Is $v(x,y) = x^3 - 3xy^2 + 2y$ a harmonic function? Prove your claim, if yes find its conjugate harmonic function $u(x,y)$ and hence obtain the analytic function whose real and imaginary parts are $u$ and $v$ respectively

$\frac{\partial v}{\partial x} = 3x^2 - 3y^2$

$\frac{\partial^2 v}{\partial x^2} = 6x$

$\frac{\partial v}{\partial y} = - 6xy + 2$

$\frac{\partial^2 v}{\partial y^2} = - 6x$

Now, $\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 6x - 6x = 0$

Thus the given function $v(x,y)$ is harmonic.

Let the harmonic conjugate be $u(x,y)$. Then,

$\frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y} \implies \frac{\partial u}{\partial y} = - 3x^2 + 3y^2 \implies u = -3x^2y + y^3 + c_1(x)$

Also, $\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} \implies \frac{\partial u}{\partial x} = - 6xy + 2 \implies u = - 3x^2y + 2x + c_2(y)$

Hence, $u(x,y) = - 3x^2y + y^3 + 2x$

Thus, the analytic function with real and imaginary parts as $u$ and $v$ respectively is,

$f(u,v) = u + iv = (-3x^2y + y^3 + 2x) + i (x^3-3xy^2+2y)$
answered Aug 26, 2017 by (1,920 points)