# Fourier Transform of one variable in a two variable function.

I have a function in two variables, that satisfies the following PDE: $$\frac{x-x_0}{x-x_1}\Psi_{xx}+\Psi_{yy}=0$$

Initially I did use Fourier series \begin{equation*} \Psi=‎ \sum_{-\infty}^{\infty} a_k(x) e^{iky}dy, 0 \leq y \leq 2 \pi \end{equation*}

and obtained the following ODE $$\frac{x-x_0}{x-x_1}\Psi_{xx}-k^2 \Psi=0$$ However, this limits k to a discrete set, but in my case it needs to be any real number, so my Professor adviced me to use Fourier Transform on y, and obtain a function in terms of x and k, which should give me exactly same ODE as a result, yet k will be a real number.

My attempt: Using the Fourier Transform \begin{equation*} \Psi=‎ \int_{-\infty}^{\infty} \hat{\Psi}(y) e^{2 \pi ik xy}dy, \ -\infty < y < \infty \end{equation*}

However, I am not able to obtain the same ODE as before (Which I need specifically). I am not so familiar with Fourier Transforms, and I do not understand what mistake am I making. Any advice would be highly appriciated. Thank you!

• The Fourier series is something you do to a periodic function, so if you use Fourier series then the solutions you get will all be periodic. – Qiaochu Yuan Jun 28 '13 at 22:27
• Do you know that $F(\phi') = 2 \pi i k F(\phi)$, where $F(\phi)$ is he Fourier transform of $\phi$ (using your convention) and $k$ is the Fourier variable? – Christopher A. Wong Jun 29 '13 at 0:20
• This is how I would define things: $$\Psi(x,y) = \int_{-\infty}^{\infty} \widehat\Psi(x,k)e^{2\pi i k x y}dk.$$ – Tunococ Jun 29 '13 at 1:03
• If I define things in this way:\begin{equation*} \Psi(x,y)=‎ \displaystyle\int_{-\infty}^{\infty} \hat{\Psi}(x,k) e^{2\pi ik xy}dk, \ -\infty < y < \infty, k \in \mathbb{R} \end{equation*}then I do not get required ODE: $$\frac{x-x_0}{x-x_1}\Psi_{xx}-k^2 \Psi=0$$, as when I derivate wrt y twice, I get some extra things, while I only need $-k^2 \Psi$. If I could define $$\Psi(x,y)=‎ \displaystyle\int_{-\infty}^{\infty} \hat{\Psi}(x,k) e^{iky}dk$$ that would be good, yet then it is not a Fourier Transform, so I am not sure what to do. – Ekushkebi Jul 1 '13 at 0:16