Дифференциальные характеристики векторного поля

Дивергенция векторного поля

Пусть в некоторой системе координат $\bar {a}(M)=P(M)\bar {i}+Q(M)\bar {j}+R(M)\bar {k}$. Скалярная величина {скалярное поле} $\left( {\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}} \right)(M)$ называется дивергенцией поля в точке $\mathbf{\textit{M}}$ и обозначается $div\bar {a}(M)$:

$div\bar {a}(M) = \left( {\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}} \right)(M)$

С помощью оператора набла дивергенция определяется как скалярное произведение $\nabla \cdot \bar {a}=\left( {\frac{\partial }{\partial x}\bar {i}+\frac{\partial }{\partial y}\bar {j}+\frac{\partial }{\partial z}\vec {k}} \right)\cdot \left( {P\bar {i}+Q\bar {j}+R\vec {k}} \right)=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial Q}{\partial z}$.

В дальнейшем мы увидим, что дивергенция инвариантна относительно системы координат и обозначает плотность источников поля, а сейчас сформулируем

Свойства дивергенции:

  1. $\bar {a}(\mathbf{\textit{M}})$ - постоянное векторное поле, то $div\bar {a}=0$;
  2. $div\left( {С_1 \bar {a}_1 +С_2 \bar {a}_2 } \right)=С_1 div\bar {a}_1 +С_2 div\bar {a}_2 $ {или $\nabla \left( {С_1 \bar {a}_1 +С_2 \bar {a}_2 } \right)=С_1 \nabla \bar {a}_1 +С_2 \nabla \bar {a}_2 )$;
  3. Если $\mathbf{\textit{u}}$ - скалярное поле, то $div\left( {u\cdot \bar {a}} \right)=\bar {a}\cdot gradu+udiv\bar {a}$ {или $\nabla \left( {u\bar {a}} \right)=\bar {a}\nabla u+u\nabla \bar {a})$. В частности, если $\bar {a}(\mathbf{\textit{M}})$ - постоянное векторное поле, то $\nabla \left( {u\bar {a}} \right)=\bar {a}\nabla u=\bar {a}\cdot gradu$.

Докажем третье свойство

$div\left( {u\cdot \bar {a}} \right)=div(\left( {(uP)\bar {i}+(uQ)\bar {j}+(uR)\bar {k}} \right) = \frac{\partial (Pu)}{\partial x}+\frac{\partial (Qu)}{\partial y}+\frac{\partial (Ru)}{\partial z}=\left( {\frac{u\partial P}{\partial x}+\frac{P\partial u}{\partial x}} \right)+\left( {\frac{u\partial Q}{\partial y}+\frac{Q\partial u}{\partial y}} \right)+\left( {\frac{u\partial R}{\partial z}+\frac{R\partial u}{\partial z}} \right) = \\ = u\left( {\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}} \right)+ \frac{P\partial u}{\partial x}+\frac{Q\partial u}{\partial y}+\frac{R\partial u}{\partial z} = udiv\bar {a}+\bar {a}\cdot gradu$.

Пример 1

Если $\bar {a}=(x^3-yz+\cos (xyz))\bar {i}-xy^2z^3\bar {j}+arctg\frac{xy}{z}\bar {k}$, то $div\bar {a}=\frac{\partial (x^3-yz+\cos (xyz))}{\partial x}-\frac{\partial \left( {xy^2z^3} \right)}{\partial y}+\frac{\partial \left( {arctg\frac{xy}{z}} \right)}{\partial z}=3x^2-yz\sin (xyz)-2xyz^3-\frac{xy}{x^2y^2+z^2}$.

Ротор векторного поля

Ротором векторного поля $\bar {a}(\mathbf{\textit{M}})$ в точке $M\in V$ называется векторная величина {векторное поле} $rot\bar {a}(M)=\left( {\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}} \right)\bar {i}+\left( {\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}} \right)\bar {j}+\left( {\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}} \right)\bar {k}$.

Запомнить эту формулу очень легко, если выразить $rot\bar {a}(M)$ через оператор Гамильтона набла: $rot\bar {a}(M)$ равен векторному произведению $\nabla \times \bar {a}$. Действительно,

$rot\bar {a}(M)=\nabla \times \bar {a}=\left| {\begin{array}{l} \bar {i}\bar {j}\bar {k} \\ \frac{\partial }{\partial x}\frac{\partial }{\partial y}\frac{\partial }{\partial z} \\ PQR \\ \end{array}} \right|$.

