Ladies and gentleman,
I use Lyx 1.1.5fix1 and found some bugs:
- The latex fault: double Superscrips should be prevented!
(You often get: a^{b}^{b} and don't see it, because the b is written
over b; possible Solution would be ^{b}{}^{b} or better prevent double
superscript)
- It seems to me like Enumerate and Math-Displays don't work together.
An example for this error is line 4707ff /layout Enumerate in the
attached file. (don't look at contents - pre-alpha state!) Simply try it,
my lyx crashes by DVI-Viewing!
Greetings
Georg Wild
#LyX 1.1 created this file. For more info see http://www.lyx.org/
\lyxformat 2.16
\textclass amsart-seq
\begin_preamble
\usepackage {amssymb}
\usepackage {amssymb}
\usepackage {stmaryrd}
\usepackage {eufrak}
\usepackage [mathcal] {euscript}
\usepackage {floatflt}
\end_preamble
\language german
\inputencoding latin1
\fontscheme default
\graphics default
\paperfontsize default
\spacing single
\papersize Default
\paperpackage widemarginsa4
\use_geometry 0
\use_amsmath 0
\paperorientation portrait
\footskip
\secnumdepth 3
\tocdepth 3
\paragraph_separation indent
\defskip medskip
\quotes_language english
\quotes_times 2
\papercolumns 1
\papersides 2
\paperpagestyle plain
\layout Problem
\begin_inset LatexCommand \label{xTAx}
\end_inset
Sei
\begin_inset Formula \( f:\mathbb R^{n}\rightarrow \mathbb R;\, x\mapsto x^{T}Ax \)
\end_inset
mit einer
\begin_inset Formula \( n\times n-Matrix \)
\end_inset
A.
Beweisen Sie, daß f differenzierbar ist und berechnen Sie das Differential
\begin_inset LatexCommand \index{Differential}
\end_inset
\begin_inset Formula \( df\left( a\right) \)
\end_inset
für
\begin_inset Formula \( a\in \mathbb R^{n}. \)
\end_inset
\lang american
33
\layout Standard
\begin_inset Formula \( f:x\mapsto Ax\, und\, f:x\mapsto x^{T} \)
\end_inset
sind linear.
\begin_inset Formula \begin{eqnarray*}
f\left( x+h\right) & = & \left( x+h\right) ^{T}A\left( x+h\right)
=x^{T}Ax+h^{T}Ax+h^{T}Ax+h^{T}Ah\\
& = & \begin{array}{ccccccc}
\underbrace{x^{T}Ax} & + & \underbrace{x^{T}Ah} & + & \underbrace{h^{T}Ax} & + &
\underbrace{h^{T}Ah}\\
f\left( x\right) & & linear & & \left( x^{T}A^{T}h\right) ^{T} & & R\left(
h\right)
\end{array}\\
& = & \begin{array}{ccccc}
\underbrace{x^{T}Ax} & + & x^{T}\left( A+A^{T}\right) h & + & h^{T}Ah\\
f\left( x\right) & & Df\left( x\right) h & & R\left( h\right)
\end{array}
\end{eqnarray*}
\end_inset
\layout Standard
f ist differenzierbar und es gilt:
\begin_inset Formula \( df\left( a\right) =a^{T}\left( A+A^{T}\right) \)
\end_inset
\layout Problem
\line_top
Gegeben ist die Funktion
\begin_inset Formula \( f:\mathbb R^{3}\rightarrow \mathbb R;\, \left( x,y,z\right)
\mapsto x^{2}y+z. \)
\end_inset
\layout Enumerate
Finden Sie den Gradienten
\begin_inset LatexCommand \index{Gradient}
\end_inset
von f im Punkt
\begin_inset Formula \( \left( 2,3,5\right) \)
\end_inset
bezüglich des euklidischen Skalarproduktes!
\layout Enumerate
Finden Sie eine lineare Approximation
\begin_inset LatexCommand \index{Approximation}
\end_inset
von f im Punkt
\begin_inset Formula \( \left( 2,3,5\right) \)
\end_inset
!
\layout Enumerate
Berechnen Sie die Richtungsableitung
\begin_inset Formula \( \partial _{\left( 1,0,-1\right) }f\left( 2,3,5\right) \)
\end_inset
zum einen direkt und zum anderem unter Verwendung des Ergebnisses von Teil
a, d.h.
verifizieren Sie an diesem Beispiel die bekannte Formel
\begin_inset Formula \( \partial _{h}f\left( x,y,z\right) =\left\langle h,grad\,
f\left( x,y,z\right) \right\rangle \)
\end_inset
!
\layout Standard
\begin_inset Formula \begin{eqnarray*}
grad\, f\left( 2,3,5\right) & = & \left( \begin{array}{c}
\partial _{1}f\left( x\right) \\
\partial _{2}f\left( x\right) \\
\partial _{3}f\left( x\right)
\end{array}\right) =\left( \begin{array}{c}
2xy\\
x^{2}\\
1
\end{array}\right) =\left( \begin{array}{c}
12\\
4\\
1
\end{array}\right) \\
f\left( x+h\right) & = & f\left( x\right) +\nabla f\left( x\right) \cdot
h+....=f\left( x\right) +\left( \begin{array}{c}
12\\
4\\
1
\end{array}\right) \cdot h\\
\partial _{\left( 1,0,-1\right) }f\left( 2,3,5\right) & = & \lim _{t\rightarrow
0}\frac{f\left( 2+t,3,5-t\right) -f\left( 2,3,5\right) }{t}=\lim _{t\rightarrow
0}\frac{\left( 2+t\right) ^{2}\cdot 3+\left( 5-t\right) -\left( 2^{2}\cdot 3+5\right)
}{t}=\\
& = & \lim _{t\rightarrow 0}\frac{12+12t+3t^{2}+5-t-17}{t}=\lim _{t\rightarrow
0}\frac{11t+3t^{2}}{t}=11\\
\partial _{\left( 1,0,-1\right) }f\left( 2,3,5\right) & = & \left\langle \left(
\begin{array}{c}
1\\
0\\
-1
\end{array}\right) ,\left( \begin{array}{c}
12\\
4\\
1
\end{array}\right) \right\rangle =12-1=11
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Gegeben ist die Funktion
\begin_inset Formula \( f:\mathbb R^{2}\rightarrow \mathbb R \)
\end_inset
mit
\begin_inset Formula \( f\left( 0,0\right) =0 \)
\end_inset
und
\layout Standard
\begin_inset Formula \[
f\left( x,y\right) =\frac{x^{3}}{\sqrt{x^{2}+y^{2}}};\left( x,y\right) \neq \left(
0,0\right) .\]
\end_inset
\layout Standard
\series bold
\bar under
Zeigen Sie:
\series default
\bar default
f ist überall differenzierbar.
\layout Standard
\begin_inset Formula \begin{eqnarray*}
Df\left( x,y,z\right) h & = & \lim _{t\rightarrow 0}\frac{f\left(
x+th_{1},x+th_{2}\right) -f\left( x,y\right) }{t}=\lim _{t\rightarrow
0}\frac{\frac{\left( x+th_{1}\right) ^{3}}{\sqrt{\left( x+th_{1}\right) ^{2}+\left(
y+th_{2}\right) ^{2}}}-\frac{x^{3}}{\sqrt{x^{2}+y^{2}}}}{t}\\
Df\left( 0,0,0\right) h & = & \lim _{t\rightarrow 0}\frac{\frac{\left( th_{1}\right)
^{3}}{\sqrt{\left( th_{1}\right) ^{2}+\left( th_{2}\right) ^{2}}}}{t}=\lim
_{t\rightarrow 0}\frac{t^{2}h_{1}^{3}}{\sqrt{t^{2}\cdot \left(
h_{1}^{2}+h_{2}^{2}\right) }}=\lim _{t\rightarrow 0}\frac{th^{3}_{1}}{\left\Vert
h\right\Vert _{2}}=0
\end{eqnarray*}
\end_inset
\begin_inset Formula \( \Longrightarrow \)
\end_inset
Die Richtungs-Ableitung von f in
\begin_inset Formula \( \left( \begin{array}{c}
0\\
0\\
0
\end{array}\right) \)
\end_inset
ist in jede Richtung 0.
In den anderen Punkten ist die Ableitung als Komposition differenzierbarer
Abbildungen differenzierbar.
\layout Problem
\line_top
Es seien
\begin_inset Formula \( \)
\end_inset
-Funktionen auf einer offenen Teilmenge des
\begin_inset Formula \( \mathbb R^{n}. \)
\end_inset
Dann gilt:
\begin_inset LatexCommand \index{Gradient, Produkt}
\end_inset
\begin_inset Formula \[
\triangle \left( fg\right) =f\cdot \triangle g+2\left\langle grad\, g,grad\,
g\right\rangle +g\cdot \triangle f\]
\end_inset
\layout Standard
\begin_inset Formula \begin{eqnarray*}
\triangle \left( fg\right) & = & \left( \begin{array}{c}
\partial _{1}^{2}\left( fg\right) \\
\partial _{2}^{2}\left( fg\right) \\
.\\
.\\
\partial _{n}^{2}\left( fg\right)
\end{array}\right) =\left( \begin{array}{c}
\partial _{1}^{2}f\cdot g+\partial _{1}f\cdot \partial _{1}g+\partial _{1}^{2}g\cdot
f+\partial _{1}f\cdot \partial _{1}g\\
\partial _{2}^{2}f\cdot g+\partial _{2}f\cdot \partial _{2}g+\partial _{2}^{2}g\cdot
f+\partial _{2}f\cdot \partial _{2}g\\
.\\
.\\
\partial _{n}^{2}f\cdot g+\partial _{n}f\cdot \partial _{n}g+\partial _{n}^{2}g\cdot
f+\partial _{n}f\cdot \partial _{n}g
\end{array}\right) =\\
& = & \left( \begin{array}{c}
\partial _{1}^{2}f\cdot g+2\cdot \partial _{1}f\cdot \partial _{1}g+\partial
_{1}^{2}g\cdot f\\
\partial _{2}^{2}f\cdot g+2\cdot \partial _{2}f\cdot \partial _{2}g+\partial
_{2}^{2}g\cdot f\\
.\\
.\\
\partial _{n}^{2}f\cdot g+2\cdot \partial _{n}f\cdot \partial _{n}g+\partial
_{n}^{2}g\cdot f
\end{array}\right) =\\
& = & \left( \begin{array}{c}
\partial _{1}^{2}f\\
\partial _{2}^{2}f\\
.\\
.\\
\partial _{n}^{2}f
\end{array}\right) +2\left( \begin{array}{c}
\partial _{1}f\cdot \partial _{1}g\\
\partial _{2}f\cdot \partial _{2}g\\
.\\
.\\
\partial _{n}f\cdot \partial _{n}g
\end{array}\right) +\left( \begin{array}{c}
\partial _{1}^{2}g\\
\partial _{2}^{2}g\\
.\\
.\\
\partial _{n}^{2}g
\end{array}\right) \\
& = & f\cdot \triangle g+2\left\langle grad\, g,grad\, g\right\rangle +g\cdot
\triangle f
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Seien
\begin_inset Formula \( I_{1},...,I_{n}\subset \mathbb R \)
\end_inset
offene Intervalle und
\begin_inset Formula \( Q:=I_{1}\times ...\times I_{n}\subset \mathbb R^{n} \)
\end_inset
.
Sei
\begin_inset Formula \( f:\mathbb K\rightarrow \mathbb R \)
\end_inset
eine partiell differenzierbare Funktion mit
\begin_inset LatexCommand \index{Ableitung, beschränkt partielle}
\end_inset
beschränkten partiellen Ableitungen:
\begin_inset Formula \( \left| \partial _{i}f\right| \leqq C\, \forall _{i\in \left\{
1..n\right\} } \)
\end_inset
.
\layout Enumerate
Man beweise für beliebige Punkte
\begin_inset Formula \( x,y\in Q \)
\end_inset
die Fehlerabschätzung:
\begin_inset Formula \[
\left| f\left( y\right) -f\left( x\right) \right| \leqq C\cdot \sum _{i=1}^{n}\left|
y_{i}-x_{i}\right| \]
\end_inset
\layout Enumerate
Man zeige, daß f stetig ist.
\layout Enumerate
Finden Sie ein Gegenbeispiel zu der Abschätzung in Teil 1, wenn an Stelle
der Kugel nur noch ein wegzusammenhängender Definitionsbereich vorausgesetzt
wird.
\layout Proof
Definiere:
\begin_inset Formula \[
a_{0}:=\left( \begin{array}{c}
x_{1}\\
x_{2}\\
.\\
.\\
x_{n}
\end{array}\right) =x;\, a_{1}:=\left( \begin{array}{c}
y_{1}\\
x_{2}\\
.\\
.\\
x_{n}
\end{array}\right) ;\, ....\, a_{n}:=\left( \begin{array}{c}
y_{1}\\
y_{2}\\
.\\
.\\
y_{n}
\end{array}\right) =y\]
\end_inset
\newline
Es gilt:
\begin_inset Formula \[
\left\{ a_{i-1}+t\left( a_{i}-a_{i-1}\right) ,t\in \left[ 0,1\right] \right\} \subset
Q\]
\end_inset
\newline
\layout Proof
\begin_inset Formula \begin{eqnarray*}
\left| f\left( a_{i}\right) -f\left( a_{i-1}\right) \right| & = & \left| f\left(
y_{1},...,y_{i},x_{i+1},...,x_{n}\right) -f\left(
y_{1},...,y_{i-1},x_{i},...,x_{n}\right) \right| =\\
& = & \left| \int _{x_{i}}^{y_{i}}\begin{array}{c}
\underbrace{\partial _{i}f\left( y_{1},...,y_{i-1},t,x_{i+1},...,x_{n}\right) }\\
\leqq C
\end{array}dt\right| \\
& & \\
& \leqq & C\cdot \left| y_{i}-x_{i}\right| \\
& & \\
\Rightarrow \left| \begin{array}{c}
\underbrace{f\left( y\right) }\\
f\left( a_{n}\right)
\end{array}-\begin{array}{c}
\underbrace{f\left( x\right) }\\
f\left( a_{0}\right)
\end{array}\right| & \leqq & \left| f\left( a_{n}\right) -f\left( a_{n-1}\right)
+f\left( a_{n-1}\right) -f\left( a_{n-2}\right) +...+f\left( a_{1}\right) -f\left(
a_{0}\right) \right| \leqq \\
& \leqq & \sum _{i=1}^{n}\left| f\left( a_{i}\right) -f\left( a_{i-1}\right) \right|
\leqq \sum _{i=1}^{n}C\cdot \begin{array}{c}
\underbrace{\left| y_{i}-x_{i}\right| }\\
\left\Vert x-y\right\Vert _{1}
\end{array}
\end{eqnarray*}
\end_inset
\layout Standard
Wegen
\begin_inset Formula \( \left| f\left( y\right) -f\left( x\right) \right| \leqq C\cdot
\left\Vert x-y\right\Vert _{1} \)
\end_inset
und da auf endlich dimensionalen Vektorräumen alle Normen äquivalent sind,
ist f Lipschitz-stetig.
\layout Standard
Wird an Stelle einer Kugel nur ein wegzusammenhängendes Gebiet vorausgesetzt,
so sind die obigen Aussagen nicht mehr gültig; sei etwa:
\newline
\begin_inset Formula \( K=\mathbb R^{2}\smallsetminus \mathbb R^{-}\times \left\{
0\right\} \)
\end_inset
\newline
\begin_inset Formula \( f\left( x_{1},x_{2}\right)
=arccos\frac{x_{1}}{\sqrt{x_{1}^{2}+x_{2}^{2}}}\cdot sgn\, x_{2} \)
\end_inset
\newline
Bei Betrachtung der Folgen
\begin_inset Formula \( x_{n}=\left( -1,-\frac{1}{n}\right) ;\, y_{n}=\left(
-1,+\frac{1}{n}\right) \)
\end_inset
folgt für
\begin_inset Formula \( n\rightarrow \infty \)
\end_inset
\begin_inset Formula \( \left\Vert y-x\right\Vert _{1}\rightarrow 0 \)
\end_inset
, aber
\begin_inset Formula \( \left| f\left( x\right) -f\left( y\right) \right| \rightarrow
2\pi \)
\end_inset
.
\layout Problem
\line_top
Man berechne
\begin_inset Formula \( d^{\left( 2\right) }f \)
\end_inset
und
\begin_inset Formula \( f'' \)
\end_inset
für
\begin_inset Formula \( f:\mathbb R_{+}\times \mathbb R\rightarrow \mathbb R,\left(
x,y\right) \mapsto x^{y}. \)
\end_inset
\layout Standard
\begin_inset Formula \begin{eqnarray*}
h^{T}d^{\left( 2\right) }f\left( x,y\right) h & = & h^{T}\left( \begin{array}{cc}
\partial _{1}^{2}f(x,y) & \partial _{12}f\left( x,y\right) \\
\partial _{21}f\left( x,y\right) & \partial _{2}^{2}f(x,y)
\end{array}\right) h=\\
& = & h^{T}\left( \begin{array}{cc}
y\left( y-1\right) x^{y-2} & x^{y-1}+y\cdot \ln x\cdot x^{y-1}\\
x^{y-1}+y\cdot \ln x\cdot x^{y-1} & ln^{2}x\cdot x^{y}
\end{array}\right) h\\
f''(x,y) & = & \left( \begin{array}{cc}
y\left( y-1\right) x^{y-2} & x^{y-1}+y\cdot \ln x\cdot x^{y-1}\\
x^{y-1}+y\cdot \ln x\cdot x^{y-1} & ln^{2}x\cdot x^{y}
\end{array}\right)
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Gegeben sei die Potenzfunktion
\begin_inset Formula \( f:\mathbb C\rightarrow \mathbb C;\, z\mapsto z^{k};k\in
\mathbb N \)
\end_inset
.
Man identifiziere den Definitionsbereich mit
\begin_inset Formula \( \mathbb R^{2} \)
\end_inset
.
Man zeige, daß f harmonisch
\begin_inset LatexCommand \index{harmonisch}
\end_inset
ist (d.h.
\begin_inset Formula \( \triangle f=0 \)
\end_inset
)
\layout Enumerate
unter Verwendung von carthesischen Koordinaten x,y
\layout Enumerate
unter Verwendung von Polarkoordinaten
\begin_inset LatexCommand \index{Polarkoordinaten}
\end_inset
r,
\begin_inset Formula \( \varphi \)
\end_inset
\layout Standard
\begin_inset Formula \begin{eqnarray*}
\triangle f\left( x,y\right) & = & \partial _{x}^{2}f\left( x,y\right) +\partial
_{y}^{2}f\left( x,y\right) =\partial _{x}^{2}\left( x+iy\right) ^{k}+\partial
_{y}^{2}\left( x+iy\right) ^{k}=\\
& = & \left[ k\left( k-1\right) +i^{2}\cdot k\left( k+1\right) \right] \left(
x+iy\right) =0\cdot \left( x+iy\right) =0
\end{eqnarray*}
\end_inset
\layout Standard
Sei nun
\begin_inset Formula \( r=\left| z^{k}\right| =\left| z\right| ^{k};\, \varphi
=arctan\left( \frac{Re\left( \left( x+iy\right) ^{k}\right) }{Im\left( \left(
x+iy\right) ^{k}\right) }\right) . \)
\end_inset
Bekanntlich gilt dann:
\begin_inset Formula \( \triangle f\left( x,y\right) =\left(
F_{rr}+\frac{1}{r^{2}}F_{\varphi \varphi }+\frac{1}{r}F_{r}\right) \left( r,\varphi
\right) ;\, F\left( r,\varphi \right) =f\left( r\, cos\varphi ,r\, sin\varphi \right)
\)
\end_inset
.
\begin_inset Formula \begin{eqnarray*}
F\left( r,\varphi \right) & = & f\left( r\, cos\varphi ,r\, sin\varphi \right)
=\left( r\, cos\varphi +i\, r\, sin\varphi \right) ^{k}=\\
& = & r^{k}\cdot \left( cos\varphi +i\, sin\varphi \right) ^{k}=r^{k}\cdot
e^{ik\varphi }\\
F_{rr}\left( r,\varphi \right) & = & e^{ik\varphi }\cdot k\left( k+1\right) r^{k-2}\\
F_{\varphi \varphi }\left( r,\varphi \right) & = & r^{k}\cdot \left( ik\right)
^{2}\cdot e^{ik\varphi }=-r^{k}k^{2}e^{ik\varphi }\\
F_{r}\left( r,\varphi \right) & = & k\cdot r^{k-1}\cdot e^{ik\varphi }\\
\triangle f\left( x,y\right) & = & e^{ik\varphi }\cdot k\left( k+1\right)
r^{k-2}+\frac{1}{r^{2}}\cdot \left( -r^{k}k^{2}e^{ik\varphi }\right)
+\frac{1}{r}k\cdot r^{k-1}\cdot e^{ik\varphi }=\\
& = & e^{ik\varphi }\cdot k\cdot r^{k-2}\cdot \left( \left( k+1\right) -k-1\right) =0
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Man ermittle alle rotationssymetrischen, harmonischen Funktionen auf einer
Kugelschale im
\begin_inset Formula \( \mathbb R^{n} \)
\end_inset
und insbesondere auf einer Kugel.
\layout Standard
\SpecialChar ~
\layout Problem
\line_top
Man untersuche die Funktion
\begin_inset Formula \( f:\mathbb R^{2}\rightarrow \mathbb R;\, f\left( x,y\right)
=x^{3}+y^{3}+3xy \)
\end_inset
auf Extrema
\begin_inset LatexCommand \index{Extrema}
\end_inset
.
\layout Standard
Für ein Extrema ist notwendig:
\begin_inset Formula \begin{eqnarray*}
grad\, f\left( x,y\right) & = & \left( \begin{array}{c}
\partial _{x}f\left( x,y\right) \\
\partial _{y}f\left( x,y\right)
\end{array}\right) =\left( \begin{array}{c}
3x^{2}+3y\\
3y^{2}+3x
\end{array}\right) \doteq 0\\
\Longrightarrow & & 3x^{2}+3y=3\left( x^{2}+y\right) \doteq 0\Leftrightarrow
x^{2}+y=0\\
\Longrightarrow & & 3y^{2}+3x=3\left( y^{2}+x\right) \doteq 0\Leftrightarrow
y^{2}+x=0\\
& \Rightarrow & \left( x^{2}+y\right) -x\left( y^{2}+x\right) =y-xy^{2}=y\left(
1-xy\right) \doteq 0\\
L\ddot{o}sung\, 1 & \Rightarrow & y=0\Rightarrow x=0\\
L\ddot{o}sung\, 2 & \Rightarrow & 1-xy=0\Rightarrow x=\frac{1}{y}\Rightarrow
y^{2}+\frac{1}{y}=0\Rightarrow \frac{1}{y}\left( y^{3}+1\right) =0\Rightarrow \\
\left( y\neq 0\right) & \Rightarrow & y=-1\Rightarrow x=-1
\end{eqnarray*}
\end_inset
\layout Standard
Es gibt also 2 stationäre Punkte:
\begin_inset Formula \( P_{1}\left( 0,0\right) ;\, P_{2}\left( -1,-1\right) \)
\end_inset
.
\begin_inset Formula \begin{eqnarray*}
H_{f}\left( x,y\right) & = & \left( \begin{array}{cc}
6x & 3\\
3 & 6y
\end{array}\right) =3\left( \begin{array}{cc}
2x & 1\\
1 & 2y
\end{array}\right) =\\
det\, H_{f}\left( x,y\right) & = & 4xy-1\\
det\, H_{f}\left( 0,0\right) & = & -1<0\Longrightarrow Sattelpunkt\\
det\, H_{f}\left( -1,-1\right) & = & 3>0\, Maximum,\, da\, H_{f}(-1,-1)_{\nwarrow }<0
\end{eqnarray*}
\end_inset
\layout Note
\line_top
Kriterien für pos.Definitheit
\begin_inset LatexCommand \index{Definitheit}
\end_inset
einer
\begin_inset Formula \( n\times n \)
\end_inset
-Matrix:
\layout Enumerate
alle Eigenwerte
\begin_inset LatexCommand \index{Eigenwerte}
\end_inset
>0
\layout Enumerate
Hurwitzkriterium
\begin_inset LatexCommand \index{Hurwitzkriterium}
\end_inset
:
\begin_inset Formula \( det\, \left( a_{ij}\right) _{1\leqq i,j\leqq k}>0 \)
\end_inset
\layout Enumerate
Charakteristisches
\begin_inset LatexCommand \index{Charakteristisches Polynom}
\end_inset
Polynom hat n pos.
