# How do I solve a certain differential equation?

I need to find out a solution to the equation:

z r[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


subject to the boundary conditions: r == R and r'[z0] = ∞.

It's known that a solution exists and it is: r^2 + z^2 == R^2, I tried:

DSolve[
{z r[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,
r == R, r'[z0] == ∞},
r[z], z]


The equation and the solution can be found in arXiv: hep-th/0605073 page 43, eqn. 7.8

• You already have the answer:r^2 + z^2 = R^2, so what's the problem.? – Mariusz Iwaniuk May 12 '19 at 9:37
• what is zr[z] ? Is this supposed to be z*r[z] and what is r'[SubStar[z]] == is supposed to be? if SubStar[z] is supposed to be some constant, why not write r'[z0] or such so it is clear? – Nasser May 12 '19 at 9:38
• @MariuszIwaniuk I have the solution but don't know how to solve the equation. – Sabyasachi Maulik May 12 '19 at 11:02
• You believe the solution is independent of d? Why? – m_goldberg May 12 '19 at 12:45
• @Michael E2 Yes z0 is equal to R. – Sabyasachi Maulik May 12 '19 at 14:17

d = 2; sol =
DSolve[{z r[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}, r[z], z]

(*Out[]= {{r[z] ->
0}, {r[z] -> -I E^-C Sqrt[-1 + E^(2 C) z^2] + C}, {r[z] ->
I E^-C Sqrt[-1 + E^(2 C) z^2] + C}}*)
s = r[z] /. sol[] /. C -> 0

(*Out[]= I E^-C Sqrt[-1 + E^(2 C) z^2]*)
Solve[s^2 + z^2 == R^2, C]

(*Out[]= {{C ->
ConditionalExpression[1/2 (2 I \[Pi] C + Log[1/R^2]),
C \[Element] Integers]}}*)


Consequently $$r(z)=\pm \sqrt {R^2-z^2}$$

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 $$d$$.

Let $$\rho^2 = z^2+r^2\quad\text{and}\quad \left({ds\over dz}\right)^2 = 1 + \left({dr \over dz}\right)^2\,.$$ Then the ODE is equivalent to $$z\,{d^2(\rho^2) \over dz^2} = (d-1)\left({ds\over dz}\right)^2 {d(\rho^2) \over dz} \,.$$ By inspection $$\rho^2 = \text{constant}$$ is a solution.

Check (there's a factor of two difference between the equations):

ode = z r[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;
With[{ds = Sqrt[1 + r'[z]^2], ρ = Sqrt[z^2 + r[z]^2]},
2 ode - (z D[ρ^2, {z, 2}] - (d - 1) ds^2 D[ρ^2, z]) //
Simplify
]
(*  0  *)