week2.tex

\pagebreak
\section*{Tydzień 2}
Portrety fazowe systemów liniowych\\
\\
Ciągły system dynamiczny jest asymptotycznie stabilny, gdy części rzeczywiste jego wartości własnych są  ujemne.\\
Ciągły system dynamiczny jest stabilny, gdy części rzeczywiste jego wartości własnych są  niedodatnie oraz klatki Jordana macierzy $J$ odpowiadające wartością własnym macierzy $A$ położonym na osi urojonej mają wymiary $1\times 1$.
\subsection*{Portrety fazowe}
\textbf{1. Wyznaczyć wielomian charakterystyczny macierzy A i wartości własne}\\
\textbf{2. Wyznaczyć wektory własne macierzy A i narysować je w układzie współrzędnych}\\
Jeśli wektor własny jest związany z $\lambda<0$ to wzdłuż tego wektora trajektorie schodzą do zera.\\
Jeśli z $\lambda>0$, uciekają w nieskończoność\\
\textbf{3. Sprawdzić kierunek trajektorii}\\
Wybieramy punkt np $(1,0)$. Mnożymy macierz A przez ten punkt. Otrzymujemy wektor, który wskazuje kierunek z tego punktu.
\begin{figure}[!h]
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	\draw[color=red](-3,-3)--(3,3);
	\draw[color=red](-3,3)--(3,-3);

	\draw[color=red][->](0,0)--(2,2);
	\draw[color=red][->](0,0)--(-2,-2);
	\draw[color=red][->](-3,3)--(-2,2);
	\draw[color=red][->](3,-3)--(2,-2);

	\draw[very thick][->](0,0)--(1,1);
	\draw[very thick][->](0,0)--(1,-1);

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	\node[color=red] at(2.2,-.5){$\lambda=-1$};

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\end{figure}
\\
\textbf{Klatki Jordana}\\
$\left[\begin{array}{cc}a&0\\0&b\end{array}\right], \lambda=a,b\ \ \ \ \left[\begin{array}{cc}a&1\\0&a\end{array}\right], \lambda=a \ \ \ \ \left[\begin{array}{cc}a&b\\-b&a\end{array}\right], \lambda=a \pm bi$\\
\\
\\
$\left[\begin{array}{cc}-1&0\\0&-1\end{array}\right]$ gwiazda, $\left[\begin{array}{cc}-1&0\\0&-2\end{array}\right]$ węzeł, $\left[\begin{array}{cc}-1&0\\0&0\end{array}\right]$ poziome, $\left[\begin{array}{cc}-1&1\\0&-1\end{array}\right]$ węzeł zdegenerowany, $\left[\begin{array}{cc}-1&0\\0&1\end{array}\right]$ siodło\\\\
$\left[\begin{array}{cc}-1&1\\-1&-1\end{array}\right]$ ognisko, $\left[\begin{array}{cc}0&1\\-1&0\end{array}\right]$ kółka\\
\\
\textbf{Frobenius}\\
\fbox{\parbox{.5\linewidth}{
$Frobenius_{2 \times 2}=\left[\begin{array}{cc}0&1\\-c_{m-2}&-c_{m-1}\end{array}\right]$\\
$m=2$\\\\
$A=\left[\begin{array}{cc}0&1\\1&0\end{array}\right] =\left[\begin{array}{cc}0&1\\-c_0&-c_1\end{array}\right]$\\
$-c_0=1, \ \ -c_1 = 0, \boxed{\text{zawsze } c_2=1 }$\\
i z tego mamy wielomian charakterystyczny:\\
$c_2\lambda^2+c_1\lambda+c_0=0$\\
tutaj : $\lambda^2-1=0$
}}
\pagebreak\\
\textbf{Dla jakich parametrów system będzie asymptotycznie stabilny}\\
1. Sprawdzamy czy macierz jest w postaci Frobeniusa\\
2. Jeśli nie wyliczamy wielomian charakterystyczny ($|A-\lambda I|=0$)\\
3. Na podstawie charakterystycznego tworzymy macierz Hurwitza\\
4. Jeżeli czy wszystkie minory są większe od 0 to system jest asymptotycznie stabilny\\
5. Jeżeli jeden minor jest równy 0 to system jest stabilny, przeciwnie niestabilny\\
\\
\textbf{Zbadaj charakter pracy układu w zależności od parametrów}\\
1. Podstawiamy $u(t)$ do pierwszego równania\\
2. Grupujemy współczynniki\\
3. Tworzymy wielomian charakterystyczny\\
4. Tworzymy macierz Hurwitza\\
5. Sprawdzamy czy minory są większe od 0\\
6. Rysujemy wykres i zaznaczamy obszary stabilności. Wewnątrz obszaru system jest \textbf{asymptotycznie stabilny}, na prostych granicznych jest \textbf{stabilny}, na przecięciach i w pozostałych obszarach jest \textbf{niestabilny}\\
7. Delta wielomianu charakterystycznego <0\\
8. Obliczamy nierówność kwadratową w zależności od $k_2$\\
9. Rysujemy tą funkcję\\
10. Powyżej funkcji będą występować oscylacje, poniżej zanikanie wykładnicze\\
\\
\textbf{Liczenie $e^{At}$}\\
1. równanie charakterystyczne\\
2. Wektory własne\\
3. Macierze $P$, $P^{-1}$, $J$\\
4. $e^{Jt}=e^\lambda \cdot J$\\
5. $e^{At}=P\cdot e^{Jt}\cdot P^{-1}$\\

