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

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

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

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.

1st problem

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 $v(x,y)=4x^{3}y-4x{y}^3$ 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 $x=\frac{z+\overline{z}}{2}$ and $y=\frac{z-\overline{z}}{2i}$.

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

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

2nd problem

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

3rd problem

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 $u_y(x,y)=-v_x(x,y)$.

4th problem

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 $v(x,y)=4x^{3}y-4x{y}^3$ and $v_y(x,y)=4x^3-12x{y}^2$ 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 $u_x(x,y)=v_y(x,y)$.