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Thread: Complex Analysis, harmonic functions

  1. #1
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    Complex Analysis, harmonic functions

    Here is the problem:

    If f= u+ iv is analytic in a region, show that uv is harmonic in the region but that u^2 need not be harmonic.

    Thanks
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  2. #2
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    Quote Originally Posted by taypez View Post
    Here is the problem:

    If f= u+ iv is analytic in a region, show that uv is harmonic in the region but that u^2 need not be harmonic.

    Thanks
    $\displaystyle f(x+iy)=u+iv$
    The function is analyic, hence homolorphic.
    It satisfies the Cauchy-Riemann Equations.
    $\displaystyle u_x=v_y$
    $\displaystyle u_y=-v_x$
    Thus,
    $\displaystyle u_xv_x+u_yv_y=0$

    We need to show,
    $\displaystyle uv$
    Solves the PDE,
    $\displaystyle f_{xx}+f_{yy}=0$.

    Now,
    $\displaystyle (uv)_{xx}=(u_xv+uv_x)_x=u_{xx}v+2u_xv_x+uv_{xx}$
    $\displaystyle (uv)_{yy}=u_{yy}v+2u_yv_y+uv_{yy}$
    Add them,
    $\displaystyle (uv)_{xx}+(uv)_{yy}=u_{xx}v+2(u_xv_x+u_yv_y)+uv_{y y}$
    But from Cauchy-Riemann Equations,
    $\displaystyle (uv)_{xx}+(uv)_{yy}=u_{xx}v+uv_{yy}$
    Maybe there is some theorem that tell the right hand side is zero. I do not know of one.
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  3. #3
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    The function is analyic, hence both u and v are harmonic.
    It satisfies the Cauchy-Riemann Equations.
    $\displaystyle u_x=v_y$ & $\displaystyle u_y=-v_x$
    Now,
    $\displaystyle (uv)_{xx}=(u_xv+uv_x)_x=u_{xx}v+2u_xv_x+uv_{xx}$
    $\displaystyle (uv)_{yy}=u_{yy}v+2u_yv_y+uv_{yy}$
    Add them,
    $\displaystyle (uv)_{xx}+(uv)_{yy}=2(u_xv_x+u_yv_y) $ because
    $\displaystyle u_{xx}+ u_{yy}=0$ & $\displaystyle v_{xx}+ v_{yy}=0.$

    But $\displaystyle u_x v_x + u_y v_y = u_x \left( { - u_y } \right) + u_y \left( { u_x } \right) = 0$
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  4. #4
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    Quote Originally Posted by Plato View Post
    The function is analyic, hence both u and v are harmonic.
    Slow down, I did not get that. You are saying that if a complex function is analytic (meaning it has a power series expansion, right?) then it must be harmonic, (satisfy the Laplace equation). So that was the extra equation you used to complete the proof?
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    a complex function is analytic (meaning it has a power series expansion, right?) then it must be harmonic, (satisfy the Laplace equation).
    For the complex analyst the definition is: a function $\displaystyle f$ said to be analytic at $\displaystyle z_0$ if its derivative exists at each point in some neighborhood of $\displaystyle z_0$.

    Moreover, a function $\displaystyle u$ is said to be harmonic if $\displaystyle u_{xx} + u_{yy} = 0$, that is the Laplace equation as you have noted.

    Theorem: If $\displaystyle f(z) = u(x,y) + iv(x,y)$ is analytic is the domain $\displaystyle D$, then both $\displaystyle u(x,y)$ and $\displaystyle v(x,y)$ are harmonic in $\displaystyle D$.
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