q4-Orientiation-Equivalence.md 14.2 KB
 Felix Hilsky committed Jan 13, 2022 1 2 3 4 5 6 7 # Orientability and Energy Minimization in Liquid Crystal Models ## Overview - *uniaxial nematic liquid crystals* werden modelliert - *Oseen-Frank*: unit vector field $n$ → nicht als $ℝP^2$ gesehen, sondern mit $n ≠ -n$ (ignoriert Symmetrie) - wird als Standardsichtweise dargestellt  Felix Hilsky committed Jan 19, 2022 8 - *Landau-de Gennes*: $Q = s(n ⊗ n - \tfrac13\operatorname{Id})$  Felix Hilsky committed Jan 13, 2022 9 10 11 12 13 14 - Resultat: - Theorien sind gleich für einfach zusammenhängende Urbilder und $W^{1,2}$ - Unterschiede in anderen Fällen sind beschrieben: - for simple energy functional, holes, various boundary conditions, difference cases are characterised (i.e. $⇔$) ## de Gennes  Felix Hilsky committed Jan 18, 2022 15   Felix Hilsky committed Jan 13, 2022 16 17 - closer to physics reality - $Q$-tensors are generally more [complex](https://arxiv.org/pdf/1409.3542.pdf), in our case just of the simple form („constrained”)  Felix Hilsky committed Jan 15, 2022 18 19 20  ### Derivation of $Q$-Tensors (from chapter 1 Introduction)  Felix Hilsky committed Jan 18, 2022 21   Felix Hilsky committed Jan 15, 2022 22 23 24 - each point has preferred direction but can have any direction → probability measure $μ(x, ·) : ℒ(𝕊^2) → [0,1]$ ($ℒ$ being all Lebesque sets) modelling very small area around $x$ - symmetry modeled as $μ(x, A) = μ(x, -A) ⇒ ⟨p⟩ = ∫_{𝕊^2} p dμ(p) = 0$ (first moment or average) - Tensor of second moments: $M_{ij} = ∫_{𝕊^2} p_ip_j dμ(p)$ ($i, j = 1, 2, 3$)  Felix Hilsky committed Jan 18, 2022 25  - $M = M^T$, $\operatorname{Tr} M = Σ_{i=1}^3 ∫_{𝕊^2} p_i^2 dμ(p) = ∫_{𝕊^2} 1 dμ(p) = 1$ ($μ$ is probability measure)  Felix Hilsky committed Jan 15, 2022 26 27  - $e · M e = ∫_{𝕊^2} (e · p)^2 dμ(p) = ⟨\cos^2(θ)⟩$ ($θ$ = angle between $p$ and $e$) for $|e| =1$) - if $μ$ is isotropic (no preferred direction): $μ_0$ with $dμ_0(p) = (4π)^{-1}dA$ with  Felix Hilsky committed Jan 19, 2022 28  $M_0 = (4π)^{-1} ∫_{𝕊^2} p ⊗ p dA = \frac 13 \operatorname{Id}$  Felix Hilsky committed Jan 15, 2022 29  - $∫_{𝕊^2} p_1 p_2 dμ(p) = 0$ since what happens on one half-sphere is countered by the other half-sphere ($p_2$, $p_3$ are same but $p_1$ is of opposite sign)  Felix Hilsky committed Jan 18, 2022 30  - $∫_{𝕊^2} p_i^2 dμ(p)$ are equal ($i=1,2,3$) and in sum $= \operatorname{Tr} M_0 = 1$  Felix Hilsky committed Jan 19, 2022 31 32 - Def. *de Gennes order-parameter tensor $Q$*: difference of second moment tensor to isotropic case $Q = M - M_0 = ∫_{𝕊^2}(p ⊗ p - \frac13 \operatorname{Id}) dμ(p)$ - $Q$ symmetric, $\operatorname{Q} = 0$ $⇒$ (spectral theorem) $Q = λ_1 ê_1 ⊗ ê_2 + λ_2 ê_2 ⊗ ê_2 - (λ_1 + λ_2)ê_3 ⊗ ê_3$ ($ê_{1,2,3}$ orthonormal eigenvector basis, $λ_1, λ_2$ eigenvectors)  Felix Hilsky committed Jan 15, 2022 33 34 35 36 37  - Eigenvectors for different non-zero eigenvalues are orthogonal: $(λ_1e_1) · (λ_2^{-1} e_2) = (Q e_1)^T (Q^{-1}e_2) = e_1 · Q^T Q^{-1} e_2 = e_1 · e_2 ⇒ e_1 ⊥ e_2$ - two eigenvalues equal and non-zero: *uniaxial* (considered here) - otherwise *biaxial* (more complicated) - Constraint on $s$: - $s$ defined by limiting process $L → 0$ and the constants $a, b, c$  Felix Hilsky committed Jan 19, 2022 38  - To make spectral representation of $Q$ fit to $Q = s(n ⊗ n - \tfrac13 \operatorname{Id})$, we need $s = -3 λ_1 = -3 λ_2, n = ê_3$. With $s$ depending on $a, b, c$ (see formula (3) on page 3 (=495)), $Q$ is further reduced to the choice of $n$. That is what we assumed as our model at the beginning.  