$\textcolor{Red}{\textbf{Diferenciación de funciones $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{m}$}}$

$\textbf{Definición.-}$ Considere la función $f:A\subset\mathbb{R}^{n}\rightarrow\mathbb{R}^{m}$ definida en un conjunto abierto A de $\mathbb{R}^{n}$ y sea $x_{0}\in A$. Se dice que esta función es diferenciable si

$$f(x_{0}+h)=f(x_{0})+f'(x_{0})\cdot h+r(h)$$
cumple
$$\lim_{h\rightarrow0}\frac{r(h)}{|h|}=\hat{0}$$

$\textbf{Ejemplo:}$ Compruebe que la función $f:\mathbb{R}^{2}\rightarrow \mathbb{R}^{3}$ definida por
$$f(x,y)=\left(e^{xy},x^{2}+y,2x^{3}y^{2}\right)$$ es diferenciable en $(1,3)$
$\textbf{Solución}$ En este caso
$$\lim_{h\rightarrow0}\frac{r(h)}{|h|}=$$
$$\lim_{(h_{1},h_{2})\rightarrow(0,0)}\frac{f(1+h_{1},3+h_{2})-f(1,3)-\left((3e^{3},e^{3})\cdot\binom{h_{1}}{h_{2}},(2,1)\cdot\binom{h_{1}}{h_{2}},(54,12)\cdot\binom{h_{1}}{h_{2}}\right)}{|(h_{1},h_{2})|}$$
$$=\lim_{(h_{1},h_{2})\rightarrow(0,0)}\frac{\left(e^{(1+h_{1})(3+h_{2})},(1+h_{1})^{2}+(3+h_{2}),2(1+h_{1})^{3}(3+h_{2})^{2}\right)-\left(e^{3},4,18\right)}{|(h_{1},h_{2})|}$$
$$\frac{-\left((3e^{3},e^{3})\cdot\binom{h_{1}}{h_{2}},(2,1)\cdot\binom{h_{1}}{h_{2}},(54,12)\cdot\binom{h_{1}}{h_{2}}\right)}{|(h_{1},h_{2})|}$$
$$=\left(\lim_{(h_{1},h_{2})\rightarrow(0,0)}\frac{e^{(1+h_{1})(3+h_{2})}-e^{3}-3e^{3}h_{1}-e^{3}h_{2}}{|(h_{1},h_{2})|},\lim_{(h_{1},h_{2})\rightarrow(0,0)}\frac{(1+h_{1})^{2}+(3+h_{2})-4-2h_{1}-h_{2}}{|(h_{1},h_{2})|},\right.$$
$$\left.\lim_{(h_{1},h_{2})\rightarrow(0,0)}\frac{2(1+h_{1})^{3}(3+h_{2})^{2}-18-54h_{1}-12h_{2}}{|(h_{1},h_{2})|}\right)$$
$$=(0,0,0)$$
por lo que la función es diferenciable.

En el ejemplo anterior se tiene que
$$\lim_{(h_{1},h_{2})\rightarrow(0,0)}\frac{f(1+h_{1},3+h_{2})-f(1,3)-\left((3e^{3},e^{3})\cdot\binom{h_{1}}{h_{2}},(2,1)\cdot\binom{h_{1}}{h_{2}},(54,12)\cdot\binom{h_{1}}{h_{2}}\right)}{|(h_{1},h_{2})|}$$
se puede expresar
$$\lim_{(h_{1},h_{2})\rightarrow (0,0)}\frac{f(1+h_{1},3+h_{2})-f(1,3)-\left( \begin{matrix}
3e^{3}&e^{3}\\
2&1\\ 54&12
\end{matrix}\right)\left[ \begin{array}{ll}
h_{1}\\ h_{2} \end{array}\right]}{\sqrt{h_{1}^{2}+h_{2}^{2}}}$$
lo que nos lleva a la siguiente definición.

$\textbf{Definición.}$ A la matriz de $m\times n$ se le llama Matriz Jacobiana de la función $f$ en $x_{0}$ y se le denota $Jf(x_{0})$.

$\textbf{Definición}$ Sea $f:\mathbb{R}^{n}\rightarrow \mathbb{R}^{m}$ definida en el abierto $\Omega$ de $\mathbb{R}^{n}$ y $x_{0}\in \Omega$. Se dice que $f$ es diferenciable en $x_{0}\in\Omega$ si y solo si existe una matriz T de $m\times n$ tal que $$\lim_{h\rightarrow 0}\frac{f(x_{0}+h)-f(x_{0})-T(x_{0})\cdot h}{|h|}=0$$donde $T(x_{0})$ es la matriz jacobiana denotada por $Jf(x_{0})$ ó $Df(x_{0})$. En notación matricial:
$$T(x_{0})\cdot h=\left( \begin{matrix}
\frac{\partial f_{1}(x_{0})}{\partial x_{1}}&\cdot\cdot\cdot& \frac{\partial f_{1}(x_{0})}{\partial x_{n}}\\
\cdot&\cdot&\cdot\\
\cdot&\cdot&\cdot\\
\cdot&\cdot&\cdot\\
\frac{\partial f_{m}(x_{0})}{\partial x_{1}}&\cdot\cdot\cdot&\frac{\partial f_{m}(x_{0})}{\partial x_{n}}
\end{matrix}\right)\cdot \left( \begin{matrix}
h_{1}\\
\cdot\\
\cdot\\
\cdot\
h_{n}
\end{matrix}\right)$$

