Cálculo vectorial

Decimos que una función ${\displaystyle \mathbf {f} }$ es continua en ${\displaystyle \mathbf {a} \Leftrightarrow \lim _{\mathbf {x} \to \mathbf {a} }\mathbf {f} {\big (}\mathbf {x} {\big )}=\mathbf {f} {\big (}\mathbf {a} {\big )}}$

${\displaystyle \lim _{\mathbf {x} \to \mathbf {a} }\mathbf {f} {\big (}\mathbf {x} {\big )}=\mathbf {b} ,\lim _{\mathbf {x} \to \mathbf {a} }\mathbf {g} {\big (}\mathbf {x} {\big )}=\mathbf {c} \Rightarrow }$

a) ${\displaystyle \lim _{\mathbf {x} \to \mathbf {a} }{\big [}\mathbf {f} +\mathbf {g} {\big ]}{\big (}\mathbf {x} {\big )}=\mathbf {b} +\mathbf {c} }$

b) ${\displaystyle \lim _{\mathbf {x} \to \mathbf {a} }\lambda \mathbf {f} {\big (}\mathbf {x} {\big )}=\lambda \mathbf {b} \quad \forall \lambda \in \mathbb {R} }$

c) ${\displaystyle \lim _{\mathbf {x} \to \mathbf {a} }{\big (}\mathbf {f} \cdot \mathbf {g} {\big )}{\big (}\mathbf {x} {\big )}=\mathbf {b} \cdot \mathbf {c} }$

(producto escalar de ${\displaystyle \mathbf {b} }$ con ${\displaystyle \mathbf {c} }$).

d) ${\displaystyle \lim _{\mathbf {x} \to \mathbf {a} }{\Big \|}\mathbf {f} {\big (}\mathbf {x} {\big )}{\Big \|}={\big \|}\mathbf {b} {\big \|}}$

${\displaystyle f}$

${\displaystyle g}$

${\displaystyle \mathbf {b} ={\big (}b_{1},\ldots ,b_{m}{\big )},\mathbf {c} ={\big (}c_{1},\ldots ,c_{m}{\big )}}$ tenemos

${\displaystyle {\begin{array}{rl}a)&\mathbf {f} {\big (}\mathbf {x} )={\big [}f_{1}{\big (}\mathbf {x} {\big )},\ldots ,f_{m}{\big (}\mathbf {x} {\big )}{\big ]},\mathbf {g} {\big (}\mathbf {x} )={\Big [}g_{1}{\big (}\mathbf {x} {\big )},\ldots ,g_{m}{\big (}\mathbf {x} {\big )}{\Big ]}\\&\lim _{\mathbf {x} \to \mathbf {a} }{\big (}\mathbf {f} +\mathbf {g} {\big )}{\big (}\mathbf {x} {\big )}=\lim _{\mathbf {x} \to \mathbf {a} }{\Big [}{\big (}f_{1}+g_{1}{\big )}{\big (}\mathbf {x} {\big )},\ldots ,{\big (}f_{m}+g_{m}{\big )}{\big (}\mathbf {x} {\big )}{\Big ]}=\\&{\Big [}\lim _{\mathbf {x} \to \mathbf {a} }{\big (}f_{1}+g_{1}{\big )}{\big (}\mathbf {x} {\big )},\ldots ,\lim _{\mathbf {x} \to \mathbf {a} }{\big (}f_{m}+g_{m}{\big )}{\big (}\mathbf {x} {\big )}{\Big ]}=\\&{\Big [}\lim _{\mathbf {x} \to \mathbf {a} }f_{1}{\big (}\mathbf {x} {\big )}+\lim _{\mathbf {x} \to \mathbf {a} }g_{1}(\mathbf {x} {\big )},\ldots ,\lim _{\mathbf {x} \to \mathbf {a} }f_{m}{\big (}\mathbf {x} {\big )}+\lim _{\mathbf {x} \to \mathbf {a} }g_{m}{\big (}\mathbf {x} {\big )}{\Big ]}=\\&{\big (}b_{1}+c_{1},\ldots ,b_{m}+c_{m}{\big )}={\big (}b_{1},\ldots ,b_{m}{\big )}+{\big (}c_{1},\ldots ,c_{m}{\big )}=\mathbf {b} +\mathbf {c} \end{array}}}$

