Identidad vectorial de Green

La siguiente pagina, es una traducción de la pagina anterior, "Green's vector identity"

La segunda derivada de dos funciones vectoriales, esta relacionada con la divergencia de las funciones vectoriales con los operadores de primer orden, es decir:

$\mathbf {P} \cdot \nabla ^{2}\mathbf {Q} -\mathbf {Q} \cdot \nabla ^{2}\mathbf {P} =\nabla \cdot \left[\mathbf {P} \left(\nabla \cdot \mathbf {Q} \right)-\mathbf {Q} \left(\nabla \cdot \mathbf {P} \right)+\mathbf {P} \times \nabla \times \mathbf {Q} -\mathbf {Q} \times \nabla \times \mathbf {P} \right].$

Ver también wikipedia: Green's vector identity

Introducción

La segunda identidad de Green establece una relación entre las derivadas de segundo y (la divergencia de) primer orden de dos funciones escalares

$p_{m}\nabla ^{2}q_{m}-q_{m}\nabla ^{2}p_{m}=\nabla \cdot \left(p_{m}\nabla q_{m}-q_{m}\nabla p_{m}\right),$

Donde $p_{m}$ y $q_{m}$ son dos campos escalares arbitrarios. Esta identidad es de gran importancia en física porque así se pueden establecer ecuaciones de continuidad para campos escalares como la masa o la energía. .

Aunque la segunda identidad de Green siempre se presenta en el análisis vectorial, solo se encuentra una versión escalar en los libros de texto. Incluso en la literatura especializada, no es fácil encontrar una versión vectorial. En la teoría de la difracción vectorial, se introducen dos versiones de la segunda identidad de Green. Una variante invoca la divergencia de un producto cruz y establece una relación en términos del rotacional-rotacional de un campo

$\mathbf {P} \cdot \left(\nabla \times \nabla \times \mathbf {Q} \right)-\mathbf {Q} \cdot \left(\nabla \times \nabla \times \mathbf {P} \right)=\nabla \cdot \left(\mathbf {Q} \times \nabla \times \mathbf {P} -\mathbf {P} \times \nabla \times \mathbf {Q} \right)$. Esta ecuación puede ser escrita en términos del Laplaciano como:

$\mathbf {P} \cdot \nabla ^{2}\mathbf {Q} -\mathbf {Q} \cdot \nabla ^{2}\mathbf {P} +\mathbf {Q} \cdot \left[\nabla \left(\nabla \cdot \mathbf {P} \right)\right]-\mathbf {P} \cdot \left[\nabla \left(\nabla \cdot \mathbf {Q} \right)\right]=\nabla \cdot \left(\mathbf {P} \times \nabla \times \mathbf {Q} -\mathbf {Q} \times \nabla \times \mathbf {P} \right).$

Sin embargo, los términos $\mathbf {Q} \cdot \left[\nabla \left(\nabla \cdot \mathbf {P} \right)\right]-\mathbf {P} \cdot \left[\nabla \left(\nabla \cdot \mathbf {Q} \right)\right]$, no podía escribirse fácilmente en términos de una divergencia. El otro enfoque introduce bi-vectores, esta formulación requiere una función de Green diádica .

Divergencia de dos campos vectoriales

Considere que los campos escalares en la segunda identidad de Green son los componentes cartesianos de los campos vectoriales, es decir $\mathbf {P} =\sum \limits _{m}p_{m}{\hat {\mathbf {e} }}_{m}$ and $\mathbf {Q} =\sum \limits _{m}q_{m}{\hat {\mathbf {e} }}_{m}$. Sumando la ecuación para cada componente, obtenemos:

$\sum \limits _{m}\left[p_{m}\nabla ^{2}q_{m}-q_{m}\nabla ^{2}p_{m}\right]=\sum \limits _{m}\left[\nabla \cdot \left(p_{m}\nabla q_{m}-q_{m}\nabla p_{m}\right)\right].$

El lado izquierdo, de acuerdo con la definición de producto punto, puede ser escrito de forma vectorial como:

$\sum \limits _{m}\left[p_{m}\nabla ^{2}q_{m}-q_{m}\nabla ^{2}p_{m}\right]=\mathbf {P} \cdot \nabla ^{2}\mathbf {Q} -\mathbf {Q} \cdot \nabla ^{2}\mathbf {P} .$