En términos $\epsilon-\delta$ se tiene que si $0<|h|<\delta$ entonces $$\frac{|f(x_{0}+h)-f(x_{0})-T(x_{0})\cdot h|}{|h|}<\epsilon$$

$\textbf{Teorema. 1}$ Supónga que $f:\Omega\subset\mathbb{R}^{n}\rightarrow \mathbb{R}^{m}$ es diferenciable en $x_{0}\in \mathbb{R}^{n}$. Entonces la matriz $T$ es única

$Demostración$ Supongamos que existen $T_{1}$ y $T_{2}$ que cumplen $$\lim_{h\rightarrow 0}\frac{f(x_{0}+h)-f(x_{0})-T_{1}(x_{0})\cdot h}{|h|}=0\quad y\quad \lim_{h\rightarrow 0}\frac{f(x_{0}+h)-f(x_{0})-T_{2}(x_{0})\cdot h}{|h|}=0$$
$\therefore$
$$\lim_{h\rightarrow 0}\frac{f(x_{0}+h)-f(x_{0})-T_{1}(x_{0})\cdot h}{|h|}-\frac{f(x_{0}+h)-f(x_{0})-T_{2}(x_{0})\cdot h}{|h|}=0$$
$$\Rightarrow \lim_{h\rightarrow 0}\frac{T_{2}(x_{0})\cdot h-T_{1}(x_{0})\cdot h}{|h|}=0$$
Sea $x$ un vector unitario en la dirección del vector $h$ y hacemos $h=tx$ con $t\in\mathbb{R}$
$\therefore$
$$ 0=\lim_{h\rightarrow 0}\frac{T_{2}(x_{0})\cdot h-T_{1}(x_{0})\cdot h}{|h|}= \lim_{t\rightarrow 0}\frac{T_{2}(x_{0})\cdot tx-T_{1}(x_{0})\cdot tx}{|tx|}=\lim_{t\rightarrow 0}t\frac{T_{2}(x_{0})\cdot x-T_{1}(x_{0})\cdot x}{|t||x|}$$
$$ \Rightarrow 0=T_{2}(x_{0})\cdot x-T_{1}(x_{0})\cdot x \Rightarrow T_{2}(x_{0})\cdot x=T_{1}(x_{0})\cdot x\Rightarrow T_{2}(x_{0})=T_{1}(x_{0})\Rightarrow T_{2}=T_{1}$$ $\square$

$\textcolor{Red}{\textbf{Operadores: Divergencia, Rotacional y Laplaciano}}$

Considere la función $f:\mathbb{R}^{3}\rightarrow\mathbb{R}^{3}$ dada por
$$f(x,y,z)=\left(f_{1}(x,y,z),f_{2}(x,y,z),f_{3}(x,y,z)\right)$$
cuya matriz jacobiana es
$$\left( \begin{matrix}
\frac{\partial f_{1}}{\partial x}&\frac{\partial f_{1}}{\partial y}&\frac{\partial f_{1}}{\partial z}\\
\frac{\partial f_{2}}{\partial x}&\frac{\partial f_{2}}{\partial y}&\frac{\partial f_{2}}{\partial z}\\
\frac{\partial f_{3}}{\partial x}&\frac{\partial f_{3}}{\partial y}&\frac{\partial f_{3}}{\partial z}\
\end{matrix}\right)$$
Con los elementos de esta matriz se forman importantes combinaciones que son la divergencia y el rotacional, conocidos también como invariantes de primer orden de esta matriz. La razón por la que se llaman invariantes es porque el valor de dichas combinaciones no se altera al efectuar un cambio de coordenadas.