Nullstellen
\layout Enumerate
\begin_inset Formula \( 2\times 2 \)
\end_inset
-Matrix Sp>0; det
\begin_inset Formula \( H_{f} \)
\end_inset
>0
\newline
\begin_inset Formula \( Tr\left( SAS^{-1}\right) =Tr\left( A\right) , \)
\end_inset
da
\begin_inset Formula \( Tr\left( AB\right) =Tr(BA) \)
\end_inset
\newline
\begin_inset Formula \( det\left( SAS^{-1}\right) =det\left( A\right) , \)
\end_inset
da
\begin_inset Formula \( det\left( AB\right) =det(BA) \)
\end_inset
\layout Problem
\line_top
Sei f eine
\begin_inset Formula \( \)
\end_inset
-Funktion auf einer Umgebung der Halbebene
\begin_inset LatexCommand \index{Halbebene}
\end_inset
\begin_inset Formula \( \mathbb H:=\mathbb R\times \left[ 0,\infty \right) \)
\end_inset
.
Man zeige: Wenn
\begin_inset Formula \( f|_{\mathbb H} \)
\end_inset
im Randpunkt
\begin_inset LatexCommand \index{Randpunkt}
\end_inset
\begin_inset Formula \( p=\left( \textrm{x},0\right) \)
\end_inset
ein lokales Extremum hat, so ist der Gradient von f im Punkt p senkrecht
\begin_inset LatexCommand \index{senkrecht}
\end_inset
zum Rand, d.h.
parallel zur
\begin_inset Formula \( x_{2} \)
\end_inset
-Achse.
\layout Proof
Angenommen
\begin_inset Formula \( \varphi '\left( 0\right) >0\Longrightarrow \exists \tilde{t} \)
\end_inset
zwischen t und 0 mit
\begin_inset Formula \( \frac{\varphi \left( t\right) -\varphi \left( 0\right)
}{t-0}=\varphi '\left( \tilde{t}\right) \)
\end_inset
nach MWS.
Wegen
\begin_inset Formula \( \)
\end_inset
gilt damit:
\begin_inset Formula \( \exists \varepsilon >0:\frac{\varphi \left( \xi \right)
-\varphi \left( 0\right) }{\xi }>0\forall \left| \xi \right| <\varepsilon \)
\end_inset
.
Für
\begin_inset Formula \( \xi <0 \)
\end_inset
folgt damit
\begin_inset Formula \( \varphi \left( \xi \right) <\varphi \left( 0\right) \)
\end_inset
, für
\begin_inset Formula \( \xi >0 \)
\end_inset
folgt damit
\begin_inset Formula \( \varphi \left( \xi \right) >\varphi \left( 0\right) \)
\end_inset
im Wiederspruch zur Voraussetzung, daß ein Extremum existiert.
\begin_inset Formula \( \Longrightarrow \varphi '\left( 0\right) \leqq 0 \)
\end_inset
\newline
Ebenso werde
\begin_inset Formula \( \varphi '\left( 0\right) <0 \)
\end_inset
ausgeschlossen.
\begin_inset Formula \( \Longrightarrow \varphi '\left( 0\right) =0\Longrightarrow
\nabla f\left( p\right) =\left( 0,f_{y}\left( p\right) \right) \)
\end_inset
\layout Problem
\line_top
Konstruktion einer Funktion auf
\begin_inset Formula \( \mathbb R^{2}, \)
\end_inset
die im Ursprung in jeder Richtung lokal konstant ist, aber in keiner Umgebung
des Ursprungs beschränkt:
\newline
Sei
\begin_inset Formula \( \varphi :\mathbb R\rightarrow \mathbb R \)
\end_inset
eine
\begin_inset Formula \( \)
\end_inset
-Funktion, die im Intervall
\begin_inset Formula \( \left( -1,1\right) \)
\end_inset
positive Werte besitzt und auf
\begin_inset Formula \( \mathbb R\smallsetminus \left( -1,1\right) \)
\end_inset
konstant gleich Null ist, z.B.
\begin_inset Formula \[
\varphi \left( t\right) =\{\begin{array}{c}
e^{\frac{-1}{1-t^{2}}}\\
0
\end{array}\begin{array}{l}
f\ddot{u}r\, t\in \left( -1,1\right) \\
sonst
\end{array}\]
\end_inset
Definiere
\begin_inset Formula \[
f:\mathbb R^{2}\rightarrow \mathbb R;f\left( x\right) :=\{\begin{array}{c}
\frac{1}{x}\varphi \left( \frac{y}{x^{2}}-2\right) \\
0
\end{array}\begin{array}{l}
f\ddot{u}r\, x\neq 0\\
f\ddot{u}r\, x=0
\end{array}\]
\end_inset
Zeige:
\begin_deeper
\layout Enumerate
\begin_inset Formula \( f\equiv 0 \)
\end_inset
in den Gebieten
\begin_inset Formula \( \left\{ \left( x,y\right) \in \mathbb R^{2}:y<x^{2}\right\} \)
\end_inset
und
\begin_inset Formula \( \left\{ \left( x,y\right) \in \mathbb R^{2}:y>3x^{2}\right\}
\)
\end_inset
.
\layout Enumerate
\begin_inset Formula \( \)
\end_inset
.
\layout Enumerate
f ist in keiner Umgebung von
\begin_inset Formula \( \left( 0,0\right) \)
\end_inset
beschränkt.
\layout Enumerate
Für jedes
\begin_inset Formula \( \left( a,b\right) \in \mathbb R^{2} \)
\end_inset
gibt es ein
\begin_inset Formula \( \varepsilon >0 \)
\end_inset
, do daß
\begin_inset Formula \( f\left( \left( 0,0\right) +t\left( a,b\right) \right) =0 \)
\end_inset
für alle
\begin_inset Formula \( t\in \left( -\varepsilon ,\varepsilon \right) \)
\end_inset
.
(Insbesondere sind alle Richtungsableitungen im Ursprung gleich Null.)
\end_deeper
\layout Enumerate
\added_space_top 1cm
\begin_inset LatexCommand \label{Nullstellen}
\end_inset
\begin_inset Formula \( f\left( x,y\right) =0, \)
\end_inset
falls x=0 oder
\begin_inset Formula \( x\neq 0\wedge \left| \frac{y}{x^{2}}-2\right| >1 \)
\end_inset
.
\newline
\begin_inset Formula \( \frac{y}{x^{2}}-2>1\Rightarrow y>3x^{2}\, \wedge \,
-\frac{y}{x^{2}}+2>1\Rightarrow y<x^{2} \)
\end_inset
\lang american
\layout Enumerate
\begin_inset Formula \( \)
\end_inset
\newline
\begin_inset Formula \( x\neq 0: \)
\end_inset
f in Umgebung von
\begin_inset Formula \( \left( x,y\right) \)
\end_inset
Verkettung von
\begin_inset Formula \( \mathbb C^{\infty } \)
\end_inset
-Funktionen
\newline
x=0,\SpecialChar ~
aber\SpecialChar ~
\begin_inset Formula \( y\neq 0 \)
\end_inset
: aus (
\begin_inset LatexCommand \ref{Nullstellen}
\end_inset
) folgt, daß f in einer Umgebung von
\begin_inset Formula \( \left( 0,y\right) \)
\end_inset
gleich 0 ist.
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
f\left( t,2t^{2}\right) & = & \frac{1}{t}\varphi \left( \frac{2t^{2}}{t^{2}}-2\right)
=\frac{1}{t}\varphi \left( 0\right) =\frac{1}{e^{t}}=e^{-t}\\
& &
\end{eqnarray*}
\end_inset
Sei V vorg.
Umgebung von
\begin_inset Formula \( \left( 0,0\right) ;t_{n}=\frac{1}{n} \)
\end_inset
\newline
\begin_inset Formula \[
\left( t_{n},2t_{n}^{2}\right) \begin{array}{c}
\\
\longrightarrow \\
n\rightarrow \infty
\end{array}\left( 0,0\right) \Longrightarrow \exists N\in \mathbb N:\left(
t_{n},2t_{n}^{2}\right) \in V\forall n\geqq N;f\left( t_{n},2t_{n}^{2}\right)
=e^{-t_{n}}\begin{array}{c}
\\
\longrightarrow \\
n\rightarrow \infty
\end{array}\infty \]
\end_inset
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
\left( x,y\right) & = & \left( 0,0\right) +t\left( a,b\right) =\left( ta,tb\right) \\
a=0 & \Rightarrow & f\left( ta,tb\right) =f\left( 0,tb\right) =0\forall t\\
a\neq 0 & \Rightarrow & \Rightarrow \left( x,y\right) =\left( ta,tb\right)
=\frac{tb}{t^{2}a^{2}}-2=\frac{b}{ta^{2}}-2;\, t\neq 0\\
& t>0: & \frac{b}{ta^{2}}-2>1\Leftrightarrow t<\frac{b}{3a^{2}}\\
& t<0: & \frac{b}{ta^{2}}-2<1\\
\Rightarrow f\left( ta,tb\right) & = & 0\forall \left| t\right| <\varepsilon
=\frac{b}{3a^{2}}\\
&
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Unabhängigkeit
\begin_inset LatexCommand \index{Unabhängigkeit}
\end_inset
der Integration
\begin_inset LatexCommand \index{Integration}
\end_inset
von der Reihenfolge:
\newline
Es sei
\begin_inset Formula \( f:\left[ a,b\right] \times \left[ c,d\right] \rightarrow
\mathbb C \)
\end_inset
eine stetige Funktion.
Dann gilt:
\begin_inset Formula \[
\int _{c}^{d}\left( \int _{a}^{b}f\left( x,y\right) dx\right) dy=\int _{a}^{b}\left(
\int _{c}^{d}f\left( x,y\right) dy\right) dx\]
\end_inset
Mit Hilfe dieser Vertauschungsregel berechne man das Integral
\layout Problem
\begin_inset Formula \[
\int _{0}^{1}\frac{x^{t}-1}{ln\, x}dx\, f\ddot{u}r\, t\geqq 0.\]
\end_inset
\layout Proof
\begin_inset Formula \begin{eqnarray*}
H_{1}\left( y\right) & = & \int _{c}^{y}\left( \int _{a}^{b}f\left( x,t\right)
dx\right) dt\\
H_{2}\left( y\right) & = & \int _{d}^{b}\left( \int _{c}^{y}f\left( x,t\right)
dt\right) dx\\
\frac{\partial }{\partial y}H_{1}\left( y\right) & = & \int _{a}^{b}f\left(
x,y\right) dx\\
&
\end{eqnarray*}
\end_inset
\begin_float footnote
\layout Standard
Beweis der glm.
Stetigkeit:
\layout Standard
\begin_inset Formula \begin{eqnarray*}
\left| \frac{\int _{y_{0}}^{y}f\left( x,t\right) dt-\int _{y_{0}}^{y}f\left(
x,y_{0}\right) dt}{y-y_{0}}\right| & = & \left| \frac{\int _{y_{0}}^{y}f\left(
x,t\right) dt-\left( y-y_{0}\right) f\left( x,y_{0}\right) }{y-y_{0}}\right| =\\
& = & \left| \frac{\int _{y_{0}}^{y}\left[ f\left( x,t\right) -f\left( x,y_{0}\right)
\right] dt}{y-y_{0}}\right| \leqq \\
& \leqq & \int _{y_{0}}^{y}\left| \frac{f\left( x,t\right) -f\left( x,y_{0}\right)
}{y-y_{0}}\right| dt\leqq \\
& \leqq & \frac{y-y_{0}}{y-y_{0}}\max \left| f\left( x,t\right) -f\left(
x,y_{0}\right) \right| =\\
& = & \max \left| f\left( x,t\right) -f\left( x,y_{0}\right) \right|
\end{eqnarray*}
\end_inset
\layout Standard
\begin_inset Formula \( \forall y\in \forall \varepsilon >0\exists \delta >0:max\left|
...\right| \leqq \varepsilon \, falls\, \left| y-y_{0}\right| <\delta \)
\end_inset
, da
\begin_inset Formula \( \left| f\left( x,t\right) -f\left( x,y_{0}\right) \right| \)
\end_inset
stetige Funktion auf Kompaktum
\begin_inset Formula \( \left[ a,b\right] \)
\end_inset
, und jede Funktion auf Kompaktum gleichmäßig stetig.
\end_float
\newline
\begin_inset Formula \begin{eqnarray*}
\frac{\partial }{\partial y}H_{2}\left( y_{0}\right) & = & H_{2}'\left( y_{0}\right)
=\lim _{y\rightarrow y_{0}}\frac{H_{2}\left( y\right) -H_{2}\left( y_{0}\right)
}{y-y_{0}}=\\
& = & \lim _{y\rightarrow y_{0}}\frac{1}{y-y_{0}}\left[ \int _{a}^{b}\left( \int
_{y_{0}}^{y}f\left( x,t\right) dt\right) dx\right] =\\
& = & \lim _{y\rightarrow y_{0}}\left[ \int _{a}^{b}f\left( x,y_{0}\right) dx\right]
\end{eqnarray*}
\end_inset
\layout Proof
\begin_inset Formula \begin{eqnarray*}
F\left( t\right) & = & \int _{0}^{1}\frac{x^{t}-1}{ln\, x}dx\\
\frac{\partial }{\partial t}\frac{x^{t}}{ln\, x} & = & \frac{\partial }{\partial
t}\frac{e^{t\, ln\, x}}{ln\, x}=e^{t\, ln\, x}=x^{t}\\
\int _{0}^{t}x^{s}ds & = & \left[ \frac{x^{s}}{ln\, x}\right]
_{0}^{t}=\frac{x^{t}-1}{ln\, x}\\
F\left( t\right) & = & \int _{0}^{1}\left( \int _{0}^{t}x^{s}ds\right) dx=\int
_{0}^{t}\left( \int _{0}^{1}x^{s}dx\right) ds=\int _{0}^{t}\left[
\frac{1}{s+1}x^{s+1}\right] _{0}^{1}ds=\\
& = & \int _{0}^{t}\frac{1}{s+1}ds=\left[ ln\left| s+1\right| \right]
_{0}^{t}=ln\left| t+1\right|
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Sei
\begin_inset Formula \( \rho \)
\end_inset
eine stetige Funktion auf dem Rechteck
\begin_inset Formula \( Q:=\left[ a,b\right] \times \left[ c,d\right] \subset \mathbb
R^{2} \)
\end_inset
.
Man zeige, daß die auf
\begin_inset Formula \( \mathbb R^{2}\smallsetminus \mathbb Q \)
\end_inset
durch
\begin_inset Formula \[
u\left( x,y\right) :=\int _{a}^{b}\left( \int _{c}^{d}ln\left( \left( x-s\right)
^{2}+\left( y-t\right) ^{2}\right) \cdot \rho \left( s,t\right) dt\right) ds\]
\end_inset
definierte Funktion harmonisch ist.
\layout Standard
Differenziationssatz für Faltung
\begin_inset LatexCommand \index{Faltung, Differentiationssatz}
\end_inset
:
\begin_inset Formula \[
\triangle U\left( x,y\right) =\int _{a}^{b}\left( \int _{c}^{d}\triangle \left(
x,y\right) ln\left( \left( x-s\right) ^{2}+\left( y-t\right) ^{2}\right) \zeta \left(
s,t\right) dt\right) ds\]
\end_inset
\layout Problem
\line_top
\begin_inset LatexCommand \label{Lagrange}
\end_inset
Man ermittle zu der Lagrange-Funktion
\begin_inset Formula \[
L\left( t,y,p\right) =\frac{1}{2}mp^{2}-mgy\, \, \, \, \, \, \left( m,g>0\right) \]
\end_inset
die Eulersche Differentialgleichung.
\layout Standard
\begin_inset Formula \begin{eqnarray*}
\frac{d}{dx}L_{p}\left( x,\varphi \left( x\right) ,\varphi '\left( x\right) \right) &
= & L_{y}\left( x,\varphi \left( x\right) ,\varphi '\left( x\right) \right) \\
\frac{d}{dt}\left( m\varphi '\left( t\right) \right) & = & -mg\\
\varphi ''\left( t\right) & = & g\\
\varphi '\left( t\right) & = & gt+v_{0}\\
y\left( t\right) & = & \frac{1}{2}gt^{2}+v_{0}t+y_{0}\, \, \, \left(
Wurfparabel\right)
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Man zeige allgemein: Hängt die Lagrange-Funktion L
\begin_inset Formula \( \left( t,y,p\right) \)
\end_inset
nicht von t ab, d.h.
\begin_inset Formula \( L\left( t,y,p\right) =\hat{L}\left( y,p\right) \)
\end_inset
, so ist für jede Lösung
\begin_inset Formula \( \varphi \)
\end_inset
der Eulerschen Differentialgleichung
\begin_inset Formula \[
\frac{d}{dt}\hat{L}_{p}\left( \varphi \left( t\right) ,\varphi '\left( t\right)
\right) =\hat{L}_{y}\left( \varphi \left( t\right) ,\varphi '\left( t\right) \right) \]
\end_inset
die Funktion
\begin_inset LatexCommand \index{Lagrange, zeitunabhängig}
\end_inset
\begin_inset Formula \[
E_{\varphi }:=\hat{L}_{p}\left( \varphi ,\varphi '\right) \varphi '-\hat{L}\left(
\varphi ,\varphi '\right) \]
\end_inset
(als Funktion von t) konstant.
\newline
Man berechne
\begin_inset Formula \( E_{\varphi } \)
\end_inset
für den Fall der Lagrangefunktion von (
\begin_inset LatexCommand \ref{Lagrange}
\end_inset
).
\layout Standard
\begin_inset Formula \begin{eqnarray*}
E_{\varphi } & = & \hat{L}_{p}\left( \varphi ,\varphi '\right) \varphi '-\hat{L}\left(
\varphi ,\varphi '\right) \\
\frac{d}{dt}E_{\varphi } & = & \frac{d}{dt}\hat{L}_{p}\left( \varphi ,\varphi '\right)
\varphi '+\hat{L}_{p}\left( \varphi ,\varphi '\right) \varphi
''-\frac{d}{dt}\hat{L}\left( \varphi ,\varphi '\right) =\\
& = & \hat{L}_{y}\left( \varphi ,\varphi '\right) \varphi '+\hat{L}_{p}\left( \varphi
,\varphi '\right) \varphi ''-\hat{L}_{y}\left( \varphi ,\varphi '\right) \varphi
'-\hat{L}_{p}\left( \varphi ,\varphi '\right) \varphi '=0\\
z.B.\, E_{\varphi } & = & mp^{2}-\frac{1}{2}mp^{2}+mgy=\frac{1}{2}my'^{2}+mgy=\\
& = & \frac{1}{2}m\left( gt+v_{0}\right) ^{2}+mg\left(
\frac{1}{2}gt^{2}+v_{0}t+y_{0}\right) =\frac{1}{2}mv_{0}^{2}+mgy_{0}=const
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Man berechne das Differential
\begin_inset Formula \( df \)
\end_inset
für die Abbildung
\begin_inset Formula \[
f:\mathbb K^{n\times n}\rightarrow \mathbb K^{n\times n},\, f\left( A\right) =A^{T}A\]
\end_inset
direkt und unter Verwendung der Produktregel.
\newline
Man berechne die Differentiale
\begin_inset Formula \( dg \)
\end_inset
und
\begin_inset Formula \( dh \)
\end_inset
für die Abbildungen
\begin_inset Formula \begin{eqnarray*}
g: & \mathbb K^{n\times n}\rightarrow \mathbb K^{n\times n} & f\left( A\right) =A^{3}\\
h: & \mathbb R^{n}\rightarrow \mathbb R & h\left( x\right) =e^{x^{T}Ax};A\in \mathbb
R^{n\times n}
\end{eqnarray*}
\end_inset
\begin_inset LatexCommand \label{expXTAx}
\end_inset
\begin_inset Formula \begin{eqnarray*}
df\left( A\right) h & = & \lim _{k\rightarrow 0}\frac{f\left( A+hk\right) -f\left(
A\right) }{k}=\lim _{k\rightarrow 0}\frac{\left( A+hk\right) ^{T}\left( A+hk\right)
-A^{T}A}{k}=\\
& = & \lim _{k\rightarrow 0}\frac{A^{T}A+A^{T}hk+\left( hk\right)
^{T}A+(hk)^{T}\left( hk\right) -A^{T}A}{k}=\\
& = & \lim _{k\rightarrow 0}\frac{A^{T}hk+\left( hk\right) ^{T}A+(hk)^{T}\left(
hk\right) }{k}=\\
& = & \lim _{k\rightarrow 0}\frac{A^{T}hk+\left( A^{T}\left( hk\right) \right)
^{T}+(hk)^{T}\left( hk\right) }{k}=\\
& = & \lim _{k\rightarrow 0}\frac{A^{T}hk+\left( A^{T}\left( hk\right) \right)
^{T}+(hk)^{T}\left( hk\right) }{k}=\\
& = & A^{T}h+\left( A^{T}\left( h\right) \right) ^{T}=A^{T}h+h^{T}A\\
d\left( A^{T}\right) h & = & \lim _{k\rightarrow 0}\frac{\left( A+hk\right)
^{T}-A^{T}}{k}=h^{T}\\
d\left( A\right) h & = & \lim _{k\rightarrow 0}\frac{\left( A+hk\right) -A}{k}=h\\
df\left( A\right) h & = & d\left( A^{T}\right) h\cdot A+A^{T}d\left( A\right)
h=h^{T}A+A^{T}h
\end{eqnarray*}
\end_inset
\layout Standard
\begin_inset Formula \begin{eqnarray*}
dg\left( A\right) h & = & \lim _{k\rightarrow 0}\frac{f\left( A+hk\right) -f\left(
A\right) }{k}=\lim _{k\rightarrow 0}\frac{\left( A+hk\right) ^{3}-A^{3}}{k}=\\
& = & \lim _{k\rightarrow 0}\frac{\left( A+hk\right) \left( A+hk\right) \left(
A+hk\right) -A^{3}}{k}=\\
& = & \lim _{k\rightarrow 0}\frac{1}{k}(A^{3}+AA\left( hk\right) +A\left( hk\right)
A+A\left( hk\right) ^{2}+\left( hk\right) AA+\\
& & +\left( hk\right) A\left( hk\right) +\left( hk\right) ^{2}A+\left( hk\right)
^{3}-A^{3}=\\
& = & AAh+Aha+hAA\\
d\left( e^{x}\right) h & = & \lim _{k\rightarrow 0}\frac{e^{x+k}-e^{x}}{k}=\lim
_{k\rightarrow 0}\frac{e^{x}\left( e^{k}-e^{0}\right) }{k-0}=e^{k}\cdot e^{0}=e^{k}\\
d\left( x^{T}Ax\right) h & = & \lim _{k\rightarrow 0}\frac{\left( x+hk\right)
^{T}A\left( x+hk\right) -x^{T}Ax}{k}=\\
& = & \lim _{k\rightarrow 0}\frac{x^{T}A\left( hk\right) +\left( hk\right)
^{T}Ax+\left( hk\right) ^{T}A\left( hk\right) }{k}=x^{T}Ah+h^{T}Ax\\
dh_{a}\left( B\right) h & = & d\left( e^{x}\right) \left( a^{T}Aa\right) \circ d\left(
x^{T}Ax\right) \left( a\right) h=e^{a^{T}Aa}\left( a^{T}Ah+h^{T}Aa\right) \\
&
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Für die Polarkoordinatenabbildung
\begin_inset LatexCommand \index{Polarkoordinaten, Abbildung}
\end_inset
\begin_inset Formula \[
P_{3}:\mathbb R^{3}\rightarrow \mathbb R^{3},\, P_{3}\left( \begin{array}{c}
r\\
\varphi _{1}\\
\varphi _{2}
\end{array}\right) :=\left( \begin{array}{c}
r\, cos\varphi _{1}\, cos\, \varphi _{2}\\
r\, sin\varphi _{1}\, cos\, \varphi _{2}\\
r\, sin\, \varphi _{2}
\end{array}\right) \]
\end_inset
berechne man die Ableitung
\begin_inset Formula \( P_{3}', \)
\end_inset
das Differential
\begin_inset Formula \( dP_{3} \)
\end_inset
und die Determinante
\begin_inset Formula \( det\, P_{3}' \)
\end_inset
.
