# Integration term by term and orthogonality

• February 23rd 2011, 02:21 PM
dwsmith
Integration term by term and orthogonality
$\displaystyle\int_0^L\varphi_m(x)\varphi_n(x) \ dx=0 \ \ \ \text{if} \ m\neq n$

$\displaystyle\int_0^L\varphi^2_n(x) \ dx>0$

$\displaystyle\int_0^L f(x)\sin\left(\frac{m\pi x}{L}\right) \ dx=\sum_{n=1}^{\infty}b_n\int_0^L\sin\left(\frac{n \pi x}{L}\right)\sin\left(\frac{m\pi x}{L}\right) \ dx$

Every term is 0 except for when m = n.

$\displaystyle\int_0^L f(x)\sin\left(\frac{m\pi x}{L}\right) \ dx=b_m\int_0^L\left[\sin\left(\frac{m\pi x}{L}\right)\right]^2 \ dx$

$\displaystyle f(x)=\sin\left(\frac{3\pi x}{L}\right)$

$\displaystyle\int_0^L \sin\left(\frac{3\pi x}{L}\right)\sin\left(\frac{m\pi x}{L}\right) \ dx=b_m\int_0^L\left[\sin\left(\frac{m\pi x}{L}\right)\right]^2 \ dx$

Looking at the LHS:

$\displaystyle b_m\int_0^L\left[\sin\left(\frac{m\pi x}{L}\right)\right]^2 \ dx=b_m\frac{1}{2}\int_0^L \left[1-\cos\left(\frac{2m\pi x}{L}\right)\right] \ dx$

$\displaystyle b_m\frac{1}{2}\left[x-\frac{L\sin\left(\frac{2m\pi x}{L}\right)}{2m\pi x}\right]_0^L\Rightarrow\frac{L}{2} b_m$

$\displaystyle b_n=\frac{2}{L}\int_0^L\sin\left(\frac{3\pi x}{L}\right)\sin\left(\frac{m\pi x}{L}\right) \ dx$

What do I do now? My solution I am obtain is way wrong.
• February 23rd 2011, 03:35 PM
zzzoak
In Fourier series the integral is

$
\displaystyle
\int_{-L}^L
$

If

$
\displaystyle
\int_{0}^L
$

other orthogonality conditions may be.
• February 23rd 2011, 03:36 PM
dwsmith
Is this the solution:

$\displaystyle b_n=\begin{cases}0 & \text{if} \ m\neq 3\\ 1 & \text{if} \ m=3\end{cases}$