Analytic function from u = y³ - 3x²y

Harmonic function

A real function u(x, y) with continuous second partial derivatives is harmonic if it satisfies Laplace's equation: u_xx + u_yy = 0. The real and imaginary parts of an analytic function are harmonic.

Solution

u = y³ − 3x²y.

u_x  = −6xy         u_xx = −6y
u_y  = 3y² − 3x²    u_yy = 6y

u_xx + u_yy = 0, so u is harmonic and a conjugate v exists.

Milne-Thomson method. f′(z) = u_x − i u_y. Putting x = z, y = 0:

f′(z) = 0 − i(0 − 3z²) = 3i z²

Integrating: f(z) = i z³ + c.

Find v

Expand z³ = (x + iy)³ = (x³ − 3xy²) + i(3x²y − y³).

i z³ = −(3x²y − y³) + i(x³ − 3xy²) = (y³ − 3x²y) + i(x³ − 3xy²)

The real part is y³ − 3x²y = u ✓, and v = x³ − 3xy² (+ constant).

Answer: f(z) = i z³ + c, where v = x³ − 3xy² + constant.