### read a little bit further

parent 33c73e7f
 ... ... @@ -83,10 +83,12 @@ Q = s(n \otimes n - \tfrac13 \operatorname{Id}) \quad (s = -3 λ_1 = -3 λ_2, n - $𝒬_2 := \{Q = s((n_1, n_2, 0) \otimes (n_1, n_2, 0) - \frac13 \operatorname{ID}) | n = (n_1, n_2, 0) ∈ 𝕊^2\}$ - $C^k$ and Lipschitz domains are defined as having a graph as the boundary (as usual) - $W^{1,p}$ defined via embedding - $W^{1,p}_{φ} = W^{1,p}$ with $φ$ on the boundary (use $\operatorname{tr}$-Operator to define boundary value) - $W^{1,p}_{φ} = W^{1,p}$ with $φ$ on the boundary (use $\operatorname{Tr}$-Operator to define boundary value) - $P : 𝕊^2 → 𝒬$ removes orientation. Orientable = in the image of $P$. Same for $Q$ only defined on $∂Ω$ - 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 - $v_{,k}$ 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. - 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 $p < 2$ ... ... @@ -114,11 +116,24 @@ Hence $∇τ = 0$ (since $x_2$, $x_3$ are analogous) weakly but this means --- 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) $W^{1-1/p, p}$. 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) $W^{1-1/p, p}$. *Proposition 3*: *Proposition 3*: If $Q = P(n)$ (orientable) on $Ω$, then $\operatorname{Tr} Q P(\operatorname{Tr} n)$ ($Q$ on $∂Ω$ is orientable). *Proof sketch*: *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 $𝕊^2$. --- *Proposition 4*: Orientability is preserved by weak convergence. For that: *Lemma 1*: (regularity is preserved) ($Ω$ bounded) * $n ∈ W^{1,p}(Ω, 𝕊^2) ⇒ Q = P(n) ∈ W^{1,p}(Ω, 𝒬)$ * $Q ∈ W^{1,p}(Ω, 𝒬) ∧ Q = P(n)$ and $n$ is measurable and continous along almost every line parallel to the coordinate axes. Then $n ∈ W^{1,p}(Ω, 𝒬)$ and $\sum_{j=1}^3 Q_{ij,k} n_j = sn_{i,k}$. *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 $gh ∈ W^{1,1} ∩ L^{∞}$. With math Q ∈  ## Energy functionals - specific energy functionals are regarded (p. 11/503) because they were looked at before and Oseen-Frank was successful for them ... ... @@ -136,3 +151,5 @@ p.5 (=497) Proposition 1 (iii): sign of $\operatorname{det} Q$ is wrong. Must 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 $n(x) = ± m(x)$ ~~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.~~ ($\overline{Ω} ⊆ B!$)
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