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Thread: [SOLVED] Harmonic conjugate

  1. #1
    Bop
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    [SOLVED] Harmonic conjugate

    Hi! I have this function:

    $\displaystyle u(x,y)=y^{3}-3x^{2}y$

    And I want it harmonic conjugate f(z) where:

    $\displaystyle f(z)=u(z)+iv(z)$


    so I use cauchy-riemann equations:

    $\displaystyle u_{x}=v_{y} and u_{y}=-v_{x}$

    So:

    $\displaystyle u_{x}=-6yx and u_{y}=3y^{2}-3x^{2}$

    $\displaystyle \int -6yx dy= -3xy^{2}+C$

    $\displaystyle \int -(3y^{2}-3x^{2}) dx= -3y^{2}x+x^{3}+C$

    Why have i two diferents values for $\displaystyle v(z)$?


    Thanks!
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  2. #2
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    $\displaystyle u_{x}(x,y) = -6xy $

    $\displaystyle u_{y}(x,y) = 3y^{2} -3x^{2} $


    $\displaystyle v(x,y) = \int u_{x} (x,y)dy$

    $\displaystyle v(x,y) = -6 \int xy \ dy = -3xy^{2} + C(x)$


    $\displaystyle v_{x}(x,y) = -3y^{2} + C'(x) = -u_{y}(x,y) = -3y^{2}+3x^{2} $

    so $\displaystyle C'(x) = 3x^{2} $

    then $\displaystyle C(x) = x^{3} $

    and $\displaystyle v(x,y) = -3xy^{2} + x^{3} + C$
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  3. #3
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    Quote Originally Posted by Bop View Post
    Hi! I have this function:
    Why have i two diferents values for $\displaystyle v(z)$?
    Thanks!
    Because when you partially integrate with respect to $\displaystyle x$ you forgot to consider an arbitrary function of $\displaystyle y$ that was lost (since any function of $\displaystyle y$ differentiated w.r.t. $\displaystyle x$ is 0) and vice versa.

    Another way of solving your problem:

    $\displaystyle u_x = -6xy, \quad u_y=3y^2-3x^2$

    $\displaystyle v_y = -6xy$
    $\displaystyle v_x = -u_y = 3x^2-3x^2$

    $\displaystyle v = \int -6xy\,dy=-3y^2x+g(x)+C$ (integrate w.r.t. y)
    $\displaystyle v = \int(3x^2-3y^2)\,dx=x^3-3y^2x+h(y)+C$ (integrate w.r.t. x)

    Therefore, by inspection, $\displaystyle g(x)$ must be $\displaystyle x^3$, and $\displaystyle h(y) = 0$.

    So $\displaystyle v = -3y^2x + x^3 + C$
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