin^3\pi l((-1)^n-1)\dfrac{l}{n\pi}~~~~~~-(2)$

Comparing eqs.(1) and eqs.(2)--

we get

$sin^3\pi l((-1)^n-1)\dfrac{l}{n\pi}$ $=ce^{-\alpha^2\omega^2t}$

from here we get ,

$c=((-1)^n-1)sin^3\pi l\dfrac{l}{n\pi}e^{\alpha^2\omega^2t}$

Put this value in eqs.(1)-

$\Rightarrow u_s=((-1)^n-1)sin^3\pi l\dfrac{l}{n\pi}e^{\alpha^2\omega^2t}e^{-\alpha^2\omega^2t}$

$\Rightarrow u_s=((-1)^n-1)sin^3\pi l\dfrac{l}{n\pi}$

Taking Inverse fourier sine transform and we get the temprature as-

$u=\int_0^{\infty}((-1)^n-1)sin^3\pi l\dfrac{l}{n\pi}sinwt dt$

This is the required temprature distribution.