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\title[Matematika 4]
{Matematika 4}

\subtitle{6. vaja}

\author[Borut] % 
{B.~Jur\v{c}i\v{c}~Zlobec\inst{1}}

\institute[Univerza v Ljubljani Slovenia] % 
{
  \inst{1}%
    Univerza v Ljubljani,\\Fakulteta za Elektrotehniko\\ 
    1000 Ljubljana, Tr\v{z}a\v{s}ka 25, Slovenija
}


\date[\today] % 
{Matematika FE, Ljubljana, Slovenija \today}

\subject{Matematika}

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\section{Variacijski ra\v{c}un}
%
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\subsection{Eulerjeva ena\v{c}ba}
\begin{frame}{Eulerjeva ena"cba}
\begin{itemize}
	\item Minimum funkcionala 
	$\displaystyle A[y(x)]=\int_{x_0}^{x_1} L(x, y(x),y'(x))dx$.
	\item Eulerjeva ena"cba: 
	\formulas{$\displaystyle 
	\frac{\partial L}{\partial y}-
	\frac{d}{dx}\frac{\partial L}{\partial y'}=0$}.
	\item "Ce je $\displaystyle \frac{\partial L}{\partial x}=0$, potem
	\formulas{$\displaystyle L-y'\frac{\partial L}{\partial y'}=C$}.
	\item Minimum funkcionala pri danem pogoju:
	$\displaystyle A[y(x)]=\int_{x_0}^{x_1} F(x, y,y')dx$, pogoj 
	$\displaystyle \int_{x_0}^{x_1} G(x, y,y')dx=\ell$.\\
	\formulas{ 
	$\displaystyle 
L(x, y,y')=F(x, y,y')+\lambda G(x, y,y')$}.
\end{itemize}
\end{frame}
%
\subsection{Funkciolnal}
\begin{frame}{Izra"cunaj vrednost funkcionala $I(f(x))=\int_a^bL(x,f(x),f'(x))dx$}
	\formulas{$\displaystyle I(f(x))=\int_0^1\sqrt{1+f'(x)^2}dx,\quad f(x)=x^2$}
	\begin{itemize}
		\item $\displaystyle \int_0^1\sqrt{1+4x^2}dx=\rightarrow$
		\item $\displaystyle \left.\frac{1}{2} \sqrt{4 x^2+1} x+\frac{1}{4} 
			\log\left(2x+\sqrt{1+4x^2}\right)\right|_0^1=\rightarrow$
		\item $\displaystyle \frac{1}{4}\left(2\sqrt{5}+
			\log\left(2+\sqrt{5}\right) \right)$.
	\end{itemize}
\end{frame}

