# Index Concordia

## Solving equations of motions in some gravity background

Posted in AdS/CFT, String Theory by Index Guy on August 31, 2008

I would like to consider the gravity background:

$ds^2 = A dx^2 + B dr^2 ,$

with the case that $A = B^{-1}.$ We saw previously that the equations of motion were given by:

$\partial_{i}\left(A\partial_{i}x^a\right) = 0$     and     $-2\partial_{i}^{2}r = \displaystyle\frac{d \left(\ln{B}\right)}{d r}\left(-\eta_{ab}\partial_{i}y^{a}\partial_{i}y^{b} +\partial_{i}r\partial_{i}r\right).$

Now we take the warp factor to have the form $B = r^{\beta}$. Then we have

$-2 r \partial_{i}^{2}r = \beta\displaystyle\left(-\eta_{ab}\partial_{i}y^{a}\partial_{i}y^{b} +\partial_{i}r\partial_{i}r\right).$

We will introduce a new symbol,

$\eta_{ab}\partial_{i}y^{a}\partial_{i}y^{b} = -W^{2}$     with $\displaystyle W(u_{1}, u_{2})$ a function of the worldsheet coordinates.

The differential equation now looks like:

$2 r \partial_{i}^{2}r + \beta \partial_{i}r\partial_{i}r = \beta W^{2}.$

We now assume that $r$ can be factored into functions of each coordinate, $r = T(u_{1}) S(u_{2})$. Then we can solve this PDE when

$k^2 = \displaystyle\frac{\beta W^2}{2 T^{2} S^{2}}$     with $k$ a constant.

Separation of variables gives the following differential equation:

$\displaystyle\frac{d^2 Y}{du_{i}^2} + \alpha \left(\displaystyle\frac{d Y}{d u_{i}}\right)^{2} = k_{i}^{2},$

with

$\alpha = 1 + \displaystyle\frac{\beta}{2}$     and     $T = \exp{Y}.$

Inserting this into Mathematica gives:

$Y(u) = C_{1} + \displaystyle\frac{1}{\alpha}\ln{\left(\cosh{\left(k_{i}\sqrt{\alpha}\left[u_{i} + C_{2}\right]\right)}\right)}.$     which means     $T(u) = C_{1} \left(\cosh{\left(k_{i}\sqrt{\alpha}\left[u_{i} + C_{2}\right]\right)} \right)^{1/\alpha}.$

Notice that the case $\beta = -2$ is interesting: for the constraints we have used the solution is an exponential function of a quadratic polynomial.

This should be more correct if one takes $W$ to be a constant and keeps all the $r$ dependence explicit. In that case the PDE looks interesting, but hard to solve.