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\title[Matematika 4]
{Matematika 4}

\subtitle{4. 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[2013] % 
{Matematika FE, Ljubljana, Slovenija \today}

\subject{Matematika}

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\section{Parcialne diferencialne ena\v{c}be}
\subsection{Splo\v{s}ne re\v{s}itve.}
\begin{frame}{Re"si parcialno diferencialno ena"cbo 
$\frac{\partial^2}{\partial x^2}u(x,y)=0$}
\begin{itemize}
	\item $\displaystyle \frac{\partial^2}{\partial x^2}u(x,y)=0$.
	\item $\frac{\partial}{\partial x}u(x,y)=f(y)$.
	\item \formulas{$u(x,y)=f(y)x+g(y)$}.
\end{itemize}
\end{frame}
\begin{frame}{Re"si parcialno diferencialno ena"cbo 
$\frac{\partial^2u(x,y)}{\partial{x}\partial{y}}=0$}
\begin{itemize}
        \item $\displaystyle \frac{\partial^2}{\partial x\partial
y}u(x,y)=0$.
        \item $\frac{\partial}{\partial x}u(x,y)=f'(x)$.
        \item \formulas{$u(x,y)=f(x)+g(y)$}.
\end{itemize}
\end{frame}


\begin{frame}{Re"si parcialno diferencialno ena"cbo 
$\frac{\partial^2u(x,y)}
{\partial x\partial y}+\frac{\partial u(x,y)}{\partial x}=0$}
\begin{itemize}
        \item $\displaystyle \frac{\partial^2}
	{\partial x\partial y}u(x,y)+\frac{\partial}{\partial x}u(x,y)=0$.
	\item Uvedemo novo spremenljivko $u(x,y)_x=v(x,y)$.
        \item $v(x,y)_y+v(x,y)=0\rightarrow v(x,y)=f'(x)e^{-y}$.
	\item $u_x(x,y)=f'(x)e^y\rightarrow u(x,y)=f(x)e^{-y}+g(y)$.
	\item \formulas{$\displaystyle u(x,y)=f(x)e^{-y}+g(y)$}.
\end{itemize}
\end{frame}

\begin{frame}{Re"si parcialno diferencialno ena"cbo 
$\frac{\partial^2u(x,y)}{\partial x^2}+u(x,y)=0$}
\begin{itemize}
        \item $\displaystyle \frac{\partial^2}
        {\partial x^2}u(x,y)+u(x,y)=0$.
        \item \formulas{$u(x,y)=f(y)\cos x+g(y)\sin x$}.
\end{itemize}
\end{frame}

\begin{frame}{Re"si parcialno diferencialno ena"cbo 
$\frac{\partial u(x,y)}{\partial y}=x$}
\begin{itemize}
        \item $\displaystyle \frac{\partial}
        {\partial y}u(x,y)=x$
        \item \formulas{$u(x,y)=x\,y+f(x)$}.
\end{itemize}
\end{frame}

\begin{frame}{Re"si parcialno diferencialno ena"cbo
$\frac{\partial^2 u(x,y)}{\partial x \partial y}+\frac{\partial u(x,y)}{\partial x}+x+y=0$}
\begin{itemize}
        \item $\displaystyle \frac{\partial^2 u}
        {\partial x\partial y}u(x,y)+\frac{\partial u}
	{\partial x}u(x,y)+x+y=0$.
	\item Uvedemo novo spremenljivko $u(x,y)_x=v(x,y)\rightarrow 
	v(x,y)_y+v=-x-y$.
	\item Re"sitev homogene ena"cbe ${v_h}_y+v_h=0$ je $v_h=f'(x)e^{-y}$.
	\item Partikularna re"sitev $\overline{v}(x,y)=C(y)e^y$.
	\item $\overline{v}(x,y)_y+\overline{v}(x,y)=-x-y\rightarrow 
	C'(y)e^{-y}-C(y)e^{-y}+C(y)e^{-y}=-x-y\rightarrow
	C'(y)=(-x-y)e^{y}\rightarrow$
	\item $C(y)=-xe^{y}-ye^{y}+e^{y}\rightarrow
	\overline{v}(x,y)=-x-y+1$.
	\item $u(x,y)_x=v(x,y)=f'(x)e^{-y}-x-y+1\rightarrow$
        \item \formulas{$\displaystyle
	u(x,y)=f(x)e^{-y}-\frac{x^2}{2}-x\,y-x+g(y)$}.
\end{itemize}
\end{frame}

\begin{frame}{Re"si parcialno diferencialno ena"cbo
$\frac{\partial u(x,y)}{\partial x}=2x\,y\,u(x,y)$}
\begin{itemize}
	\item Vzamemo, da je $y$ konstanta. 
	\item Navadna diferencialna 
	ena"cba ima lo"cljive spremenljivke. 
	\item Pi"semo $v(x)=u(x,y)\rightarrow$
	\item $v'=2x\,y\,v\rightarrow \frac{dv}{v}=2x\,y\,dx$.
	\item $\ln |v|=x^2y+\ln|C|\rightarrow v(x)=C\,e^{x^2y}$.
	\item \formulas{$\displaystyle u(x,y)=C(y)e^{x^2y}$}.
\end{itemize}
\end{frame}