\begin{frame}{Paradoks dvoj"ckov}
Lastni "cas sistema, ki v prostoru "casu opisuje pot $x(t)$,  
dolo"ca funkcional 
\textcolor{darkred}{
$\displaystyle\tau=\int_{t_0}^{t_1}\sqrt{1-x'(t)^2}dt$}. 
Prostor "cas ima v na"sem primeru eno "casovno in eno 
prostorsko dimenzijo. 
Vzeli smo, da je svetlobna hitrost enaka 1. 
Na za"cetku, v "casu ni"c, 
se dvoj"cka nahajata v koordinatnem izhodi"s"cu. Eden ostane tam 
drugi pa odpotuje, njegovo pot v prostoru "casu opi"se funkcija 
$x(t)=t(2-t)/2$. 
V "casu $t=2$ se zopet sre"cata v koordinatnem izhodi"s"cu.
Izra"cunaj lastna "casa mirujo"cega in gibajo"cega dvoj"cka. 
Pot mirujo"cega dvoj"cka opi"se funkcija $x(t)=0$. 
\begin{enumerate}
	\item \formulas{$\displaystyle\tau_1=\int_0^2\sqrt{1}dt=2$}
	\item \formulas{$\displaystyle\tau_2=\int_0^2\sqrt{1-(1-t)^2}dt=
	\pi/2$}.\\ Polovica plo"s"cine kroga s polmerom 1. 
	\item \formulas{$\tau_1>\tau_2$}. 
\end{enumerate}
\end{frame}
\begin{frame}
	\begin{center}
		\includegraphics[height=0.9\textheight]{dvojcka.pdf}
	\end{center}
\end{frame}
\subsection{Geometrijski primeri}
\begin{frame}{Poi"s"ci najkraj"so pot med dvema to"ckama v ravnini}
\formulas{$T_0(x_0,y_0),\quad T_1(x_1,y_1)$}
\begin{itemize}
	\item I"s"cemo funkcijo $y=f(x)$, $y_0=f(x_0)$ in $y_1=f(x_1)$,
	za katero funkcional\\
	$\displaystyle A(f(x))=\int_{x_0}^{x_1}\sqrt{1+f'(x)^2}dx$
	dose"ze minimum.
	\item $\displaystyle L=\sqrt{1+y'^2}\rightarrow$, 
	$\frac{\partial L}{\partial x}=0\rightarrow$, 
	$\displaystyle L-y'\frac{\partial L}{\partial y'}=C\rightarrow$,
	\item $\displaystyle\sqrt{1+y'^2}-y'\frac{y'}{\sqrt{1+y'^2}}=
	C\rightarrow$,
	$\displaystyle \frac{1}{\sqrt{1+y'^2}}=C\rightarrow$, 
	\item $\displaystyle y'=k\rightarrow y=kx+m$.
	\item Funkcija je \formulas{$\displaystyle f(x)=kx+m$}, iz pogojev 
	$y_i=kx_i+m$, $i=0,1$, dolo"cimo koeficienta $k$ in $m$. 
	\item Najkraj"sa pot je ravna "crta. 
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci optimalno pot med dvema 
dogodkoma v prostoru "casu (svetlobna hitrost je enaka 1).}
\formulas{$T_0(x_0,t_0),\quad T_1(x_1,t_1)$}
\begin{itemize}
	\item I"s"cemo funkcijo $x=x(t)$, $x_0=x(t_0)$ in $x_1=x(t_1)$,
	za katero funkcional (lastni "cas) \\
	$\displaystyle A(x(t))=\int_{t_0}^{t_1}\sqrt{1-x'(t)^2}dt$
	dose"ze optimum, (lasni "cas je najdalj"si).  
	\item $\displaystyle L=\sqrt{1-x'^2}\rightarrow$, 
	$\frac{\partial L}{\partial t}=0\rightarrow$, 
	$\displaystyle L-x'\frac{\partial L}{\partial x'}=C\rightarrow$,
	\item $\displaystyle\sqrt{1-x'^2}+x'\frac{x'}{\sqrt{1-x'^2}}=
	C\rightarrow$,
	$\displaystyle \frac{1}{\sqrt{1-x'^2}}=C\rightarrow$, 
	\item $\displaystyle x'=k\rightarrow x=k\,t+m$.
	\item Funkcija je \formulas{$\displaystyle x(t)=k\,t+m$}, 
	iz pogojev 
	$x_i=k\,t_i+m$, $i=0,1$, dolo"cimo koeficienta $k$ in $m$. 
\end{itemize}
\end{frame}

\begin{frame}{Rotacijska ploskev z najmanj"so povr"sino}
\formulas{I"s"cemo funkcijo $f(x)$, $f(x_0)=a$, $f(x_1)=b$, 
$\displaystyle P=2\pi\int_{x_0}^{x_1}f(x)\sqrt{1+f'(x)^2}dx$
tako, da je $P$ minimalna.}
\begin{itemize}
	\item $\displaystyle L=y\sqrt{1+y'^2}\rightarrow$, 
	\item $\displaystyle L-y'\frac{\partial L}{\partial y'}=C_1$,
	\item $\displaystyle\frac{y}{\sqrt{1-y'^2}}=C_1\rightarrow$,
	$y=C_1\sqrt{1+y'^2}\rightarrow$,$\displaystyle\frac{dy}
	{\sqrt{\left(\frac{y}{C_1}\right)^2-1}}=dx$.
	\item $\displaystyle y=C_1\mbox{ch}\,\frac{x-C_2}{C_1}$.
	\item Iz pogojev dolo"cimo konstanti $C_1$ in $C_2$. 
\end{itemize}
\end{frame}

