Conjugation of Complex Functions in Partial Differential Equations

In summary, the conversation discusses a problem with a partial differential equation, which is part of a system of PDEs. The equation is simplified to 2A*(z*) - 2A(z) + (z-z*) [partial derivative of A* with respect to z* + partial derivative of A with respect to z] = 0. The conversation then delves into confusion over notation and the possibility of solving the equation. Finally, it is determined that the equation can be solved by interpreting the bar notation as conjugation of the entire function, not just the variables.
  • #1
L0r3n20
36
2
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?
 
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  • #2
L0r3n20 said:
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
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
L0r3n20 said:
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
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
L0r3n20 said:
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
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!
 

1. What is a PDE with complex argument?

A PDE with complex argument is a partial differential equation that involves complex-valued functions as arguments. This means that the independent variables and coefficients in the equation can have complex values.

2. How are PDEs with complex argument different from regular PDEs?

PDEs with complex argument are more general than regular PDEs, as they allow for complex-valued functions and coefficients. They also tend to have more complex solutions compared to regular PDEs.

3. What are some applications of PDEs with complex argument?

PDEs with complex argument are commonly used in physics, engineering, and mathematics to model phenomena that involve complex quantities, such as electromagnetic fields and quantum mechanics.

4. How are PDEs with complex argument solved?

Similar to regular PDEs, there are several methods for solving PDEs with complex argument, including separation of variables, Fourier series, and numerical methods such as finite difference and finite element methods.

5. Are there any challenges in solving PDEs with complex argument?

Yes, solving PDEs with complex argument can be more challenging compared to regular PDEs due to the complex nature of the solutions. It also requires a good understanding of complex analysis and advanced mathematical techniques.

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