harmonic conjugate for the $u(x,y)=x^3 +2xy-4xy^2$

Question

If $$u(x,y)=x^3 +2xy-4xy^2$$ find the harmonic conjugate $v(x,y)$ and explain why the function is entire!

So I tried to solve it and that's what I got

$$ux=3x^2+2y-4y^2$$

$$uy=2x-8xy$$

after applying Cauchy-Riemann Equations

$$vy=ux=3x^2+2y-4y^2$$

and after integration

$$v(x,y)=3x^2y+y^2-(4/3) y^3+\beta(X)$$

and after trying to solve for $\beta$ I found it equal to

$$\beta=x^2y-x^2$$

and after applying it to the

$$v(x,y)= 4x^2+y^2=(4/3)y^3-x^2$$

which in obviously is not applying C-R equations if we want to prove the solution is correct and ux is not equal to vy !!!

So does this function is not analytic or entire at all or I am doing something wrong in somewhere!

regards .

Answer 1

$u$ is not harmonic because $u_{xx}+u_{yy}\neq0$ so it won't have a harmonic conjugate.

Answer 2

The function is not harmonic, so it cannot have a conjugate.

On the other hand, if we fix the example as $$ u(x,y)=x^3+2xy-\color{red}{3}xy^2, $$ then \begin{alignat}{2} u_x&=3x^2+2y-3y^2 &\qquad\qquad u_{xx}&=6x \\[4px] u_y&=2x-6xy & u_{yy}&=-6x \end{alignat} and this $u$ is harmonic.

The harmonic conjugate $v$ must satisfy $$ v_x=-u_y \qquad\qquad v_y=u_x $$ Thus $$ v_x=-2x+6xy \qquad\qquad v_y=3x^2+2y-3y^2 $$ Hence $$ v=-x^2+3x^2y+f(y) $$ and $$ v_y=3x^2+\beta'(y) $$ that gives $\beta'(y)=2y-3y^2$, so $\beta(y)=y^2-y^3$ (plus an arbitrary constant).

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

You can show that $$ f(z)=f(x+iy)=u(x,y)+iv(x,y)=z^3-iz^2 $$