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Calculation of Potential Functions


A B C D E F G H I J K L M N O P Q R S T U V W X Y Z overview

Assume that the vector field $ \Phi: D \longrightarrow \mathbb{R}^3 $ given by

$\displaystyle \par
\left(\begin{array}{c}
x \\
y \\
z
\end{array}\right)
\par...
...eft(\begin{array}{c}
p(x,y,z) \\
q(x,y,z) \\
r(x,y,z)
\end{array}\right) \ .
$

has a potential function $ u(x,y,z) .$ Then using parameter integrals the potential function $ u$ may be determined as follows.

Start with the integration of $ p(x,y,z)$ with respect ot the first variable $ x . $

$\displaystyle u(x,y,z) = \int p(x,y,z) dx =
u_1(x,y,z)+c_1(y,z)\,.
$

From $ q(x,y,z) = u_y = \frac{\partial}{\partial y} u_1 +
\frac{\partial}{\partial y} c_1 $ it follows that

$\displaystyle c_1(y,z) = \int \left(q(x,y,z) - \frac{\partial}{\partial y} u_1 \right)\,dy =
u_2(y,z) + c_2(z) \ .
$

Finally $ r(x,y,z) = u_z = \frac{\partial}{\partial z} u_1 +
\frac{\partial}{\partial z} u_2
+ \frac{\partial}{\partial z} c_2$ yields

$\displaystyle c_2(z) = \int \left(r(x,y,z) - \frac{\partial}{\partial z}u_1 -
\frac{\partial}{\partial z} u_2\right)\,dz
= u_3(z) + c \ .
$

Therefore

$\displaystyle u(x,y,z) = u_1(x,y,z)+u_2(y,z)+u_3(z)+c\,.
$

()

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  automatically generated 6/15/2005