HP 49g+, Руководство пользователя, Страница: 474 / 862

Page 15-7
produces the vector potential function
Φ
2
= [0, ZYX-2YX, Y-(2ZX-X)], which is
different from
Φ
1
. The last command in the screen shot shows that indeed F =
∇× Φ
2
. Thus, a vector potential function is not uniquely determined.
The components of the given vector field,
F (x,y,z) = f(x,y,z) i +g(x,y,z) j
+h(x,y,z) k , and those of the vector potential function,
Φ
(x,y,z) =
φ
(x,y,z) i +
ψ
(x,y,z) j +
η
(x,y,z) k , are related by f =
∂η
/
∂
y -
∂ψ
/
∂
x, g =
∂φ
/
∂
z -
∂η
/
∂
x, and h =
∂ψ
/
∂
x -
∂φ
/
∂
y.
A condition for function
Φ
(x,y,z) to exists is that div F =
∇•
F = 0, i.e.,
∂
f/
∂
x +
∂
g/
∂
y +
∂
f/
∂
z = 0. Thus, if this condition is not satisfied, the vector potential
function
Φ
(x,y,z) does not exist. For example, given F = [X+Y,X-Y,Z^2],
function VPOTENTIAL returns an error message, since function F does not
satisfy the condition
∇•
F = 0:
The condition
∇•
F
≠
0 is verified in the following screen shot: