# Heat equation

1. Apr 18, 2007

### catcherintherye

1. The problem statement, all variables and given/known data

I am solving the heat equation $$1/a^2\theta_t = \theta_x$$

boundary conditions are $$\theta(0,t) = \theta(L,t) = 0 t > 0$$

initial conditions are $$\theta(x,0) = T_0sin(x\pi/L)$$

now I have derived the steady solution to be 0 and I have derived that the general solution will be of the form

$$\theta(x,t) = \sum_{n=1}^\infty\B_n sin\frac{n\pix}{L}exp\frac{-a^2n^\pi^2}{L}t$$

I am next required to determine the sequence $${B_n} \\ \mbox{now usually I would proceed in the following manner, using the initial conditions}\\ \theta(x,0) = \sum_{n=1}^\infty B_n sinn\pix/L = T_0sin(\frac{\pix}{L})$$

I would then multiply both sides by $$\int_{0}^{L}sin\frac{mx\pi}{L}$$

but instead i decided first to separted term $$B_1$$ and cancel term sin(mx\pi/L)

so I proceeded using the cancellation by the orthogonality conditions and derived the following

$$B_mL/2 = (T_0 -B_1)\int_{0}^{L}sin\frac{mx\pi}{L}\\ = \frac{(B_1 -T_0)}{m\pix}\left[cosmx\pi/L\right]_{0}^{L}\\ = \frac{B_1-T_0}{m\pi}\left[(-1)^m -1]$$

you see the problem is I end up with $$B_1$$ undetermined
$$B_1 = \frac{-4T_0}{m\pi -4}$$

where did I go wrong/ what should I have done differently??

2. Relevant equations

3. The attempt at a solution

Last edited: Apr 18, 2007
2. Apr 18, 2007

### HallsofIvy

You didn't do anything wrong except that you guarenteed you would not be able to determine B1 when you "decided first to separted term $B_1$ and cancel term sin(mx\pi/L)"!

There are many different ways to do this but I would probably expand $T_0 sin(\pi x/L)$ in a Fourier sine series in $\pi x/L$ and equate coefficients.

3. Apr 18, 2007

### catcherintherye

mmm...I am not familiar with this method, you see the reason I took out term B1 was because I knew it would allow me to proceed by multplying by

$$sin\frac{m\pix}{L}$$and then integrating from 0 to L. Without taking out term B1 this would eventually lead me to do the following integral
$$L/2B_m = T_0\int_{0}^{L} sin(\frac{m\pix}{L})sin(\frac{pix}{L})dx$$

is this also a valid method of solution and if so How do I do the integral? Do I have to use some sort of trig identity

Last edited: Apr 18, 2007