Solving Heat Equation PDE

Question

I need some help on solving this heat PDE :

Question : Consider a bar length L. The face at x=0 is insulated so that the heat flow across is zero, and the face at x=L is held at temperature u=0. The temperature distribution is governed by heat equation

$$ {\partial u \over \partial t} - k {\partial ^2 u \over \partial x^2} =0$$

Show that the normal modes of u(x,t) are

$$\large U_n(x,t)=B_n\cos[(2n-1)\pi x/2L]e^{[-(2n-1)^2\pi^2k^2t]/4L^2}$$

Given BCs are : u(0,t)=0 and u(L,t)=0

I already solve this PDE

and I got my answer in this form :

$$\large U_n(x,t)=B_n\sin[n\pi x/2L] e^{[-(n)^2\pi^2k^2t]/4L^2}$$

How am I need to change my answer into the form they asking ??? Anyone willing to help me ?

Answer 1

You need to express the problem as

$$u_t = k u_{xx} $$

$$ u_x(0,t) = 0 ,\;\;\; u(L,t) = 0 $$

Use separation of variables, i.e. $u(x,t) = X(x)T(t)$ and get

$$X'' + \frac{\lambda}{k} X = 0 $$

where $-\lambda$ is the separation constant. The solution to this equation is

$$ X(x) = A \cos{\left (\sqrt{\frac{\lambda}{k}} x \right)} + B \sin{\left (\sqrt{\frac{\lambda}{k}} x \right)}$$

The condition at $x=0$ implies that $B=0$:

$$ X'(x) = -\sqrt{\frac{\lambda}{k}} A \sin{\left (\sqrt{\frac{\lambda}{k}} x \right)} + B \sqrt{\frac{\lambda}{k}}\cos{\left (\sqrt{\frac{\lambda}{k}} x \right)}$$

$$X'(0) = B \sqrt{\frac{\lambda}{k}} = 0$$

The condition at $x=L$ implies that

$$\cos{\left (\sqrt{\frac{\lambda}{k}} L \right)} = 0$$

so that

$$\lambda = \left [ (2 n-1) \frac{\pi}{2 L} \right ]^2 k $$

You should be able to take it from there with the time equation.