From ce72c8b27b09ecbf98a454f3b37019aaa948a57e Mon Sep 17 00:00:00 2001 From: Andrea Mozzini Vellen Date: Mon, 2 May 2022 16:02:40 +0200 Subject: Intro chapters --- buch/papers/kreismembran/main.tex | 23 ++---- buch/papers/kreismembran/teil0.tex | 16 +---- buch/papers/kreismembran/teil1.tex | 142 +++++++++++++++++++++++++------------ buch/papers/kreismembran/teil2.tex | 79 ++++++++++++--------- 4 files changed, 148 insertions(+), 112 deletions(-) diff --git a/buch/papers/kreismembran/main.tex b/buch/papers/kreismembran/main.tex index 67b436c..eafec18 100644 --- a/buch/papers/kreismembran/main.tex +++ b/buch/papers/kreismembran/main.tex @@ -3,28 +3,19 @@ % % (c) 2020 Hochschule Rapperswil % -\chapter{Thema\label{chapter:kreismembran}} -\lhead{Thema} +\chapter{Schwingungen einer kreisförmligen Membran\label{chapter:kreismembran}} +\lhead{Schwingungen einer kreisförmligen Membran} \begin{refsection} -\chapterauthor{Hans Muster} - -Ein paar Hinweise für die korrekte Formatierung des Textes +\chapterauthor{Andrea Mozzini Vellen und Tim Tönz} \begin{itemize} \item -Absätze werden gebildet, indem man eine Leerzeile einfügt. -Die Verwendung von \verb+\\+ ist nur in Tabellen und Arrays gestattet. +Tim ist ein snitch \item -Die explizite Platzierung von Bildern ist nicht erlaubt, entsprechende -Optionen werden gelöscht. -Verwenden Sie Labels und Verweise, um auf Bilder hinzuweisen. +ich dachte wir sind gute Freunden \item -Beginnen Sie jeden Satz auf einer neuen Zeile. -Damit ermöglichen Sie dem Versionsverwaltungssysteme, Änderungen -in verschiedenen Sätzen von verschiedenen Autoren ohne Konflikt -anzuwenden. +du schuldest mir ein bier \item -Bilden Sie auch für Formeln kurze Zeilen, einerseits der besseren -Übersicht wegen, aber auch um GIT die Arbeit zu erleichtern. +auch ein gin tonic \end{itemize} \input{papers/kreismembran/teil0.tex} diff --git a/buch/papers/kreismembran/teil0.tex b/buch/papers/kreismembran/teil0.tex index e4b1711..1552259 100644 --- a/buch/papers/kreismembran/teil0.tex +++ b/buch/papers/kreismembran/teil0.tex @@ -3,20 +3,8 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Teil 0\label{kreismembran:section:teil0}} -\rhead{Teil 0} -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua \cite{kreismembran:bibtex}. -At vero eos et accusam et justo duo dolores et ea rebum. -Stet clita kasd gubergren, no sea takimata sanctus est Lorem ipsum -dolor sit amet. +\section{Einleitung\label{kreismembran:section:teil0}} +\rhead{Einleitung} -Lorem ipsum dolor sit amet, consetetur sadipscing elitr, sed diam -nonumy eirmod tempor invidunt ut labore et dolore magna aliquyam -erat, sed diam voluptua. -At vero eos et accusam et justo duo dolores et ea rebum. Stet clita -kasd gubergren, no sea takimata sanctus est Lorem ipsum dolor sit -amet. diff --git a/buch/papers/kreismembran/teil1.tex b/buch/papers/kreismembran/teil1.tex index b715075..29a47a6 100644 --- a/buch/papers/kreismembran/teil1.tex +++ b/buch/papers/kreismembran/teil1.tex @@ -2,54 +2,102 @@ % teil1.tex -- Beispiel-File für das Paper % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -% -\section{Teil 1 -\label{kreismembran:section:teil1}} -\rhead{Problemstellung} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. -Nemo enim ipsam voluptatem quia voluptas sit aspernatur aut odit -aut fugit, sed quia consequuntur magni dolores eos qui ratione -voluptatem sequi nesciunt + +\section{Die Hankel Transformation \label{kreismembran:section:teil1}} +\rhead{Die Hankel Transformation} + +Hermann Hankel (1839-1873) war ein deutscher Mathematiker, der für seinen Beitrag zur mathematischen Analyse und insbesondere für seine namensgebende Transformation bekannt ist. +Diese Transformation tritt bei der Untersuchung von funktionen auf, die nur von der Enternung des Ursprungs abhängen. +Er studierte auch funktionen, jetzt Hankel- oder Bessel- Funktionen genannt, der dritten Art. +Die Hankel Transformation mit Bessel Funktionen al Kern taucht natürlich bei achsensymmetrischen Problemen auf, die in Zylindrischen Polarkoordinaten formuliert sind. +In diesem Kapitel werden die Theorie der Transformation und einige Eigenschaften der Grundoperationen erläutert. + +Wir führen die Definition der Hankel Transformation aus der zweidimensionalen Fourier Transformation und ihrer Umkehrung ein, die durch: +\begin{align} + \mathscr{F}\{f(x,y)\} & = F(k,l)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-i( \bm{\kappa}\cdot \mathbf{r})}f(x,y) dx dy,\label{equation:fourier_transform}\\ + \mathscr{F}^{-1}\{F(x,y)\} & = f(x,y)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{i(\bm{\kappa}\cdot \mathbf{r}))}F(k,l) dx dy \label{equation:inv_fourier_transform} +\end{align} +wo $\mathbf{r}=(x,y)$ und $\bm{\kappa}=(k,l)$. Wie bereits erwähnt, sind Polarkoordinaten für diese Art von Problemen am besten geeignet, also mit, $(x,y)=r(\cos\theta,\sin\theta)$ und $(k,l)=\kappa(\cos\phi,\sin\phi)$, findet man $\bm{\kappa}\cdot\mathbf{r}=\kappa r(\cos(\theta-\phi))$ und danach: +\begin{align} + F(k,\phi)=\frac{1}{2\pi}\int_{0}^{\infty}r dr \int_{0}^{2\pi}e^{-ikr\cos(\theta-\phi)}f(r,\theta) d\phi. + \label{equation:F_ohne_variable_wechsel} +\end{align} +Dann wird angenommen dass, $f(r,\theta)=e^{in\theta}f(r)$, was keine strenge Einschränkung ist, und es wird eine Änderung der Variabeln vorgenommen $\theta-\phi=\alpha-\frac{\pi}{2}$, um \ref{equation:F_ohne_variable_wechsel} zu reduzieren: +\begin{align} + F(k,\phi)=\frac{1}{2\pi}\int_{0}^{\infty}rf(r) dr \int_{\phi_{0}}^{2\pi+\phi_{0}}e^{in(\phi-\frac{\pi}{2})+i(n\alpha-kr\sin\alpha)} d\alpha, + \label{equation:F_ohne_bessel} +\end{align} +wo $\phi_{0}=(\frac{\pi}{2}-\phi)$. + +Unter Verwendung der Integral Darstellung der Besselfunktion vom Ordnung n +\begin{align} + J_n(\kappa r)=\frac{1}{2\pi}\int_{\phi_{0}}^{2\pi + \phi_{0}}e^{i(n\alpha-\kappa r \sin \alpha)} d\alpha + \label{equation:bessel_n_ordnung} +\end{align} +\eqref{equation:F_ohne_bessel} wird sie zu: +\begin{align} + F(k,\phi)&=e^{in(\phi-\frac{\pi}{2})}\int_{0}^{\infty}rJ_n(\kappa r) f(r) dr \label{equation:F_mit_bessel_step_1} \\ + &=e^{in(\phi-\frac{\pi}{2})}\tilde{f}_n(\kappa), + \label{equation:F_mit_bessel_step_2} +\end{align} +wo $\tilde{f}_n(\kappa)$ ist die \textit{Hankel Transformation} von $f(r)$ und ist formell definiert durch: +\begin{align} + \mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)=\int_{0}^{\infty}rJ_n(\kappa r) f(r) dr. + \label{equation:hankel} +\end{align} + +Ähnlich verhält es sich mit der inversen Fourier Transformation in Form von polaren Koordinaten unter der Annahme $f(r,\theta)=e^{in\theta}f(r)$ mit \ref{equation:F_mit_bessel_step_2}, wird die inverse Fourier Transformation \ref{equation:inv_fourier_transform}: + +\begin{align*} + e^{in\theta}f(r)&=\frac{1}{2\pi}\int_{0}^{\infty}\kappa d\kappa \int_{0}^{2\pi}e^{i\kappa r \cos (\theta - \phi)}F(\kappa,\phi) d\phi\\ + &= \frac{1}{2\pi}\int_{0}^{\infty}\kappa \tilde{f}_n(\kappa) d\kappa \int_{0}^{2\pi}e^{in(\phi - \frac{\pi}{2})- i\kappa r \cos (\theta - \phi)} d\phi, +\end{align*} +was durch den Wechsel der Variablen $\theta-\phi=-(\alpha+\frac{\pi}{2})$ und $\theta_0=-(\theta+\frac{\pi}{2})$, + +\begin{align} + &= \frac{1}{2\pi}\int_{0}^{\infty}\kappa \tilde{f}_n(\kappa) d\kappa \int_{\theta_0}^{2\pi+\theta_0}e^{in(\theta + \alpha - i\kappa r \sin\alpha)} d\alpha \nonumber \\ + &= e^{in\theta}\int_{0}^{\infty}\kappa J_n(\kappa r) \tilde{f}_n(\kappa) d\kappa,\quad \text{von \eqref{equation:bessel_n_ordnung}} +\end{align} + +Also, die inverse \textit{Hankel Transformation} ist so definiert: +\begin{align} + \mathscr{H}^{-1}_n\{\tilde{f}_n(\kappa)\}=f(r)=\int_{0}^{\infty}\kappa J_n(\kappa r) \tilde{f}_n(\kappa) d\kappa. + \label{equation:inv_hankel} +\end{align} + +Anstelle von $\tilde{f}_n(\kappa)$, wird häufig für die Hankel Transformation verwendet, indem die Ordnung angegeben wird. +\eqref{equation:hankel} und \eqref{equation:inv_hankel} Integralen existieren für eine grosse Klasse von Funktionen, die normalerweise in physikalischen Anwendungen benötigt werden. +Alternativ kann auch die berühmte Hankel Transformationsformel verwendet werden, + +\begin{align} + f(r) = \int_{0}^{\infty}\kappa J_n(\kappa r) d\kappa \int_{0}^{\infty} p J_n(\kappa p)f(p) dp, + \label{equation:hankel_integral_formula} +\end{align} +um die Hankel Transformation \eqref{equation:hankel} und ihre Inverse \eqref{equation:inv_hankel} zu definieren. +Insbesondere die Hankel Transformation der nullten Ordnung ($n=0$) und der ersten Ordnung ($n=1$) sind häufig nützlich, um Lösungen für Probleme mit der Laplace Gleichung in einer achsensymmetrischen zylindrischen Geometrie zu finden. + +\subsection{Operative Eigenschaften der Hankel Transformation\label{sub:op_properties_hankel}} +In diesem Kapitel werden die operativen Eigenschaften der Hankel Transformation aufgeführt. Der Beweis für ihre Gültigkeit wird jedoch nicht analysiert. + +\subsubsection{Skalierung \label{subsub:skalierung}} +Wenn $\mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)$, dann: + +\begin{equation} + \mathscr{H}_n\{f(ar)\}=\frac{1}{a^{2}}\tilde{f}_n \left(\frac{\kappa}{a}\right), \quad a>0. +\end{equation} + +\subsubsection{Persevalsche Relation \label{subsub:perseval}} +Wenn $\tilde{f}(\kappa)=\mathscr{H}_n\{f(r)\}$ und $\tilde{g}(\kappa)=\mathscr{H}_n\{g(r)\}$, dann: + \begin{equation} -\int_a^b x^2\, dx -= -\left[ \frac13 x^3 \right]_a^b -= -\frac{b^3-a^3}3. -\label{kreismembran:equation1} + \int_{0}^{\infty}rf(r) dr = \int_{0}^{\infty}\kappa\tilde{f}(\kappa)\tilde{g}(\kappa) d\kappa. \end{equation} -Neque porro quisquam est, qui dolorem ipsum quia dolor sit amet, -consectetur, adipisci velit, sed quia non numquam eius modi tempora -incidunt ut labore et dolore magnam aliquam quaerat voluptatem. - -Ut enim ad minima veniam, quis nostrum exercitationem ullam corporis -suscipit laboriosam, nisi ut aliquid ex ea commodi consequatur? -Quis autem vel eum iure reprehenderit qui in ea voluptate velit -esse quam nihil molestiae consequatur, vel illum qui dolorem eum -fugiat quo voluptas nulla pariatur? - -\subsection{De finibus bonorum et malorum -\label{kreismembran:subsection:finibus}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga \eqref{000tempmlate:equation1}. - -Et harum quidem rerum facilis est et expedita distinctio -\ref{kreismembran:section:loesung}. -Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil -impedit quo minus id quod maxime placeat facere possimus, omnis -voluptas assumenda est, omnis dolor repellendus -\ref{kreismembran:section:folgerung}. -Temporibus autem quibusdam et aut officiis debitis aut rerum -necessitatibus saepe eveniet ut et voluptates repudiandae sint et -molestiae non recusandae. -Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis -voluptatibus maiores alias consequatur aut perferendis doloribus -asperiores repellat. +\subsubsection{Hankel Transformationen von Ableitungen \label{subsub:ableitungen}} +Wenn $\tilde{f}_n(\kappa)=\mathscr{H}_n\{f(r)\}$, dann: +\begin{align} + &\mathscr{H}_n\{f'(r)\}=\frac{\kappa}{2n}\left[(n-1)\tilde{f}_{n+1}(\kappa)-(n+1)\tilde{f}_{n-1}(\kappa)\right], \quad n\geq1, \\ + &\mathscr{H}_1\{f'(r)\}=-\kappa \tilde{f}_0(\kappa), +\end{align} +bereitgestellt dass $[rf(r)]$ verschwindet als $r\to0$ und $r\to\infty=0$. \ No newline at end of file diff --git a/buch/papers/kreismembran/teil2.tex b/buch/papers/kreismembran/teil2.tex index 7ed217f..45357f2 100644 --- a/buch/papers/kreismembran/teil2.tex +++ b/buch/papers/kreismembran/teil2.tex @@ -1,40 +1,49 @@ % -% teil2.tex -- Beispiel-File für teil2 -% % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Teil 2 -\label{kreismembran:section:teil2}} -\rhead{Teil 2} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum -\label{kreismembran:subsection:bonorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. Nam libero tempore, cum soluta nobis -est eligendi optio cumque nihil impedit quo minus id quod maxime -placeat facere possimus, omnis voluptas assumenda est, omnis dolor -repellendus. Temporibus autem quibusdam et aut officiis debitis aut -rerum necessitatibus saepe eveniet ut et voluptates repudiandae -sint et molestiae non recusandae. Itaque earum rerum hic tenetur a -sapiente delectus, ut aut reiciendis voluptatibus maiores alias -consequatur aut perferendis doloribus asperiores repellat. +\section{Lösung der partiellen Differentialgleichung + \label{kreismembran:section:teil2}} +\rhead{Lösung der partiellen Differentialgleichung} + +Wie im vorherigen Kapitel gezeigt, lautet die partielle Differentialgleichung, die die Schwingungen einer Membran beschreibt: +\begin{equation*} + \frac{1}{c^2}\frac{\partial^2u}{\partial t^2} = \Delta u +\end{equation*} +Da es sich um eine Kreisscheibe handelt, werden Polarkoordinaten verwendet, so dass sich der Laplaceoperator ergibt: +\begin{equation*} + \Delta + = + \frac{\partial^2}{\partial r^2} + + + \frac1r + \frac{\partial}{\partial r} + + + \frac{1}{r 2} + \frac{\partial^2}{\partial\varphi^2}. + \label{buch:pde:kreis:laplace} +\end{equation*} + +Es wird eine runde elastische Membran berücksichtigt, die den Gebietbereich $\Omega$ abdeckt und am Rand $\Gamma$ befestigt ist. +Es wird daher davon ausgegangen, dass die Membran aus einem homogenen Material von vernachlässigbarer Dicke gefertigt ist. +Die Membran kann verformt werden, aber innere elastische Kräfte wirken den Verformungen entgegen. Es wirken keine äusseren Kräfte. Es handelt sich somit von einer kreisförmligen eigespannten homogenen schwingenden Membran. + +Daher ist die Membranabweichung im Punkt $(r,\theta)$ $\in$ $\overline{\rm \Omega}$ zum Zeitpunkt $t$: +\begin{align*} + u: \overline{\rm \Omega} \times \mathbb{R}_{\geq 0} &\longrightarrow \mathbb{R}\\ + (r,\theta,t) &\longmapsto u(r,\theta,t) +\end{align*} +Da die Membran am Rand befestigt ist, kann es keine Schwingungen geben, so dass die \textit{Dirichlet-Randbedingung} gilt: +\begin{equation*} + u\big|_{\Gamma} = 0 +\end{equation*} + + +Um eine eindeutige Lösung bestimmen zu können, werden die folgenden Anfangsbedingungen festgelegt: + +\begin{align*} + u(r,\theta, 0) &:= f(x,y)\\ + \frac{\partial}{\partial t} u(r,\theta, 0) &:= g(x,y) +\end{align*} +An dieser Stelle könnte man zum Beispiel die bereits in Kapitel (TODO:refKAPITEL) vorgestellte Methode der Separation anwenden. Da es sich in diesem Fall jedoch um einem achsensymmetrischen Problem handelt, das in Polarkoordinaten formuliert ist, wird man die Transformationsmethode verwenden, insbesondere die Hankel Transformation. -- cgit v1.2.1 From 955047b8a63a3b08b27d9203030e2b5193e21dab Mon Sep 17 00:00:00 2001 From: Andrea Mozzini Vellen Date: Wed, 18 May 2022 13:55:56 +0200 Subject: Ersten Entwurf --- buch/papers/kreismembran/main.tex | 10 -- buch/papers/kreismembran/teil1.tex | 181 ++++++++++++++++++------------------- buch/papers/kreismembran/teil2.tex | 128 +++++++++++++++++++------- buch/papers/kreismembran/teil3.tex | 102 ++++++++++++++------- 4 files changed, 255 insertions(+), 166 deletions(-) diff --git a/buch/papers/kreismembran/main.tex b/buch/papers/kreismembran/main.tex index eafec18..e63a118 100644 --- a/buch/papers/kreismembran/main.tex +++ b/buch/papers/kreismembran/main.tex @@ -7,16 +7,6 @@ \lhead{Schwingungen einer kreisförmligen Membran} \begin{refsection} \chapterauthor{Andrea Mozzini Vellen und Tim Tönz} -\begin{itemize} -\item -Tim ist ein snitch -\item -ich dachte wir sind gute Freunden -\item -du schuldest mir ein bier -\item -auch ein gin tonic -\end{itemize} \input{papers/kreismembran/teil0.tex} \input{papers/kreismembran/teil1.tex} diff --git a/buch/papers/kreismembran/teil1.tex b/buch/papers/kreismembran/teil1.tex index 29a47a6..aef5b79 100644 --- a/buch/papers/kreismembran/teil1.tex +++ b/buch/papers/kreismembran/teil1.tex @@ -3,101 +3,98 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil -\section{Die Hankel Transformation \label{kreismembran:section:teil1}} -\rhead{Die Hankel Transformation} - -Hermann Hankel (1839-1873) war ein deutscher Mathematiker, der für seinen Beitrag zur mathematischen Analyse und insbesondere für seine namensgebende Transformation bekannt ist. -Diese Transformation tritt bei der Untersuchung von funktionen auf, die nur von der Enternung des Ursprungs abhängen. -Er studierte auch funktionen, jetzt Hankel- oder Bessel- Funktionen genannt, der dritten Art. -Die Hankel Transformation mit Bessel Funktionen al Kern taucht natürlich bei achsensymmetrischen Problemen auf, die in Zylindrischen Polarkoordinaten formuliert sind. -In diesem Kapitel werden die Theorie der Transformation und einige Eigenschaften der Grundoperationen erläutert. - -Wir führen die Definition der Hankel Transformation aus der zweidimensionalen Fourier Transformation und ihrer Umkehrung ein, die durch: -\begin{align} - \mathscr{F}\{f(x,y)\} & = F(k,l)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-i( \bm{\kappa}\cdot \mathbf{r})}f(x,y) dx dy,\label{equation:fourier_transform}\\ - \mathscr{F}^{-1}\{F(x,y)\} & = f(x,y)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{i(\bm{\kappa}\cdot \mathbf{r}))}F(k,l) dx dy \label{equation:inv_fourier_transform} -\end{align} -wo $\mathbf{r}=(x,y)$ und $\bm{\kappa}=(k,l)$. Wie bereits erwähnt, sind Polarkoordinaten für diese Art von Problemen am besten geeignet, also mit, $(x,y)=r(\cos\theta,\sin\theta)$ und $(k,l)=\kappa(\cos\phi,\sin\phi)$, findet man $\bm{\kappa}\cdot\mathbf{r}=\kappa r(\cos(\theta-\phi))$ und danach: -\begin{align} - F(k,\phi)=\frac{1}{2\pi}\int_{0}^{\infty}r dr \int_{0}^{2\pi}e^{-ikr\cos(\theta-\phi)}f(r,\theta) d\phi. - \label{equation:F_ohne_variable_wechsel} -\end{align} -Dann wird angenommen dass, $f(r,\theta)=e^{in\theta}f(r)$, was keine strenge Einschränkung ist, und es wird eine Änderung der Variabeln vorgenommen $\theta-\phi=\alpha-\frac{\pi}{2}$, um \ref{equation:F_ohne_variable_wechsel} zu reduzieren: -\begin{align} - F(k,\phi)=\frac{1}{2\pi}\int_{0}^{\infty}rf(r) dr \int_{\phi_{0}}^{2\pi+\phi_{0}}e^{in(\phi-\frac{\pi}{2})+i(n\alpha-kr\sin\alpha)} d\alpha, - \label{equation:F_ohne_bessel} -\end{align} -wo $\phi_{0}=(\frac{\pi}{2}-\phi)$. - -Unter Verwendung der Integral Darstellung der Besselfunktion vom Ordnung n -\begin{align} - J_n(\kappa r)=\frac{1}{2\pi}\int_{\phi_{0}}^{2\pi + \phi_{0}}e^{i(n\alpha-\kappa r \sin \alpha)} d\alpha - \label{equation:bessel_n_ordnung} -\end{align} -\eqref{equation:F_ohne_bessel} wird sie zu: -\begin{align} - F(k,\phi)&=e^{in(\phi-\frac{\pi}{2})}\int_{0}^{\infty}rJ_n(\kappa r) f(r) dr \label{equation:F_mit_bessel_step_1} \\ - &=e^{in(\phi-\frac{\pi}{2})}\tilde{f}_n(\kappa), - \label{equation:F_mit_bessel_step_2} -\end{align} -wo $\tilde{f}_n(\kappa)$ ist die \textit{Hankel Transformation} von $f(r)$ und ist formell definiert durch: -\begin{align} - \mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)=\int_{0}^{\infty}rJ_n(\kappa r) f(r) dr. - \label{equation:hankel} -\end{align} - -Ähnlich verhält es sich mit der inversen Fourier Transformation in Form von polaren Koordinaten unter der Annahme $f(r,\theta)=e^{in\theta}f(r)$ mit \ref{equation:F_mit_bessel_step_2}, wird die inverse Fourier Transformation \ref{equation:inv_fourier_transform}: +\section{Lösungsmethode 1: Separationsmethode  + \label{kreismembran:section:teil1}} +\rhead{Lösungsmethode 1: Separationsmethode} +An diesem Punkt bleibt also nur noch die Lösung der partiellen Differentialgleichung. In diesem Kapitel wird sie mit Hilfe der Separationsmetode gelöst. + +Wie im vorherigen Kapitel gezeigt, lautet die partielle Differentialgleichung, die die Schwingungen einer Membran beschreibt: +\begin{equation*} + \frac{1}{c^2}\frac{\partial^2u}{\partial t^2} = \Delta u +\end{equation*} +Da es sich um eine Kreisscheibe handelt, werden Polarkoordinaten verwendet, so dass sich der Laplaceoperator ergibt: +\begin{equation*} + \Delta + = + \frac{\partial^2}{\partial r^2} + + + \frac1r + \frac{\partial}{\partial r} + + + \frac{1}{r 2} + \frac{\partial^2}{\partial\varphi^2}. + \label{buch:pde:kreis:laplace} +\end{equation*} + +Es wird eine runde elastische Membran berücksichtigt, die den Gebietbereich $\Omega$ abdeckt und am Rand $\Gamma$ befestigt ist. +Es wird daher davon ausgegangen, dass die Membran aus einem homogenen Material von vernachlässigbarer Dicke gefertigt ist. +Die Membran kann verformt werden, aber innere elastische Kräfte wirken den Verformungen entgegen. Es wirken keine äusseren Kräfte. Es handelt sich somit von einer kreisförmligen eigespannten homogenen schwingenden Membran. + +Daher ist die Membranabweichung im Punkt $(r,\varphi)$ $\in$ $\overline{\rm \Omega}$ zum Zeitpunkt $t$: \begin{align*} - e^{in\theta}f(r)&=\frac{1}{2\pi}\int_{0}^{\infty}\kappa d\kappa \int_{0}^{2\pi}e^{i\kappa r \cos (\theta - \phi)}F(\kappa,\phi) d\phi\\ - &= \frac{1}{2\pi}\int_{0}^{\infty}\kappa \tilde{f}_n(\kappa) d\kappa \int_{0}^{2\pi}e^{in(\phi - \frac{\pi}{2})- i\kappa r \cos (\theta - \phi)} d\phi, + u: \overline{\rm \Omega} \times \mathbb{R}_{\geq 0} &\longrightarrow \mathbb{R}\\ + (r,\varphi,t) &\longmapsto u(r,\varphi,t) \end{align*} -was durch den Wechsel der Variablen $\theta-\phi=-(\alpha+\frac{\pi}{2})$ und $\theta_0=-(\theta+\frac{\pi}{2})$, - -\begin{align} - &= \frac{1}{2\pi}\int_{0}^{\infty}\kappa \tilde{f}_n(\kappa) d\kappa \int_{\theta_0}^{2\pi+\theta_0}e^{in(\theta + \alpha - i\kappa r \sin\alpha)} d\alpha \nonumber \\ - &= e^{in\theta}\int_{0}^{\infty}\kappa J_n(\kappa r) \tilde{f}_n(\kappa) d\kappa,\quad \text{von \eqref{equation:bessel_n_ordnung}} -\end{align} - -Also, die inverse \textit{Hankel Transformation} ist so definiert: -\begin{align} - \mathscr{H}^{-1}_n\{\tilde{f}_n(\kappa)\}=f(r)=\int_{0}^{\infty}\kappa J_n(\kappa r) \tilde{f}_n(\kappa) d\kappa. - \label{equation:inv_hankel} -\end{align} - -Anstelle von $\tilde{f}_n(\kappa)$, wird häufig für die Hankel Transformation verwendet, indem die Ordnung angegeben wird. -\eqref{equation:hankel} und \eqref{equation:inv_hankel} Integralen existieren für eine grosse Klasse von Funktionen, die normalerweise in physikalischen Anwendungen benötigt werden. -Alternativ kann auch die berühmte Hankel Transformationsformel verwendet werden, - -\begin{align} - f(r) = \int_{0}^{\infty}\kappa J_n(\kappa r) d\kappa \int_{0}^{\infty} p J_n(\kappa p)f(p) dp, - \label{equation:hankel_integral_formula} -\end{align} -um die Hankel Transformation \eqref{equation:hankel} und ihre Inverse \eqref{equation:inv_hankel} zu definieren. -Insbesondere die Hankel Transformation der nullten Ordnung ($n=0$) und der ersten Ordnung ($n=1$) sind häufig nützlich, um Lösungen für Probleme mit der Laplace Gleichung in einer achsensymmetrischen zylindrischen Geometrie zu finden. - -\subsection{Operative Eigenschaften der Hankel Transformation\label{sub:op_properties_hankel}} -In diesem Kapitel werden die operativen Eigenschaften der Hankel Transformation aufgeführt. Der Beweis für ihre Gültigkeit wird jedoch nicht analysiert. - -\subsubsection{Skalierung \label{subsub:skalierung}} -Wenn $\mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)$, dann: - -\begin{equation} - \mathscr{H}_n\{f(ar)\}=\frac{1}{a^{2}}\tilde{f}_n \left(\frac{\kappa}{a}\right), \quad a>0. -\end{equation} - -\subsubsection{Persevalsche Relation \label{subsub:perseval}} -Wenn $\tilde{f}(\kappa)=\mathscr{H}_n\{f(r)\}$ und $\tilde{g}(\kappa)=\mathscr{H}_n\{g(r)\}$, dann: - +Da die Membran am Rand befestigt ist, kann es keine Schwingungen geben, so dass die \textit{Dirichlet-Randbedingung} gilt: +\begin{equation*} + u\big|_{\Gamma} = 0 +\end{equation*} +Um eine eindeutige Lösung bestimmen zu können, werden die folgenden Anfangsbedingungen festgelegt: +\begin{align*} + u(r,\varphi, 0) &= f(r,\varphi)\\ + \frac{\partial}{\partial t} u(r,\varphi, 0) &= g(r,\varphi) +\end{align*} +Daher muss an dieser Stelle von einer Separation der Variablen ausgegangen werden: +\begin{equation*} + u(r,\varphi, t) = F(r)G(\varphi)T(t) +\end{equation*} +Dank der Randbedingungen kann also gefordert werden, dass $F(R)=0$ ist, und natürlich, dass $G(\varphi)$ $2\pi$ periodisch ist. Eingesetz in der Differenzialgleichung ergibt: +\begin{equation*} + \frac{1}{c^2}\frac{T''(t)}{T(t)}=\frac{F''(r)}{F(r)}+\frac{1}{r}\frac{F'(r)}{F(r)}+\frac{1}{r^2}\frac{G''(\varphi)}{G(\varphi)} +\end{equation*} +Da die linke Seite nur von $t$ und die rechte Seite nur von $r$ und $\varphi$ abhängt, müssen sie gleich einer reellen Zahl sein. Aus physikalischen Grunden suchen wir nach Lösungen, die weder exponentiell in der Zeit wachsen noch exponentiell abklingen. Dies bedeutet, dass die Konstante negativ sein muss, also schreibt man $k=-k^2$. Daraus ergeben sich die folgenden zwei Gleichungen: +\begin{gather*} + T''(t) + c^2\kappa^2T(t) = 0\\ + r^2\frac{F''(r)}{F(r)} + r \frac{F'(r)}{F(r)} +\kappa^2 r^2 = - \frac{G''(\varphi)}{G(\varphi)} +\end{gather*} +In der zweiten Gleichung hängt die linke Seite nur von $r$ ab, während die rechte Seite nur von $\varphi$ abhängt. Sie müssen also wiederum gleich einer reellen Zahl $\nu$ sein. Also das: +\begin{gather*} + r^2F''(r) + rF'(r) + (\kappa^2 r^2 - \nu)F(r) = 0 \\ + G''(\varphi) = \nu G(\varphi) +\end{gather*} +$G$ kann in einer Fourierreihe entwickelt werden, so dass man sieht, dass $\nu$ die Form $n^2$ mit einer positiven ganzen Zahl sein muss, also: +\begin{equation*} + G(\varphi) = C_n \cos(\varphi) + D_n \sin(\varphi) +\end{equation*} +Die Gleichung $F$ hat die Gestalt +\begin{equation*} + r^2F''(r) + rF'(r) + (\kappa^2 r^2 - n^2)F(r) = 0 \quad (*) +\end{equation*} +Wir bereits in der Vorlesung von Prof. Müller gezeigt, sind die Besselfunktionen +\begin{equation*} + J_{\nu}(x) = r^\nu \displaystyle\sum_{m=0}^{\infty} \frac{(-1)^m x^{2m}}{2^{2m+\nu}m! \Gamma (\nu + m+1)} +\end{equation*} +Lösungen der "Besselschen Differenzialgleichung" +\begin{equation*} + x^2 y'' + xy' + (x^2 - \nu^2)y = 0 +\end{equation*} +Die Funktionen $F(r) = J_n(\kappa r)$ lösen also die Differentialgleichung $(*)$. Die +Randbedingung $F(R)=0$ impliziert, dass $\kappa R$ eine Nullstelle der Besselfunktion +$J_n$ sein muss. Man kann zeigen, dass die Besselfunktionen $J_n, n \geq 0$, alle unendlich +viele Nullstellen +\begin{equation*} + \alpha_{1n} < \alpha_{2n} < ... +\end{equation*} +haben, und dass $\underset{\substack{m\to\infty}}{\text{lim}} \alpha_{mn}=\infty$. Somit ergit sich, dass $\kappa = \frac{\alpha_{mn}}{R}$ für ein $m\geq 1$, und dass +\begin{equation*} + F(r) = J_n (\kappa_{mn}r) \quad mit \quad \kappa_{mn}=\frac{\alpha_{mn}}{R} +\end{equation*} +Die Differenzialgleichung $T''(t) + c^2\kappa^2T(t) = 0$, wird auf ähnliche Weise gelöst wie $G(\varphi)$. Durch Überlagerung aller Ergebnisse erhält man die Lösung \begin{equation} - \int_{0}^{\infty}rf(r) dr = \int_{0}^{\infty}\kappa\tilde{f}(\kappa)\tilde{g}(\kappa) d\kappa. + u(r, \varphi, t) = \displaystyle\sum_{m=1}^{\infty}\displaystyle\sum_{n=0}^{\infty} J_n (k_{mn}r)\cos(n\varphi)[a_{mn}\cos(c \kappa_{mn} t)+b_{mn}\sin(c \kappa_{mn} t)] \end{equation} +Dabei sind m und n ganze Zahlen, wobei m für die Anzahl der Knotenkreise und n +für die Anzahl der Knotenlinien steht. Es gibt bestimmte Bereiche auf der Membran, in denen es keine Bewegung oder Vibration gibt. Wenn der nicht schwingende Bereich ein Kreis ist, nennt man ihn einen Knotenkreis, und wenn er eine Linie ist, nennt man ihn ebenfalls eine Knotenlinie. $Jn(\kappa_{mn}r)$ ist die Besselfunktion $n$-ter Ordnung, wobei kmn die Wellenzahl und $r$ der Radius ist. $a_{mn}$ und $b_{mn}$ sind die zu bestimmenden Konstanten. -\subsubsection{Hankel Transformationen von Ableitungen \label{subsub:ableitungen}} -Wenn $\tilde{f}_n(\kappa)=\mathscr{H}_n\{f(r)\}$, dann: - -\begin{align} - &\mathscr{H}_n\{f'(r)\}=\frac{\kappa}{2n}\left[(n-1)\tilde{f}_{n+1}(\kappa)-(n+1)\tilde{f}_{n-1}(\kappa)\right], \quad n\geq1, \\ - &\mathscr{H}_1\{f'(r)\}=-\kappa \tilde{f}_0(\kappa), -\end{align} -bereitgestellt dass $[rf(r)]$ verschwindet als $r\to0$ und $r\to\infty=0$. \ No newline at end of file +An diesem Punkt stellte sich die Frage, ob es möglich wäre, die partielle Differentialgleichung mit einer anderen Methode als der der Trennung der Variablen zu lösen. Nach einer kurzen Recherche und Diskussion mit Prof. Müller wurde festgestellt, dass die beste Methode die Transformationsmethode ist, genauer gesagt die Anwendung der Hankel-Transformation. Im nächsten Kapitel wird daher diese Integraltransformation vorgestellt und entwickelt, und es wird erläutert, warum sie für diese Art von Problem geeignet ist. diff --git a/buch/papers/kreismembran/teil2.tex b/buch/papers/kreismembran/teil2.tex index 45357f2..8afe817 100644 --- a/buch/papers/kreismembran/teil2.tex +++ b/buch/papers/kreismembran/teil2.tex @@ -2,48 +2,112 @@ % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Lösung der partiellen Differentialgleichung - \label{kreismembran:section:teil2}} -\rhead{Lösung der partiellen Differentialgleichung} +\section{Die Hankel Transformation \label{kreismembran:section:teil2}} +\rhead{Die Hankel Transformation} + +Hermann Hankel (1839-1873) war ein deutscher Mathematiker, der für seinen Beitrag zur mathematischen Analyse und insbesondere für seine namensgebende Transformation bekannt ist. +Diese Transformation tritt bei der Untersuchung von funktionen auf, die nur von der Enternung des Ursprungs abhängen. +Er studierte auch funktionen, jetzt Hankel- oder Bessel- Funktionen genannt, der dritten Art. +Die Hankel Transformation mit Bessel Funktionen al Kern