finish ana7
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\documentclass[uebung]{../../../lecture}
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\title{Analysis II: Übungsblatt 7}
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\author{Leon Burgard, Christian Merten}
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\begin{document}
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\punkte
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\begin{aufgabe}
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\begin{enumerate}[(a)]
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\item $f\colon R^2 \to \R$, $f(x,y) = e^{x} \cos(y) + \ln(1+y^2)$. Dann gilt
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\[
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\nabla f = \begin{pmatrix} e^{x} \cos(y) \\ -e^{x} \sin(y) + \frac{2y}{1+y^2} \end{pmatrix}
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.\]
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\item $f\colon R^2 \to R$ mit
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\[
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f(x,y) = \begin{cases}
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xy \frac{x^2 - y^2}{x^2 + y^2} & (x,y) \neq (0,0) \\
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0 & (x,y) = (0,0)
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\end{cases}
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.\]
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Beh.: $f$ überall zweimal partiell differenzierbar, aber
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\[
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\frac{\partial^2f(0,0)}{\partial x \partial y} \neq \frac{\partial^2f(0,0)}{\partial y \partial x}
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.\]
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\begin{proof}
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Für $(x,y) \neq (0,0)$ gilt
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\begin{salign*}
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\frac{\partial f}{\partial x} &= \frac{\partial}{\partial x}
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\left( \frac{x^{3}y - xy^{3}}{x^2 + y^2} \right) \\
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&= \frac{yx^{4} + 4x^2y^{3} - y^{5}}{(x^2 + y^2)^2} \\
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\frac{\partial f}{\partial y} &=
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\frac{\partial}{\partial y} \left( \frac{x^{3}y - xy^{3}}{x^2 + y^2} \right) \\
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&= \frac{x^{5} - 4y^2 x^{3} - y^{4}x}{(x^2+y^2)^2}
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.\end{salign*}
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Diese partiellen Ableitungen sind als Quotient von Polynomen mit $(x^2+ y^2)^2 \neq 0$
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wieder partiell differenzierbar.
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$f$ ist im Punkt $(0,0)$ partiell differenzierbar, denn
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\begin{align*}
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\frac{\partial f}{\partial x}(0,0)
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&= \lim_{h \to 0} \frac{f((0,0) + (h, 0)) - f(0,0)}{h}
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= \lim_{h \to 0} \frac{\overbrace{f(h,0)}^{= 0}}{h} = 0 \\
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\frac{\partial f}{\partial y}(0,0) &=
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\lim_{h \to 0} \frac{\overbrace{f(0,h)}^{= 0}}{h} = 0
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.\end{align*}
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Für die zweiten partiellen Ableitungen in $(0,0)$ gilt
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\begin{salign*}
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\frac{\partial^2f(0,0)}{\partial x \partial y}
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&= \lim_{h \to 0} \frac{\frac{\partial f}{\partial y}(h,0) - \frac{\partial f}{\partial y}(0)}{h}
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= \lim_{h \to 0} \frac{\frac{\partial f}{\partial y}(h,0)}{h}
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= \frac{h^{5}}{h h^{4} } = 1 \\
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\frac{\partial^2f(0,0)}{\partial y \partial x}
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&= \lim_{h \to 0} \frac{\frac{\partial f}{\partial x}(0,h) - \frac{\partial f}{\partial x}(0)}{h}
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= \lim_{h \to 0} \frac{\frac{\partial f}{\partial y}(0,h)}{h}
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= - \frac{h^{5}}{h h^{4} } = -1
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.\end{salign*}
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Also existieren die zweiten partiellen Ableitungen auf ganz $\R^2$, aber es gilt
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\[
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\frac{\partial^2 f(0,0)}{\partial x \partial y} \neq
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\frac{\partial^2 f(0,0)}{\partial y \partial x}
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.\]
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\end{proof}
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Beh.: Die partielle Ableitung $\frac{\partial f}{\partial x \partial y}$
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ist unstetig in $(0,0)$.
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\begin{proof}
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Es gilt für $(x,y) \neq (0,0)$:
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\begin{align*}
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\frac{\partial f}{\partial x \partial y}
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= \frac{x^{6} + 9x^{4}y^2 - 9x^2y^{4} - y^{6}}{(x^2 + y^2)^{3}}
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.\end{align*}
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Mit $(x,y)_n = \left( \frac{1}{n}, \frac{1}{n} \right) $ gilt
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$(x,y)_n \xrightarrow{n \to \infty} (0,0)$, aber
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\begin{align*}
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\frac{\partial f}{\partial x \partial y}(x,y)_n
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= \frac{\frac{1}{n^{6}} + \frac{9}{n^{6}} - \frac{9}{n^{6}} - \frac{1}{n^{6}}}{\frac{8}{n^{6}}} = 0 \neq 1 = \frac{\partial f}{\partial x \partial y} (0,0)
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.\end{align*}
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\end{proof}
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Der Satz von Schwarz für ein $x \in D$ gilt nur, wenn $f$ 2-mal stetig
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partiell differenzierbar ist in $x$, dies ist
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hier für $x = (0,0)$ nicht der Fall.
