\input /home/pasky/school/fastex/lib.tex


\subchapter{Implicitn� funkce}{

	M�jme soustavu funkc�
	$$ \eqalign{F_1(x_1,\ldots,x_m,y_1,\ldots,y_n) &= 0\cr
	F_2(x_1,\ldots,x_m,y_1,\ldots,y_n) &= 0\cr
	\vdots &\cr
	F_n(x_1,\ldots,x_m,y_1,\ldots,y_n) &= 0} $$
	a bod $(x_0,y_0) \in \real^{m+n}$, $F_i(x_0,y_0) = 0$ (pro $1 \le i \le n$).

	To v�e m��eme m�t, pr�v� kdy�
	$$y_i = f_i(x_1,\ldots,x_m)\qquad 1 \le i \le n$$
	(lok�ln� v okol� $(x_0,y_0)$).

	Zave�me si zna�en�:
	$$F_i=F_i(x_1,\ldots,x_m,y_1,\ldots,y_n) \qquad f_i=f_i(x_1,\ldots,x_m)$$
	$$F=(F_1,\ldots,n) \qquad f=(f_1,\ldots,n)$$
	$$x=(x_1,\ldots,x_m) \qquad y=(y_1,\ldots,y_n)$$
	$$F_x'(x,y) = \pmatrix{\diffi{F_1}{x_1} & \cdots & \diffi{F_1}{x_m} \cr
				\vdots & & \vdots \cr
				\diffi{F_n}{x_1} & \cdots & \diffi{F_n}{x_m}}(x,y)$$
				\smallskip
	$$f'(x) = \pmatrix{\diffi{f_1}{x_1} & \cdots & \diffi{F_1}{x_m} \cr
				\vdots & & \vdots \cr
				\diffi{f_n}{x_1} & \cdots & \diffi{f_n}{x_m}}(x)$$
				\smallskip
	$$F_y'(x,y) = \pmatrix{\diffi{F_1}{y_1} & \cdots & \diffi{F_1}{y_n} \cr
				\vdots & & \vdots \cr
				\diffi{F_n}{y_1} & \cdots & \diffi{F_n}{y_n}}(x,y)$$

\theorem{14}{o implicitn�ch funkc�ch}{

	M�jme $F=(F_1,\ldots,F_n): W \to \real^n$
	jako zobrazen� definovan� na okol� $W \subset \real^{m+n}$
	bodu $(x_0,y_0)$ (kde $x_0 \in \real^m$, $y_0 \in \real^n$)
	spl�uj�c� n�sleduj�c� podm�nky:
	\list{
	\listItem{$F_i \in C^1(W)$ pro $1 \le i \le n$}
	\listItem{$F_i(x_0,y_0) = 0$ pro $1 \le i \le n$}
	\listItem{$\det F_y'(x_0,y_0) \ne 0$}
	}

	Potom existuj� okol� $x_0 \in U \subset \real^m$, $y_0 \in V \subset \real^n$
	takov�, �e $U \times V \subset W$ a z�rove�
	$$ \forall x \in U\ \onexists y \in V: F_i(x,y) = 0 \qquad 1 \le i \le n $$

	\medskip

	Tj. existuje zobrazen� $f=(f_1,\ldots,f_n)\colon U \to V$ takov�, �e
	$$ \forall (x,y) \in U \times V: F(x,y) = \overline{0}
		\Leftrightarrow y = f(x) $$
	a nav�c $f_i \in C^1(U)$ pro $1 \le i \le n$.
	Tedy zobrazen� $f$ je diferencovateln� v $\forall x \in U$ a z�rove�
	$$ f'(x) = -F_y'(x,f(x))^{-1} \cdot F_x'(x,f(x)) $$

}

	\penalty-200

	\TODO{n�jak hezky nav�zat, tady v�ta samotn� kon��... co to tu vlastn� je? ;-)}

	$$ F_k(x_1,\ldots,x_m,f_1(x),\ldots,f_n(x)) = 0 \qquad 1 \le k \le n,\ x \in U $$
	$$ \diff{F_k}{x_i} + \sum_{j=1}^n \diff{F_k}{y_j}\diff{f_j}{x_i} = 0 $$

	$$ \undernote{\pmatrix{\diffi{F_1}{y_1} & \cdots & \diffi{F_1}{y_n} \cr
		\vdots & & \vdots \cr
		\diffi{F_n}{y_1} & \cdots & \diffi{F_n}{y_n}}}{A}
		\pmatrix{\diffi{f_1}{x_i} \cr \vdots \cr \diffi{f_n}{x_i}}
		= -\pmatrix{\diffi{F_1}{x_i} \cr \vdots \cr \diffi{F_n}{x_i}} $$
	$$ \diffi{f_j}{x_i} = -{\pmatrix{\cdots & \diffi{F_1}{x_i} & \cdots \cr & \vdots & \cr & \diffi{F_n}{x_i} &} \over \det(F_y'(x,f(x))) } = - {\det(A') \over \det(A)}$$
	(v �itateli zam�n�me $j$-t� sloupec).

