Introducción

Teorema de la Función Implícita (sistemas de ecuaciones

Teorema 1. Considere las funciones $z_1=F(x,y,u,v)$ y $z_2=G(x,y,u,v)$. Sea $P=(x,y,u,v)\in {\mathbb{R}}^4$ un punto tal que $F(P)=G(P)=0$. Suponga que en una bola $\mathit{\text{B}}\in {\mathbb{R}}^4$ de centro $P$ las funciones $F$ y $G$ tienen (sus cuatro) derivadas parciales continuas. Si el Jacobiano ${\displaystyle \frac{\partial (F,G)}{\partial (u,v)}(P)\ne 0}$ entonces las expresiones $F(x,y,u,v)=0$ y $G(x,y,u,v)=0$ definen funciones (implícitas) $u={\phi }_1(x,y)$ y $v={\phi }_2(x,y)$ definidas en una vecindad $v$ de $(x,y)$ las cuales tienen derivadas parciales continuas en $v$

Dadas las funciones $F$ y $G$ de las variables $u,v,x,y$ nos preguntamos cuando de las expresiones

$F(x,y,u,v)=0$
$G(x,y,u,v)=0$

podemos despejar a $u$ y $v$ en términos de $x$ y $y$ en caso de ser posible diremos que las funciones $u={\phi }_1(x,y)$ y $v={\phi }_2(x,y)$ son funciones implícitas dadas. Se espera que ${\mathrm{\exists }}^{\prime }$n funciones $u={\phi }_1(x,y)$ y
$v={\phi }_2(x,y)$ en

$F(x,y,{\phi }_1(x,y),{\phi }_2(x,y)$
$G(x,y,{\phi }_1(x,y),{\phi }_2(x,y)$

con $(x,y)$ en alguna vecindad $V$

Suponiendo que existen ${\phi }_1$ y ${\phi }_2$ veamos sus derivadas

${\displaystyle \frac{\partial F}{\partial x}{\displaystyle \frac{\partial x}{\partial x}+{\displaystyle \frac{\partial F}{\partial y}{\displaystyle \frac{\partial y}{\partial x}+{\displaystyle \frac{\partial F}{\partial u}{\displaystyle \frac{\partial u}{\partial x}+{\displaystyle \frac{\partial F}{\partial v}{\displaystyle \frac{\partial v}{\partial x}=0}}}}}}}}$ $$ $\Rightarrow$ $$ ${\displaystyle \frac{\partial F}{\partial u}{\displaystyle \frac{\partial u}{\partial x}+{\displaystyle \frac{\partial F}{\partial v}{\displaystyle \frac{\partial v}{\partial x}=-{\displaystyle \frac{\partial F}{\partial x}}}}}}$

${\displaystyle \frac{\partial G}{\partial x}{\displaystyle \frac{\partial x}{\partial x}+{\displaystyle \frac{\partial G}{\partial y}{\displaystyle \frac{\partial y}{\partial x}+{\displaystyle \frac{\partial G}{\partial u}{\displaystyle \frac{\partial u}{\partial x}+{\displaystyle \frac{\partial G}{\partial v}{\displaystyle \frac{\partial v}{\partial x}=0}}}}}}}}$ $$ $\Rightarrow$ $$ ${\displaystyle \frac{\partial G}{\partial u}{\displaystyle \frac{\partial u}{\partial x}+{\displaystyle \frac{\partial G}{\partial v}{\displaystyle \frac{\partial v}{\partial x}=-{\displaystyle \frac{\partial G}{\partial x}}}}}}$

Lo anterior se puede ver como un sistema de 2 ecuaciones con 2 incógnitas ${\displaystyle \frac{\partial u}{\partial x}}$ y ${\displaystyle \frac{\partial v}{\partial x}}$. Aquí se ve que para que el sistema tenga solución.

$det\left|\begin{array}{cc}{\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial F}{\partial v}}\\ {\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial G}{\partial v}}\end{array}\right|\ne 0$ en $(P)$ (el $det$ Jacobiano) y según la regla de Cramer

${\displaystyle \frac{\partial u}{\partial x}=\frac{det\left|\begin{array}{cc}{\displaystyle -\frac{\partial F}{\partial x}}& {\displaystyle \frac{\partial F}{\partial v}}\\ {\displaystyle -\frac{\partial G}{\partial x}}& {\displaystyle \frac{\partial G}{\partial v}}\end{array}\right|}{det\left|\begin{array}{cc}{\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial F}{\partial v}}\\ {\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial G}{\partial v}}\end{array}\right|}=-\frac{\frac{\partial (F,G)}{\partial (x,v)}}{\frac{\partial (F,G)}{\partial (u,v)}}}$, ${\displaystyle \frac{\partial v}{\partial x}=\frac{det\left|\begin{array}{cc}{\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle -\frac{\partial F}{\partial x}}\\ {\displaystyle \frac{\partial G}{\partial u}}& {\displaystyle -\frac{\partial G}{\partial x}}\end{array}\right|}{det\left|\begin{array}{cc}{\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial F}{\partial v}}\\ {\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial G}{\partial v}}\end{array}\right|}=-\frac{\frac{\partial (F,G)}{\partial (u,x)}}{\frac{\partial (F,G)}{\partial (u,v)}}}$.

