Solving Coupled PDEs Numerically with Unknown Functions u(x,y) and v(x,y)

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This discussion focuses on solving two coupled partial differential equations (PDEs) with unknown functions u(x,y) and v(x,y) numerically, particularly when one equation is non-linear. The recommended method is a finite-timestep approach, ensuring that the timestep Δx is small relative to the expected fluctuations in the solutions. The iterative form of the equations is emphasized, allowing for the generation of solutions using boundary values. The problem is also linked to complex analysis, suggesting that the equations may relate to the analyticity condition for complex variables.

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I have two unknown function namely u(x,y) and v(x,y). These functions are part of two coupled partial differential equations. I realize that it will be almost impossible to get a general solution seeing as one on the PDEs is non-linear. But given a set of boundary conditions I wish to solve for these unknown functions numerically. I don’t quite know how to go about this though, so any help would be appreciated. The equations are attached to this thread
 

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If I wanted to solve this numerically, I'd use a finite-timestep approach.

Remember:

[tex]\frac{\partial f(x)}{\partial x} \equiv \lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}[/tex]

So: just make sure that your timestep [itex]\Delta x[/itex] is small compared to the (expected) fluctuations in your solutions. In that case, you can re-write your equations in an iterative form (note that I used [itex]\Delta x=1[/itex]):

[tex]f(n+1) = {\rm some\; function\; of}\; f(n)[/tex]

which you can do for both of your functions. Now, you can start with your boundary values (for time [itex]n=0[/itex]) and generate the solutions for [itex]n>0[/itex] with a computer. Computationally intensive, but that should be no problem for your equations...

Succes!
 
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It looks to me as though this is a problem from complex analysis. Your second equation is the analyticity condition for a function of a complex variable. Perhaps the first equation is simply expressed in terms of that function.

dhris
 

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