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Proving a complex harmonic function must be constant if the square is also constant

  1. Jan 29, 2012 #1
    Let f(x,y)=u+iv be a complex harmonic function such that (f(x,y))2 is also harmonic (both on a domain D). Show that f(x,y) must be a constant

    I've attempted to show this through brute force, but perhaps there is something more elegant? I know that the Laplace equations must hold, so what I have done (I'll spare the readers the details) is to calculate out f(x,y)2, separate the real and imaginary parts in terms of u and v, and calculate the second order partial derivatives for the Laplace equations. After some calculation and cancellation (in particular of the second partials using the fact that f itself is harmonic), I get:

    ux2+uy2-vx2-vy2=0

    and

    uxvy+uxvy=0

    I need to somehow get from these two equations to the fact that f is constant. Looking at this, it seems like I'm just a step away, but I am missing something. Any suggestions on what I might be overlooking?
     
  2. jcsd
  3. Jan 29, 2012 #2

    Dick

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    Re: Proving a complex harmonic function must be constant if the square is also consta

    Did you think about trying to use the maximum principle?
     
  4. Jan 29, 2012 #3
    Re: Proving a complex harmonic function must be constant if the square is also consta

    unfortunately, we haven't covered the maximum principle yet. Of course, I could try to include the proof as part of my solution, but the fact that it hasn't been discussed yet suggests to me that there must be another way of solving.
     
  5. Jan 29, 2012 #4

    Dick

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    Re: Proving a complex harmonic function must be constant if the square is also consta

    Judging more by the title thread than what you actually stated in it, if f(x,y)^2=C a constant, then f(x,y) has a choice of at most two possible values in D. The two square roots of C. If f(x,y) is harmonic then it must be continuous. So it can't switch values. Is it that simple?
     
  6. Jan 29, 2012 #5
    Re: Proving a complex harmonic function must be constant if the square is also consta

    No, sorry, my error. That should have said "if the square is also harmonic."
     
  7. Jan 29, 2012 #6

    Dick

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    Re: Proving a complex harmonic function must be constant if the square is also consta

    Well, if f(x,y)=x+iy (which is complex harmonic, I think) then f(x,y)^2=(x^2-y^2)+2xyi. f(x,y)^2 is complex harmonic, isn't it? Now I really don't get what the question is. Do you mean f(x,y) to be real harmonic?
     
    Last edited: Jan 29, 2012
  8. Jan 30, 2012 #7
    Re: Proving a complex harmonic function must be constant if the square is also consta

    If what you are saying in your last post is correct, then the problem itself wouldn't make sense. But what are you using to determine that f^2 is harmonic? The product of harmonic functions need not be harmonic.

    Also, just to clarify with my confusing title:

    Assumptions: f and f^2 are harmonic
    Need to show: f is constant
     
  9. Jan 30, 2012 #8

    morphism

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    Re: Proving a complex harmonic function must be constant if the square is also consta

    What is your definition of a complex harmonic function? Is it just a function f:R^2->C such that u(x,y)=Re(f(x,y)) and v(x,y)=Im(f(x,y)) are real harmonic functions, i.e. satisfy the Laplace equation u_xx+u_yy=v_xx+v_yy=0?

    If so, then Dick's f(x,y)=x+iy is a counterexample to the claim "f and f^2 harmonic => f is constant".
     
  10. Jan 30, 2012 #9

    Dick

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    Re: Proving a complex harmonic function must be constant if the square is also consta

    I'm checking that the laplacian of x^2-y^2 and xy are zero. But I didn't really need to. f(x,y)=x+iy is analytic. So f(x,y)^2 is analytic. Your problem makes sense if f:R^2->R. I.e. if f(x,y) is real (not complex) harmonic.
     
  11. Jan 30, 2012 #10
    Re: Proving a complex harmonic function must be constant if the square is also consta

    Maybe I am having some deeper conceptual problem. I am now confused on several points (I think Morphism may have hit on it in that I am most likely misunderstanding what it means to be a complex harmonic function):

    1. How is the function f(x,y)=x+iy harmonic? u would be x and v would be y, and neither has a laplacian equal to zero. u_x+u_y=1+0=1. So this can't be a counterexample because it contradicts the hypotheses.

    2. Harmonic implies analytic but not vice versa, right? So I'm not sure what that comment meant.

    3. Back to morphism, let's say for a moment that this were a real function. The problem wouldn't even make sense then right? If f(x,y) is harmonic then f_x+f_y=0. The Laplacian of f(x,y)^2 would be 2*f(x,y)*f_x+2*f(x,y)*f_y=2f(x,y)(f_x+f_y)=0, so it would always be harmonic, and there would be nothing to force it to be constant. I can't find any other definition of a complex harmonic function though, so I am completely lost.

    I hope all this helps point out where my misunderstanding is!!
     
  12. Jan 30, 2012 #11

    morphism

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    Re: Proving a complex harmonic function must be constant if the square is also consta

    The Laplacian is [itex]\partial_{xx}+\partial_{yy}[/itex] and not [itex]\partial_x+\partial_y[/itex]. Maybe this is what's tripping you up!

    (And no: harmonic doesn't imply analytic. What is true is that analytic implies real and imaginary parts are (real) harmonic - this follows from the Cauchy-Riemann equations.)
     
  13. Jan 30, 2012 #12
    Re: Proving a complex harmonic function must be constant if the square is also consta

    oh okay, just figured part of my confusion out. If the whole function is analytic then each part is harmonic and they have to be harmonic conjugates. Still confused on the other portions.
     
  14. Jan 30, 2012 #13
    Re: Proving a complex harmonic function must be constant if the square is also consta

    ha, that's embarrassing (I really did know that as you can see from my first post). I've been looking at this one too long. Thanks! So it must be that it is supposed to be a real function, not complex. I'll have to check with the professor on that. So then it should be simpler. Now I'm getting the laplacian is 2((f_x)^2+f_xx+(f_y)^2+f_yy), which reduces because fxx+fyy must be zero since f is harmonic. So I am just left with (f_x)^2+(f_y)^2=0. From there I should be able to treat this as a differential equation? (That is another thing that bothers me, we keep integrating though we haven't proven the integrals exist!)
     
  15. Jan 30, 2012 #14

    Dick

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    Re: Proving a complex harmonic function must be constant if the square is also consta

    If (f_x)^2+(f_y)^2=0 then f_x=0 and f_y=0.
     
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