diff options
Diffstat (limited to 'buch/papers/lambertw')
| -rw-r--r-- | buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png | bin | 0 -> 124329 bytes | |||
| -rw-r--r-- | buch/papers/lambertw/main.log | 686 | ||||
| -rw-r--r-- | buch/papers/lambertw/teil1.tex | 37 | ||||
| -rw-r--r-- | buch/papers/lambertw/teil4.tex | 116 |
4 files changed, 167 insertions, 672 deletions
diff --git a/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png b/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png Binary files differnew file mode 100644 index 0000000..53eb2f9 --- /dev/null +++ b/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png diff --git a/buch/papers/lambertw/main.log b/buch/papers/lambertw/main.log index 4b0af4d..754563d 100644 --- a/buch/papers/lambertw/main.log +++ b/buch/papers/lambertw/main.log @@ -1,14 +1,12 @@ -This is pdfTeX, Version 3.141592653-2.6-1.40.23 (TeX Live 2021/W32TeX) (preloaded format=pdflatex 2021.11.16) 15 MAR 2022 13:23 +This is pdfTeX, Version 3.141592653-2.6-1.40.23 (MiKTeX 21.8) (preloaded format=pdflatex 2021.9.21) 20 JUL 2022 18:38 entering extended mode - restricted \write18 enabled. - %&-line parsing enabled. -**main.tex -(./main.tex -LaTeX2e <2021-11-15> -L3 programming layer <2021-11-12> +**./main.tex +(main.tex +LaTeX2e <2021-06-01> patch level 1 +L3 programming layer <2021-08-27> ! Undefined control sequence. l.6 \chapter - {Thema\label{chapter:lambertw}} + {Verfolgungskurven\label{chapter:lambertw}} The control sequence at the end of the top line of your error message was never \def'ed. If you have misspelled it (e.g., `\hobx'), type `I' and the correct @@ -22,16 +20,28 @@ See the LaTeX manual or LaTeX Companion for explanation. Type H <return> for immediate help. ... -l.6 \chapter{T - hema\label{chapter:lambertw}} +l.6 \chapter{V + erfolgungskurven\label{chapter:lambertw}} You're in trouble here. Try typing <return> to proceed. If that doesn't work, type X <return> to quit. -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! +Missing character: There is no V in font nullfont! 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Undefined control sequence. l.7 \lhead {Thema} @@ -61,666 +71,46 @@ or <return> to continue without it. ! Undefined control sequence. l.9 \chapterauthor - {Hans Muster} + {David Hugentobler und Yanik Kuster} The control sequence at the end of the top line of your error message was never \def'ed. If you have misspelled it (e.g., `\hobx'), type `I' and the correct spelling (e.g., `I\hbox'). 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Missing character: There is no r in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (5.55557pt too wide) in paragraph at lines 26--28 -\/cmr/m/n/10 ^^?u - [] - -Overfull \hbox (7.50002pt too wide) in paragraph at lines 26--28 -[]\/cmr/m/n/10 U +Overfull \hbox (20.0pt too wide) in paragraph at lines 6--10 +[][] [] @@ -734,16 +124,16 @@ Enter file name: <read *> l.30 \input{papers/lambertw/teil0.tex} - ^^M + *** (cannot \read from terminal in nonstop modes) Here is how much of TeX's memory you used: - 36 strings out of 478371 - 593 string characters out