\layout Standard
\begin_inset Formula \begin{eqnarray*}
P_{3}' & = & \left( \begin{array}{ccc}
cos\varphi _{1}\, cos\, \varphi _{2} & -r\, sin\, \varphi _{1}\, cos\, \varphi _{2} &
-r\, cos\, \varphi _{1}\, sin\, \varphi _{2}\\
sin\varphi _{1}\, cos\, \varphi _{2} & r\, cos\, \varphi _{1}\, cos\, \varphi _{2} &
-r\, cos\, \varphi _{1}\, sin\, \varphi _{2}\\
sin\, \varphi _{2} & 0 & r\, cos\, \varphi _{2}
\end{array}\right) \\
dP_{3}'h & = & P_{3}'h\\
det\, P_{3}' & = & r^{2}\, cos\, \varphi _{2}
\end{eqnarray*}
\end_inset
\layout Standard
\begin_float margin
\layout Standard
\align center \pextra_type 3
\begin_inset Figure size 89 168
file G1.eps
width 3 15.00
height 3 20.00
flags 11
\end_inset
\layout Standard
\align center \pextra_type 3 \pextra_widthp 50
\begin_inset Figure size 89 168
file G2.eps
width 3 15.00
height 3 20.00
flags 11
\end_inset
\layout Standard
\noindent \align center \pextra_type 3 \pextra_widthp 40
\begin_inset Figure size 89 168
file G3.eps
width 3 15.00
height 3 20.00
flags 11
\end_inset
\end_float
\layout Problem
\line_top
Gegeben sei die Funktion
\begin_inset Formula \[
f:\mathbb R^{2}\rightarrow \mathbb R^{2};\, \left( x,y\right) \rightarrow \left(
x^{2}-y^{2},2xy\right) .\]
\end_inset
Außerdem seien die Scharen parametrisierter Kurven
\begin_inset Formula \begin{eqnarray*}
\gamma _{1}^{\left( a\right) }:\mathbb R\rightarrow \mathbb R^{2} & ; & t\mapsto
\left( a,t\right) \\
\gamma _{2}^{\left( b\right) }:\mathbb R\rightarrow \mathbb R^{2} & ; & t\mapsto
\left( t,b\right) \, und\\
\gamma _{3}^{\left( r\right) }:\mathbb R\rightarrow \mathbb R^{2} & ; & t\mapsto
\left( r\, cos\, t,r\, sin\, t\right)
\end{eqnarray*}
\end_inset
\begin_inset Formula \( \left( a,b,r\in \mathbb R,r>0\right) \)
\end_inset
im Urbildraum von f gegeben.
\begin_deeper
\layout Enumerate
\pextra_type 3 \pextra_widthp 50
Man skizziere die Bilder der Kurven
\begin_inset Formula \( f\circ \gamma _{1}^{\left( a\right) },f\circ \gamma
_{2}^{\left( b\right) }\, und\, f\circ \gamma _{3}^{\left( r\right) } \)
\end_inset
für einige Werte a,b,r.
\layout Enumerate
\pextra_type 3 \pextra_widthp 50
Man berechne die Ableitung und das Differential von f im Punkt
\begin_inset Formula \( \left( a,b\right) \)
\end_inset
.
Man bermerke, daß
\begin_inset Formula \( f'\left( a,b\right) \)
\end_inset
eine Ähnlichkeitsmatrix (d.h.
\begin_inset Formula \( c\cdot A \)
\end_inset
mit einer orthogonalen Matrix A und
\begin_inset Formula \( c\in \mathbb R\smallsetminus \left\{ 0\right\} \)
\end_inset
) ist.
\layout Enumerate
\pextra_type 3 \pextra_widthp 50
Man berechne die Bilder der Tangentialvektoren
\begin_inset Formula \( v_{1}:=\left( \gamma _{1}^{\left( a\right) }\right) '\left(
b\right) \)
\end_inset
und
\begin_inset Formula \( v_{2}:=\left( \gamma _{2}^{\left( b\right) }\right) '\left(
a\right) \)
\end_inset
im Punkt
\begin_inset Formula \( \left( a,b\right) \)
\end_inset
unter
\begin_inset Formula \( df \)
\end_inset
.
Welcher Winkel liegt zwischen den Tangentialvektoren?
\layout Enumerate
\pextra_type 3 \pextra_widthp 50
Für zwei Kurven
\begin_inset Formula \( \alpha \, \beta :\left( -\varepsilon ;\varepsilon \right)
\rightarrow \mathbb R^{n} \)
\end_inset
mit
\begin_inset Formula \( \alpha \left( 0\right) =\beta \left( 0\right) \neq \left(
0,0\right) \)
\end_inset
beweise man:
\lang american
\lang german
\begin_inset Formula \begin{eqnarray*}
\frac{\left\langle df_{\left( a,b\right) }\alpha '\left( 0\right) ,df_{\left(
a,b\right) }\beta '\left( 0\right) \right\rangle }{\left\Vert df_{\left( a,b\right)
}\alpha '\left( 0\right) \right\Vert \cdot \left\Vert df_{\left( a,b\right) }\beta
'\left( 0\right) \right\Vert } & = & \frac{\left\langle \alpha '\left( 0\right) ,\beta
'\left( 0\right) \right\rangle }{\left\Vert \alpha '\left( 0\right) \right\Vert \cdot
\left\Vert \beta '\left( 0\right) \right\Vert },\\
& &
\end{eqnarray*}
\end_inset
wobei
\begin_inset Formula \( \left\langle \cdot ,\cdot \right\rangle \)
\end_inset
das euklidsche Skalarprodukt und
\begin_inset Formula \( \left\Vert \cdot \right\Vert \)
\end_inset
die euklidsche Norm auf
\begin_inset Formula \( \)
\end_inset
bedeuten.
\end_deeper
\layout Standard
\begin_inset Formula \begin{eqnarray*}
df_{\left( a,b\right) }\left( x,y\right) & = & \left( \begin{array}{cc}
2a & -2b\\
2b & 2a
\end{array}\right) \left( \begin{array}{c}
x\\
y
\end{array}\right) \\
df_{\left( 1,0\right) }\left( x,y\right) & = & \left( \begin{array}{cc}
2 & 0\\
0 & 2
\end{array}\right) \left( \begin{array}{c}
x\\
y
\end{array}\right) =2\left( \begin{array}{c}
x\\
y
\end{array}\right) \\
f'(a,b) & = & \left( \begin{array}{cc}
2a & -2b\\
2b & 2a
\end{array}\right) =2\sqrt{a^{2}+b^{2}}\left( \begin{array}{cc}
\frac{a}{\sqrt{a^{2}+b^{2}}} & \frac{-b}{\sqrt{a^{2}+b^{2}}}\\
\frac{b}{\sqrt{a^{2}+b^{2}}} & \frac{a}{\sqrt{a^{2}+b^{2}}}
\end{array}\right) =\\
& = & 2\sqrt{a^{2}+b^{2}}\left( \begin{array}{cc}
cos\, \varphi & -sin\, \varphi \\
sin\, \varphi & cos\, \varphi
\end{array}\right) \in 2\sqrt{a^{2}+b^{2}}\cdot EO\left( 2\right) \\
& &
\end{eqnarray*}
\end_inset
\layout Standard
\begin_inset Formula \( f'_{\left( a,b\right) }=A \)
\end_inset
hat Eigenschaft
\begin_inset Formula \( A^{T}A=c\cdot \mathbb 1 \)
\end_inset
mit geeignetem
\begin_inset Formula \( c=4\left( a^{2}+b^{2}\right) >0. \)
\end_inset
\layout Standard
\begin_inset Formula \begin{eqnarray*}
\alpha '\left( 0\right) =:v & ; & \beta '\left( 0\right) =:w\\
\frac{\left\langle Av,Aw\right\rangle }{\left\Vert Av\right\Vert \cdot \left\Vert
Aw\right\Vert } & \begin{array}{c}
!\\
=\\
\end{array} & \frac{\left\langle v,w\right\rangle }{\left\Vert v\right\Vert \cdot
\left\Vert w\right\Vert }\Leftrightarrow \\
\Leftrightarrow \frac{v^{T}A^{T}Aw}{\sqrt{v^{T}A^{T}Av}\cdot \sqrt{w^{T}A^{T}Aw}} & =
& \frac{v^{T}w}{\sqrt{v^{T}v}\sqrt{w^{T}w}}\Leftrightarrow \\
\Leftrightarrow \frac{c\cdot v^{T}w}{c\cdot \sqrt{v^{T}v}\sqrt{w^{T}w}} & = &
\frac{v^{T}w}{\sqrt{v^{T}v}\sqrt{w^{T}w}}\, \, \, \surd
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Die Joukowski-Abbildung
\begin_inset Formula \( f:\mathbb C\smallsetminus \mathbb C,\, z\mapsto
\frac{1}{2}\left( z+\frac{1}{z}\right) \)
\end_inset
je3des der beiden Gebiete
\begin_inset Formula \( D_{1}:=\left\{ z\in \mathbb C:\left| z\right| <1\right\} \)
\end_inset
bijektiv und konform auf
\begin_inset Formula \( G:=\mathbb C\smallsetminus \left\{ t\in \mathbb R:-1\leqq
t\leqq 1\right\} \)
\end_inset
ab.
\newline
Man zeige: Jede der beiden Einschränkungen
\begin_inset Formula \( f|_{D_{1}} \)
\end_inset
und
\begin_inset Formula \( f_{D_{2}} \)
\end_inset
ist sogar ein Diffeomorphismus von
\begin_inset Formula \( D_{1} \)
\end_inset
bzw.
\begin_inset Formula \( D_{2} \)
\end_inset
auf G und die inversen Abbildungen sind auch wieder konform.
\layout Standard
\begin_inset Formula \begin{eqnarray*}
\frac{1}{2}\left( z+\frac{1}{z}\right) & = & w\\
z^{2}-2wz+1 & = & 0\\
z_{1|2} & = & w\pm \sqrt{w^{2}-1}
\end{eqnarray*}
\end_inset
Bekannt: f bijektiv (injektiv würde genügen); konform (stetig würde genügen)
\newline
\bar under
Satz von der Diffeomorphie:
\bar default
f differenzierbar genau dann, wenn
\begin_inset Formula \( f' \)
\end_inset
nicht singulär, d.h.
kein Eigenwert = 0, d.
h.
f' invertierbar.
\layout Standard
\begin_inset Formula \begin{eqnarray*}
f'\left( x,y\right) & = & \left( \begin{array}{cc}
a & -b\\
b & a
\end{array}\right) \\
a & = & \frac{1}{2}+\frac{y^{2}-x^{2}}{2\left( x^{2}-y^{2}\right) }\\
b & = & \frac{xy}{\left( x^{2}+y^{2}\right) ^{2}}
\end{eqnarray*}
\end_inset
\layout Standard
\begin_inset Formula \begin{eqnarray*}
f'\left( x,y\right) \, invertierbar & \Leftrightarrow & det\, f'\left( x,y\right)
=a^{2}+b^{2}\neq 0\Leftrightarrow \\
\Leftrightarrow \left( a,b\right) \neq \left( 0,0\right) & \Leftrightarrow & \left\{
nicht\, y=0\left( \rightarrow b=0\right) \, und\, nicht\, a=0\Leftrightarrow
x^{4}-x^{2}=0\right\} \Leftrightarrow \\
\Leftrightarrow \left\{ nicht\, y=0\, und\, x\in \left\{ -1,0,1\right\} \right\} &
\Rightarrow & f'\, \ddot{u}berall\, auf\, D_{1}\, und\, D_{2}\, invertierbar\\
f' & = & \left( \begin{array}{cc}
a & b\\
-b & a
\end{array}\right) \\
\left( f'\right) ^{-1} & = & \frac{1}{a^{2}+b^{2}}\left( \begin{array}{cc}
a & b\\
-b & a
\end{array}\right) ;f\, konform
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Gegeben sei das Gleichungssystem
\begin_inset Formula \begin{eqnarray*}
f_{1}\left( x,y_{1},y_{2}\right) & = & x^{3}+y_{1}^{3}+y_{2}^{3}-7=0\\
f_{2}\left( x,y_{1},y_{2}\right) & = & xy_{1}+y_{1}y_{2}+y_{2}x+2=0
\end{eqnarray*}
\end_inset
und die Nullstelle
\begin_inset Formula \( \left( 2,-1,0\right) \)
\end_inset
.
Man zeige, daß das Gleichungssystem in der Nähe dieser Nullstelle nach
\begin_inset Formula \( y_{1} \)
\end_inset
und
\begin_inset Formula \( y_{2} \)
\end_inset
aufgelöst werden kann, d.h.
lokal um
\begin_inset Formula \( \left( 2,-1,0\right) \)
\end_inset
gilt für alle Lösungen des Gleichungssystems:
\begin_inset Formula \[
\]
\end_inset
Man berechne
\begin_inset Formula \( g'\left( 2\right) . \)
\end_inset
\layout Standard
\begin_inset Formula \begin{eqnarray*}
F\left( x,g_{1}\left( x\right) ,g_{2}\left( x\right) \right) & = & \left(
\begin{array}{c}
f_{1}\\
f_{2}
\end{array}\right) \left( x,g_{1}\left( x\right) ,g_{2}\left( x\right) \right) =\\
& = & \left( \begin{array}{c}
x^{3}+y_{1}^{3}+y_{2}^{3}-7\\
xy_{1}+y_{1}y_{2}+y_{2}x+2
\end{array}\right) \left( x,g_{1}\left( x\right) ,g_{2}\left( x\right) \right) =0\\
&
\end{eqnarray*}
\end_inset
Kurve darf nicht senkrecht auf der y-Achse stehen
\begin_inset Formula \( \Rightarrow \)
\end_inset
x-Achse
\begin_inset Formula \( \in \)
\end_inset
Ebene der beiden Normalenvektoren (=Gradienten).
\begin_inset Formula \begin{eqnarray*}
0=F_{1}'\left( x\right) & = & \partial _{1}f_{1}\left( x,g_{1}\left( x\right)
,g_{2}\left( x\right) \right) \cdot 1+\partial _{2}f_{1}\cdot g_{1}'\left( x\right)
+\partial _{3}f_{2}\cdot g_{2}'\left( x\right) \\
0=F_{2}'\left( x\right) & = & \partial _{1}f_{2}+d_{2}f_{2}\cdot g_{1}'\left(
x\right) +\partial _{3}f_{2}\cdot g_{2}'\left( x\right) \\
f\ddot{u}r\, x=2: & & g_{1}\left( x\right) =-1;\, g_{2}\left( x\right) =0\\
\partial _{1}f_{1}\left( 2,-1,0\right) =12 & & \partial _{1}f_{2}\left( 2,-1,0\right)
=-1\\
\partial _{2}f_{1}\left( 2,-1,0\right) =3 & \times & \partial _{2}f_{2}\left(
2,-1,0\right) =2\\
\partial _{3}f_{1}\left( 2,-1,0\right) =0 & & \partial _{3}f_{2}\left( 2,-1,0\right)
=1\\
F_{1}'\left( 2\right) & = & 12+3g_{1}'\left( 2\right) +0=0\Rightarrow g_{1}'\left(
2\right) =-4\\
F_{2}'\left( 2\right) & = & -1+2g_{1}'\left( 2\right) +g_{2}'\left( 2\right)
=0\Rightarrow g_{2}\left( 2\right) =9\\
\Rightarrow & & g'\left( 2\right) =\left( \begin{array}{c}
-4\\
9
\end{array}\right)
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Die Gleichung
\begin_inset Formula \[
z^{3}+z+xy=2\]
\end_inset
hat für jedes
\begin_inset Formula \( \left( x,y\right) \in \mathbb R^{2} \)
\end_inset
genau eine reelle Lösung
\begin_inset Formula \( z=g\left( x,y\right) \)
\end_inset
.
Man zeige, daß
\begin_inset Formula \( g:\mathbb R^{2}\rightarrow \mathbb R \)
\end_inset
differenzierbar ist, und berechne
\begin_inset Formula \( g'\left( 1,2\right) \)
\end_inset
.
Man untersuche g auf Extrema.
\layout Standard
\begin_inset Formula \begin{eqnarray*}
\nabla f & = & \left( \begin{array}{c}
y\\
x\\
1+3z^{2}
\end{array}\right) ;\, z-Komponente>0\\
& & \Rightarrow lokal\, differenzierbar\, und\, nach\, z\, aufl\ddot{o}sbar\\
& & \, \, \left( Satz\, \ddot{u}ber\, implizite\, Funktionen\right) \\
z & \mapsto & z^{3}+z\, ist\, bijektiv\\
& \Rightarrow & global\, differenzierbar\, h\ddot{a}ngt\, z\, von\, x,y\, ab\\
\left[ d_{\left( x,y\right) }f\right] v & = & \left( y,x\right) \cdot v\, \left( v\in
\mathbb R^{2}\right) \\
\left[ d_{\left( z\right) }f\right] w & = & \left( 3z^{2}+1\right) \cdot w\, \left(
v\in \mathbb R\right) \\
&
\end{eqnarray*}
\end_inset
Formel für g'
\begin_inset Formula \( \left( x,y\right) \)
\end_inset
:
\begin_inset Formula \begin{eqnarray*}
dg_{\left( x,y\right) }v & = & -\left[ d_{\left( z\right) }f\left( x,y,z\right)
\right] ^{-1}\cdot \left[ d_{\left( x,y\right) }f\left( x,y,z\right) \right] v=\\
& = & -\frac{1}{3z^{2}+1}\left( y,x\right) v
\end{eqnarray*}
\end_inset
Speziell für
\begin_inset Formula \( \left( x,y\right) =\left( 1,2\right) \)
\end_inset
:
\begin_inset Formula \begin{eqnarray*}
z^{3}+z+2 & = & 2\, \Rightarrow z=0\\
dg_{\left( 1,2\right) }v & = & -\frac{1}{3\cdot 0^{2}+1}\left( 2,1\right) v=\left(
-2,-1\right) v;\, v\in \mathbb R^{2}
\end{eqnarray*}
\end_inset
Extrema:
\begin_inset Formula \begin{eqnarray*}
\left( x,y\right) & = & \left( 0,0\right) \, einziger\, Kandidat\\
F\left( x,y\right) & := & f\left( x,y,g\left( x,y\right) \right) =2\\
0 & = & \partial _{x}F\left( x,y\right) =\partial _{1}f\cdot 1+\partial _{2}f\cdot
0+\partial _{3}f\cdot g_{x}\left( x,y\right) \\
0 & = & \partial _{y}F\left( x,y\right) =\partial _{1}f\cdot 0+\partial _{2}f\cdot
1+\partial _{3}f\cdot g_{y}\left( x,y\right) \\
0 & = & 1y+0x+\left( 3z^{2}+1\right) g_{x}\left( x,y\right) \\
0 & = & 0y+1x+\left( 3z^{2}+1\right) g_{y}\left( x,y\right) \\
g_{x} & = & g_{y}=0\Rightarrow x=y=0;\, z^{3}+z+0=0\Rightarrow z=1\\
0 & = & \partial _{x}^{2}F\left( x,y\right) =\\
& = & \partial _{1}^{2}f+\partial _{2}\partial _{1}f+\partial _{3}\partial _{1}f\cdot
g_{x}\left( x,y\right) +\partial _{1}\partial _{3}f\cdot g_{x}\left( x,y\right)
+\partial _{2}\partial _{3}f\cdot 0\cdot g_{x}\left( x,y\right) +\\
& & +\partial _{3}\partial _{3}f\cdot \left[ g_{x}\left( x,y\right) \right]
^{2}+\partial _{3}f\cdot g_{xx}\left( x,y\right) \\
\partial _{xy}f\left( 0,0\right) & = & 0\\
\partial _{3}f\left( 0,0,1\right) & = & 4\\
\partial _{1}^{2} & = & 0\\
g_{xx}\left( 0,0\right) & = & 0\\
aus\, Symmetrie: & & g_{yy}\left( 0,0\right) =0\\
0 & = & \partial _{y}\partial _{x}F=\\
& = & \partial _{1}^{2}\cdot 0+\partial _{2}\partial _{1}f+\partial _{3}\partial
_{1}f\cdot g_{y}+\left[ \partial _{1}\partial _{3}f+\partial _{2}\partial
_{3}f+\partial _{3}\partial _{3}f\cdot g_{y}\right] g_{x}+\partial _{3}f\cdot g_{xy}\\
0 & = & 4g_{xy}\left( 0,0\right) +1\\
g_{xy}\left( 0,0\right) & = & -\frac{1}{4}\\
g''\left( 0,0\right) & = & \left( \begin{array}{cc}
0 & -\frac{1}{4}\\
-\frac{1}{4} & 0
\end{array}\right) \\
det\, g''\left( 0,0\right) & = & -\frac{1}{16}\Rightarrow g''\, indefinit\Rightarrow
\left( 0,0\right) \, ist\, Sattelpunkt
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Sei
\begin_inset Formula \( f:\mathbb R^{3}\rightarrow \mathbb R \)
\end_inset
eine
\begin_inset Formula \( \)
\end_inset
-Funktion und
\begin_inset Formula \( A=\left( p_{0},T_{0},V_{0}\right) \)
\end_inset
eine Nullstelle von f mit
\begin_inset Formula \( \partial _{p}f\left( A\right) \neq 0 \)
\end_inset
,
\begin_inset Formula \( \partial _{T}f\left( A\right) \neq 0 \)
\end_inset
und
\begin_inset Formula \( \partial _{V}f\left( A\right) \neq 0. \)
\end_inset
Thermodynamiker verwenden die Relation
\begin_inset Formula \[
\frac{\partial p}{\partial T}\cdot \frac{\partial T}{\partial V}\cdot \frac{\partial
V}{\partial p}=-1.\]
\end_inset
Wie ist eine solche Relation zu interpretieren?
\layout Standard
\begin_inset Formula \begin{eqnarray*}
\partial _{p}f\left( p_{0},V_{0},T_{0}\right) & \neq & 0\\
& &
\end{eqnarray*}
\end_inset
\begin_inset Formula \( \Rightarrow \)
\end_inset
Man kann f in a lokal nach
\begin_inset Formula \( p_{1} \)
\end_inset
auflösen und
\begin_inset Formula \( dp\left( T_{0},V_{0}\right) =-\left[ d_{\left( p\right)
}f\left( p_{0},T_{0},V_{0}\right) \right] ^{-1}\cdot d_{\left( T,V\right) }f\left(
p_{0},T_{0},V_{0}\right) \)
\end_inset
.
\newline
Hier ist
\begin_inset Formula \( d_{\left( p\right) }f\left( p_{0},T_{0},V_{0}\right) :\mathbb
R\rightarrow \mathbb R;\, h\mapsto \partial _{p}f\left( p_{0},T_{0},V_{0}\right) \cdot
h. \)
\end_inset
\newline
\begin_inset Formula \begin{eqnarray*}
d_{\left( T,V\right) }f\left( p_{0},T_{0},V_{0}\right) : & \mathbb R^{2} & \Rightarrow
\mathbb R\\
& k & \mapsto \left( \partial _{T}f,\partial _{V}f\right) _{\left(
p_{0},T_{0},V_{0}\right) }\cdot k
\end{eqnarray*}
\end_inset
Speziell:
\begin_inset Formula \begin{eqnarray*}
\partial _{T}p\left( T_{0},V_{0}\right) & = & -\frac{\partial _{V}f\left(
p_{0},T_{0},V_{0}\right) }{\partial _{p}f\left( p_{0},T_{0},V_{0}\right) }\\
genauso:\partial _{V}p\left( T_{0},V_{0}\right) & = & -\frac{\partial _{V}f\left(
p_{0},T_{0},V_{0}\right) }{\partial _{T}f\left( p_{0},T_{0},V_{0}\right) }\\
\partial _{p}p\left( T_{0},V_{0}\right) & = & -\frac{\partial _{p}f\left(
p_{0},T_{0},V_{0}\right) }{\partial _{V}f\left( p_{0},T_{0},V_{0}\right) }\\
& \Rightarrow & \partial _{T}p\cdot \partial _{V}T\cdot \partial _{p}V=-1
\end{eqnarray*}
\end_inset
ideales Gas:
\begin_inset Formula \begin{eqnarray*}
f\left( p,T,V\right) & = & \frac{pV}{T}-c\\
p\left( T,V\right) & = & \frac{cT}{V};\, \partial _{R}p\left(
p_{0},T_{0},V_{0}\right) =\frac{c}{V_{0}}
\end{eqnarray*}
\end_inset
Tatsächlich ist auch
\begin_inset Formula \[
-\frac{\partial _{T}f}{\partial
_{p}f}=-\frac{-\frac{pV_{0}}{T_{0}^{2}}}{\frac{V_{0}}{T}}=\frac{p_{0}}{T_{0}}=\frac{c}{V_{0}}\]
\end_inset
\layout Problem
\line_top
Es sei M und N Untermannigfalten des
\begin_inset Formula \( \mathbb R^{m} \)
\end_inset
bzw.
