Finding Y Component for Vector Field F with Zero Divergence

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stunner5000pt
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Homework Statement


Suppose we have the vector field F whose x component is given by [itex]F_{x}=Ax[/itex] and whose divergence is known to be zero [itex]\vec{\nabla}\cdot\vec{F}=0[/itex], then find a possible y component for this field. How many y components are possible?

2. The attempt at a solution

So the divergence in cartesian coordinates is given by
[tex]\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y} = 0[/tex]

Using the fact that [tex]F_{x}=Ax[/tex]
[tex]A+\frac{\partial F}{\partial y} = 0[/tex]
[tex]\frac{\partial F}{\partial y} = -A[/tex]
integrate both sides with respect to y we get

[tex]F_{y}=-Ay+B[/tex]

where B is a constant
is that sufficient for a possible y component? For the question with howm any are possible... arent there infinite possibilities since B could be anything. But they are all parallel to each... linearly dependent on the above answer?
 
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stunner5000pt said:

Homework Statement


Suppose we have the vector field F whose x component is given by [itex]F_{x}=Ax[/itex] and whose divergence is known to be zero [itex]\vec{\nabla}\cdot\vec{F}=0[/itex], then find a possible y component for this field. How many y components are possible?

2. The attempt at a solution

So the divergence in cartesian coordinates is given by
[tex]\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y} = 0[/tex]

Using the fact that [tex]F_{x}=Ax[/tex]
[tex]A+\frac{\partial F}{\partial y} = 0[/tex]
[tex]\frac{\partial F}{\partial y} = -A[/tex]
integrate both sides with respect to y we get

[tex]F_{y}=-Ay+B[/tex]

where B is a constant
is that sufficient for a possible y component? For the question with howm any are possible... arent there infinite possibilities since B could be anything. But they are all parallel to each... linearly dependent on the above answer?

As far as I'm concerned, you're good to go.
 
asleight said:
As far as I'm concerned, you're good to go.

but.. infinitely many solutions because of B or finite because they are all linearly dependent on the solution given?
 
stunner5000pt said:
but.. infinitely many solutions because of B or finite because they are all linearly dependent on the solution given?

There is an infinite amount of parallel solutions.