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Math Help - [SOLVED] Harmonic conjugate

  1. #1
    Bop
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    [SOLVED] Harmonic conjugate

    Hi! I have this function:

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

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

    f(z)=u(z)+iv(z)


    so I use cauchy-riemann equations:

    u_{x}=v_{y} and u_{y}=-v_{x}

    So:

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

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

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

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


    Thanks!
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  2. #2
    Super Member Random Variable's Avatar
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     u_{x}(x,y) = -6xy

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


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

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


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

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

    then  C(x) = x^{3}

    and  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 v(z)?
    Thanks!
    Because when you partially integrate with respect to x you forgot to consider an arbitrary function of y that was lost (since any function of y differentiated w.r.t. x is 0) and vice versa.

    Another way of solving your problem:

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

    v_y = -6xy
    v_x = -u_y = 3x^2-3x^2

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

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

    So v = -3y^2x + x^3 + C
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