4
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

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

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.