I have an action given by,
\begin{equation} S = \int^{\tau_f}_{\tau_0} d\tau \left(\frac{1}{z^3}\sqrt{-f(u,z) \dot{u}^2 + 2 \dot{u} \dot{z} + \dot{x}^2} + \frac{2 \dot{z}}{z^3} \right), \end{equation}
where the Lagrangian is,
\begin{equation} L = \frac{1}{z^3}\sqrt{-f(u,z) \dot{u}^2 + 2 \dot{u} \dot{z} + \dot{x}^2} + \frac{2 \dot{z}}{z^3}, \qquad f(u,z) = 1 - m e^{-g u} z^4, \end{equation}
$m$ and $g$ are constants.
The boundary term that arises from this is,
\begin{equation} \left( -f(u,z) \dot{u} + \dot{z} \right) \delta u + \left( \dot{u} + 2 \sqrt{-f(u,z) \dot{u}^2 + 2 \dot{u}\dot{z} + \dot{x}^2} \right) \delta z + \dot{x} \delta x \Biggr|^{\tau_f}_{\tau_0} = 0. \end{equation}
The point at $\tau_f$ is fixed and that is not the important issue here, while the point at $\tau_0$ can move freely in the $u-z$ plane but at a fixed $x$ so that I have a constraint $\delta x = 0$, while I don't require $\delta u$ and $\delta z$ to be zero so that,
\begin{equation} -f(u_0,z_0) \dot{u}_0 + \dot{z}_0 = 0, \qquad \dot{u}_0 + 2 \sqrt{-f(u_0,z_0) \dot{u}_0^2 + 2 \dot{u}_0 \dot{z}_0 + \dot{x}_0^2} = 0. \end{equation}
In the action, I could have parameterized in terms of $x$ so that the boundary condition becomes,
\begin{equation} -f(u_0,z_0) u'_0 + z'_0 = 0, \qquad u'_0 + 2 \sqrt{-f(u_0,z_0) u'^2_0 + 2 u'_0 z'_0 + 1} = 0. \end{equation}
This leads to a boundary condition that is in the form of a derivative,
\begin{equation} u'_0 = \pm 2 \sqrt{\frac{1}{1-4 f(u_0,z_0)}}, \qquad z'_0 = \pm 2 f(u,z) \sqrt{\frac{1}{1-4 f(u_0,z_0)}}. \end{equation}
I'm using finite difference numerical method to solve the differential equation obtained from the Euler-Lagrange equation, I need to impose the derivative boundary condition given some test boundary values $u_0$ and $z_0$. After injecting the test values, the finite difference numerical method will randomly move the boundary points and settle to the best approximate solution. The issue is, since $u_0$ and $z_0$ are inside the square-root, sometimes there will be values $u_0$ and $z_0$ where the inside of the square-root becomes negative hence giving imaginary values.
I would like to ask if there is any way to go around this problem? One possibility is if there is an identity to make the quantity in the square-root come out, although I'm not yet sure what it is. Another possibility is to make the Lagrangian the quadratic Lagrangian, although there are subtleties that could possibly not make it work.
Specifically, as in the post, it says that as long as $dL/ds = 0$ where $s$ is some parameter, then $L^2/2$ gives the same equation of motion as $L$. I wanted to verify if that applies in my case. In my case, we can choose $x$ as a parameter, and $dL/dx = 0$, so does that mean I can always use $L^2/2$?
Any help would be appreciated.
Edit: I just noticed that even $L^2/2$ would still contain the square root, so it seems like this method also does not work.
For what it's worth, OP's square root action $$\begin{align} S[u,x,z]~=~& \int^{\tau_f}_{\tau_0} d\tau~ L,\cr L~=~&z^{-3}\left(\sqrt{G}+2 \dot{z}\right), \cr G ~=~& -f(u,z) \dot{u}^2 + 2 \dot{u} \dot{z} + \dot{x}^2,\cr f(u,z) ~=~& 1 - m e^{-g u} z^4, \end{align}\tag{1}$$ is equivalent to the following non-square root action $$\begin{align} \widetilde{S}[u,x,z,e]~=~& \int^{\tau_f}_{\tau_0} d\tau~\widetilde{L},\cr \widetilde{L}~=~&z^{-3}\left(\frac{G}{2e}+\frac{e}{2}+2 \dot{z}\right),\cr e~>~&0, \end{align}\tag{2}$$ because the Euler-Lagrange (EL) equation for the auxiliary variable $e$ is $$e~=~\sqrt{G},\tag{3}$$ and $$ \widetilde{S}[u,x,z,e\!=\!\sqrt{G}]~=~S[u,x,z].\tag{4}$$