taucht natürlich bei achsensymmetrischen Problemen auf, die in Zylindrischen Polarkoordinaten formuliert sind. +In diesem Kapitel werden die Theorie der Transformation und einige Eigenschaften der Grundoperationen erläutert. + + +Wir führen die Definition der Hankel Transformation aus der zweidimensionalen Fourier Transformation und ihrer Umkehrung ein, die durch: +\begin{align} + \mathscr{F}\{f(x,y)\} & = F(k,l)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-i( \bm{\kappa}\cdot \mathbf{r})}f(x,y) dx dy,\label{equation:fourier_transform}\\ + \mathscr{F}^{-1}\{F(x,y)\} & = f(x,y)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{i(\bm{\kappa}\cdot \mathbf{r}))}F(k,l) dx dy \label{equation:inv_fourier_transform} +\end{align} +wo $\mathbf{r}=(x,y)$ und $\bm{\kappa}=(k,l)$. Wie bereits erwähnt, sind Polarkoordinaten für diese Art von Problemen am besten geeignet, also mit, $(x,y)=r(\cos\theta,\sin\theta)$ und $(k,l)=\kappa(\cos\phi,\sin\phi)$, findet man $\bm{\kappa}\cdot\mathbf{r}=\kappa r(\cos(\theta-\phi))$ und danach: +\begin{align} + F(k,\phi)=\frac{1}{2\pi}\int_{0}^{\infty}r dr \int_{0}^{2\pi}e^{-ikr\cos(\theta-\phi)}f(r,\theta) d\phi. + \label{equation:F_ohne_variable_wechsel} +\end{align} +Dann wird angenommen dass, $f(r,\theta)=e^{in\theta}f(r)$, was keine strenge Einschränkung ist, und es wird eine Änderung der Variabeln vorgenommen $\theta-\phi=\alpha-\frac{\pi}{2}$, um \eqref{equation:F_ohne_variable_wechsel} zu reduzieren: +\begin{align} + F(k,\phi)=\frac{1}{2\pi}\int_{0}^{\infty}rf(r) dr \int_{\phi_{0}}^{2\pi+\phi_{0}}e^{in(\phi-\frac{\pi}{2})+i(n\alpha-kr\sin\alpha)} d\alpha, + \label{equation:F_ohne_bessel} +\end{align} +wo $\phi_{0}=(\frac{\pi}{2}-\phi)$. + +Unter Verwendung der Integral Darstellung der Besselfunktion vom Ordnung n +\begin{align} + J_n(\kappa r)=\frac{1}{2\pi}\int_{\phi_{0}}^{2\pi + \phi_{0}}e^{i(n\alpha-\kappa r \sin \alpha)} d\alpha + \label{equation:bessel_n_ordnung} +\end{align} +\eqref{equation:F_ohne_bessel} wird sie zu: +\begin{align} + F(k,\phi)&=e^{in(\phi-\frac{\pi}{2})}\int_{0}^{\infty}rJ_n(\kappa r) f(r) dr \label{equation:F_mit_bessel_step_1} \\ + &=e^{in(\phi-\frac{\pi}{2})}\tilde{f}_n(\kappa), + \label{equation:F_mit_bessel_step_2} +\end{align} +wo $\tilde{f}_n(\kappa)$ ist die \textit{Hankel Transformation} von $f(r)$ und ist formell definiert durch: +\begin{align} + \mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)=\int_{0}^{\infty}rJ_n(\kappa r) f(r) dr. + \label{equation:hankel} +\end{align} + +Ähnlich verhält es sich mit der inversen Fourier Transformation in Form von polaren Koordinaten unter der Annahme $f(r,\theta)=e^{in\theta}f(r)$ mit \eqref{equation:F_mit_bessel_step_2}, wird die inverse Fourier Transformation \eqref{equation:inv_fourier_transform}: + +\begin{align} + e^{in\theta}f(r)&=\frac{1}{2\pi}\int_{0}^{\infty}\kappa d\kappa \int_{0}^{2\pi}e^{i\kappa r \cos (\theta - \phi)}F(\kappa,\phi) d\phi\\ + &= \frac{1}{2\pi}\int_{0}^{\infty}\kappa \tilde{f}_n(\kappa) d\kappa \int_{0}^{2\pi}e^{in(\phi - \frac{\pi}{2})- i\kappa r \cos (\theta - \phi)} d\phi, +\end{align} +was durch den Wechsel der Variablen $\theta-\phi=-(\alpha+\frac{\pi}{2})$ und $\theta_0=-(\theta+\frac{\pi}{2})$, + +\begin{align} + &= \frac{1}{2\pi}\int_{0}^{\infty}\kappa \tilde{f}_n(\kappa) d\kappa \int_{\theta_0}^{2\pi+\theta_0}e^{in(\theta + \alpha - i\kappa r \sin\alpha)} d\alpha \nonumber \\ + &= e^{in\theta}\int_{0}^{\infty}\kappa J_n(\kappa r) \tilde{f}_n(\kappa) d\kappa,\quad \text{von \eqref{equation:bessel_n_ordnung}} +\end{align} + +Also, die inverse \textit{Hankel Transformation} ist so definiert: +\begin{align} + \mathscr{H}^{-1}_n\{\tilde{f}_n(\kappa)\}=f(r)=\int_{0}^{\infty}\kappa J_n(\kappa r) \tilde{f}_n(\kappa) d\kappa. + \label{equation:inv_hankel} +\end{align} + +Anstelle von $\tilde{f}_n(\kappa)$, wird häufig für die Hankel Transformation verwendet, indem die Ordnung angegeben wird. +\eqref{equation:hankel} und \eqref{equation:inv_hankel} Integralen existieren für eine grosse Klasse von Funktionen, die normalerweise in physikalischen Anwendungen benötigt werden. +Alternativ kann auch die berühmte Hankel Transformationsformel verwendet werden, + +\begin{align} + f(r) = \int_{0}^{\infty}\kappa J_n(\kappa r) d\kappa \int_{0}^{\infty} p J_n(\kappa p)f(p) dp, + \label{equation:hankel_integral_formula} +\end{align} +um die Hankel Transformation \eqref{equation:hankel} und ihre Inverse \eqref{equation:inv_hankel} zu definieren. +Insbesondere die Hankel Transformation der nullten Ordnung ($n=0$) und der ersten Ordnung ($n=1$) sind häufig nützlich, um Lösungen für Probleme mit der Laplace Gleichung in einer achsensymmetrischen zylindrischen Geometrie zu finden. + +\subsection{Operative Eigenschaften der Hankel Transformation\label{sub:op_properties_hankel}} +In diesem Kapitel werden die operativen Eigenschaften der Hankel Transformation aufgeführt. Der Beweis für ihre Gültigkeit wird jedoch nicht analysiert. + +\subsubsection{Theorem 1: Skalierung \label{subsub:skalierung}} +Wenn $\mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)$, dann: -Wie im vorherigen Kapitel gezeigt, lautet die partielle Differentialgleichung, die die Schwingungen einer Membran beschreibt: \begin{equation*} - \frac{1}{c^2}\frac{\partial^2u}{\partial t^2} = \Delta u + \mathscr{H}_n\{f(ar)\}=\frac{1}{a^{2}}\tilde{f}_n \left(\frac{\kappa}{a}\right), \quad a>0. \end{equation*} -Da es sich um eine Kreisscheibe handelt, werden Polarkoordinaten verwendet, so dass sich der Laplaceoperator ergibt: + +\subsubsection{Theorem 2: Persevalsche Relation \label{subsub:perseval}} +Wenn $\tilde{f}(\kappa)=\mathscr{H}_n\{f(r)\}$ und $\tilde{g}(\kappa)=\mathscr{H}_n\{g(r)\}$, dann: + \begin{equation*} - \Delta - = - \frac{\partial^2}{\partial r^2} - + - \frac1r - \frac{\partial}{\partial r} - + - \frac{1}{r 2} - \frac{\partial^2}{\partial\varphi^2}. - \label{buch:pde:kreis:laplace} + \int_{0}^{\infty}rf(r) dr = \int_{0}^{\infty}\kappa\tilde{f}(\kappa)\tilde{g}(\kappa) d\kappa. \end{equation*} -Es wird eine runde elastische Membran berücksichtigt, die den Gebietbereich $\Omega$ abdeckt und am Rand $\Gamma$ befestigt ist. -Es wird daher davon ausgegangen, dass die Membran aus einem homogenen Material von vernachlässigbarer Dicke gefertigt ist. -Die Membran kann verformt werden, aber innere elastische Kräfte wirken den Verformungen entgegen. Es wirken keine äusseren Kräfte. Es handelt sich somit von einer kreisförmligen eigespannten homogenen schwingenden Membran. +\subsubsection{Theorem 3: Hankel Transformationen von Ableitungen \label{subsub:ableitungen}} +Wenn $\tilde{f}_n(\kappa)=\mathscr{H}_n\{f(r)\}$, dann: -Daher ist die Membranabweichung im Punkt $(r,\theta)$ $\in$ $\overline{\rm \Omega}$ zum Zeitpunkt $t$: \begin{align*} - u: \overline{\rm \Omega} \times \mathbb{R}_{\geq 0} &\longrightarrow \mathbb{R}\\ - (r,\theta,t) &\longmapsto u(r,\theta,t) + &\mathscr{H}_n\{f'(r)\}=\frac{\kappa}{2n}\left[(n-1)\tilde{f}_{n+1}(\kappa)-(n+1)\tilde{f}_{n-1}(\kappa)\right], \quad n\geq1, \\ + &\mathscr{H}_1\{f'(r)\}=-\kappa \tilde{f}_0(\kappa), \end{align*} -Da die Membran am Rand befestigt ist, kann es keine Schwingungen geben, so dass die \textit{Dirichlet-Randbedingung} gilt: -\begin{equation*} - u\big|_{\Gamma} = 0 -\end{equation*} +bereitgestellt dass $[rf(r)]$ verschwindet als $r\to0$ und $r\to\infty$. +\subsubsection{Theorem 