A Variant of the Heat Equation on the First Quadrant

Subjects: Fourier Analysis, Partial Differential Equations
Links: The Heat Equation, Fourier Transform in R, The Heat Equation on the Real line

The equation $$x^2 \frac{\partial^2 u}{\partial x^2}+ ax\frac{\partial u}{\partial x} = \frac{\partial u}{\partial t}$$with $u(x,0)=f(x)$, for $0<x<\mathrm{\infty }$ and $t>0$ is a variant of the heat equation which occurs in a number of applications. We can see some similarity with a Cauchy-Euler Differential Equation, and thus use a similar technique by making the substitution $x=e^{-y}$, so that $-\mathrm{\infty }<y<\mathrm{\infty }$.

We set $U(y,t)=u(e^{-y},t)$ and $F(y)=f(e^{-y})$. Then The problem reduces to the equation $$\frac{\partial^2 U}{\partial y^2} + (1-a) \frac{\partial U}{\partial y} = \frac{\partial U}{\partial t}$$with $U(y,0)=F(y)$.

With this in mind, we can apply the Fourier transform on $y$, and get that $$-4\pi^2\omega^2 \hat U + (1-a) 2\pi i \omega \hat U = \frac{\partial \hat U}{\partial t}$$Fixing $\omega$, then we have that $$\hat U (\omega, t)= A(\omega) \exp(-4\pi^2\omega^2 t)\exp(2\pi i \omega (1-a)t)$$
Given the initial condition, then $$\hat U (\omega, t)= \hat F(\omega) \exp(-4\pi^2\omega^2 t)\exp(2\pi i \omega (1-a)t)$$
We can calculate the inverse Fourier transform of $\exp (-4{\pi }^2{\omega }^{2}t)\exp (2\pi i\omega (1-a)t)$, if we do that we get that $$\int_{-\infty}^\infty \exp(-4\pi^2\omega^2 t) e^{2\pi i \omega(x+(1-a)t)}, d\omega = \mathcal H_t(x+(1-a)t)=g_t(x)$$
Then we can get that $$U(y, t) = (F* g_t)(x)$$
expanding the integrals and returning to the original variables we get that $$u(x,t) = \frac1{\sqrt{4\pi t}} \int_0^\infty f(v) \exp\left(\frac{-(\ln(v/x)+(1-a)t)^2}{4t}\right) \frac{dv}{v}$$