# Harmonic function and harmonic conjugate

Let $$u:G\subset\mathbb{R} \rightarrow \mathbb{R}$$ a harmonic function $$v:G\rightarrow \mathbb{R}$$ the harmonic conjugate function, with $$G$$ a domain. Prove that $$u^2-v^2$$ and $$uv$$ are harmonic without derivatives.

Before this, I proved that $$u^2$$ is harmonic, if $$u$$ is an harmonic function. Then, I thought that $$u^2$$ and $$v^2$$ are harmonic functions, and I wanted to conclude that $$u^2-v^2$$ is a harmonic function.

Nonetheless, this interpretation is wrong.

Since $$v$$ is a harmonic conjugate to $$u$$, this means that the function $$f=u+iv$$ is holomorphic in the domain $$G$$ specified. However it is easy to show that if $$f$$ is holomorphic, then $$f^2$$ is holomorphic and by the Cauchy-Riemann equations, it's real and it's imaginary part are harmonic functions. Expanding shows that

$$f^2=(u+iv)^2=u^2-v^2+i(2uv)\equiv \Re f+ i \Im f$$

and therefore the functions $$\Re f=u^2-v^2$$ and $$\Im f=2uv$$ and are harmonic (obviously if $$2uv$$ is harmonic then also $$uv$$ is).

If $$v$$ is the harmonic conjugate of $$u$$, then

$$f(z) = u(z) + iv(z) \tag 1$$

is holomorphic; thus so is $$(f(z))^2$$; now,

$$(f(z))^2 = (u(z) + iv(z))^2 = u^2(z) - v^2(z) + 2iu(z)v(z); \tag 2$$

since $$u^2 - v^2$$ and $$2uv$$ are the real and imaginary parts of $$f^2(z)$$, they are harmonic; $$2uv$$ harmonic implies that $$uv$$ is.

Nota Bene: if $$u$$ is harmonic, then $$u^2$$ is only harmonic if and only if $$\nabla u = 0$$, for we have

$$\nabla u^2 = \nabla \cdot (\nabla u^2) = \nabla \cdot (2u\nabla u)$$ $$= 2\nabla u \cdot \nabla u + 2u\nabla \cdot \nabla u = 2\nabla u \cdot \nabla u + 2u\nabla^2 u = 2\nabla u \cdot \nabla u, \tag 3$$

since

$$\nabla^2 u = 0; \tag 4$$

we are left with

$$\nabla u^2 = 2\nabla u \cdot \nabla u; \tag 5$$

thus for harmonic $$u$$, $$u^2$$ harmonic is equivalent to

$$\nabla u = 0, \tag 6$$

that is, $$u$$ is constant on connected components of $$G$$.

In this argument, I have used the well-known identity

$$\nabla \cdot (u \nabla u) = \nabla u \cdot \nabla u + u \nabla \cdot \nabla u, \tag 7$$

which is found in many sources on the gradient operator $$\nabla$$ and vector calculus identities; the reader my check out wikipedia or simply google around for more. End of Note.