\begin_inset Formula \( \mathbb R^{n} \)
\end_inset
.
Dann ist das direkte Produkt
\begin_inset Formula \( M\times N \)
\end_inset
eine Untermannigfaltigkeit des
\begin_inset Formula \( \mathbb R^{m\times n} \)
\end_inset
der Dimension
\begin_inset Formula \( dim\, M+dim\, N \)
\end_inset
.
\layout Standard
Offenbar
\begin_inset Formula \( M\times N\subseteq \mathbb R^{m+n} \)
\end_inset
.
Bezeichne
\begin_inset Formula \( d|_{M}:=dim\, M;\, d|_{N}:=dim\, N. \)
\end_inset
Sei
\begin_inset Formula \( \left( a,b\right) \in M\times N\Rightarrow \exists
Umgebungen\, U\subseteq \mathbb R^{m}\, von\, a;\, V\subseteq \mathbb R^{n}\, von\, b
\)
\end_inset
und Diffeomorphismous
\begin_inset Formula \( \varphi ,\psi ;\, \varphi :U\rightarrow \hat{U}\in \mathbb
R^{m};\, \psi :V\rightarrow \hat{V}\in \mathbb R^{n} \)
\end_inset
mit
\begin_inset Formula \( \varphi \left( M\cap U\right) =\mathbb R_{0}^{d_{M}}\cap
\hat{U} \)
\end_inset
und
\begin_inset Formula \( \varphi \left( N\cap V\right) =\mathbb R_{0}^{d_{N}}\cap
\hat{V} \)
\end_inset
.
Dann ist
\begin_inset Formula \( \eta :U\times V\rightarrow \hat{U}\times \hat{V};\, \left(
p,y\right) \mapsto \left( \varphi \left( p\right) ,\psi \left( p\right) \right) \)
\end_inset
ein Diffeomorphismus
\begin_inset Formula \( \eta '\left( p,q\right) =\left( \begin{array}{cc}
\varphi '_{\left( p\right) }0 & \\
0 & \psi '_{\left( q\right) }
\end{array}\right) ;\, \eta \left( \left( M\times N\right) \cap \left( U\times
V\right) \right) =\left( \mathbb R_{0}^{d_{M}}\cap \hat{U}\right) \times \left(
\mathbb R_{0}^{d_{N}}\cap \hat{V}\right) \subset \mathbb R^{m+n} \)
\end_inset
\layout Problem
\line_top
Gegeben sei ein Gelenk
\begin_inset LatexCommand \index{Gelenk}
\end_inset
in der Ebene, bestehend aus zwei Stangen
\begin_inset LatexCommand \index{Stangen}
\end_inset
der Längen
\begin_inset Formula \( l_{1},l_{2}. \)
\end_inset
Man interpretiere die Menge aller möglichen Positionen
\begin_inset Formula \( \left( a_{1},a_{2},a_{3}\right) \)
\end_inset
der Eckpunkte als eine 4-dimensionale Untermannigfaltigkeit M des
\begin_inset Formula \( \mathbb R^{6}. \)
\end_inset
Man finde eine geeignete Parametrisierung
\begin_inset Formula \( \mathbb R^{2}\times \mathbb S^{1}\times \mathbb
S^{1}\rightarrow M \)
\end_inset
!
\layout Standard
Nullstellengebilde von
\begin_inset Formula \( f:\mathbb R^{6}\rightarrow \mathbb R^{2};\, \left(
\begin{array}{c}
x_{1}\\
y_{1}\\
x_{2}\\
y_{2}\\
x_{3}\\
y_{3}
\end{array}\right) \mapsto \left( \begin{array}{c}
\left( x_{1}-x_{2}\right) ^{2}+\left( y_{1}-y_{2}\right) ^{2}-l_{1}^{2}\\
\left( x_{2}-x_{3}\right) ^{2}+\left( y_{2}-y_{3}\right) ^{2}-l_{2}^{2}
\end{array}\right) \)
\end_inset
.
f ist
\begin_inset Formula \( \)
\end_inset
-Funktion.
\begin_inset Formula \[
f'\left( x_{1},y_{1},x_{2},y_{2},x_{3},y_{3}\right) ^{T}=2\left( \begin{array}{cccccc}
x_{1}-x_{2} & y_{1}-y_{2} & x_{2}-x_{1} & y_{2}-y_{1} & 0 & 0\\
0 & 0 & x_{2}-x_{3} & y_{2}-y_{3} & x_{3}-x_{2} & y_{3}-y_{2}
\end{array}\right) \]
\end_inset
\begin_inset Formula \( \Rightarrow f'\left( p\right) \, regul\ddot{a}r\, f\ddot{u}r\,
alle\, p\in M\Rightarrow M\, Mannigfaltigkeit\, der\, Dimension\, 6-2=4. \)
\end_inset
\layout Problem
\line_top
Rotationsflächen im
\begin_inset Formula \( \mathbb R^{3}. \)
\end_inset
\newline
Es sei
\begin_inset Formula \( M=f^{-1}\left( 0\right) \)
\end_inset
eine 1-dimensionale Untermannigfaltigkeit des
\begin_inset Formula \( \mathbb R^{2} \)
\end_inset
, wobei
\begin_inset Formula \( f:\mathbb R_{+}\times \mathbb R\rightarrow \mathbb R \)
\end_inset
eine stetig differenzierbare Funktion sei mit 0 als regulärem Wert.
Man zeige:
\begin_inset Formula \[
R:=\left\{ \left( x,y,z\right) \in \mathbb R^{3}|f\left( \sqrt{x^{2}-y^{2}},z\right)
=0\right\} \]
\end_inset
ist eine 2-dimensionale Untermannigfaltigkeit des
\begin_inset Formula \( \mathbb R^{3}. \)
\end_inset
\layout Standard
\begin_inset Formula \begin{eqnarray*}
g & : & \mathbb R^{3}\smallsetminus \left\{ \left( 0,0\right) \right\} \times \mathbb
R\rightarrow \mathbb R^{+}\times \mathbb R\\
\left( \begin{array}{c}
x\\
y\\
z
\end{array}\right) & \mapsto & \left( \begin{array}{c}
\sqrt{x^{2}+y^{2}}\\
z
\end{array}\right)
\end{eqnarray*}
\end_inset
Dann ist
\begin_inset Formula \( R=\left( f\circ g\right) ^{-1}\left( 0\right) \)
\end_inset
.
\newline
Zu zeigen: 0 ist regulärer Wert von
\begin_inset Formula \( f\circ g \)
\end_inset
.
\begin_inset Formula \begin{eqnarray*}
\left( f\circ g\right) '\left( x,y,z\right) & = & \left( f_{x},f_{y}\right) _{\left(
\begin{array}{c}
x\\
y\\
z
\end{array}\right) }\circ \left( \begin{array}{ccc}
\frac{x}{\sqrt{x^{2}+y^{2}}} & \frac{y}{\sqrt{x^{2}+y^{2}}} & 0\\
0 & 0 & 1
\end{array}\right) =\\
& = & \left(
f_{x}\frac{x}{\sqrt{x^{2}+y^{2}}},f_{x}\frac{y}{\sqrt{x^{2}+y^{2}}},f_{y}\right)
\end{eqnarray*}
\end_inset
\begin_inset Formula \( f_{x},f_{y} \)
\end_inset
nicht beide 0 (regulärer Wert)
\begin_inset Formula \( \Rightarrow \)
\end_inset
mindestens ein Eintrag nicht 0
\begin_inset Formula \( \Rightarrow \)
\end_inset
für
\begin_inset Formula \( \left( \begin{array}{c}
x\\
y\\
z
\end{array}\right) \in \mathbb R \)
\end_inset
ist
\begin_inset Formula \( \left( f\circ g\right) '_{\left( \begin{array}{c}
x\\
y\\
z
\end{array}\right) } \)
\end_inset
surjektiv
\begin_inset Formula \( \Rightarrow \)
\end_inset
R Mannigfaltigkeit der Dimension 3-1=2
\layout Problem
\line_top
Man zeige, daß jeder Punkt
\begin_inset Formula \( \left( p,q\right) \in \mathbb R^{2} \)
\end_inset
mit
\begin_inset Formula \( pq\neq 0 \)
\end_inset
auf genau zwei Parabeln
\begin_inset Formula \( P_{a_{1}},P_{a_{2}} \)
\end_inset
der Schar
\begin_inset Formula \[
P_{a}:=\left\{ \left( x,y\right) \in \mathbb R^{2}:ax^{2}-y-\frac{1}{4a}=0\right\} \]
\end_inset
liegt und daß sich diese beiden Parabeln im Punkt
\begin_inset Formula \( \left( p,q\right) \)
\end_inset
senkrecht schneiden.
\layout Standard
\begin_inset Formula \begin{eqnarray*}
ap^{2}-q-\frac{1}{4a} & = & 0\\
4a^{2}p^{2}-4aq-1 & = & 0\\
a_{1|2} & = & \frac{4q\pm \sqrt{16q^{2}+4\cdot 4p^{2}}}{2\cdot
4p^{2}}=\frac{q}{2p^{2}}\pm \frac{\sqrt{p^{2}+q^{2}}}{2p^{2}}
\end{eqnarray*}
\end_inset
Da
\begin_inset Formula \( pq\neq 0 \)
\end_inset
existieren zwei Lösungen.
Diese Schneiden sich in
\begin_inset Formula \( \left( p,q\right) \)
\end_inset
.
\layout Problem
\line_top
\SpecialChar ~
\layout Enumerate
Man formuliere den Satz vom regulären Wert für den Spezialfall einer Hyperfläche
(d.h.
Untermannigfaltigkeit der Dimension n-1) im
\begin_inset Formula \( \mathbb R^{n} \)
\end_inset
.
\layout Enumerate
Ist
\begin_inset Formula \( M:=\left\{ \left( x,y\right) \in \mathbb R^{2}:x^{2}=0\right\}
\)
\end_inset
eine Untermannigfaltigkeit der
\begin_inset Formula \( \mathbb R^{2} \)
\end_inset
?
\layout Enumerate
Die Neilsche Parabel:
\newline
Es sei
\begin_inset Formula \( N:=\left\{ \left( x,y\right) \in \mathbb
R^{2}:x^{3}=y^{2}\right\} \)
\end_inset
.
Man zeige: Für
\begin_inset Formula \( A\in N,\, a\neq \left( 0,0\right) \)
\end_inset
, hat
\begin_inset Formula \( T_{a}N \)
\end_inset
die Dimension 1, dagegen hat
\begin_inset Formula \( T_{\left( 0,0\right) }N \)
\end_inset
die Dimension 0.
Man folgere, daß N keine Untermannigfaltigkeit des
\begin_inset Formula \( \mathbb R^{2} \)
\end_inset
ist.
\layout Problem
\line_top
\SpecialChar ~
\layout Enumerate
Man bestimme alle Extrema der Funktion
\begin_inset Formula \( f:\mathbb R^{3}\rightarrow \mathbb R;\, \left( x,y,z\right)
\mapsto xyz \)
\end_inset
auf der Sphäre
\begin_inset Formula \( \mathbb S^{2}. \)
\end_inset
\layout Enumerate
Gegeben sei die Funktion
\begin_inset Formula \[
f:\mathbb R^{3}\rightarrow \mathbb R;\, \left( x,y,z\right) \mapsto
z+\frac{1}{3}\left( x^{2}+y^{2}\right) .\]
\end_inset
Man zeige, daß f auf der Sphäre
\begin_inset Formula \( \mathbb S^{2} \)
\end_inset
im Punkt p=
\begin_inset Formula \( \left( 0,0,1\right) \)
\end_inset
ein Maximum hat, daß aber die Einschränkung des Differenzials zweiter Ordnung
auf den Tangentialraum
\begin_inset Formula \( T_{p}M \)
\end_inset
positiv definit ist.
Die Einschränkung der Hessematrix auf den Tangentialraum liefert also kein
Kriterium dafür, ob an einem kritischen Punkt ein Maximum oder Minimum
oder Sattelpunkt vorliegt.
\layout Standard
\SpecialChar ~
\layout Standard
\begin_inset Formula \begin{eqnarray*}
\nabla f\left( x,y,z\right) & = & \left( \begin{array}{c}
yz\\
xz\\
xy
\end{array}\right) \\
\nabla g_{\mathbb S}\left( x,y,z\right) & = & 2\left( \begin{array}{c}
x\\
y\\
z
\end{array}\right) \\
Lagrange: & & \\
Fall\, 1 & : & xyz\neq 0\\
\left( \begin{array}{c}
yz\\
xz\\
xy
\end{array}\right) & = & \lambda \left( \begin{array}{c}
x\\
y\\
z
\end{array}\right) ;\, \lambda \in \mathbb R\\
\Rightarrow \lambda & = & \frac{yz}{x}=\frac{xz}{y}=\frac{xy}{z}\Rightarrow \\
\Rightarrow \lambda xyz & = & y^{2}z^{2}=x^{2}z^{2}=x^{2}y^{2}\Rightarrow \\
& \Rightarrow & x^{2}=y^{2}=z^{2}=\frac{1}{3}\\
& \Rightarrow & \left| x\right| =\left| y\right| =\left| z\right|
=\frac{\sqrt{3}}{3}\\
Fall\, 2 & : & Sonderf\ddot{a}lle\, xyz=0\\
f\left( 0,0,\pm 1\right) & = & 0\\
Wegen\, f\left( \varepsilon ,\varepsilon ,\sqrt{1-2\varepsilon ^{2}}\right) & > & 0\\
und\, f\left( -\varepsilon ,\varepsilon ,\sqrt{1-2\varepsilon ^{2}}\right) & < & 0\,
liegt\, kein\, Extrema\, vor.\\
& & (andere\, Sonderf\ddot{a}lle\, werden\, durch\, Symmetrie\, ausgeschlossen)\\
Da\, \mathbb S^{2}\, kompakt: & & \\
f\left( x,y,z\right) & & Extemum\, falls\, \left| x\right| =\left| y\right| =\left|
z\right| =\frac{\sqrt{3}}{3}\\
Minimum & & falls\, sgn\left( xyz\right) <0\\
Maximum & & falls\, sgn\left( xyz\right) >0
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Man skizziere das Geschwindigkeitsfeld
\begin_inset Formula \( v:\mathbb R^{3}\rightarrow \mathbb R^{3},x\mapsto w\times x \)
\end_inset
einer starren Drehung im
\begin_inset Formula \( \mathbb R^{3} \)
\end_inset
mit der verktoriellen Drehgeschwindigkeit
\begin_inset Formula \( \omega \in \mathbb R^{3} \)
\end_inset
und berechne die Divergenz
\begin_inset Formula \[
div\, v:=\sum ^{3}_{i=1}\frac{\partial v_{i}}{\partial x_{i}}.\]
\end_inset
\layout Standard
\begin_inset Formula \begin{eqnarray*}
v\left( x\right) & = & \left( \begin{array}{c}
w_{1}\\
w_{2}\\
w_{3}
\end{array}\right) \times \left( \begin{array}{c}
x_{1}\\
x_{2}\\
x_{3}
\end{array}\right) =\left( \begin{array}{c}
w_{2}x_{3}-w_{3}x_{2}\\
w_{3}x_{1}-w_{1}x_{3}\\
w_{1}x_{2}-w_{2}x_{1}
\end{array}\right) \\
div\, v & = & \sum ^{3}_{i=1}\frac{\partial v_{i}}{\partial x_{i}}=0
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Man löse das Anfangswertproblem
\begin_inset Formula \[
\left( \begin{array}{c}
\dot{x}\\
\dot{y}
\end{array}\right) =\left( \begin{array}{c}
-y\\
x
\end{array}\right) ;\, \left( \begin{array}{c}
x\left( 0\right) \\
y\left( 0\right)
\end{array}\right) =\left( \begin{array}{c}
1\\
0
\end{array}\right) \]
\end_inset
mittels Picard-Lindelöf-Iteration.
\layout Standard
\begin_inset Formula \[
F\left( \left( x,y\right) ,t\right) =\left( \begin{array}{c}
-y\\
x
\end{array}\right) \, Lipschitzstetig\, bzgl\, \left( x,y\right) \]
\end_inset
\begin_inset Formula \begin{eqnarray*}
\varphi _{0}(t) & = & \left( \begin{array}{c}
x_{0}\left( t\right) \\
y_{0}\left( t\right)
\end{array}\right) =\left( \begin{array}{c}
1\\
0
\end{array}\right) \\
\varphi _{k+1}(t) & = & \left( \begin{array}{c}
x_{k+1}\left( t\right) \\
y_{k+1}\left( t\right)
\end{array}\right) =\int _{0}^{t}\left( \begin{array}{c}
-y_{k}\left( s\right) \\
x_{k}\left( s\right)
\end{array}\right) ds+\left( \begin{array}{c}
x_{0}\\
y_{0}
\end{array}\right) \\
\varphi _{1}(t) & = & \int _{0}^{t}\left( \begin{array}{c}
-0\\
1
\end{array}\right) ds+\left( \begin{array}{c}
1\\
0
\end{array}\right) =\left( \begin{array}{c}
1\\
t
\end{array}\right) \\
\varphi _{2}(t) & = & \int _{0}^{t}\left( \begin{array}{c}
-t\\
1
\end{array}\right) ds+\left( \begin{array}{c}
1\\
0
\end{array}\right) =\left( \begin{array}{c}
1-\frac{1}{2}t^{2}\\
t
\end{array}\right) \\
\varphi _{3}(t) & = & \int _{0}^{t}\left( \begin{array}{c}
-t\\
1-\frac{1}{2}t^{2}
\end{array}\right) ds+\left( \begin{array}{c}
1\\
0
\end{array}\right) =\left( \begin{array}{c}
1-\frac{1}{2}t^{2}\\
t-\frac{1}{6}t^{3}
\end{array}\right) \\
\varphi _{2n}(t) & = & \left( \begin{array}{c}
1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}-\frac{t^{6}}{6!}+...\pm \frac{1}{\left( 2n\right)
!}t^{2n}\\
t-\frac{t^{3}}{3!}+\frac{t^{5}}{5!}-\frac{t^{7}}{7!}+...\pm \frac{1}{\left(
2n-1\right) !}t^{2n-1}
\end{array}\right) \\
\lim _{n\rightarrow \infty }\varphi _{2n}(t) & = & \left( \begin{array}{c}
cos\, t\\
sin\, t
\end{array}\right)
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
\noun on
Das Lemma von Morse
\noun default
\begin_inset LatexCommand \index{Morse}
\end_inset
\newline
Sei f eine reelle
\begin_inset Formula \( \)
\end_inset
-Funktion in einer Umgebung U von
\begin_inset Formula \( 0\in \mathbb R^{n} \)
\end_inset
mit
\begin_inset Formula \( f\left( 0\right) =0,f'\left( 0\right) =0 \)
\end_inset
und nicht ausgearteter Hessematrix
\begin_inset Formula \( f''\left( 0\right) . \)
\end_inset
Dann existiert ein Diffeomorphismus
\begin_inset Formula \( \phi :U_{0}\rightarrow V \)
\end_inset
einer Umgebung
\begin_inset Formula \( U_{0}\subset U \)
\end_inset
von 0 auf eine Umgebung V von 0 so, daß
\begin_inset Formula \[
f\circ \phi ^{-1}\left( y\right) =\left( y_{1}^{2}+...+y_{k}^{2}\right) -\left(
y_{k+1}^{2}+...+y_{n}^{2}\right) .\]
\end_inset
\layout Enumerate
Worin besteht der Unterschied zur Taylor-Formel?
\layout Enumerate
Man veranschauliche und beweise die Aussage am Beispiel n=1.
\layout Enumerate
Man beweise das Lemma von Morse:
\newline
Zunächst zeige man, daß es eine Nullumgebung Û und eine
\begin_inset Formula \( \)
\end_inset
-Funktion
\begin_inset Formula \( A:\hat{U}\rightarrow \mathbb R^{n\times n} \)
\end_inset
gibt mit
\begin_inset Formula \[
f\left( x\right) =x^{T}A\left( x\right) x.\]
\end_inset
\layout Enumerate
Wie sehen im Fall einer positiv definiten Hessematrix die Niveauflächen
in der Nähe von
\begin_inset Formula \( 0\in U \)
\end_inset
aus?
\layout Standard
\SpecialChar ~
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
f\left( 0\right) & = & 0\in \mathbb R\\
f'\left( 0\right) & = & 0\in \mathbb R^{n\times n}\\
f\left( x\right) & = & x^{T}Ax+R\left( x\right)
\end{eqnarray*}
\end_inset
Durch Diagonalisierung und Normierung wird
\begin_inset Formula \( x^{T}Ax=\left( x_{1}^{2}+...+x_{k}^{2}\right) -\left(
x_{k+1}^{2}+...+x_{n}^{2}\right) \)
\end_inset
(lineare Koordinatentransformation!).
\newline
\bar under
Unterschied:
\bar default
Man läßt nicht nur lineare Koordinatentransformationen zu, sondern auch
einen Diffeomorphismus
\begin_inset Formula \( \Rightarrow \)
\end_inset
R
\begin_inset Formula \( \left( x\right) \)
\end_inset
fällt weg.
