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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 $y(x,t)$?

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  • $\begingroup$ @ John Smith: what physical system is this for? $\endgroup$ – Robert Lewis Dec 3 '13 at 19:45
  • $\begingroup$ @RobertLewis: its diffusion in a cylindrical system with power-laws for density and diffusivity. $\endgroup$ – John Smith Dec 3 '13 at 19:47
  • $\begingroup$ @ John Smith: thanks for the info! $\endgroup$ – Robert Lewis Dec 3 '13 at 19:49
  • $\begingroup$ for $\gamma = 1$ it is the generator of a squared bessel process. these were studied by fourier methods in the 50's by Feller, and are well know today as the CIR interest rate processes. $\endgroup$ – mike Dec 3 '13 at 19:49
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\partiald{{\rm y}\pars{x,t}}{t} = Ax^{\gamma}\,\partiald[2]{{\rm y}\pars{x,t}}{x} + Ax^{\gamma - 1}\,\partiald{{\rm y}\pars{x,t}}{x}}.$

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}
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  • $\begingroup$ @ Felix Marin: You seem to have made a mistake when you insert the definitions into the PDE (first multi-line equation). Correcting for this, I get the following ODE for $f(\xi)$:$$ f''(\xi)+f'(\xi)\left(\frac{\xi}{2 A (1-\gamma/2)^2}+\frac{1}{\xi}\right)=0.$$This gives me a solution of $$y(x,t) = f(0) + B \mathrm{Ei}\left(-\frac{x^{2-\gamma}}{A\,t\,(2-\gamma)^2}\right).$$ However, I fail to see where the initial condition comes in. Suppose you have a dirac delta function at some point $x_0$ at $t_0=0$. How can you calculate the time evolution? Can you help me out here? $\endgroup$ – John Smith Dec 4 '13 at 20:16
  • $\begingroup$ @JohnSmith I'll check later since now I'm busy. It happens frequently with long calculations. Usually, I make the calculations directly on the screen. I call this $\tt CalcuTEX$. Thanks. $\endgroup$ – Felix Marin Dec 4 '13 at 20:19
  • $\begingroup$ @JohnSmith You were right. I just rechecked every thing. $\endgroup$ – Felix Marin Dec 6 '13 at 2:24
  • $\begingroup$ I accepted the answer as it solves the original question. But do you know how this can use a delta function as initial condition? $\endgroup$ – John Smith Dec 7 '13 at 6:17

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