\begin{frame}{Izoperimetri"cni problem}
	\formulas{Poi"s"ci funkcijo $y=f(x)$, ki maksimizira 
plo"s"cino $\displaystyle\int_a^b f(x) dx$ pri dani lo"cni dol"zini 
$\displaystyle\int_a^b\sqrt{1+f'(x)^2}dx=\ell$ in $f(a)=f(b)=0$.}
\begin{itemize}
	\item $\displaystyle L=y-\lambda\sqrt{1+y'^2}$.
	\item Ker je $\displaystyle\frac{\partial L}{\partial x}=0$, velja
	$L-y'\frac{\partial L}{\partial y'}=
	y-\frac{\lambda}{\sqrt{1+y'^2}}=C_2$.
	\item $\displaystyle y'=\frac{\sqrt{\lambda^2-(y-C_2)^2}}{y-C_2}$.
	\item Lo"cimo spremenljivke in dobimo $(x-C_1)^2+(y-C_2)^2=\lambda^2$.
	Dolo"cimo konstante $C_1$, $C_2$ in $\lambda$ tako, da bo 
	$y(a)=y(b)=0$ in dol"zina loka enaka $\ell$. 
\end{itemize}	
\end{frame}

\begin{frame}{Maksimalna entropija in normalna porazdelitev.
}
\begin{itemize}
        \item  Gostota porazdelitve slu"cajne spremenljivke
        \textcolor{darkblue}{$\rho(x)\ge0$ na $\mathbb{R}$}.
        \begin{enumerate}
        \item \textcolor{darkred}{
                $\displaystyle \int_{-\infty}^\infty\rho(x)\,dx=1$},
	\item \textcolor{darkred}{
                $\displaystyle \mu=\int_{-\infty}^\infty x\rho(x)\,dx$},
                matemati"cno upanje,
        \item \textcolor{darkred}{$\displaystyle \sigma^2=
		\int_{-\infty}^\infty (x-\mu)^2\rho(x)\,dx$}, disperzija in
        \end{enumerate}
        \item Entropija je enaka
        \textcolor{darkblue}{$\displaystyle S[\rho]=
        -\int_{-\infty}^{\infty}\rho(x)\log\rho(x)\,dx$}.
        \item \textcolor{blue}{Poi"s"ci gostoto porazdelitve,
        ki ima pri dani disperziji $\sigma^2$ in matemati"cnem
        upanju $\mu$, najve"cjo entropijo.}
\end{itemize}
\end{frame}

\begin{frame}
        \begin{itemize}
                \item $\displaystyle 
			L= \rho\log \rho + \lambda_1\rho + \lambda_2 x \rho + \lambda_3 (x-\mu)^2 \rho$,
                \item $\displaystyle \frac{\partial L}{\partial \rho}=
			\log\rho + \lambda_1 + \lambda_2 x + \lambda_3 (x-\mu)^2=0$,
		\item $\displaystyle \rho=e^{-\lambda_1-\lambda_2 x - \lambda_3 (x-\mu)^2}$
		\item $\displaystyle e^{-\lambda_1}=\frac{1}{\sqrt{2\pi}\sigma},\quad \lambda_2=0,\quad 
				\lambda_3=\frac{1}{2\sigma^2}$.
		\item \textcolor{darkblue}{
			$\displaystyle \rho=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$}
        \end{itemize}
\end{frame}

