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 $z=x+iy$ and ${\bar {z}}\ =x-iy$ where $x$ and $y$ are real.

Let $f(z)=u(x,y)+iv(x,y)$ be any holomorphic function.

Example 1: $z^{4}=(x^{4}-6x^{2}y^{2}+y^{4})+i(4x^{3}y-4xy^{3})$

Example 2: $\exp(iz)=\cos(x)\exp(-y)+i\sin(x)\exp(-y)$

In his article, Milne-Thomson considers the problem of finding $f(z)$ when 1. $u(x,y)$ and $v(x,y)$ are given, 2. $u(x,y)$ is given and $f(z)$ is real on the real axis, 3. only $u(x,y)$ is given, 4. only $v(x,y)$ 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.

Given both real and imaginary parts

Problem: $u(x,y)$ and $v(x,y)$ are known; what is $f(z)$?

Answer: $f(z)=u(z,0)+iv(z,0)$

In words: the holomorphic function $f(z)$ can be obtained by putting $x=z$ and $y=0$ in $u(x,y)+iv(x,y)$.

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

Example 2: with $u(x,y)=\cos(x)\exp(-y)$ and $v(x,y)=\sin(x)\exp(-y)$ we obtain $f(z)=\cos(z)+i\sin(z)=\exp(iz)$.

Proof:

From the first pair of definitions $x={\frac {z+{\bar {z}}}{2}}$ and $y={\frac {z-{\bar {z}}}{2i}}$.

Therefore $f(z)=u\left({\frac {z+{\bar {z}}}{2}}\ ,{\frac {z-{\bar {z}}}{2i}}\right)+iv\left({\frac {z+{\bar {z}}}{2}}\ ,{\frac {z-{\bar {z}}}{2i}}\right)$.

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

Given only real part

Problem: $u(x,y)$ is known, $v(x,y)$ is unknown, $f(x+i0)$ is real; what is $f(z)$?

Answer: $f(z)=u(z,0)$.

Only example 1 applies here: with $u(x,y)=x^{4}-6x^{2}y^{2}+y^{4}$ we obtain $f(z)=z^{4}$.

Proof: "$f(x+i0)$ is real" means $v(x,0)=0$. In this case the answer to problem 1 becomes $f(z)=u(z,0)$.

General version

Problem: $u(x,y)$ is known, $v(x,y)$ is unknown; what is $f(z)$?

Answer: $f(z)=u(z,0)-i\int u_{y}(z,0)dz$ (where $u_{y}(x,y)$ is the partial derivative of $u(x,y)$ with respect to $y$).

Example 1: with $u(x,y)=x^{4}-6x^{2}y^{2}+y^{4}$ and $u_{y}(x,y)=-12x^{2}y+4y^{3}$ we obtain $f(z)=z^{4}+iC$ with real but undetermined $C$.

Example 2: with $u(x,y)=\cos(x)\exp(-y)$ and $u_{y}(x,y)=-\cos(x)\exp(-y)$ we obtain $f(z)=\cos(z)+i\int \cos(z)dz=\cos(z)+i(\sin(z)+C)=\exp(iz)+iC$.

Proof: This follows from $f(z)=u(z,0)+i\int v_{x}(z,0)dz$ and the 2nd Cauchy-Riemann equation $u_{y}(x,y)=-v_{x}(x,y)$.

Given only imaginary part

Problem: $u(x,y)$ is unknown, $v(x,y)$ is known; what is $f(z)$?

Answer: $f(z)=\int v_{y}(z,0)dz+iv(z,0)$.

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

Example 2: with $v(x,y)=\sin(x)\exp(-y)$ and $v_{y}(x,y)=-\sin(x)\exp(-y)$ we obtain $f(z)=-\int \sin(z)dz+i\sin(z)=\cos(z)+C+i\sin(z)=\exp(iz)+C$.

Proof: This follows from $f(z)=\int u_{x}(z,0)dz+iv(z,0)$ and the 1st Cauchy-Riemann equation $u_{x}(x,y)=v_{y}(x,y)$.