### Author Topic: problem involving fourier series  (Read 2661 times)

#### Shaghayegh A

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• Karma: 0 ##### problem involving fourier series
« on: December 11, 2016, 10:31:05 PM »
Problem: Solve by Fourier method \begin{align} & u_{tt}-u_{xx}=0\qquad -\frac{\pi}{2}<x<\frac{\pi}{2},\label{1-1}\\ & u_x|_{x=-\pi/2}=u_x|_{x=\pi/2}=0,\label{1-2}\\ &u| _{t=0}=x^2,\qquad u_t|_{t=0}=0.\label{1-3} \end{align}

I have $X(x) = A\sin(\sqrt\lambda x) +B\cos(\sqrt\lambda x)$ Using (2), I can either take

$X(x) = A\sin(nx), \lambda = n^2$ OR $X(x) = B\cos(2nx), \lambda = 4n^2$ Is it correct to take either one? Or is there a right one?

#### Victor Ivrii ##### Re: problem involving fourier series
« Reply #1 on: December 13, 2016, 07:11:19 AM »
You make your conclusion basing on the equation and the boundary conditions. The easiest way is to change variables $x'=x+\frac{\pi}{2}$; then $\lambda_n=n^2$ and $X_n(x')=\cos (nx')$ from the standard problem; returning to old coordinates
\begin{equation}
X_n(x)=\cos (nx+\frac{\pi 2}{2})=\left\{\begin{aligned}
&(-1)^m \cos (2mx) && n=2m,\\
&(-1)^{m+1}\sin ((2m+1)x) && n=2m+1.
\end{aligned}\right.
\end{equation}
We can drop sign at our wish. So, you got correctly only half of the eigenfunctions, but in this particular problem it would lead to a correct solution. Indeed, since "initial" functions $x^2$ and $0$ are even only $\cos(2mx)$ would be in the end. However, it will cost you points!

If we observe that the functions $x^2$ and $0$ are even, and thus solution must be even, we can reduce interval to $(0,\frac{\pi}{2})$ and set conditions $u_{x=0}=u_{x=\pi/2}=0$ which would lead to  $\lambda_m=4m^2$ and $X_m=\cos (2mx)$ from the beginning.