$\textbf{Definición}$ Dada la matriz Jacobiana $$\left( \begin{matrix}
\textcolor{Red}{\frac{\partial f_{1}}{\partial x}}&\frac{\partial f_{1}}{\partial y}&\frac{\partial f_{1}}{\partial z}\\
\frac{\partial f_{2}}{\partial x}&\textcolor{Red}{\frac{\partial f_{2}}{\partial y}}&\frac{\partial f_{2}}{\partial z}\\
\frac{\partial f_{3}}{\partial x}&\frac{\partial f_{3}}{\partial y}&\textcolor{Red}{\frac{\partial f_{3}}{\partial z}}\
\end{matrix}\right)$$
Se define la divergencia de f como
$$div~(f)=\textcolor{Red}{\frac{\partial f_{1}}{\partial x}+\frac{\partial f_{2}}{\partial y}+\frac{\partial f_{3}}{\partial z}}$$
Se puede ver que es igual a la suma de los elementos de la diagonal principal, es decir, que constituye la traza de la matriz jacobiana de f. La defnición de divergencia puede darse también mediante el operador $\nabla$ (nabla)
$$\nabla\cdot f=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\cdot \left(f_{1},f_{2},f_{3}\right)=\frac{\partial f_{1}}{\partial x}+\frac{\partial f_{2}}{\partial y}+\frac{\partial f_{3}}{\partial z}$$

Dada la matriz Jacobiana
$$\left( \begin{matrix}
\frac{\partial f_{1}}{\partial x}&\textcolor{Green}{\frac{\partial f_{1}}{\partial y}}&\textcolor{Red}{\frac{\partial f_{1}}{\partial z}}\
\textcolor{Green}{\frac{\partial f_{2}}{\partial x}}&\frac{\partial f_{2}}{\partial y}&\textcolor{Blue}{\frac{\partial f_{2}}{\partial z}}\
\textcolor{Red}{\frac{\partial f_{3}}{\partial x}}&\textcolor{Blue}{\frac{\partial f_{3}}{\partial y}}&\frac{\partial f_{3}}{\partial z}\
\end{matrix}\right)$$
se define el rotacional como
$$rot~(f)=\left(\textcolor{Blue}{\frac{\partial f_{3}}{\partial y}-\frac{\partial f_{2}}{\partial z}},\textcolor{Red}{\frac{\partial f_{3}}{\partial x}-\frac{\partial f_{1}}{\partial z}},\textcolor{Green}{\frac{\partial f_{2}}{\partial x}-\frac{\partial f_{1}}{\partial y}}\right)$$
Y podemos ver que las componentes del rotacional están definidas por las diferencias de los elementos situados simétricamente con respecto a la diagonal principal de la matriz Jacobiana.
La defnición de rotacional puede darse también mediante el operador $\nabla$ (nabla)
$$\nabla\times f=\left|\begin{matrix}\hat{i}&\hat{j}&\hat{k}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\f_{1}&f_{2}&f_{3}\end{matrix}\right|=\left(\frac{\partial f_{3}}{\partial y}-\frac{\partial f_{2}}{\partial z},\frac{\partial f_{3}}{\partial x}-\frac{\partial f_{1}}{\partial z},\frac{\partial f_{2}}{\partial x}-\frac{\partial f_{1}}{\partial y}\right)$$

$\textbf{Definición.}$ Sea $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$ definida en el abierto $\Omega$ de $\mathbb{R}^{n}$ tal que $f$ es de clase $C^{2}$ en $\Omega$. La expresión
$$\nabla^{2}f=\frac{\partial^{2}f}{\partial x^{2}}+\frac{\partial^{2}f}{\partial y^{2}}$$
es llamada Laplaciano de f. La ecuación
$$\frac{\partial^{2}f}{\partial x^{2}}+\frac{\partial^{2}f}{\partial y^{2}}=0$$
es llamada la ecuación de Laplace. Las funciones f de clase $C^{2}$ que cumplen la ecuación de Laplace se llaman funciones Armónicas.

$\textbf{Ejercicio.}$ Sean $f,g:A\subset\mathbb{R}^{3}\rightarrow\mathbb{R}^{3}$ dos funciones diferenciables en una región $A\subset\mathbb{R}^{3}$ y $\varphi:\mathbb{R}^{3}\rightarrow\mathbb{R}$
Pruebe que

(a) $div~(f+g)=div~f+div~g$

(b) $div~(\varphi f)=\varphi~div~(f)+grad~(\varphi)\cdot f$
(c) $rot~(f+g)=rot~(f)+rot~g$
(d) Sean $\phi,\psi:A\subset \mathbb{R}^{2}\rightarrow\mathbb{R}$ dos funciones $\psi=\psi(x,y),~\phi=\phi(x,y)$ tales que $$\frac{\partial \phi}{\partial x}=\frac{\partial \psi}{\partial y},~~\frac{\partial \phi}{\partial y}=-\frac{\partial \psi}{\partial x}$$
Demuestre que $\phi,~\psi$ son armónicas.

