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Incompressiblitiy and unit length implies constants

  1. Nov 19, 2009 #1
    Let [tex]u,v\in C^1(R^2)[/tex], and
    [tex]
    (u)^1+(v)^2=1,\partial_x u+ \partial_y v=0[/tex].
    Then [tex]u,v[/tex] are constants.

    Is it right? How to prove?
     
    Last edited: Nov 19, 2009
  2. jcsd
  3. Nov 21, 2009 #2
    Any guys here knows? Thank you...
     
  4. Nov 21, 2009 #3

    HallsofIvy

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    It's hard to tell what you are asking. "[itex]C^1(R^2)[/itex]" is the set of continuously differentiable functions in two variables but what are "(u)1" and "(v)2"?
     
  5. Nov 21, 2009 #4
    Thank HallsofIvy for reminding...And (u)^2 is just the square of u, same as the meaning of (v)^2...
     
  6. Nov 22, 2009 #5
    But your question has [itex](u)^1[/itex] not [itex](u)^2[/itex] ... was that an error?
     
  7. Nov 24, 2009 #6
    What about (u,v) = (cos t, sin t) over all R^2?
     
  8. Nov 24, 2009 #7

    HallsofIvy

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    Then what is "t"? Do you mean the [itex]\theta[/itex] coordinate in polar coordinates for the point (x,y)? In that case, [itex]\partial u/\partial v+ \partial v/\partial y[/itex] would not be equal to 0.
     
  9. Nov 25, 2009 #8
    No, time. I am thinking in terms of fluid mechanics, where (u,v) is the velocity of the fluid (incompressibility suggested this to me). So the fluid is incompressible and has speed 1.
    Otherwise I don't understand how the question makes any sense, "Then u,v are constants?"... constant with respect to what, if not time?
     
  10. Jan 4, 2010 #9
    I mean if X=(u,v) =(u(x,y),v(x,y)) satisifies
    1. |X|=1;
    2. div X=0;
    then X is a constant, i.e. idependent of x,y.
     
  11. Jan 4, 2010 #10

    LCKurtz

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    Consider:

    [tex] u(x,y) = \frac {-y}{\sqrt{x^2+y^2}},\ v(x,y) = \frac {x}{\sqrt{x^2+y^2}}[/tex]

    [Edit] I just noticed your original post required continuity and this example isn't continuous at (0,0).
     
    Last edited: Jan 4, 2010
  12. Jan 6, 2010 #11
    Year, the question is originated from vector analysis, then all functions are smooth, i.e. infinitely differentiable.Thank you.
     
  13. Jan 7, 2010 #12
    Well, it's not always true (although I don't have a counter-example),
    But it resembles the fact that analytic complex function with constant absolute values,
    must be constant.
    Here u and v play the roll of the real & imaginary parts of the complex function, and
    the condition about the derivatives is one of the cauchy reimann equations. Problem is
    you need the second one to complete the proof.

    Let me show how I would start the proof, and then when I get stuck:

    First of all, [tex]u^{2}+v^{2}=1[/tex] So at every point, either u or v are not zero.
    No let's differentiate the equality wrt x and wrt y:
    (I also divided the equations by 2)
    [tex]uu_{x}+vv_{x}=0[/tex]
    [tex]uu_{y}+vv_{y}=0[/tex]

    So we have a system of equation wrt to u and v. But we also know that we have at each point a non-trivial solution, so the determinant must equal zero:

    [tex]det(A)=u_{x}v_{y}-v_{x}u_{y}[/tex]

    Now I only know that [tex]u_{x}+v_{y}=0[/tex], but it's not enought to complete the proof, beacuse I'd also need the other CR equation: [tex]u_{y}-v_{x}=0[/tex]

    Then I could substitute this into the determinant condition and get:

    [tex]u^{2}_{x}+u^{2}_{y}=0[/tex]
    [tex]v^{2}_{x}+v^{2}_{y}=0[/tex]
    Which implies

    [tex]u_{x}=u_{y}=v_{x}=v_{y}=0[/tex]

    So u & v are constant.

    But you don't have the extra information, so generally it's not true.
     
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