# Stuck with a PDE system

1. Jul 5, 2012

### L0r3n20

Hi all! I'm stuck with a system of PDE. I'm not sure I want to write it here in full, so l'll write just one of them. I've found a solution to this equation but I'm not sure it's the most general one since when I plug this solution in to the other eqs, I get a trivility condition for the coefficients
$$2\bar{k}^1\left(\bar{s},\bar{t},\bar{u}\right)-2 k^1\left(s,t,u\right)+\left(s-\bar{s}\right)\left(\partial_s k^1\left(s,t,u\right) + \bar{\partial}_{\bar{s}}\bar{k}^1\left(\bar{s}, \bar{t}, \bar{u}\right)\right) =0$$
Can someone help?

2. Jul 5, 2012

### haruspex

Not sure I've understood the equation. Is this simplification valid:
2u(x) - 2v(y) + (y-x)(∂v/∂y + ∂u/∂x) = 0
?
If so:
∂v/∂y + ∂u/∂x = 2(u-v)/(x-y)
Consider (u, v+v') is also a solution. So
∂v'/∂y = 2v'/(y-x)
v' = (y-x)2f(x)
where f is an arbitrary function of x.
Does that help?

3. Jul 6, 2012

### L0r3n20

Thank you for the answer. Ok that was useful, at least a bit. In fact, although it is correct, I need a u and a v depending ONLY from one of the two variables (x and y). In fact I'm dealing with (anti)holomorphic functions and I need them to respect the holomorphicity condition.

4. Jul 7, 2012

### haruspex

OK, so make f(x) constant. You can do likewise for (u+u', v).

5. Jul 8, 2012

### L0r3n20

Even thought I set f(x) constant I get the x-dependence from y-x, right?