spherically symmetric configurations

Question

$$\Delta S -S +S^3=0$$ How this Differential equation can be written in this form:

\begin{equation} \frac{d^2S}{d\rho^2}+\frac{D-1}{\rho}\,\frac{dS}{d\rho} -S+S^3=0 \end{equation} Which is spherically symmetric configurations. D= dimensions. Details in the paper equations (21, 44)

Answer 1

If $S$ is speherically symmetric, say $S(x) = S(\rho)$ with $\rho = |x|$, we have \begin{align*} \Delta S &= \sum_i \partial_i^2S\\ &= \sum_i \partial_i \bigl(\partial_\rho S\cdot \partial_i\rho\bigr)\\ &= \sum_i \left(\partial_\rho^2 S \cdot (\partial_i \rho)^2 + \partial_\rho S \cdot \partial_i^2\rho\right) \end{align*} Now $$ \partial_i\rho(x) = \frac{x_i}{\rho(x)} $$ and hence $$ \partial_i^2\rho(x) = \frac{\rho(x) - x_i\frac{x_i}{\rho(x)}}{\rho^2(x)} = \frac{\rho^2(x) - x_i^2}{\rho^3(x)} $$ So $$\sum_i (\partial_i \rho)^2 = \frac{1}{\rho^2(x)}\sum_i x_i^2 = 1$$ and $$ \sum_i \partial_i^2\rho = \frac{D\rho^2 - \sum_i x_i^2}{\rho^3} = \frac{D-1}{\rho} $$ and hence $$ \Delta S = \partial_\rho^2 S + \frac{D-1}\rho \cdot \partial_\rho S $$

Answer 2

The expression for Laplacian operator in the $D$-dimensional spherical coordinates is $$\Delta=\frac{\partial^2}{\partial\rho^2}+\frac{D-1}{\rho}\frac{\partial}{\partial\rho}+\frac{1}{\rho^2}\Delta_{S^{D-1}},$$ where $\Delta_{S^{D-1}}$ is a differential operator called Laplace-Beltrami operator on the $(D-1)$-dimensional sphere $S^{D-1}$. It depends only on the angular variables (in particular, all the derivatives are with respect to them). Therefore, if you are looking for rotationally symmetric solutions of your PDE, angular derivatives are all zero and you end up with the ODE you have written.

For example, as $D=2$, set $x=\rho\cos\varphi$, $y=\rho\sin\varphi$ and this angular part of Laplacian will be equal to $$\Delta_{S^1}=\frac{\partial^2}{\partial\varphi^2}.$$

Similarly, for $D=3$, introducing the spherical coordinates $$\begin{cases}x=\rho\cos\varphi\sin\theta,\\ y=\rho\sin\varphi\sin\theta,\\ z=\rho\cos\theta,\end{cases}$$ we have the expression $$\Delta_{S^2}=\frac{\partial^2}{\partial\theta^2}+\cot\theta\frac{\partial}{\partial\theta}+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}.$$