Integrating Vector Derivatives

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Q: Given that G(x,y,z)=(6xz+x3, 3x2y+y2, 4x+2yz-3z2). Find F such that curl F = G.

Solution:
...
A particular solution is
Fo=(-3x2yz-y2z, 2x2-3xz3-x3z)

And then my textbook says that the general solution is F=Fo + grad f where f is an arbitrary C1 function.
===============================

Now my questions:

If f is C1 function, why must F=Fo+gradf be a solution to curl F=G?
I believe that curl(grad f)=0 for f a C2 (not C1) function. Why does C1 work as well?

Secondly, why can we be sure that F=Fo+gradf is the general solution to curl F=G? (i.e. why is every solution contained in it?)


I would really appreciate if someone could explain.
 

Answers and Replies

  • #2
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Just to give some background, here we're trying to find the "vector potential function"

Can somebody help?
 
  • #3
Tom Mattson
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If f is C1 function, why must F=Fo+gradf be a solution to curl F=G?
I believe that curl(grad f)=0 for f a C2 (not C1) function. Why does C1 work as well?
I think you've actually found an error here. You do in fact need the second partials to be continuous, as that's the only way you are guaranteed to be able to switch the order of second mixed partials. Check it out with your prof to make sure we're not missing something subtle here.

Secondly, why can we be sure that F=Fo+gradf is the general solution to curl F=G? (i.e. why is every solution contained in it?)
It's because you haven't specified the f. It's just like saying that the general antiderivative of [itex]x[/itex] is [itex]\frac{1}{2}x^2+C[/itex]. Since you haven't committed to a particular C, the family of functions is still as general as possible.
 
  • #4
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I think you've actually found an error here. You do in fact need the second partials to be continuous, as that's the only way you are guaranteed to be able to switch the order of second mixed partials. Check it out with your prof to make sure we're not missing something subtle here.
Yes, I think it has to be C2.
But it is written in my textbook as C1, it must be a typo or something...

Just one more question:
We can add an arbitrary function f of class C2, is it because that
curl(Fo+gradf) = curl(Fo) + curl(gradf) = curl(Fo) + 0 = curl(Fo) ?
I am not too sure about the first equal sign. In general, is it true that curl(H1+H2) = curl(H1) + curl(H2)


It's because you haven't specified the f. It's just like saying that the general antiderivative of [itex]x[/itex] is [itex]\frac{1}{2}x^2+C[/itex]. Since you haven't committed to a particular C, the family of functions is still as general as possible.
For f a C2 function,
I can see that every vector field of the form F=Fo+gradf is a solution to curl F=G.
But is every solution to curl F=G of the form F=Fo+gradf? Can there exist a solution not of this form? How do you know?


Thanks!!
 
  • #5
Tom Mattson
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Just one more question:
We can add an arbitrary function f of class C2, is it because that
curl(Fo+gradf) = curl(Fo) + curl(gradf) = curl(Fo) + 0 = curl(Fo) ?
I am not too sure about the first equal sign. In general, is it true that curl(H1+H2) = curl(H1) + curl(H2)
Yes, certainly. The curl is just a formal linear combination of partial derivatives (with unit vector coefficients). Since differentiation is itself linear, the whole shebang is linear.

For f a C2 function,
I can see that every vector field of the form F=Fo+gradf is a solution to curl F=G.
But is every solution to curl F=G of the form F=Fo+gradf? Can there exist a solution not of this form? How do you know?
There's a basic theorem from vector calculus (can't recall the name) that states that the curl of a vector field [itex]\vec{F}[/itex] vanishes if and only if [itex]\vec{F}=\nabla{f}[/itex] for some scalar function [itex]f[/itex].
 

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