\pagebreak
%###################                 2.1.1              #################################%
\subsection*{Zadanie 2.1.1} {\color{darkgray}
	Naszkicować portrety fazowe systemów dynamicznych\\
	$\dot{x}(t)=\left[\begin{array}{cc}0&1\\-1&0\end{array}\right]x(t)\ \ \ $ i $\ \ \ \dot{x}(t)=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]x(t)$\\
	i opisać czym się różnią.\\
}\lineh
\\\\
\begin{multicols}{2}\noindent
$\lambda^2+1=0$\\
$\lambda = \pm i$\\
$J=A$\\
$\left[\begin{array}{cc}-i&1\\-1&-i\end{array}\right]\left[\begin{array}{c}\omega_1\\ \omega_2\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right]$\\
$-i\omega_1+\omega_2 =0$\\
$-\omega_1-i\omega_2=0$\\
$\omega_1=-i\omega_2$
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
$\lambda^2=1$\\
$\lambda = \pm 1$\\
$J=\left[\begin{array}{cc}\lambda_1&0\\0&\lambda_2\end{array}\right]=\left[\begin{array}{cc}1&0\\0&-1\end{array}\right] \neq A$\\
$\boxed{\lambda=1}$\\
$\left[\begin{array}{cc}-1&1\\1&-1\end{array}\right]\left[\begin{array}{c}\omega_1\\ \omega_2\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right]$\\
$ \begin{cases} -\omega_1+\omega_2 =0\\ \omega_1-\omega_2=0\\\end{cases}\Rightarrow \omega_1=\omega_2$\\
$\boxed{\lambda=-1}$\\
$\left[\begin{array}{cc}1&1\\1&1\end{array}\right]\left[\begin{array}{c}\omega_1\\ \omega_2\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right]$\\
$ \omega_1+\omega_2 = 0\Rightarrow \omega_1=-\omega_2$\\
\\
wektory własne:\\
$\left[\begin{array}{c}1\\1\end{array}\right] \wedge \left[\begin{array}{c}-1\\1\end{array}\right] \ $ wyznaczają osie\\\\
kierunek strzałek:\\
$\left[\begin{array}{cc}0&1\\1&0\end{array}\right]\left[\begin{array}{c}1\\1\end{array}\right]=\left[\begin{array}{c}1\\1\end{array}\right]$ taki sam wektor więc strzałki $+\infty \ \ -\infty$\\\\
$\left[\begin{array}{cc}0&1\\1&0\end{array}\right]\left[\begin{array}{c}-1\\1\end{array}\right]=\left[\begin{array}{c}1\\-1\end{array}\right]$ inny więc strzałki do 0\\

\end{multicols}
\begin{figure}[!h]
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\hspace*{3cm}
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\draw [color=blue](3.0,2.0)--(2.93,1.9)--(2.87,1.8)--(2.81,1.7)--(2.75,1.6)--(2.68,1.49)--(2.62,1.37)--(2.56,1.25)--(2.5,1.11)--(2.43,0.97)--(2.37,0.8)--(2.31,0.58)--(2.25,0.25)--(2.25,-0.25)--(2.31,-0.58)--(2.37,-0.8)--(2.43,-0.97)--(2.5,-1.11)--(2.56,-1.25)--(2.62,-1.37)--(2.68,-1.49)--(2.75,-1.6)--(2.81,-1.7)--(2.87,-1.8)--(2.93,-1.9)--(3.0,-2.0);
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\draw [color=blue](-3.0,1.41)--(-2.93,1.27)--(-2.87,1.12)--(-2.81,0.95)--(-2.75,0.75)--(-2.68,0.47)--(-2.68,-0.47)--(-2.75,-0.75)--(-2.81,-0.95)--(-2.87,-1.12)--(-2.93,-1.27)--(-3.0,-1.41);
\draw [color=blue](3.0,1.41)--(2.93,1.27)--(2.87,1.12)--(2.81,0.95)--(2.75,0.75)--(2.68,0.47)--(2.68,-0.47)--(2.75,-0.75)--(2.81,-0.95)--(2.87,-1.12)--(2.93,-1.27)--(3.0,-1.41);
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	\draw[color=blue,very thick][<-](-.7,-1.2)--(-.5,-1.1);
	\draw[color=blue,very thick][<-](.7,1.2)--(.5,1.1);
	\draw[color=blue,very thick][<-](1.2,.7)--(1.1,.5);
	\draw[color=blue,very thick][<-](-1.2,-.7)--(-1.1,-.5);

	\draw[dashed](-3,-3)--(3,3);
	\draw[dashed](-3,3)--(3,-3);

	\draw[very thick][->](-3,0)--(3,0) node[right=.2] {$x_1$};
	\draw[very thick][->](0,-2.625)--(0,2.75) node[above=.2] {$x_2$};
\end{tikzpicture}
\end{figure}
Pierwszy portret fazowy to środek, a drugi to siodło.

%###################                 2.2.1              #################################%
\pagebreak
\subsection*{Zadanie 2.2.1} {\color{darkgray}
	Naszkicować portrety fazowe systemów dynamicznych\\
	$\begin{array}{l}\dot{x}_1(t)=x_2(t)\\\dot{x}_2(t)=-x_1(t)\end{array}$ i $\begin{array}{l}\dot{x}_1(t)=10x_2(t)\\\dot{x}_2(t)=-10x_1(t)\end{array}$\\
	i opisać czym się różnią.\\
}\lineh
\\\\
\begin{multicols}{2}\noindent
$\begin{cases}\dot{x}_1(t)=x_2(t) \\ \dot{x}_2(t)=-x_1(t)\end{cases}$\\
$\dot{x}(t)=\left[\begin{array}{cc}0&1\\-1&0\end{array}\right]x(t)$\\
$J=A$
\\
$\begin{cases}\dot{x}_1(t)=10x_2(t) \\ \dot{x}_2(t)=-10x_1(t)\end{cases}$\\
$\dot{x}(t)=\left[\begin{array}{cc}0&10\\-10&0\end{array}\right]x(t)$\\
$J=A$\\
\end{multicols}

\begin{figure}[!h]
\begin{tikzpicture}
	\draw[color=blue] (0,0) circle(.5);
	\draw[color=blue] (0,0) circle(1);
	\draw[color=blue] (0,0) circle(1.5);
	\draw[color=blue] (0,0) circle(2);
	\draw[color=blue] (0,0) circle(2.5);

