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Math Help - [Complex Analysis] Am I heading the right direction with this proof? Feel like it's m

  1. #1
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    [Complex Analysis] Am I heading the right direction with this proof? Feel like it's m

    In this problem, you are guided through the start of the proof of the proposition:
    If u(x,y) is a harmonic function and v(x,y) is its harmonic conjugate, then the function φ(x,y) = u(x,y)v(x,y) is harmonic.
    Proof Suppose f(z) = u(x,y) +iv(x,y) is analytic in a domain D. We saw in Section 3.2 that the product of two analytic functions is analytic. Hence [f(z)]2 is analytic. Now examine [f(z)]2 and finish the proof.

    This is what I have so far: If u(x,y) is a harmonic function and v(x,y) is its harmonic conjugate, then the function φ(x,y) = u(x,y)v(x,y) is harmonic. Since f(z) = u(x,y) +iv(x,y) is an analytic function then [f(z)]2 is also analytic. Therefore:
    [f(z)]2 = [u(x,y) + iv(x,y)]2 = [u(x,y)]2 - [v(x,y)]2 + i2u(x,y)(v(x,y)
    Sine the product of a harmonic function is harmonic, then u(x,y)v(x,y) is harmonic.
    I feel like I'm not saying anything, though! Any insight?

    Another attempt: Suppose f(z) = u(x,y) +iv(x,y) is analytic in a domain D. We saw in Section 3.2 that the product of two analytic functions is analytic. Hence [f(z)]2 is analytic. Thus:
    [f(z)]2 = [u(x,y) + iv(x,y)]2 = [u(x,y)]2 - [v(x,y)]2 + i2u(x,y)(v(x,y)
    Since u(x,y) is a harmonic equation then [u(x,y)]2 is also harmonic. Similarly, since v(x,y) is the harmonic conjugate of u(x,y) then [v(x,y)]2 is also the harmonic conjugate of [u(x,y)]2.
    Therefore, focusing on the term i2u(x,y)(v(x,y) and finding the partial derivatives with respect to x and respect to y respectively: Here I do the math.
    When put into LaPlace's Formula it simplified to 4i[VxUx + UyVy]. Since the function is analytic the CR equations hold and with substitution the function is 4i[Vx(Vy) + (-Vx)Vy) = 4i(o0 =0. Showing the product of conjugate harmonic components is also harmonic.

    Is this too big of a jump? Can I just assume since u(x,y) is a harmonic equation then [u(x,y)]2 is also harmonic?
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  2. #2
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    Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

    just show that $\nabla^2 \varphi(x,y) = 0$

    $\varphi(x,y)=uv$

    $\varphi_x=\dfrac {\partial}{\partial x} \varphi = u_x v + u v_x$

    $\varphi_y=\dfrac {\partial}{\partial y} \varphi = u_y v + u v_y$

    $\varphi_{xx}=\dfrac {\partial}{\partial x}\left(u_x v + u v_x\right)=u_{xx}v + 2 u_x v_x + u v_{xx}$

    $\varphi_{yy}=\dfrac {\partial}{\partial y}\left( u_y v + u v_y \right) = u_{yy} v + 2 u_y v_y + u v_{yy}$

    $\varphi_{xx}+\varphi_{yy}=v(u_{xx}+u_{yy}) + u(v_{xx}+ v_{yy}) + 2(u_x v_x + u_y v_y) = 0 + 0 + u_x (-u_y) + u_y (u_x)=0$

    the Laplacians of u and v are each 0 as they are each harmonic functions.

    the last equality is from the CR equations of u and v which are satisfied as they are harmonic conjugates of one another.
    Last edited by romsek; April 21st 2014 at 08:22 PM.
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    Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

    Let me put your suggestions into proof form and I'll post it in a jiffy. Thanks!
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    Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

    Quote Originally Posted by romsek View Post
    just show that $\nabla^2 \varphi(x,y) = 0$

    $\varphi(x,y)=uv$

    $\varphi_x=\dfrac {\partial}{\partial x} \varphi = u_x v + u v_x$

    $\varphi_y=\dfrac {\partial}{\partial y} \varphi = u_y v + u v_y$

    $\varphi_{xx}=\dfrac {\partial}{\partial x}\left(u_x v + u v_x\right)=u_{xx}v + 2 u_x v_x + u v_{xx}$

    $\varphi_{yy}=\dfrac {\partial}{\partial y}\left( u_y v + u v_y \right) = u_{yy} v + 2 u_y v_y + u v_{yy}$

    $\varphi_{xx}+\varphi_{yy}=v(u_{xx}+u_{yy}) + u(v_{xx}+ v_{yy}) + 2(u_x v_x + u_y v_y) = 0 + 0 + u_x (-u_y) + u_y (u_x)=0$

    the Laplacians of u and v in each 0 as they are each harmonic functions.

    the last equality is from the CR equations of u and v which are satisfied as they are harmonic conjugates of one another.
    I'm looking through my Complex Analysis notes and I'm not seeing why $u_x (-u_y) + u_y (u_x)=0$. Is this a known theorem?
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    Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

    Quote Originally Posted by WingardiumLeviOsa View Post
    I'm looking through my Complex Analysis notes and I'm not seeing why $u_x (-u_y) + u_y (u_x)=0$. Is this a known theorem?
    look at it more closely

    $\Large u_x (-u_y) + u_y (u_x) = u_x u_y - u_x u_y = 0$
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    Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

    Quote Originally Posted by romsek View Post
    look at it more closely

    $\Large u_x (-u_y) + u_y (u_x) = u_x u_y - u_x u_y = 0$
    Wow... I know I'm blind but I didn't know I was that blind! Thanks! It's been a long night.
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    Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

    Quote Originally Posted by romsek View Post
    look at it more closely

    $\Large u_x (-u_y) + u_y (u_x) = u_x u_y - u_x u_y = 0$
    [Complex Analysis] Am I heading the right direction with this proof? Feel like it's m-2014-04-22-00.09.07.jpg
    Does that look great or do I need to explain myself in other parts? Thank you so much for helping me see this.
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    Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

    Quote Originally Posted by WingardiumLeviOsa View Post
    Click image for larger version. 

Name:	2014-04-22 00.09.07.jpg 
Views:	4 
Size:	714.4 KB 
ID:	30723
    Does that look great or do I need to explain myself in other parts? Thank you so much for helping me see this.
    looks fine to me
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  9. #9
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    Re: [Complex Analysis] Am I heading the right direction with this proof? Feel like it

    Quote Originally Posted by romsek View Post
    looks fine to me
    Thanks!
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