of 5852527 - 296836 words of memory out of 5000000 - 18242 multiletter control sequences out of 15000+600000 - 403598 words of font info for 28 fonts, out of 8000000 for 9000 + 22 strings out of 478927 + 609 string characters out of 2852535 + 290175 words of memory out of 3000000 + 17980 multiletter control sequences out of 15000+600000 + 403430 words of font info for 27 fonts, out of 8000000 for 9000 1141 hyphenation exceptions out of 8191 - 23i,1n,32p,120b,183s stack positions out of 5000i,500n,10000p,200000b,80000s + 16i,0n,26p,94b,28s stack positions out of 5000i,500n,10000p,200000b,80000s ! ==> Fatal error occurred, no output PDF file produced! diff --git a/buch/papers/lambertw/teil1.tex b/buch/papers/lambertw/teil1.tex index 3415c45..aa7f226 100644 --- a/buch/papers/lambertw/teil1.tex +++ b/buch/papers/lambertw/teil1.tex @@ -25,25 +25,25 @@ Wir verwenden die Hergeleiteten Gleichungen \frac{1}{4}\left(\left(y_0+r_0\right)\eta+\left(r_0-y_0\right)ln\left(\eta\right)-r_0+3y_0\right) \\ \chi &= - \frac{r_0+y_0}{r_0-y_0}; \cdot\chi \\ + \frac{r_0+y_0}{r_0-y_0}\\ \eta &= \left(\frac{x}{x_0}\right)^2 - \:;\: + \\ r_0 - = + &= \sqrt{x_0^2+y_0^2} \\ \end{align*} Wir definieren einen Treffer wenn die Koordinaten des Verfolgers mit denen des Ziels übereinstimmen bei einem diskreten Zeitpunkt $t_1$. Aus dem vorangegangenem Beispiel, sind die Gleichungen zu den x- und y-Koordinaten des Verfolgers bekannt. Die Des Ziels sind \begin{equation} - \overrightarrow{Z}(t) + \vec{Z}(t) = \left( \begin{array}{c} 0 \\ v \cdot t \end{array} \right) = \left( \begin{array}{c} 0 \\ t \end{array} \right) ;\quad - \overrightarrow{V}(t) + \vec{V}(t) = \left( \begin{array}{c} x(t) \\ y(t) \end{array} \right) \label{lambertw:Anfangspunkte} @@ -52,10 +52,10 @@ Wir definieren einen Treffer wenn die Koordinaten des Verfolgers mit denen des Z Somit gilt es \begin{equation*} - \overrightarrow{Z}(t_1)=\overrightarrow{V}(t_1) + \vec{Z}(t_1)=\vec{V}(t_1) \end{equation*} -zu lösen. Da die $y(t)$ viel komplexer ist als $x(t)$ wird das Problem in zwei einzelne Teilprobleme zerlegt. Wobei die Bedingung der x- und y-Koordinaten einzeln überprüft werden. +zu lösen. Da $y(t)$ viel komplexer ist als $x(t)$ wird das Problem in zwei einzelne Teilprobleme zerlegt. Wobei die Bedingung der x- und y-Koordinaten einzeln überprüft werden. \begin{align*} 0 @@ -72,7 +72,10 @@ zu lösen. Da die $y(t)$ viel komplexer ist als $x(t)$ wird das Problem in zwei \\ \end{align*} -Zuerst wird die Bedingung der x-Koordinate betrachtet. Diese kann durch quadrieren und anschliessendes multiplizieren von $\chi$ vereinfacht werden. +Zuerst wird die Bedingung der x-Koordinate betrachtet. +Diese kann durch quadrieren und anschliessendes multiplizieren von $\chi$ vereinfacht werden. +Es ist zu beachten, dass $W(x)$ die Lambert W-Funktion ist, welche im Kapitel \eqref{buch:section:lambertw} behandelt wurde. +Die Gleichung \begin{equation} 0 @@ -80,7 +83,8 @@ Zuerst wird die Bedingung der x-Koordinate betrachtet. Diese kann durch quadrier W\left(\chi\cdot e^{\chi-\frac{4t}{r_0-y_0}}\right) \end{equation} -Dies entspricht genau den Nullstellen