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Milne-Thomson method for finding a holomorphic function

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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

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Let z=x+iy and z¯ =x−iy where x and y are real.

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

Example 1: z4=(x4−6x2y2+y4)+i(4x3y−4xy3)

Example 2: exp⁡(iz)=cos⁡(x)exp⁡(−y)+isin⁡(x)exp⁡(−y)

In his article,[1] 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

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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)=x4−6x2y2+y4 and v(x,y)=4x3y−4xy3 we obtain f(z)=z4.

Example 2: with u(x,y)=cos⁡(x)exp⁡(−y) and v(x,y)=sin⁡(x)exp⁡(−y) we obtain f(z)=cos⁡(z)+isin⁡(z)=exp⁡(iz).

Proof:

From the first pair of definitions x=z+z¯2 and y=z−z¯2i.

Therefore f(z)=u(z+z¯2 ,z−z¯2i)+iv(z+z¯2 ,z−z¯2i).

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

Given only real part

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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)=x4−6x2y2+y4 we obtain f(z)=z4.

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

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

Answer: f(z)=u(z,0)−i∫uy(z,0)dz (where uy(x,y) is the partial derivative of u(x,y) with respect to y).

Example 1: with u(x,y)=x4−6x2y2+y4 and uy(x,y)=−12x2y+4y3 we obtain f(z)=z4+iC with real but undetermined C.

Example 2: with u(x,y)=cos⁡(x)exp⁡(−y) and uy(x,y)=−cos⁡(x)exp⁡(−y) we obtain f(z)=cos⁡(z)+i∫cos⁡(z)dz=cos⁡(z)+i(sin⁡(z)+C)=exp⁡(iz)+iC.

Proof: This follows from f(z)=u(z,0)+i∫vx(z,0)dz and the 2nd Cauchy-Riemann equation uy(x,y)=−vx(x,y).

Given only imaginary part

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

Answer: f(z)=∫vy(z,0)dz+iv(z,0).

Example 1: with v(x,y)=4x3y−4xy3 and vy(x,y)=4x3−12xy2 we obtain f(z)=∫4z3dz+i0=z4+C with real but undetermined C.

Example 2: with v(x,y)=sin⁡(x)exp⁡(−y) and vy(x,y)=−sin⁡(x)exp⁡(−y) we obtain f(z)=−∫sin⁡(z)dz+isin⁡(z)=cos⁡(z)+C+isin⁡(z)=exp⁡(iz)+C.

Proof: This follows from f(z)=∫ux(z,0)dz+iv(z,0) and the 1st Cauchy-Riemann equation ux(x,y)=vy(x,y).

References

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  1. 1 2 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.