${\displaystyle {\begin{array}{rl}b)&\lim _{\mathbf {x} \to \mathbf {a} }\lambda \mathbf {f} {\big (}\mathbf {x} {\big )}=\lim _{\mathbf {x} \to \mathbf {a} }\lambda {\Big [}f_{1}{\big (}\mathbf {x} {\big )},\ldots ,f_{m}{\big (}\mathbf {x} {\big )}{\Big ]}=\lim _{\mathbf {x} \to \mathbf {a} }{\Big [}\lambda f_{1}{\big (}\mathbf {x} {\big )},\ldots ,\lambda f_{m}{\big (}\mathbf {x} {\big )}{\Big ]}=\\&{\Big [}\lim _{\mathbf {x} \to \mathbf {a} }\lambda f_{1}{\big (}\mathbf {x} {\big )},\ldots ,\lim _{\mathbf {x} \to \mathbf {a} }\lambda f_{m}{\big (}\mathbf {x} {\big )}{\Big ]}={\Big [}\lambda \lim _{\mathbf {x} \to \mathbf {a} }f_{1}{\big (}\mathbf {x} {\big )},\ldots ,\lambda \lim _{\mathbf {x} \to \mathbf {a} }f_{m}{\big (}\mathbf {x} {\big )}{\Big ]}=\\&\lambda {\Big [}\lim _{\mathbf {x} \to \mathbf {a} }f_{1}{\big (}\mathbf {x} {\big )},\ldots ,\lim _{\mathbf {x} \to \mathbf {a} }f_{m}{\big (}\mathbf {x} {\big )}{\Big ]}=\lambda {\big (}b_{1},\ldots ,b_{m}{\big )}=\lambda \mathbf {b} \end{array}}}$

${\displaystyle c)\quad {\big (}\mathbf {f} \cdot \mathbf {g} {\big )}{\big (}\mathbf {x} {\big )}-\mathbf {b} \cdot \mathbf {c} ={\Big [}\mathbf {f} {\big (}\mathbf {x} {\big )}-\mathbf {b} {\Big ]}\cdot {\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}-\mathbf {c} {\Big ]}+\mathbf {b} \cdot {\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}-\mathbf {c} {\Big ]}+\mathbf {c} \cdot {\Big [}\mathbf {f} {\big (}\mathbf {x} {\big )}-\mathbf {b} {\Big ]}}$

Aplicando la desigualdad triangular y la desigualdad de Cauchy-Schwarz tenemos

${\displaystyle {\begin{array}{l}{\Big |}{\big (}\mathbf {f} \cdot \mathbf {g} {\big )}{\big (}\mathbf {x} {\big )}-\mathbf {b} \cdot \mathbf {c} {\Big |}\leqslant {\Big \|}\mathbf {f} {\big (}\mathbf {x} {\big )}-\mathbf {b} {\Big \|}\cdot {\Big \|}\mathbf {g} {\big (}\mathbf {x} {\big )}-\mathbf {c} {\Big \|}+{\big \|}\mathbf {b} {\big \|}\cdot {\Big \|}\mathbf {g} {\big (}\mathbf {x} {\big )}-\mathbf {c} {\Big \|}+{\big \|}\mathbf {c} {\big \|}\cdot {\Big \|}\mathbf {f} {\big (}\mathbf {x} {\big )}-\mathbf {b} {\Big \|}\Rightarrow \\0\leqslant \lim _{{\big \|}\mathbf {x} -\mathbf {a} {\big \|}\to 0}{\Big |}{\big (}\mathbf {f} \cdot \mathbf {g} {\big )}{\big (}\mathbf {x} {\big )}-\mathbf {b} \cdot \mathbf {c} {\Big |}\leqslant \lim _{{\big \|}\mathbf {x} -\mathbf {a} {\big \|}\to 0}{\Big \|}\mathbf {f} {\big (}\mathbf {x} {\big )}-\mathbf {b} {\Big \|}\cdot \lim _{{\big \|}\mathbf {x} -\mathbf {a} {\big \|}\to 0}{\Big \|}\mathbf {g} {\big (}\mathbf {x} {\big )}-\mathbf {c} {\Big \|}+\\{\big \|}\mathbf {b} {\big \|}\cdot \lim _{{\big \|}\mathbf {x} -\mathbf {a} {\big \|}\to 0}{\Big \|}\mathbf {g} {\big (}\mathbf {x} {\big )}-\mathbf {c} {\Big \|}+{\big \|}\mathbf {c} {\big \|}\lim _{{\big \|}\mathbf {x} -\mathbf {a} {\big \|}\to 0}{\Big \|}\mathbf {f} {\big (}\mathbf {x} {\big )}-\mathbf {b} {\Big \|}=0\cdot 0+{\big \|}\mathbf {b} {\big \|}\cdot 0+{\big \|}\mathbf {c} {\big \|}\cdot 0=\\0\end{array}}}$

