Milne-Thomson method for finding a holomorphic function explained

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

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} ^{2y} ^{2+y} ^{4)} + i( 4 x ^{3y-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.

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} ^{2y} ^{2+y} ^{4} and v(x,y) =4 x ^{3y-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 = (z + \bar {z)/() } and y = (z - \bar {z)/() }.

Therefore

f(z) = u \left( (z + \bar {z)/() }\, \frac\right) + iv\left(\frac\, \frac\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 ) .

2nd problem

Problem:

u(x,y) is known, v(x,y) is unknown, f(x + i 0 ) is real; what is f(z) ?

Answer:

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

Only example 1 applies here: with

u(x,y) = x ^{4-6x} ^{2y} ^{2+y} ^{4} we obtain f(z) = z ^{4} .

Proof: "

f(x + i 0 ) 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} ^{2y} ^{2+y} ^{4} and u ^{2 y +4 y} _{y(x,y) =-12 x} ^{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+i} v(z, 0 ) .

Example 1: with

v(x,y) =4 x ^{3y-4xy} ^{3} and v ^{3-12 xy} _{y(x,y) =4 x} ^{2} we obtain f(z) = \int 4 z ^{3dz+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 + i v(z, 0 ) and the 1st Cauchy-Riemann equation u _{x(x,y)=v} _{y(x,y)} .

Notes and References

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