Felix Hilsky committed Jan 15, 2022 39 40 41 42 43 44  - but also $Q$ has an original definition via $μ$. Together: math Qn · n = n^T s (n n^T - \frac 13 \operatorname{Id}) n = s (n^T n n^T n - \frac13 n^T \operatorname{Id}n) = s(1 · 1 - \frac13 · 1) \frac23 s  Felix Hilsky committed Jan 18, 2022 45 46   math  Felix Hilsky committed Jan 15, 2022 47 48 49 50 51 52 53  = ∫_{𝕊^2} n^T p p^T n - \frac13 n^T n dμ(p) = ∫_{𝕊^2} (n · p)^2 - \frac 13 dμ(p) = ⟨\cos^2 θ - \frac13⟩ ⇒ s = \frac32 ⟨\cos^2 θ - \frac13⟩  - $⇒ -\frac12 \leq s \leq 1$: $s$ says how much the molecules agree, are „in order“ (max: $s = 1$: perfectly ordered $p∥n$; $s = -\frac12$: $p ⊥ n$; $s=0$: $Q = 0$, isotropic) - $⇒$ s is *scalar order parameter associated to the tensor $Q$*  Felix Hilsky committed Jan 13, 2022 54 55  - usual assumption: $s$ is constant $⇒$ space of $Q$-Tensors "is" $ℝP^2$  Felix Hilsky committed Jan 15, 2022 56 57 58 ### Simpliest energy functional (p. 495 = 3) math  Felix Hilsky committed Jan 18, 2022 59 ℱ_{\text{LG}}[Q] = ∫_{Ω} \left(\sum_{i,j,k = 1}^3 \frac L2 Q_{ij,k} Q_{ij,k} - \frac a2 \operatorname{Tr} Q^2 - \frac b3 \operatorname{Tr} Q^3 + \frac c4 (\operatorname{Tr} Q^2)^2 \right) dx  Felix Hilsky committed Jan 15, 2022 60 61 62  $a, b, c$ constants, $L$ „*elastic constant*”. Physics is interested in $L → 0$  Felix Hilsky committed Jan 13, 2022 63 64 ## Oseen-Frank - simpler, but sometimes wrong  Felix Hilsky committed Jan 15, 2022 65 66 67 68 69 70 71 72 73 74 - (here:) with orientation ($n ∈ 𝕊^2$, not $n ∈ ℝP^2$) - problem: *„fake defects”*: „non-orientable line field” (places where vector field has to be rough but would be OK if we took away the orientation ### Simpliest energy functional minimizers of $ℱ_{\text{LG}}$ for $L → 0$ are "suitably approximated" by minimizers of math ℱ_{\text{OF}}[Q] = ∫_{Ω} \sum_{i, j, k = 1}^3 Q_{ij,k} Q_{ij,k} dx  *if* $Q ∈ W^{1,2}$ with $Q$ uniaxial almost everywhere, i.e. math  Felix Hilsky committed Jan 19, 2022 75 Q = s(n ⊗ n - \tfrac13 \operatorname{Id}) \quad (s = -3 λ_1 = -3 λ_2, n = ê_3)  Felix Hilsky committed Jan 15, 2022 76   Felix Hilsky committed Jan 13, 2022 77   Felix Hilsky committed Jan 17, 2022 78 79 ### Notation - $P : 𝕊^2 → 𝒬$ removes orientation. Orientable = in the image of $P$. Same for $Q$ only defined on $∂Ω$  Felix Hilsky committed Jan 19, 2022 80 81 - $𝒬 := \{Q = s(n ⊗ n - \frac13 \operatorname{ID}) | n ∈ 𝕊^2\}$ - $𝒬_2 := \{Q = s((n_1, n_2, 0) ⊗ (n_1, n_2, 0) - \frac13 \operatorname{ID}) | n = (n_1, n_2, 0) ∈ 𝕊^2\}$  Felix Hilsky committed Jan 17, 2022 82 83 - $C^k$ and Lipschitz domains are defined as having a graph as the boundary (as usual) - $W^{1,p}$ defined via embedding  Felix Hilsky committed Jan 18, 2022 84 - $W^{1,p}_{φ} = W^{1,p}$ with $φ$ on the boundary (use $\operatorname{Tr}$-Operator to define boundary value)  Felix Hilsky committed Jan 15, 2022 85 - $P : 𝕊^2 → 𝒬$ removes orientation. Orientable = in the image of $P$. Same for $Q$ only defined on $∂Ω$  