\begin{frame}{Maksimalna entropija in eksponentna porazdelitev}
\textcolor{darkblue}{Gostota porazdelitve 
$\displaystyle\begin{cases}\rho(x)>0,&x>0\\\rho(x)=0,&x\le 0\end{cases}$  
z maksimalno entropijo pri danem matemati"cnem upanju $\mu>0$.}
\begin{itemize} 
\item Pogoji: $\displaystyle \int_0^\infty\rho(x)dx=1$, 
	$\displaystyle\mu=\int_0^\infty x\rho(x)$. 
\item Funkcional: $\displaystyle 
	S[\rho(x)]=-\int_0^\infty\rho(x)\log(\rho(x))dx$. 
\item Lagrange: $\displaystyle L=\rho\log(\rho)+\lambda_1\rho+\lambda_2 x \rho$
\item Eulerjeva ena"cba: $\displaystyle \log\rho + \lambda_1 + \lambda_2 x =0$,
\item Re"sitev: $\displaystyle \rho(x)=e^{-\lambda_1-\lambda_2 x}$,  
	$e^{-\lambda_1}=\frac{1}{\mu}$, $\lambda_2=\frac{1}{\mu}$.
\item \textcolor{darkblue}{
	$\displaystyle \rho(x)=\frac{1}{\mu}e^{-\frac{1}{\mu}}$}.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci ekstremalo funkcionala pri danem pogoju}
\textcolor{darkblue}{
	$\displaystyle\int_0^2 x'(t)^2dt$, $x(0)=0$, $x(2)=4$ in  
	$\displaystyle\int_0^2 \left(x(t)t^2+x'(t)\right)dt=8$.
}
\begin{itemize}
	\item $\displaystyle L(t,x(t),x'(t))=x'(t)^2+\lambda \left(x(t) t^2+x'(t)\right)$.
	\item Eulerjeva ena"cba $\displaystyle \lambda t^2-2x''(t)=0$,
	\item $\displaystyle x(t)=\frac{1}{24}\left( 48 t-8t\lambda+t^4\lambda\right)$.
	\item Re"sitev ena"cbe $\displaystyle \int_0^2\left(x(t) t^2+x'(t)\right)dt=8$, $\lambda=7$.
	\item \textcolor{darkblue}{$\displaystyle x(t)=\frac{1}{24}\left(-8t+7t^4\right) $}.
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci ekstremalo funkcionala pri danem pogoju}
\textcolor{darkblue}{
	$\displaystyle\int_0^\pi x'(t)^2 dt$, $x(0)=0$, $x(\pi)=0$ in  
	$\displaystyle\int_0^\pi x(t)^2dt=\frac{\pi}{2}$.
}
\begin{itemize}
	\item $\displaystyle L(t,x(t),x'(t))=x'(t)^2+\lambda x(t)^2$.
	\item Eulerjeva ena"cba $\displaystyle \lambda x(t)-x''(t)=0$.
	\item Upo"stevajo"c robne pogoje $\lambda=-n^2$, $n\in \mathbb{Z}$. 
	\item Re"sitev Eulerjeve ena"cbe z upo"stevanjem robnih pogojev: 
	\item $x(t)=A \sin(n t)$,
	\item $\displaystyle A \int_0^\pi\sin(n t)^2dt=\frac{\pi}{2}$,
	\item \textcolor{darkblue}{$\displaystyle x(t)=\sin(n t),
		\quad n\in \mathbb{Z},\quad n\ne0$.}
\end{itemize}
\end{frame}


\subsection{Fizikalni primeri}
\begin{frame}{Princip najmanj"se akcije}
\begin{itemize}
	\item Lagrangeeva funkcija $L(t,x,\dot{x})=T-U$, razlika 
	kineti"cne in potencialne energije.
	\item Kineti"cna $T=\frac{m\dot{x}^2}{2}$ in 
		potencialna energija $U=U(x)$. 
	\item Re"sitev je trajektorija $x=x(t)$ za katero zavzame akcija 
	\formulas{$\displaystyle S=\int_{t_0}^{t_1}L(x,\dot{x})dt$}
	najmanj"so vrednost.
	\item Ker je $\frac{\partial L}{\partial t}=0$, potem 
	$L-\dot{x}\frac{\partial L}{\partial \dot{x}}=C=-E$.
	\item Zakon o ohranitvi energije 	
	$\displaystyle \frac{m\dot{x}^2}{2}+U(x)=E$. 
	\item Vsota kineti"cne in potencialne energije je konstantna. 
	\item Newtonov zakon: $\frac{\partial L}{\partial x}-
	\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}=0\rightarrow$,
	$m\ddot{x}=-\frac{\partial U}{\partial x}$. 
\end{itemize}
\end{frame}


