Question

3. Consider the following Neumann problem for the heat equation: 14(0,t)=14(L,t)=0, t>0 u(x,0)- f(x),0<x<L (a) Give a short physical interpretation of this problem. (b) Given the following initial condition, 2 *2 2 solve the initial boundary value problem for u(x,t.

Answer 1

(a) The problem can be thought of a initial temperature distribution u(x,0) given for a rod of length L where the ends are insulated, resulting in neumann type boundary conditions.

(b) Lets us write in variable separable form the solution as

Substituting back into the equation

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

$\implies g(x)h'(t)= kg''(x)h(t)$

where

$h'(t)= \frac{dh}{dt} \quad g'(x) =\frac{dg}{dx} \quad g''(x) = \frac{d^2g}{dx^2}$

The boundary conditions thus become

$u_x(0,t) =g'(0)h(t) = 0 \implies g'(0) = 0$

$u_x(L,t) =g'(L)h(t) = 0 \implies g'(L) = 0$

Similarly the initial conditions are

Looking at the modifiedheat equation &

rearranging we have,

$\frac{g''(x)}{g(x)} = \frac{h'(t)}{kh(t)}$

Since the LHS is dependent upon x, only & RHS dependent on t only that means both must be equalling a constant say $-\lambda^2$

The minus sign is chosen so as to get a periodic solution satisfying periodic neumann conditions at all times.

Hence we have two sets of ordinary differential equations, viz.

$g''(x) + \lambda^2g(x) =0$

$h'(t)+ k\lambda^2h(t)=0$

The gene solution for the first one is

$g(x) = c_1\cos{\lambda x} +c_2\sin{\lambda x}$

where $c_1$ & $c_2$ are constants satisfying boundary conditions. Now,

$g'(x) = -\lambda\left(c_1 \sin{\lambda x} + c_2 \cos{\lambda x}\right )$

$g'(0) = 0 \implies c_2 = 0$

$g'(L) = 0 \implies \sin{\lambda L} = 0 \implies \lambda L = n\pi \quad n = 0, \pm 1,\pm 2, \pm 3, \ldots$

Thus,

$\lambda = \frac{n\pi}{L}$

For , the solution is obtained by multiplying $e^{k\lambda^2t}$ on both sides of equation as in

$e^{k\lambda^2t} h'(t) + {k\lambda^2}e^{k\lambda^2t} h(t) = 0$

$\implies \frac{d\left(e^{k\lambda^2t} h(t) \right )}{dt} = 0$

$\implies e^{k\lambda^2t} h(t) = \Lambda \implies h(t) = \Lambda e^{-k\lambda^2t}$

Since $\lambda$ can assume many values based on , many such $g_n$ & $h_n$ exist

Thus ,

$u(x,t) = \sum_{n=-\infty}^{n=\infty}A_n g_n(x)h_n(t)=\sum_{n=-\infty}^{n=\infty}A_n \cos{\left( \frac{n\pi x}{L}\right )}e^{-k\lambda_{n}^{2}t}$

where $A_n$ is the amplitude of the $n^{th}$ mode and has absorbed the other constants like $c_1$ & $\Lambda$ .

Moreover $\lambda_n = \frac{n\pi}{L}$

Now applying initial conditions we have

$u(x,0) = \sum_{n=-\infty}^{n=\infty}A_n \cos{\frac{n\pi x}{L}} = f(x)$

Since the above is a fourier series we can find $A_n$ as

$A_n = \frac{2}{L}\int_{0}^{L}f(x)\cos{\left(\frac{n\pi x}{L} \right )}dx$

Now given that

$f(x) = \frac{L}{2} \quad \forall x \in \left[0\ \frac{L}{2}\right]$

$f(x) = x-\frac{L}{2} \quad \forall x \in \left[ \frac{L}{2}\ L\right]$

$\implies A_n = \frac{2}{L}\left[ \int_{0}^{L/2}\frac{L}{2}\cos{\frac{n\pi x}{L}}dx + \int_{L/2}^{L}\left(x-\frac{L}{2}\right)\cos{\frac{n\pi x}{L}}dx\right]$