\newline
\layout Enumerate
\bar under
Spezialfall n=1:
\newline
\bar default
Taylor:
\begin_inset Formula \( f\left( x\right) =ax^{2}+R\left( x\right) =\pm \left(
\sqrt{\left| a\right| }\times \sqrt{1+\frac{R\left( x\right) }{ax^{2}}}\right)
^{2}=\pm \phi ^{2}\left( x\right) \)
\end_inset
\newline
\bar under
Behauptung
\bar default
:
\begin_inset Formula \( \phi \left( x\right) \)
\end_inset
ist Diffeomorphismus
\newline
\begin_inset Formula \begin{eqnarray*}
\phi \left( x\right) & = & \sqrt{\left| a\right| }x\cdot \sqrt{1+\frac{R\left(
x\right) }{ax^{2}}}\, differenzierbar\, \forall x\neq 0\\
\phi '\left( 0\right) & = & \lim _{x\rightarrow 0}\frac{\phi \left( x\right) -\phi
\left( 0\right) }{x-0}=\lim _{x\rightarrow 0}\sqrt{\left| a\right| }\cdot
\sqrt{1+\frac{R\left( x\right) }{ax^{2}}}=\sqrt{\left| a\right| }\, existiert\, in\,
\mathbb R\\
\phi '\left( x\right) & & stetig\, f\ddot{u}r\, x\neq 0\\
& & Stetigkeit\, in\, x=0\\
R\left( x\right) & = & \frac{1}{2}\int _{0}^{y}\left( x-t\right) ^{2}f^{\left(
3\right) }\left( t\right) dt\\
F\left( x,y\right) & := & \frac{1}{2}\int _{0}^{y}\left( x-t\right) ^{2}f^{\left(
3\right) }\left( t\right) dt\\
\partial _{y}F & = & \frac{1}{2}\left( x-y\right) ^{2}f^{\left( 3\right) }\left(
y\right) \\
\partial _{x}F & = & \int _{0}^{y}\left( x-t\right) f^{\left( 3\right) }\left(
t\right) dt\\
R\left( x\right) & = & F\left( x,x\right) \\
R'\left( x_{0}\right) & = & \partial _{x}F\left( x_{0},x_{0}\right) +\partial
_{y}F\left( x_{0},x_{0}\right) =\int _{0}^{x_{0}}\left( x-t\right) f^{\left( 3\right)
}\left( t\right) dt\\
\lim _{x\rightarrow 0}\frac{R\left( x\right) }{x^{2}} & = & 0\, (Taylor)\\
\lim _{x\rightarrow 0}\frac{R'\left( x\right) }{x} & = & \lim _{x\rightarrow 0}\int
_{0}^{x}\frac{x-t}{x}f^{\left( 3\right) }\left( t\right) dt=0\\
& & Damit\, durch\, Rechnung\\
\lim _{x\rightarrow 0}\phi '\left( x\right) & = & \sqrt{\left| a\right| }=\phi
'\left( 0\right) \\
f\circ \phi ^{-1}\left( x\right) & = & \pm x^{2}
\end{eqnarray*}
\end_inset
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
f\left( x\right) & = & \sum _{i=1}^{n}a_{i}\left( x\right) x_{i}+f\left( 0\right) \\
a_{i}\left( x\right) & = & \int _{0}^{1}\partial _{i}f\left( tx\right) dt\leftarrow
a_{i}\left( 0\right) =\partial _{i}f\left( 0\right) =0;\, \partial _{j}a_{i}\left(
0\right) =\partial _{j}\partial _{i}f\left( 0\right) \\
a_{i}\left( x\right) & = & \sum _{j=1}^{n}a_{ij}\left( x\right) x_{j}+a_{i}\left(
0\right) =\sum ^{n}_{j=1}a_{ij}\left( x\right) x_{j}\\
f\left( x\right) & = & \sum ^{n}_{i,j=1}a_{ij}\left( x\right) x_{i}x_{j}=x^{T}A\left(
x\right) \cdot x=\frac{1}{2}x^{T}\left( A\left( x\right) +A\left( x\right) ^{T}\right)
\cdot x\, \, \, \left( oBdA\, symmetrisch\right) \\
&
\end{eqnarray*}
\end_inset
\layout Enumerate
Eigenwerte positiv
\newline
\begin_inset Formula \( f\left( x\right) =x_{1}^{2}+...+x_{n}^{2} \)
\end_inset
\hfill
Paraboilid
\hfill
\layout Standard
\pagebreak_top
\SpecialChar ~
\layout Note
\line_top
\bar under
Exponentialfunktion:
\layout Note
\begin_inset Formula \begin{eqnarray*}
\Lambda & = & \left( \begin{array}{ccc}
\lambda _{1} & \cdots & 0\\
\vdots & \ddots & \vdots \\
0 & \cdots & \lambda _{n}
\end{array}\right) \\
\Lambda ^{k} & = & \left( \begin{array}{ccc}
\lambda _{1}^{k} & \cdots & 0\\
\vdots & \ddots & \vdots \\
0 & \cdots & \lambda _{n}^{k}
\end{array}\right) \\
e^{\left( \begin{array}{ccc}
\lambda _{1} & \cdots & 0\\
\vdots & \ddots & \vdots \\
0 & \cdots & \lambda _{n}
\end{array}\right) } & = & \sum _{k=0}^{\infty }\frac{\Lambda ^{k}}{k!}=\left(
\begin{array}{ccc}
\sum _{k=0}^{\infty }\frac{1}{k!}\lambda _{1}^{k} & \cdots & 0\\
\vdots & \ddots & \vdots \\
0 & \cdots & \sum _{k=0}^{\infty }\frac{1}{k!}\lambda _{n}^{k}
\end{array}\right) =\left( \begin{array}{ccc}
e^{\lambda _{1}} & \cdots & 0\\
\vdots & \ddots & \vdots \\
0 & \cdots & e^{\lambda _{n}}
\end{array}\right) \\
\mathbb I & = & \left( \begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right) \\
\mathbb I^{2} & = & \left( \begin{array}{cc}
-1 & 0\\
0 & -1
\end{array}\right) =-1\cdot \mathbb E\\
\left( \mathbb I^{2}\right) ^{k} & = & I^{2k}=\left( -1\right) ^{k}\mathbb E\\
e^{t\mathbb I} & = & \sum _{k=0}^{\infty }\frac{\left( t\mathbb I\right)
^{k}}{k!}=E+\mathbb I+\frac{t^{2}\left( -1\right) ^{2}}{2!}\mathbb E+\frac{t^{3}\left(
-1\right) ^{3}}{3!}\mathbb I+....=\\
& = & \left( \sum _{k=0}^{\infty }\left( -1\right) ^{k}\frac{t^{2k}}{\left( 2k\right)
!}\right) E+\left( \sum _{k=0}^{\infty }\left( -1\right) ^{k}\frac{t^{2k-1}}{\left(
2k-1\right) !}\right) I=\left( \begin{array}{c}
cos\, t\\
sin\, t
\end{array}\right) \\
\mathbb I^{*} & = & \left( \begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{array}\right) \\
\mathbb I^{*}\, ^{2} & = & \left( \begin{array}{ccc}
0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right) \\
e^{\mathbb I^{*}} & = & \mathbb E+t\left( \begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{array}\right) +\frac{t^{2}}{2!}\left( \begin{array}{ccc}
0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right)
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
\SpecialChar ~
\layout Enumerate
Man berechne
\begin_inset Formula \( e^{At} \)
\end_inset
aufgrund der Potenzreihe für
\begin_inset Formula \( A_{1}:=\left( \begin{array}{cc}
a & -b\\
b & a
\end{array}\right) \)
\end_inset
und
\begin_inset Formula \( A_{2}:=\left( \begin{array}{ccc}
\lambda & 1 & 0\\
0 & \lambda & 1\\
0 & 0 & \lambda
\end{array}\right) \)
\end_inset
.
\layout Enumerate
Man ermittle ein Fundamentalsystem für die Lösungen von
\begin_inset Formula \( \dot{x}=Ax \)
\end_inset
mit den Matrizen aus Teil (1).
\layout Enumerate
Man zeige: Das Additionstheorem
\begin_inset LatexCommand \index{Additionstheorem}
\end_inset
für die Exponentialfunktion
\begin_inset LatexCommand \index{Exponentialfunktion}
\end_inset
von Matrizen gilt nicht allgemein: Berechne
\begin_inset Formula \( e^{A}\cdot e^{B} \)
\end_inset
und
\begin_inset Formula \( e^{A+B} \)
\end_inset
für die Matrizen
\begin_inset Formula \( A:=\left( \begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right) \)
\end_inset
und
\begin_inset Formula \( B:=\left( \begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right) \)
\end_inset
.
\layout Standard
\SpecialChar ~
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
A_{1} & = & \left( \begin{array}{cc}
a & -b\\
b & a
\end{array}\right) =a\mathbb E+b\mathbb I\\
e^{tA_{1}} & = & e^{ta\mathbb E+tb\mathbb I}=\left( \begin{array}{ccc}
e^{t_{a}} & \cdots & 0\\
\vdots & \ddots & \vdots \\
0 & \cdots & e^{t_{a}}
\end{array}\right) \left( \begin{array}{cc}
cos\, tb & -sin\, tb\\
sin\, tb & cos\, tb
\end{array}\right) =e^{t_{a}}\left( \begin{array}{cc}
cos\, tb & -sin\, tb\\
sin\, tb & cos\, tb
\end{array}\right) \\
A_{2} & = & \left( \begin{array}{ccc}
\lambda & 1 & 0\\
0 & \lambda & 1\\
0 & 0 & \lambda
\end{array}\right) \\
e^{tA_{2}} & = & e^{t\lambda \mathbb E}\cdot e^{t\mathbb I}=e^{t\lambda }\mathbb
E\left( \mathbb E+t\left( \begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{array}\right) +\frac{t^{2}}{2!}\left( \begin{array}{ccc}
0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right) \right) =e^{t\lambda }\left( \begin{array}{ccc}
1 & t & \frac{t^{2}}{2}\\
0 & 1 & t\\
0 & 0 & 1
\end{array}\right)
\end{eqnarray*}
\end_inset
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
\phi _{1}\left( t\right) & = & k_{1}e^{at}\left( \begin{array}{c}
cos\, bt\\
sin\, bt
\end{array}\right) +k_{2}e^{at}\left( \begin{array}{c}
-sin\, bt\\
cos\, bt
\end{array}\right) \, \, f\ddot{u}r\, A_{1}\\
\phi _{2}\left( t\right) & = & k_{1}e^{\lambda t}+k_{2}e^{\lambda t}\left(
\begin{array}{c}
t\\
1\\
0
\end{array}\right) +k_{3}e^{\lambda t}\left( \begin{array}{c}
\frac{1}{2}t^{2}\\
t\\
1
\end{array}\right)
\end{eqnarray*}
\end_inset
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
e^{A}=e^{\left( \begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right) } & = & \sum _{k=0}^{\infty }\frac{\left( \begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right) ^{k}}{k!}=\mathbb E+\left( \begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right) +0=\left( \begin{array}{cc}
1 & 1\\
0 & 1
\end{array}\right) \\
e^{B}=e^{\left( \begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right) } & = & \sum _{k=0}^{\infty }\frac{\left( \begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right) ^{k}}{k!}=\mathbb E+\left( \begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right) \sum _{k=1}^{\infty }\frac{1}{k!}=\left( \begin{array}{cc}
e^{1} & 0\\
0 & 1
\end{array}\right) \\
e^{A}\cdot e^{B} & = & \left( \begin{array}{cc}
e & 1\\
0 & 1
\end{array}\right) \\
e^{B}\cdot e^{A} & = & \left( \begin{array}{cc}
e & e\\
0 & 1
\end{array}\right) \\
e^{A+B} & = & e^{\left( \begin{array}{cc}
1 & 1\\
0 & 0
\end{array}\right) }=\sum _{k=0}^{\infty }\frac{\left( \begin{array}{cc}
1 & 1\\
0 & 0
\end{array}\right) ^{k}}{k!}=\left( \begin{array}{cc}
e & e-1\\
0 & 1
\end{array}\right)
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Man zeige: Das früher definierte charakteristische
\begin_inset LatexCommand \index{Polynom, charakteristisches}
\end_inset
Polynom einer lienearen Differentialgleichung n-ter Ordnung mit konstanten
Koeffizienten
\begin_inset Formula \[
x^{\left( n\right) }+a_{n-1}x^{\left( n-1\right) }+...+a_{1}x'+a_{0}x=0\]
\end_inset
ist bis auf einen konstanten Faktor das charakteristische Polynom der Matrix
des assoziierten Systems erster Ordnung.
\layout Standard
\begin_inset Formula \begin{eqnarray*}
\left( \begin{array}{c}
x\\
x'\\
\vdots \\
\vdots \\
\vdots \\
x^{\left( n-1\right) }
\end{array}\right) & = & \left( \begin{array}{cccccc}
0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & \ddots & 0 & 0\\
0 & 0 & 0 & \ddots & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 1\\
-a_{0} & -a_{1} & -a_{2} & 0 & -a_{n-2} & -a_{n-1}
\end{array}\right) \left( \begin{array}{c}
x\\
x'\\
\vdots \\
\vdots \\
\vdots \\
x^{\left( n-1\right) }
\end{array}\right) \\
det\left( A-\lambda \mathbb E\right) & = & \left( \begin{array}{cccccc}
-\lambda & 1 & 0 & 0 & 0 & 0\\
0 & -\lambda & 1 & 0 & 0 & 0\\
0 & 0 & -\lambda & \ddots & 0 & 0\\
0 & 0 & 0 & \ddots & 1 & 0\\
0 & 0 & 0 & 0 & -\lambda & 1\\
-a_{0} & -a_{1} & -a_{2} & 0 & -a_{n-2} & -a_{n-1}-\lambda
\end{array}\right) =\\
& = & \left( -\lambda \right) ^{n-1}\left( -a_{n-1}-\lambda \right) +a_{n-2}\left(
-\lambda \right) ^{n-2}+a_{n-3}\left( -\lambda \right) ^{n-3}+...=\\
& = & \left( -1\right) ^{n}\left[ \left( a_{n-1}+\lambda \right) \lambda
^{n-1}+a_{n-2}\lambda ^{n-2}+...+a_{1}\lambda +a_{0}\right]
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Set
\begin_inset Formula \( v:\mathbb R^{n}\rightarrow \mathbb R^{n} \)
\end_inset
ein
\begin_inset Formula \( \)
\end_inset
-Vektorfeld.
Man zeige:
\layout Enumerate
Ist
\begin_inset Formula \( x_{0} \)
\end_inset
eine Nullstelle von
\begin_inset Formula \( v \)
\end_inset
und
\begin_inset Formula \( \varphi \)
\end_inset
eine maximale Integralkurve von v mit
\begin_inset Formula \( \varphi \left( t_{0}\right) =x_{0} \)
\end_inset
, so ist
\begin_inset Formula \( \varphi \)
\end_inset
auf ganz
\begin_inset Formula \( \mathbb R \)
\end_inset
definiert und
\begin_inset Formula \( \varphi \left( t\right) =x_{0} \)
\end_inset
für alle
\begin_inset Formula \( t\in \mathbb R \)
\end_inset
.
\layout Enumerate
Jede maximale Integralkurve zu
\begin_inset Formula \( \tilde{v}:\mathbb R^{n}\rightarrow \mathbb R^{n},\,
\tilde{v}\left( x\right) :=sin\left( \left\Vert x\right\Vert _{2}^{2}\right) \cdot
v\left( x\right) \)
\end_inset
ist auf ganz
\begin_inset Formula \( \mathbb R \)
\end_inset
definiert.
Wo verläuft sie?
\layout Standard
Lösung:
\layout Enumerate
ist Lösung des AWP.
Da
\begin_inset Formula \( \varphi \)
\end_inset
auf ganz
\begin_inset Formula \( \mathbb R \)
\end_inset
definiert ist, ist es bereits die maximale Lösung.
\layout Enumerate
\begin_inset Formula \[
\tilde{v}\left( x\right) :=sin\left( \left\Vert x\right\Vert _{2}^{2}\right) \cdot
v\left( x\right) \]
\end_inset
Maximale Lösung
\begin_inset LatexCommand \index{Lösung, maximale}
\end_inset
örtlich
\begin_inset LatexCommand \index{örtlich}
\end_inset
beschränkt
\begin_inset Formula \( \Rightarrow \)
\end_inset
zeitlich
\begin_inset LatexCommand \index{zeitlich}
\end_inset
auf ganz
\begin_inset Formula \( \mathbb R \)
\end_inset
unbeschränkt
\begin_inset LatexCommand \index{unbeschränkt}
\end_inset
!
\layout Problem
\line_top
\SpecialChar ~
\layout Enumerate
Man zeige: Der Tangentialraum
\begin_inset Formula \( T_{\mathbb 1}O\left( n\right) \)
\end_inset
besteht aus allen schiefsymmetrischen reellen
\begin_inset Formula \( n\times n \)
\end_inset
-Matritzen.
\layout Enumerate
Man zeige: Sei
\begin_inset Formula \( H\in T_{\mathbb 1}O\left( n\right) \)
\end_inset
.
Dann ist
\begin_inset Formula \[
\gamma :\mathbb R\rightarrow O\left( n\right) ,\, t\mapsto e^{Ht}\]
\end_inset
eine differenzierbare Kurve in
\begin_inset Formula \( O\left( n\right) \)
\end_inset
mit
\begin_inset Formula \( \)
\end_inset
\layout Standard
\SpecialChar ~
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
f & : & \mathbb R^{n\times n}\rightarrow R^{n\times n}\\
& & A\mapsto A^{T}A\\
O\left( n\right) & = & f^{-1}\left( \mathbb 1\right)
\end{eqnarray*}
\end_inset
\begin_inset Formula \( \mathfrak 1 \)
\end_inset
ist regulärer Wert:
\newline
Sei
\begin_inset Formula \( A\in f^{-1}\left( \mathfrak 1\right) \)
\end_inset
, d.h.
\begin_inset Formula \( A^{T}A=1 \)
\end_inset
.
\begin_inset Formula \( df_{A}H=A^{T}H+H^{T}A\, \forall H\in T_{A}\mathbb R^{n\times
n} \)
\end_inset
.
Sei
\begin_inset Formula \( y\in T_{\mathbb 1}\mathbb R_{s}^{n\times n}\approxeq
R_{s}^{n\times n} \)
\end_inset
.
\newline
Wähle:
\begin_inset Formula \begin{eqnarray*}
H & := & \frac{1}{2}\left( A^{T}\right) ^{-1}Y=\frac{1}{2}AY\\
df_{A}H & = & \frac{1}{2}\left( A^{T}AY+Y^{T}A^{T}A\right) =\frac{1}{2}\left(
Y+Y^{T}\right) =Y
\end{eqnarray*}
\end_inset
\begin_inset Formula \( \Rightarrow df_{A} \)
\end_inset
surjektiv
\begin_inset Formula \( \Rightarrow \)
\end_inset
Satz über implizite
\begin_inset LatexCommand \index{implizite}
\end_inset
Funktionen
\newline
\begin_inset Formula \( O\left( n\right) \)
\end_inset
ist Mannigfaltigkeit der Dimension
\begin_inset Formula \( n^{2}-\frac{n\left( n+1\right) }{2} \)
\end_inset
.
\newline
\begin_inset Formula \( T_{\mathbb 1}O\left( n\right) =Kern\, df_{\mathbb 1}\mathbb
R^{n\times n}=\left\{ X\in \mathbb R^{n\times n}:\mathbb 1^{T}x+x^{T}\mathbb
1=0\right\} =\left\{ X\in \mathbb R^{n\times n}:x+x^{T}=0\right\} \)
\end_inset
\layout Enumerate
Sei
\begin_inset Formula \( H\in \left\{ X\in \mathbb R^{n\times n}:x+x^{T}=0\right\} \)
\end_inset
.
\newline
\begin_inset Formula \begin{eqnarray*}
f_{1} & : & \mathbb R\rightarrow \mathbb R^{n\times n};\, t\mapsto e^{Ht}\,
differenzierbar\\
f_{1}\left( 0\right) & = & e^{0H}=\mathfrak 1;\, \gamma _{1}'\left( t\right)
=He^{Ht}\\
\gamma _{1}'\left( 0\right) & = & He^{0H}=H
\end{eqnarray*}
\end_inset
Behauptung:
\begin_inset Formula \( e^{Ht}\in O\left( n\right) \, \forall t. \)
\end_inset
\newline
zu Zeigen:
\begin_inset Formula \( \left( e^{Ht}\right) ^{T}e^{Ht}=\mathfrak 1 \)
\end_inset
\newline
\begin_inset Formula \( \left( e^{Ht}\right) ^{T}e^{Ht}=e^{-Ht}e^{Ht}=e^{\mathfrak
0}=\mathfrak 1 \)
\end_inset
\layout Problem
\line_top
Man bestimme durch Reihenentwicklung zwei unabhängige Lösungen von
\begin_inset Formula \( \ddot{x}=tx \)
\end_inset
.
Sind die Reihen für numerische Zwecke gut geeignet?
\layout Standard
\begin_inset Formula \begin{eqnarray*}
x\left( t\right) & = & tx=a_{0}+a_{1}t+a_{2}t^{2}+a_{3}t^{3}+...\\
\dot{x}\left( t\right) & = & a_{1}+2a_{2}t+3a_{3}t^{2}+4a_{4}t^{3}+....\\
\ddot{x}\left( t\right) =tx\left( t\right) & = & 0+2a_{2}+6a_{2}t+12a_{4}t^{2}\\
\Rightarrow a_{2} & = & 0\\
a_{n+3} & = & \frac{1}{\left( n+3\right) \left( n+2\right) }a_{n}\forall n\in \mathbb
N\\
a_{2} & = & a_{5}=a_{8}=...=0\\
x\left( t\right) & = & a_{0}\left( 1+\frac{t^{3}}{3\cdot 2}+\frac{t^{6}}{\left(
6\cdot 5\right) \left( 3\cdot 2\right) }+...\right) +\\
& & +a_{1}\left( t+\frac{t^{4}}{3\cdot 4}+\frac{t^{7}}{\left( 6\cdot 7\right) \left(
3\cdot 4\right) }+...\right)
\end{eqnarray*}
\end_inset
Vergleich mit Potenzreihe
\begin_inset LatexCommand \index{Potenzreihe}
\end_inset
: Konvergenz sehr gut, besonders für kleine t.
\newline
Für negative T: Leipnitzkriterium:
\begin_inset Formula \( \sum _{n=0}^{\infty }\left( -1\right) ^{n}a_{n};\, a_{n}\geq
a_{n+1}\rightarrow 0;\, \left| R_{n}\right| =\left| \sum _{i=n+1}^{\infty }\left(
-1\right) ^{i}a_{i}\right| \leq a_{n+1};\, sgn\, R_{n}=\left( -1\right) ^{n+1}. \)
\end_inset
\begin_inset Formula \begin{eqnarray*}
x_{1}\left( t\right) & = & 1+\frac{t^{3}}{3\cdot 2}+\frac{t^{6}}{\left( 6\cdot
5\right) \left( 3\cdot 2\right) }+...\\
x_{2}\left( t\right) & = & 1+\frac{t^{4}}{3\cdot 4}+\frac{t^{7}}{\left( 6\cdot
7\right) \left( 3\cdot 4\right) }+...
\end{eqnarray*}
\end_inset
Zwei linear unabhängige Lsg.
bilden eine Basis.
\begin_inset Formula \begin{eqnarray*}
x\left( t\right) & = & a_{0}x_{1}\left( t\right) +a_{1}x_{2}\left( t\right) \\
& & \forall a_{0},a_{1}\in \mathbb R
\end{eqnarray*}
\end_inset
\layout Note
\line_top
Ist
\begin_inset Formula \( \phi \)
\end_inset
eine Fundamentalmatrix
\begin_inset LatexCommand \index{Fundamentalmatrix}
\end_inset
der homogenen
\begin_inset LatexCommand \index{homogenen}
\end_inset
Gleichung
\begin_inset Formula \( \dot{x}=Ax, \)
\end_inset
so ist
\begin_inset Formula \[
x_{p}\left( t\right) :=\phi \left( t\right) \cdot c\left( t\right) \, mit\, c:=\int
\phi \left( s\right) ^{-1}b\left( s\right) ds\]
\end_inset
eine Lösung der inhomogenen
\begin_inset LatexCommand \index{inhomogenen}
\end_inset
Gleichung
\begin_inset Formula \( \dot{x}=Ax+b \)
\end_inset
.
