Fourier sine series in solution to 1D Heat Equation

Question

I have reduced my solution of a 1D heat equation boundary value problem to the following:

$$W(z, t) = \sum_{n=1}^\infty b_n \sin(\lambda_n z) e^{-\lambda_n^2 \alpha t}$$

To get the coefficients $b_n$, I apply the initial condition that: $W(z, 0) = T_0$, which gives the Fourier Sine Series:

$$\sum_{n=1}^\infty b_n \sin(\lambda_n z) = T_0$$

My question is how to obtain the coefficients $b_n$ for my problem here using the integral formula for the Fourier Sine Series? Namely, if

$$f(x) = \sum_{n = 1}^\infty b_n \sin(nx)$$

Then:

$$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx)$$

The argument in the sine function for my problem is $\lambda_n$ = (some function of n, and not explicitly equal to $n$ as in the integral formula above. Is there a way that I am suppose to transform the argument such that the formula can be applied? Thanks kindly in advance,

Answer 1

Hint

1) Multiply both sides of your second equation by ${\sin {\lambda _m}z}$ and integrate from $a$ to $b$.

2) Use this property of your sin functions called orthogonality

$$\eqalign{ & \int\limits_a^b {\sin {\lambda _n}z\sin {\lambda _m}z{\rm{d}}z} = {\delta _{mn}}\int\limits_a^b {{{\sin }^2}{\lambda _n}z{\rm{d}}z} \cr & {\delta _{mn}} = \left\{ {\matrix{ 1 & {m = n} \cr 0 & {m \ne n} \cr } } \right. \cr} $$

where $a \le z \le b$ is your domain of interest.

Answer 2

Your formula isn't precise. To get the coefficients of $$f(x)=\sum b_n X_n(\lambda_n x),\quad b\in[a,b]$$ you have to evaluate $$b_n = \frac{1}{b-a} \int_{a}^{b} f(x) X_n(\lambda_n x)dx$$ where $X_n$ is an eigenfunction that belongs to the eigenvalue $\lambda_n$, which you have found by solving Storm Leuvil equation (in your example $X_n(x)=\sin(\lambda_n x)$). You can figure it out by yourself using H.R.'s hint.

Note that $$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} T_0 \sin(\lambda_n x)dx=0$$ which suggest that you have a mistake, for example in finding eigenfunctions\value .