Are linear, CTCS codes always stable?

I would like to solve some equations which basically look like this $$\frac{\partial u}{\partial x}=F\left(v,\frac{\partial v}{\partial y},\frac{\partial^2 v}{\partial y^2}\right),$$ $$\frac{\partial v}{\partial x}=G\left(u,\frac{\partial u}{\partial y},\frac{\partial^2 u}{\partial y^2}\right).$$ My current plan is to solve this by replacing every derivative with a centred-difference equation and so the numerical scheme looks like this $$u_{i+1,j}=u_{i,j} + \Delta x\,F\left(v_{i+1/2,j},v_{i+1/2,j+1},v_{i+1/2,j-1}\right)$$ $$v_{i+3/2,j}=v_{i+1/2,j} + \Delta x\,G\left(u_{i+1/2,j},u_{i+1/2,j+1},u_{i+1/2,j-1}\right),$$ where I define each $$v_{i+1/2,j}$$ to be located halfway between $$u_{i,j}$$ and $$u_{i+1,j}$$ in the $$x$$-direction. I initialisise $$u$$ and $$v$$ at $$x=x_{min}$$ and then iterate along in $$x$$. Is this scheme stable? Do you know if a scheme for solving such a problem is documented anywhere. I have coded it up and it seems to be stable for certain values of $$\Delta x$$ and $$\Delta y$$. However, I do not know if there is a fundamental problem with my scheme or if I have simply made a silly coding error somewhere.

Edit:

I have linked a pdf showing the expressions for F and G. They are highlighted in red text on page 3.

https://www.dropbox.com/s/43q5kkkqrrkbvrq/Normal_mode_code.pdf?dl=0

• Without knowing the details of your implementation it’s impossible for us to know if something is wrong or not. Also, if it is possible, specify $F$ and $G$ operators explicitly. – Alone Programmer Feb 22 '20 at 3:30
• I suppose that CTCS refers to centred in time and centred in space. If that's the case, then the answer is no. One counterexample is the advection equation. – nicoguaro Feb 22 '20 at 13:06
• @nicoguaro I would say that's the answer. – Anton Menshov Feb 22 '20 at 19:41
• @AntonMenshov, I'm not that sure. I just calculated the stability condition using von Neumann analysis and it seemed to be stable (if CFL≤1). – nicoguaro Feb 22 '20 at 23:57
• Thank you for your comments. The expressions for F and G are quite complicated. They involve complex numbers. I have attached a pdf showing the equations. – Peanutlex Feb 24 '20 at 10:29