View source for 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.<ref name=TMG>{{cite journal|last1=Milne-Thomson|first1=L. M.|title=1243. On the relation of an analytic function of z to its real and imaginary parts|journal=The Mathematical Gazette|date=July 1937|volume=21|issue=244|pages=228–229|doi=10.2307/3605404|jstor=3605404|s2cid=125681848 }}</ref> It is named after [[Louis Melville Milne-Thomson]].

==Introduction==

Let <math>z = x + iy</math> and <math> \bar {z}\ = x - iy</math> where <math>x</math> and <math>y</math> are [[real number|real]].

Let <math>f(z) = u(x,y) + iv(x,y)</math> be any [[holomorphic function]].

Example 1: <math>z^4  = (x^4-6x^2y^2+y^4) +i(4x^3y-4xy^3)</math>

Example 2: <math>\exp(iz)=\cos(x)\exp(-y)+i\sin(x)\exp(-y)</math>

In his article,<ref name="TMG" /> Milne-Thomson considers the problem of finding <math>f(z)</math> when 1. <math>u(x,y)</math> and <math>v(x,y)</math> are given, 2. <math>u(x,y)</math> is given and <math>f(z)</math> is real on the real axis, 3. only <math>u(x,y)</math> is given, 4. only <math>v(x,y)</math> 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'': <math>u(x,y)</math> and <math>v(x,y)</math> are known; what is <math>f(z)</math>?

''Answer'': <math>f(z)=u(z,0)+iv(z,0)</math>

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

Example 1: with <math>u(x,y)=x^4-6x^2y^2+y^4</math> and <math>v(x,y)=4x^3y-4xy^3</math> we obtain <math>f(z)=z^4</math>.

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

''Proof'':

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

Therefore <math>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) </math>.

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

==2nd problem==

''Problem'': <math>u(x,y)</math> is known, <math>v(x,y)</math> is unknown, <math>f(x+i0)</math> is real; what is <math>f(z)</math>?

''Answer'': <math>f(z)=u(z,0)</math>.

Only example 1 applies here: with <math>u(x,y)=x^4-6x^2y^2+y^4</math> we obtain <math>f(z)=z^4</math>.

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

==3rd problem==

''Problem'': <math>u(x,y)</math> is known, <math>v(x,y)</math> is unknown; what is <math>f(z)</math>?

''Answer'': <math>f(z)=u(z,0)-i \int u_y(z,0) dz</math> (where <math>u_y(x,y)</math> is the partial derivative of <math>u(x,y)</math> with respect to <math>y</math>).

Example 1: with <math>u(x,y)=x^4-6x^2y^2+y^4</math> and <math>u_y(x,y)=-12x^2y+4y^3</math> we obtain <math>f(z)=z^4+iC </math> with real but undetermined <math>C</math>.

Example 2: with <math>u(x,y)=\cos(x)\exp(-y)</math> and <math>u_y(x,y)=-\cos(x)\exp(-y)</math> we obtain <math>f(z)=\cos(z)+i\int\cos(z)dz=\cos(z)+i(\sin(z)+C)=\exp(iz)+iC</math>.

''Proof'': This follows from <math>f(z)=u(z,0)+i \int v_x(z,0) dz</math> and the [[Cauchy–Riemann equations|2nd Cauchy-Riemann equation]]  <math>u_y(x,y)=-v_x(x,y)</math>.

==4th problem==

''Problem'': <math>u(x,y)</math> is unknown, <math>v(x,y)</math> is known; what is <math>f(z)</math>?

''Answer'': <math>f(z)=\int v_y(z,0)dz+i v(z,0)</math>.

Example 1: with <math>v(x,y)=4x^3y-4xy^3</math> and <math>v_y(x,y)=4x^3-12xy^2</math> we obtain <math>f(z)=\int 4z^3dz+i0=z^4+C</math> with real but undetermined <math>C</math>.

Example 2: with <math>v(x,y)=\sin(x)\exp(-y)</math> and <math>v_y(x,y)=-\sin(x)\exp(-y)</math> we obtain <math>f(z)=-\int\sin(z)dz+i\sin(z)=\cos(z)+C+i\sin(z)=\exp(iz)+C</math>.

''Proof'': This follows from <math>f(z)=\int u_x(z,0) dz +i v(z,0)</math> and the [[Cauchy–Riemann equations|1st Cauchy-Riemann equation]] <math>u_x(x,y)=v_y(x,y)</math>.

== References ==
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[[Category:Analytic functions]]