der Lambert W-Funktion. Da die Lambert W-Funktion genau eine Nullstelle bei + +entspricht genau den Nullstellen der Lambert W-Funktion. Da die Lambert W-Funktion genau eine Nullstelle bei \begin{equation*} W(0)=0 @@ -100,5 +104,20 @@ Dies nützt nicht viel, da unendlich viel Zeit vergehen müsste damit ein Treffe Somit kann nach den Gestellten Bedingungen das Ziel nie getroffen werden. Dieses Resultat ist aber eher akademischer Natur, weil der Verfolger und das Ziel als Punkt betrachtet wurden. Wobei aber in Realität nicht von Punkten sondern von Objekten mit einer räumlichen Ausdehnung gesprochen werden kann. +Somit wird in einer nächsten Betrachtung untersucht, ob der Verfolger dem Ziel näher kommt als ein definierter Trefferradius. +Falls dies stattfinden sollte, wird dies als Treffer interpretiert. +Mathematisch kann dies mit + +\begin{equation} + |\vec{V}-\vec{Z]|<a_min \quad a_min\in\mathbb{R}>0 +\end{equation} + +beschrieben werden, wobei $a_min$ dem Trefferradius entspricht. +Diese Gleichung wird noch quadriert, um die Wurzeln des Betrages loszuwerden. +Da sowohl der Betrag als auch $a_min$ grösser null sind, bleibt die Aussage unverändert. + +\begin{equation} + |\vec{V}-\vec{Z]|^2<a_min^2 \quad a_min\in\mathbb{R}>0 +\end{equation} diff --git a/buch/papers/lambertw/teil4.tex b/buch/papers/lambertw/teil4.tex index 598a57e..6184369 100644 --- a/buch/papers/lambertw/teil4.tex +++ b/buch/papers/lambertw/teil4.tex @@ -6,24 +6,22 @@ \section{Beispiel Verfolgungskurve \label{lambertw:section:teil4}} \rhead{Beispiel Verfolgungskurve} -In diesem Abschnitt wird rechnerisch das Beispiel einer Verfolgungskurve beschreiben. +In diesem Abschnitt wird rechnerisch das Beispiel einer Verfolgungskurve mit der Verfolgungsstrategie 1 beschreiben. -\subsection{Ziel bewegt sich auf einer Gerade -\label{lambertw:subsection:malorum}} -Das zu verfolgende Ziel \(A\) wandert auf einer Gerade, wobei diese Gerade der \(y\)-Achse entspricht. Der Verfolger \(P\) startet auf einem beliebigen Punkt auf dem ersten Quadrant.Um die Rechnungen zu vereinfachen wir die Geschwindigkeit \(v\) auf 1 gesetzt. Diese Anfangspunkte oder Anfangsbedingungen können wie folgt formuliert werden: +Das zu verfolgende Ziel \(\overrightarrow{Z}\) wandert auf einer Gerade mit konstanter Geschwindigkeit \(v = 1\), wobei diese Gerade der \(y\)-Achse entspricht. Der Verfolger \(\overrightarrow{V}\) startet auf einem beliebigen Punkt im ersten Quadrant und bewegt sich auch mit konstanter Geschwindigkeit. Diese Anfangspunkte oder Anfangsbedingungen können wie folgt formuliert werden: \begin{equation} - A + \overrightarrow{Z} = \left( \begin{array}{c} 0 \\ v \cdot t \end{array} \right) = \left( \begin{array}{c} 0 \\ t \end{array} \right) ; - P + \overrightarrow{V} = \left( \begin{array}{c} x \\ y \end{array} \right) \label{lambertw:Anfangspunkte} \end{equation} -Wenn man diese Startpunkte in die Gleichung der Verfolgungskurve einfügt ergibt sich folgender Ausdruck: +Wenn man diese Startpunkte in die Gleichung der Verfolgungskurve \eqref{lambertw:pursuerDGL} einfügt ergibt sich folgender Ausdruck: \begin{equation} \frac{\left( \begin{array}{c} 