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Laplace's Eqn and Cauchy's Integral Formula

  1. Oct 21, 2005 #1
    Is there a connection between Laplace's Equation and Cauchy's integral formula? There seems to be quite a similarity, eg, solutions of Laplaces Eqn are determined by their values at the boundary.
  2. jcsd
  3. Oct 21, 2005 #2


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    Yes, there is a connection. Cauchy's integral formula assumes that the function in question is analytic. A function is analytic if and only if it satisfies the Cauchy-Riemann equations:
    If f(z)=u(x,y)+iv(x,y), then
    [tex]\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}[/tex]
    [tex]\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}[/tex]
    Since the function is analytic, then u and y have continous partial derivatives of all orders, so we may differentiate the above expressions to obtain:
    [tex]\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 v}{\partial x \partial y}[/tex]
    [tex]\frac{\partial^2 u}{\partial y^2}=-\frac{\partial^2 v}{\partial y \partial x}[/tex]
    Since these derivatives are continuos, then:
    [tex]\frac{\partial^2 v}{\partial y \partial x}=\frac{\partial^2 v}{\partial x \partial y}[/tex]
    [tex]\frac{\partial^2 u}{\partial x^2}=-\frac{\partial^2 u}{\partial y^2}[/tex]
    [tex]\rightarrow \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0[/tex]
    Which is Laplace's equation. It can be proven similarly that the imaginary part of f also satisfies Laplace's equation.
    Last edited: Oct 21, 2005
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