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The following is the well-known nonlinear differential equation for director's distribution at static equilibrium in liquid crystal displays(LCD). I want to obtain weak form of the given differential equation for FEM simulation.

$(\epsilon_{\parallel} - \epsilon_{\perp}) E^2 sin(u) cos(u) + (k_{3} - k_{1})sin(u)cos(u) (\frac{d u}{d z})^2 + ( k_{1} cos^2(u) + k_{3} sin^2(u) ) \frac{d^2 u}{dz^2} = 0$

where $\epsilon_{\perp}$, $\epsilon_{\parallel}$, $k_{1}$ and $k_{2}$ are constants and E is the input parameter that is to be set.

Can anyone tell what would be a correct weak form of the equation?

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First notice that $$ (k_3 - k_1)\sin(u)\cos(u)(u')^2 + (k_1\cos^2(u) + k_3\sin^2(u))u'' = ((k_1\cos^2(u) + k_3\sin^2(u))u')' + (k_1 - k_3)\sin(u)\cos(u)(u')^2. $$ I choose to rearrange our equation so that the elliptic part is nonnegative:

$$-((k_1\cos^2(u) + k_3\sin^2(u))u')' = (k_1 - k_3)\sin(u)\cos(u)(u')^2 + (\epsilon_{\parallel} - \epsilon_{\perp})E^2\sin(u)\cos(u)$$

For brevity, rename these functions $D(u)$ and $F(u)$ such that this equation reads $$-(D(u)u')' = F(u).$$ We can then multiply through by a test function and apply integration by parts to the divergence term:

$$ \begin{aligned} \int_a^b-(D(u)u')v \ dz &= \int_a^bF(u)v \ dz \\ \implies \int_a^b D(u)u'v' \ dz &= -\left[D(u)u'v\right]_a^b + \int_a^bF(u)v \ dz. \end{aligned} $$

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  • $\begingroup$ What's the elliptic part? I apologize if the question is too silly. $\endgroup$
    – Pu Zhang
    May 10, 2023 at 11:42
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    $\begingroup$ For a linear equation, the bilinear form $\int D \nabla u\cdot \nabla v \ dx$ where $D > 0$ is tyuupically termed "elliptic" $\endgroup$
    – whpowell96
    May 10, 2023 at 16:38

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