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}
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\begin{enumerate}[(a)]
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\item $f\colon R^2 \to R$ mit
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\[
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f(x,y) = \begin{cases}
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\frac{xy^2}{x^2 + y^{4}} & (x,y) \neq (0,0) \\
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0 & (x,y) = (0,0)
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\end{cases}
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.\] Beh.: $f$ ist im Punkt $(x,y) = (0,0)$ nicht total differenzierbar.
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\begin{proof}
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Mit dem Hinweis g.z.z., dass $f$ in $(0,0)$ nicht stetig ist. Mit
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$(x,y)_n = \left( \frac{1}{n^2}, \frac{1}{n} \right) $ gilt
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$(x,y)_n \xrightarrow{n \to \infty} (0,0)$ aber, es ist
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\[
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f\left( \frac{1}{n^2}, \frac{1}{n} \right) = \frac{\frac{1}{n^{4}}}{\frac{1}{n^{4}} + \frac{1}{n^{4}}} = \frac{1}{2} \neq 0 = f(0, 0)
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.\] Also $f$ unstetig in $(0,0)$.
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\end{proof}
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Beh.: $f$ besitzt Richtungsableitungen in alle Richtungen in $(x,y) = (0,0)$.
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\begin{proof}
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Sei $v \in \R^2$ mit $\Vert v \Vert_2 = 1$ beliebig. Dann sei
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$v = \begin{pmatrix} v_x \\ v_y \end{pmatrix}$. Dann gilt
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\begin{salign*}
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\frac{\partial f}{\partial v} (0, 0) = \lim_{t \searrow 0}
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\frac{f(tv) - f(0)}{t}
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= \lim_{t \searrow 0} \frac{f(tv)}{t}
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= \lim_{t \searrow 0} \frac{v_xv_y^2}{v_x^2 + \frac{1}{t^2}v_y^{4}} = \frac{v_y^2}{v_x}
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.\end{salign*}
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Also existieren alle Richtungsableitungen in $(x,y) = (0,0)$.
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\end{proof}
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\item $f\colon R^2 \to \R$ mit
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\[
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f(x,y) = \begin{cases}
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(x^2 + y^2) \sin\left( \frac{1}{\sqrt{x^2 + y^2} } \right) & (x,y) \neq (0,0) \\
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0 & (x,y) = (0,0)
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\end{cases}
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.\]
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Beh.: $D f(0,0) = \begin{pmatrix} 0 \\ 0 \end{pmatrix}^{T}$, also $f$
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im Punkt $(x,y) = (0,0)$ total differenzierbar.
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\begin{proof}
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Sei $h \in \R^2$ mit $h \neq 0$ beliebig. Dann gilt
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\begin{salign*}
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\left| \frac{f((0,0) + h) - f(0,0) - \begin{pmatrix} 0 & 0 \end{pmatrix} \cdot h }{\Vert h \Vert_2} \right|
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&= \left| \frac{f(h)}{\Vert h \Vert_2} \right| \\
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&= \frac{\Vert h \Vert_2^2 \sin\left( \frac{1}{\Vert h \Vert_2} \right) }{\Vert h \Vert_2} \\
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&= \left| \Vert h \Vert_2 \sin\left( \frac{1}{\Vert h \Vert_2} \right) \right| \\
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&\le \Vert h \Vert_2 \xrightarrow{h \to 0} 0
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.\end{salign*}
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Also folgt
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\begin{align*}
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\lim_{h \to 0} \frac{f((0,0) + h) - f(0,0) - \begin{pmatrix} 0 & 0 \end{pmatrix} h}{\Vert h \Vert_2} = 0
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.\end{align*}
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Also $f$ total differenzierbar in $(x,y) = (0,0)$ mit
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Differential $D f(0, 0) = \begin{pmatrix} 0 & 0 \end{pmatrix} $.
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\end{proof}
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Beh.: $\frac{\partial f}{\partial x}$ nicht stetig in $(x,y) = (0,0)$.
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\begin{proof}
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Mit der (a) folgt, dass $J_f(0,0) = D f(0,0)$, also insbesondere
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$\frac{\partial f}{\partial x} = 0$.