$Solución$ Para el inciso (a) se tiene
$$div~(f+g)=\nabla\cdot (f+g)$$
$$=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\cdot (f+g)$$
$$=\frac{\partial}{\partial x}(f+g)+\frac{\partial}{\partial y}(f+g)+\frac{\partial}{\partial z}(f+g)$$
$$=\frac{\partial}{\partial x}f+\frac{\partial}{\partial x}g+\frac{\partial}{\partial y}f+\frac{\partial}{\partial y}g+\frac{\partial}{\partial z}f+\frac{\partial}{\partial z}g$$
$$=\frac{\partial}{\partial x}f+\frac{\partial}{\partial y}f+\frac{\partial}{\partial z}f+\frac{\partial}{\partial x}g+\frac{\partial}{\partial y}g+\frac{\partial}{\partial z}g$$
$$=\nabla\cdot f+\nabla\cdot g$$

Para el inciso (a) se tiene
$$div~(f+g)=\nabla\cdot (f+g)$$
$$=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\cdot (f+g)$$
$$=\frac{\partial}{\partial x}(f+g)+\frac{\partial}{\partial y}(f+g)+\frac{\partial}{\partial z}(f+g)$$
$$=\frac{\partial}{\partial x}f+\frac{\partial}{\partial x}g+\frac{\partial}{\partial y}f+\frac{\partial}{\partial y}g+\frac{\partial}{\partial z}f+\frac{\partial}{\partial z}g$$
$$=\frac{\partial}{\partial x}f+\frac{\partial}{\partial y}f+\frac{\partial}{\partial z}f+\frac{\partial}{\partial x}g+\frac{\partial}{\partial y}g+\frac{\partial}{\partial z}g$$
$$=\nabla\cdot f+\nabla\cdot g$$

Para el inciso (c) se tiene
$$\nabla\times (f+g)=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\times(f_{1}+g_{1},f_{2}+g_{2},f_{3}+g_{3})$$
$$=\left(\frac{\partial (f_{3}+g_{3})}{\partial y}-\frac{\partial (f_{2}+g_{2})}{\partial z},\frac{\partial (f_{3}+g_{3})}{\partial x}-\frac{\partial (f_{1}+g_{1})}{\partial z},\frac{\partial (f_{2}+g_{2})}{\partial x}-\frac{\partial (f_{1}+g_{1})}{\partial y}\right)$$
$$=\left[\left(\frac{\partial f_{3}}{\partial y}-\frac{\partial f_{2}}{\partial z}\right)+\left(\frac{\partial g_{3}}{\partial y}-\frac{\partial g_{2}}{\partial z}\right),\left(\frac{\partial f_{3}}{\partial x}-\frac{\partial f_{1}}{\partial z}\right)+\left(\frac{\partial g_{3}}{\partial x}-\frac{\partial g_{1}}{\partial z}\right),\left(\frac{\partial f_{2}}{\partial x}-\frac{\partial f_{1}}{\partial y}\right)+\left(\frac{\partial g_{2}}{\partial x}-\frac{\partial g_{1}}{\partial y}\right)\right]$$
$$=\nabla\times f+\nabla\times g$$

Para el inciso (d) se tiene
$$\frac{\partial \phi}{\partial x}=\frac{\partial \psi}{\partial y},~~~~\frac{\partial \phi}{\partial y}=-\frac{\partial \psi}{\partial x}$$
por lo que
$$\frac{\partial^{2} \phi}{\partial x^{2}}=\frac{\partial}{\partial x}\left(\frac{\partial \psi}{\partial y}\right)$$
$$=\frac{\partial^{2}\psi}{\partial x\partial y}$$
por otro lado
$$\frac{\partial^{2} \phi}{\partial y^{2}}=\frac{\partial}{\partial y}\left(-\frac{\partial \psi}{\partial x}\right)$$
$$=-\frac{\partial^{2}\psi}{\partial y\partial x}$$
por lo tanto
$$\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}=\frac{\partial^{2}\psi}{\partial x\partial y}-\frac{\partial^{2}\psi}{\partial y\partial x}=0$$
Analogamente se tiene
$$\frac{\partial^{2} \psi}{\partial x^{2}}=\frac{\partial}{\partial x}\left(-\frac{\partial \phi}{\partial y}\right)$$
$$=-\frac{\partial^{2}\phi}{\partial x\partial y}$$
por otro lado
$$\frac{\partial^{2} \psi}{\partial y^{2}}=\frac{\partial}{\partial y}\left(\frac{\partial \phi}{\partial x}\right)$$
$$=\frac{\partial^{2}\phi}{\partial y\partial x}$$
por lo tanto
$$\frac{\partial^{2} \psi}{\partial x^{2}}+\frac{\partial^{2} \psi}{\partial y^{2}}=-\frac{\partial^{2}\phi}{\partial x\partial y}+\frac{\partial^{2}\phi}{\partial y\partial x}=0$$
como las funciones $\psi,~\phi$ satisfacen la ecuación de Laplace entonces ambas funciones son armónicas.