Harmonic Conjugate

If the real part of a complex function is harmonic, then there exists a harmonic imaginary part, so that the function is analytic. The imaginary part is known as the harmonic conjugate of the real part. In other words, the harmonic conjugate to a real valued function is an imaginary function, such that the whole function is differentiable.

A harmonic conjugate satisfies both the Cauchy–Riemann equations and Laplace’s equation.

Theorem

Let’s say the complex function is f(z) and the real part is u(x,y) and the imaginary part is v(x,y). Then we can write the function as,

f(z) = u(x,y) + vi(x,y)

If u(x,y) of the function is harmonic, then there exists a harmonic v(x,y) part, so f(z) is differentiable. To prove a function is harmonic, you need to prove that the complex function is differentiable.

To continue, we need to understand how can a complex function be differentiable.

Conditions to be a differentiable complex function

A complex function is only differentiable when it satisfies two following conditions.

#1 The complex function should be continues.

#2 It should satisfy Cauchy-Reimann (CR) Equations. If you need to learn about those equations, read MechCollege’s article on Cauchy-Reimann Equations.

∂u/∂x =  ∂v/∂y ——–(CR Eq. 1)

∂u/∂y = – ∂v/∂x ——–(CR Eq. 2)

Solved Problems

Example #1 | Basic example for finding harmonic conjugate

Find the harmonic conjugate of u(x,y) = x2 – y2 – x + y . And u(x,y) is a harmonic function and f(z) = u(x,y) + vi(x,y).

So, we’re given the real part of the equation. So, now we need to find the imaginary part v(x,y), in way that the complex function f(z) is differentiable.

f(z) = u(x,y) + vi(x,y)

To be the f(z) differentiable, according to Cauchy-Reimann equations, the real part should satisfy the following equations.

∂u/∂x =  ∂v/∂y

∂u/∂y = – ∂v/∂x

Lets get the partial differentiation of u(x,y) with respect to x and y.

∂u/∂x = 2x – 1

∂u/∂y = -2y + 1

So,

∂v/∂y = ∂u/∂x = 2x – 1

∂v/∂x = – ∂u/∂y = 2y – 1

There are two methods to solve the rest of the problem.

Method #1

Partially integrating ∂v/∂x with respect to x gives,

v = 2xy – x + h(x)

Partially integrating ∂v/∂y with respect to y gives,

v = 2xy – y + g(x)

So, the  h(x) = – y and g(x) = – x

Therefore, the harmonic conjugate v(x,y) is,

v(x,y) =  2xy – x – y

Method #2

Partially integrating ∂v/∂x with respect to y gives,

v = 2xy – y + g(x)

By finding the partial differentiation of v with respect to x,

∂v/∂x = 2y + g'(x)

Since ∂v/∂x = 2y – 1,

2y – 1 = 2y + g'(x)

g'(x) = -1

Partially integrating g'(x) with respect to x gives,

g(x) = – x

Therefore, the harmonic conjugate v(x,y) is,

v(x,y) =  2xy – x – y

Example #2 | The harmonic conjugate of an exponential function

Show that u(x,y) = exsin(y) is a harmonic function and find the harmonic conjugate v(x,y) such that f(z) = u(x,y) + vi(x,y) is analytic.

In this question, we can see something we didn’t talk about earlier. How to prove a function is harmonic? To that, you need to prove that the function has continuous second order partial derivatives.

Lets get the partial differentiation of u(x,y) with respect to x and y.

∂u/∂x = exsin(y)

∂u/∂y = excos(y)

Since we need to prove u(x,y) = exsin(y) is a harmonic function, lets take the second degree partial derivatives.

2u/∂x2 = excos(y)

2u/∂y2 = – exsin(y)

Since it has continues second order partial derivatives, u(x,y) = exsin(y) is a harmonic function.

Now, let’s find the harmonic conjugate. According Cauchy-Reimann equations,

∂v/∂y = ∂u/∂x = exsin(y)

∂v/∂x = – ∂u/∂y = – excos(y)

Partially integrating ∂v/∂y with respect to y gives,

v = – excos(y) + g(x)

By finding the partial differentiation of v with respect to x,

∂v/∂x = – excos(y) + g'(x)

Since ∂v/∂x =- excos(y),

– excos(y) =  – excos(y) + g'(x)

g'(x) = 0

Partially integrating g'(x) with respect to x gives,

g(x) = C

Therefore, the harmonic conjugate v(x,y) is,

v(x,y) = – excos(y) + C