, como queríamos demostrar.

${\displaystyle d)\quad \mathbf {g} {\big (}\mathbf {x} {\big )}=\mathbf {f} {\big (}\mathbf {x} {\big )},\mathbf {c} =\mathbf {b} \Rightarrow \lim _{\mathbf {x} \to \mathbf {a} }{\Big \|}\mathbf {f} {\big (}\mathbf {x} {\big )}{\Big \|}^{2}={\big \|}\mathbf {b} {\big \|}^{2}}$, como queríamos demostrar.

Sean ${\displaystyle \mathbf {f} }$ y ${\displaystyle \mathbf {g} }$ dos funciones tales que la función compuesta ${\displaystyle \mathbf {f} \circ \mathbf {g} }$ está definida en ${\displaystyle \mathbf {a} }$, siendo

${\displaystyle {\big (}\mathbf {f} \circ \mathbf {g} {\big )}{\big (}\mathbf {x} {\big )}=\mathbf {f} {\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}{\Big ]}}$

${\displaystyle \mathbf {g} }$ es continua en ${\displaystyle \mathbf {a} }$ y ${\displaystyle \mathbf {f} }$ es continua en ${\displaystyle \mathbf {g} {\big (}\mathbf {a} {\big )}\Rightarrow {\big (}\mathbf {f} \circ \mathbf {g} {\big )}}$ es continua en ${\displaystyle \mathbf {a} }$.

${\displaystyle \mathbf {y} =\mathbf {g} {\big (}\mathbf {x} {\big )}}$

${\displaystyle \mathbf {b} =\mathbf {g} {\big (}\mathbf {a} {\big )}}$

${\displaystyle {\begin{array}{l}\lim _{{\big \|}\mathbf {x} -\mathbf {a} {\big \|}\to 0}{\Big \|}\mathbf {f} {\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}{\Big ]}-\mathbf {f} {\Big [}\mathbf {g} {\big (}\mathbf {a} {\big )}{\Big ]}{\Big \|}=\lim _{{\big \|}\mathbf {y} -\mathbf {b} {\big \|}\to 0}{\Big \|}\mathbf {f} {\big (}\mathbf {y} {\big )}-\mathbf {f} {\big (}\mathbf {b} {\big )}{\Big \|}=0\Rightarrow \\\lim _{\mathbf {x} \to \mathbf {a} }\mathbf {f} {\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}{\Big ]}=\mathbf {f} {\Big [}\mathbf {g} {\big (}\mathbf {a} {\big )}{\Big ]}\end{array}}}$

como queríamos demostrar.

Sea ${\displaystyle f:S\subseteq \mathbb {R} ^{n}\longrightarrow \mathbb {R} }$. Sea ${\displaystyle \mathbf {x} }$ un vector cuyo origen es el origen de coordenadas y cuyo extremo ${\displaystyle \in S,}$ e ${\displaystyle \mathbf {y} }$ un vector arbitrario de ${\displaystyle \mathbb {R} ^{n}}$. Definimos la derivada de f en ${\displaystyle \mathbf {x} }$ respecto a ${\displaystyle \mathbf {y} }$ como