	\draw[color=blue,very thick][<-](-.3,-.45)--(-.1,-.5);
	\draw[color=blue,very thick][<-](-.3,-.95)--(-.1,-1);
	\draw[color=blue,very thick][<-](-.3,-1.45)--(-.1,-1.5);
	\draw[color=blue,very thick][<-](-.3,-1.95)--(-.1,-2);
	\draw[color=blue,very thick][<-](-.3,-2.45)--(-.1,-2.5);

	\draw[very thick][->](-3,0)--(3,0) node[right=.2] {$x_1$};
	\draw[very thick][->](0,-2.625)--(0,2.75) node[above=.2] {$x_2$};
\end{tikzpicture}
\hspace*{3cm}
\begin{tikzpicture}
	\draw[color=blue] (0,0) circle(.5);
	\draw[color=blue] (0,0) circle(1);
	\draw[color=blue] (0,0) circle(1.5);
	\draw[color=blue] (0,0) circle(2);
	\draw[color=blue] (0,0) circle(2.5);

	\draw[color=blue,very thick][<-](-.3,-.45)--(-.1,-.5);
	\draw[color=blue,very thick][<-](-.3,-.95)--(-.1,-1);
	\draw[color=blue,very thick][<-](-.3,-1.45)--(-.1,-1.5);
	\draw[color=blue,very thick][<-](-.3,-1.95)--(-.1,-2);
	\draw[color=blue,very thick][<-](-.3,-2.45)--(-.1,-2.5);

	\draw[very thick][->](-3,0)--(3,0) node[right=.2] {$x_1$};
	\draw[very thick][->](0,-2.625)--(0,2.75) node[above=.2] {$x_2$};
\end{tikzpicture}
\end{figure}
Portrety są identyczne. Jedyną różnicą jest szybkość poruszania się trajektorii w dziedzinie czasu. Dla 1 mamy $t$, a dla 2 $10t$



%###################                 2.3.1              #################################%
\pagebreak
\subsection*{Zadanie 2.3.1} {\color{darkgray}
	Podać wartości własne, jakie mogą odpowiadać poniższemu portretowi fazowemu.
\begin{figure}[!h]
\begin{tikzpicture}
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\draw [color=blue](0.37,2.66)--(0.5,2.0)--(0.62,1.6)--(0.75,1.33)--(0.87,1.14)--(1.0,1.0)--(1.12,0.88)--(1.25,0.8)--(1.37,0.72)--(1.5,0.66)--(1.62,0.61)--(1.75,0.57)--(1.87,0.53)--(2.0,0.5)--(2.12,0.47)--(2.25,0.44)--(2.37,0.42)--(2.5,0.4)--(2.62,0.38)--(2.75,0.36)--(2.87,0.34)--(3.0,0.33);
\draw [color=blue](-3.0,-0.33)--(-2.87,-0.34)--(-2.75,-0.36)--(-2.62,-0.38)--(-2.5,-0.4)--(-2.37,-0.42)--(-2.25,-0.44)--(-2.12,-0.47)--(-2.0,-0.5)--(-1.87,-0.53)--(-1.75,-0.57)--(-1.62,-0.61)--(-1.5,-0.66)--(-1.37,-0.72)--(-1.25,-0.8)--(-1.12,-0.88)--(-1.0,-1.0)--(-0.87,-1.14)--(-0.75,-1.33)--(-0.62,-1.6)--(-0.5,-2.0)--(-0.37,-2.66);
\draw [color=blue](0.37,-2.66)--(0.5,-2.0)--(0.62,-1.6)--(0.75,-1.33)--(0.87,-1.14)--(1.0,-1.0)--(1.12,-0.88)--(1.25,-0.8)--(1.37,-0.72)--(1.5,-0.66)--(1.62,-0.61)--(1.75,-0.57)--(1.87,-0.53)--(2.0,-0.5)--(2.12,-0.47)--(2.25,-0.44)--(2.37,-0.42)--(2.5,-0.4)--(2.62,-0.38)--(2.75,-0.36)--(2.87,-0.34)--(3.0,-0.33);
\draw [color=blue](-3.0,0.33)--(-2.87,0.34)--(-2.75,0.36)--(-2.62,0.38)--(-2.5,0.4)--(-2.37,0.42)--(-2.25,0.44)--(-2.12,0.47)--(-2.0,0.5)--(-1.87,0.53)--(-1.75,0.57)--(-1.62,0.61)--(-1.5,0.66)--(-1.37,0.72)--(-1.25,0.8)--(-1.12,0.88)--(-1.0,1.0)--(-0.87,1.14)--(-0.75,1.33)--(-0.62,1.6)--(-0.5,2.0)--(-0.37,2.66);