Felix Hilsky committed Jan 17, 2022 86 87  - for $Q ∈ W^{1,p}(Ω, 𝒬)$ $ℒ^d$-almost everywhere - for $Q ∈ W^{1-\frac1p, p}(∂Ω, 𝒬)$ orientable to $n ∈ W^{1-\frac1p, p}(Ω, 𝕊^2)$ $ℋ^{d-1}$-almost everywhere  Felix Hilsky committed Jan 18, 2022 88 89 - $v_{,k}$ is the variable $v$ differentiated in the direction $k$. $Q_{ij,k}$ is the $i-j$'th component of the matrix field $Q$ differentiated in the direction $k$. This is not mentioned anywhere!! - Indeces which appear twice, are summed over, without mentioning it once!! Even when both are lower indeces.  Felix Hilsky committed Jan 19, 2022 90 91 92 93 94 95 96 - $b:𝒬 → ℝP^2$ (and $b:𝒬_2 → ℝP^1$) is isometry between $𝒬$ and $RP^2$ by math b(s(n ⊗ n - \tfrac13 \operatorname{Id})) = \{n, -n\} ∈ ℝP^2  $⇒ 𝒬$ is Riemannian manifold (p. 12 (=504) section "Orientability Issues" With this identification, $P$ is a covering map (details p. 12)  Felix Hilsky committed Jan 14, 2022 97 98 99 100 101 - in chapter 3.1 firstly for continous $Q$ since that's standard from topology - *Theorem 1*: $𝒬_2$ (one-dim), orientable on boundary of holes, $Q$ continous $⇒$ orientable (long proof with a lot of fiddling!) - chapter 3.2 Theorem 2: $Ω$ simply connected, $Q ∈ W^{1, p}$, $p \geq 2$ (!) $⇒$ orientable with Sobolev-Seminorm estimate. Counterexample for $p < 2$ - chapter 4: different cases of $Q$ orientable on boundary $⇒$/$⇔$/$⇐$ on $Ω$. Sometimes $Q: → 𝒬_2$, sometimes $Q$ continous (on boundary)  Felix Hilsky committed Jan 17, 2022 102 103 104 105 106 107 108 109 ### Chapter 2 Propositions (choose better subsection title!) *Proposition 2*: $Q ∈ W^{1,p}(Ω, 𝒬)$ ($1 \leq p \leq ∞$) can have only two orientations. *Proof*: In one dimension weakly differentiable means that $f(x) = f(a) + ∫_a^x f'(y) dy$. (Which is in turn equivalent to absolutely continous.)  Felix Hilsky committed Jan 18, 2022 110 In several dimensions this is [true in each direction](https://math.aalto.fi/~jkkinnun/files/sobolev_spaces.pdf#section.1.14) but only for almost all $x_2$, $x_3$.  Felix Hilsky committed Jan 17, 2022 111 112 113 114 115 116 117 118 119 120 121 122 123 So: $n$, $m$ have representatives such that for almost everywhere $x_2$, $x_3$: $n$, $m=τn$ are absolutely continous, so also $τ = τ·1 = τn·n = m·n$. Absolute continuity implies continuity, so $τ$ has a representative that is for those almost all $x_2, x_3$, constant. Now use this in Fubini on a small ball in $Ω$ around $y$: math ∫_{B_{ε}(y)} ∂_1 τ(x) φ(x) dx = ∫_{B_{ε}(y)} τ(x) ∂_1 φ(x) dx = ∫_{B_{ε}(y)} ∂_1(τφ)(x) dx = 0  (first =: Fubini, second =: $τ$ is constant along $x_1$ for almost all $x_2, x_3$, third =: first integrated over $x_1$ direction, this is $± φ(x_{1, \max}, x_2, x_3) - (± φ(x_{1, \min}, x_2, x_3)) = 0 - 0 = 0$, then integrated over $x_2$, $x_3$, it stays 0) Hence $∇τ = 0$ (since $x_2$, $x_3$ are analogous) weakly but this means that $τ$ is constant on $B_{ε}(y)$ for every $y ∈ Ω$ and therefore constant on $Ω$. Done. ---  Felix Hilsky committed Jan 18, 2022 124 Trace operator is introduced by Evans but with continuity estimate for $\operatorname{Tr} : W^{1,p} → L^p$. Here we need the stronger statement that $\operatorname{Tr}$ maps into the [Sobolev-Slobodeckij space](https://en.wikipedia.org/wiki/Trace_operator#Characterization_using_Sobolev%E2%80%93Slobodeckij_spaces) $W^{1-1/p, p}$.  