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Cauchy Riemann Conditions Question

  1. Oct 15, 2004 #1
    Ok, I am told in a complex analysis book that the gradient squared of u is equal to the gradient squared of v which is equal to 0.

    We know the derivate of w exists, and w(z)=u(x,y) + iv(x,y)

    Thus the Cauchy Riemann conditions must hold. (When I use d assume that it refers to a partial derivative)

    So du/dx=dv/dy and du/dy=-dv/dx by Cauchy Riemann

    We know the gradient squared of u is equal to d^2u/dx^2 + d^2u/dy^2 is equal to d^2v/dy^2 - d^2v/dx^2

    We know the gradient squared of v is equal to d^2v/dx^2 + d^2v/dy^2 which is equal to d^2u/dx^2 - d^2u/dy^2

    Am I correct in assuming this? And if I am, I dont see how the gradient squared of u is equal to the gradient squared of v. Any suggestions?
     
  2. jcsd
  3. Oct 16, 2004 #2

    HallsofIvy

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    Cauchy-Riemann conditions:
    (a) du/dx= dv/dy and (b) du/dy= - dv/dx

    Differentiate both sides of (a) with respect to x:
    d2/dx2= d2v/dydx

    Differentiate both sides of (b) with respect to y:
    d2u/dy2= -d2v/dydx

    Adding: d2/dx2+ d2u/dy2= d2v/dydx -d2v/dydx= 0.

    Differentiate (a) with respect to y and (b) with respect to x and subtract to get the formula for v.

    The "Laplacian" (what you are calling "gradient squared") of u and v are not just equal, they are both 0. The real and imaginary parts of an analytic function are always "harmonic functions", i.e. they satisfy
    [itex]\Delta u=0[/itex] and [itex]\Delta v= 0[/itex].
     
    Last edited: Oct 16, 2004
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