# Piecewise initial condition heat equation

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1. Feb 2, 2016

1. The problem statement, all variables and given/known data
I have the solution to the heat equation, with the BC's and everything but the IC applied. So I am just trying to solve for the coefficients, the solution without the coefficients is
$$u(x,t) = \sum_{n=1}^{\infty} A_n\sin(nx)e^{-n^2t}$$
If the initial condition is $u(x,0) = f(x)$ such that $$f(x) = \begin{cases} 0 & 0 < x < \frac{\pi}{3} \\ 100 & \frac{\pi}{3} < x < \frac{2\pi}{3} \\ 0 & \frac{2\pi}{3} < x < \pi \end{cases}$$
I used the formula $$A_m = \frac{2}{\pi}\int_0^\pi f(x)\sin(mx)dx=\frac{200}{m\pi}\bigg[\cos(\frac{\pi}{3}m) - cos(\frac{2\pi}{3}m)\bigg]$$

I couldn't find a pattern in the coefficients other than all the even indices go to $0$. Also is this even correct? When I try to graph this at $t=0$ it isn't giving me the piecewise function $f(x)$. Is it just that I didn't use enough terms to make it noticeable? Also how would I show that at any time $t>0$ the temperature distribution in the rod achieves a local maximum at $x=\pi/2$?

Last edited: Feb 2, 2016
2. Feb 4, 2016

### RUber

That's right. Use more terms. I used 100, and it was still pretty rough. Fourier sums like your solution work well for smooth functions, abrupt changes like that require a lot of terms to reproduce.

3. Feb 4, 2016

### Staff: Mentor

$$\cos(\frac{\pi}{3}m) - cos(\frac{2\pi}{3}m)=\cos\left(\frac{\pi}{3}m\right)-2\cos^2\left(\frac{\pi}{3}m\right)+1$$
m=1: $\frac{1}{2}-\frac{1}{2}+1=1$
m = 2: $-\frac{1}{2}-\frac{1}{2}+1=0$
m=3: $-1-2+1=-2$
m=4: $-\frac{1}{2}-\frac{1}{2}+1=0$
m=5: $1$
m=6: $0$
m=7: $1$
m=8: $0$
m=9: $-2$

There are 3 separate summations.

For m=6k+1 (k=0,1,2,...), the coeff is 1
For m = 6k+3 (k=0,1,2,..), the coeff is -2
For m = 6k+5 (k=0,1,2,...), the coeff is 1