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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?

(For further clarification, I have tried to draw a single iteration of what I am attempting to describe.)

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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.

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  • $\begingroup$ 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.) $\endgroup$
    – Mathews24
    Commented Oct 31, 2016 at 2:37
  • $\begingroup$ The difference between the time variable and the space variable is that because of causality, you typically only have initial conditions and then you can advance the solution one time step to the next, only taking information from the past into the future. On the other hand, for spatial variables, you typically have information going both left to right hand right to left, and so you can't advance in one direction only. $\endgroup$ Commented Oct 31, 2016 at 19:18
  • $\begingroup$ 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. $\endgroup$
    – Mathews24
    Commented Nov 1, 2016 at 4:23
  • $\begingroup$ I kept doing this until I got to the final i (M) for k = 0, then I would go to k = 1 at i = 0, and continue the process to k = 1, i = M, and then go back to k =2, i = 0 and so forth. And my solution matched exactly what we expect using the method I described above. It's almost like the coupling of terms in these equations updates E in t. since I am only computing values at points at the beginning (i.e. z = 0) and then moving forward by one increment, where I assume I am in a mirror-less environment and the wave propagates forward and thus does not affect points behind me. $\endgroup$
    – Mathews24
    Commented Nov 1, 2016 at 4:23
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    $\begingroup$ Yes, if you have one-way transport (in both time and space) then you can use these marching techniques. $\endgroup$ Commented Nov 1, 2016 at 14:06

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