\layout Problem
\line_top
Man bestimme die Lösung des Anfangswertproblems
\begin_inset LatexCommand \index{Anfangswertproblems}
\end_inset
:
\begin_inset Formula \begin{eqnarray*}
\left( \begin{array}{c}
x\\
y
\end{array}\right) ^{*} & = & \left( \begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right) \cdot \left( \begin{array}{c}
x\\
y
\end{array}\right) +\left( \begin{array}{c}
0\\
3t+1
\end{array}\right) \\
\left( \begin{array}{c}
x\\
y
\end{array}\right) \left( 0\right) & = & \left( \begin{array}{c}
0\\
3
\end{array}\right)
\end{eqnarray*}
\end_inset
\layout Standard
Fundamentalmatrix
\begin_inset LatexCommand \index{Fundamentalmatrix}
\end_inset
:
\begin_inset Formula \begin{eqnarray*}
\phi \left( t\right) & = & e^{t\left( \begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right) }=\left( \begin{array}{cc}
cos\, t & -sin\, t\\
sin\, t & cos\, t
\end{array}\right) \\
c & = & \int _{0}^{s}\left( \begin{array}{cc}
cos\, t & sin\, t\\
-sin\, t & cos\, t
\end{array}\right) \left( \begin{array}{c}
0\\
3t+1
\end{array}\right) dt=\int _{0}^{s}\left( 3t+1\right) \left( \begin{array}{c}
sin\, t\\
cos\, t
\end{array}\right) dt=\\
& = & 3\int _{0}^{s}\left( \begin{array}{c}
t\, sin\, t\\
t\, cos\, t
\end{array}\right) dt+\int _{0}^{s}\left( \begin{array}{c}
sin\, t\\
cos\, t
\end{array}\right) dt=3\left[ \left( \begin{array}{c}
sin\, t-t\, cos\, t\\
cos\, t+t\, sin\, t
\end{array}\right) \right] _{0}^{s}+\left[ \left( \begin{array}{c}
-cos\, t\\
sin\, t
\end{array}\right) \right] _{0}^{s}=\\
& = & 3\left( \begin{array}{c}
sin\, s-s\, cos\, s\\
cos\, s+s\, sin\, s-1
\end{array}\right) +\left( \begin{array}{c}
-cos\, s+1\\
sin\, s
\end{array}\right) =\left( \begin{array}{c}
3sin\, s-3s\, cos\, s-cos\, s+1\\
3cos\, s+3s\, sin\, s+sin\, s-3
\end{array}\right) =\\
& = & \left( 3s+1\right) \left( \begin{array}{c}
-cos\, s\\
sin\, s
\end{array}\right) +\left( \begin{array}{c}
3sin\, s+1\\
3cos\, s-3
\end{array}\right) \\
x_{p}\left( t\right) & = & \left( \begin{array}{cc}
cos\, s & -sin\, s\\
sin\, s & cos\, s
\end{array}\right) \left( \begin{array}{c}
3sin\, s-3s\, cos\, s-cos\, s+1\\
3cos\, s+3s\, sin\, s+sin\, s-3
\end{array}\right) =\\
& = & \left( \begin{array}{c}
3sin\, s\, cos\, s-3s\, cos^{2}s-cos^{2}s+cos\, s-3sin\, s\, cos\, s-3s\,
sin^{2}s-sin^{2}s+3sin\, s\\
3sin^{2}s-3s\, sin\, s\, cos\, s-sin\, s\, cos\, s+sin\, s+3cos^{2}s+3s\, sin\, s\,
cos\, s+sin\, s\, cos\, s-3cos\, s
\end{array}\right) =\\
& = & \left( \begin{array}{c}
+cos\, s-3s+3sin\, s-1\\
+sin\, s-3cos\, s+3
\end{array}\right)
\end{eqnarray*}
\end_inset
Die allgemeine Lösung des AWP ist
\begin_inset Formula \begin{eqnarray*}
\left( \begin{array}{c}
x\\
y
\end{array}\right) & = & x_{p}+\phi \left( t\right) \left( \begin{array}{c}
c_{1}\\
c_{2}
\end{array}\right) =\\
& = & \left( \begin{array}{c}
cos\, s-3s+3sin\, s-1\\
sin\, s-3cos\, s+3
\end{array}\right) +c_{1}\left( \begin{array}{c}
cos\, s\\
sin\, s
\end{array}\right) +c_{2}\left( \begin{array}{c}
-sin\, s\\
cos\, s
\end{array}\right) =\\
& = & \left( \begin{array}{c}
cos\, s-3s+3sin\, s-1\\
sin\, s-3cos\, s+3
\end{array}\right) +3\left( \begin{array}{c}
-sin\, s\\
cos\, s
\end{array}\right) =\left( \begin{array}{c}
cos\, s-3s-1\\
sin\, s+3
\end{array}\right) =\\
\left( \begin{array}{c}
x\\
y
\end{array}\right) \left( 0\right) & = & \left( \begin{array}{c}
0\\
3
\end{array}\right) \, \surd \\
\left( \begin{array}{c}
cos\, s-3s-1\\
sin\, s+3
\end{array}\right) ^{*} & = & \left( \begin{array}{cc}
0 & -1\\
1 & 0
\end{array}\right) \cdot \left( \begin{array}{c}
cos\, s-3s-1\\
sin\, s+3
\end{array}\right) +\left( \begin{array}{c}
0\\
3t+1
\end{array}\right) \, \surd
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Sei
\begin_inset Formula \( A\in \mathbb C^{n\times n} \)
\end_inset
und
\begin_inset Formula \( \)
\end_inset
.
Man zeige:
\newline
Falls
\begin_inset Formula \( \alpha \)
\end_inset
kein Eigenwert von A ist, hat die Differentialgleichung
\begin_inset Formula \[
\dot{x}=Ax+e^{at}\cdot c\]
\end_inset
genau eine Lösung der Form
\begin_inset Formula \( x\left( t\right) =e^{at}\cdot d \)
\end_inset
mit
\begin_inset Formula \( d\in \mathbb C^{n} \)
\end_inset
.
\layout Standard
\begin_inset Formula \begin{eqnarray*}
e^{at}d & = & ae^{at}\cdot d\\
A\left( e^{at}d\right) +e^{at}\cdot c & = & e^{at}\left( Ad-c\right) \\
e^{at}ad & = & e^{at}\left( Ad-c\right) \\
ad & = & Ad+c,\, da\, e^{at}\left( Ad+c\right) \\
-c & = & Ad-qd\\
-c & = & \left( A-a\mathbb E\right) d
\end{eqnarray*}
\end_inset
\begin_inset Formula \( \alpha \)
\end_inset
kein Eigenwert
\begin_inset Formula \( \Rightarrow d:=\left( A-a\mathbb E\right) ^{-1}\cdot c \)
\end_inset
eindeutig.
\newline
\begin_inset Formula \( \alpha \)
\end_inset
ein Eigenwert
\layout Problem
\line_top
Man diskutiere die Gleichung zweiter Ordnung
\begin_inset Formula \[
\ddot{x}=-\left( x-\frac{1}{2}x^{2}\right) .\]
\end_inset
\layout Enumerate
Wie lautet das assoziierte
\begin_inset LatexCommand \index{assoziierte}
\end_inset
System erster Ordnung?
\layout Enumerate
Man ermittle dazu ein erstes Integral und diskutiere qualitativ dessen Niveaulin
ien in der nähe der kritischen Stellen.
\layout Enumerate
Man berechne explizit die Niveaulinien.
\layout Standard
\SpecialChar ~
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
\left\{ \begin{array}{c}
x_{0}:=x\\
x_{1}:=\dot{x}
\end{array}\right\} & \Rightarrow & \left( \begin{array}{c}
x_{0}\\
x_{1}
\end{array}\right) ^{*}=\left( \begin{array}{c}
\dot{x}_{0}\\
\dot{x}_{1}
\end{array}\right) =\left( \begin{array}{c}
\dot{x}_{1}\\
-\left( x-\frac{1}{2}x^{2}\right)
\end{array}\right) \\
bezeichne:\, \left( \begin{array}{c}
x_{0}\\
x_{1}
\end{array}\right) =\left( \begin{array}{c}
x\\
y
\end{array}\right) & \Rightarrow & \left( \begin{array}{c}
x\\
y
\end{array}\right) ^{*}=\left( \begin{array}{c}
y\\
-\left( x-\frac{1}{2}x^{2}\right)
\end{array}\right)
\end{eqnarray*}
\end_inset
\layout Enumerate
E sei erstes Integral
\begin_inset Formula \( E\left( x\left( t\right) ,y\left( t\right) \right) =const \)
\end_inset
.
\begin_inset Formula \begin{eqnarray*}
\frac{d}{dt}E & \Rightarrow & Ex\dot{x}+Ey\dot{y}=0\\
grad\, E\cdot \left( \begin{array}{c}
x\\
y
\end{array}\right) & = & 0\\
\nabla E & \perp & \left( \begin{array}{c}
y\\
-\left( x-\frac{1}{2}x^{2}\right)
\end{array}\right) \\
Ansatz:\, E_{x} & = & \left( x-\frac{1}{2}x^{2}\right) \\
E_{y} & = & y
\end{eqnarray*}
\end_inset
kritischer Punkt für E;
\begin_inset Formula \( grad\, E=0 \)
\end_inset
\hfill
\begin_inset Formula \( E\left( x,y\right)
=\frac{1}{2}x^{2}-\frac{1}{6}x^{3}+\frac{1}{2}y^{2} \)
\end_inset
\begin_inset Formula \begin{eqnarray*}
E_{y} & = & y=0\\
E_{x} & = & x\frac{1}{2}x^{2}=0\\
\Rightarrow 2\, krit.\, Punkte & & \left( x,y\right) =\{\begin{array}{c}
\left( 0,0\right) \\
\left( 2,0\right)
\end{array}
\end{eqnarray*}
\end_inset
Lösungskurven verlaufen in den Niveumengen
\begin_inset Formula \( E^{-1}\left( \left\{ \left( 0,0\right) \right\} \right) \)
\end_inset
und
\begin_inset Formula \( E^{-1}\left( \left\{ \left( 2,0\right) \right\} \right) \)
\end_inset
.
Mit dem Lemma von Morse folgt wegen
\begin_inset Formula \begin{eqnarray*}
\left( x,y\right) & = & \left( \begin{array}{cc}
1-x & 0\\
0 & 1
\end{array}\right) \, \, \, \, \, \, \, \, Hessematrix\\
E''\left( 0,0\right) & = & \left( \begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right) \, \, \, \, \, \, positiv\, definit,
\end{eqnarray*}
\end_inset
daß die Lösungskurven in einer Umgebung von
\begin_inset Formula \( \left( 0,0\right) \)
\end_inset
auf
\begin_inset Formula \( \)
\end_inset
-Deformierten Kreisen verlaufen.
E ist äquivalent zu
\begin_inset Formula \( \tilde{E}\left( u,v\right) =u^{2}+v^{2} \)
\end_inset
in Umgebung von
\begin_inset Formula \( \left( 0,0\right) . \)
\end_inset
\begin_inset Formula \begin{eqnarray*}
E''\left( 2,0\right) & = & \left( \begin{array}{cc}
-1 & 0\\
0 & 1
\end{array}\right) \\
\tilde{E}\left( u,v\right) & = & u^{2}+v^{2}
\end{eqnarray*}
\end_inset
Niveaulinie heißt nicht Lösungskurve!, denn die Lösung läuft in unendlich
langer Zeit zu
\begin_inset Formula \( \left( 2,0\right) \)
\end_inset
.
In
\begin_inset Formula \( \left( 2,0\right) \)
\end_inset
steht konstante Lösung.
\begin_inset Formula \begin{eqnarray*}
E\left( x,y\right) & = & \frac{1}{2}x^{2}-\frac{1}{6}x^{3}+\frac{1}{2}y^{2}\\
E^{-1}\left( \left\{ \left( 0,0\right) \right\} \right) & = & \left\{ \left(
x,y\right) :y=\pm \sqrt{\frac{1}{3}x^{3}-x^{2}}\right\} \\
E^{-1}\left( \left\{ \left( 2,0\right) \right\} \right) & = & \left\{ \left(
x,y\right) :y=\pm \sqrt{\frac{4}{3}+\frac{1}{3}x^{3}-x^{2}}\right\}
\end{eqnarray*}
\end_inset
\layout Enumerate
\begin_inset Formula \[
y=\pm \left| x\right| \sqrt{\frac{1}{3}x-1}\]
\end_inset
\layout Problem
\line_top
Sei A eine reelle
\begin_inset Formula \( 3\times 3 \)
\end_inset
-Matrix mit Eigenwerten
\begin_inset Formula \( -\lambda \pm i\mu \)
\end_inset
und
\begin_inset Formula \( -\lambda \)
\end_inset
, wobei
\begin_inset Formula \( \lambda ,\gamma ,\mu \in \mathbb R_{+}. \)
\end_inset
Man skizziere die Integralkurven des Vektorfeldes
\begin_inset Formula \( v:\mathbb R^{3}\rightarrow \mathbb R^{3},\, v\left( x\right)
=Ax \)
\end_inset
, und untersuche sie für
\begin_inset Formula \( t\rightarrow \infty \)
\end_inset
.
\layout Standard
\SpecialChar ~
\layout Problem
\line_top
Man zeige für
\begin_inset Formula \( A\in \mathbb C^{n\times n} \)
\end_inset
:
\begin_inset Formula \[
det\, e^{A}=e^{Spur\, A}\]
\end_inset
\layout Enumerate
direkt unter Verwendung der Beziehung
\begin_inset Formula \( e^{T^{-1}AT}=T^{-1}e^{A}T \)
\end_inset
.
\layout Enumerate
mit Hilfe der Formel von Liouville
\begin_inset LatexCommand \index{Liouville}
\end_inset
.
\layout Standard
\SpecialChar ~
\layout Enumerate
Wähle
\begin_inset Formula \( T\in \mathbb C^{n\times n} \)
\end_inset
invertierbar, so daß
\begin_inset Formula \( T^{-1}AT \)
\end_inset
Jordan-Normalform.
\begin_inset Formula \begin{eqnarray*}
T^{-1}AT & = & \left( \begin{array}{ccc}
\lambda _{1} & & *\\
& \ddots & \\
0 & & \lambda _{n}
\end{array}\right) =X\\
X^{2} & = & \left( \begin{array}{ccc}
\lambda _{1}^{2} & & *\\
& \ddots & \\
0 & & \lambda _{n}^{2}
\end{array}\right) \\
e^{X} & = & \left( \begin{array}{ccc}
e^{\lambda _{1}} & & *\\
& \ddots & \\
0 & & e^{\lambda _{n}}
\end{array}\right) \\
det\, e^{A} & = & det\, e^{X}=\prod _{k=1}^{n}e^{\lambda _{k}}=e^{\sum
_{k=0}^{n}\lambda _{k}}=e^{tr\, X}=e^{tr\, A}
\end{eqnarray*}
\end_inset
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
\frac{d}{dt}vol_{n}\left( \phi _{t}\left( M\right) \right) & = & \int _{M}div\,
v\left( \phi _{t}\left( x\right) \right) \left[ det\, \phi '\left( x\right) \right]
dx\\
& &
\end{eqnarray*}
\end_inset
Wähle v so, daß
\begin_inset Formula \( \phi _{t}\left( x\right) =e^{At}x, \)
\end_inset
d.h.
wähle
\begin_inset Formula \( v\left( x\right) =Ax \)
\end_inset
.
\begin_inset Formula \begin{eqnarray*}
\frac{d}{dt}vol_{M}\left( e^{At}M\right) & = & \int _{M}div\, v\left( e^{At}x\right)
\cdot det\left( e^{At}\cdot \mathbb 1\right) dx\\
\frac{d}{dt}\left( det\, e^{At}\cdot vol\left( M\right) \right) & = & \int
_{M}tr\left( A\right) \cdot det\, e^{At}dx\\
\frac{d}{dt}\left( det\, e^{At}\right) \cdot vol\left( M\right) & = & tr\, A\, det\,
e^{At}\int _{M}dx\\
\frac{d}{dt}det\, e^{At} & = & tr\, A\, det\, e^{At}\\
\Rightarrow f\left( t\right) & = & const\cdot e^{tr\, A\cdot t}
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
\SpecialChar ~
\layout Enumerate
Man beschreibe die Zustandsabbildung
\begin_inset Formula \( \phi _{t} \)
\end_inset
für das Vektorfeld
\begin_inset Formula \( v\left( \begin{array}{c}
x\\
y
\end{array}\right) =\left( \begin{array}{cc}
1 & -1\\
1 & 1
\end{array}\right) \left( \begin{array}{c}
x\\
y
\end{array}\right) . \)
\end_inset
\layout Enumerate
Wie sehen die Bildmengen des Quadrates
\begin_inset Formula \( Q=\left\{ \left( x,y\right) :\left| x-2\right| \leq 1,\left|
y\right| \leq 1\right\} \)
\end_inset
unter
\begin_inset Formula \( \phi _{1} \)
\end_inset
und
\begin_inset Formula \( \phi _{2} \)
\end_inset
aus?
\layout Enumerate
Verifizieren Sie, daß die Divergenz des Vektorfeldes v konstant ist.
Was folgt daraus für die Flächeninhalte von
\begin_inset Formula \( \phi _{t}\left( A\right) \)
\end_inset
für eine Teilmenge
\begin_inset Formula \( a\subset \mathbb R^{2} \)
\end_inset
? Verifizieren Sie Ihr Ergebnis am Beispiel des obigen Quadrats Q.
\layout Problem
\line_top
Sei
\begin_inset Formula \( \alpha :I\rightarrow \mathbb R \)
\end_inset
eine stetige Funktion ohne Nullstelle auf einem Intervall
\begin_inset Formula \( I\subset \left[ 0,\infty \right) \)
\end_inset
.
Man ermittle für das Vektorfeld
\begin_inset Formula \[
v\left( \begin{array}{c}
x\\
y
\end{array}\right) =\alpha \left( r\right) \cdot \left( \begin{array}{c}
-y\\
x
\end{array}\right) \, mit\, r:=\sqrt{x^{2}+y^{2}}\]
\end_inset
auf dem Kreisring
\begin_inset Formula \( K_{1}:=\left\{ \left( x,y\right) \in \mathbb
R^{2}:\sqrt{x^{2}+y^{2}}\in I\right\} \)
\end_inset
einen globalen Fluß
\begin_inset Formula \( \phi :\mathbb R\times K_{I}\rightarrow \mathbb R^{2}. \)
\end_inset
\layout Standard
\SpecialChar ~
\layout Problem
\line_top
Seien
\begin_inset Formula \( \omega ,\omega _{1},\omega _{2} \)
\end_inset
1-Formen und f eine Funktion auf
\begin_inset Formula \( U\subset \mathbb R^{n}. \)
\end_inset
Man definiert 1-Formen auf
\begin_inset Formula \( \omega _{1}+\omega _{2}\, und\, f\omega \)
\end_inset
durch
\begin_inset Formula \begin{eqnarray*}
\left( \omega _{1}+\omega _{2}\right) & := & \omega _{1}\left( x\right) h+\omega
_{2}\left( x\right) h\\
\left( f\omega \right) \left( x\right) h & := & f\left( x\right) \omega \left(
x\right) h.
\end{eqnarray*}
\end_inset
\layout Enumerate
Wie erhält man aus den Koordinatendarstellungen von
\begin_inset Formula \( \omega ,\omega _{1}\, und\, \omega _{2} \)
\end_inset
diejenigen von
\begin_inset Formula \( \omega _{1}+\omega _{2} \)
\end_inset
und
\begin_inset Formula \( f\omega \)
\end_inset
?
\layout Enumerate
Wenn
\begin_inset Formula \( \omega _{1} \)
\end_inset
und
\begin_inset Formula \( \omega _{2} \)
\end_inset
exakt sind, dann ist es auch
\begin_inset Formula \( \omega _{1}+\omega _{2} \)
\end_inset
.
Gilt analoges für das Produkt
\begin_inset Formula \( f\omega \)
\end_inset
?
\layout Problem
\line_top
\series bold
Techniken zur Berechnung von Stammfunktionen
\begin_inset LatexCommand \index{Stammfunktionen}
\end_inset
\layout Enumerate
Durch Integration längs radialer Wege ermittle man eine Stammfunktion zu
\begin_inset Formula \[
\omega =\left( x^{2}-yz\right) dx+\left( y^{2}-xz\right) dy-xy\, dz.\]
\end_inset
\layout Enumerate
Es sei
\begin_inset Formula \( \omega =f_{1}dx_{1}+f_{2}dx_{2} \)
\end_inset
eine
\begin_inset Formula \( \)
\end_inset
auf einem Rechteck
\begin_inset Formula \( R=I\times J. \)
\end_inset
Man zeige:
\begin_inset Formula \( \omega \)
\end_inset
besitzt höchsten dann eine Stammfunktion auf R, wenn die sogenannte Integrabili
tätsbedingung
\begin_inset Formula \( \partial _{1}f_{2}=\partial _{2}f_{1} \)
\end_inset
ist.
Gegebenenfalls ermittle man eine Kandidatin f für eine Stammfunktion durch
achsenparallele Integration und zeige, daß
\begin_inset Formula \( df=\omega . \)
\end_inset
\layout Enumerate
Anwendung: Man verifiziere, daß die Windungsform
\begin_inset Formula \( \omega _{W}=\frac{-y\, dx+x\, dy}{x^{2}+y^{2}} \)
\end_inset
die Integrabilitätsbedingung auf
\begin_inset Formula \( \mathbb R^{2}\smallsetminus \left\{ 0\right\} \)
\end_inset
erfüllt und berechne eine Stammfunktion in der rechten Halbebene.
\newline
Warum kann
\begin_inset Formula \( \omega _{W} \)
\end_inset
auf
\begin_inset Formula \( \mathbb R^{2}\smallsetminus \left\{ 0\right\} \)
\end_inset
keine Stammfunktion haben?
\layout Enumerate
Man berechne eine Stammfunktion auf der längs der negativen x-Achse geschlitzen
Ebenen
\begin_inset Formula \( \mathbb R^{2}\smallsetminus S,\, S=\left\{ \left( x,0\right)
:x\leq 0\right\} \)
\end_inset
, durch Integration.
\layout Problem
\line_top
Für eine 1-Form
\begin_inset Formula \( \omega =\sum _{i=1}^{n}a_{i}dx_{i} \)
\end_inset
definiere man
\begin_inset Formula \( \left\Vert \omega \left( x\right) \right\Vert _{2}:=\sqrt{\sum
_{i=1}^{n}\left| a_{i}\left( x\right) \right| ^{2}} \)
\end_inset
.
Man zeige: Ist
\begin_inset Formula \( \omega \)
\end_inset
eine stetige 1-Form auf U und
\begin_inset Formula \( \gamma :\left[ a,b\right] \rightarrow U \)
\end_inset
ein Integrationsweg, so gilt:
\begin_inset Formula \[
\left| \int _{\gamma }\omega \right| \leq \max _{t\in \left[ a,b\right] }\left\Vert
\omega \left( \gamma \left( t\right) \right) \right\Vert _{2}\cdot L\left( \gamma
\right) ,\]
\end_inset
wobei
\begin_inset Formula \( L\left( \gamma \right) \)
\end_inset
die Bogenlänge\SpecialChar \@.
von
\begin_inset Formula \( \gamma \)
\end_inset
bezeichnet.
\layout Standard
\SpecialChar ~
\layout Problem
\line_top
Es sei F ein
\begin_inset Formula \( \)
\end_inset
- bzw.
\begin_inset Formula \( \)
\end_inset
-Vertorfeld auf einer offenen Menge
\begin_inset Formula \( U\subset \mathbb R^{3} \)
\end_inset
.
Man zeige:
\layout Enumerate
Für jeden Vektor
\begin_inset Formula \( v\in \mathbb R^{3} \)
\end_inset
gilt
\begin_inset Formula \( rot\left( F\times v\right) =\partial _{v}F-div\, F\cdot v \)
\end_inset
.
\layout Enumerate
\begin_inset Formula \( div\, rot\, F=0 \)
\end_inset
\layout Problem
\line_top
Berchnen Sie das Differential und die Ableitung von
\begin_inset Formula \( f:\mathbb R^{n}\smallsetminus \left\{ 0\right\} \rightarrow
\mathbb R \)
\end_inset
für
\layout Enumerate
\begin_inset Formula \[
f\left( x\right) =\frac{x^{T}Ax}{\left\Vert x\right\Vert _{2}}\]
\end_inset
\layout Enumerate
\begin_inset Formula \[
f\left( x\right) =e^{-1/\left\Vert x\right\Vert _{2}^{2}}\]
\end_inset
\layout Standard
\SpecialChar ~
\layout Enumerate
Siehe auch Problem
\begin_inset LatexCommand \ref{xTAx}
\end_inset
!
\begin_inset Formula \begin{eqnarray*}
\partial _{k}\frac{1}{\left\Vert x\right\Vert _{2}} & = & \partial
_{k}\frac{1}{\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+..+x_{n}^{2}}}=-\frac{x_{k}}{\left\Vert
x\right\Vert _{2}^{3}}\\
\left( \frac{1}{\left\Vert x\right\Vert _{2}}\right) '\left( x\right) & = &
-\frac{x^{T}}{\left\Vert x\right\Vert _{2}}\\
df\left( x\right) h & = & d\left( x^{T}Ax\cdot \frac{1}{\left\Vert x\right\Vert
_{2}}\right) \left( x\right) h=d\left( x^{T}Ax\right) h\cdot \frac{1}{\left\Vert
x\right\Vert _{2}}-x^{T}Ax\cdot \frac{x^{T}}{\left\Vert x\right\Vert _{2}^{3}}h=\\
& = & x^{T}\left( A+A^{T}\right) h\cdot \frac{1}{\left\Vert x\right\Vert
_{2}}+x^{T}Ax\cdot \frac{x^{T}}{\left\Vert x\right\Vert _{2}}h=\frac{1}{\left\Vert
x\right\Vert _{2}}\left( x^{T}Ah+x^{T}A^{T}h-\frac{x^{T}Axx^{T}}{\left\Vert
x\right\Vert _{2}^{2}}h\right) =\\
& = & \frac{x^{T}}{\left\Vert x\right\Vert _{2}}\left(
A+A^{T}-\frac{Axx^{T}}{\left\Vert x\right\Vert _{2}^{2}}\right) h\\
f'\left( x\right) & = & \frac{x^{T}}{\left\Vert x\right\Vert _{2}}\left(
A+A^{T}-\frac{Axx^{T}}{\left\Vert x\right\Vert _{2}^{2}}\right)
\end{eqnarray*}
\end_inset
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
f\left( x\right) & = & e^{-1/\left\Vert x\right\Vert _{2}^{2}}\\
df\left( x\right) h & = & e^{-1/\left\Vert x\right\Vert _{2}^{2}}\cdot \left(
-\frac{2}{\left\Vert x\right\Vert _{2}^{3}}\right) \cdot \frac{x^{T}}{\left\Vert
x\right\Vert _{2}}h=\left( -2e^{-1/\left\Vert x\right\Vert _{2}^{2}}\right)
\frac{x^{T}}{\left\Vert x\right\Vert _{2}^{4}}h\\
f''\left( x\right) & = & \left( -2e^{-1/\left\Vert x\right\Vert _{2}^{2}}\right)
\frac{x^{T}}{\left\Vert x\right\Vert _{2}^{4}}h
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Die Funtionen
\begin_inset Formula \( f:\mathbb R^{2}\rightarrow \mathbb R \)
\end_inset
mit
\begin_inset Formula \( f\left( 0,0\right) =0 \)
\end_inset
und
\begin_inset Formula \[
f\left( x,y\right) =\frac{xy^{2}}{x^{2}+y^{4}}\]
\end_inset
ist im Nullpunkt unstetig, hat dort aber Ableitungen in jeder Richtung.