0-x \\ t-y \end{array} \right)}{\sqrt{x^2 + (t-y)^2}} \circ @@ -79,12 +77,12 @@ Wenn man nun beidseitig die Quadratwurzel zieht, dann ergibt sich im Vergleich z = 0 \label{lambertw:equation5} \end{equation} -Um die Ableitung nach der Zeit wegzubringen wird beidseitig mit \(\dot{x}\) dividiert, wobei \(\frac{\dot{y}}{\dot{x}} = \frac{dy}{dt}/\frac{dx}{dt} = \frac{dy}{dx}\) entspricht. +Um die Ableitung nach der Zeit wegzubringen, wird beidseitig mit \(\dot{x}\) dividiert, wobei \(\frac{\dot{y}}{\dot{x}} = \frac{dy}{dt}/\frac{dx}{dt} = \frac{dy}{dx}\) entspricht. \[ x \frac{\dot{y}}{\dot{x}} + (t-y) \frac{\dot{x}}{\dot{x}} = 0 \] -Nach dem kürzen ergibt sich folgende DGL: +Nach dem Kürzen und Vereinfachen ergibt sich folgende DGL: \begin{equation} x y^{\prime} + t - y = 0 @@ -146,21 +144,109 @@ Diese kann mit den selben Methoden gelöst werden, diesmal in Kombination mit de &= \int \frac{1}{2} (e^{ln(x)+C} - e^{-(ln(x)+C)}) \\ &= - C_1 + C_2 x^2 - C_3 ln(x) + \frac{e^C}{4} x^2 - \frac{ln(x)}{2 \cdot e^C} + C_1 \\ + &= + C_1 + C_2 x^2 - \frac{ln(x)}{8 \cdot C_2} \end{align*} -Das Resultat wie ersichtlich ist folgende Funktion welche mittels Anfangsbedingungen parametrisiert werden kann: + +\begin{figure} + \centering + \includegraphics{papers/lambertw/Bilder/VerfolgungskurveBsp.png} + \caption[Graph der Verfolgungskurve]{Graph der Verfolgungskurve wobei, ({\color{red}rot}) die Funktion \ensuremath{y(x)} ist, ({\color{darkgreen}grün}) der quadratische Teil und ({\color{blue}blau}) dem \ensuremath{ln(x)}-Teil entspricht. + \label{lambertw:BildFunkLoes} + } +\end{figure} + +Das Resultat, wie ersichtlich, ist folgende Funktion \eqref{lambertw:funkLoes} welche mittels Anfangsbedingungen parametrisiert werden kann: \begin{equation} - y(x) + {\color{red}{y(x)}} = - C_1 + C_2 x^2 - C_3 ln(x) + C_1 + C_2 {\color{darkgreen}{x^2}} {\color{blue}{-}} \frac{\color{blue}{ln(x)}}{8 \cdot C_2} \label{lambertw:funkLoes} \end{equation} -Für die Koeffizienten \(C_1, C_2\) und \(C_3\) ergibt sich ein Anfangswertproblem, welches für deren Bestimmung gelöst werden muss. Zuerst soll aber eine qualitative Intuition, oder Idee für das Aussehen der Funktion \(\bf{y(x)}\) geschaffen werden: +Für die Koeffizienten \(C_1\) und \(C_2\) ergibt sich ein Anfangswertproblem, welches für deren Bestimmung gelöst werden muss. Zuerst soll aber eine qualitative Intuition, oder Idee für das Aussehen der Funktion \(\bf{y(x)}\) geschaffen werden: \begin{itemize} \item Für grosse \(x\)-Werte welche in der Regel in der Nähe von \(x_0\) sein sollten, ist der quadratisch Term in der Funktion dominant und somit für immer kleiner werdende \(x\) geht der Verfolger in Richtung \(y\)-Achse wobei seine Steigung stetig sinkt, was Sinn macht wenn der Verfolgte entlang der \(y\)-Achse steigt. \item Für \(x\)-Werte in der Nähe von \(0\) ist das asymptotische Verhalten des Logarithmus dominant, dies macht auch Sinn da sich der Verfolgte auf der \(y\)-Achse bewegt und der Verfolger im nachgeht. \item - Aufgrund des Monotoniewechsels in der Kurve muss die Kurve auch ein Minimum aufweisen. Es stellt sich nun die Frage: Wo