Chemistry - Cauchy-Riemann equations

In mathematics, the Cauchy-Riemann differential equations in complex analysis, named after Augustin Cauchy and Bernhard Riemann, are two partial differential equations which provide a necessary and sufficient condition for a function to be holomorphic.

Let f(x+iy) = u + iv be a function from an open subset of the complex numbers C to C, and regard u and v as real-valued functions defined on an open subset of R². Then f is holomorphic if and only if u and v are differentiable and their partial derivatives satisfy the Cauchy-Riemann equations, which are:

${ \partial u \over \partial x } = { \partial v \over \partial y}$

and

${ \partial u \over \partial y } = -{ \partial v \over \partial x}.$

It follows from the equations that u and v must be harmonic functions. The equations can therefore be seen as the conditions on a given pair of harmonic functions to come as real and imaginary parts of a complex-analytic function.

Derivation

Consider a function f(z) = u(x, y) + i v(x, y) over C, and we wish to calculate its derivative at some point, z₀. We can essentially approach z₀ along the real axis towards 0, or down the imaginary axis towards 0.

If we take the first path:

$f'(z)=\lim_{h\rightarrow 0} {f(z+h)-f(z) \over h}=\lim_{h\rightarrow 0}{u(x+h,y)+iv(x+h,y)-(u(x,y)+iv(x,y))\over h}$

$=\lim_{h\rightarrow 0}{(u(x+h,y)-u(x,y))+i(v(x+h,y)-v(x,y))\over h}=\lim_{h\rightarrow 0}{u(x+h,y)-u(x,y) \over h}+i{v(x+h,y)-v(x,y) \over h}.$

This is now in the form of two difference quotients, so now

$f'(z)={\partial u \over \partial x} + i {\partial v \over \partial x}.$

Taking the second path:

$f'(z)=\lim_{h\rightarrow 0} {f(z+ih)-f(z) \over ih}=\lim_{h\rightarrow 0}{u(x,y+h)+iv(x,y+h)-(u(x,y)+iv(x,y))\over ih}$

$=\lim_{h\rightarrow 0}{u(x,y+h)-u(x,y) \over ih} +i{v(x,y+h)-v(x,y) \over ih}$

$=\lim_{h\rightarrow 0}-i{u(x,y+h)-u(x,y) \over h}+{v(x,y+h)-v(x,y) \over h}=\lim_{h\rightarrow 0}{v(x,y+h)-v(x,y) \over h}-i{u(x,y+h)-u(x,y) \over h}.$

Again, this is now in the form of two difference quotients, so

$f'(z)={\partial v \over \partial y} - i {\partial u \over \partial y}.$

Equating these two we get

${\partial u \over \partial x} + i {\partial v \over \partial x} = {\partial v \over \partial y} - i {\partial u \over \partial y}.$

Equating real and imaginary parts, then

${\partial u \over \partial x} = {\partial v \over \partial y}$

${\partial v \over \partial x} = - {\partial u \over \partial y}, {\partial u \over \partial y} = -{\partial v \over \partial x}.\quad\square$

Polar representation

Considering the polar representation $z = r e i θ$ , the equations take the form

${ \partial u \over \partial r } = {1 \over r}{ \partial v \over \partial \theta},$

${ \partial v \over \partial r } = -{1 \over r}{ \partial u \over \partial \theta}.$

Categories: Partial differential equations | Complex analysis | Equations