\draw [color=blue](0.75,2.66)--(0.87,2.28)--(1.0,2.0)--(1.12,1.77)--(1.25,1.6)--(1.37,1.45)--(1.5,1.33)--(1.62,1.23)--(1.75,1.14)--(1.87,1.06)--(2.0,1.0)--(2.12,0.94)--(2.25,0.88)--(2.37,0.84)--(2.5,0.8)--(2.62,0.76)--(2.75,0.72)--(2.87,0.69)--(3.0,0.66);
\draw [color=blue](-3.0,-0.66)--(-2.87,-0.69)--(-2.75,-0.72)--(-2.62,-0.76)--(-2.5,-0.8)--(-2.37,-0.84)--(-2.25,-0.88)--(-2.12,-0.94)--(-2.0,-1.0)--(-1.87,-1.06)--(-1.75,-1.14)--(-1.62,-1.23)--(-1.5,-1.33)--(-1.37,-1.45)--(-1.25,-1.6)--(-1.12,-1.77)--(-1.0,-2.0)--(-0.87,-2.28)--(-0.75,-2.66);
\draw [color=blue](0.75,-2.66)--(0.87,-2.28)--(1.0,-2.0)--(1.12,-1.77)--(1.25,-1.6)--(1.37,-1.45)--(1.5,-1.33)--(1.62,-1.23)--(1.75,-1.14)--(1.87,-1.06)--(2.0,-1.0)--(2.12,-0.94)--(2.25,-0.88)--(2.37,-0.84)--(2.5,-0.8)--(2.62,-0.76)--(2.75,-0.72)--(2.87,-0.69)--(3.0,-0.66);
\draw [color=blue](-3.0,0.66)--(-2.87,0.69)--(-2.75,0.72)--(-2.62,0.76)--(-2.5,0.8)--(-2.37,0.84)--(-2.25,0.88)--(-2.12,0.94)--(-2.0,1.0)--(-1.87,1.06)--(-1.75,1.14)--(-1.62,1.23)--(-1.5,1.33)--(-1.37,1.45)--(-1.25,1.6)--(-1.12,1.77)--(-1.0,2.0)--(-0.87,2.28)--(-0.75,2.66);

\draw [color=blue](1.12,2.66)--(1.25,2.4)--(1.37,2.18)--(1.5,2.0)--(1.62,1.84)--(1.75,1.71)--(1.87,1.6)--(2.0,1.5)--(2.12,1.41)--(2.25,1.33)--(2.37,1.26)--(2.5,1.2)--(2.62,1.14)--(2.75,1.09)--(2.87,1.04)--(3.0,1.0);
\draw [color=blue](-3.0,-1.0)--(-2.87,-1.04)--(-2.75,-1.09)--(-2.62,-1.14)--(-2.5,-1.2)--(-2.37,-1.26)--(-2.25,-1.33)--(-2.12,-1.41)--(-2.0,-1.5)--(-1.87,-1.6)--(-1.75,-1.71)--(-1.62,-1.84)--(-1.5,-2.0)--(-1.37,-2.18)--(-1.25,-2.4)--(-1.12,-2.66);
\draw [color=blue](1.12,-2.66)--(1.25,-2.4)--(1.37,-2.18)--(1.5,-2.0)--(1.62,-1.84)--(1.75,-1.71)--(1.87,-1.6)--(2.0,-1.5)--(2.12,-1.41)--(2.25,-1.33)--(2.37,-1.26)--(2.5,-1.2)--(2.62,-1.14)--(2.75,-1.09)--(2.87,-1.04)--(3.0,-1.0);
\draw [color=blue](-3.0,1.0)--(-2.87,1.04)--(-2.75,1.09)--(-2.62,1.14)--(-2.5,1.2)--(-2.37,1.26)--(-2.25,1.33)--(-2.12,1.41)--(-2.0,1.5)--(-1.87,1.6)--(-1.75,1.71)--(-1.62,1.84)--(-1.5,2.0)--(-1.37,2.18)--(-1.25,2.4)--(-1.12,2.66);

\draw [color=blue](1.87,2.66)--(2.0,2.5)--(2.12,2.35)--(2.25,2.22)--(2.37,2.1)--(2.5,2.0)--(2.62,1.9)--(2.75,1.81)--(2.87,1.73)--(3.0,1.66);
\draw [color=blue](-3.0,-1.66)--(-2.87,-1.73)--(-2.75,-1.81)--(-2.62,-1.9)--(-2.5,-2.0)--(-2.37,-2.1)--(-2.25,-2.22)--(-2.12,-2.35)--(-2.0,-2.5)--(-1.87,-2.66);
\draw [color=blue](1.87,-2.66)--(2.0,-2.5)--(2.12,-2.35)--(2.25,-2.22)--(2.37,-2.1)--(2.5,-2.0)--(2.62,-1.9)--(2.75,-1.81)--(2.87,-1.73)--(3.0,-1.66);
\draw [color=blue](-3.0,1.66)--(-2.87,1.73)--(-2.75,1.81)--(-2.62,1.9)--(-2.5,2.0)--(-2.37,2.1)--(-2.25,2.22)--(-2.12,2.35)--(-2.0,2.5)--(-1.87,2.66);

	\draw[color=blue,very thick][<-](-1.41,-1.41)--(-1.42,-1.4);
	\draw[color=blue,very thick][<-](-1.41,1.41)--(-1.42,1.4);
	\draw[color=blue,very thick][<-](1.41,1.41)--(1.42,1.4);
	\draw[color=blue,very thick][<-](1.41,-1.41)--(1.42,-1.4);


	\draw[very thick][->](-3,0)--(3,0)node[right=.2] {$x_1$};
	\draw[very thick][->](0,-2.625)--(0,2.75)node[above=.2] {$x_2$};
\end{tikzpicture}
\end{figure}
\\
}\lineh
\\\\
Dla siodła: dwie wartości własne rzeczywiste przeciwnych znaków np 1 i -1\\
$\left[\begin{array}{cc}-1&0\\0&1\end{array}\right]$