\begin{frame}{Re"si parcialno diferencialno ena"cbo 
$\frac{\partial u(x,y)}{\partial x}+
x \frac{\partial^2 u(x,y)}{\partial x^2}=y$}
\begin{itemize}
	\item Uvedemo novo spremenljivko in vzamemo, da je y konstanta. 
	\item $v(x)=u(x,y)_x\rightarrow v+xv'=y\rightarrow x\frac{dv}{dx}=y-v$.
	\item Lo"cimo spremenljivke $\frac{dv}{y-v}=\frac{dx}{x}$.
	\item $\ln|y-v|=\ln|x|+\ln|C|\rightarrow y-v=C\,x
	\rightarrow v=y-Cx$.
	\item $u(x,y)_x=y+f(y)x\rightarrow 
		u(x,y)=x\,y+f(y)\frac{x^2}{2}+g(y)$.
	\item \formulas{$\displaystyle u(x,y)=x\,y+f(y)\frac{x^2}{2}+g(y)$}.
\end{itemize}
\end{frame}
\begin{frame}{Re"si parcialno diferencialno ena"cbo
$\frac{\partial^2u(x,y)}{\partial y^2}+
\frac{\partial u(x,y)}{\partial y}=xy$}
\begin{itemize}
\item Uvedemo novo spremenljivko in vzamemo, da je x konstanta.
\item $v(y)=u(x,y)_y\rightarrow v'+v=x\,y$.
\item Re"sitev homogene ena"cbe je $v(y)=C e^{-y}$. 
\item Nastavek za partikularno re"sitev $\overline{v}(y)=C(y)e^{-y}$.
\item $C'(y)e{-y}=x\,y\rightarrow C'(y)=x\,y\,e^{y}\rightarrow
	C(y)=x (ye^y-e^{y})$
\item $\overline{v}(y)=x\,y-x\rightarrow v(y)=x\,y-x+C e^{-y}$.
\item $u(x,y)_x=x\,y-x+f(x) e^{-y}\rightarrow 
	u(x,y)=\frac{1}{2}x\,y^2-x\,y+f(x)e^{-y}+g(x)$.
\item \formulas{$\displaystyle
	u(x,y)=\frac{1}{2}x\,y^2-x\,y+f(x)e^{-y}+g(x)$}.
\end{itemize}
\end{frame}

\subsection{Uvedba novih spremenljivk}
\begin{frame}{Vpelji nove spremenljivke in re"si ena"cbo
$\frac{\partial^2u(x,y)}{\partial x^2} -
	2\frac{\partial^2 u(x,y)}{\partial x\partial y} + 
\frac{\partial^2u(x,y)}{\partial y^2}=0$, $t=x$, $z=x+y$.}
\begin{itemize}
	\item $\displaystyle\frac{\partial u}{\partial x}=
	\frac{\partial u}{\partial t}\frac{\partial t}{\partial x}
	+\frac{\partial u}{\partial z}\frac{\partial z}{\partial x}$.
	\item $\frac{\partial u}{\partial y}=\frac{\partial u}{\partial t}
	\frac{\partial t}{\partial y}
        +\frac{\partial u}{\partial z}\frac{\partial z}{\partial y}$.
	\item $u_x=u_t+u_z\quad u_y=u_z\rightarrow 
	u_{xx}=(u_t+u_z)_t+(u_t+u_z)_z=u_{tt}+2u_{tz}+u_{zz},
	\quad u_{yy}=u_{zz}
	\rightarrow u_{xy}=(u_t+u_z)_z=u_{tz}+u_zz$.
	\item $u_{xx} -2u_{xy} + u_{yy}=
	u_{tt}+2u_{tz}+u_{zz}-2u_{tz}-2u_{zz}+u_{zz}=u_{tt}=0$.
	\item $u=f(z) t+g(z)\rightarrow u(x,y)=f(x+y) x + g(x+y)$.
	\item \formulas{$\displaystyle u(x,y)=f(x+y) x + g(x+y) $}.
\end{itemize}
\end{frame}


\begin{frame}{Vpelji nove spremenljivke in re"si ena"cbo
$x\frac{\partial u(x,y)}{\partial x}-
	y\frac{\partial u(x,y)}{\partial y}=2u(x,y)$,
	$t=x^2$, $z=x\,y$}
\begin{itemize}
	\item $\frac{\partial u}{\partial x}=2x\frac{\partial u}{\partial x}+
	y\frac{\partial u}{z}\rightarrow$
        $\frac{\partial u}{\partial y}=x\frac{\partial u}{z}$
	\item $\frac{\partial u}{\partial x}=
	2\sqrt{t}\frac{\partial u}{\partial x}+
        \frac{z}{\sqrt{t}}\frac{\partial u}{z}\rightarrow$        
        $\frac{\partial u}{\partial y}=
	\sqrt{t}\frac{\partial u}{\partial z}$.
	\item $\sqrt{t}2\sqrt{t}u_t+\sqrt{t}\frac{z}{\sqrt{t}}u_z-
	\frac{z}{\sqrt{t}}\sqrt{t}u_z=2u$
	\item $2tu_t=2u\rightarrow \frac{du}{u}=
	\frac{dt}{t}\rightarrow u=tf(z)$
	\item \formulas{$\displaystyle u(x,y)=x^2f(xy)$}.
\end{itemize}
\end{frame}


