Let $u = u(t,x)$ satisfy the PDE $$ \frac{\partial u}{\partial t} = \frac{1}{2}c^2\frac{\partial^2 u}{\partial x^2} + (a + bx)\frac{\partial u}{\partial x} + f u, $$ where $a,b,c,f \in \mathbb{R}$ are constant.

I'm aware of solution methods for when $c \propto x^2$ (so not constant) and $a = 0$, for which I would make the change of variables $x \mapsto \log x$ to make it constant coefficient, use the Fourier transform to make it an ODE and solve from there. This seemingly easier PDE has got me stumped, though, and I would appreciate a push in the right direction!


The general solution is given by: $$u(x,t) = e^{- \lambda t} \, e^{-\frac{x}{2c}(2a + b x)} H_{\alpha}\left(\frac{a + bx}{\sqrt{2b}} \right)$$ where $\alpha = \frac{f+\lambda}{b} - 1$ and $H_{n}(z)$ is the Hermite polynomial.

The solution is obtained by the following.

Let $u(x,t) = e^{- \lambda t} \, F(x)$ in the equation \begin{align} u_{t} = c u_{xx} + (a + b x) u_{x} + d u \end{align} to obtain \begin{align} F'' + \left(\frac{a}{c} + \frac{b}{c} \, x \right) F' + \left( \frac{d + \lambda}{c} \right) \, F = 0. \end{align} Now, from Wolfram Alpha, the equation $y'' + (\alpha + \beta x) y' + \gamma y = 0$ has the solution \begin{align} y = e^{- \frac{x}{2} (2 \alpha + \beta x)} \, H_{p}\left(\frac{\alpha + \beta x}{\sqrt{2 \beta}} \right) \end{align} where $p = \frac{\gamma}{\beta} - 1$. From this the solution above is obtained.

  • $\begingroup$ Thanks for the solution. Do you know a good reference on this technique? $\endgroup$ – bcf Jun 8 '15 at 21:27
  • $\begingroup$ It doesn't seem like $p \in \mathbb{N}$ necessarily. What is meant by non-integral polynomials? $\endgroup$ – bcf Jun 9 '15 at 13:45
  • $\begingroup$ Since the Hermite polynomials have an equivalent hypergeometric representation one can replace one form for the other. If $p$, identified in the solution, is not well formed for the Hermite polynomial form then the hypergeometric form will be utilized more efficiently. $\endgroup$ – Leucippus Jun 9 '15 at 14:07

You can directly apply separation of variables, but in fact this PDE can get the more simplified form when applying separation of variables by applying the following change of variables:

Let $\begin{cases}x_1=a+bx\\t_1=t\end{cases}$ ,

Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial x_1}\dfrac{\partial x_1}{\partial x}+\dfrac{\partial u}{\partial t_1}\dfrac{\partial t_1}{\partial x}=b\dfrac{\partial u}{\partial x_1}$

$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(b\dfrac{\partial u}{\partial x_1}\right)=\dfrac{\partial u}{\partial x_1}\left(b\dfrac{\partial u}{\partial x_1}\right)\dfrac{\partial x_1}{\partial x}+\dfrac{\partial u}{\partial t_1}\left(b\dfrac{\partial u}{\partial x_1}\right)\dfrac{\partial t_1}{\partial x}=b^2\dfrac{\partial^2u}{\partial x_1^2}$

$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial x_1}\dfrac{\partial x_1}{\partial t}+\dfrac{\partial u}{\partial t_1}\dfrac{\partial t_1}{\partial t}=\dfrac{\partial u}{\partial t_1}$

$\therefore\dfrac{\partial u}{\partial t_1}=\dfrac{b^2c^2}{2}\dfrac{\partial^2u}{\partial x_1^2}+bx_1\dfrac{\partial u}{\partial x_1}+fu$

With reference to Change variables into Fokker-Planck PDE,

Let $\begin{cases}x_2=x_1e^{bt_1}\\t_2=t_1\end{cases}$ ,

Then $\dfrac{\partial u}{\partial x_1}=\dfrac{\partial u}{\partial x_2}\dfrac{\partial x_2}{\partial x_1}+\dfrac{\partial u}{\partial t_2}\dfrac{\partial t_2}{\partial x_1}=e^{bt_1}\dfrac{\partial u}{\partial x_2}=e^{bt_2}\dfrac{\partial u}{\partial x_2}$

$\dfrac{\partial^2u}{\partial x_1^2}=\dfrac{\partial}{\partial x_1}\left(e^{bt_2}\dfrac{\partial u}{\partial x_2}\right)=\dfrac{\partial u}{\partial x_2}\left(e^{bt_2}\dfrac{\partial u}{\partial x_2}\right)\dfrac{\partial x_2}{\partial x_1}+\dfrac{\partial u}{\partial t_2}\left(e^{bt_2}\dfrac{\partial u}{\partial x_2}\right)\dfrac{\partial t_2}{\partial x_1}=e^{2bt_2}\dfrac{\partial^2u}{\partial x_2^2}$

$\dfrac{\partial u}{\partial t_1}=\dfrac{\partial u}{\partial x_2}\dfrac{\partial x_2}{\partial t_1}+\dfrac{\partial u}{\partial t_2}\dfrac{\partial t_2}{\partial t_1}=bx_1e^{bt_1}\dfrac{\partial u}{\partial x_2}+\dfrac{\partial u}{\partial t_2}$

$\therefore bx_1e^{bt_1}\dfrac{\partial u}{\partial x_2}+\dfrac{\partial u}{\partial t_2}=\dfrac{b^2c^2e^{2bt_2}}{2}\dfrac{\partial^2u}{\partial x_2^2}+bx_1e^{bt_1}\dfrac{\partial u}{\partial x_2}+fu$

$\dfrac{\partial u}{\partial t_2}-fu=\dfrac{b^2c^2e^{2bt_2}}{2}\dfrac{\partial^2u}{\partial x_2^2}$


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