Milne-Thomson method for finding a holomorphic function

In mathematics, the Milne-Thomson method is a method for finding a holomorphic function whose real or imaginary part is given. It is named after Louis Melville Milne-Thomson.

Introduction

Let [math]\displaystyle{ z = x + iy }[/math] and [math]\displaystyle{ \bar {z}\ = x - iy }[/math] where [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math] are real.

Let [math]\displaystyle{ f(z) = u(x,y) + iv(x,y) }[/math] be any holomorphic function.

Example 1: [math]\displaystyle{ z^4 = (x^4-6x^2y^2+y^4) +i(4x^3y-4xy^3) }[/math]

Example 2: [math]\displaystyle{ \exp(iz)=\cos(x)\exp(-y)+i\sin(x)\exp(-y) }[/math]

In his article, Milne-Thomson considers the problem of finding [math]\displaystyle{ f(z) }[/math] when 1. [math]\displaystyle{ u(x,y) }[/math] and [math]\displaystyle{ v(x,y) }[/math] are given, 2. [math]\displaystyle{ u(x,y) }[/math] is given and [math]\displaystyle{ f(z) }[/math] is real on the real axis, 3. only [math]\displaystyle{ u(x,y) }[/math] is given, 4. only [math]\displaystyle{ v(x,y) }[/math] is given. He is really interested in problems 3 and 4, but the answers to the easier problems 1 and 2 are needed for proving the answers to problems 3 and 4.

1st problem

Problem: [math]\displaystyle{ u(x,y) }[/math] and [math]\displaystyle{ v(x,y) }[/math] are known; what is [math]\displaystyle{ f(z) }[/math]?

Answer: [math]\displaystyle{ f(z)=u(z,0)+iv(z,0) }[/math]

In words: the holomorphic function [math]\displaystyle{ f(z) }[/math] can be obtained by putting [math]\displaystyle{ x = z }[/math] and [math]\displaystyle{ y = 0 }[/math] in [math]\displaystyle{ u(x,y)+iv(x,y) }[/math].

Example 1: with [math]\displaystyle{ u(x,y)=x^4-6x^2y^2+y^4 }[/math] and [math]\displaystyle{ v(x,y)=4x^3y-4xy^3 }[/math] we obtain [math]\displaystyle{ f(z)=z^4 }[/math].

Example 2: with [math]\displaystyle{ u(x,y)=\cos(x)\exp(-y) }[/math] and [math]\displaystyle{ v(x,y)=\sin(x)\exp(-y) }[/math] we obtain [math]\displaystyle{ f(z)=\cos(z)+i\sin(z)=\exp(iz) }[/math].

Proof:

From the first pair of definitions [math]\displaystyle{ x = \frac{z + \bar {z}}{2} }[/math] and [math]\displaystyle{ y = \frac{z - \bar {z}}{2i} }[/math].

Therefore [math]\displaystyle{ f(z)= u \left( \frac{z + \bar {z}}{2}\ , \frac{z - \bar {z}}{2i}\right) + iv\left( \frac{z + \bar {z}}{2}\ , \frac{z - \bar {z}}{2i}\right) }[/math].

This is an identity even when [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math] are not real, i.e. the two variables [math]\displaystyle{ z }[/math] and [math]\displaystyle{ \bar {z}\ }[/math] may be considered independent. Putting [math]\displaystyle{ \bar {z} =z }[/math] we get [math]\displaystyle{ f(z) = u(z,0) + iv(z,0) }[/math].

2nd problem

Problem: [math]\displaystyle{ u(x,y) }[/math] is known, [math]\displaystyle{ v(x,y) }[/math] is unknown, [math]\displaystyle{ f(x+i0) }[/math] is real; what is [math]\displaystyle{ f(z) }[/math]?

Answer: [math]\displaystyle{ f(z)=u(z,0) }[/math].

Only example 1 applies here: with [math]\displaystyle{ u(x,y)=x^4-6x^2y^2+y^4 }[/math] we obtain [math]\displaystyle{ f(z)=z^4 }[/math].

Proof: "[math]\displaystyle{ f(x+i0) }[/math] is real" means [math]\displaystyle{ v(x,0)=0 }[/math]. In this case the answer to problem 1 becomes [math]\displaystyle{ f(z)=u(z,0) }[/math].

3rd problem

Problem: [math]\displaystyle{ u(x,y) }[/math] is known, [math]\displaystyle{ v(x,y) }[/math] is unknown; what is [math]\displaystyle{ f(z) }[/math]?

Answer: [math]\displaystyle{ f(z)=u(z,0)-i \int u_y(z,0) dz }[/math] (where [math]\displaystyle{ u_y(x,y) }[/math] is the partial derivative of [math]\displaystyle{ u(x,y) }[/math] with respect to [math]\displaystyle{ y }[/math]).

Example 1: with [math]\displaystyle{ u(x,y)=x^4-6x^2y^2+y^4 }[/math] and [math]\displaystyle{ u_y(x,y)=-12x^2y+4y^3 }[/math] we obtain [math]\displaystyle{ f(z)=z^4+iC }[/math] with real but undetermined [math]\displaystyle{ C }[/math].

Example 2: with [math]\displaystyle{ u(x,y)=\cos(x)\exp(-y) }[/math] and [math]\displaystyle{ u_y(x,y)=-\cos(x)\exp(-y) }[/math] we obtain [math]\displaystyle{ f(z)=\cos(z)+i\int\cos(z)dz=\cos(z)+i(\sin(z)+C)=\exp(iz)+iC }[/math].

Proof: This follows from [math]\displaystyle{ f(z)=u(z,0)+i \int v_x(z,0) dz }[/math] and the 2nd Cauchy-Riemann equation [math]\displaystyle{ u_y(x,y)=-v_x(x,y) }[/math].

4th problem

Problem: [math]\displaystyle{ u(x,y) }[/math] is unknown, [math]\displaystyle{ v(x,y) }[/math] is known; what is [math]\displaystyle{ f(z) }[/math]?

Answer: [math]\displaystyle{ f(z)=\int v_y(z,0)dz+i v(z,0) }[/math].

Example 1: with [math]\displaystyle{ v(x,y)=4x^3y-4xy^3 }[/math] and [math]\displaystyle{ v_y(x,y)=4x^3-12xy^2 }[/math] we obtain [math]\displaystyle{ f(z)=\int 4z^3dz+i0=z^4+C }[/math] with real but undetermined [math]\displaystyle{ C }[/math].

Example 2: with [math]\displaystyle{ v(x,y)=\sin(x)\exp(-y) }[/math] and [math]\displaystyle{ v_y(x,y)=-\sin(x)\exp(-y) }[/math] we obtain [math]\displaystyle{ f(z)=-\int\sin(z)dz+i\sin(z)=\cos(z)+C+i\sin(z)=\exp(iz)+C }[/math].

Proof: This follows from [math]\displaystyle{ f(z)=\int u_x(z,0) dz +i v(z,0) }[/math] and the 1st Cauchy-Riemann equation [math]\displaystyle{ u_x(x,y)=v_y(x,y) }[/math].