I am trying to solve the CDR-Equation in 2D:
$$\frac{\partial c(x,y)}{\partial t} + \nabla \cdot ( -d\nabla c(x,y) + \vec{v}(x,y) c(x,y))+ a c(x,y)=0\,,$$ with Boundary Conditions (length of square is $L$): $$c(0,y)=0$$ $$-d\nabla c(L,y) + v(L,y) (c(L,y))=0\,.$$. $$ c(x,y=0)=0$$ $$ c(x,y=L)=0$$
1.) Why exactly is this equation stiff? Does it depend on the reaction term $ac(x,y)$?
2.) Can I solve the equation with the Crank-Nicholson method? Is the error huge? If yes, what is the best method?
Mathematically speaking, stiffness is meaningless for a single differential equation, and is rather attributed to a set of differential equations that have different time-scales (e.g. when trying to solve two coupled equations with time-scales of 1 second and 1 day, respectively). However, a single equation can also be referred to as stiff if certain numerical integration methods are numerically unstable. Source/sink terms can give rise to stiffness, for instance: $$\frac{dy}{dt}=-1000y$$ is a good example where sink term has given rise to stiffness. That being said, $ac(x,y)$ might be causing the stiffness in your PDE.
Theoretically, you can solve any time-dependent PDE with Crank-Nicolson method. However, you should use adaptive time-steps when solving a stiff equation, otherwise the computational cost is huge. The naive Crank-Nicolson formulation is based on fixed time-step, however, you can derive your customized version with adaptive time-step. But, I do not recommend as it is cumbersome. Your best alternative, in my opinion, is to use method of lines. You can find useful information about this method here. To give you a rough idea, in the method of lines, you discretize your PDE with respect to spatial variables (and not time). This eventually gives you a set of first order ODEs (ordinary differential equation) with respect to time. Now you can simply use a stiff ODE solver (e.g. Adam's Bashforth method) to integrate the set of ODEs.
