Does anyone know a solution to this PDE? I ran into this PDE:
$$\frac{\partial y(x,t)}{\partial t} = A\,x^{\gamma-1} \left(\frac{\partial y(x,t)}{\partial x} + x \, \frac{\partial^2 y(x,t)}{\partial x^2}\right)$$
If it helps in any way, this is for a physical system, where $x,y,t$ and $A$ are positive real numbers and we just care about solutions in $x>0$.
For starters, a solution for $\gamma=1$ would already be enough:
$$\frac{\partial y(x,t)}{\partial t} = A \left(\frac{\partial y(x,t)}{\partial x} + x \, \frac{\partial^2 y(x,t)}{\partial x^2}\right)$$
I got as far as to find the solution in Fourier space for a Dirac delta distribution as initial condition $y(x,0)= F \, \delta(x-\hat x)$, which is:
$$ y(k,t) = \frac{F}{2\pi}\,\exp\left(-\frac{\mathrm{i} \, \hat x \, k}{1-A\, \mathrm{i} \, k\, t}\right) \, \left(1-A\, \mathrm{i} \, k\, t\right).$$
But now I fail to get a nice closed-form expression for the inverse Fourier transform of that solution. Can anyone find that, or find another way to solve for $y(x,t)$?
 A: $\newcommand{\+}{^{\dagger}}%
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$\ds{\partiald{{\rm y}\pars{x,t}}{t}
=
Ax^{\gamma}\,\partiald[2]{{\rm y}\pars{x,t}}{x}
+
Ax^{\gamma - 1}\,\partiald{{\rm y}\pars{x,t}}{x}}.$
With the scaling $x = \alpha\tilde{x}$ and $t = \beta\tilde{t}$, we have

\begin{align}
{1 \over \beta}\,\partiald{y}{\tilde{t}}
&=A\alpha^{\gamma}\tilde{x}^{\gamma}\,{1 \over \alpha^{2}}\,
\partiald[2]{y}{\tilde{x}}
+ A\alpha^{\gamma -1}\tilde{x}^{\gamma - 1}\,{1 \over \alpha}\,\partiald{y}{x}
\\[3mm]
\partiald{y}{\tilde{t}}
&=A\pars{\beta\alpha^{\gamma - 2}}\tilde{x}^{\gamma}\,\partiald[2]{y}{\tilde{x}}
+ A\pars{\beta\alpha^{\gamma - 2}}\tilde{x}^{\gamma - 1}\,\partiald{y}{x}
\end{align}
such that the equation is invariant whenever $\beta\alpha^{\gamma - 2} = 1$:
$$
1 = {t \over \tilde{t}}\,\pars{x \over \tilde{x}}^{\gamma - 2}\quad\imp\quad
{x^{1 - \gamma/2} \over t^{1/2}} = {\tilde{x}^{1 - \gamma/2} \over t^{1/2}}
$$
Let $\ds{\xi \equiv {x^{1 - \gamma/2} \over t^{1/2}}}$ and
${\rm y}\pars{x,t} \equiv \fermi\pars{\xi}
 = \fermi\pars{x^{1 - \gamma/2} \over t^{1/2}}$:
\begin{align}
\partiald{{\rm y}\pars{x,t}}{t}
&=
\fermi'\pars{\xi}\,\pars{-\,{1 \over 2}\,{x^{1 - \gamma/2} \over t^{3/2}}}
\\[3mm]
\partiald{{\rm y}\pars{x,t}}{x}
&=
\fermi'\pars{\xi}\,{\pars{1 - \gamma/2}x^{-\gamma/2} \over t^{1/2}}
\\[3mm]
\partiald[2]{{\rm y}\pars{x,t}}{x}
&=
\fermi''\pars{\xi}\,{\pars{1 - \gamma/2}^{2}x^{-\gamma} \over t}
-
\fermi'\pars{\xi}\,{\gamma \over 2}\,
{\pars{1 - \gamma/2}x^{-\gamma/2 - 1} \over t^{1/2}}
\end{align}

\begin{align}
-\,{x^{1 - \gamma/2} \over 2t^{3/2}}\,\fermi'\pars{\xi}
&=
Ax^{\gamma}\bracks{%
{\pars{1 - \gamma/2}^{2}x^{-\gamma} \over t}\,\fermi''\pars{\xi}
-
{\gamma\pars{1 - \gamma/2}x^{-\gamma/2 - 1} \over 2t^{1/2}}\,\fermi'\pars{\xi}}
\\[3mm]&\phantom{=}\mbox{}+
Ax^{\gamma -1}\bracks{%
{\pars{1 - \gamma/2}x^{-\gamma/2} \over t^{1/2}}\,\fermi'\pars{\xi}}
\end{align}
Multiplying both members by the factor $\ds{2t^{3/2}x^{\gamma/2 - 1}}$
\begin{align}
-\fermi'\pars{\xi}
&=\bracks{%
2A\pars{1 - {\gamma \over 2}}^{2}\,\overbrace{t^{1/2}x^{\gamma/2 - 1}}
^{\ds{\xi^{-1}}}\,\fermi''\pars{\xi}
-
A\gamma\pars{1 - {\gamma \over 2}}\,
\overbrace{tx^{\gamma - 2}}^{\ds{\xi^{-2}}}\fermi'\pars{\xi}}
+
2A\pars{1 - {\gamma \over 2}}\,\overbrace{tx^{\gamma - 2}}^{\ds{\xi^{-2}}}
\fermi'\pars{\xi}
\\[3mm]&=
{2A\pars{1 - \gamma/2}^{2} \over \xi}\,\fermi''\pars{\xi}
+
{2A\pars{1 - \gamma/2}^{2} \over \xi^{2}}\,\fermi'\pars{\xi}
\end{align}
$\fermi\pars{\xi}$ obeys the equation:
$$
\fermi''\pars{\xi}
+
\bracks{%
{1 \over \xi} + {\xi \over 2A\pars{1 - \gamma/2}^{2}}}\fermi'\pars{\xi} = 0
$$
Now, we multiply both members by the integrating factor
$\ds{\braces{\xi\exp\pars{{\xi^{2} \over 4A\bracks{1 - \gamma/2}^{2}}}}}$. It leads to:
$$
\totald{}{\xi}\bracks{%
{\xi\exp\pars{\xi^{2} \over 4A\pars{1 - \gamma/2}^{2}}\fermi'\pars{\xi}}} = 0
\quad\imp\quad
\fermi'\pars{\xi}
=
{B \over \xi}\,\exp\pars{-\xi^{2} \over 4A\pars{1 - \gamma/2}^{2}}
$$
where $B$ is a $\it constant$.

\begin{align}
\fermi\pars{\xi}
&=\fermi\pars{\xi_{0}}
+
B\int_{\xi_{0}}^{\xi}\exp\pars{-z^{2} \over 4A\pars{1 - \gamma/2}^{2}}
\ {\dd z \over z}
\end{align}

