# Laplace's Eqn and Cauchy's Integral Formula

1. Oct 21, 2005

### Euclid

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. Oct 21, 2005

### LeonhardEuler

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
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$
$$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$
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:
$$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 v}{\partial x \partial y}$$
$$\frac{\partial^2 u}{\partial y^2}=-\frac{\partial^2 v}{\partial y \partial x}$$
Since these derivatives are continuos, then:
$$\frac{\partial^2 v}{\partial y \partial x}=\frac{\partial^2 v}{\partial x \partial y}$$
Therefore:
$$\frac{\partial^2 u}{\partial x^2}=-\frac{\partial^2 u}{\partial y^2}$$
$$\rightarrow \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$$
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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