\begin{frame}{Navpi"cni met navzgor}
\formulas{Orientirajmo os x navpi"cno navzgor. 
Kamen vr"zemo navpi"cno, z za"cetno hitrostjo $v_0$, 
	iz polo"zaja $x_0=0$.}
\begin{itemize}
	\item Kineti"cna energija $T=\frac{m\dot{x}^2}{2}$,
	potencialna energija $U=mgx$. 
	\item Lagrangeeva funkcija 
	$\displaystyle L=T-U=\frac{m\dot{x}^2}{2}-mgh$.
	\item Eulerjeva ena"cba: 
	$m\ddot{x}=-mg\rightarrow$, 
	$\ddot{x}=-g\rightarrow$, $x(t)=C_2+C_1t-\frac{gt^2}{2}$.
	\item Re"sitev: 
	\formulas{$\displaystyle x(t)=v_0t-\frac{gt^2}{2}$}.
\end{itemize}
\end{frame}
\begin{frame}{Nihanje vzmetnega nihala}
\formulas{Os $x$ orientirajmo vodoravno.
Ute"z z maso $m$ je pripeta na vzmet s konstanto vzmeti $k$.
Ravnovesna lega je $x_0=0$. Na za"cetku nihalo miruje. 
Ute"z zmaknemo iz mirovne lege za $\Delta x=\ell$}.
\begin{itemize}
	\item Kineti"cna energija je $T=\frac{m\dot{x}^2}{2}$ medtem, ko je 
        potencialna energija enaka $U=\frac{k x^2}{2}$. 
	\item Lagrangeeva funkcija 
	$\displaystyle L=T-U=\frac{m\dot{x}^2}{2}-\frac{k x^2}{2}$.
	\item Eulerjeva ena"cba:
        $m\ddot{x}=-k x\rightarrow$, $\ddot{x}+\frac{k}{m}x=0\rightarrow$, 
	re"sitev 
	$x(t)=A\cos(\omega t)+B\sin(\omega t)$, $\omega^2=\frac{k}{m}$. 
	\item \formulas{$x(t)=\ell\cos(\sqrt\frac{k}{m} t)$}.
\end{itemize}
\end{frame}


\begin{frame}{Oblika prosto vise"ce verige}
\formulas{Veriga vpeta v to"ckah $T_0(x_0,y_0)$ in $T_1(x_1,y_1)$, 
dol"zine $\ell$. 
}
\begin{itemize}
	\item Oblika verige je podana s funkcijo $y=f(x)$, kjer je 
	$y_i=f(x_i)$, $i=0,1$ in 
	dol"zina je $\int_{x_0}^{x_1}\sqrt{1+y'^2} dx=\ell$.
	\item Oblika verige je taka, da je njena potencialna 
	energija minimalna.
	\item $\displaystyle dU=yg\,dm=yg\rho\,ds=g\rho y\sqrt{1+y'^2}dx$.
	\item $\displaystyle U=g\rho\int_{x_0}^{x_1} y\sqrt{1+y'^2}dx$.
	\item Lagrangeeva funkcija 
	$L= y\sqrt{1+y'^2}+\lambda\sqrt{1+y'^2}$.
	\item Prvi integral Eulerjeve ena"cbe: 
	$\displaystyle\frac{y+\lambda}{\sqrt{1+y'^2}}=C$.
	\item $y'=\sqrt{C^2(y-\lambda)^2-1}$, 
		\formulas{$y=f(x)=C\mbox{ch}\frac{x+C_1}{C}-\lambda$}.
	\item Dolo"cimo konstante $C$, $C_1$ in $\lambda$, da bo\\ 
	$y_i=f(x_i)$, $i=0,1$ in 
	$\displaystyle\int_{x_0}^{x_1}\sqrt{1+y'^2}dx=\ell$.   
\end{itemize}
\end{frame}

