# Classification of method for solving PDEs

If I have a system of equations as follows (where $i = \sqrt{-1}$):

$$\frac {\partial A}{\partial t} = iA^*B - A \tag{1} \\$$ $$\frac {\partial B}{\partial z} = AB^* - B \tag{2}$$

Using the method of lines, in general, I discretize space and solve the system for all time for a particular $z_k$—this transforms the PDEs into a system of ODEs. But in the above case, I have a spatial derivative and also a time derivative that determines the evolution of my system. Is the particular method of discretizing both time and space and solving over such a grid with different partial derivatives (e.g. $t, z$) given a particular name? Is such a method invalid for any particular reason when solving PDEs? I realize I will not have any intermediate values for $B$ when advancing in $t$, and lacking intermediate values for $A$ when advancing in $z$ when using a method such as a 4th order Runge-Kutta. This, I assume, will result in a poor approximation, but is such a method invalid for any reason? Does such a method have a particular name?

It is easier to see what is happening if you put the $\partial B/\partial z$ term also on the right hand side: Then, you have one equation with a time derivative and one that does not. This is what we typically call a "Differential Algebraic Equation" (DAE), and it is the same as what is happening, for example, in the time dependent Stokes and Navier-Stokes equations. There are a good number of methods that implement efficient schemes for DAEs, such as projection methods.
• Thank you for the insight. I'll certainly look into projection methods. Although a part of my initial question still remains: is advancing $B$ in $z$ invalid for any reason? (I can open up a new question for this.) – Mathews24 Oct 31 '16 at 2:37
• That makes sense although I don't know if my method violates anything you wrote. And luckily, I do have both boundary and initial conditions, and it appears the method did work.The main reason I am insistent on clarifying the matter is because I did actually solve the following system by stepping N and P in t, while E in z by starting at $z_{k=0}$ and $t_{i=0}$, and then computing E at i=0,k=1, and P and N at i=1,k=0. I then went to $z_{k=0}$ and $t_{i=1}$ and computed E at i=1,k=1, and P and N at i=2,k=0. – Mathews24 Nov 1 '16 at 4:23