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Showing scalar functions u(x,y,z) and v(x,y,z) are related

  1. Jun 16, 2013 #1
    1.a. Show that ∇F[u(x,y,z),v(x,y,z)] = Fu∇v + Fv∇u
    1.b. Show that a necessary and sufficient condition that u and v are functionally related by the equation F(u,v) = 0 is ∇u x ∇v = 0

    2. Relevant equations
    ∇ = [itex]\frac{\partial}{\partial x}[/itex][itex]\widehat{i}[/itex] + [itex]\frac{\partial}{\partial y}[/itex][itex]\widehat{j}[/itex] + [itex]\frac{\partial}{\partial z}[/itex][itex]\widehat{k}[/itex]

    3. The attempt at a solution 1.a
    ∇F[u(x,y,z),v(x,y,z)] = (Fuux + Fvvx)[itex]\widehat{i}[/itex] + (Fuuy + Fvvy)[itex]\widehat{j}[/itex] + (Fuuz + Fvvz)[itex]\widehat{k}[/itex] = Fu∇u + Fv∇v


    4. The attempt at solution 1.b
    I'm honestly stuck. The necessary and sufficient condition throws me. If I work from the assumption that F[u,v] = 0, I can get:

    ∇F x ∇v = Fu∇u x ∇v = 0
    ∇F x ∇u = Fv∇v x ∇u = 0

    Either of which lead to ∇u x ∇v = 0

    But this seems to show neither necessity nor sufficiency.

    I know that this leads to developing the Jacobian and I have an inkling that the delta function may help, but can't get anywhere with that. Any pointers would be greatly appreciated.
     
  2. jcsd
  3. Jun 16, 2013 #2
    On the contrary, your work so far demonstrates necessity. You proved [itex]F(u,v) = 0 \implies \nabla u \times \nabla v = 0[/itex]. In other words, you proved [itex]\nabla u \times \nabla v = 0[/itex] is necessary follows from [itex]F(u,v) = 0[/itex].

    Now prove [itex]\nabla u \times \nabla v = 0 \implies F(u,v) = 0[/itex] to demonstrate sufficiency.
     
    Last edited: Jun 16, 2013
  4. Jun 16, 2013 #3
    Let F[u,v] = uv

    ∇F = u∇v + v∇u

    Repeat the cross products? I guess I'm lost on the starting point to demonstrate sufficiency.
     
  5. Jun 17, 2013 #4
    My previous post was in error, which I will proceed to correct with a detailed explanation.

    Assume [itex]F(u,v) = 0[/itex].

    [tex]F(u,v) = 0[/tex]

    [tex]\implies \begin{cases} \frac{dF}{dx} = 0 \\\\ \frac{dF}{dy} = 0 \\\\ \frac{dF}{dz} = 0 \end{cases}[/tex]

    [tex]\implies \begin{cases} \frac{\partial F}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial F}{\partial v}\frac{\partial v}{\partial x} = 0 \\\\ \frac{\partial F}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial F}{\partial v}\frac{\partial v}{\partial y} = 0 \\\\ \frac{\partial F}{\partial u}\frac{\partial u}{\partial z} + \frac{\partial F}{\partial v}\frac{\partial v}{\partial z} = 0 \end{cases}[/tex]

    [tex]\implies \begin{cases} \frac{\partial F}{\partial u}\frac{\partial u}{\partial x} = -\frac{\partial F}{\partial v}\frac{\partial v}{\partial x} \\\\ \frac{\partial F}{\partial u}\frac{\partial u}{\partial y} = -\frac{\partial F}{\partial v}\frac{\partial v}{\partial y} \\\\ \frac{\partial F}{\partial u}\frac{\partial u}{\partial z} = -\frac{\partial F}{\partial v}\frac{\partial v}{\partial z} \end{cases}[/tex]

    [tex]\implies \begin{cases} \frac{\partial u}{\partial x} \frac{\partial v}{\partial y} = \frac{\partial u}{\partial y}\frac{\partial v}{\partial x} \\\\ \frac{\partial u}{\partial x} \frac{\partial v}{\partial z} = \frac{\partial u}{\partial z}\frac{\partial v}{\partial x} \\\\ \frac{\partial u}{\partial y} \frac{\partial v}{\partial z} = \frac{\partial u}{\partial z}\frac{\partial v}{\partial y} \end{cases}[/tex]

    Now, compute [itex]\nabla u \times \nabla v[/itex].

    [tex]\nabla u \times \nabla v = (\frac{\partial u}{\partial y}\frac{\partial v}{\partial z} - \frac{\partial u}{\partial z}\frac{\partial v}{\partial y})\mathbf{i} - (\frac{\partial u}{\partial x}\frac{\partial v}{\partial z} - \frac{\partial u}{\partial z}\frac{\partial v}{\partial x})\mathbf{j} + (\frac{\partial u}{\partial x}\frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}\frac{\partial v}{\partial x})\mathbf{k} = 0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} = \mathbf{0}[/tex]

    We have proved [itex]\nabla u \times \nabla v = \mathbf{0}[/itex] necessarily follows from [itex]F(u,v) = 0[/itex]. In other words, [itex]\nabla u \times \nabla v = \mathbf{0}[/itex] is necessary for [itex]F(u,v) = 0[/itex].

    Reversing the direction of the proof shows [itex]\nabla u \times \nabla v = \mathbf{0} \implies F(u,v) = 0[/itex]. Therefore, [itex]\nabla u \times \nabla v = \mathbf{0}[/itex] is sufficient for [itex]F(u,v) = 0[/itex].

    You will notice we did not directly use the result from 1.a. in our answer to 1.b. I will investigate further and report any connection I discover. Cheers!
     
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