\begin{frame}{Sisetmi Eurejevih ena"cb}
\begin{itemize}
	\item \textcolor{darkblue}{
	$\displaystyle I[x,y]=\int_a^b L(t,x(t),y(t),
	\dot{x}(t),\dot{y}(t)]dt$}.
	\item Sistem Eulejevih ena"cb\\
	$\displaystyle \frac{\partial L}{\partial x}-
		  \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=0$\\
	$\displaystyle \frac{\partial L}{\partial y}-
                  \frac{d}{dt}\frac{\partial L}{\partial \dot{y}}=0$\\ 
\end{itemize}
\end{frame}

\begin{frame}{Po"sevni met}
\begin{itemize}
	\item $\displaystyle L = \frac{m}{2}(\dot{x}^2+\dot{y}^2)-m g y$.
	\item $\displaystyle x(0)=x_0,\quad y(0)=y_0,\quad 
	\dot{x}(0)=v_{x0},\quad \dot{y}(0)=v_{y0}$.
	\item Sistem Eulerjevih ena"cb:\\
	$\displaystyle \ddot{x}(t)=0,\quad \ddot{y}(t)=-g$.
	\item Re"sitev sistema:\\
	\textcolor{darkblue}{$\displaystyle x(t)=x_0+v_{x0}t,\quad 
	y(t)=-\frac{g}{2}t^2+ y_0+v_{y0}t$}.
\end{itemize}
\end{frame}

\begin{frame}{Eulerjeva ena"cba za funkcije ve"c spremenljivk}
\textcolor{darkblue}{$\displaystyle 
	I[u(x,y)]=\iint_\mathcal{D}L\left(x,y,u(x,y),
	\frac{\partial u}{\partial x},
	\frac{\partial u}{\partial y}\right)dxdy$, 
	$\displaystyle u(x,y)|_{\partial\mathcal{D}}=f(x,y)$}
\begin{itemize}
	\item Ozna"cimo $\displaystyle 
	p=\frac{\partial u}{\partial x}, \quad 
	q=\frac{\partial u}{\partial y}.$
	\item $L=L(x,y,u,p,q)$, Eulerjeva ena"cba:\\
	\item $\displaystyle \frac{\partial L}{\partial u}-
	\frac{d}{dx}\frac{\partial L}{\partial p}-
	\frac{d}{dy}\frac{\partial L}{\partial q}=0$.
\end{itemize}
\end{frame}

\begin{frame}{Laplaceova diferencialna ena"cba}
\textcolor{darkblue}{$\displaystyle 
	I[u]=\iint_{\mathcal{D}}\left|\nabla{u}\right|^2dS,\quad 
	u(x,y)_{\partial\mathcal{D}}=f(x,y)$}.
\begin{itemize}
	\item $\displaystyle L=\frac{1}{2}(p^2+q^2)$.
	\item Eulerjeva ena"cba: 
	$\displaystyle -\frac{d}{dx}p-\frac{d}{dy}q=0$.
	\item $\displaystyle -\Delta u(x,y)=0,\quad 
		u(x,y)|_{\partial\mathcal{D}}=f(x,y)$. 
\end{itemize}	
\end{frame}

\begin{frame}{Ploskev minimalne povr"sine}
\textcolor{darkblue}{
$\displaystyle I[u]=\iint_\mathcal{D}
	\sqrt{1+
           \left(\frac{\partial u}{\partial x}\right)^2+
           \left(\frac{\partial u}{\partial y}\right)^2}
	dxdy,\quad u(x,y)|_{\partial\mathcal{D}}=f(x,y)$}.
\begin{itemize}
	\item Poiskali bomo pribli"zno re"sitev. 
	\item Za mahne vrednosti odvodov (polo"zna ploskev):
	\item $\displaystyle \sqrt{1+x}\approx 1+\frac{1}{2}x$,
	\item $\displaystyle L=1+\frac{1}{2}(p^2+q^2)$.
	\item Eulerjeva ena"cba $\displaystyle -\Delta u(x,y)=0$.
\end{itemize}	
\end{frame}


\end{document}


