Solving $\partial P(x,t)/\partial t = \left( D \partial^2 /\partial x^2-\mu \partial /\partial x \right) P(x,t)$ with no-flux boundary conditions

45 Views Asked by At

In a mathematical physical problem, I would like to solve the following partial differential equation of type Fokker-Planck: $$ \frac{\partial P(x,t)}{\partial t} = \hat{H} P(x,t) \, , $$ with the operator $$ \hat{H}:= D \, \frac{\partial^2}{\partial x^2} -\mu \, \frac{\partial}{\partial x} $$ where $D$ and $\mu$ are strictly positive numbers. This equation is subject to the boundary conditions of zero flux at boundaries, namely $$ \left. D \, \frac{\partial P(x,t)}{\partial x} - \mu P(x,t) \right|_{x= \pm a} = 0 \, , $$ in addition to the initial condition $P(x,t=0) = \delta(x-x_0)$, where $x_0 \in (-a,a)$.

For the solution of the above PDE, it can readily be noticed that the operator $\hat{H}$ is self-adjoint and has eigenfunctions $\phi_n(x)$ of the form $$ \phi_n(x) = A_n \exp \left( \frac{\mu + \sqrt{\mu^2-4Dn}}{D} \, x\right) + B_n \exp \left( \frac{\mu - \sqrt{\mu^2-4Dn}}{D} \, x\right) \, , $$ for $ N = 0,1,2, \cdots, \infty$.

Then, we have $\hat{H}\phi_n(x) = -n\phi_n(x)$ so that the $n$th eigenvalue is $\lambda_n = -n$. Therefore, the spectral decomposition of $P(x,t)$ can now be given as $$ P(x,t) = \sum_{n=0}^\infty \phi_n(x) \exp \left( -nt \right) \, . $$

However, when trying to apply the prescribed boundary conditions at $x = \pm a$, i guess that we should impose that $$ \left. D \phi_n'(x) - \mu \phi_n(x) \right|_{x=\pm a} = 0, $$ leading to a homogeneous linear system of equations whose solution is trivial, i.e. $A_n = B_n =0$. This does not seem to be correct!

It would be great if someone here could let me know if something went wrong in my development. Any hint or help is highly appreciated. Thank you,