El lado derecho es un poco más difícil de expresar en términos de operadores vectoriales. Debido a la distributividad del operador de divergencia sobre la suma, la suma de la divergencia es igual a la divergencia de la suma, es decir $\sum \limits _{m}\left[\nabla \cdot \left(p_{m}\nabla q_{m}-q_{m}\nabla p_{m}\right)\right]=\nabla \cdot \left(\sum \limits _{m}p_{m}\nabla q_{m}-\sum \limits _{m}q_{m}\nabla p_{m}\right)$. Recordar que la identidad vectorial para el gradiente de un producto escalar $\nabla \left(\mathbf {P} \cdot \mathbf {Q} \right)=\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} +\left(\mathbf {Q} \cdot \nabla \right)\mathbf {P} +\mathbf {P} \times \nabla \times \mathbf {Q} +\mathbf {Q} \times \nabla \times \mathbf {P}$,la cual, escrita en componentes vectoriales viene dada por $\nabla \left(\mathbf {P} \cdot \mathbf {Q} \right)=\nabla \sum \limits _{m}p_{m}q_{m}=\sum \limits _{m}p_{m}\nabla q_{m}+\sum \limits _{m}q_{m}\nabla p_{m}.$ Este resultado es similar a lo que deseamos evidenciar en términos vectoriales 'excepto' por el signo menos. Dado que los operadores diferenciales en cada término actúan sobre un vector (es decir $p_{m}$’s) o el otro ($q_{m}$’s) , la contribución a cada término debe ser

$\sum \limits _{m}p_{m}\nabla q_{m}=\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} +\mathbf {P} \times \nabla \times \mathbf {Q} ,$ $\sum \limits _{m}q_{m}\nabla p_{m}=\left(\mathbf {Q} \cdot \nabla \right)\mathbf {P} +\mathbf {Q} \times \nabla \times \mathbf {P} .$

Se puede demostrar rigurosamente que estos resultados son correctos mediante evaluación de los componentes del vector. Por lo tanto, el lado derecho se puede escribir en forma vectorial como

$\sum \limits _{m}p_{m}\nabla q_{m}-\sum \limits _{m}q_{m}\nabla p_{m}=\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} +\mathbf {P} \times \nabla \times \mathbf {Q} -\left(\mathbf {Q} \cdot \nabla \right)\mathbf {P} -\mathbf {Q} \times \nabla \times \mathbf {P} .$

Putting together these two results, a theorem for vector fields analogous to Green’s theorem for scalar fields is obtained

$\color {green}{\mathbf {P} \cdot \nabla ^{2}\mathbf {Q} -\mathbf {Q} \cdot \nabla ^{2}\mathbf {P} =\nabla \cdot \left[\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} +\mathbf {P} \times \nabla \times \mathbf {Q} -\left(\mathbf {Q} \cdot \nabla \right)\mathbf {P} -\mathbf {Q} \times \nabla \times \mathbf {P} \right].}$

Reassuringly, from the vector relationship we can go back to the scalar case as shown in the scalar limit. The curl of a cross product can be written as $\nabla \times \left(\mathbf {P} \times \mathbf {Q} \right)=\left(\mathbf {Q} \cdot \nabla \right)\mathbf {P} -\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} +\mathbf {P} \left(\nabla \cdot \mathbf {Q} \right)-\mathbf {Q} \left(\nabla \cdot \mathbf {P} \right)$; Green’s vector identity can then be rewritten as

$\mathbf {P} \cdot \nabla ^{2}\mathbf {Q} -\mathbf {Q} \cdot \nabla ^{2}\mathbf {P} =\nabla \cdot \left[\mathbf {P} \left(\nabla \cdot \mathbf {Q} \right)-\mathbf {Q} \left(\nabla \cdot \mathbf {P} \right)-\nabla \times \left(\mathbf {P} \times \mathbf {Q} \right)+\mathbf {P} \times \nabla \times \mathbf {Q} -\mathbf {Q} \times \nabla \times \mathbf {P} \right].$

Since the divergence of a curl is zero, the third term vanishes and Green’s vector identity is