\layout Standard
\begin_inset Formula \begin{eqnarray*}
\partial _{h}f\left( 0,0\right) & = & \lim _{t\rightarrow 0}\frac{f\left( \left(
0,0\right) +t\left( h_{1},h_{2}\right) \right) -f\left( \left( 0,0\right) \right)
}{t}=\lim _{t\rightarrow 0}\frac{\left( th_{1}\right) \left( th_{2}\right)
^{2}}{\left[ \left( \left( th_{1}\right) ^{2}+\left( th_{2}\right) ^{4}\right) \right]
t}=\\
& = & \lim _{t\rightarrow 0}\frac{t^{3}h_{1}h_{2}^{2}}{\left[
h_{1}^{2}+t^{2}h_{2}^{4}\right] t^{3}}=\lim _{t\rightarrow
0}\frac{h_{1}h_{2}^{2}}{\left[ h_{1}^{2}+t^{2}h_{2}^{4}\right]
}=\frac{h_{1}h_{2}^{2}}{h_{1}^{2}}=\frac{h^{2}}{h_{1}}\, falls\, h_{2}\neq 0\\
& = & \lim _{t\rightarrow 0}h_{1}\cdot \frac{h_{2}^{2}}{\left[ t^{2}h_{2}^{4}\right]
}=0\, falls\, h_{2}=0\\
& = & 0\, f\ddot{u}r\, h=\left( 0,0\right) \, nV
\end{eqnarray*}
\end_inset
f ist unstetig bei
\begin_inset Formula \( \left( 0,0\right) \)
\end_inset
:
\begin_inset Formula \begin{eqnarray*}
\left( \varepsilon ^{2},\varepsilon \right) & \mapsto & f\left( \varepsilon
^{2},\varepsilon \right) =\frac{\left( \varepsilon ^{2}\right) ^{2}}{2\varepsilon
^{4}}=\frac{1}{2}\\
\lim _{\varepsilon \rightarrow 0}\left( \varepsilon ^{2},\varepsilon \right) & = &
\left( 0,0\right) \\
\lim _{\varepsilon \rightarrow 0}f\left( \varepsilon ^{2},\varepsilon \right) & = &
\frac{1}{2}
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
\SpecialChar ~
\layout Enumerate
Bestimmen Sie den Gradienten der Funktion
\begin_inset Formula \( f:\mathbb R^{3}\rightarrow \mathbb R,\, \left( x,y,z\right)
\mapsto x^{2}y+z \)
\end_inset
im Punkt
\begin_inset Formula \( \left( 2,3,5\right) \)
\end_inset
bezüglich des Skalarproduktes
\begin_inset Formula \[
\left\langle v,w\right\rangle :=v^{T}\left( \begin{array}{ccc}
1 & 0 & 0\\
0 & 3 & 0\\
0 & 0 & 2
\end{array}\right) w.\]
\end_inset
\layout Enumerate
Sei
\begin_inset Formula \( \left( G_{ij}\right) \)
\end_inset
eine positiv definite, symmetrische
\begin_inset Formula \( n\times n \)
\end_inset
-Matrix.
Dann definiert
\begin_inset Formula \( \left\langle v,w\right\rangle _{G}:=v^{T}\cdot G\cdot w \)
\end_inset
ein Skalarprodukt auf
\begin_inset Formula \( \mathbb R^{n} \)
\end_inset
.
Bestimmen Sie für eine allgemeine Funktion
\begin_inset Formula \( g\in \mathbb C^{1}\left( \mathbb R^{n},\mathbb R\right) \)
\end_inset
den Gradienten
\begin_inset Formula \( grad_{G}g\left( a\right) \)
\end_inset
in einem Punkt a bezüglich dieses Skalarproduktes!
\layout Standard
\SpecialChar ~
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
\partial _{x}f\left( 2,3,5\right) & = & 12\\
\partial _{y}f\left( 2,3,5\right) & = & 4\\
\partial _{z}f\left( 2,3,5\right) & = & 1\\
df\left( a\right) h=12h_{1}+4h_{2}+h_{3} & = & \left\langle grad\, f\left( a\right)
,h\right\rangle =\left( \alpha ,\beta ,\gamma \right) \left( \begin{array}{ccc}
1 & 0 & 0\\
0 & 3 & 0\\
0 & 0 & 2
\end{array}\right) \left( \begin{array}{c}
h_{1}\\
h_{2}\\
h_{3}
\end{array}\right) =\\
& = & \alpha h_{1}+3\beta h_{2}+2\gamma h_{3}\\
\Rightarrow & & \alpha =12;\, \beta =\frac{4}{3};\, \gamma =\frac{1}{2}\\
\Rightarrow & & grad\, f\left( a\right) =\left( \begin{array}{c}
12\\
\frac{4}{3}\\
\frac{1}{2}
\end{array}\right)
\end{eqnarray*}
\end_inset
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
\left\langle grad\, g\left( a\right) ,h\right\rangle & = & dg\left( a\right) \cdot h\\
\left[ grad\, g\left( a\right) \right] ^{T}Gh & = & dg\left( a\right) h=\left(
\partial _{1}g,\partial _{2}g,...,\partial _{n}g\right) h\\
\left[ grad\, g\left( a\right) \right] ^{T}G & = & \left( \partial _{1}g,\partial
_{2}g,...,\partial _{n}g\right) \left( a\right) \\
\left[ grad\, g\left( a\right) \right] ^{T} & = & \left( \partial _{1}g,\partial
_{2}g,...,\partial _{n}g\right) G^{-1}=g'\left( a\right) \cdot G^{-1}\\
grad\, g\left( a\right) & = & \left( G^{-1}\right) ^{T}\left( g'\left( a\right)
\right) ^{T}
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Es sei
\begin_inset Formula \( f:\mathbb R^{2}\rightarrow \mathbb R \)
\end_inset
eine
\begin_inset Formula \( \)
\end_inset
-Funktion.
Durch Komposition mit einer Polarkoordinatenabbildung
\begin_inset Formula \( P_{2} \)
\end_inset
bilde man
\begin_inset Formula \( F:=f\circ P_{2}, \)
\end_inset
\begin_inset Formula \[
F\left( r,\varphi \right) =f\left( r\, cos\, \varphi ,r\, sin\, \varphi \right) .\]
\end_inset
Man zeige: In jedem Punkt
\begin_inset Formula \( \left( x,y\right) :=\left( r\, cos\, \varphi ,r\, sin\,
\varphi \right) \)
\end_inset
mit
\begin_inset Formula \( r\neq 0 \)
\end_inset
gilt:
\begin_inset Formula \[
\nabla f\left( x,y\right) =\left( F_{rr}+\frac{1}{r^{2}}F_{\varphi \varphi
}+\frac{1}{r}F_{r}\right) \left( r,\varphi \right) \]
\end_inset
\layout Standard
\begin_inset Formula \begin{eqnarray*}
F_{r}\left( r,\varphi \right) & = & \frac{\partial }{\partial r}f\left( r\, cos\,
\varphi ,r\, sin\, \varphi \right) =\frac{\partial f}{\partial x}\frac{\partial
x}{\partial v}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial v}=f_{x}cos\,
\varphi +f_{y}sin\, \varphi \\
F_{rr}\left( r,\varphi \right) & = & f_{xx}cos^{2}\varphi +f_{xy}sin\, \varphi \,
cos\, \varphi +f_{yx}sin\, \varphi \, cos\, \varphi +f_{yy}sin^{2}\varphi =\\
& = & f_{xx}cos^{2}\varphi +2f_{xy}sin\, \varphi \, cos\, \varphi
+f_{yy}sin^{2}\varphi \\
F_{\varphi }\left( r,\varphi \right) & = & -f_{x}r\, sin\, \varphi +f_{y}r\, cos\,
\varphi \\
F_{\varphi \varphi }\left( r,\varphi \right) & = & \left[ f_{xx}\left( -r\, sin\,
\varphi \right) +f_{xy}r\, cos\, \varphi \right] \left( -r\, sin\, \varphi \right)
-f_{x}r\, cos\, \varphi +\\
& & +\left[ f_{yx}\left( -r\, sin\, \varphi \right) +f_{yy}r\, cos\, \varphi \right]
\left( r\, cos\, \varphi \right) -f_{y}r\, sin\, \varphi \\
F_{rr}+\frac{1}{r}F_{r}+\frac{1}{r^{2}}F_{\varphi \varphi } & = & ...=f_{xx}\left(
cos^{2}\varphi +sin^{2}\varphi \right) +f_{yy}\left( sin^{2}\varphi +cos^{2}\varphi
\right) =\nabla f
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Sei
\begin_inset Formula \( \)
\end_inset
Dann ist
\begin_inset Formula \( \psi :\mathbb R^{n}\times \mathbb R\rightarrow \mathbb R, \)
\end_inset
\begin_inset Formula \[
\psi \left( x,t\right) :=f\left( \left\langle v,x\right\rangle -c\left\Vert
v\right\Vert _{2}t\right) \]
\end_inset
eine Lösung der Wellengleichung
\begin_inset Formula \[
\triangle _{x}\psi =\frac{1}{c^{2}}\psi _{tt}.\]
\end_inset
\layout Standard
\begin_inset Formula \begin{eqnarray*}
\frac{\partial }{\partial x_{i}}\psi & = & f'\cdot v_{i}\qquad \triangle _{x}\psi
=f''\left( ...\right) \left\Vert v\right\Vert _{2}^{2}\\
\frac{\partial ^{2}}{\partial x_{i}^{2}}\psi & = & f''\cdot v_{i}^{2}\\
\frac{\partial }{\partial t}\psi & = & f'\cdot \left( -c\left\Vert v\right\Vert
_{2}\right) \\
\frac{\partial ^{2}}{\partial t^{2}}\psi & = & f''\cdot c^{2}\left\Vert v\right\Vert
_{2}^{2}=\psi _{tt}=c^{2}\triangle _{x}\psi
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Beispiel für
\begin_inset Formula \( \partial _{ij}\neq \partial _{ji}: \)
\end_inset
\newline
Es sei
\begin_inset Formula \( f:\mathbb R^{2}\rightarrow \mathbb R \)
\end_inset
gegeben durch
\begin_inset Formula \( f\left( 0,0\right) =0 \)
\end_inset
und
\begin_inset Formula \[
f\left( x,y\right) :=\frac{x^{3}y-xy^{3}}{x^{2}+y^{2}}\, f\ddot{u}r\, \left(
x,y\right) \neq \left( 0,0\right) .\]
\end_inset
Man zeige:
\layout Enumerate
f ist eine
\begin_inset Formula \( \)
\end_inset
-Funktion auf
\begin_inset Formula \( \)
\end_inset
\layout Enumerate
\begin_inset Formula \( \partial _{xy}f \)
\end_inset
und
\begin_inset Formula \( \partial _{yx}f \)
\end_inset
existieren auf
\begin_inset Formula \( \)
\end_inset
und sind stetig auf
\begin_inset Formula \( \mathbb R^{2}\smallsetminus \left\{ 0\right\} \)
\end_inset
.
\layout Enumerate
\begin_inset Formula \( \partial _{xy}f\left( 0,0\right) =1 \)
\end_inset
und
\begin_inset Formula \( \partial _{yx}\left( 0,0\right) =-1 \)
\end_inset
.
\layout Standard
\SpecialChar ~
\layout Standard
\begin_inset Formula \begin{eqnarray*}
\partial _{x}f\left( x,y\right) & = & \frac{x^{4}y+4x^{2}y^{3}-y^{4}}{\left(
x^{2}+y^{2}\right) ^{2}}\, f\ddot{u}r\, \left( x,y\right) \neq \left( 0,0\right) \\
\partial _{x}f\left( 0,0\right) & = & 0\, f\ddot{u}r\, x=\left( 0,0\right) \\
\partial _{y}f\left( x,y\right) & = & \frac{x\left( x^{4}-4x^{2}y^{2}-y^{4}\right)
}{\left( x^{2}+y^{2}\right) ^{2}}
\end{eqnarray*}
\end_inset
für Stetigkeit von
\begin_inset Formula \( \partial _{x}f \)
\end_inset
verwende man
\begin_inset Formula \( \frac{x^{2}}{x^{2}+y^{2}}\leq 1,\frac{y^{2}}{x^{2}+y^{2}}\leq
1 \)
\end_inset
für
\begin_inset Formula \( \left( x,y\right) \neq 0 \)
\end_inset
.
\begin_inset Formula \( \left| \partial _{x}f\left( x,y\right) -\partial _{x}f\left(
0,0\right) \right| =\frac{\left| x^{4}y\right| +\left| 4x^{2}y^{2}y\right| +\left|
y^{4}y\right| }{\left( x^{2}+y^{2}\right) ^{2}}\leq \left| y\right| +4\left| y\right|
+\left| y\right| =6\left| y\right| \leq 6\left\Vert \left( x,y\right) \right\Vert \)
\end_inset
.
Sei
\begin_inset Formula \( \varepsilon >0 \)
\end_inset
.
Wähle:
\begin_inset Formula \( \delta =\frac{\varepsilon }{6}\Rightarrow \forall \left(
x,y\right) \in \mathbb R^{2}:\left\Vert \left( x,y\right) -\left( 0,0\right)
\right\Vert <\delta \)
\end_inset
.
\begin_inset Formula \begin{eqnarray*}
\left\Vert \partial _{x}f\left( x,y\right) -\partial _{x}f\left( 0,0\right)
\right\Vert & < & \varepsilon \\
\partial _{y}\partial _{x}f & = & \frac{x^{6}+9x^{4}y^{2}-9x^{2}y^{4}-y^{6}}{\left(
x^{2}+y^{2}\right) ^{3}}\, f\ddot{u}r\, \left( x,y\right) \neq \left( 0,0\right) \\
\left( \partial _{x}f\right) \left( 0,y\right) & = & -y\, f\ddot{u}r\, y\neq 0\,
und\, auch\, \partial _{x}f\left( 0,0\right) =0\\
\partial _{y}\partial _{x}f\left( 0,0\right) & = & -1,\, analog\, \partial
_{x}\partial _{y}f=1
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
\SpecialChar ~
\layout Enumerate
Man beweise die folgende Version einer allgemeinen Kettenregel: Seien
\begin_inset Formula \( f:U\rightarrow \mathbb R,\, U\subset \mathbb R^{n} \)
\end_inset
und
\begin_inset Formula \( \)
\end_inset
-Funktionen.
Dann ist auch
\begin_inset Formula \( g\circ f \)
\end_inset
eine
\begin_inset Formula \( \)
\end_inset
-Funktion und es gilt:
\begin_inset Formula \[
\left( g\circ f\right) '=\left( g'\circ f\right) \cdot f'.\]
\end_inset
\layout Enumerate
Man zeige, daß
\begin_inset Formula \( f:K\rightarrow \mathbb R,\, x\mapsto e^{\frac{-1}{1-\left\Vert
x\right\Vert _{2}^{2}}} \)
\end_inset
eine
\begin_inset Formula \( \)
\end_inset
-Funktion ist, wobei
\begin_inset Formula \( K:=\left\{ x\in \mathbb R^{N}:\left\Vert x\right\Vert
_{2}<1\right\} \)
\end_inset
.
\layout Standard
\SpecialChar ~
\layout Enumerate
Induktion nach k:
\newline
\begin_inset Formula \begin{eqnarray*}
k=1 & : & \widehat{f_{i}}\left( t\right) =f\left( a+te_{i}\right) \textrm{
differenzierbar}\\
& & \partial _{i}\left( g\circ f\right) \left( a\right) =\frac{d}{dt}\left( g\circ
\widehat{f_{i}}\right) \left( a\right) =g'\left( \widehat{f_{i}}\left( a\right)
\right) \cdot \widehat{f_{i}}'\left( a\right) =g'\left( f\left( a\right) \right)
\partial _{i}f\left( a\right) \in \mathcal C^{1}\\
& & \left( g\circ f\right) \left( a\right) =\left( g'\circ f\right) \left( a\right)
\cdot f'\left( a\right) \textrm{ stetig partiell differenzierbar }\Rightarrow \mathcal
C^{1}\\
k\rightarrow k+1 & : & \textrm{Seien }f,g\in \mathcal C^{k+1}\textrm{ und die
Behauptung bewiesen f}\ddot{\textrm{u}}\textrm{r k}.\\
& & \textrm{Zun}\ddot{\textrm{a}}\textrm{chst f},g\in \mathcal C^{1}\\
& & \Rightarrow \partial _{1}\left( g\circ f\right) =\left( g'\circ f\right)
\partial _{i}f\in \mathcal C^{k}\\
& & \Rightarrow \textrm{jedes }\partial _{i_{1}},...,\partial _{i_{k}}f\textrm{
stetig partiell differenzierbar }\Rightarrow \\
& & \Rightarrow \textrm{stetig differenzierbar}\Rightarrow f\in \mathcal C^{k+1}
\end{eqnarray*}
\end_inset
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
f:\mathbb R^{n}\rightarrow \mathbb R & ; & x\mapsto \left\Vert x\right\Vert _{2}\\
g:\mathbb R\rightarrow \mathbb R & ; & x\mapsto \left\{ \begin{array}{c}
e^{\frac{-1}{1-x}}\\
0
\end{array}\begin{array}{c}
\textrm{auf }\left( -1,1\right) \\
sonst
\end{array}\right\} \\
\hat{f}:\mathbb R^{n}\rightarrow \mathbb R & ; & x\mapsto \left\Vert x\right\Vert
_{2}^{2}\\
\hat{g}:\mathbb R\rightarrow \mathbb R & ; & x\mapsto \left\{ \begin{array}{c}
e^{\frac{-1}{1-x}}\\
0
\end{array}\begin{array}{c}
\textrm{auf }\left( -\infty ,1\right) \\
sonst
\end{array}\right\} \\
\left( g\circ \hat{f}\right) & \in & \mathcal C^{\infty }
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
\SpecialChar ~
\layout Enumerate
Bestimmen Sie die Differentiale
\begin_inset Formula \( d^{\left( 2\right) }\left( f\right) ,d^{\left( 3\right)
}\left( f\right) \)
\end_inset
und die zweiten Ableitungen
\begin_inset Formula \( f'' \)
\end_inset
für die Funktionen
\begin_inset Formula \begin{eqnarray*}
f:\mathbb R^{2}\rightarrow \mathbb R & ; & \left( x,y\right) \mapsto sin\left(
xy\right) \textrm{ im Punkt }\left( \pi ,1\right) \\
g:\mathbb R^{3}\times \mathbb R_{+}\rightarrow \mathbb R & ; & \left( x,y,z\right)
\mapsto xy\, ln\left( x\right) \textrm{ im Punkt }\left( 2,-3,1\right)
\end{eqnarray*}
\end_inset
\layout Enumerate
Berechnen Sie das Taylorpolynom dritter Ordnung für die Funktion
\begin_inset Formula \[
\]
\end_inset
um den Entwicklungspunkt
\begin_inset Formula \( \left( \frac{\pi }{8},\frac{\pi }{16}\right) \)
\end_inset
.
\layout Standard
\SpecialChar ~
\layout Enumerate
\SpecialChar ~
\layout Enumerate
\SpecialChar ~
\layout Problem
\line_top
Charakterisierung des Laplace-Operators durch die Drehinvarianz:
\newline
Der Differentialoperator
\begin_inset Formula \( \)
\end_inset
,
\begin_inset Formula \[
P\left( D\right) f:=\sum _{i,j=1}^{n}c_{ik}\partial _{i}\partial _{k}f\]
\end_inset
mit
\begin_inset Formula \( c_{ik}\in \mathbb R \)
\end_inset
habe die folgende Eigenschaft:
\newline
Für jede
\begin_inset Formula \( \)
\end_inset
-Funktion f auf
\begin_inset Formula \( \mathbb R^{n} \)
\end_inset
und jede orthogonale Matrix
\begin_inset Formula \( A\in \mathbb R^{n\times n} \)
\end_inset
gilt mit der durch
\begin_inset Formula \( x\mapsto f\left( Ax\right) \)
\end_inset
erklärten Funktion
\begin_inset Formula \( f_{A}: \)
\end_inset
\begin_inset Formula \[
\left( P\left( D\right) f_{A}\right) \left( x\right) =\left( P\left( D\right) f\right)
\left( Ax\right) \]
\end_inset
Man zeige:
\begin_inset Formula \( P\left( D\right) =c\triangle \)
\end_inset
mit einer Konstanten
\begin_inset Formula \( \)
\end_inset
.
\layout Proof
oBdA
\begin_inset Formula \( c_{ij}=c_{ji} \)
\end_inset
für alle i,j, sonst ersetze beide durch
\begin_inset Formula \( \frac{c_{ij}+cji}{2}. \)
\end_inset
\newline
Seien nun
\begin_inset Formula \( i,j\in \left\{ 1..n\right\} ;\, i\neq j \)
\end_inset
.
Wähle
\begin_inset Formula \( f\left( x\right) =x_{i}x_{j} \)
\end_inset
;
\begin_inset Formula \[
A:=\left( \begin{array}{ccccc}
1 & & & & 0\\
& \ddots & & & \\
& & \left( -1\right) & & \\
& & & \ddots & \\
0 & & & & 1
\end{array}\right) \]
\end_inset
\begin_inset Formula \( \partial _{ij}f=\partial _{ji}f=1 \)
\end_inset
auf
\begin_inset Formula \( \mathbb R^{n},\, \partial _{k,l}f=0\, sonst \)
\end_inset
\newline
\begin_inset Formula \( P\left( D\right) f=2c_{ij} \)
\end_inset
auf ganz
\begin_inset Formula \( \)
\end_inset
\newline
Andererseits gilt für
\begin_inset Formula \( f_{A}\left( x\right) =f\left( Ax\right) =\left( Ax\right)
_{i}\left( Ax\right) _{j}=-x_{i}x_{j} \)
\end_inset
.