%###################                 2.4.1              #################################%
\pagebreak
\subsection*{Zadanie 2.4.1} {\color{darkgray}
	Dla systemu\\
	$\begin{array}{rcl}x(t)+4\ddot{x}(t)+\dot{x}(t)&=&u(t) \\ u(t)&=&k_1\dot{x}(t)-k_2x(t)\end{array}$\\
	zbadać zachowanie się ukłądu w zależności od $k_1$ i $k_2$. Zaznaczyć odpowiednie obszary na płaszczyźnie $k_1 \times k_2$\\
}\lineh
\\\\
$x+4\ddot{x}+\dot{x}=k_1\dot{x}-k_2x$\\
$4\ddot{x}+(1-k_1)\dot{x}+(1+k_2)x=0$\\
wielomian charakterystyczny:\\
$x=e^{\lambda t} \ \ \ \dot{x}=\lambda e^{\lambda t} \ \ \ \ddot{x}=\lambda^2 e^{\lambda t}$\\
$4\lambda^2+(1-k_1)\lambda+1+k_2=0$\\
macierz Hurwitza dla wielomianu stopnia drugiego:
$a_0x^2+a_1x+a_2=0$\\
$\left[\begin{array}{cc}a_1&0\\a_0&a_2\end{array}\right] \ \ $ czyli $\ \ 
\left[\begin{array}{cc}1-k_1&0\\4&1+k_2\end{array}\right]$\\
żeby układ był stabilny to $|a_1|>0$ i $\left|\begin{array}{cc}a_1&0\\a_0&a_2\end{array}\right|>0$\\
więc:\\
$1-k_1>0 \Rightarrow\boxed{ k_1<1}$\\
$(1-k_1)(1+k_2)>0 \Rightarrow 1+k_2>0 \Rightarrow \boxed{k_2>-1}$\\
dla $\Delta<0$ występują oscylacje, więc:\\
$\Delta=(1-k_1)^2-4\cdot4(1-k_2)=(1-k_1)^2-16-16k_2<0$\\
$k_2>\frac{1}{16}(1-k_1)^2-1$\\
\begin{figure}[!h]
\begin{tikzpicture}
\fill[fill=blue, opacity=.2](-6,3)--(1,3)--(1,-1)--(-6,-1);
\draw [pattern=my north east lines, line space=8pt, draw=green!50!black](-6,3)--(-6.0,2.06)--(-5.75,1.84)--(-5.5,1.64)--(-5.25,1.44)--(-5.0,1.25)--(-4.75,1.06)--(-4.5,0.89)--(-4.25,0.72)--(-4.0,0.56)--(-3.75,0.41)--(-3.5,0.26)--(-3.25,0.12)--(-3.0,0.0)--(-2.75,-0.12)--(-2.5,-0.23)--(-2.25,-0.33)--(-2.0,-0.43)--(-1.75,-0.52)--(-1.5,-0.6)--(-1.25,-0.68)--(-1.0,-0.75)--(-0.75,-0.8)--(-0.5,-0.85)--(-0.25,-0.9)--(0.0,-0.93)--(0.25,-0.96)--(0.5,-0.98)--(0.75,-0.99)--(1.0,-1.0)--(1.25,-0.99)--(1.5,-0.98)--(1.75,-0.96)--(2.0,-0.93)--(2.25,-0.9)--(2.5,-0.85)--(2.75,-0.8)--(3.0,-0.75)--(3.25,-0.68)--(3.5,-0.6)--(3.75,-0.52)--(4.0,-0.43)--(4.25,-0.33)--(4.5,-0.23)--(4.75,-0.12)--(5.0,0.0)--(5.25,0.12)--(5.5,0.26)--(5.75,0.41)--(6.0,0.56)--(6,3);
\draw [color=blue](-6.0,2.06)--(-5.75,1.84)--(-5.5,1.64)--(-5.25,1.44)--(-5.0,1.25)--(-4.75,1.06)--(-4.5,0.89)--(-4.25,0.72)--(-4.0,0.56)--(-3.75,0.41)--(-3.5,0.26)--(-3.25,0.12)--(-3.0,0.0)--(-2.75,-0.12)--(-2.5,-0.23)--(-2.25,-0.33)--(-2.0,-0.43)--(-1.75,-0.52)--(-1.5,-0.6)--(-1.25,-0.68)--(-1.0,-0.75)--(-0.75,-0.8)--(-0.5,-0.85)--(-0.25,-0.9)--(0.0,-0.93)--(0.25,-0.96)--(0.5,-0.98)--(0.75,-0.99)--(1.0,-1.0)--(1.25,-0.99)--(1.5,-0.98)--(1.75,-0.96)--(2.0,-0.93)--(2.25,-0.9)--(2.5,-0.85)--(2.75,-0.8)--(3.0,-0.75)--(3.25,-0.68)--(3.5,-0.6)--(3.75,-0.52)--(4.0,-0.43)--(4.25,-0.33)--(4.5,-0.23)--(4.75,-0.12)--(5.0,0.0)--(5.25,0.12)--(5.5,0.26)--(5.75,0.41)--(6.0,0.56);



	\draw[very thick][->](-6,0)--(6,0)node[right=.2] {$k_1$};
	\draw[very thick][->](0,-3)--(0,3)node[above=.2] {$k_2$};

	\draw (-0.1,-1) -- (0.1,-1) node [left=3pt]{{-1}};
	\draw (5,-0.1) -- (5,0.1) node [below=4pt]{{5}};
	\draw (-3,-0.1) -- (-3,0.1) node [below=4pt]{{-3}};
	\draw (1,-0.1) -- (1,0.1) ;
	\node at (1.2,-0.2) {1};

	\draw[dashed, color=red](-6,-1)--(6,-1);
	\draw[dashed, color=red](1,-3)--(1,3);
	\draw[thick](1,-1) circle(.15);
\end{tikzpicture}
\end{figure}
\\
wewnątrz niebieskiego obszaru asymptotycznie stabilny $k_1<1 \wedge k_2>-1)$\\
na czerwonych prostych granicznych stabilny $(k_1=1 \vee k_2=-1)$ bez punktu wspólnego\\
niestabilny na przecięciu prostych i w pozostałych obszarach\\
oscylacje dla zakreskowanego $k_2>\frac{1}{16}(1-k_1)^2-1$