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Für $(x,y) \neq (0,0)$ folgt
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\begin{salign*}
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\frac{\partial f}{\partial x} &= 2x \sin\left( \frac{1}{\sqrt{x^2 + y^2} } \right)
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- \frac{x}{\sqrt{x^2 + y^2}} \cos\left( \frac{1}{\sqrt{x^2 + y^2} } \right)
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.\end{salign*}
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Mit $(x, y)_n = \left( \frac{1}{2\pi n}, 0 \right) $ gilt
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$x_n \xrightarrow{n \to \infty} (0,0)$, aber
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\begin{salign*}
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\frac{\partial f}{\partial x}\left( \frac{1}{2\pi n}, 0 \right)
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= \frac{1}{\pi n} \underbrace{\sin\left( 2 \pi n \right)}_{= 0} - \cos (2\pi n) = -1 \neq 0
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= \frac{\partial f}{\partial x}(0)
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.\end{salign*}
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\end{proof}
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}
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Sei $D \coloneqq \R^2 \setminus \{ (x_1, x_2)^{T} \mid x_2 \le 0 \text{ oder } x_1x_2 = k\pi + \frac{\pi}{2}, k \in \Z\} $, $f\colon D \to \R^2$ und $g \colon R^2 \to \R^2$ mit
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\[
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f(x_1, x_2) = \begin{pmatrix} x_1 \ln(x_2) \\ \tan(x_1, x_2) \end{pmatrix},
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\quad g(y_1, y_2) = \begin{pmatrix} y_1^2 \\ y_2^2 \end{pmatrix}
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.\] Weiter sei $h = g \circ f \colon D \to \R^2$.
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$D_h(x)$ ohne Kettenregel:
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\begin{salign*}
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h(x) = g(f(x)) = \begin{pmatrix} x_1^2 \ln^2(x_2) \\ \tan^2(x_1x_2) \end{pmatrix}
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.\end{salign*}
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Damit folgt direkt
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\begin{salign*}
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D_h(x) = \begin{pmatrix} \frac{\partial h_1}{\partial x_1} & \frac{\partial h_1}{\partial x_2} \\
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\frac{\partial h_2}{\partial x_1} & \frac{\partial h_2}{\partial x_2}
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\end{pmatrix}
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=
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\begin{pmatrix}
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2x_1 \ln^2(x_2) & 2 \frac{x_1^2}{x_2} \ln(x_2) \\
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\frac{2x_2 \tan(x_1x_2)}{\cos^2(x_1x_2)} & \frac{2x_1 \tan(x_1x_2)}{\cos^2(x_1x_2)}
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\end{pmatrix}
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.\end{salign*}
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Mit Kettenregel ist zunächst
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\begin{salign*}
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D_g(y) = \begin{pmatrix} 2y_1 & 0 \\ 0 & 2 y_2 \end{pmatrix},
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\quad
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D_f(x) = \begin{pmatrix} \ln(x_2) & \frac{x_1}{x_2} \\
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\frac{x_2}{\cos^2(x_1x_2)} & 2 \frac{x_1y_2}{\cos^2(x_1x_2)}\end{pmatrix}
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.\end{salign*}
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Damit folgt
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\begin{salign*}
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D_h(x) &= D_g(f(x)) D_f(x) \\
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&= \begin{pmatrix} 2 x_1 \ln(x_2) & 0 \\ 0 & 2 \tan(x_1 x_2) \end{pmatrix}
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\begin{pmatrix} \ln(x_2) & \frac{x_1}{x_2} \\
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\frac{x_2}{\cos^2(x_1x_2)} & 2 \frac{x_1y_2}{\cos^2(x_1x_2)}\end{pmatrix} \\
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&=
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\begin{pmatrix}
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2x_1 \ln^2(x_2) & 2 \frac{x_1^2}{x_2} \ln(x_2) \\
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\frac{2x_2 \tan(x_1x_2)}{\cos^2(x_1x_2)} & \frac{2x_1 \tan(x_1x_2)}{\cos^2(x_1x_2)}
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\end{pmatrix}
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.\end{salign*}
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Das Einsetzen von $x_0 = (1,e)^{T} \in D$ ist dem Lesenden als Aufgabe überlassen.
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\end{aufgabe}
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\begin{aufgabe}
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$f\colon \R \to \R^2$ mit
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\[
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f(x) = \begin{pmatrix} \cos(x) \\ \sin(x) \end{pmatrix}
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.\] Weiter seien $a = 0$ und $b = 2\pi$.
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Beh.: Es ex. kein $\xi \in (a,b)$ mit $f(b) - f(a) = D_f(\xi)(b - a)$.
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\begin{proof}
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Es ist zunächst
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\[
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D_f(x) = \begin{pmatrix} - \sin(x) \\ \cos(x) \end{pmatrix}
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.\] Da $\sin(x)$ und $\cos(x)$ keine gemeinsamen Nullstellen in $\R$ besitzen, folgt $\forall x \in \R$:
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\[
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D_f(x) \neq 0
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.\] Damit folgt direkt $\forall \xi \in (a,b)$:
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\begin{salign*}
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f(b) - f(a) = \begin{pmatrix} \cos(2\pi) \\ \sin(2\pi) \end{pmatrix}
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- \begin{pmatrix} \cos(0) \\ \sin(0) \end{pmatrix}
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= \begin{pmatrix} 1 \\ 0 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \end{pmatrix}
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= 0 \neq 2\pi \begin{pmatrix} - \sin(\xi) \\ \cos(\xi) \end{pmatrix}
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= D_f(\xi)
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.\end{salign*}
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\end{proof}
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\end{aufgabe}
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\end{document}
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