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PDE with complex argument |
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| Sep11-12, 03:10 PM | #1 |
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PDE with complex argument
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? |
| Sep11-12, 03:40 PM | #2 |
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[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] |
| Sep11-12, 04:39 PM | #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! :) |
| Sep12-12, 07:51 AM | #4 |
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PDE with complex argument[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. |
| Sep12-12, 12:01 PM | #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.. |
| Sep12-12, 01:58 PM | #6 |
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[tex]A(z)=iz[/tex] [tex]A(\overline{z})=i\overline{z}[/tex] [tex]\overline{A(\overline{z})}=-iz[/tex] Not sure though ok? |
| Sep20-12, 01:34 AM | #7 |
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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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