%###################                 2.5.1              #################################%
\pagebreak
\subsection*{Zadanie 2.5.1} {\color{darkgray}
	Zbadać charakter pracy układu\\
	$\begin{array}{rcl}\ddot{x}(t)+\dot{x}(t)+x(t)&=&u(t) \\ u(t)&=&Kx(t)\end{array}$\\
	w zależności od parametru $K$. Zaznaczyć wszystkie istotne rodzaje zachowań na osi liczbowej.\\
}\lineh
\\\\
$\ddot{x}+\dot{x}+x=Kx$\\
$\ddot{x}+\dot{x}+x(1-K)=0$\\
$\lambda^2+\lambda+1-K=0$ wielomian charakterystyczny\\
$\left[\begin{array}{cc}1&0\\1&1-K\end{array}\right]$ macierz Hurwitza\\
$1-K>0 \Rightarrow K<1$\\
$\Delta=1-4(1-K)=-3+4K<0\Rightarrow K<\frac 34$\\
\begin{figure}[!h]
\begin{tikzpicture}
	\draw[thick][->](-2,0)--(3,0)node[right=.2] {$K$};

	\draw (0,-0.1) -- (0,0.1) node [below=4pt]{{0}};
	\draw (2,-0.1) -- (2,0.1) node [below=4pt]{{1}};
	\draw (1.5,-0.1) -- (1.5,0.1) node [below=4pt]{$\frac34$};

	\draw(-2,1)--(1.7,1)--(2,0);
	\node at (0,.75){asymptotycznie stabilny};
	\draw[->](2.5,-1)--(2.1,-0.1);
	\node at (3.1,-1.1){stabilny};
	\draw(4,1)--(2.3,1)--(2,0);
	\node at (3.2,.75){niestabilny};
	\draw(-2,.5)--(1.2,.5)--(1.5,0);
	\node at (0,.25){oscylacje};
	\draw(4,.5)--(1.8,.5)--(1.5,0);
	\node at (3.3,.25){brak oscylacji};

\end{tikzpicture}
\end{figure}
\\

%###################                 2.6.1              #################################%
\pagebreak
\subsection*{Zadanie 2.6.1} {\color{darkgray}
	Dla jakich wartości parametru $k$ system opisany równaniami:\\
	$\begin{array}{rcl}4\dot{x}_1&=&12x_1-0.25kx_2  \\  0.5\dot{x}_2&=&\frac 1kx_1+kx_2\end{array}$\\
	będzie niestabilny.\\
}\lineh
\\\\
$\begin{array}{rcl}\dot{x}_1&=&3x_1-\frac{1}{16}kx_2  \\  \dot{x}_2&=&\frac 2kx_1+2kx_2\end{array}$\\
$\dot{x}=\left[\begin{array}{cc}3&-\frac{1}{16}k\\\frac{2}{k}&2k\end{array}\right]x$\\
$\left|\begin{array}{cc}3-\lambda&-\frac{1}{16}k\\\frac{2}{k}&2k-\lambda\end{array}\right|=(3-\lambda)(2k-\lambda)+(\frac{1}{16}k)(\frac{2}{k})=\lambda^2-(3+2k)\lambda+6k+\frac{1}{8}$ wielomian charakterystyczny\\
$\left[\begin{array}{cc}-(3+2k) &0\\1&6k+\frac 18\end{array}\right]$ macierz Hurwitz'a\\
$-3-2k>0\Rightarrow k<-\frac 32$\\
$-(3+2k)(6k+\frac 18)>0 \Rightarrow 6k+\frac 18>0\Rightarrow k>-\frac 18$\\
stabilny dla $k<-\frac 32 \wedge k>-\frac 18 \Rightarrow k\in \varnothing$\\
niestabilny dla $k \in \mathbb{R}$




%###################                 2.7.1              #################################%
\pagebreak
\subsection*{Zadanie 2.7.1} {\color{darkgray}
	Wyznaczyć macierz $e^{At}$ dla macierzy\\
	$\left[\begin{array}{cc}-2&1\\-2&0\end{array}\right]$\\
}\lineh
\\\\
$e^{At}=P\cdot e^{Jt}\cdot P^{-1}$\\
$e^{Jt}=e^\lambda \cdot J$\\\\
$\left|\begin{array}{cc}-2-\lambda&1\\-2&-\lambda\end{array}\right|=\lambda(2+\lambda)+2=\lambda^2+2\lambda+2=0$\\
$\sqrt{\Delta}=2i$\\
$\lambda_1=\frac{-2+2i}{2}=-1+i \ \ \ \ \ \ \ \lambda_2=-1-i$\\
\\
$\boxed{ \lambda_1=-1+i}$\\
{\color{lightgray}
$\left[\begin{array}{cc}-2+1-i&1\\-2&1-i\end{array}\right]\left[\begin{array}{c}\omega_1\\ \omega_2\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right]$\\
$\begin{cases}-(1+i)\omega_i+\omega_2=0 \Rightarrow \omega_2=\omega_1+i\omega_1 \\-2\omega_1+(1-i)\omega_2=0\end{cases}$\\
$-2\omega_1+(1-i)(1+i)\omega_1=0$\\
$-2\omega_1+2\omega_1=0$\\
}
$\boxed{\begin{aligned}
\text{dla }\lambda=a \pm ib\\
J=\left[\begin{array}{cc}a&b\\-b&a\end{array}\right]\\
e^{tJ}=e^{a t}\left[\begin{array}{cc}\cos bt&\sin bt\\-\sin bt& \cos t\end{array}\right]
\end{aligned}}$\\\\
$J=\left[\begin{array}{cc}-1&1\\-1&-1\end{array}\right]$\\
$e^{tJ}=e^{-t}\left[\begin{array}{cc}\cos t & \sin t \\ -\sin t & \cos t\end{array}\right]$\\
$W=\left[\begin{array}{c}1\\1+i\end{array}\right]\omega=s\left[\begin{array}{c}1\\1\end{array}\right]+pi\left[\begin{array}{c}0\\1\end{array}\right]$\\
$P=\left[\begin{array}{cc}1&0\\1&1\end{array}\right]$\\
$P^{-1}=\left[\begin{array}{cc}1&0\\-1&1\end{array}\right]$\\\\
$\boxed{\left[\begin{array}{cc}a&b\\c&d\end{array}\right]^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d&-b\\-c&a\end{array}\right]}$\\\\\\
$e^{At}=\left[\begin{array}{cc}1&0\\1&1\end{array}\right]\cdot e^{-t}\cdot\left[\begin{array}{cc}\cos t&\sin t \\-\sin t &\cos t\end{array}\right]\cdot \left[\begin{array}{cc}1&0\\-1&1\end{array}\right]=e^{-t}\left[\begin{array}{cc}\cos t & \sin t \\ \cos t-\sin t & \sin t +\cos t\end{array}\right]\cdot \left[\begin{array}{cc}1&0\\-1 &1\end{array}\right]=$\\
$=e^{-t}\left[\begin{array}{cc}\cos t - \sin t & \sin t \\-2\sin t &\sin t +\cos t\end{array}\right]$\\