Felix Hilsky committed Jan 17, 2022 125   Felix Hilsky committed Jan 18, 2022 126 *Proposition 3*: If $Q = P(n)$ (orientable) on $Ω$, then $\operatorname{Tr} Q P(\operatorname{Tr} n)$ ($Q$ on $∂Ω$ is orientable).  Felix Hilsky committed Jan 17, 2022 127   Felix Hilsky committed Jan 18, 2022 128 129 130 131 *Proof sketch*: Approximate $n$ with differentable functions on and around $Ω$, then use continuity of $\operatorname{Tr}$ and $P$ (show it!) in $W^{1,p}(Ω)$ and $L^p(∂Ω)$ to show $P(\operatorname{Tr} Q) = \operatorname{Tr} P(Q)$. Show with integral approximation property that $\operatorname{Tr} n$ lies in $𝕊^2$. ---  Felix Hilsky committed Jan 19, 2022 132 For Proposition 4: *Lemma 1*: (regularity is preserved) ($Ω$ bounded)  Felix Hilsky committed Jan 18, 2022 133 134 135 * $n ∈ W^{1,p}(Ω, 𝕊^2) ⇒ Q = P(n) ∈ W^{1,p}(Ω, 𝒬)$ * $Q ∈ W^{1,p}(Ω, 𝒬) ∧ Q = P(n)$ and $n$ is measurable and continous along almost every line parallel to the coordinate axes. Then $n ∈ W^{1,p}(Ω, 𝒬)$ and $\sum_{j=1}^3 Q_{ij,k} n_j = sn_{i,k}$.  Felix Hilsky committed Jan 18, 2022 136 137 138 139 140 *Proof of Lemma 1*: Since $Ω$ is bounded, $W^{1,p}(Ω) ⊆ W^{1,1}(Ω)$, so from $n ∈ W^{1,p}$ we get $n ∈ W^{1,1}$. Also $|n(x)|=1$ for all $x$, so $n ∈ L^{∞}$. For $g,h∈W^{1,1} ∩ L^{∞}$, we have $gh ∈ W^{1,1} ∩ L^{∞}$. (Here used with $g = n_i$, $h=n_{j,k}$) With $Q = P(n)$ we get $Q ∈ W^{1,1} ∩ L^{∞}$ and also $∇Q ∈ L^p$ since components of $∇Q$ are products (and sums) of $n_i$ (bounded by $1$) and $n_{j,k} ∈ L^p$. Formula is straight calculation (!! $n_j n_j$ means $\sum_{j=1}^3 n_jn_j$!!). For converse ($Q ∈ W^{1,p}$ given), using the formula to define a $n_{i, k}$ candidate doesn't help since when trying to use this, we need the derivative of $n_j$ again which we want to show that it exists:  Felix Hilsky committed Jan 18, 2022 141 math  Felix Hilsky committed Jan 19, 2022 142 ñ_{ik} := \sum_{j} Q_{ij,k} n_j  Felix Hilsky committed Jan 18, 2022 143   Felix Hilsky committed Jan 18, 2022 144 math  Felix Hilsky committed Jan 19, 2022 145 \text{For } φ ∈ C^{∞}_c: ∫_{Ω} ñ_{ik} φ  Felix Hilsky committed Jan 18, 2022 146 147 148 149 150 151 152 = ∫ \sum_j s^{-1} Q_{ij,k} n_j φ = ∫ - \sum_j Q_{ij} \underbrace{(n_j φ)_{,k}}_{\text{exists?!??}}  That's why the authors take an ugly route via normal differentiability almost everywhere (using the assumption about continuity line-wise). Aditionally the continuity is necessary because otherwise the direction of $n$ can swap everywhere funnily. For the theorems about differentiability almost everywhere parallel to the axis, see [ACL characterization from Nikodym](https://math.aalto.fi/~jkkinnun/files/sobolev_spaces.pdf#page=40).  Felix Hilsky committed Jan 19, 2022 153 154 155 156 157 158 The set of lines parallel to axes is a Lebesgue-zero set (assumption) and the set of points where $Q$ is not differentable in the axes directions is a zero set (Nikodym). Hence for almost all $x ∈ Ω$, $n$ is continous along the line $(x + ℝe_k) ∩ Ω$ and $Q$ is differentable at $x$ in the direction $e_k$. Then calculate a lot (start with difference quotient $Δ_t Q_{ij,k}$, multiply with $\frac12(n_j(x+te_k)+n_j(x))$ and sum over $j$) and reach math s · \lim_{t → 0} \frac{n_i (x+te_k) - n_i(x)}{t} = Q_{ij, k}(x) n_j(x)  $∇n ∈ L^p$ since $∇Q ∈ L^p$ and $n ∈ L^{∞}$.  