${\displaystyle f'{\big (}\mathbf {x} ;\mathbf {y} {\big )}=\lim _{h\to 0}{\cfrac {f{\big (}\mathbf {x} +h\mathbf {y} {\big )}-f{\big (}\mathbf {x} {\big )}}{h}}}$

${\displaystyle {\cfrac {\partial f}{\partial x_{k}}}=\lim _{h\to 0}{\cfrac {f{\big (}x_{1},\ldots ,x_{k}+h,\ldots ,x_{n}{\big )}-f{\big (}x_{1},\ldots ,x_{k},\ldots ,x_{n}{\big )}}{h}}}$

Decimos que f es diferenciable en ${\displaystyle \mathbf {a} \Leftrightarrow }$

${\displaystyle \exists f_{L}:\mathbb {R} ^{n}\longrightarrow \mathbb {R} {\Big |}\lim _{{\big \|}\mathbf {v} {\big \|}\to \mathbf {0} }f{\big (}\mathbf {a} +\mathbf {v} {\big )}=f{\big (}\mathbf {a} {\big )}+f_{L}{\big (}\mathbf {v} {\big )}}$.

${\displaystyle f_{L}}$ ha de ser una aplicación lineal, que definimos como la diferencial de f en a.

${\displaystyle f}$ es diferenciable en ${\displaystyle \mathbf {x} }$ con diferencial ${\displaystyle f_{L}{\big (}\mathbf {y} {\big )}\Rightarrow }$

a) ${\displaystyle \exists f'{\big (}\mathbf {x} ;\mathbf {y} {\big )}\quad \forall \mathbf {y} \in \mathbb {R} ^{n}}$

b) ${\displaystyle f'{\big (}\mathbf {x} ;\mathbf {y} {\big )}=\sum _{k=1}^{n}y_{k}{\cfrac {\partial f}{\partial x_{k}}}}$

${\displaystyle {\begin{array}{rl}a)&\mathbf {v} =h\mathbf {y} ,\quad h\in \mathbb {R} ,\\&\lim _{{\big \|}\mathbf {v} {\big \|}\to \mathbf {0} }f{\big (}\mathbf {x} +\mathbf {v} {\big )}=\lim _{{\big \|}\mathbf {v} {\big \|}\to \mathbf {0} }f{\big (}\mathbf {x} +h\mathbf {y} {\big )}=f{\big (}\mathbf {x} {\big )}+f_{L}{\big (}h\mathbf {y} {\big )}=\\&f{\big (}\mathbf {x} {\big )}+hf_{L}{\big (}\mathbf {y} {\big )}\Rightarrow \\&\lim _{h\to 0}{\cfrac {f{\big (}\mathbf {x} +h\mathbf {y} {\big )}-f{\big (}\mathbf {x} {\big )}}{h}}=f'{\big (}\mathbf {x} ;\mathbf {y} {\big )}=f_{L}{\big (}\mathbf {y} {\big )}\end{array}}}$

como queríamos demostrar.

${\displaystyle b)}$  Expresando ${\displaystyle y}$ en función de sus componentes en la base

${\displaystyle {\begin{array}{l}{\big \{}\mathbf {e} _{1},\ldots ,\mathbf {e} _{n}{\big \}},f_{L}{\big (}\mathbf {y} {\big )}=f_{L}{\big (}\sum _{k=1}^{n}y_{k}\mathbf {e} _{k}{\big )}=\sum _{k=1}^{n}y_{k}f_{L}{\big (}\mathbf {e} _{k}{\big )}=\sum _{k=1}^{n}y_{k}f'{\big (}\mathbf {x} ;\mathbf {e} _{k}{\big )}=\\\sum _{k=1}^{n}y_{k}{\cfrac {\partial f}{\partial x_{k}}}\end{array}}}$

como queríamos demostrar.