%###################                 2.8.1              #################################%
\pagebreak
\subsection*{Zadanie 2.8.1} {\color{darkgray}
	Wyznaczyć rozwiązanie $x(t), t\geqslant 0$ równania\\
	$\ddot{x}(t)+\dot{x}(t)+3x(t)=0$\\
	$x(0)=1, \ \ \dot{x}(0)=0$\\
}\lineh
\\\\
$x=e^{\lambda t} \ \ \ \dot{x}=\lambda e^{\lambda t} \ \ \ \ddot{x}=\lambda^2 e^{\lambda t}$\\
$\lambda^2e^{\lambda t}+\lambda e^{\lambda t}+3e^{\lambda t}=0$\\
$\lambda^2+\lambda+3=0$\\
$\Delta=1-12=-11<0$\\
$\lambda_1=\alpha+i\beta \ \ \ \ \ \alpha=\frac{-b}{2a} \ \ \ \ \ \alpha=\frac{-1}{2}$\\
$\lambda_2=\alpha-i\beta \ \ \ \ \ \beta=\frac{\sqrt{|\Delta|}}{2a} \ \ \ \ \ \beta=\frac{\sqrt{11}}{2}$\\
\\
$x(t)=Ae^{\alpha t}\cos \beta t+Be^{\alpha t}\sin \beta t$\\
$x(t)=Ae^{-\frac{t}{2}}\cos(\frac{\sqrt{11}}{2}t)+Be^{-\frac t2}\sin(\frac{\sqrt{11}}{2}t)$\\
$x(0)=A\cdot e^0\cdot \cos 0+B\cdot e^0 \cdot  \underbrace{\sin 0}_{=0}=\boxed{A=1}$\\
$\dot{x}(t)=-\beta Ae^{\alpha t}\sin \beta t+\alpha A e^{\alpha t}\cos \beta t+\beta Be^{\alpha t}\cos\beta t+\alpha B e^{\alpha t}\sin \beta t$\\
$\dot{x}(0)=\underbrace{-\frac{\sqrt{11}}{2}\cdot e^0\cdot\sin 0}_{=0}-\frac12\cdot e^0\cos 0+\frac{\sqrt{11}}{2}B\cdot e^0\cdot\cos 0 -\underbrace{\frac12Be^0\sin0}_{=0}=$\\
$=-\frac12+\frac{\sqrt{11}}{2}B=0\Rightarrow\boxed{B=\frac12\cdot\frac{2}{\sqrt{11}}=\frac{\sqrt{11}}{11}}$\\
$\boxed{x(t)=e^{-\frac t2}\cos(\frac{\sqrt{11}}{2}t)+\frac{\sqrt{11}}{11}e^{-\frac t2}\sin(\frac{\sqrt{11}}{2}t)}$\\


%###################                 2.9.1              #################################%
\pagebreak
\subsection*{Zadanie 2.9.1} {\color{darkgray}
	Na gładkim stole leży sznur o długości 0.3 m i masie 50g, przy czym część sznura zwisa ze stołu jak na rysunku. Zamodelować ruch sznura po stole za pomocą równania różniczkowego. Naszkicować portret fazowy systemu opisanego tym równaniem\\}
\begin{figure}[!h]
\begin{tikzpicture}
	
\draw[very thick,pattern=my north east lines, line space=5pt, draw=black] (0,0) -- (4,0) -- (4,-.5) -- (3.2,-.5)--(3.2,-2.5)--(2.7,-2.5)--(2.7,-.5)--(0,-.5);
\filldraw[fill=black] (1,0.05)--(1,.4)--(4,.4)--(4.2,.37)  --(4.3,.3)--(4.37,.2) --(4.4,0)--(4.4,-1.7)--(4.1,-1.7)--(4.1,0.05);

\draw[color=black] (4.1,0)--(5.2,0);
\draw[color=black] (4.1,-1.7)--(5.2,-1.7);
\draw[<->][color=black] (5,0)--(5,-1.7);
\draw[->][color=black] (4.25,-1.7)--(4.25,-2.5);

