Commit 045aa20a by Harry Fuchs

### 2019-06-04

parent f26338a1
Pipeline #2371 passed with stage
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 ... ... @@ -3095,7 +3095,7 @@  $$\frac{2}{\pi} \operatorname{arctan}(\lVert y \rVert)y \reflectbox{\rotatebox[origin=c]{0}{\mapsto}} y \frac{2}{\pi} \operatorname{arctan}(\lVert y \rVert)y \mathrel{ \reflectbox{\rotatebox[origin=c]{0}{ \mapsto }} } y$$ ** Satz ... ... @@ -3130,7 +3130,195 @@ Sei $H\colon N\times[0,1]\to M$ eine Homotopie zwischen $i_1$ und $i_2$ $$0&=& \int_{N\times [0,1]} H^*(\diffd \omega) \\&=& \int_{N\times [0,1]} \intd (H^*\omega) \\&\overset{Stokes}=& \int_{\partial (N\times [0,1])} H^* \omega \\&\overset{\text{Stokes}}=& \int_{\partial (N\times [0,1])} H^* \omega \\&=& \int_N H^*_1 \omega - \int_N H^*_0 \omega \\&=& \int_{i_1(N)}\omega - \int_{i_0(N)} \omega$$ %2019-06-04 * Gaußscher Integralsatz $M\subset \mathbb R^3$ beschränktes Gebiet, $\partial M\subset \mathbb R^3$ glatt orientiert durch $\nu\colon \partial M \to \mathbb R^3$ Normalenfeld $$L &\colon& M\to \mathbb R^3 \text{ glattes Vektorfeld } \\ L&=&CL_x, Ly, Lz$$ $$\int_{\partial M}\langle L, \nu\rangle \intd \underbrace{S}_{\text{ Flächenelement } } = \int_M \operatorname{div} L \intd x\intd y\intd z$$ $$\operatorname{div} L = \frac{\partial L_x}{\partial x} + \frac{\partial L_y}{\partial y} + \frac{\partial L_z}{\partial z}$$ Wenn $\partial M$ parametrisiert ist: $$x &=& x(u,v) \\ y &=& y(u,v) \\ z &=& z(u,v)$$ $$\int_{\partial M} f(x,y,z) \intd S := \int_v f(x(u,v), y(u,v), z(u,v)) \cdot \sqrt{\det G(u,v)} \intd u \intd v$$ $$G(u,v) = (D_{(u,v)}\psi )^T D_{(u,v)} \psi$$ die Gram-Matrix der Koordinatenbasis ($\psi\colon U \to \partial M$, $(u,v)\mapsto (x(u,v)\ldots )$) $$= \int f(x(u,v), y(u,v), z(u,v) ) \lVert e_u \times e_v \rVert \intd u \intd v$$ Stokes: $$\int_{\partial M} \omega = \int_M \intd \omega$$ wollen: $$\diffd \omega = \operatorname{div} L \intd x \wedge \diffd y \wedge \diffd z$$ $$\omega = L_x \intd y\wedge \diffd z + L_y \intd z\wedge \diffd z + L_z \diffd x \wedge \diffd y$$ $$\diffd \omega &=& \frac{\partial L_x}{\partial x} \intd x\wedge \diffd y \wedge \diffd z + \frac{\partial L_y}{\partial y} \intd y\wedge \diffd z \wedge \diffd x + \frac{\partial L_z}{\partial z} \intd z\wedge \diffd x \wedge \diffd y + 0 + \ldots + 0 \\&=& (\operatorname{div}L) \intd x \wedge \diffd y \wedge \diffd z$$ $$\int_{partial M} \omega = \int_{\partial M} L_x \intd y\wedge \diffd z + L_y \intd z\wedge \diffd x + L_z \intd x\wedge \diffd y \\ &\overset{?}=& \int_{\partial M}\langle L, \nu \rangle\intd S$$ $$\\&& \int_{\partial M} L_x \intd y\wedge \diffd z \\&=& \int_U L_x (x(u,v), y(u,v), z(u,v) ) \left[\frac{\partial y}{\partial u} \intd u + \frac{\partial y}{\partial v} \intd v \right]\wedge \wedge \left(\frac{\partial z}{\partial u} \intd u + \frac{\partial z}{\partial v}\intd v\right) \\&=& \int_U L_x(u,v) \underbrace{ \left[ \frac{\partial y}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial y}{\partial v} \frac{\partial z}{\partial u} \right] }_{(e_u\times e_v)_x} \intd u\wedge \intd v \\&=& \int_U L_x (u,v) \frac{(e_u\times e_v)_x}{\underbrace{ \lVert e_u \times e_v \rVert}_{\nu_x} } \lVert e_u \times e_v \rVert \intd u \wedge \diffd v \\&=& \int_{\partial M} L_x \nu_x \intd S$$ $M= \mathbb R^2 \setminus \{ 0 \}$ $$A = A_x \intd x + A_y \intd y \in \Omega^1 (M)$$ $$F &:=& \diffd A \\&=& \frac{\partial Ay}{\partial x} \intd x \wedge \intd y - \frac{\partial Ax}{\partial y} \intd x \wedge \intd y \\&=& \left[ \frac{\partial Ay}{\partial x} -\frac{\partial A x}{\partial y} \right] \intd x \wedge \diffd y$$ Wenn $F=0$ ($\Rightarrow A$ geschlossen) Frage: - Ist $A$ exakt? - Äquivalente Fragestellung: $$H^1(M) &=& \{\text{geschlossene 1-Form }\} / \{\text{ exakt 1-Form } \} \\&=& Z' / B' \\&=& 0?$$ $$H^1(M) = 0 \Leftrightarrow \text{jede geschlossene 1-Form ist exakt}$$ %TOOD Bildchen (2) $\Rightarrow H^1(\mathbb R^2 \setminus \{0\}) = H^1(S^1) = \mathbb R$ ** Behauptung $\forall r> 0$ gilt: $$\int_{\underbrace{\partial B(0,r)}_{r \cdot S^1} } A = \int_{\partial B(0,1)} A$$ %TODO Bilchen (3) $$N := \{ x\in \mathbb R^2 \mathrel| 1\leqslant \lVert x \rVert \leqslant r\}\quad (\text{bzw. } r\leqslant \lVert x \rVert \leqslant 1)$$ $$\Rightarrow \partial N = (r \cdot S^1) \cup (-S^1)$$ Stokes: $$0 \overset{\diffd A = 0}= \int_{N} \intd A = \int_{r\cdot S'} A - \int_{S'} A$$ Wenn $A= \diffd f$ : $\int_{S'} A = \int_{S'} \intd f \overset{\text{Stokes} }= \int_{\partial S'} f = \int_{\emptyset} f = 0$ $\Rightarrow \int_{S^1} A \neq 0$ $\Rightarrow A$ nicht exakt. (z.B. $A = \frac{x\intd y - y\intd x}{x^2 + y^2}$) ist geschlossen, aber nicht exakt $$T^* \mathbb R^n \cong \mathbb R^n \times \mathbb R^n \ni (\underbrace{q^1, \ldots, q^n}_{\text{in } \mathbb R^n}, \underbrace{p^1, \ldots, p^n}_{\text{in } T_q^* \mathbb R^n})$$ $$\omega = \sum_{i=1}^n \diffd q^i \wedge \diffd p^i \in \Omega^2 (T^* \mathbb R^n)$$ 1) $\diffd \omega = 0$ 2) $\omega (\underbrace{ v}_{\equiv (a,p)\in T^* R^n}) \in \bigwedge ^2 T^*_v \mathbb R^n = \{ \alpha \mathrel| \alpha \colon T_v \mathbb R^n \times T_v \mathbb R^n \to \mathbb R \text{ bilinear, schiefsymetrisch }\}$ $$\frac{\partial}{\partial q^1}, \ldots, \frac{\partial}{\partial q^n}, \frac{\partial}{\partial p^1}, \ldots, \frac{\partial}{\partial p^n}$$ ist eine Basis von $T_v \mathbb R^n$. Wie sieht die Matrix von $\omega$ in dieser Basis aus? $$\omega &=& \sum_{i=1}^n(e^q_i)\wedge(e^p_i)^* \\ \omega(e_i^q, e_j^q) &=& 0 = \omega(e_i^p, e_j^p) \\ \omega(e_i^q, e_j^p) &=& \partial_{ij}$$ $\Rightarrow \omega$ ist nicht ausgeartet (an jedem Punkt!) $\omega$ nicht ausgeartet $\Rightarrow$ definiert an jedem Punkt einen Isomorphismus $$\hat \omega(v) \colon T_v^* \mathbb R^n \to T_v\mathbb R^n$$ $$\hat \omega^{-1}(v) \colon T_v\mathbb R^n &\to& T^*_v \mathbb R^n \\ \chi &\mapsto& (\eta \mapsto \omega(v)(\chi , \eta) ) \\ \frac{}{q^i} e^q_i &\mapsto& (\eta \mapsto \omega(v) (e_q^i, \eta)) = \left(e_{i}^p\right)^* = \diffd p^i \\ e_i^p &\mapsto& - \diffd q^1$$ $$\hat \omega\colon \Omega^1(T^* \mathbb R^n) &\to& \Gamma(T^* \mathbb R^n) \\ \reflectbox{\rotatebox[origin=c]{90}{\in}} && \\ \alpha &\mapsto& (v\mapsto \hat \omega (\alpha(v)) )$$ Für jede Funktion $H \rightsquigarrow X_H := \hat \omega (\diffd H) -$ Vektorfeld auf $T^* \mathbb R^n$ ** Behauptung Die Differentialgleichungen für den Fluss von $X_H$ $$\dot q_i &=& \frac{\partial H}{\partial p^i} \\ \dot p_i &=& - \frac{\partial H}{\partial q^i}$$
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