= Solution
The three contributions to the <Frank elastic energy> penalize distinct slow distortions of the <nematic director>. <Nematic splay> measures local spreading or convergence and is governed by $K_1(\nabla\cdot\mathbf n)^2/2$. <Nematic twist> measures rotation around an axis perpendicular to the director, with cost $K_2(\mathbf n\cdot\nabla\times\mathbf n)^2/2$. <Nematic bend> measures curvature of director integral lines; for a unit vector, $(\mathbf n\cdot\nabla)\mathbf n=-\mathbf n\times(\nabla\times\mathbf n)$, giving cost $K_3|\mathbf n\times\nabla\times\mathbf n|^2/2$. These terms respect the head-tail symmetry of a <nematic liquid crystal>. Positive elastic constants oppose these distortions.
For the one-dimensional reduction assume lateral uniformity, so $\theta$ and $\phi$ depend only on $z$. A prime below means $d/dz$. Directly differentiating the <nematic director> gives
$$
\nabla\cdot\mathbf n=\cos\theta\,\theta',\qquad \nabla\times\mathbf n=(-n_y',n_x',0),\qquad \mathbf n\cdot\nabla\times\mathbf n=-\cos^2\theta\,\phi'.
$$
The squared <curl> is $\sin^2\theta\,\theta'^2+\cos^2\theta\,\phi'^2$. Subtracting the squared twist invariant gives
$$
|\mathbf n\times\nabla\times\mathbf n|^2=\sin^2\theta\,\theta'^2+\sin^2\theta\cos^2\theta\,\phi'^2.
$$
Consequently the <one-dimensional twisted nematic energy> has elastic coefficients
$$
\boxed{f(\theta)=K_1\cos^2\theta+K_3\sin^2\theta,\qquad g(\theta)=K_2\cos^4\theta+K_3\sin^2\theta\cos^2\theta.}
$$
A normal <electric field> has $\mathbf E\cdot\mathbf n=E_0\sin\theta$. There is a genuine factor-of-two inconsistency in the source: its initial functional places the electric term inside the overall factor $1/2$, whereas its subsequent density and threshold use the full electric term. For the requested density and threshold, use the usual orientational coupling $-\epsilon E_0^2\sin^2\theta/(8\pi)$. Here $\epsilon$ is the <dielectric anisotropy of a nematic>, in the printed Gaussian-unit convention. If the first functional is instead read literally, this coupling is halved and the resulting threshold voltage is multiplied by $\sqrt2$.
Put $c=\epsilon E_0^2/(8\pi)$ and $\mathcal L=f\theta'^2/2+g\phi'^2/2-c\sin^2\theta$. Apply the <Euler-Lagrange equations for two fields>. Variation in $\phi$, which does not appear explicitly, gives
$$
\boxed{(g(\theta)\phi')'=0.}
$$
Variation in $\theta$ gives $(f\theta')'-f_\theta\theta'^2/2-g_\theta\phi'^2/2+2c\sin\theta\cos\theta=0$, or
$$
\boxed{f\theta''+\frac12f_\theta\theta'^2-\frac12g_\theta\phi'^2+\frac{\epsilon E_0^2}{4\pi}\sin\theta\cos\theta=0.}
$$
At $\theta=0$, $f=K_1$, $g=K_2$, and $f_\theta=g_\theta=0$. With zero field, $\theta'=0$ and $\phi'=\pi/(2L)$ satisfy both equations and the imposed azimuthal endpoints. Thus the quarter-turn planar texture is a stationary solution. Stationarity alone is not a claim of stability for arbitrary elastic constants.
To find the <Fréedericksz transition>, impose the additional strong planar anchoring $\theta(0)=\theta(L)=0$. Write $\alpha=\pi/(2L)$. The small-angle expansions are
$$
f=K_1+(K_3-K_1)\theta^2+O(\theta^4),\qquad g=K_2+(K_3-2K_2)\theta^2+O(\theta^4).
$$
The conserved $g\phi'$ and fixed total twist imply $\phi'=\alpha+O(\theta^2)$ for a pure tilt perturbation. Linearizing the tilt <Euler-Lagrange equation> therefore gives
$$
K_1\theta''+\left[\frac{\epsilon E_0^2}{4\pi}-(K_3-2K_2)\alpha^2\right]\theta=0.
$$
Substitute $\theta=\theta_0\sin(\pi z/L)$. A nonzero neutral amplitude requires
$$
\frac{\epsilon E_c^2}{4\pi}=\frac{\pi^2}{L^2}\left[K_1+\frac{K_3-2K_2}{4}\right].
$$
Since $V_c=E_cL$, the <quarter-turn nematic instability threshold> is
$$
\boxed{V_c=2\pi^{3/2}\sqrt{K_1+\frac14(K_3-2K_2)}\,\epsilon^{-1/2}.}
$$
This is the first zero-endpoint mode: higher sine modes replace $K_1$ by $n^2K_1$ and cost more elastic energy. The quadratic energy change is $\frac12\int[K_1\theta'^2+\{(K_3-2K_2)\alpha^2-\epsilon E_0^2/(4\pi)\}\theta^2]dz$, so the first mode has negative stiffness above this voltage. The result assumes $\epsilon>0$ and $K_1+(K_3-2K_2)/4>0$. If the latter is negative the planar state is already unstable at zero field; for negative <dielectric anisotropy of a nematic>, a normal field opposes this tilt. The literal initial electric prefactor would give $V_{c,\mathrm{literal}}=\sqrt2V_c$.
For reflective operation of the <twisted nematic field effect>, orient each <polarizer>'s transmission axis along the adjacent planar <nematic director>. Light entering through the top <polarizer> initially has $y$ <linear polarization>. At low voltage, <adiabatic optical following in a twisted nematic> rotates it through a quarter turn to $x$ as it travels downward, so it passes the bottom <polarizer> and reaches the mirror. On returning, it follows the reverse twist back to $y$ and exits the top <polarizer>: the pixel appears bright. At sufficiently high voltage, positive <dielectric anisotropy of a nematic> aligns most of the director with the normal field. The bulk no longer produces the quarter-turn optical rotation, so the crossed bottom <polarizer> blocks the incident light and the pixel appears dark. Elastic relaxation restores the bright twisted state after voltage is removed. The adiabatic requirement is more precisely $2\pi|\Delta n|/\lambda\gg|\phi'|$, involving <birefringence>, not thickness alone. \b[Voltage switches director orientation and hence the reflected brightness; it does not require the liquid crystal to emit light.]
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