# How to solve this differential equation with Fourier Transform?

Consider the differential equation

$$\dfrac{\partial w}{\partial t} = -\alpha \dfrac{\partial w}{\partial x} + D \dfrac{\partial ^2 w}{\partial x^2}$$

together with the boundary conditions that

$$\lim_{x\to \pm \infty} w(x,t) = \lim_{x\to \pm \infty} \dfrac{\partial w}{\partial x}(x,t) = \lim_{x\to \pm \infty} \dfrac{\partial^2 w}{\partial x^2}(x,t) = 0$$

and the initial condition

$$w(x,0)=\delta(x-x_0).$$

I need to solve this using Fourier Transform. For that, let $\mathcal{F}_x$ be the Fourier Transform operator on the $x$ variable. Applying it to the equation and using the relation for the Fourier transform of a derivative we get

$$\dfrac{\partial}{\partial t}\mathcal{F}_x(w)(k,t) = -\alpha (-ik) \mathcal{F}_x(w)(k,t) + D(-ik)^2 \mathcal{F}_x(w)(k,t),$$

This is equivalent to

$$\dfrac{\partial}{\partial t}\mathcal{F}_x(w)(k,t)- (\alpha ik +Dk^2) \mathcal{F}_x(w)(k,t)=0$$

Considering then $k$ fixed, the solution can be easily be obtained as

$$\mathcal{F}_x(w)(k,t)= A \exp\left[(\alpha ik + Dk^2)t\right].$$

But then we need to invert this. Using the inversion formula and some manipulations I've obtained

$$w(x,t) = \dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \mathcal{F}_x(w)(k,t)e^{-ikx}dx = \dfrac{A}{\sqrt{2Dt}}\exp\left[\dfrac{-x^2-2\alpha t - \alpha^2 t^2}{4Dt}\right].$$

Now, I don't know how to use the boundary conditions and the initial condition on this problem. It seems the boundary condition is automatically satisfied, but I'm not sure. Also the initial condition I have no idea on how to use.

How can I finish this by applying the boundary conditions and the initial condition in the right way?

• I think your "vanishing at infinity" BCs are built into the assumption that the FT exists as a function (rather than as a distribution). – Ian Sep 27 '15 at 20:28
• But I think you need to incorporate the initial condition into your solution of the ODE in Fourier space. For that you will need to know how to compute the Fourier transform of the Dirac delta. – Ian Sep 27 '15 at 20:31