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Homework Help: Technical question in multi-variable differentiation

  1. Feb 2, 2008 #1
    Let [tex]f(x+iy)=u(x,y)+iv(x,y)[/tex]. Suppose we know [tex]|f|^2=u^2+v^2[/tex] is a constant function. Then we are allowed to say that [tex](u^2+v^2)_x=(u^2+v^2)_y=0[/tex]. But are we allowed to differentiate u by x and v by y? IE, are we allowed to make the following statement:
    [tex](u^2)_x+(v^2)_y=0[/tex]

    I'm guessing 'no', but I'm not too sure why. Intuitively, I would guess that you could change u and v in such a way that those changes balance each other out? (Very unclear way to say it...)
     
  2. jcsd
  3. Feb 2, 2008 #2
    no youre not allowed.

    add the functions together and differentiate it with respect to whatever subscipt it is
     
  4. Feb 2, 2008 #3

    quasar987

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    Not you're not allowed. Consider for instance the functions u²=2x+3y and v²=-2x-3y. Then [tex](u^2+v^2)_x=(u^2+v^2)_y=0[/tex], but [tex](u^2)_x+(v^2)_y=-1[/tex]
     
  5. Feb 3, 2008 #4

    HallsofIvy

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    Perhaps you could but that has nothing to do with the derivative. My question is why on earth would you even consider that [itex]u^2_x+ v^2_y= 0[/itex]?
     
  6. Feb 3, 2008 #5
    My friend and I used that in a complex analysis proof that, for a function [tex]f(x+iy) = u(x,y)+iv(x,y)[/tex] that is holomorphic on an open set, if |f| is constant then f is constant.

    If |f| is constant, then [tex]|f|^2 = u^2+v^2[/tex] is constant. Then the derivatives [tex](u^2+v^2)_x[/tex] and [tex](u^2+v^2)_y = 0[/tex].

    We then also used the fact that [tex](u^2)_x+(v^2)_y = 0[/tex] and manipulated the three equations using the Cauchy-Riemann equations to show that all partials were equal to zero ([tex]u_x = u_y = v_x = v_y = 0[/tex]).

    Will have to rethink the proof...
     
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