In mathematics, the Milne-Thomson method is a method of finding a holomorphic function, whose real or imaginary part is given. The method greatly simplifies the process of finding the holomorphic function whose real or imaginary or any combination of the two parts is given. It is named after Louis Melville Milne-Thomson.
Let f ( z ) = u ( x , y ) + i v ( x , y ) be any holomorphic function.
Let z = x + i y and z ¯ = x − i y where x and y are real.
Hence,
x = z + z ¯ 2 and y = z − z ¯ 2 i Therefore, f ( z ) = u ( x , y ) + i v ( x , y ) is equal to
f ( z ) = u ( z + z ¯ 2 , z − z ¯ 2 i ) + i v ( z + z ¯ 2 , z − z ¯ 2 i ) This can be regarded as an identity in two independent variables z and z ¯ . We can therefore, put z = z ¯ and get f ( z ) = u ( z , 0 ) + i v ( z , 0 )
So, f ( z ) can be obtained in terms of z simply by putting x = z and y = 0 in f ( z ) = u ( x , y ) + i v ( x , y ) when f ( z ) is a holomorphic function.
Now, f ′ ( z ) = ∂ u ∂ x + i ∂ v ∂ x .
Since, f ( z ) is holomorphic, hence Cauchy–Riemann equations are satisfied. Hence, f ′ ( z ) = ∂ u ∂ x − i ∂ u ∂ y .
Let ∂ u ∂ x = Φ ( x , y ) and ∂ u ∂ y = Ψ ( x , y ) .
Then
f ′ ( z ) = ∂ u ∂ x − i ∂ u ∂ y f ′ ( z ) = Φ ( x , y ) − i Ψ ( x , y ) Now, putting x = z and y = 0 in the above equation, we get
f ′ ( z ) = Φ ( z , 0 ) − i Ψ ( z , 0 ) . Integrating both sides of the above equation we get
∫ f ′ ( z ) d z = ∫ Φ ( z , 0 ) d z − i ∫ Ψ ( z , 0 ) d z or
f ( z ) = ∫ f ′ ( z ) d z = ∫ Φ ( z , 0 ) d z − i ∫ Ψ ( z , 0 ) d z + c which is the required holomorphic function.
Let u ( x , y ) = x 4 − 6 x 2 y 2 + y 4 , and let the desired holomorphic function be f ( z ) = u ( x , y ) + i v ( x , y )
Then as per the above process we know that
f ′ ( z ) = ∂ u ( x , y ) ∂ x + i ∂ v ( x , y ) ∂ x . But as f ( z ) is holomorphic, so it satisfies Cauchy–Riemann equations.
Hence, ∂ u ( x , y ) ∂ x = ∂ v ( x , y ) ∂ y and ∂ u ( x , y ) ∂ y = − ∂ v ( x , y ) ∂ x
Or u x = v y and u y = − v x .
Substituting these values in f ′ ( z ) we get,
f ′ ( z ) = ∂ u ( x , y ) ∂ x − i ∂ u ( x , y ) ∂ y Hence,
f ′ ( z ) = ( 4 x 3 − 12 x y 2 ) − i ( − 12 x 2 y + 4 y 3 ) This can be written as f ′ ( z ) = Φ ( x , y ) − i Ψ ( x , y )
where, Φ ( x , y ) = ( 4 x 3 − 12 x y 2 ) and Ψ ( x , y ) = − 12 x 2 y + 4 y 3 .
Rewriting f ′ ( z ) = Φ ( x , y ) − i Ψ ( x , y ) using x = z and y = 0
f ′ ( z ) = 4 z 3 − i ( 0 ) Integrating both sides w.r.t d z we get,
∫ f ′ ( z ) d z = ∫ 4 z 3 d z + ∫ 0 d z Hence, f ( z ) = z 4 + c is the required holomorphic function.