$\color {green}{\mathbf {P} \cdot \nabla ^{2}\mathbf {Q} -\mathbf {Q} \cdot \nabla ^{2}\mathbf {P} =\nabla \cdot \left[\mathbf {P} \left(\nabla \cdot \mathbf {Q} \right)-\mathbf {Q} \left(\nabla \cdot \mathbf {P} \right)+\mathbf {P} \times \nabla \times \mathbf {Q} -\mathbf {Q} \times \nabla \times \mathbf {P} \right].}$

With a similar porcedure, the Laplacian of the dot product can be expressed in terms of the Laplacians of the factors

$\nabla ^{2}\left(\mathbf {P} \cdot \mathbf {Q} \right)=\mathbf {P} \cdot \nabla ^{2}\mathbf {Q} -\mathbf {Q} \cdot \nabla ^{2}\mathbf {P} +2\nabla \cdot \left[\left(\mathbf {Q} \cdot \nabla \right)\mathbf {P} +\mathbf {Q} \times \nabla \times \mathbf {P} \right].$

corollary

As a corollary, the awkward terms in the introduction can now be written in terms of a divergence by comparison with the vector Green equation

$\mathbf {P} \cdot \left[\nabla \left(\nabla \cdot \mathbf {Q} \right)\right]-\mathbf {Q} \cdot \left[\nabla \left(\nabla \cdot \mathbf {P} \right)\right]=\nabla \cdot \left[\mathbf {P} \left(\nabla \cdot \mathbf {Q} \right)-\mathbf {Q} \left(\nabla \cdot \mathbf {P} \right)\right].$

This result can be verified by expanding the divergence of a scalar times a vector on the RHS.

derivation by components

In order to evaluate $\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} +\mathbf {P} \times \nabla \times \mathbf {Q}$ consider the first term in three dimensional Cartesian components $\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} =\left(p_{x}{\frac {\partial }{\partial x}}+p_{y}{\frac {\partial }{\partial y}}+p_{z}{\frac {\partial }{\partial z}}\right)\left(q_{x}{\hat {\mathbf {e} }}_{x}+q_{y}{\hat {\mathbf {e} }}_{y}+q_{z}{\hat {\mathbf {e} }}_{z}\right)$ that may be written as

$\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} =\left(p_{x}{\frac {\partial q_{x}}{\partial x}}+p_{y}{\frac {\partial q_{x}}{\partial y}}+p_{z}{\frac {\partial q_{x}}{\partial z}}\right){\hat {\mathbf {e} }}_{x}+\left(p_{x}{\frac {\partial q_{y}}{\partial x}}+p_{y}{\frac {\partial q_{y}}{\partial y}}+p_{z}{\frac {\partial q_{y}}{\partial z}}\right){\hat {\mathbf {e} }}_{y}+\left(p_{x}{\frac {\partial q_{z}}{\partial x}}+p_{y}{\frac {\partial q_{z}}{\partial y}}+p_{z}{\frac {\partial q_{z}}{\partial z}}\right){\hat {\mathbf {e} }}_{z}.$

The curl in the second term is $^{R}\mathbf {Q} =\nabla \times \mathbf {Q} =\left({\frac {\partial q_{z}}{\partial y}}-{\frac {\partial q_{y}}{\partial z}}\right){\hat {\mathbf {e} }}_{x}+\left({\frac {\partial q_{x}}{\partial z}}-{\frac {\partial q_{z}}{\partial x}}\right){\hat {\mathbf {e} }}_{y}+\left({\frac {\partial q_{y}}{\partial x}}-{\frac {\partial q_{x}}{\partial y}}\right){\hat {\mathbf {e} }}_{z}.$ The cross product is $\mathbf {P} \times {}^{R}\mathbf {Q} =\left(p_{y}{}^{R}q_{z}-p_{z}{}^{R}q_{y}\right){\hat {\mathbf {e} }}_{x}+\left(p_{z}{}^{R}q_{x}-p_{x}{}^{R}q_{z}\right){\hat {\mathbf {e} }}_{y}+\left(p_{x}{}^{R}q_{y}-p_{y}{}^{R}q_{x}\right){\hat {\mathbf {e} }}_{z};$ The second term is then

