Heat equation solution using Fourier transform

Question

I want to solve the equation $$x^2\frac{\partial^2 u(x,t)}{\partial x^2}+ax\frac{\partial u(x,t)}{\partial x}=\frac{\partial u(x,t)}{\partial t}$$ with $u(x,0)=f(x)$ for $0<x<\infty$ and $t>0$. Substituting $U(y,t)=u(e^{-y},t)$ and $F(y)=f(e^{-y})$ we get $$ \frac{\partial^2 U(y,t)}{\partial y^2}+(1-a)\frac{\partial U(y,t)}{\partial y}=\frac{\partial U(y,t)}{\partial t},$$ with the solution $$\hat U(\xi,t)=\hat F(\xi)e^{(-4\pi\xi^2+(1-a)2\pi i\xi)t},$$ since $\hat U(\xi,0)=\hat F(\xi)$. Taking the Fourier transform in the y variable (assuming that u satisfies the necessary conditions and $\hat {\frac{\partial U}{\partial t}}$=$\frac{\partial}{\partial t}\hat U$), and using $$\hat F(\xi)=\int_{-\infty}^{\infty} F(x)e^{-2\pi xi\xi} dx=\int_{0}^{\infty} \frac{f(y)}{y}e^{2\pi i\log(y)\xi} dy,$$ we are supposed to get $$u(x,t)=\frac{1}{\sqrt{4\pi t}}\int_0^{\infty}e^{-(\log(v/x)+(1-a)t)^2/(4t)}f(v) \frac{dv}{v},$$ whereas I get $$\int_{-\infty}^{\infty} \int_0^{\infty} e^{(-4\pi^2\xi^2+(1-a)2\pi i\xi)t}e^{2\pi\log(y/x)i\xi} \frac{f(y)}{y} dyd\xi$$ and don't know how to simplify it.

Answer 1

Use $$-4\pi\xi^2t+(1-a)2\pi i\xi t+ 2\pi i\xi\log(y/x)=\frac{t}{\pi}\left(2\pi i\xi+\frac{(1-a)t+\log(y/x)}{2t} \pi\right)^2-\frac{((1-a)t+\log(y/x))^2}{4t}\pi$$ and $$\int_{-\infty}^{\infty}e^{(2\pi i \xi+a)^2} d\xi=\frac{1}{2\sqrt{\pi}}$$ for any $a\in\mathbb{R}$. The integral above you can calculate by transforming it to the case $a=0$ by substitution, and for the case $a=0$ you have a gaussian integral which is well known. By this and some further elementary calculation you should get your solution.