Análogamente si derivamos con respecto a $y$ obtenemos

${\displaystyle \frac{\partial F}{\partial u}{\displaystyle \frac{\partial u}{\partial y}+{\displaystyle \frac{\partial F}{\partial v}{\displaystyle \frac{\partial v}{\partial y}={\displaystyle \frac{\partial F}{\partial y}}}}}}$

${\displaystyle \frac{\partial G}{\partial u}{\displaystyle \frac{\partial u}{\partial y}+{\displaystyle \frac{\partial G}{\partial v}{\displaystyle \frac{\partial v}{\partial y}={\displaystyle \frac{\partial G}{\partial y}}}}}}$

de donde

${\displaystyle \frac{\partial u}{\partial y}=-\frac{det\left|\begin{array}{cc}{\displaystyle -\frac{\partial F}{\partial y}}& {\displaystyle \frac{\partial F}{\partial v}}\\ {\displaystyle -\frac{\partial G}{\partial y}}& {\displaystyle \frac{\partial G}{\partial v}}\end{array}\right|}{det\left|\begin{array}{cc}{\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial F}{\partial v}}\\ {\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial G}{\partial v}}\end{array}\right|}=-\frac{\frac{\partial (F,G)}{\partial (y,v)}}{\frac{\partial (F,G)}{\partial (u,v)}}}$, ${\displaystyle \frac{\partial v}{\partial y}=-\frac{det\left|\begin{array}{cc}{\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle -\frac{\partial F}{\partial y}}\\ {\displaystyle \frac{\partial G}{\partial u}}& {\displaystyle -\frac{\partial G}{\partial y}}\end{array}\right|}{det\left|\begin{array}{cc}{\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial F}{\partial v}}\\ {\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial G}{\partial v}}\end{array}\right|}=-\frac{\frac{\partial (F,G)}{\partial (u,y)}}{\frac{\partial (F,G)}{\partial (u,v)}}}$.

Al determinante $det\left|\begin{array}{cc}{\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial F}{\partial v}}\\ {\displaystyle \frac{\partial G}{\partial u}}& {\displaystyle \frac{\partial G}{\partial v}}\end{array}\right|$ lo llamamos Jacobiano y lo denotamos por ${\displaystyle \frac{\partial (F,G)}{\partial (u,v)}}$.

Ejemplo. Analizar la solubilidad del sistema
$$e^u+e^v=x+ye$$
$$u{e}^u+v{e}^v=xye$$
$Solución$ En este caso definimos
$$F(x,y,u,v)=e^u+e^v-x-ye=0$$
$$G(x,y,u,v)=u{e}^u+v{e}^v-xye=0$$
por lo que el sistema tendra solución si ${\displaystyle det\left|\begin{array}{cc}{\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial F}{\partial v}}\\ {\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial G}{\partial v}}\end{array}\right|\ne 0}$

En este caso
$$det\left|\begin{array}{cc}{\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial F}{\partial v}}\\ {\displaystyle \frac{\partial F}{\partial u}}& {\displaystyle \frac{\partial G}{\partial v}}\end{array}\right|=det\left|\begin{array}{cc}{\displaystyle e^u}& {\displaystyle e^v}\\ u{e}^u+e^{e^u}& v{e}^v+e^v\end{array}\right|=e^u(v{e}^v+e^v)-e^v(u{e}^u+e^u)=v{e}^{u+v}-u{e}^{v+u}\ne 0$$
por lo tanto u y v se pueden ver en términos de x,y $\therefore$ se pueden calcular sus parciales en $u=0,v=1,x=1,y=1$ que es este caso dan
$${\displaystyle \frac{\partial u}{\partial x}=-\frac{det\left|\begin{array}{cc}-1& -ye\\ e^v& v{e}^v+e^v\end{array}\right|}{v{e}^{u+v}-u{e}^{v+u}}=-\frac{-(v{e}^v+e^v)+e^{v}ye}{v{e}^{u+v}-u{e}^{v+u}}|(0,1,1,1)=\frac{2e-e^2}{e}=2-e}$$ $${\displaystyle \frac{\partial v}{\partial x}=-\frac{det\left|\begin{array}{cc}{e}^u& u{e}^u+e^u\\ -1& -ye\end{array}\right|}{v{e}^{u+v}-u{e}^{v+u}}=-\frac{-y{e}^{u}e+u{e}^u+e^u}{v{e}^{u+v}-u{e}^{v+u}}|(0,1,1,1)=\frac{e-1}{e}=1-e^{-1}}$$
$${\displaystyle \frac{\partial u}{\partial y}=-\frac{det\left|\begin{array}{cc}-e& -xe\\ e^v& v{e}^v+e^v\end{array}\right|}{v{e}^{u+v}-u{e}^{v+u}}=-\frac{-e(v{e}^v+e^v)+e^{v}xe}{v{e}^{u+v}-u{e}^{v+u}}|(0,1,1,1)=\frac{e^2+e^2-e^2}{e}=e}$$ $${\displaystyle \frac{\partial v}{\partial y}=-\frac{det\left|\begin{array}{cc}{e}^u& u{e}^u+e^u\\ -e& -xe\end{array}\right|}{v{e}^{u+v}-u{e}^{v+u}}=-\frac{-e^{u}xe+e(u{e}^u+e^u)}{v{e}^{u+v}-u{e}^{v+u}}|(0,1,1,1)=\frac{e-e}{e}=0}$$

