= Solution
For unit mass, the <Lagrangian> in <cylindrical coordinates> is
$$
L=\frac12\left(\dot r^2+r^2\dot\varphi^2+\dot z^2\right)-\Phi(r,z).
$$
Introduce <shearing sheet> coordinates by $r=r_0+x$ and $\varphi=\Omega_0t+y/r_0$. Evaluate all derivatives of $\Phi$ at $(r_0,0)$. The reference <circular orbit> obeys $\Phi_r=r_0\Omega_0^2$, while midplane symmetry gives $\Phi_z=\Phi_{rz}=0$. Its <Taylor expansion> is
$$
\Phi=\Phi_0+\Phi_rx+\frac12\Phi_{rr}x^2+\frac12\Phi_{zz}z^2+O(3).
$$
Expanding the kinetic energy to the same order gives
$$
\frac12r^2\dot\varphi^2
=\frac12r_0^2\Omega_0^2+r_0\Omega_0^2x+r_0\Omega_0\dot y
+\frac12\Omega_0^2x^2+2\Omega_0x\dot y+\frac12\dot y^2+O(3).
$$
The terms linear in $x$ cancel by circular-orbit balance. Discard the constant and the term $r_0\Omega_0\dot y$ by <total-time-derivative invariance of a Lagrangian>; these do not change the <Euler-Lagrange equations>. The <particle Lagrangian in a shearing sheet> is therefore
$$
\boxed{L_2=\frac12(\dot x^2+\dot y^2+\dot z^2)+2\Omega_0x\dot y-\Phi_t},
$$
with <shearing-sheet tidal potential>
$$
\boxed{\Phi_t=\frac12(\Phi_{rr}-\Omega_0^2)x^2+\frac12\Phi_{zz}z^2
=-q_0\Omega_0^2x^2+\frac12\Omega_z^2z^2}.
$$
The second form follows from $\Phi_r=r\Omega^2$, the <orbital shear parameter> $q_0=-d\log\Omega/d\log r$ and the <vertical epicyclic frequency> $\Omega_z^2=\Phi_{zz}$.
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