$\mathbf {P} \times \nabla \times \mathbf {Q} =\left(p_{y}\left({\frac {\partial q_{y}}{\partial x}}-{\frac {\partial q_{x}}{\partial y}}\right)-p_{z}\left({\frac {\partial q_{x}}{\partial z}}-{\frac {\partial q_{z}}{\partial x}}\right)\right){\hat {\mathbf {e} }}_{x}+\left(p_{z}\left({\frac {\partial q_{z}}{\partial y}}-{\frac {\partial q_{y}}{\partial z}}\right)-p_{x}\left({\frac {\partial q_{y}}{\partial x}}-{\frac {\partial q_{x}}{\partial y}}\right)\right){\hat {\mathbf {e} }}_{y}+\left(p_{x}\left({\frac {\partial q_{x}}{\partial z}}-{\frac {\partial q_{z}}{\partial x}}\right)-p_{y}\left({\frac {\partial q_{z}}{\partial y}}-{\frac {\partial q_{y}}{\partial z}}\right)\right){\hat {\mathbf {e} }}_{z}$

that expands to

$\mathbf {P} \times \nabla \times \mathbf {Q} =\left(p_{y}{\frac {\partial q_{y}}{\partial x}}-p_{y}{\frac {\partial q_{x}}{\partial y}}-p_{z}{\frac {\partial q_{x}}{\partial z}}+p_{z}{\frac {\partial q_{z}}{\partial x}}\right){\hat {\mathbf {e} }}_{x}+\left(p_{z}{\frac {\partial q_{z}}{\partial y}}-p_{z}{\frac {\partial q_{y}}{\partial z}}-p_{x}{\frac {\partial q_{y}}{\partial x}}+p_{x}{\frac {\partial q_{x}}{\partial y}}\right){\hat {\mathbf {e} }}_{y}+\left(p_{x}{\frac {\partial q_{x}}{\partial z}}-p_{x}{\frac {\partial q_{z}}{\partial x}}-p_{y}{\frac {\partial q_{z}}{\partial y}}+p_{y}{\frac {\partial q_{y}}{\partial z}}\right){\hat {\mathbf {e} }}_{z}.$

Evaluate $\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} +\mathbf {P} \times \nabla \times \mathbf {Q}$ in the x direction

$\left[\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} +\mathbf {P} \times \nabla \times \mathbf {Q} \right]_{x}=\left(p_{x}{\frac {\partial q_{x}}{\partial x}}+p_{y}{\frac {\partial q_{x}}{\partial y}}+p_{z}{\frac {\partial q_{x}}{\partial z}}+p_{y}{\frac {\partial q_{y}}{\partial x}}-p_{y}{\frac {\partial q_{x}}{\partial y}}-p_{z}{\frac {\partial q_{x}}{\partial z}}+p_{z}{\frac {\partial q_{z}}{\partial x}}\right){\hat {\mathbf {e} }}_{x},$

canceling out terms $\left[\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} +\mathbf {P} \times \nabla \times \mathbf {Q} \right]_{x}=\left(p_{x}{\frac {\partial q_{x}}{\partial x}}+p_{y}{\frac {\partial q_{y}}{\partial x}}+p_{z}{\frac {\partial q_{z}}{\partial x}}\right){\hat {\mathbf {e} }}_{x}.$

Analogous results are obtained in the other directions so that

$\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} +\mathbf {P} \times \nabla \times \mathbf {Q} =\left(p_{x}{\frac {\partial q_{x}}{\partial x}}+p_{y}{\frac {\partial q_{y}}{\partial x}}+p_{z}{\frac {\partial q_{z}}{\partial x}}\right){\hat {\mathbf {e} }}_{x}+\left(p_{x}{\frac {\partial q_{x}}{\partial y}}+p_{y}{\frac {\partial q_{y}}{\partial y}}+p_{z}{\frac {\partial q_{z}}{\partial y}}\right){\hat {\mathbf {e} }}_{y}+\left(p_{x}{\frac {\partial q_{x}}{\partial z}}+p_{y}{\frac {\partial q_{y}}{\partial z}}+p_{z}{\frac {\partial q_{z}}{\partial z}}\right){\hat {\mathbf {e} }}_{z}.$

that may be written out in vector form as $\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} +\mathbf {P} \times \nabla \times \mathbf {Q} =\mathbf {P} \cdot {\frac {\partial \mathbf {Q} }{\partial x}}+\mathbf {P} \cdot {\frac {\partial \mathbf {Q} }{\partial y}}+\mathbf {P} \cdot {\frac {\partial \mathbf {Q} }{\partial z}}.$ However, the terms can be rearranged as