Teorema de la Función Implícita (n-sistemas de ecuaciones

Considere las n-funciones
$$u_i=F_i(x_1,\dots ,x_m,y_1,\dots ,y_n),i=1,\dots ,n$$ Sea $P=(\bar{x}1,\dots ,\bar{x}m,\bar{y}1,\dots ,\bar{y}n)\in {\mathbb{R}}^{n+m}$ un punto tal que $F_i(P)=0$. Suponga que en una bola $\mathit{\text{B}}\in {\mathbb{R}}^{n+m}$ de centro $P$ las funciones $F_i$ tienen (sus $m+n$) derivadas parciales continuas. Si el Jacobiano $$\frac{\partial (F_1,F_2,\dots ,F_n)}{\partial (y_1,y_2,\dots ,y_n)}=\left|\begin{array}{cccc}\frac{\partial F_1}{\partial y_1}& \frac{\partial F_1}{\partial y_2}& \cdots & \frac{\partial F_1}{\partial y_n}\\ \frac{\partial F_2}{\partial y_1}& \frac{\partial F_2}{\partial y_2}& \cdots & \frac{\partial F_2}{\partial y_n}\\ ⋮& ⋮& \ddots & ⋮\\ \frac{\partial F_n}{\partial y_1}& \frac{\partial F_n}{\partial y_2}& \cdots & \frac{\partial F_n}{\partial y_n}\end{array}\right|\ne 0enP$$

entonces las expresiones
$F_i(x_1,\dots ,x_m,y_1,\dots ,y_n)=0$ y $G(x,y,u,v)=0$ definen funciones (implícitas)
$y_i={\phi }_i(x_1,\dots ,x_m),i=1,\dots ,n$ definidas en una vecindad $v$ de $(\bar{x}1,\dots ,\bar{x}m)$ las cuales tienen derivadas parciales
continuas en $v$ que se pueden calcular como
$$\frac{\partial y_i}{\partial x_j}=\frac{\frac{\partial (F_1,F_2,\dots ,F_n)}{\partial (y_1,\dots ,y_{i-1},x_j,y_{i+1},\dots ,y_n)}}{\frac{\partial (F_1,F_2,\dots ,F_n)}{\partial (y_1,y_2,\dots ,y_n)}}$$

Ejemplo. Considere las ecuaciones
$$\begin{array}{c}F(x,y,u,v,w)=x+y+u+v+w=0\\ G(x,y,u,v,w)=x^2-y^2+u^2-2v^2+w^2+1=0\\ H(x,y,u,v,w)=x^3+y^3+u^4-3v^4+8w^4+2=0\end{array}$$

En el punto $P=(1,-1,1,-1,0)$, se tiene $F(P)=G(P)=H(P)=0$. Todas las derivadas parciales de F, G, H son continuas. Se tiene además que
$$\frac{\partial (F,G,H)}{\partial (u,v,w)}=det{\left|\begin{array}{ccc}1& 1& 1\\ 2u& -4v& 2w\\ 4u^3& -12v^3& 32w^2\end{array}\right|}_{\begin{array}{c}u=1\\ v=-1\\ w=0\end{array}}=8\ne 0$$
Entonces el teorema asegura que en torno a P podemos despejar $u,v,w$ en términos de $x,y$ y establecer funciones
$$u=u(x,y),v=v(x,y),w=w(x,y)$$
las cuales tienen derivadas parciales continuas en una vecindad de $(1,-1)$ que se pueden calcular

$$\frac{\partial u}{\partial x}=-\frac{\frac{\partial (F,G,H)}{\partial (x,v,w)}}{\frac{\partial (F,G,H)}{\partial (u,v,w)}},\frac{\partial u}{\partial y}=-\frac{\frac{\partial (F,G,H)}{\partial (y,v,w)}}{\frac{\partial (F,G,H)}{\partial (u,v,w)}}$$

$$\frac{\partial v}{\partial x}=-\frac{\frac{\partial (F,G,H)}{\partial (u,x,w)}}{\frac{\partial (F,G,H)}{\partial (u,v,w)}},\frac{\partial v}{\partial y}=-\frac{\frac{\partial (F,G,H)}{\partial (u,y,w)}}{\frac{\partial (F,G,H)}{\partial (u,v,w)}}$$

$$\frac{\partial w}{\partial x}=-\frac{\frac{\partial (F,G,H)}{\partial (u,v,x)}}{\frac{\partial (F,G,H)}{\partial (u,v,w)}},\frac{\partial w}{\partial y}=-\frac{\frac{\partial (F,G,H)}{\partial (u,v,y)}}{\frac{\partial (F,G,H)}{\partial (u,v,w)}}$$