Если теперь раскрыть этот определитель по первой строке, получим $ rot\bar {a}(M)=\nabla \times \bar {a}=\left| {\begin{array}{l} \bar {i}\bar {j}\bar {k} \\ \frac{\partial }{\partial x}\frac{\partial }{\partial y}\frac{\partial }{\partial z} \\ PQR \\ \end{array}} \right|=\bar {i}\left( {\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}} \right)-\bar {j}\left( {\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}} \right)+ \bar {k}\left( {\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}} \right)=\\=\left( {\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}} \right)\bar {i}+\left( {\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}} \right)\bar {j}+\left( {\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}} \right)\bar {k}. $

Пример 2

Если $\bar {a}=(x^3-yz+\cos (xyz))\bar {i}-xy^2z^3\bar {j}+arctg\frac{xy}{z}\bar {k}$, то $rot\bar {a}=\left| {\begin{array}{l} \bar {i}\bar {j}\bar {k} \\ \frac{\partial }{\partial x}\frac{\partial }{\partial y}\frac{\partial }{\partial z} \\ x^3-yz+\cos (xyz)-xy^2z^3arctg\frac{xy}{z} \\ \end{array}} \right|=\bar {i}\left( {\frac{\partial \left( {arctg\frac{xy}{z}} \right)}{\partial y}-\frac{\partial \left( {-xy^2z^3} \right)}{\partial z}} \right)-\bar {j}\left( {\frac{\partial \left( {arctg\frac{xy}{z}} \right)}{\partial x}-\frac{\partial \left( {x^3-yz+\cos (xyz)} \right)}{\partial z}} \right)+ \\ + \bar {k}\left( {\frac{\partial \left( {-xy^2z^3} \right)}{\partial x}-\frac{\partial \left( {x^3-yz+\cos (xyz)} \right)}{\partial y}} \right)=\left( {\frac{xz}{x^2y^2+z^2}+3xy^2z^2} \right)\bar {i}-\left( {\frac{yz}{x^2y^2+z^2}+y+xy\sin (xyz)} \right)\bar {j}-\left( {y^2z^3-z-xz\sin (xyz)} \right)\bar {k}$

Свойства ротора

  1. Если $\bar {a}(\mathbf{\textit{M}})$ - постоянное векторное поле, то $rot\bar {a}=\bar {0}$;
  2. $rot\left( {С_1 \bar {a}_1 +С_2 \bar {a}_2 } \right)=С_1 rot\bar {a}_1 +С_2 rot\bar {a}_2 $ {или $\nabla \times \left( {С_1 \bar {a}_1 +С_2 \bar {a}_2 } \right)=С_1 \nabla \times \bar {a}_1 +С_2 \nabla \times \bar {a}_2 )$;
  3. Если $\mathbf{\textit{u}}$ - скалярное поле, то $rot\left( {u\cdot \bar {a}} \right)=gradu\times \bar {a}+urot\bar {a}$ {или $\nabla \times \left( {\bar {a}u} \right)=\nabla u\times \bar {a}+u\left( {\nabla \times \bar {a}} \right))$. В частности, если $\bar {a}(\mathbf{\textit{M}})$ - постоянное векторное поле, то $\nabla \times \left( {u\bar {a}} \right)=\nabla u\times \bar {a}=gradu\times \bar {a}$.

Докажем третье свойство

$rot\left( {u\cdot \bar {a}} \right)=\left| {\begin{array}{l} \bar {i}\bar {j}\bar {k} \\ \frac{\partial }{\partial x}\frac{\partial }{\partial y}\frac{\partial }{\partial z} \\ uPuQuR \\ \end{array}} \right|=\left( {\frac{\partial (uR)}{\partial y}-\frac{\partial (uQ)}{\partial z}} \right)\bar {i}-\left( {\frac{\partial (uR)}{\partial x}-\frac{\partial (uP)}{\partial z}} \right)\bar {j}+\left( {\frac{\partial (uQ)}{\partial x}-\frac{\partial (uP)}{\partial y}} \right)\bar {k}=\left( {R\frac{\partial u}{\partial y}-Q\frac{\partial u}{\partial z} + \\+u\left( {\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}} \right)} \right)\bar {i}-\left( {R\frac{\partial u}{\partial x}-P\frac{\partial u}{\partial z}+u\left( {\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}} \right)} \right)\bar {j}+\left( {Q\frac{\partial u}{\partial x}-P\frac{\partial u}{\partial y}+u\left( {\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}} \right)} \right)\bar {k}= \\ =\left( {R\frac{\partial u}{\partial y}-Q\frac{\partial u}{\partial z}} \right)\bar {i}-\left( {R\frac{\partial u}{\partial x}-P\frac{\partial u}{\partial z}} \right)\bar {j}+\left( {Q\frac{\partial u}{\partial x}-P\frac{\partial u}{\partial y}} \right)\bar {k}+u\left[ {\left( {\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}} \right)\bar {i}-\left( {\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z}} \right)\bar {j}+\left( {\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}} \right)\bar {k}} \right]= \\ = \left| {\begin{array}{l} \bar {i}\bar {j}\bar {k} \\ \frac{\partial u}{\partial x}\frac{\partial u}{\partial y}\frac{\partial u}{\partial z} \\ PQR \\ \end{array}} \right|+urota=gradu\times \bar {a}+urota$