Answer to Question #169937 in Differential Equations for Ezekiel Berunio

Question

A thin, homogenous bar of length L has insulated ends and initial temperature f(x) = x. Determine the temperature distribution u(x,t) in the bar.

Answer 1

We know one dimensional equation for a rod is given as,

$\dfrac{\delta u}{\delta t} = c^2\dfrac{\delta^2 u}{\delta x^2}$

It's solution can be given as $u(x,t) = (C_1cos(px) + C_2sin(px))e^{-c^2p^2t}$

At x = 0 ,$u(x,t) = 0 = u(0,t) = C_1e^{-c^2p^2t}$

Hence, $C_1 = 0$

$u(x,t) = C_2sin(px)e^{-c^2p^2t}$

Using condition u(a,t) = 0

Hence, $C_2sin(px)e^{-c^2p^2t} = 0$

But actually it is not equal to zero to get the desired solution

then, $sin(px) = 0$

$pa = n\pi$

Hence, $p = \dfrac{n\pi}{a}$

Putting the value of p in above equation, we get

$u(x,t) = C_2sin{\dfrac{n\pi x}{a}}e^-\dfrac{n^2\pi^2kt}{a^2}$

where, n = 0,1,2,3.....

$u(x,t) = \sum_{1}^{\infin}b_nsin{\dfrac{n\pi x}{a}}e^-{\dfrac{n^2\pi^2kt}{a^2}}$ .......(A)

It is given that u(x,0) = f(x)

So, $f(x) = \sum_{1}^{\infin}b_nsin{\dfrac{n\pi x}{a}}$

then $b_n = \dfrac{2}{a}\int _{0}^{a}f(x)sin{\dfrac{n\pi x}{a}}dx$ (Using Fourier Sine Series)

Putting this value of $b_n$ in equation (A) we get the desired equation.

$u(x,t) = \sum_{1}^{\infin} \dfrac{2}{a}\int _{0}^{a}f(x)sin{\dfrac{n\pi x}{a}}dx (sin{\dfrac{n\pi x}{a}}e^-{\dfrac{n^2\pi^2kt}{a^2}})$