# Find a harmonic conjugate for the function $u(x,y)=x^{3}+Axy^{2}$ if one exists - $u(x,y)$ might not be harmonic?

For $u(x,y) = x^{3} + Axy^{2}$, where $A$ is a real number, I need to find a harmonic conjugate, or if one does not exist, show that it does not exist.

I began my approach to this problem by first checking whether $u(x,y)$ is even harmonic.

However, I got that $\displaystyle \frac{\partial ^{2} u}{\partial x^{2}} = 6x$ and $\displaystyle \frac{\partial ^{2}u}{\partial y^{2}} = 2Ax$.

So, $u(x,y)$ appears to satisfy Laplace's Equation only when $A = -3$ or at $x = 0$ for any $A \in \mathbb{R}$.

I believe that in order for $u(x,y)$ to be harmonic, however, it would need to be like analyticity, where Laplace's Equation would need to hold in at least a neighborhood of $x = 0$, the imaginary axis, and not just on the axis itself. But, that is only my intuition. I'm not sure if this is actually true. If so, then we would need $A = -3$.

So, I suppose then I could go ahead and try to find a harmonic conjugate in the case where $A$ must be $-3$?

• I think you got it right: In my opinion it must be $\;A=-3\;$ . Commented Mar 25, 2016 at 22:25
In general you can solve for $v$ from its derivatives by the Cauchy-Riemann equations. This problem can be eyeballed pretty quickly, though: $u$ is the real part of $f(z) = z^3.$
• but $A$ must be $-3$, right?
• Yes, the condition $\Delta u = 0$ has to be true everywhere so your argument above is fine. Commented Mar 25, 2016 at 22:50