Felix Hilsky committed Jan 18, 2022 159 160 161 --- *Proposition 4*: Orientability is preserved by weak convergence.  Felix Hilsky committed Jan 17, 2022 162   Felix Hilsky committed Jan 19, 2022 163 164 165 166 167 168 169 170 171 172 *Proof of Proposition 4*: [Uniform boundedness principle](https://en.wikipedia.org/wiki/Uniform_boundedness_principle#Theorem) implies that from $Q^{(k)} ⇀ Q$ follows $\lVert Q^{(k)} \rVert$ is bounded and with $|n(x)|$ bounded by $1$ and the formula $\rVert n^{(k)} \rVert_{W^{1,p}}$ is bounded. From boundedness we get a weakly convergent subsequence $n^{(k_l)} ⇀ n$, which also has $n^{(k_l)}(x) → n(x)$ almost everywhere, so $P(n) = Q$. --- *Lemma 2*: non-orientability is a stable property with respect to the $W^{1,p)(Ω, ℝ^9)$ norm: Let $Q ∈ W^{1,p}(Ω, 𝒬), 1 ≤ p ≤ ∞$ be non-orientable. Then there exists $ε > 0$, depending on $Q$, so that for all $Q̃ ∈ W^{1,p}(Ω, 𝒬) with $\lVert Q̃−Q \rVert_{W^{1,p}(M,ℝ^9)} < ε$the line field Q̃ is also non-orientable. *Proof of Lemma 2*: obvious, given Proposition 4  Felix Hilsky committed Jan 14, 2022 173 ## Energy functionals  Felix Hilsky committed Jan 19, 2022 174   Felix Hilsky committed Jan 14, 2022 175 176 - specific energy functionals are regarded (p. 11/503) because they were looked at before and Oseen-Frank was successful for them - conversion between energy functionals possible  Felix Hilsky committed Jan 15, 2022 177   Felix Hilsky committed Jan 19, 2022 178 179 180 181 182 183 184 185 186 187  ## Compatibility - Oseen-Frank and de Gennes are compatible if the oriented line field (= unit vector field) can be oriented (without changing regularity) - otherwise Oseen-Frank might miss a global minimizer because it is not orientable - Question if orientable calculable with integer programming problem (p. 4 = 496) (= (linear) optimization problem with only integer coefficiants) - In chapter 3, orienting a line field is translated into lifting$Q$from$ℝP^2$to covering space$𝕊^2$. - Classical algebraic topology tools only consider continous maps, but we have$W^{1,p}(Ω)$maps and want to preserve this regularity  Felix Hilsky committed Jan 15, 2022 188 ## Questions  Felix Hilsky committed Jan 17, 2022 189 190 191 192 193 194  ## Errors in the paper p.5 (=497) Proposition 1 (iii): sign of$\operatorname{det} Q$is wrong. Must be positive, since exactly two of the eigenvalues are negative p. 7 (=499) Proposition 2: it says$Q(x) ∈ W^{1,p}(Ω, 𝒬)$but$(x)$must be erased ($Q(x) ∈ 𝒬$) Also the statement of the proposition is trivial and not what the colloquial formulation says: it should be "... with$P(n) = P(m) = Q$and$n ≠ m$[as$W^{1,p}$-generalized functions, so on a non-0-Lebesgue set], we have$m = -n$almost everywhere in$Ω$. If$n ≠ m$almost everywhere, it follows straight away that$m = -n$almost everywhere since pointwise$n(x) = ± m(x)$ Felix Hilsky committed Jan 18, 2022 195 196  ~~p. 8 (=500) Proof Proposition 3: "Mollify$\overline n$o get$\overline n^{(j)} ∈ C^1(\overline{Ω}, ℝ^3)$should be$∈ C^1(B, ℝ^3)$. By mollifying we cannot get a bigger domain and it would be weird to first extend$n$to then ignore this extension when mollifying again.~~ ($\overline{Ω} ⊆ B!\$)