$\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} +\mathbf {P} \times \nabla \times \mathbf {Q} =\left(p_{x}{\frac {\partial q_{x}}{\partial x}}{\hat {\mathbf {e} }}_{x}+p_{x}{\frac {\partial q_{x}}{\partial y}}{\hat {\mathbf {e} }}_{y}+p_{x}{\frac {\partial q_{x}}{\partial z}}{\hat {\mathbf {e} }}_{z}\right)+\left(p_{y}{\frac {\partial q_{y}}{\partial x}}{\hat {\mathbf {e} }}_{x}+p_{y}{\frac {\partial q_{y}}{\partial y}}{\hat {\mathbf {e} }}_{y}+p_{y}{\frac {\partial q_{y}}{\partial z}}{\hat {\mathbf {e} }}_{z}\right)+\left(p_{z}{\frac {\partial q_{z}}{\partial x}}{\hat {\mathbf {e} }}_{x}+p_{z}{\frac {\partial q_{z}}{\partial y}}{\hat {\mathbf {e} }}_{y}+p_{z}{\frac {\partial q_{z}}{\partial z}}{\hat {\mathbf {e} }}_{z}\right),$

and thus $\left(\mathbf {P} \cdot \nabla \right)\mathbf {Q} +\mathbf {P} \times \nabla \times \mathbf {Q} =\left(p_{x}\nabla q_{x}+p_{y}\nabla q_{y}+p_{z}\nabla q_{z}\right).$ An equivalent procedure for $\left(\mathbf {Q} \cdot \nabla \right)\mathbf {P} +\mathbf {Q} \times \nabla \times \mathbf {P}$ gives $\left(\mathbf {Q} \cdot \nabla \right)\mathbf {P} +\mathbf {Q} \times \nabla \times \mathbf {P} =q_{x}\nabla p_{x}+q_{y}\nabla p_{y}+q_{z}\nabla p_{z}.$

scalar case

If we take one component vectors, for example, $\mathbf {P} \rightarrow p_{x}{\hat {\mathbf {e} }}_{x},\mathbf {Q} \rightarrow q_{x}{\hat {\mathbf {e} }}_{x}$, the vector relationship ([eq:vec green]) becomes $p_{x}\nabla ^{2}q_{x}-q_{x}\nabla ^{2}p_{x}=\nabla \cdot \left[p_{x}{\frac {\partial q_{x}}{\partial x}}{\hat {\mathbf {e} }}_{x}-q_{x}{\frac {\partial p_{x}}{\partial x}}{\hat {\mathbf {e} }}_{x}+\mathbf {P} \times \nabla \times \mathbf {Q} -\mathbf {Q} \times \nabla \times \mathbf {P} \right].$ Since $\nabla \times \mathbf {Q} ={\frac {\partial q_{x}}{\partial z}}{\hat {\mathbf {e} }}_{y}-{\frac {\partial q_{x}}{\partial y}}{\hat {\mathbf {e} }}_{z}$,

$\mathbf {P} \times \nabla \times \mathbf {Q} =\left(p_{x}{\frac {\partial q_{x}}{\partial y}}\right){\hat {\mathbf {e} }}_{y}+\left(p_{x}{\frac {\partial q_{x}}{\partial z}}\right){\hat {\mathbf {e} }}_{z}$

and $\mathbf {Q} \times \nabla \times \mathbf {P} =\left(q_{x}{\frac {\partial p_{x}}{\partial y}}\right){\hat {\mathbf {e} }}_{y}+\left(q_{x}{\frac {\partial p_{x}}{\partial z}}\right){\hat {\mathbf {e} }}_{z}$. Therefore $p_{x}\nabla ^{2}q_{x}-q_{x}\nabla ^{2}p_{x}=\nabla \cdot \left[p_{x}\nabla q_{x}-q_{x}\nabla p_{x}\right],$ we recover Green’s second identity for the functions $p_{x},\,q_{x}$.

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