In order to compute the holographic entanglement entropy for a spherical subsystem in AdS using the Ryu-Takayanagi conjecture, one needs to solve the following second order nonlinear differential equation:

$$zr(z)r''(z) - (d-1)r(z)(r'(z))^3 - (d-2)z(r'(z))^2 - (d-1)r(z)r'(z) - (d-2)z = 0$$

and obtain the solution $r^2 + z^2 = l^2$, I need help in obtaining this solution.

N.B.: The equation (in a mildly different form), and its solution are given in hep-th/0605073


1 Answer 1


In the case of a disk $d=2$, the equation is simplified: $$zr(z)r''(z) - r(z)(r'(z))^3 - r(z)r'(z) = 0$$ We discard the obvious solution of the equation $r = 0$, then putting $r' = y$ we find $$zy'=y^3+y$$ This equation is easily integrated, and using the boundary condition $y(R) =\infty $, we find $$\frac {r'}{\sqrt {1+r'^2}}=\frac {z}{R}$$

Solving the equation for $r`$, we have $$r'=\pm \frac {z}{\sqrt {R^2-z^2}}$$ And find finally $$r=\pm \sqrt {R^2-z^2}$$ Note that in the article Aspects of Holographic Entanglement Entropy the authors cited another equation on p.34 $$ rzz′′ + (d−1)z(z′)^3 + (d−1)zz′ + dr(z′)^2 + dr = 0. $$ For this equation, the solution $z^2+r^2=R^2$ exists for any $d$.

  • $\begingroup$ Thank you, any suggestion on how to generalize to higher dimensions? $\endgroup$ Commented May 12, 2019 at 17:12
  • $\begingroup$ @SabyasachiMaulik See update to my answer $\endgroup$ Commented May 12, 2019 at 20:12
  • $\begingroup$ Can you elaborate on that please? $\endgroup$ Commented May 13, 2019 at 2:47
  • $\begingroup$ @SabyasachiMaulik Use code r*z*z'' + (d \[Minus] 1) z (z')^3 + (d \[Minus] 1) z*z' + d*r (z')^2 + d*r /. {z -> Sqrt[R^2 - r^2], z' -> -(r/Sqrt[-r^2 + R^2]), z'' -> -(r^2/(-r^2 + R^2)^(3/2)) - 1/Sqrt[-r^2 + R^2]} // Simplify $\endgroup$ Commented May 13, 2019 at 15:48

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