\begin_inset Formula \( P\left( D\right) f_{A}=-2c_{ij} \)
\end_inset
auf ganz
\begin_inset Formula \( \mathbb R^{n}. \)
\end_inset
Insbesondere
\begin_inset Formula \begin{eqnarray*}
\left( P\left( D\right) f\right) \left( Ax\right) & = & 2c_{ij}\\
\left( P\left( D\right) f_{A}\right) \left( x\right) & = & -2c_{ij}\qquad
\Longrightarrow c_{ij}=0
\end{eqnarray*}
\end_inset
\begin_inset Formula \( P\left( D\right) f=\sum _{i=1}^{n}c_{ii}\partial _{ii}f \)
\end_inset
\newline
Wähle
\begin_inset Formula \( f\left( x\right) =x_{1}^{2};\, A\left( x\right) =\left(
\begin{array}{ccccc}
0 & & 1 & & \\
& 1 & & & \\
1 & & 0 & & \\
& & & \ddots & \\
& & & & 1
\end{array}\right) \, \left( i-te\, Zeile\right) \)
\end_inset
\newline
\begin_inset Formula \( \partial _{1}\partial _{1}f=2\textrm{ auf ganz }\mathbb R;\,
\partial _{i}\partial _{j}f=0\textrm{ f}\ddot{\textrm{u}}\textrm{r j}\neq 1 \)
\end_inset
\newline
\begin_inset Formula \( P\left( D\right) f=2c_{11} \)
\end_inset
insbesondere
\begin_inset Formula \( \left( P\left( D\right) f\right) \left( Ax\right) =2c_{11} \)
\end_inset
\newline
\begin_inset Formula \( f_{A}\left( x\right) =f\left( Ax\right) =\left[ \left(
Ax\right) _{1}\right] ^{2}=x_{i}^{2} \)
\end_inset
\newline
\begin_inset Formula \( \left( P\left( D\right) f_{A}\right) \left( x\right) =2c_{ii}
\)
\end_inset
\newline
\begin_inset Formula \( \left( P\left( D\right) \left( f_{A}\right) \right) \left(
x\right) =2c_{ii}=2c_{11} \)
\end_inset
\newline
\begin_inset Formula \( \Rightarrow c_{ii}=c\, \forall i\in \left\{ 1..n\right\}
\Rightarrow P\left( D\right) =c\triangle \)
\end_inset
\layout Problem
\line_top
Man zeige: Wenn
\begin_inset Formula \( f:\mathbb R^{2}\smallsetminus \left\{ 0\right\} \rightarrow
\mathbb R \)
\end_inset
harmonisch ist, dann auch
\begin_inset Formula \( g:\mathbb R^{2}\smallsetminus \left\{ 0\right\} \rightarrow
\mathbb R,x\mapsto f\left( \frac{x}{\left\Vert x\right\Vert _{2}^{2}}\right) \)
\end_inset
.
\layout Standard
\begin_inset Formula \begin{eqnarray*}
g\left( x\right) & = & f\left( h\left( x\right) \right) \textrm{ mit h}\left(
\textrm{x}\right) =\frac{x}{\left\Vert x\right\Vert _{2}^{2}}=\frac{\left(
x_{1},x_{2}\right) ^{T}}{x_{1}^{2}+x_{2}^{2}}\\
h_{1} & = & \frac{\left( 1,0\right) \left( x_{1}^{2}+x_{2}^{2}\right) -\left(
x_{1},x_{2}\right) \cdot 2x_{1}}{\left( x_{1}^{2}+x_{2}^{2}\right) ^{2}}=\frac{\left(
x_{2}^{2}-x_{1}^{2},-2x_{1}x_{2}\right) }{\left( x_{1}^{2}+x_{2}^{2}\right) ^{2}}\\
h_{2} & = & \frac{\left( -2x_{1}x_{2},-x_{2}^{2}+x_{1}^{2}\right) }{\left(
x_{1}^{2}+x_{2}^{2}\right) ^{2}}\, \textrm{aus Symmetrie }\\
h_{11} & = & \frac{\left( -2x_{1},-2x_{2}\right) \left( x_{1}^{2}+x_{2}^{2}\right)
^{2}-\left( x_{2}^{2}-x_{1}^{2},-2x_{1}x_{2}\right) 2\left( x_{1}^{2}+x_{2}^{2}\right)
\cdot 2x_{1}}{\left( x_{1}^{2}+x_{2}^{2}\right) ^{2}}=\\
& = & \frac{\left( 2x_{1}^{3}-6x_{1}x_{2}^{2},-2x_{2}^{3}+6x_{1}^{2}x_{2}\right)
}{\left( x_{1}^{2}+x_{2}^{2}\right) ^{3}}\\
h_{22} & = & \frac{\left(
-2x_{1}^{3}+6x_{1}x_{2}^{2},2x_{2}^{3}-6x_{1}^{2}x_{2}\right) }{\left(
x_{1}^{2}+x_{2}^{2}\right) ^{3}}\\
\Rightarrow & & Ah=h_{11}+h_{22}=0\Rightarrow \textrm{ h ist harmonisch}
\end{eqnarray*}
\end_inset
\layout Standard
\begin_inset Formula \begin{eqnarray*}
\triangle g & = & g_{11}+g_{22}\\
g_{1} & = & f_{1}h_{1}+f_{2}h_{2}\\
g_{11} & = &
f_{11}h_{1}^{2}+h_{12}h_{1}^{2}+f_{1}h_{11}+f_{12}h_{1}^{2}+f_{22}h_{1}^{2}+f_{2}h_{11}\\
g_{22} & = &
f_{22}h_{2}^{2}+f_{12}h_{2}^{2}+f_{2}h_{22}+f_{12}h_{2}^{2}+f_{11}h_{2}^{2}+f_{1}h_{22}\\
\triangle g & = &
g_{11}+g_{22}=f_{11}h_{1}^{2}+h_{12}h_{1}^{2}+f_{1}h_{11}+f_{12}h_{1}^{2}+f_{22}h_{1}^{2}+f_{2}h_{11}+\\
& &
+f_{22}h_{2}^{2}+f_{12}h_{2}^{2}+f_{2}h_{22}+f_{12}h_{2}^{2}+f_{11}h_{2}^{2}+f_{1}h_{22}=\\
& = & \left( f_{11}+f_{22}\right) \left( h_{1}^{2}+h_{2}^{2}\right) +\left(
f_{1}+f_{2}\right) \left( h_{11}^{2}+h_{22}^{2}\right) +2f_{12}\left(
h_{1}^{2}+h_{2}^{2}\right) =\\
& = & 2f_{12}\left( h_{1}^{2}+h_{2}^{2}\right)
\end{eqnarray*}
\end_inset
\begin_inset Formula \( \Rightarrow g \)
\end_inset
harmonisch
\layout Problem
\line_top
Man untersuche Folgende Funktion auf Extrema:
\begin_inset Formula \[
f\left( x,y,z\right) =x^{2}+y^{2}+z^{2}-2xyz\textrm{ auf }\mathbb R^{3}\]
\end_inset
\layout Standard
\begin_inset Formula \begin{eqnarray*}
\partial _{x}f\left( x,y,z\right) & = & 2x-2yz\\
\partial _{y}f\left( x,y,z\right) & = & 2y-2xz\\
\partial _{z}f\left( x,y,z\right) & = & 2z-2xy\\
grad\, f & = & 0\, \Leftrightarrow \left( \begin{array}{c}
2x-2yz\\
2y-2xz\\
2z-2xy
\end{array}\right) =0\Leftrightarrow \left\{ \begin{array}{c}
x=yz\\
y=xz\\
z=xy
\end{array}\right\} \Rightarrow \left\{ \begin{array}{c}
y^{2}=1\\
x^{2}=1\\
z^{2}=1
\end{array}\right\} \\
Extrema & : & P\left( x,y,z\right) \, mit\, \left| x\right| =\left| y\right| =\left|
z\right| =1;\, xyz=1\\
& & P\left( 0,0,0\right) \\
f''\left( x,y,z\right) & = & \left( \begin{array}{ccc}
2 & -2z & -2y\\
-2z & 2 & -2x\\
-2y & -2x & 2
\end{array}\right) \\
f''\left( 0,0,0\right) & = & \left( \begin{array}{ccc}
2 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 2
\end{array}\right) \textrm{ positiv definit }\Rightarrow Minimum\, \left( 0,0,0\right)
\\
f''\left( 1,1,1\right) & = & \left( \begin{array}{ccc}
2 & -2 & -2\\
-2 & 2 & -2\\
-2 & -2 & 2
\end{array}\right) \\
& &
\end{eqnarray*}
\end_inset
\layout Standard
Eigenwerte:
\begin_inset Formula \begin{eqnarray*}
det\left( f''\left( 1,1,1\right) -\lambda \mathbb E\right) & = & \left( 2-\lambda
\right) ^{3}-8\cdot 2-4\left( 2-\lambda \right) \cdot 3=\\
& = & 8-12\lambda +6\lambda ^{2}-\lambda ^{3}-16-24+12\lambda =\lambda ^{3}+6\lambda
^{2}-32=0\\
\Rightarrow \lambda _{1} & = & -2\\
\lambda _{2|3} & = & 4\\
& \Rightarrow & Sattelpunkt\, in\, \left( 1,1,1\right) \\
f''\left( -1,-1,1\right) & = & \left( \begin{array}{ccc}
2 & -2 & 2\\
-2 & 2 & 2\\
2 & 2 & 2
\end{array}\right) \\
det\left( f''\left( 1,1,1\right) -\lambda \mathbb E\right) & = & \left( 2-\lambda
\right) ^{3}-16-12\left( 2-\lambda \right) =0\\
\Rightarrow \lambda _{1} & = & -2\\
\lambda _{2|3} & = & 4\\
& \Rightarrow & Sattelpunkt\, in\, \left( 1,1,1\right)
\end{eqnarray*}
\end_inset
Da jede der drei Koordinaten in der exakt gleichen Form auftreten, gilt
analog:
\newline
\begin_inset Formula \( \left( -1,1,-1\right) \)
\end_inset
und
\begin_inset Formula \( \left( 1,-1,-1\right) \)
\end_inset
usw.
sind Sattelpunkte.
\layout Problem
\line_top
Man zeige, daß die Funktion
\begin_inset Formula \( f:\mathbb R^{2}\rightarrow \mathbb R \)
\end_inset
,
\begin_inset Formula \[
f\left( x,y\right) =\left( y-x^{2}\right) \left( y-3x^{2}\right) \]
\end_inset
in
\begin_inset Formula \( \left( 0,0\right) \)
\end_inset
kein lokales Minimum hat, daß aber ihre sämtlichen Beschränkungen auf die
Geraden durch
\begin_inset Formula \( \left( 0,0\right) \)
\end_inset
dort isolierte lokale Minima haben.
\layout Standard
\begin_inset Formula \begin{eqnarray*}
f\left( x,y\right) & = & \left( y-x^{2}\right) \left( y-3x^{2}\right)
=y^{2}-4x^{2}y+3x^{4}\\
f_{x}\left( x,y\right) & = & -8xy+12x^{3};\, f_{xx}=-8y+36x^{2}\\
f_{y}\left( x,y\right) & = & 2y-4x^{2};\, f_{yy}=2\\
f_{xy}\left( x,y\right) & = & f_{yx}\left( x,y\right) =-8x\\
\nabla f\left( 0,0\right) & = & \left( 0,0\right) \\
f''\left( 0,0\right) & = & \left( \begin{array}{cc}
0 & 0\\
0 & 2
\end{array}\right) \Rightarrow Eigenwerte\, \lambda _{1}=2;\lambda _{2}=0\Rightarrow
\textrm{keine Aussage}\\
f\left( \varepsilon ,\varepsilon ^{2}\right) & = & \left( \varepsilon
^{2}-\varepsilon ^{2}\right) \left( \varepsilon ^{2}-3\varepsilon ^{2}\right) =0;\,
f\left( 0,0\right) =0\\
\Rightarrow & & \textrm{bez}\ddot{\textrm{u}}\textrm{glich Parabel besitzt f in
}\left( 0,0\right) \textrm{ kein lokales Minimum}\\
f\left( x,ax\right) & = & a^{2}x^{2}-4ax^{3}+3x^{4}\\
f_{x}\left( x,ax\right) & = & 2a^{2}x-12ax^{2}+12x^{3}=0\, in\, \left( 0,0\right) \\
f_{xx}\left( x,ax\right) & = & 2a^{2}-24ax+36x^{2}>0\, \forall a\neq 0\, in\, \left(
0,0\right) \\
f\left( x,0\right) & = & 3x^{4}\textrm{ besitzt in }\left( 0,0\right) \textrm{
Minimum}\\
\Rightarrow & & \textrm{ f hat in }\left( 0,0\right) \textrm{ zu jeder Geraden ein
Minimum}
\end{eqnarray*}
\end_inset
\layout Problem
\line_top
Man untersuche folgende Funktionen auf Extrema (auch Randextrema):
\begin_inset Formula \[
g\left( x,y\right) =y\left( x-1\right) e^{-\left( x^{2}-y^{2}\right) }\textrm{ auf
}\left[ 0;\infty \right) \times \mathbb R\]
\end_inset
\layout Standard
\begin_inset Formula \begin{eqnarray*}
g\left( x,y\right) & = & yxe^{-\left( x^{2}+y^{2}\right) }-ye^{-x^{2}+y^{2}}\\
\partial _{x}g\left( x,y\right) & = & ye^{-\left( x^{2}+y^{2}\right)
}-2x^{2}ye^{-\left( x^{2}+y^{2}\right) }+2xye^{-\left( x^{2}+y^{2}\right) }=\left(
-2x^{2}y+2xy+y\right) e^{-\left( x^{2}+y^{2}\right) }\\
\partial _{y}g\left( x,y\right) & = & xe^{-\left( x^{2}+y^{2}\right)
}-2xy^{2}e^{-\left( x^{2}+y^{2}\right) }-e^{-\left( x^{2}+y^{2}\right)
}+2y^{2}e^{-\left( x^{2}+y^{2}\right) }=\\
& & =\left( -2xy^{2}+2y^{2}+x-1\right) e^{-\left( x^{2}+y^{2}\right) }\\
\partial _{xx}g\left( x,y\right) & = & \left( -4xy+2y\right) e^{-\left(
x^{2}+y^{2}\right) }-2x\left( -2x^{2}y+2xy+y\right) e^{-\left( x^{2}+y^{2}\right) }=\\
& = & 2y\left( 1+2x^{3}-2x^{2}-3x\right) e^{-\left( x^{2}+y^{2}\right) }\\
\partial _{yy}g\left( x,y\right) & = & \left( -4xy+4y\right) e^{-\left(
x^{2}+y^{2}\right) }-2y\left( -2xy^{2}+2y^{2}+x-1\right) e^{-\left( x^{2}+y^{2}\right)
}=\\
& = & 2y\left( 3+2xy^{2}-2y^{2}-3x\right) e^{-\left( x^{2}+y^{2}\right) }\\
\partial _{xy}g\left( x,y\right) & = & \left( -2y^{2}+1\right) e^{-\left(
x^{2}+y^{2}\right) }-2x\left( -2xy+2y^{2}+x-1\right) e^{-\left( x^{2}+y^{2}\right) }=\\
& = & \left( 1-2x^{2}-2y^{2}-4xy+2x+4x^{2}y^{2}\right) e^{-\left( x^{2}+y^{2}\right)
}=\partial _{yx}g\left( x,y\right)
\end{eqnarray*}
\end_inset
Nullstellen der 1.
Ableitung:
\begin_inset Formula \begin{eqnarray*}
-2x^{2}y+2xy+y & = & 0\\
-2xy^{2}+2y^{2}+x-1 & = & 0
\end{eqnarray*}
\end_inset
\begin_inset Formula \begin{eqnarray*}
& & \Rightarrow P_{1}=\left( 1,0\right) \\
\textrm{aus }1: & y\left( -2x^{2}+2x+1\right) =0 & \Rightarrow x_{1|2}=\frac{-2\pm
\sqrt{4+8}}{-4}=\frac{1\pm \sqrt{3}}{2}\\
\textrm{aus }2: & y^{2}\left( -2x^{2}+2\right) +x-1=0 & \Rightarrow y^{2}=\frac{1}{2}\\
& & \Rightarrow P_{2}=\left( \frac{1+\sqrt{3}}{2},\frac{\sqrt{2}}{2}\right) \\
& & \Rightarrow P_{3}=\left( \frac{1+\sqrt{3}}{2},-\frac{\sqrt{2}}{2}\right) \\
& & \Rightarrow P_{4}=\left( \frac{1-\sqrt{3}}{2},\frac{\sqrt{2}}{2}\right) \\
& & \Rightarrow P_{5}=\left( \frac{1-\sqrt{3}}{2},-\frac{\sqrt{2}}{2}\right)
\end{eqnarray*}
\end_inset
\begin_inset Formula \begin{eqnarray*}
\partial _{xx}g\left( P_{2}\right) & = & \sqrt{2}e^{-\left( \frac{1}{4}\left(
1+2\sqrt{3}\right) +\frac{1}{2}\right) }\left( 1-\frac{3}{2}\left( 1+\sqrt{3}\right)
-\frac{1}{2}\left( 4+2\sqrt{3}+\frac{1}{4}\left( 1+3+3+3+3\sqrt{3}\right) \right)
\right) =\\
& = & -\sqrt{6}e^{-\frac{1}{2}\left( 3+\sqrt{3}\right) }\\
\partial _{yy}g\left( P_{2}\right) & = & \sqrt{2}e^{-\frac{1}{2}\left(
3+\sqrt{3}\right) }\left(
3+\frac{1}{2}+\frac{1}{2}\sqrt{3}+\frac{1}{2}+\frac{1}{2}\sqrt{3}-1\right)
=\sqrt{2}\left( 1-\sqrt{3}\right) e^{-\frac{1}{2}\left( 3+\sqrt{3}\right) }\\
\partial _{xy}g\left( P_{2}\right) & = & \\
\partial _{yx}g\left( P_{2}\right) & = & e^{-\frac{1}{2}\left( 3+\sqrt{3}\right)
}\left( 1-1-\frac{1}{2}\left( 4+2\sqrt{3}\right) +1\left( 1+\sqrt{3}\right)
+\frac{1}{2}\left( 4+2\sqrt{3}\right) -\left( 1+\sqrt{3}\right) \right) =0\\
g''\left( P_{2}\right) & = & \left( \begin{array}{cc}
-\sqrt{6} & 0\\
0 & \sqrt{2}\left( 1-\sqrt{3}\right)
\end{array}\right) e^{-\frac{1}{2}\left( 3+\sqrt{3}\right) }\\
det\, g''\left( P_{2}\right) & = & \left( e^{-\frac{1}{2}\left( 3+\sqrt{3}\right)
}\right) ^{2}det\, \left( \begin{array}{cc}
\sqrt{6} & 0\\
0 & \sqrt{2}\left( 1-\sqrt{3}\right)
\end{array}\right) =e^{-\left( 3+\sqrt{3}\right) }\left( 6-2\sqrt{3}\right) >0\\
& \Rightarrow & Maximum\, P_{2}\\
& & \textrm{weil g}\left( \textrm{x},-y\right) =-g\left( x,y\right) \textrm{ ist
}P_{3}\textrm{ Minimum}
\end{eqnarray*}
\end_inset
Randextrema:
\begin_inset Formula \begin{eqnarray*}
& & x=0\, und\, \partial _{y}g=0\\
e^{-\left( x^{2}+y^{2}\right) }\left( x-1-2y^{2}x+2y^{2}\right) & = & 0\, mit\, x=0\\
-1+2y^{2} & = & 0\Rightarrow 2y^{2}=1\Rightarrow y=\pm \sqrt{\frac{1}{2}}\\
g\left( 0,\sqrt{\frac{1}{2}}\right) & = & \sqrt{\frac{1}{2}}\left( -1\right)
e^{-\frac{1}{2}}=-\sqrt{\frac{1}{2}}e^{-\frac{1}{2}}\\
& & \textrm{wegen Symmetrie }g\left( x,-y\right) =-g\left( x,y\right) \\
& \Rightarrow & g\left( 0,-\sqrt{\frac{1}{2}}\right)
=\frac{1}{2}\sqrt{2}e^{-\frac{1}{2}}\\
& \Rightarrow & \textrm{Randmaximum bei }\left( 0,-\frac{1}{2}\sqrt{2}\right) \\
& & \textrm{Randmaximum bei }\left( 0,\frac{1}{2}\sqrt{2}\right)
\end{eqnarray*}
\end_inset
\begin_inset Formula \( P_{1}\left( 1,0\right) \)
\end_inset
ist Sattelpunkt!
\layout Problem
\line_top
Eine Funktion
\begin_inset Formula \( f\in \mathcal C^{2}\left( \mathbb R^{2}\right) \)
\end_inset
genüge der partiellen Differentialgleichung
\begin_inset Formula \[
\triangle f=c\cdot f\textrm{ mit der Konstanten }c>0\]
\end_inset
in einer Umgebung der abgeschlossenen Kreisscheibe
\begin_inset Formula \( D:=\left\{ \left( x,y\right) \in \mathbb R^{2}:x^{2}+y^{2}\leq
1\right\} . \)
\end_inset
Ferner erfülle f die Randbedingung
\begin_inset Formula \[
f\left( x,y\right) =0\textrm{ f}\ddot{\textrm{u}}\textrm{r alle x}\in \partial
D=\left\{ \left( x,y\right) \in \mathbb R^{2}:x^{2}+y^{2}=0\right\} .\]
\end_inset
Man zeige:
\begin_inset Formula \( f\equiv 0 \)
\end_inset
auf ganz D.
\newline
Hinweis: Welches Vorzeichen hat
\begin_inset Formula \( \triangle f \)
\end_inset
an der Stelle eines Maximums/Minimums?
\layout Standard
\SpecialChar ~
\layout Problem
\line_top
Eine Möglichkeit zur Berechnung von
\begin_inset Formula \( \int _{\mathbb R}e^{-x^{2}}dx: \)
\end_inset
\layout Enumerate
Man zeige, daß durch
\begin_inset Formula \[
F\left( t\right) =\int _{0}^{1}\frac{e^{-\left( 1+x^{2}\right) t^{2}}}{1+x^{2}}dx\]
\end_inset
eine diefferenzierbare Funktion
\begin_inset Formula \( F:\mathbb R\rightarrow \mathbb R \)
\end_inset
erklärt wird mit
\begin_inset Formula \( F'\left( t\right) =-2e^{-t^{2}}\cdot \int
_{0}^{t}e^{-u^{2}}du. \)
\end_inset
\layout Enumerate
Man folgere daraus:
\begin_inset Formula \( F\left( t\right) =-\left( \int _{0}^{t}e^{-u^{2}}du\right)
^{2}+\frac{\pi }{4}. \)
\end_inset
\newline
Hinweis:
\begin_inset Formula \( F' \)
\end_inset
hat die Form
\begin_inset Formula \( F'\left( t\right) =-2f'\left( t\right) \cdot f\left( t\right)
\)
\end_inset
.
\layout Enumerate
Man zeige:
\begin_inset Formula \( \int _{\mathbb R}e^{-u^{2}}du=\sqrt{\pi } \)
\end_inset
.
\layout Standard
\SpecialChar ~
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
F\left( t\right) & = & \int _{0}^{1}\frac{e^{-\left( 1+x^{2}\right)
t^{2}}}{1+x^{2}}dx\textrm{ ist stetig auf }\left[ 0,1\right] \times \mathbb R\\
& & \Rightarrow \textrm{F stetig partiell differenzierbar }\Rightarrow F\textrm{
stetig differenzierbar}\\
& &
\end{eqnarray*}
\end_inset
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
F'\left( t\right) & = & \int _{0}^{1}\frac{\partial }{\partial t}\frac{e^{-\left(
1+x^{2}\right) t^{2}}}{1+x^{2}}dx=\int _{0}^{1}-2te^{-\left( 1+x^{2}\right)
t^{2}}dx=-2e^{-t^{2}}\int _{0}^{1}e^{-x^{2}t^{2}}t\, dx=\\
& = & -2e^{-t^{2}}\int _{0}^{t}e^{-u^{2}}du=-2f^{'}\left( t\right) f\left( t\right) \\
F\left( t\right) & = & -\left[ f\left( t\right) \right] ^{2}+c\\
F\left( 0\right) & = & \int _{0}^{1}\frac{1}{1+x^{2}}dx=arctan\, x|_{0}^{1}=\frac{\pi
}{4}-0\\
F\left( t\right) & = & -\left[ \int _{0}^{t}e^{-u^{2}}du\right] ^{2}+\frac{\pi }{4}
\end{eqnarray*}
\end_inset
\layout Enumerate
\begin_inset Formula \begin{eqnarray*}
\left| f\left( t\right) \right| & = & \int
_{0}^{1}\frac{e^{-t^{2}}e^{-t^{2}x^{2}}}{1+x^{2}}dx\leq \int
_{0}^{1}\frac{e^{-t^{2}\cdot 1}}{1}=e^{-t^{2}}\rightarrow 0\textrm{
f}\ddot{\textrm{u}}\textrm{r t}\rightarrow \infty \\
& \Rightarrow & \int _{0}^{\infty }e^{-u^{2}}du=\frac{\sqrt{\pi }}{2}\\
& \Rightarrow & \int _{-\infty }^{\infty }e^{-u^{2}}du=\sqrt{\pi }
\end{eqnarray*}
\end_inset
\layout Standard
\begin_inset LatexCommand \printindex{}
\end_inset
\the_end
