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.[1] It is named after Louis Melville Milne-Thomson.

Introduction

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

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

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

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

In his article,[1] Milne-Thomson considers the problem of finding ${\displaystyle f(z)}$ when 1. ${\displaystyle u(x,y)}$ and ${\displaystyle v(x,y)}$ are given, 2. ${\displaystyle u(x,y)}$ is given and ${\displaystyle f(z)}$ is real on the real axis, 3. only ${\displaystyle u(x,y)}$ is given, 4. only ${\displaystyle v(x,y)}$ 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: ${\displaystyle u(x,y)}$ and ${\displaystyle v(x,y)}$ are known; what is ${\displaystyle f(z)}$?

Answer: ${\displaystyle f(z)=u(z,0)+iv(z,0)}$

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

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

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

Proof:

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

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

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

2nd problem

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

Answer: ${\displaystyle f(z)=u(z,0)}$.

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

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

3rd problem

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

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

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

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

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

4th problem

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

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

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

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

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

References

1. Milne-Thomson, L. M. (July 1937). "1243. On the relation of an analytic function of z to its real and imaginary parts". The Mathematical Gazette. 21 (244): 228–229. doi:10.2307/3605404. JSTOR 3605404. S2CID 125681848.