Milne Thompson Method for finding Analytic Function

Milne Thompson Method is a method of finding an Analytic Function, whose real or imaginary part is given. The method greatly simplifies the process of finding the Analytic Function, whose real or imaginary or any combination of the two parts is given.

Method for finding the Analytic Function

Let $f(z) = u(x,y) + iv(x,y)$ be any Analytic Function.

Let $z = x + iy$ and $\bar {z}\ = x - iy$

Hence, $x = \frac{z + \bar {z}}{{2}}\$ and $y = \frac{z - \bar {z}}{{2i}}\$

Therefore, $f(z) = u(x,y) + iv(x,y)$ is equal to

$f(z) = u( \frac{z + \bar {z}}{{2}}\ , \frac{z - \bar {z}}{{2i}}\ ) + iv( \frac{z + \bar {z}}{{2}}\ , \frac{z - \bar {z}}{{2i}}\ )$

This can be regarded as an identity in two independent variables $z$ and $\bar {z}\$. We can therefore, put $z$ = $\bar {z}\$ and get $f(z) = u(z,0) + iv(z,0)$

So, $f(z)$ can be obtained in terms of $z$ simply by putting $x$ = $z$ and $y$ = $0$ in $f(z) = u(x,y) + iv(x,y)$ when $f(z)$ is Analytic Function.

Now, $f'(z) = {\partial u \over \partial x} + i{\partial v \over \partial x}$.

Since, $f(z)$ is Analytic, hence Cauchy-Riemann Equations are satisfied. Hence, $f'(z) = {\partial u \over \partial x} - i{\partial u \over \partial y}$.

Let ${\partial u \over \partial x}$ = $\Phi(x,y)$ and ${\partial u \over \partial y}$ = $\Psi(x,y)$.

Then,
$f'(z) = {\partial u \over \partial x} - i{\partial u \over \partial y}$
$f'(z) = \Phi(x,y) - i\Psi(x,y)$
Now, putting $x = z$ and $y = 0$ in the above equation, we get
$f'(z) = \Phi(z,0) - i\Psi(z,0)$.

Integrating the above equation we get $\int f'(z)\,dz = \int \Phi(z,0) dz - i \int \Psi(z,0) dz$

Or $f(z) = \int f'(z)\,dz = \int \Phi(z,0) dz - i \int \Psi(z,0) dz + c$
which is the required Analytic Function.

Example

Let $u(x,y) = x^4-6x^2y^2+y^4$ be any real function whose Harmonic Conjugate, and hence the Analytic Function, is to be determined.

Let, the desired Analytic Function be $f(z) = u(x,y) + i v(x,y)$
Then as per the above process we know that
$f'(z) = {\partial u(x,y) \over \partial x} + i {\partial v(x,y) \over \partial x}$

But as $f(z)$ is analytic, so it satisfies Cauchy-Riemann Equations.
Hence, ${\partial u(x,y) \over \partial x} = {\partial v(x,y) \over \partial y}$ and ${\partial u(x,y) \over \partial y} = -{\partial v(x,y) \over \partial x}$

Or $u_x = v_y$ and $u_y = -v_x$
Substituting these values in $f'(z)$ we get,
$f'(z) = {\partial u(x,y) \over \partial x} - i {\partial u(x,y) \over \partial y}$
Hence,
$f'(z) = (4x^3 - 12xy^2) - i (-12x^2y + 4y^3)$

This can be written as $f'(z) = \Phi(x,y) - i \Psi(x,y)$

Where, $\Phi(x,y) = (4x^3 - 12xy^2)$ and $\Psi(x,y) = -12x^2y + 4y^3$

Rewriting $f'(z) = \Phi(x,y) - i \Psi(x,y)$ using $x = z$ and $y = 0$
$f'(z) = 4z^3 - i (0)$

Integrating both sides w.r.t $dz$ we get,
$\int f'(z) dz = \int 4z^3 dz + \int 0 dz$

Hence, $f(z) = z^4 + c$ is the required Analytic Function.