Solving a Killing Vector Problem in General Relativity: Help with a PDE System

[Magister]

Hi
Solving a Killing vector problem, in General Relativity, I got the following PDE system:
<br /> \frac{\partial X^0}{\partial x}=0<br />
<br /> \frac{\partial X^1}{\partial y}=0<br />
<br /> \frac{\partial X^2}{\partial z}=0<br />
<br /> \frac{\partial X^0}{\partial y} + \frac{\partial X^1}{\partial x}=0<br />
<br /> \frac{\partial X^0}{\partial z} + \frac{\partial X^2}{\partial x}=0<br />
<br /> \frac{\partial X^1}{\partial z} + \frac{\partial X^2}{\partial y}=0<br />

where X^a , a=0,1,2 are the three components of the Killing vector that I am looking for. I have spend a lot of time trying to solve this system but I am not getting any solution.
Thanks for any idea.

 

[Mute]

I don't know what a Killing vector is, so assuming that is has a "nice" form, where by "nice" I mean it satisfies the continuity conditions, etc, which mean that equality between mixed partial derivatives holds, i.e., \partial_{x}\partial_{y} = \partial_{y}\partial_{x}, then it follows that

\partial_{yy}X^{0} = 0
\partial_{zz}X^{0} = 0
\partial_{xx}X^{1} = 0
\partial_{zz}X^{1} = 0
\partial_{xx}X^{2} = 0
\partial_{yy}X^{2} = 0

You get these by differentiating your coupled equations with respect to one of the variables for which the derivative of X^a is zero, then swap the order of the partials and that term goes away. You can do that twice with each equation.

I haven't thought of this much past that - my guess is that integrating those and then making sure the arbitrary functions of integration/boundary conditions work out should be sufficient.

 

[Magister]

I don't know what a Killing vector is, so assuming that is has a "nice" form, where by "nice" I mean it satisfies the continuity conditions, etc, which mean that equality between mixed partial derivatives holds, i.e., LaTeX graphic is being generated. Reload this page in a moment., then it follows that

Yes, you can assume that they are "nice".
Can you tell me how many solutions should I expect and why? The problem is that I always get lost with the vairous solutions I get.
Thanks for the replies

 

[Mute]

Well, the second order differential equations will give you two linearly independent solutions. In this case, since the derivative is equal to zero, your solutions will just be a term proportional to the variable and a constant - where you have to note that the constant is actually a function of the other two variables which you did not differentiate with respect to.

You have three differential equations for each X^a, two of which are second order and one of which is first order. Solving all three is easy, and so after that you just have to match up the arbitrary functions so that the X^a satisfies all three of its differential equations.