\node at(5.5,-.9){$x,m$};
\node at(4.3,-2.7){$Q=mg$};
\end{tikzpicture}
\end{figure}
\\
\lineh
\\\\
$l=0.3m \ \ \ m=50g$\\
$m=M\cdot\frac xl$\\
$M\cdot\ddot{x}=m\cdot g$\\
$\cancel{M}\cdot\ddot{x}=\cancel{M}\cdot\frac xl\cdot g$\\
$\ddot{x}=x\frac gl \ \ \ \ k=\frac gl$\\
$x_1=x$\\
$x_2=\dot{x}$\\
$\dot{x}_1=\dot{x}=x_2$\\
$\dot{x}_2=\ddot{x}=x_1k$\\
$\dot{x}=\left[\begin{array}{cc}0&1\\k&0\end{array}\right]x$\\
$\lambda^2-k=0$\\
$\lambda =\pm \sqrt{k}$\\
$J=\left[\begin{array}{cc}\sqrt{k}&0\\0&-\sqrt{k}\end{array}\right]$\\
\\
$\lambda=\sqrt{k}$\\
$\left[\begin{array}{cc}-\sqrt{k}&1\\k&-\sqrt{k}\end{array}\right]\left[\begin{array}{c}\omega_1\\\omega_2\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right]$\\
$-\sqrt{k}\omega_1+\omega_2=0$\\
$k\omega_1-\sqrt{k}\omega_2=0$\\
$\omega_2=\sqrt{k}\omega_1$\\
$\left[\begin{array}{c}1\\\sqrt{k}\end{array}\right]$\\
$\left[\begin{array}{cc}0&1\\k&0\end{array}\right]\left[\begin{array}{c}1\\\sqrt{k}\end{array}\right]=\left[\begin{array}{c}\sqrt{k}\\k\end{array}\right]$\\

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	\draw[color=blue,very thick][<-](-.7,-1.2)--(-.5,-1.1);
	\draw[color=blue,very thick][<-](.7,1.2)--(.5,1.1);
	\draw[color=blue,very thick][<-](1.2,.7)--(1.1,.5);
	\draw[color=blue,very thick][<-](-1.2,-.7)--(-1.1,-.5);

	\draw[dashed](-3,-3)--(3,3);
	\draw[dashed](-3,3)--(3,-3);

	\draw[very thick][->](-3,0)--(3,0) node[right=.2] {$x_1$};
	\draw[very thick][->](0,-2.625)--(0,2.75) node[above=.2] {$x_2$};
\end{tikzpicture}
\end{figure}



%###################                 2.10.1              #################################%
\pagebreak
\subsection*{Zadanie 2.10.1} {\color{darkgray}
	Dany jest system opisany równaniem\\
	$\dot{x}_1(t)=-\pi x_2(t)$\\
	$\dot{x}_2(t)=\pi x_1(t)$\\
	Naszkicować zbiór punktów powstałych z trajektorii stanu systemu w chwili $t=0.75s$ dla warunków początkowych branych ze zbioru $X=\{(x_1,x_2)\in \mathbb{R}^2:|x_1+x_2|=1\}$\\
}\lineh
\\\\
$\dot{x}=\left[\begin{array}{cc}0&-\pi\\\pi&0\end{array}\right]x$\\
$\left|\begin{array}{cc}-\lambda&-\pi\\\pi&-\lambda\end{array}\right|=\lambda^2+\pi^2=0\Rightarrow\lambda^2=-\pi^2 \ \ \ \ \ \ \ \lambda=\pm i\pi$\\
$J=\left[\begin{array}{cc}0&-\pi\\\pi&0\end{array}\right]$\\
$A = J$, $a=0, b=\pi$\\
$e^{tJ}=e^{t}\left[\begin{array}{cc}\cos(\pi t)&\sin(\pi t)\\-\sin(\pi t)&\cos(\pi t)\end{array}\right]$\\
$x(t)=e^{tJ}x(0)+\underbrace{\int_0^te^{(t-\tau)A}Bu(\tau)\ d\tau}_{=0 \text{, \ bo }u=0 \ \ B=0}$\\
$x(t)=e^{tJ}x(0)$\\\\
$x(t)=\left[\begin{array}{cc}\cos(\pi t)&\sin(\pi t)\\-\sin(\pi t)&\cos(\pi t)\end{array}\right]x(0)$\\
$t=\frac 34 s$\\
$x(\frac 34)=\left[\begin{array}{cc}-\frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}\\-\frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2}\end{array}\right]x(0)=-\frac{\sqrt{2}}{2}\left[\begin{array}{cc}1&-1\\1&1\end{array}\right]x(0)$\\
$\begin{cases}
x_1(\frac34)=-\frac{\sqrt{2}}{2}(x_1(0)-x_2(0))\\
x_2(\frac34)=-\frac{\sqrt{2}}{2}(x_1(0)+x_2(0))\\
|x_1+x_2|=1\Rightarrow \begin{array}{ccc}x_1+x_2=1&\vee&x_1+x_2=-1 \\ x_1=1-x_2&\vee& x_1=-1-x_2\end{array}
\end{cases}$\\
$\begin{cases}
x_1(\frac34)=-\frac{\sqrt{2}}{2}(-x_2(0)+1-x_2(0))=-\frac{\sqrt{2}}{2}-\sqrt{2}x_2(0)\\
x_2(\frac34)=-\frac{\sqrt{2}}{2}(-x_2(0)+1+x_2(0))=-\frac{\sqrt{2}}{2}
\end{cases}$\\
$\begin{cases}
x_1(\frac34)=-\frac{\sqrt{2}}{2}(-1-x_2(0)-x_2(0))=\frac{\sqrt{2}}{2}+\sqrt{2}x_2(0)\\
x_2(\frac34)=-\frac{\sqrt{2}}{2}(-1-x_2(0)+x_2(0))=\frac{\sqrt{2}}{2}
\end{cases}$\\

\begin{figure}[!h]
\begin{tikzpicture}
	\draw [color=blue](-3,1)--(3,1);
	\draw [color=blue](-3,-1)--(3,-1);


	\draw[very thick][->](-3,0)--(3,0) node[right=.2] {$x_1(\frac34)$};
	\draw[very thick][->](0,-2.625)--(0,2.75) node[above=.2] {$x_2(\frac34)$};
\end{tikzpicture}
\end{figure}


%  \left[\begin{array}{cc}\end{array}\right]