4 \label{subsub:thorem4}} +Wenn $\mathscr{H}_n\{f(r)\}=\tilde{f}_n(\kappa)$, dann: -Um eine eindeutige Lösung bestimmen zu können, werden die folgenden Anfangsbedingungen festgelegt: +\begin{equation*} + \mathscr{H}_n \left\{ \left( \nabla^2 - \frac{n^2}{r^2} f(r)\right)\right\}= \mathscr{H}_n\left\{\frac{1}{r}\frac{d}{dr}\left(r\frac{df}{dr}\right) - \frac{n^2}{r^2}f(r)\right\}=-\kappa^2\tilde{f}_{n}(\kappa), +\end{equation*} +bereitgestellt dass $rf'(r)$ und $rf(r)$ verschwinden als $r\to0$ und $r\to\infty$. -\begin{align*} - u(r,\theta, 0) &:= f(x,y)\\ - \frac{\partial}{\partial t} u(r,\theta, 0) &:= g(x,y) -\end{align*} -An dieser Stelle könnte man zum Beispiel die bereits in Kapitel (TODO:refKAPITEL) vorgestellte Methode der Separation anwenden. Da es sich in diesem Fall jedoch um einem achsensymmetrischen Problem handelt, das in Polarkoordinaten formuliert ist, wird man die Transformationsmethode verwenden, insbesondere die Hankel Transformation. diff --git a/buch/papers/kreismembran/teil3.tex b/buch/papers/kreismembran/teil3.tex index 73dee0f..bef8b5f 100644 --- a/buch/papers/kreismembran/teil3.tex +++ b/buch/papers/kreismembran/teil3.tex @@ -3,38 +3,76 @@ % % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil % -\section{Teil 3 +\section{Lösungsmethode 2: Transformationsmethode \label{kreismembran:section:teil3}} -\rhead{Teil 3} -Sed ut perspiciatis unde omnis iste natus error sit voluptatem -accusantium doloremque laudantium, totam rem aperiam, eaque ipsa -quae ab illo inventore veritatis et quasi architecto beatae vitae -dicta sunt explicabo. Nemo enim ipsam voluptatem quia voluptas sit -aspernatur aut odit aut fugit, sed quia consequuntur magni dolores -eos qui ratione voluptatem sequi nesciunt. Neque porro quisquam -est, qui dolorem ipsum quia dolor sit amet, consectetur, adipisci -velit, sed quia non numquam eius modi tempora incidunt ut labore -et dolore magnam aliquam quaerat voluptatem. Ut enim ad minima -veniam, quis nostrum exercitationem ullam corporis suscipit laboriosam, -nisi ut aliquid ex ea commodi consequatur? Quis autem vel eum iure -reprehenderit qui in ea voluptate velit esse quam nihil molestiae -consequatur, vel illum qui dolorem eum fugiat quo voluptas nulla -pariatur? - -\subsection{De finibus bonorum et malorum -\label{kreismembran:subsection:malorum}} -At vero eos et accusamus et iusto odio dignissimos ducimus qui -blanditiis praesentium voluptatum deleniti atque corrupti quos -dolores et quas molestias excepturi sint occaecati cupiditate non -provident, similique sunt in culpa qui officia deserunt mollitia -animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis -est et expedita distinctio. Nam libero tempore, cum soluta nobis -est eligendi optio cumque nihil impedit quo minus id quod maxime -placeat facere possimus, omnis voluptas assumenda est, omnis dolor -repellendus. Temporibus autem quibusdam et aut officiis debitis aut -rerum necessitatibus saepe eveniet ut et voluptates repudiandae -sint et molestiae non recusandae. Itaque earum rerum hic tenetur a -sapiente delectus, ut aut reiciendis voluptatibus maiores alias -consequatur aut perferendis doloribus asperiores repellat. +\rhead{Lösungsmethode 2: Transformationsmethode} +Die Hankel-Transformation wird dann zur Lösung der Differentialgleichung verwendet. Es müssen jedoch einige Änderungen an dem Problem vorgenommen werden, damit es mit den Annahmen übereinstimmt, die für die Verwendung der Hankel-Transformation erforderlich sind. Das heisst, dass die Funktion u nur von der Entfernung zum Ausgangspunkt abhängt. Wir führen also das Konzept einer unendlichen und achsensymmetrischen Membran ein: +\begin{equation*} + \frac{\partial^2u}{\partial t^2} + = + c^2 \left(\frac{\partial^2 u}{\partial r^2} + + + \frac{1}{r} + \frac{\partial u}{\partial r} \right), \quad 00 + \label{eq:PDE_inf_membane} +\end{equation*} + +\begin{align} + u(r,0)=f(r), \quad \frac{\partial}{\partial t} u(r,0) = g(r), \quad \text{für} \quad 0 Date: Wed, 18 May 2022 14:20:41 +0200 Subject: =?UTF-8?q?Dreiecke=20f=C3=BCr=20Nav?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../050-differential/uebungsaufgaben/airy.cpp | 4 +- buch/chapters/110-elliptisch/ellintegral.tex | 2 +- buch/papers/nav/images/Makefile | 10 ++- buch/papers/nav/images/common.inc | 28 +++++- buch/papers/nav/images/dreieck3d1.pdf | Bin 0 -> 90451 bytes buch/papers/nav/images/dreieck3d1.pov | 12 +-- buch/papers/nav/images/dreieck3d2.pdf | Bin 0 -> 69523 bytes buch/papers/nav/images/dreieck3d2.pov | 6 +- buch/papers/nav/images/dreieck3d3.pdf | Bin 0 -> 82512 bytes buch/papers/nav/images/dreieck3d3.pov | 8 +- buch/papers/nav/images/dreieck3d4.pdf | Bin 0 -> 85037 bytes buch/papers/nav/images/dreieck3d4.pov | 8 +- buch/papers/nav/images/dreieck3d5.pdf | Bin 0 -> 70054 bytes buch/papers/nav/images/dreieck3d5.pov | 6 +- buch/papers/nav/images/dreieck3d6.pov | 2 +- buch/papers/nav/images/dreieck3d7.pov | 10 +-- buch/papers/nav/images/dreieck3d8.jpg | Bin 0 -> 93432 bytes buch/papers/nav/images/dreieck3d8.pdf | Bin 0 -> 107370 bytes buch/papers/nav/images/dreieck3d8.pov | 96 +++++++++++++++++++++ buch/papers/nav/images/dreieck3d8.tex | 57 ++++++++++++ vorlesungen/stream/countdown.html | 2 +- 21 files changed, 217 insertions(+), 34 deletions(-) create mode 100644 buch/papers/nav/images/dreieck3d1.pdf create mode 100644 buch/papers/nav/images/dreieck3d2.pdf create mode 100644 buch/papers/nav/images/dreieck3d3.pdf create mode 100644 buch/papers/nav/images/dreieck3d4.pdf create mode 100644 buch/papers/nav/images/dreieck3d5.pdf create mode 100644 buch/papers/nav/images/dreieck3d8.jpg create mode 100644 buch/papers/nav/images/dreieck3d8.pdf create mode 100644 buch/papers/nav/images/dreieck3d8.pov create mode 100644 buch/papers/nav/images/dreieck3d8.tex diff --git a/buch/chapters/050-differential/uebungsaufgaben/airy.cpp b/buch/chapters/050-differential/uebungsaufgaben/airy.cpp index e4df8e1..eb5c6be 100644 --- a/buch/chapters/050-differential/uebungsaufgaben/airy.cpp +++ b/buch/chapters/050-differential/uebungsaufgaben/airy.cpp @@ -44,8 +44,8 @@ double h0f1(double c, double x) { double f1(double x) { // unfortunately, gsl_sf_hyperg_0F1 does not work if c<1, because // it uses a relation to the bessel functions - //return gsl_sf_hyperg_0F1(2/3, x*x*x/9.); - return h0f1(2./3., x*x*x/9.); + return gsl_sf_hyperg_0F1(2/3, x*x*x/9.); + //return h0f1(2./3., x*x*x/9.); } double f2(double x) { diff --git a/buch/chapters/110-elliptisch/ellintegral.tex b/buch/chapters/110-elliptisch/ellintegral.tex index 4cb2ba3..3acce2f 100644 --- a/buch/chapters/110-elliptisch/ellintegral.tex +++ b/buch/chapters/110-elliptisch/ellintegral.tex @@ -651,7 +651,7 @@ werden, dass $1-k'^2=k^2$ ist. \begin{definition} Ist $0\le k\le 1$ der Modul eines elliptischen Integrals, dann heisst -$k' = \sqrt{1-k^2}$ er {\em Komplementärmodul} oder {\em Komplement +$k' = \sqrt{1-k^2}$ der {\em Komplementärmodul} oder {\em Komplement des Moduls}. Es ist $k^2+k'^2=1$. \end{definition} diff --git a/buch/papers/nav/images/Makefile b/buch/papers/nav/images/Makefile index c9dcacc..bbdea2f 100644 --- a/buch/papers/nav/images/Makefile +++ b/buch/papers/nav/images/Makefile @@ -50,7 +50,8 @@ DREIECKE3D = \ dreieck3d4.pdf \ dreieck3d5.pdf \ dreieck3d6.pdf \ - dreieck3d7.pdf + dreieck3d7.pdf \ + dreieck3d8.pdf dreiecke3d: $(DREIECKE3D) @@ -106,3 +107,10 @@ dreieck3d7.jpg: dreieck3d7.png dreieck3d7.pdf: dreieck3d7.tex dreieck3d7.jpg pdflatex dreieck3d7.tex +dreieck3d8.png: dreieck3d8.pov common.inc + povray +A0.1 $(POVRAYOPTIONS) -Odreieck3d8.png dreieck3d8.pov +dreieck3d8.jpg: dreieck3d8.png + convert dreieck3d8.png -density 300 -units PixelsPerInch dreieck3d8.jpg +dreieck3d8.pdf: dreieck3d8.tex dreieck3d8.jpg + pdflatex dreieck3d8.tex + diff --git a/buch/papers/nav/images/common.inc b/buch/papers/nav/images/common.inc index 33d9384..e2a1ed0 100644 --- a/buch/papers/nav/images/common.inc +++ b/buch/papers/nav/images/common.inc @@ -97,13 +97,13 @@ union { } #end -#macro winkel(w, p, q, staerke) +#macro winkel(w, p, q, staerke, r) #declare n = vnormalize(w); #declare pp = vnormalize(p - vdot(n, p) * n); #declare qq = vnormalize(q - vdot(n, q) * n); intersection { sphere { <0, 0, 0>, 1 + staerke } - cone { <0, 0, 0>, 0, 1.2 * vnormalize(w), 0.4 } + cone { <0, 0, 0>, 0, 1.2 * vnormalize(w), r } plane { -vcross(n, qq) * vdot(vcross(n, qq), pp), 0 } plane { -vcross(n, pp) * vdot(vcross(n, pp), qq), 0 } } @@ -113,8 +113,30 @@ union { sphere { p, 1.5 * staerke } #end +#macro dreieck(p, q, r, farbe) + #declare n1 = vnormalize(vcross(p, q)); + #declare n2 = vnormalize(vcross(q, r)); + #declare n3 = vnormalize(vcross(r, p)); + intersection { + plane { n1, 0 } + plane { n2, 0 } + plane { n3, 0 } + sphere { <0, 0, 0>, 1 + 0.001 } + pigment { + color farbe + } + finish { + metallic + specular 0.4 + } + } +#end + #declare fett = 0.015; -#declare fine = 0.010; +#declare fein = 0.010; + +#declare klein = 0.3; +#declare gross = 0.4; #declare dreieckfarbe = rgb<0.6,0.6,0.6>; #declare rot = rgb<0.8,0.2,0.2>; diff --git a/buch/papers/nav/images/dreieck3d1.pdf b/buch/papers/nav/images/dreieck3d1.pdf new file mode 100644 index 0000000..015bce7 Binary files /dev/null and b/buch/papers/nav/images/dreieck3d1.pdf differ diff --git a/buch/papers/nav/images/dreieck3d1.pov b/buch/papers/nav/images/dreieck3d1.pov index 8afe60e..e491075 100644 --- a/buch/papers/nav/images/dreieck3d1.pov +++ b/buch/papers/nav/images/dreieck3d1.pov @@ -12,9 +12,9 @@ union { punkt(A, fett) punkt(B, fett) punkt(C, fett) - punkt(P, fine) - seite(B, P, fine) - seite(C, P, fine) + punkt(P, fein) + seite(B, P, fein) + seite(C, P, fein) pigment { color dreieckfarbe } @@ -25,7 +25,7 @@ union { } object { - winkel(A, B, C, fine) + winkel(A, B, C, fein, gross) pigment { color rot } @@ -36,7 +36,7 @@ object { } object { - winkel(B, C, A, fine) + winkel(B, C, A, fein, gross) pigment { color gruen } @@ -47,7 +47,7 @@ object { } object { - winkel(C, A, B, fine) + winkel(C, A, B, fein, gross) pigment { color blau } diff --git a/buch/papers/nav/images/dreieck3d2.pdf b/buch/papers/nav/images/dreieck3d2.pdf new file mode 100644 index 0000000..6b3f09d Binary files /dev/null and b/buch/papers/nav/images/dreieck3d2.pdf differ diff --git a/buch/papers/nav/images/dreieck3d2.pov b/buch/papers/nav/images/dreieck3d2.pov index c23a54c..c0625ce 100644 --- a/buch/papers/nav/images/dreieck3d2.pov +++ b/buch/papers/nav/images/dreieck3d2.pov @@ -12,9 +12,9 @@ union { punkt(A, fett) punkt(B, fett) punkt(C, fett) - punkt(P, fine) - seite(B, P, fine) - seite(C, P, fine) + punkt(P, fein) + seite(B, P, fein) + seite(C, P, fein) pigment { color dreieckfarbe } diff --git a/buch/papers/nav/images/dreieck3d3.pdf b/buch/papers/nav/images/dreieck3d3.pdf new file mode 100644 index 0000000..7d79455 Binary files /dev/null and b/buch/papers/nav/images/dreieck3d3.pdf differ diff --git a/buch/papers/nav/images/dreieck3d3.pov b/buch/papers/nav/images/dreieck3d3.pov index f2496b5..b6f64d5 100644 --- a/buch/papers/nav/images/dreieck3d3.pov +++ b/buch/papers/nav/images/dreieck3d3.pov @@ -12,9 +12,9 @@ union { punkt(A, fett) punkt(B, fett) punkt(C, fett) - punkt(P, fine) - seite(B, P, fine) - seite(C, P, fine) + punkt(P, fein) + seite(B, P, fein) + seite(C, P, fein) pigment { color dreieckfarbe } @@ -25,7 +25,7 @@ union { } object { - winkel(A, B, C, fine) + winkel(A, B, C, fein, gross) pigment { color rot } diff --git a/buch/papers/nav/images/dreieck3d4.pdf b/buch/papers/nav/images/dreieck3d4.pdf new file mode 100644 index 0000000..e1ea757 Binary files /dev/null and b/buch/papers/nav/images/dreieck3d4.pdf differ diff --git a/buch/papers/nav/images/dreieck3d4.pov b/buch/papers/nav/images/dreieck3d4.pov index bddcf7c..b6f17e3 100644 --- a/buch/papers/nav/images/dreieck3d4.pov +++ b/buch/papers/nav/images/dreieck3d4.pov @@ -6,9 +6,9 @@ #include "common.inc" union { - seite(A, B, fine) - seite(A, C, fine) - punkt(A, fine) + seite(A, B, fein) + seite(A, C, fein) + punkt(A, fein) punkt(B, fett) punkt(C, fett) punkt(P, fett) @@ -25,7 +25,7 @@ union { } object { - winkel(B, C, P, fine) + winkel(B, C, P, fein, gross) pigment { color rgb<0.6,0.4,0.2> } diff --git a/buch/papers/nav/images/dreieck3d5.pdf b/buch/papers/nav/images/dreieck3d5.pdf new file mode 100644 index 0000000..6848331 Binary files /dev/null and b/buch/papers/nav/images/dreieck3d5.pdf differ diff --git a/buch/papers/nav/images/dreieck3d5.pov b/buch/papers/nav/images/dreieck3d5.pov index 32fc9e6..188f181 100644 --- a/buch/papers/nav/images/dreieck3d5.pov +++ b/buch/papers/nav/images/dreieck3d5.pov @@ -6,9 +6,9 @@ #include "common.inc" union { - seite(A, B, fine) - seite(A, C, fine) - punkt(A, fine) + seite(A, B, fein) + seite(A, C, fein) + punkt(A, fein) punkt(B, fett) punkt(C, fett) punkt(P, fett) diff --git a/buch/papers/nav/images/dreieck3d6.pov b/buch/papers/nav/images/dreieck3d6.pov index 7611950..191a1e7 100644 --- a/buch/papers/nav/images/dreieck3d6.pov +++ b/buch/papers/nav/images/dreieck3d6.pov @@ -25,7 +25,7 @@ union { } object { - winkel(B, A, P, fine) + winkel(B, A, P, fein, gross) pigment { color rgb<0.6,0.2,0.6> } diff --git a/buch/papers/nav/images/dreieck3d7.pov b/buch/papers/nav/images/dreieck3d7.pov index fa48f5b..aae5c6c 100644 --- a/buch/papers/nav/images/dreieck3d7.pov +++ b/buch/papers/nav/images/dreieck3d7.pov @@ -10,13 +10,13 @@ union { seite(A, P, fett) seite(C, P, fett) - seite(A, B, fine) - seite(B, C, fine) - seite(B, P, fine) + seite(A, B, fein) + seite(B, C, fein) + seite(B, P, fein) punkt(A, fett) punkt(C, fett) punkt(P, fett) - punkt(B, fine) + punkt(B, fein) pigment { color dreieckfarbe } @@ -27,7 +27,7 @@ union { } object { - winkel(A, P, C, fine) + winkel(A, P, C, fein, gross) pigment { color rgb<0.4,0.4,1> } diff --git a/buch/papers/nav/images/dreieck3d8.jpg b/buch/papers/nav/images/dreieck3d8.jpg new file mode 100644 index 0000000..52bd25e Binary files /dev/null and b/buch/papers/nav/images/dreieck3d8.jpg differ diff --git a/buch/papers/nav/images/dreieck3d8.pdf b/buch/papers/nav/images/dreieck3d8.pdf new file mode 100644 index 0000000..9d630aa Binary files /dev/null and b/buch/papers/nav/images/dreieck3d8.pdf differ diff --git a/buch/papers/nav/images/dreieck3d8.pov b/buch/papers/nav/images/dreieck3d8.pov new file mode 100644 index 0000000..9e9921a --- /dev/null +++ b/buch/papers/nav/images/dreieck3d8.pov @@ -0,0 +1,96 @@ +// +// dreiecke3d8.pov +// +// (c) 2022 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +// +#include "common.inc" + +union { + seite(A, B, fett) + seite(B, C, fett) + seite(A, C, fett) + seite(A, P, fein) + seite(B, P, fett) + seite(C, P, fett) + punkt(A, fett) + punkt(B, fett) + punkt(C, fett) + punkt(P, fett) + pigment { + color dreieckfarbe + } + finish { + specular 0.95 + metallic + } +} + +object { + winkel(A, B, C, fein, klein) + pigment { + color rot + } + finish { + specular 0.95 + metallic + } +} + +object { + winkel(B, C, A, fein, klein) + pigment { + color gruen + } + finish { + specular 0.95 + metallic + } +} + +object { + winkel(C, A, B, fein, gross) + pigment { + color blau + } + finish { + specular 0.95 + metallic + } +} + +object { + winkel(A, P, C, fein/2, gross) + pigment { + color rgb<0.8,0,0.8> + } + finish { + specular 0.95 + metallic + } +} + +object { + winkel(B, P, C, fein, klein) + pigment { + color rgb<1,0.8,0> + } + finish { + specular 0.95 + metallic + } +} + +object { + winkel(B, P, A, fein/2, gross) + pigment { + color rgb<0.4,0.6,0.8> + } + finish { + specular 0.95 + metallic + } +} + +dreieck(A, B, C, White) + + diff --git a/buch/papers/nav/images/dreieck3d8.tex b/buch/papers/nav/images/dreieck3d8.tex new file mode 100644 index 0000000..c59c7b0 --- /dev/null +++ b/buch/papers/nav/images/dreieck3d8.tex @@ -0,0 +1,57 @@ +% +% dreieck3d8.tex +% +% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule +% +\documentclass[tikz]{standalone} +\usepackage{times} +\usepackage{amsmath} +\usepackage{txfonts} +\usepackage[utf8]{inputenc} +\usepackage{graphics} +\usetikzlibrary{arrows,intersections,math} +\usepackage{ifthen} +\begin{document} + +\newboolean{showgrid} +\setboolean{showgrid}{false} +\def\breite{4} +\def\hoehe{4} + +\begin{tikzpicture}[>=latex,thick] + +% Povray Bild +\node at (0,0) {\includegraphics[width=8cm]{dreieck3d8.jpg}}; + +% Gitter +\ifthenelse{\boolean{showgrid}}{ +\draw[step=0.1,line width=0.1pt] (-\breite,-\hoehe) grid (\breite, \hoehe); +\draw[step=0.5,line width=0.4pt] (-\breite,-\hoehe) grid (\breite, \hoehe); +\draw (-\breite,-\hoehe) grid (\breite, \hoehe); +\fill (0,0) circle[radius=0.05]; +}{} + +\node at (0.7,3.8) {$A$}; +\node at (-3.4,-0.8) {$B$}; +\node at (3.3,-2.1) {$C$}; +\node at (-1.4,-3.5) {$P$}; + +\node at (-1.9,2.1) {$c$}; +\node at (-0.2,-1.2) {$a$}; +\node at (2.6,1.5) {$b$}; +\node at (-0.8,0) {$l$}; + +\node at (-2.6,-2.2) {$p_b$}; +\node at (1,-2.9) {$p_c$}; + +\node at (0.7,3.3) {$\alpha$}; +\node at (0.8,2.85) {$\omega$}; +\node at (-2.6,-0.6) {$\beta$}; +\node at (2.3,-1.2) {$\gamma$}; +\node at (-2.6,-1.3) {$\beta_1$}; +\node at (-2.1,-0.8) {$\kappa$}; + +\end{tikzpicture} + +\end{document} + diff --git a/vorlesungen/stream/countdown.html b/vorlesungen/stream/countdown.html index d8ec82e..e9d7d6e 100644 --- a/vorlesungen/stream/countdown.html +++ b/vorlesungen/stream/countdown.html @@ -25,7 +25,7 @@ function checkfor(d) { console.log("time string: " + ds); let start = new Date(ds).getTime(); console.log("now: " + now); - if ((start > now) && ((start-now) < 3600*1000)) { + if ((start > now) && ((start-now) < 3300*1000)) { deadline = start; console.log("set deadline to: " + ds); } else { -- cgit v1.2.1 From 525ff82400b685dc6dd0d6376253545720471be0 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Andreas=20M=C3=BCller?= Date: Wed, 18 May 2022 14:25:26 +0200 Subject: remove bad files --- buch/buch.aux | 22 - buch/buch.bbl | 0 buch/buch.blg | 48 - buch/buch.idx | 0 buch/buch.log | 2106 --------------------------------- buch/papers/nav/images/dreieck3d5.pdf | Bin 70054 -> 70045 bytes 6 files changed, 2176 deletions(-) delete mode 100644 buch/buch.aux delete mode 100644 buch/buch.bbl delete mode 100644 buch/buch.blg delete mode 100644 buch/buch.idx delete mode 100644 buch/buch.log diff --git a/buch/buch.aux b/buch/buch.aux deleted file mode 100644 index 6730af9..0000000 --- a/buch/buch.aux +++ /dev/null @@ -1,22 +0,0 @@ -\relax -\providecommand\hyper@newdestlabel[2]{} -\providecommand\babel@aux[2]{} -\@nameuse{bbl@beforestart} -\catcode `"\active -\providecommand\HyperFirstAtBeginDocument{\AtBeginDocument} -\HyperFirstAtBeginDocument{\ifx\hyper@anchor\@undefined -\global\let\oldcontentsline\contentsline -\gdef\contentsline#1#2#3#4{\oldcontentsline{#1}{#2}{#3}} -\global\let\oldnewlabel\newlabel -\gdef\newlabel#1#2{\newlabelxx{#1}#2} -\gdef\newlabelxx#1#2#3#4#5#6{\oldnewlabel{#1}{{#2}{#3}}} -\AtEndDocument{\ifx\hyper@anchor\@undefined -\let\contentsline\oldcontentsline -\let\newlabel\oldnewlabel -\fi} -\fi} -\global\let\hyper@last\relax -\gdef\HyperFirstAtBeginDocument#1{#1} -\providecommand\HyField@AuxAddToFields[1]{} -\providecommand\HyField@AuxAddToCoFields[2]{} -\providecommand\BKM@entry[2]{} diff --git a/buch/buch.bbl b/buch/buch.bbl deleted file mode 100644 index e69de29..0000000 diff --git a/buch/buch.blg b/buch/buch.blg deleted file mode 100644 index 706b1d8..0000000 --- a/buch/buch.blg +++ /dev/null @@ -1,48 +0,0 @@ -This is BibTeX, Version 0.99d -Capacity: max_strings=200000, hash_size=200000, hash_prime=170003 -The top-level auxiliary file: buch.aux -I found no \citation commands---while reading file buch.aux -I found no \bibdata command---while reading file buch.aux -I found no \bibstyle command---while reading file buch.aux -You've used 0 entries, - 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TeX capacity exceeded, sorry [main memory size=3000000]. - \XKV@resb - -l.575 } - -If you really absolutely need more capacity, -you can ask a wizard to enlarge me. - - -Here is how much of TeX's memory you used: - 81845 strings out of 478582 - 2061696 string characters out of 2856069 - 3000001 words of memory out of 3000000 - 98832 multiletter control sequences out of 15000+600000 - 484206 words of font info for 75 fonts, out of 8000000 for 9000 - 1143 hyphenation exceptions out of 8191 - 105i,3n,99p,3369b,2422s stack positions out of 10000i,1000n,20000p,200000b,80000s -! ==> Fatal error occurred, no output PDF file produced! diff --git a/buch/papers/nav/images/dreieck3d5.pdf b/buch/papers/nav/images/dreieck3d5.pdf index 6848331..0c86d36 100644 Binary files a/buch/papers/nav/images/dreieck3d5.pdf and b/buch/papers/nav/images/dreieck3d5.pdf differ -- cgit v1.2.1 From 93bdfca3b41397e43537ee334e57883a9ef79279 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Andreas=20M=C3=BCller?= Date: Wed, 18 May 2022 14:30:12 +0200 Subject: fix nav/makefile.inc --- buch/papers/nav/Makefile.inc | 12 +++++++----- 1 file changed, 7 insertions(+), 5 deletions(-) diff --git a/buch/papers/nav/Makefile.inc b/buch/papers/nav/Makefile.inc index b30377e..24ab4ee 100644 --- a/buch/papers/nav/Makefile.inc +++ b/buch/papers/nav/Makefile.inc @@ -6,9 +6,11 @@ dependencies-nav = \ papers/nav/packages.tex \ papers/nav/main.tex \ - papers/nav/references.bib \ - papers/nav/teil0.tex \ - papers/nav/teil1.tex \ - papers/nav/teil2.tex \ - papers/nav/teil3.tex + papers/nav/einleitung.tex \ + papers/nav/flatearth.tex \ + papers/nav/geschichte.tex \ + papers/nav/nautischesdreieck.tex \ + papers/nav/sincos.tex \ + papers/nav/trigo.tex \ + papers/nav/references.bib -- cgit v1.2.1 From 88cc6d9775114c70d8723a52a869179ce806d2f7 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Andreas=20M=C3=BCller?= Date: Wed, 18 May 2022 14:43:23 +0200 Subject: fix paper offset --- buch/splitpapers | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/buch/splitpapers b/buch/splitpapers index 9ae5aae..e1b6834 100755 --- a/buch/splitpapers +++ b/buch/splitpapers @@ -16,7 +16,7 @@ then fi awk 'BEGIN { - offsetpage = 10 + offsetpage = 12 startpage = 0 identifier = "" chapterno = 0 -- cgit v1.2.1