Потенциал на ВП

[Добромир Глухаров]

Потенциално векторно поле. Скаларен потенциал

От горния линк може да се види, че за да бъде полето $\vec{a}=(1+y^2z^3)\vec{i}+(1+2xyz^3)\vec{j}+(1+3xy^2z^2)\vec{k}$ потенциално, трябва $rot\ \vec{a}=\vec{0}\Rightarrow\vec{\bigtriangledown}\times\vec{\bigtriangledown}\vec{a}=\left(\frac{\partial}{\partial x}\vec{i}+\frac{\partial}{\partial y}\vec{j}+\frac{\partial}{\partial z}\vec{k}\right)\times\left(\frac{\partial}{\partial x}\vec{i}+\frac{\partial}{\partial y}\vec{j}+\frac{\partial}{\partial z}\vec{k}\right)\ \vec{a}=$

$=\begin{array}{|ccc|}\vec{i}&\vec{j}&\vec{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ 1+y^2z^3&1+2xyz^3&1+3xy^2z^2\end{array}=$

$=\left(\frac{\partial(1+3xy^2z^2)}{\partial y}-\frac{\partial(1+2xyz^3)}{\partial z}\right)\vec{i}-\left(\frac{\partial(1+3xy^2z^2)}{\partial x}-\frac{\partial(1+y^2z^3)}{\partial z}\right)\vec{j}+\left(\frac{\partial(1+2xyz^3)}{\partial x}-\frac{\partial(1+y^2z^3)}{\partial y}\right)\vec{k}=$

$=(6xyz^2-6xyz^2)\vec{i}-(3y^2z^2-3y^2z^2)\vec{j}+(2yz^3-2yz^3)\vec{k}=\vec{0}$

Видяхме, че наистина полето е потенциално.

Остава да намерим потенциала $\varphi$ от условието $\vec{grad}\varphi=\vec{a}$

Имаме $\frac{\partial\varphi}{\partial x}\vec{i}+\frac{\partial\varphi}{\partial y}\vec{j}+\frac{\partial\varphi}{\partial z}\vec{k}=\vec{a}=(1+y^2z^3)\vec{i}+(1+2xyz^3)\vec{j}+(1+3xy^2z^2)\vec{k}$

Следователно:

$\frac{\partial\varphi}{\partial x}=1+y^2z^3\Rightarrow\varphi=x+xy^2z^3+C_1(y,z)$

$\frac{\partial\varphi}{\partial y}=1+2xyz^3\Rightarrow\varphi=y+xy^2z^3+C_2(x,z)$

$\frac{\partial\varphi}{\partial z}=1+3xy^2z^2\Rightarrow\varphi=z+xy^2z^3+C_3(x,y)$

$\Rightarrow x+xy^2z^3+C_1(y,z)=y+xy^2z^3+C_2(x,z)=z+xy^2z^3+C_3(x,y)$

$\Rightarrow x+C_1(y,z)=y+C_2(x,z)=z+C_3(x,y)=x+y+z+C$

$\Rightarrow C_1=y+z+C;\ C_2=x+z+C;\ C_3=x+y+C$

$\Rightarrow \varphi=x+y+z+xy^2z^3+C$