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.

  • $\begingroup$ What have you already tried? $\endgroup$ – razivo Jun 15 '20 at 19:32
  • $\begingroup$ Well, I know that $\int_{-\infty}^{\infty} e^{-x^2}dx=\sqrt{\pi}$ (seems to be related), and I think we could interchange the order of integration, but I can't see how to apply it. $\endgroup$ – ranger281 Jun 15 '20 at 19:35
  • $\begingroup$ The $\xi$ integral is just the Fourier transform of a Gaussian since you will treat the other variables as a constant. Apply the known formula, or feel free to complete the square and derive it yourself. $\endgroup$ – Ninad Munshi Jun 15 '20 at 19:36
  • $\begingroup$ I’d recommend trying to simplify the current expression as much as possible, migrating constants out of the integral, maybe canceling the log you got there, then, try to do some “renaming” of constants to try and find the most reduced form of the integral. $\endgroup$ – razivo Jun 15 '20 at 19:38
  • $\begingroup$ And, this seems very “Fourier integral” to me. $$\frac{1}{\pi}\int^{\infty}_0 \int^{\infty}_{-\infty} e^{-iwt’}f(t)dt’e^{iwt}dw$$ $\endgroup$ – razivo Jun 15 '20 at 19:39

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.


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