\begin{frame}{Vpelji nove spremenljivke in re"si ena"cbo
$\frac{\partial^2 u(x,y)}{\partial x^2} 
  +\frac{\partial^2u(x,y)}{\partial x\partial y} - 
	2\frac{\partial^2u(x,y)}{\partial y^2}=0$,
$t=x+y$, $z=2x-y$}.
\begin{itemize}
	\item $u_x=u_t+2u_z\quad u_y=u_t-u_z\rightarrow$ 
	\item $u_{xx}=(u_t+2u_z)_t+2(u_t+2u_z)_z=u_{tt}+4u_{tz}+4u_{zz},$
	$u_{yy}=(u_t-u_z)_t-(u_t-u_z)_z=u_{tt}-2u_{tz}+u_{zz},$ 
	$u_{xy}=(u_t+2u_z)_t-(u_t+2u_z)_z$.
	\item 
	$u_{xx} -2u_{xy} + u_{yy}=
	u_{tt}+2u_{tz}+u_{zz}-2u_{tz}-2u_zz+u_{zz}\rightarrow$
	$u_{tt}=0$.
	\item $u=f(z) t+g(z)\rightarrow u(x,y)=f(2x-y)(x+y) 
		+ g(2x-y)$.
	\item \formulas{$\displaystyle u(x,y)=f(2x-y)(x+y) + g(2x-y) $}.
\end{itemize}
\end{frame}
\begin{frame}{Poi"s"ci re"sitev diferencialne 
ena"cbe $\frac{\partial u(x,y)}{\partial x}+\frac{\partial u(x,y)}{\partial y}=0$ 
v obliki $u(x,y)=X(x)Y(y)$}
\begin{itemize}
	\item Vstavimo $u(x,y)=X(x)Y(y)$, $X'(x)Y(y)+X(x)Y'(y)=0\rightarrow$
	\item $\displaystyle \frac{X'(x)}{X(x)}+\frac{Y'(y)}{Y(y)}=0$. 
	\item Velja $\displaystyle \frac{X'(x)}{X(x)}=-\frac{Y'(y)}{Y(y)}=k$,
	\item kjer je $k$ parameter nodvisen od $x$ in $y$.
	\item $\frac{dX}{X}=k\,dx$ in $\frac{dY}{Y}=-k\,dy\rightarrow$
	\item $X(x)=Ae^{kx}$ in $Y(y)=Be^{-ky}\rightarrow$
	\item \formulas{$\displaystyle u(x,y)=C e^{k(x-y)}$}
\end{itemize}
\end{frame}


\begin{frame}{Poi"s"ci re"sitev diferencialne 
ena"cbe $x^2u_{xy}+3y^2u=0$ v obliki $u=XY$}
\begin{itemize}
	\item $x^2X'Y'+3y^2XY=0\rightarrow$
	\item $\displaystyle x^2\frac{X'}{X}\frac{Y'}{Y}+3y^2=0\rightarrow$ 
	\item $x^2\frac{X'}{X}=-3y^2\frac{Y}{Y'}=k$,
	\item kjer je $k$ parameter nodvisen od $x$ in $y$.
	\item $\frac{dX}{X}=k\frac{dx}{x^2}$ in 
	$\frac{dY}{Y}=-3y^2\frac{dy}{k}\,dy\rightarrow$
	\item $X(x)=Ae^{-\frac{k}{x}}$ in $Y(y)=Be^{-\frac{y^3}{k}}\rightarrow$
	\item \formulas{$\displaystyle u(x,y)=C e^{-\frac{k}{x}-\frac{y^3}{k}}$}
\end{itemize}
\end{frame}

\begin{frame}{Poi"s"ci re"sitev diferencialne
ena"cbe $u_x+yu_y=0$ v obliki $u=XY$, ki ustreza pogojema 
$u(1,0)=1$ in $u(0,1)=2$.}
\begin{itemize}
        \item $X'Y+yXY'=0\rightarrow$
        $\displaystyle \frac{X'}{X}+y\frac{Y'}{Y}=0\rightarrow$
        \item $\frac{X'}{X}=-y\frac{Y'}{Y}=k$,
        kjer je $k$ parameter nodvisen od $x$ in $y$.
        \item $\frac{dX}{X}=k\,dx$ in
        $\frac{dY}{Y}=-y\,dy\rightarrow$
        \item $X(x)=Ae^{kx}$ in $Y(y)=Be^{-k\frac{y^2}{2}}\rightarrow$
        \formulas{$\displaystyle u(x,y)=C e^{kx-k\frac{y^2}{2}}$}.
	\item $u(1,0)=C e^{k}=1$ in $u(0,1)=C e^{-\frac{k}{2}}=2$
        \item \formulas{$k=-\frac{1}{3}\log{4}$, $C=4^{1/3}$}. 
\end{itemize}
\end{frame}



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