befindet sich dieser Punkt? Durch eine logische Überlegung kann eine Abschätzung darüber getroffen werden und zwar, dass dieser dann entsteht, wenn \(A\) und \(P\) die gleiche \(y\)-Koordinaten besitzen. In diesem Moment ändert die Richtung der \(y\)-Komponente der Geschwindigkeit und somit auch sein Vorzeichen. + Aufgrund des Monotoniewechsels in der Kurve muss es auch ein Minimum aufweisen. Es stellt sich nun die Frage: Wo befindet sich dieser Punkt? Durch eine logische Überlegung kann eine Abschätzung darüber getroffen werden und zwar, dass dieser dann entsteht, wenn \(A\) und \(P\) die gleiche \(y\)-Koordinaten besitzen. In diesem Moment ändert die Richtung der \(y\)-Komponente der Geschwindigkeit und somit auch sein Vorzeichen. \end{itemize} +Alle diese Eigenschafte stimmen mit dem überein, was man von einer Kurve dieser Art erwarten würde, siehe \ref{lambertw:BildFunkLoes}. Nun stellt sich die Frage wie die Kurve wirklich aussieht, dies wird durch das Einsetzen folgender Anfangsbedingungen erreicht: +\begin{equation} + y(x)\big \vert_{t=0} + = + y(x_0) + = + y_0 + \:;\: + \frac{dy}{dx}\bigg \vert_{t=0} + = + y^{\prime}(x_0) + = + \frac{y_0}{x_0} +\end{equation} +Leitet man die Funktion \eqref{lambertw:funkLoes} nach x ab und setzt die Anfangsbedingungen ein, dann ergibt sich folgendes Gleichungssystem: +\begin{subequations} + \begin{align} + y_0 + &= + C_1 + C_2 x^2_0 - \frac{ln(x_0)}{8 \cdot C_2} \\ + \frac{y_0}{x_0} + &= + 2 \cdot C_2 x_0 - \frac{ln(x_0)}{8 \cdot C_2} + \end{align} +\end{subequations} +... Mit folgenden Formeln geht es weiter: +\begin{align*} + \eta + &= + \left(\frac{x}{x_0}\right)^2 + \:;\: + r_0 + = + \sqrt{x_0^2+y_0^2} \\ + y + &= + \frac{1}{4}\left(\left(y_0+r_0\right)\eta+\left(r_0-y_0\right)ln\left(\eta\right)-r_0+3y_0\right) \\ + y^\prime + &= + \frac{1}{2}\left(\left(y_0+r_0\right)\frac{x}{x_0^2}+\left(r_0-y_0\right)\frac{1}{x}\right) \\ + -4t + &= + \left(y_0+r_0\right)\left(\eta-1\right)+\left(r_0-y_0\right)ln\left(\eta\right) \\ + -4t+\left(y_0+r_0\right) + &= + \left(y_0+r_0\right)\eta+\left(r_0-y_0\right)ln\left(\eta\right) \\ + e^{-4t+\left(y_0+r_0\right)} + &= + e^{\left(y_0+r_0\right)\eta}\cdot\eta^{\left(r_0-y_0\right)} \\ + e^{\frac{-4t}{r_0-y_0}+\frac{y_0+r_0}{r_0-y_0}} + &= + e^{\frac{y_0+r_0}{r_0-y_0}\eta}\cdot\eta\ \\ + \chi + &= + \frac{y_0+r_0}{r_0-y_0}; \cdot\chi \\ + \chi\cdot e^{\chi-\frac{4t}{r_0-y_0}} + &= + \chi\eta\cdot e^{\chi\eta} \\ + W\left(\chi\cdot e^{\chi-\frac{4t}{r_0-y_0}}\right) + &= + \chi\eta \\ + \frac{W\left(\chi\cdot e^{\chi-\frac{4t}{r_0-y_0}}\right)}{\chi} + &= + \eta \\ + \frac{W\left(\chi\cdot e^{\chi-\frac{4t}{r_0-y_0}}\right)}{\chi} + &= + \left(\frac{x}{x_0}\right)^2 \\ + x\left(t\right) + &= + \sqrt{\frac{W\left(\chi\cdot e^{\chi-\frac{4t}{r_0-y_0}}\right)}{\chi}} +\end{align*} +\begin{equation} + y(t) + = + \frac{1}{4}\left(\left(y_0+r_0\right)\frac{W\left(\chi\cdot e^{\chi\ -\ \frac{4t}{r_0-y_0}}\right)}{\chi}+\left(r_0-y_0\right)\cdot\mathrm{ln}\ \left(\frac{W\left(\chi\cdot e^{\chi\ -\ \frac{4t}{r_0-y_0}}\right)}{\chi}\right)-r_0+3y_0\right) + \label{lambertw:funkNachT} +\end{equation} |
