# Setting up my Fourier series for $B_n$

Related but not necessary to know: here

Looking at the temperature distribution in an infinitely long cylinder of metal with insulated sides and initial temperature distribution f(x)= \left\{\begin{align}0,\quad|x|\lt L \\ C,\quad|x| \gt L \end{align} \right.

$C$ is constant.

Now I want to workout $B_n$ for the fourier series, and I thought that I would want:

$$B_n = \frac2L\int_L^\infty C\sin\left(\frac{n\pi x}{L}\right) dx$$

But perhaps I haven't setup the integral correctly. Thank you for listening.

I based my choice off of the general form of the solution to the heat equation: $u(x,t)=\sum \limits_{n=1}^\infty B_n e^{({-n\pi C/L})^2} \sin\left(\frac{n\pi x}{L}\right)$ With $B_n = \frac2L \int_0^L \sin\left(\frac{n\pi x}{L}\right) f(x) dx$

• It's not a periodic function, so why are you trying to compute a Fourier series to begin with? – Hans Lundmark Oct 19 '14 at 12:24
• @HansLundmark My logic is added above – Analysis Oct 19 '14 at 12:30
• That's the solution to the heat equation on an interval $[0,L]$. It's not going to help you if the interval is unbounded, as in your case. You know that a Fourier series is something that's associated to a periodic function, don't you? It just doesn't make sense to try to compute Fourier coefficients for a function that's not periodic. – Hans Lundmark Oct 19 '14 at 12:35
• @HansLundmark That is a good point. I'll have to find some other way to find $u(x,t)$ thank you – Analysis Oct 19 '14 at 12:37
• The Fourier transform should be more useful. But make the change of variables $v(x,t)=u(x,t)-C$ first, so that your initial data are zero for $|x| > L$, otherwise you're going to run into convergence problems again. – Hans Lundmark Oct 19 '14 at 12:48

It would seem you have a number of concepts confused, and I would personally suggest you refer to a textbook on this matter.

You want to use $$\frac{\partial u}{\partial t} = c^2 \frac{\partial^2 u}{\partial x^2}$$

which helps you obtain your standard $F''+p^2 F =0$ and $\dot{G} + c^2p^2 G=0$, if you can't get this you are in trouble.

f(x)= \left\{\begin{align}0,\quad|x|\lt L \\ C,\quad|x| \gt L \end{align} \right.

$C, L\lt x$ and $x\lt -L$

$u(x,t) = \frac{C}{2c\sqrt{\pi t}}\int_L^\infty$

(CURRENTLY EDITTING)