Sea ${\displaystyle f:S\subset \mathbb {R} ^{n}\longrightarrow \mathbb {R} }$ un campo escalar y ${\displaystyle \mathbf {x} :J\subset \mathbb {R} \longrightarrow S}$. Definimos la función compuesta ${\displaystyle g=f\circ \mathbf {x} }$ como ${\displaystyle g(t)=f{\Big [}\mathbf {x} {\big (}t{\big )}{\Big ]}}$, entonces ${\displaystyle \quad g'{\big (}t{\big )}=\sum _{k=1}^{n}{\cfrac {\partial f}{\partial x_{k}}}\cdot {\cfrac {dx_{k}}{dt}}}$

Sea ${\displaystyle \mathbf {f} :S\subseteq \mathbb {R} ^{n}\longrightarrow \mathbb {R} ^{m}}$ un campo vectorial. Sea ${\displaystyle \mathbf {x} \in S}$ e ${\displaystyle \mathbf {y} }$ un vector cualquiera. Definimos la derivada

${\displaystyle \mathbf {f'} {\big (}\mathbf {x} ;\mathbf {y} {\big )}=\lim _{h\to 0}{\cfrac {\mathbf {f} {\big (}\mathbf {x} +h\mathbf {y} {\big )}-\mathbf {f} {\big (}\mathbf {x} {\big )}}{h}}}$

Decimos que ${\displaystyle \mathbf {f} }$ es diferenciable ${\displaystyle \Leftrightarrow \exists \mathbf {f} _{L}:\mathbb {R} ^{n}\longrightarrow \mathbb {R} ^{m}}$, aplicación lineal que verifica:

${\displaystyle \lim _{{\big \|}\mathbf {v} {\big \|}\to 0}\mathbf {f} {\big (}\mathbf {x} +\mathbf {v} {\big )}=\mathbf {f} {\big (}\mathbf {x} {\big )}+\mathbf {f} _{L}{\big (}\mathbf {v} {\big )}}$.

Si un campo vectorial ${\displaystyle \mathbf {f} }$ es diferenciable en ${\displaystyle \mathbf {x} \Rightarrow }$ es continuo en ${\displaystyle \mathbf {x} }$.

Sea ${\displaystyle \mathbf {h} {\big (}\mathbf {x} {\big )}={\big (}\mathbf {f} \circ \mathbf {g} {\big )}{\big (}\mathbf {x} {\big )}}$ un campo vectorial definido y diferenciable en ${\displaystyle \mathbf {x} }$. Su diferencial ${\displaystyle \mathbf {h} '{\big (}\mathbf {x} {\big )}}$ resulta ser

${\displaystyle \mathbf {h} '{\big (}\mathbf {x} {\big )}=\mathbf {f} '{\Big [}\mathbf {g} {\big (}\mathbf {x} {\big )}{\Big ]}\circ \mathbf {g} '{\big (}\mathbf {x} {\big )}}$

${\displaystyle {\cfrac {\partial ^{2}f}{\partial x_{i}\partial x_{j}}}={\cfrac {\partial ^{2}f}{\partial x_{j}\partial x_{i}}}\quad \forall i\neq j\Leftrightarrow }$ ambas derivadas parciales existen y son continuas en ${\displaystyle \mathbf {x} }$.

Un campo escalar tiene un máximo en ${\displaystyle \mathbf {x} =\mathbf {a} \Leftrightarrow }$ existe una n-bola ${\displaystyle B{\big (}\mathbf {a} {\big )}{\Big |}\forall \mathbf {x} \in B{\big (}\mathbf {a} {\big )}\quad f{\big (}\mathbf {x} {\big )}\leqslant f{\big (}\mathbf {a} {\big )}}$

Un campo escalar tiene un mínimo en ${\displaystyle \mathbf {x} =\mathbf {a} \Leftrightarrow }$ existe una n-bola ${\displaystyle B{\big (}\mathbf {a} {\big )}{\Big |}\forall \mathbf {x} \in B{\big (}\mathbf {a} {\big )}\quad f{\big (}\mathbf {x} {\big )}\geqslant f{\big (}\mathbf {a} {\big )}}$

Un campo escalar tiene un punto de ensilladura ${\displaystyle \Leftrightarrow }$

${\displaystyle \forall B{\big (}\mathbf {a} {\big )}\quad \exists \mathbf {x} {\big |}f{\big (}\mathbf {x} {\big )}\leqslant f{\big (}\mathbf {a} {\big )}\land \exists \mathbf {x} {\big |}f{\big (}\mathbf {x} {\big )}\geqslant f{\big (}\mathbf {a} {\big )}}$.