A simple approach here would be to use the shooting method, by integrating from $\xi$=0 to infinity (some large value) the ODE written as

$ \frac{d}{d\xi} \phi = \sqrt{-2 S(\phi)} $

Due to the symmetry we have the constraint $d_{\xi} \phi$=0 at $\xi$=0, and normally that would be enough to integrate the ODE from zero to infinity. However, there are two solutions here that satisfy the condition $d_{\xi} \phi$=0 at $\xi$=0; one is $\phi$=const and the other one is the correct solution that dies out at $\xi \rightarrow \infty$. In other words, there are two integral curves that touch at $\xi$=0, and using the shooting method would keep us on the wrong integral curve.

To overcome this difficulty, let's manipulate the equation a little bit. Take the derivative of the equation

$ \frac{d}{d\xi} (\frac{1}{2} (\dot{\phi})^2 + S(\phi) = 0) $

That produces

$ \dot{\phi} \ddot{\phi} + S' \dot{\phi} = 0, $

where the dot stands for $d_{\xi}$ and prime stands for $d_{\phi}$.

Now we can divide by $\dot{\phi}$ to write it as

$ \ddot{\phi} = -S' $

Then introduce variable $\psi=\dot{\phi}$, and we have a system of ODEs amenable to solving by the shooting method

$ \dot{\phi} = \psi \\ \dot{\psi} = -S'(\phi) $

The initial conditions at $\xi$=0 are $\psi$=0 and $\phi$=$\phi_0$, where the parameter $\phi_0$ is found by shooting to satisfy the BC $\phi=0$ at infinity.

A simple approach here would be to use the shooting method, by integrating from $\xi$=0 to infinity (some large value) the ODE written as

$ \frac{d}{d\xi} \phi = \sqrt{-2 S(\phi)} $

Due to the symmetry we have the constraint $d_{\xi} \phi$=0 at $\xi$=0, and to find the right value of $\phi$ at $\xi=0$ we'd need to solve $S(\phi)$=0 which is not easy to do. But even if we find the right value $\phi(0)$, there are two solutions here that satisfy the condition $d_{\xi} \phi$=0 at $\xi$=0; one is $\phi$=const and the other one is the correct solution that dies out at $\xi \rightarrow \infty$. In other words, there are two integral curves that touch at $\xi$=0, and integrating the ODE from $\xi=0$ would keep us on the wrong integral curve.

To overcome this difficulty, let's manipulate the equation a little bit. Take the derivative of the equation

$ \frac{d}{d\xi} (\frac{1}{2} (\dot{\phi})^2 + S(\phi) = 0) $

That produces

$ \dot{\phi} \ddot{\phi} + S' \dot{\phi} = 0, $

where the dot stands for $d_{\xi}$ and prime stands for $d_{\phi}$.

Now we can divide by $\dot{\phi}$ to write it as

$ \ddot{\phi} = -S' $

Then introduce variable $\psi=\dot{\phi}$, and we have a system of ODEs amenable to solving by the shooting method

$ \dot{\phi} = \psi \\ \dot{\psi} = -S'(\phi) $

The initial conditions at $\xi$=0 are $\psi$=0 and $\phi$=$\phi_0$, where the parameter $\phi_0$ is found by shooting to satisfy the BC $\phi=0$ at infinity.