PDE with complex argument

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Main Question or Discussion Point

I'm in truble with a partial differential equation. Actually it is a system of PDE but It would be useful to solve at least one of them.
The most easy one is this one

[tex]
2 \bar{\xi}\left(\bar{s},\bar{t},\bar{u}\right) - 2 \xi\left(s,t,u\right) + \left(s-\bar{s}\right)\left(\bar{\partial}_\bar{s} \bar{\xi} + \partial_s \xi \right) = 0
[/tex]

This equation can be simplified to

[tex]
2 A^*\left(z^*\right) - 2 A\left(z\right) + \left(z-z^*\right)\left(\bar{\partial}_{z^*}A^*+\partial_{z}A\right)= 0
[/tex]

I further developed my computation using [tex] A(z) = u(x,y) + i v(x,y) [/tex] with [tex] u,v \in \mathbb{R}[/tex]
finding (I used Cauchy-Riemann equations)
[tex] v(x,y) = y^2 f(x+y) [/tex]
Here is where I get stucked since I cannot find a suitable form of "f(x+y)" in order to obtain "u" and satisfy Cauchy-Riemann equations...
Any ideas?
 

Answers and Replies

  • #2
1,796
53
I'm in truble with a partial differential equation. Actually it is a system of PDE but It would be useful to solve at least one of them.
The most easy one is this one

[tex]
2 \bar{\xi}\left(\bar{s},\bar{t},\bar{u}\right) - 2 \xi\left(s,t,u\right) + \left(s-\bar{s}\right)\left(\bar{\partial}_\bar{s} \bar{\xi} + \partial_s \xi \right) = 0
[/tex]

This equation can be simplified to

[tex]
2 A^*\left(z^*\right) - 2 A\left(z\right) + \left(z-z^*\right)\left(\bar{\partial}_{z^*}A^*+\partial_{z}A\right)= 0
[/tex]

Any ideas?
Consider the expression:

[tex]
\bar{\partial}_{z^*}A^*[/tex]

I assume that means:

[tex]\overline{\frac{\partial \bar{f}}{\partial\bar{s}}}[/tex]

but we know that:

[tex]\frac{\partial \overline{f}}{\partial \overline{s}}=\overline{\frac{\partial f}{\partial s}}[/tex]

which means you have:

[tex]2\overline{A}(\overline{z})-2A(z)+2(z-\overline{z})\frac{\partial A}{\partial z}=0[/tex]
 
  • #3
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I'm afraid I use the wrong notation or maybe I didn't understand at all! =)
with
[tex] \bar{\partial}_{\bar{s}} \xi^* [/tex]
I mean the derivate of xi* wrt the complex conjugate of s (i.e. \bar{s}). I use the bar over the partial derivative to point out that the derivate is made over \bar{s} and not s. Sorry about this misleading notation! :)
 
  • #4
1,796
53
I'm afraid I use the wrong notation or maybe I didn't understand at all! =)
with
[tex] \bar{\partial}_{\bar{s}} \xi^* [/tex]
I mean the derivate of xi* wrt the complex conjugate of s (i.e. \bar{s}). I use the bar over the partial derivative to point out that the derivate is made over \bar{s} and not s. Sorry about this misleading notation! :)
Ok, that's confussing. Tell you what, how about we just do it my way:

[tex]2\overline{A}(\overline{z})-2A(z)+2(z-\overline{z})\frac{d A}{d z}=0[/tex]

Can we even solve that one? The conjugate variables really hit me with a surprise though and I'm not use to working with DEs like that. I mean what do you do with something like that? Is it even well-posed? Suppose nobody could help us and we had to do something with it, a thesis or something? What do we do? Suppose we could first look at:

[tex] \frac{dy}{dz}+\overline{y}(\overline{z})=0[/tex]

Can we even do that one? Does it even make sense? Looks like another whole-semester type problem to me.
 
Last edited:
  • #5
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I have to admit I'm confused too...
My problem, I mean in its original formulation, require to find the holomorphic Killing vector of a given Kahler manifold. In order to do that I found I have to solve that equation (and many more to be honest...).
Now I wondering if by [tex] \bar{A}(\bar{z})[/tex]
they actually mean [tex] \left(A(z)\right)^*[/tex]
In that case I can set [tex] A = u(x,y)+iv(x,y) \qquad \bar{A} = u(x,y)-iv(x,y)[/tex]
For which I found this solution
[tex]
u(x,y) = \frac{1}{2} C_1 \left(x^2-y^2\right)+C_2 x + C_3 \qquad
v(x,y) = C_1 xy + C_2 y
[/tex]
Which is a bit tempting since it satisfy also Cauchy Riemann equations..
 
  • #6
1,796
53
Now I wondering if by [tex] \bar{A}(\bar{z})[/tex]
they actually mean [tex] \left(A(z)\right)^*[/tex]
I think that means the conjugate of A at the conjugate of z. So if:

[tex]A(z)=iz[/tex]

[tex]A(\overline{z})=i\overline{z}[/tex]

[tex]\overline{A(\overline{z})}=-iz[/tex]

Not sure though ok?
 
  • #7
33
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Bulletin from the front. :)

As I supposed they intended just the conjugation of the entire function not of both function and variables... So I solved, thank you anyway!
 

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