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Vector Calculus Proof: Curl V = 0 -> V = grad phi

  1. Sep 23, 2013 #1
    If ∇ x v = 0 in all of three dimensional space, show that there exists a scalar function ##\phi (x,y,z)## such that v = ∇##\phi##. (from Walter Strauss' Partial Differential Equations, 2nd edition; problem 11; pg 20.)

    I'm not really sure where to begin with this problem. I asked a few of my friends in math and they all provided me with suggestions that are above my level (ie. using something called Poincare's Lemma which we haven't covered in any of my classes yet). My difficulty lies in proving this is true in the direction required.

    I can easily show this is true in the reverse direction. If ∇##\phi## = v, then ∇ x v = 0. I can show this with relative ease by simply writing out each component explicitly. (I won't do this here, however, since that's not what the question is asking me to do). I'm not sure how to go in the reverse direction, however. Any help would be greatly appreciated

    (Also, to put the problem into context: this problem is an assigned problem near the beginning of an introductory PDEs class that assumes a familiarity with vector calculus, basic linear algebra and ODEs.)

    It also might be worth mentioning that I think we can assume ##\phi## is a "nice" function - that is, that it is continuous and twice differentiable. Thanks a bunch in advance for any nudges in the right direction.
     
  2. jcsd
  3. Sep 23, 2013 #2

    WannabeNewton

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    Hint: Stokes' Theorem.
     
  4. Sep 23, 2013 #3
    In the words of my pre-calculus teacher: "If you don't know what to do, do something!"

    What do you know about the curl (stuff you can do to curls, for example) that might even vaguely be useful? I find, often, just listing my resources gives rise to a simple proof.

    Edit: WN got there first. Too slow. :rofl:

    As a side note, Poincaré's Lemma seems like a fun way to do this, and a little more straightforward to do. I might have to try it from that angle in the morning. :biggrin:
     
  5. Sep 23, 2013 #4
    Just from the definition I can write:

    ∇ x v = <## \frac{∂ v^{z}}{∂ y} - \frac{∂ v^{y}}{∂ z}##, ##\frac{∂ v^{x}}{∂ z} - \frac{∂ v^{z}}{∂ x}##, ##\frac{∂ v^{y}}{∂ x} - \frac{∂ v^{x}}{∂ y}##>

    Where the notation ##v^{i}## refers to the ith component of v.

    For this expression to be 0 we must have each component 0, which suggests that ##\frac{∂ v^{z}}{∂ y} = \frac{∂ v^{y}}{∂ z}## (and similar for the other components). I'm not too sure where to go from here following the "do something" approach.


    The approach of using Stoke's Theorem let's me say that if the curl of a vector field is zero then the vector field must be conservative and we know that a function f exists such that the gradient of f is this vector field. This approach seems way too easy and I love it. Is this all it is?

    Suppose I didn't know Stoke's Theorem - where would I go from what I've done above to brute force out the answer (if this is possible)? Thanks a bunch guys!



    EDIT: I may as well look up Poincare's Lemma myself - this assignment isn't due until next Monday so I have plenty of time to fool around with different approaches!
     
  6. Sep 23, 2013 #5

    WannabeNewton

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    Well you have that ##\int _{S}(\nabla \times V) \cdot dA = \int _{\partial S}V\cdot dr = 0##. Thus the line integral of ##V## around any closed loop is zero. How can you then use this to construct a scalar field ##\varphi## such that ##V = \nabla \varphi##?

    The Poincare Lemma directly gives you the result but the proof of the Poincare Lemma itself is extremely non-elementary and requires algebraic topology to motivate and setup so I wouldn't worry about it.
     
  7. Sep 23, 2013 #6

    Office_Shredder

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    If you want to avoid Stoke's Theorem, then you have to construct your function f directly. Given v= (v1,v2,v3), we know that
    [tex] f + A(y,z) = \int v_1 dx [/tex]
    [tex] f +B(x,z) = \int v_2 dy [/tex]
    [tex] f + C(x,y) = \int v_3 dz [/tex]

    for some functions A, B and C (check that e.g. the first equation gives that df/dx = v1). If you can show that there exist functions A,B and C such that
    [tex] \int v_1 dx - A(y,z) = \int v_2 dy - B(x,z) = \int v_3 dz - C(x,y)[/tex]

    Then the function f is equal to all three of these. The existence of these functions will have to come about using the curl = 0 condition (and the differential equations that you wrote down from there).
     
  8. Sep 23, 2013 #7
    I recall a theorem from vector calculus that states something along the lines of: if the line integral of a vector function ##V## around any closed loop is zero then this vector function is conservative which means that it can be written as the gradient of a scalar function. I suppose you are in essence asking me to prove this part,correct?

    I would do something like this:

    Select an arbitrary point (a,b) in the domain D and compute the line integral from this point (a,b) to another arbitrary point (x,y) (where (a,b) ##\neq## (x,y)) that is composed of one or more curves, at least one of which is horizontal (call this union of curves ##C_{1}##). Do the same thign for a separate curve except with one piece being vertical (call this union of curves ##C_{2}@@). Then the line integral over the combination of these two curves will be zero because they will form a closed loop (that is ##C_{1} U -C_{2}## is a closed loop). Because this loop has components that are horizontal and vertical we know how the function changes along these horiztonal and vertical lines and we should find that their sum (which should be the gradient of the scalar function) satisfies the conditions necessary.

    I realize this is a rather rough sketch of my idea. I essentially would look at forming a closed loop that is composed of at least one part horizontal and one part vertical so that I could write v
    = P i + Q j. I'll have to review the section on line integrals in order to come up with a more precise proof.
     
  9. Sep 24, 2013 #8

    WannabeNewton

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    The idea is to show that if the line integral is zero around closed loops, then the line integral is independent of the path taken. The desired result will then follow suit.
     
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