}

\theorem{15}{o inverzn�ch funkc�ch}{

	M�jme $f\colon U \to \real^m$, otev�enou $U \subset \real^m$,
	$x_0 \in U$ (ozna�me $y_0=f(x_0)$).
	Nech� $f$ spl�uje $f \in C^1(U)$,
	$$\det(\delta_{xj} f_i(x_0))_{i,j=1}^m \ne 0$$

	Pak existuj� okol�
	$x_0 \in U' \subset \real^m$,
	$y_0 \in V \subset \real^m$
	takov�, �e $f\colon U' \to V$ je bijekce,
	$f^{-1} \in C^1(V)$ a~z�rove�
	$$ \forall x \in U', y = f(x) \in V: Df^{-1}(y) = (Df(x))^{-1} $$
	(a podobn� i pro Jacobiho matice).

	\proof{
		Plyne z \ias{}{14}{}.
		Vezm�me $x = x_1,\ldots,x_m$, $y = y_1,\ldots,y_m$:
		$$ F(x,y) = f(x) - y $$
		$$ x_i = g_i(y_1,\ldots,y_m) \Rightarrow F(x,y) = \overline{0} \qquad 1 \le i \le m $$
		P�edpoklady \ias{}{14}{} jsme tak tedy splnili
		($F$ je $C^1$, $F(x_0,y_0) = \overline{0}$, determinant je tak� nulov�).

		M�m okol� $y_0 \in V \subset \real^m$, $g\colon V \to U$
		takov�, �e $$ \forall y \in V: f(x) - y = 0 \Leftrightarrow x = g(y) $$
		$$ \Rightarrow f(g(y)) = y \Rightarrow g = f^{-1} $$

		Vezmu $U' := g(V)$. Pak $f\colon U' \to V$ a $g\colon V \to U'$.
		$U'$ je okol� $x_0$, proto�e $U'$ je $f^{-1}(V)$.
		\qed
	}

	\note{
		M�jme $f\colon U' \to V$, $U,V \in \real^m$, $f$ je bijekce,
		$f \in C^1(U')$, $f^{-1} \in C^1(V)$.
		Takov�mu zobrazen� ��k�me {\bf difeomorfismus}.
	}

}

\medskip


M�jme $U \subset \real^m$ otev�enou, $f\colon U \to \real$,
$F_i\colon U \to \real$ ($1 \le i \le n < m$).
Nech� $f,F_i \in C^1(U)$:
$$ H = \{ x \in \real^m: F_1(x) = \cdots = F_n(x) = 0 \} $$
Jak naj�t lok�ln� extr�my $f$ na $H$?


\theorem{16}{Lagrangeovy multiplik�tory}{

	$a \in H$ bu� bod lok�ln�ho extr�mu, $f$ na $H$.
	M�jme matici $(\delta_{xj}F_i(a))_{i,j=1}^{n,m}$
	maj�c� maxim�ln� hodnost
	(tj. $n$, neboli $\grad F_1(a),\ldots,\grad F_n(a)$
	jsou line�rn� nez�visl� vektory v $\real^m$).
	Potom existuj� ��sla $\lambda_1,\ldots,\lambda_n \in \real$
	takov�, �e: $$\grad f(a) - \sum_{i=1}^n \lambda_i \grad F_i(a) = \overline{0}$$
	$$ \Rightarrow \diff{f}{x_j}(a) - \sum_{i=1}^n \lambda_i\diff{F_i}{x_j}(a) = 0 \qquad 1 \le j \le m $$

	\proof{
		Bez �jmy na obecnosti nech�:
		$$ \det(\delta_{x_{m-n+1}} F(a), \delta_{x_{m-n+2}} F(a), \ldots, \delta_{x_m} F(a)) \ne 0 $$
		$$ x_1,\ldots,x_m = \undernote{x_1,\ldots,x_{m-n}}{y},\undernote{x_{m-n+1},\ldots,x_m}{z} $$
		$$ a = (\undernote{a_1,\ldots,a_{m-n}}{y_0},\undernote{a_{m-n+1},\ldots,a_m}{z_0}) $$

		Pak podle \ias{}{14}{} existuje zobrazen� $g=(g_1,\ldots,g_n)$
		definovan� na okol� $y_0$ takov�, �e (lok�ln�) plat�:
		$$ F_i(y,z) = 0 \Leftrightarrow z = g(y) \qquad 1 \le i \le n $$

		Nyn� p�edpokl�d�me, �e funkce
		$$ h(y) := f(y,g_1(y),\ldots,g_n(y)) $$
		m� v bod� $y_0$ lok�ln� extr�m, bez vazby,
		neboli $\grad h(y_0) = \overline{0}$:
		$$ \diff{f}{y_i}(y_0,g(y_0)) + \sum_{j=1}^n \diff{f}{z_j}(y_0,g(y_0))\cdot\diff{g_j}{y_i}(y_0) = 0
			\qquad 1 \le i \le m-n $$
		$$ D_yf(y_0,g(y_0)) + D_zf(y_0,g(y_0))\cdot Dg(y_0) = 0 $$

		Podle \ias{}{14}{} si ale m��eme $Dg(y_0)$ spo��tat:
		$$ Dg(y_0) = -(D_z F(y_0,g(y_0)))^{-1}\cdot D_y F(y_0,g(y_0)) $$
		$$ D_yf - \undernote{D_zf \cdot (D_zF)^{-1}}{\lambda} \cdot D_yF = \overline{0} $$
		$$ \lambda = (\lambda_1,\ldots,\lambda_n) = D_zf\cdot(D_zF)^{-1}$$
		$$ D_yf - \lambda D_yF = \overline{0} $$
		$$ D_zf - \lambda D_zF = \overline{0} $$
		$$ \Rightarrow Df - \lambda DF = 0 $$
		\qed
	}

	\notes{
	\list{
	\listItem{Jak� je geometrick� v�znam t�to v�ty?
		M�jme $a \in H$ jako bod lok�ln�ho extr�mu $f$ na $H$.
		V~tom p��pad� $$ \grad f(a) \in \mathop{\rm Lin}(\{\grad F_1(a),\ldots,\grad F_n(a)\}) $$
		(kde $\mathop{\rm Lin}$ je line�rn� obal).
		V�imn�me si, �e tohle plat� trivi�ln� pro $\grad f(a) = \overline{0}$.
		Jak je to ale v p��pad�, �e $\grad f(a) \ne \overline{0}$?
		$$ T N_a = \{ x \in \real^m: x \perp \grad f(a) \Leftrightarrow \left<\grad f(a),x\right> = 0 \} $$
		je te�n� prostor (posunut� od po��tku) k plo�e
		$$ N = \{x \in \real^m: f(x) = f(a) \} $$
		v bod� $a$ (s dimenz� $m-1$).

		\penalty-500
		
		M�jme plochu
		$$H = \{x \in \real^m: F_1(x) = \cdots = F_n(x) = 0\}$$
		$$T H_a = \{x \in \real^m: \left<\grad F_1(a),x\right> = \cdots = \left<\grad F_n(a),x\right> = 0 \} $$
		pak bude te�n� prostor k plo�e $H$ v bod� $a$ (s dimenz� $m-n$).

		\medskip

		Pak se d� \ias{}{16}{} p�eformulovat jako:
		m�-li $f$ v $a$ lok�ln� extr�m vzhledem k $H$, pak
		$$T H_a \subset T N_a$$
	}

	\listItem{Vezm�me si {\bf Lagrangovu funkci}:
		$$ L(x,\lambda) = L(x_1,\ldots,x_m,\lambda_1,\ldots,\lambda_n) = f(x) - \sum_{i=1}^n \lambda_iF_i(x) $$
		$$ \grad L = \left(\diff{f}{x_1}(x) - \sum_{i=1}^n\lambda_i\diff{F_i}{x_1}(x), \ldots,
					\diff{f}{x_m}(x) - \sum_{i=1}^n\lambda_i\diff{F_i}{x_m}(x),
					-F_1(x),\ldots,-F_n(x)
					\right)$$
		$$ \grad L(x,\lambda) = \overline{0} \Longleftrightarrow \grad f(x) = \sum_{i=1}^n \lambda_i \grad F_i(x)
				\,\et\, F_1(x) = \cdots = F_n(x) = 0 \textnote{tj. $x \in H$} $$

		Tedy \ias{}{16}{} lze p�eformulovat tak� jako:
		m�-li $f$ v $a$ lok�ln� extr�m vzhledem k $H$, pak
		$$\existss \lambda \in \real^n: \grad L(a,\lambda) = \overline{0}$$
	}
	}
	}

}

\theorem{17}{}{

	M�jme otev�enou $U \subset \real^m$,
	$f,F_i \in C^2(U)$, $1 \le i \le n < m$,
	$$ H = \{ x \in \real^m: F_1(x) = \cdots F_n(x) = 0 \} $$
	a pro $\forall x \in U$ m� matice $(\delta_{xj} F_i(x))_{i,j=1}^{n,m}$
	maxim�ln� hodnost $n$, $a \in H$ a $\grad f(a) \ne \overline{0}$.
	Pak plat�:
	\list{
	\listItem{Pokud pro $\forall \lambda \in \real^n$ m�me
		$\grad L(a,\lambda) \ne \overline{0}$,
		pak $f$ v $a$ nem� lok�ln� extr�m vzhledem k $H$.
		(D�sledek \ias{}{16}{}.)
	}
	\listItem{Nech� $\lambda \in \real^n$ spl�uje
		$\grad L(a,\lambda) = \overline{0}$.
		Pokud kvadratick� forma
		$$P(h_1,\ldots,h_m) = \sum_{i,j=1}^m {\delta^2 L \over \delta x_i \delta x_j}(a,\lambda) h_i h_j$$
		je pozitivn� (negativn�) definitn� na vektorech $h \in T H_a$,
		pak $f$ m� v $a$ ostr� lok�ln� minimum (maximum) vzhledem k $H$.
	}
	\listItem{Pokud je tato kvadratick� forma indefinitn�,
		$f$ nem� v $a$ lok�ln� extr�m vzhledem k $H$.
	}
	}

}


\bend
