Properties of harmonic functions

Question

If $u:D \mapsto \mathbb{R}$ and $v:D \mapsto \mathbb{R}$ are harmonic functions, then also function $uv:D \mapsto \mathbb{R}$ is harmonic.

Is the statement correct?

Answer 1

A harmonic function is a function $u$ such that $\Delta u=0$, where $\Delta$ is the Laplacian of $u$, i.e. $\Delta = \sum\frac{\partial^2}{\partial x_i^2}$. If $u$ and $v$ are harmonic, you have \begin{equation} \Delta(uv)=\sum_i\frac{\partial}{\partial x_i}\left(\frac{\partial(uv)}{\partial x_i}\right)= \sum_i\frac{\partial}{\partial x_i}\left(\frac{\partial u}{\partial x_i}v+\frac{\partial v}{\partial x_i}u\right)=\\ =\sum_i\frac{\partial^2u}{\partial x_i^2}v+\frac{\partial u}{\partial x_i}\frac{\partial v}{\partial x_i}+\frac{\partial^2v}{\partial x_i^2}u+\frac{\partial v}{\partial x_i}\frac{\partial u}{\partial x_i}=\\ =(\Delta u) v+(\Delta v)u +2\sum_i\frac{\partial u}{\partial x_i}\frac{\partial v}{\partial x_i} \end{equation}

Since $\Delta u = \Delta v =0$, we have \begin{equation} \Delta(uv)=2\sum_i\frac{\partial u}{\partial x_i}\frac{\partial v}{\partial x_i} \end{equation} which, in general, it is not zero. The condition for two functions to have a harmonic product is therefore $\nabla u\cdot\nabla v=0$.

Answer 2

Lauds and +1 to marco trevi; here`s an essentially the same but more coordinate-independent argument; I finished it up just as marco posted, so decided to go ahead and offer a slightly different perspective:

What needs to be done is to find out when $\nabla^2 (uv) = 0$ given that $\nabla^2 u =\nabla^2 v = 0$ individually. We have:

$\nabla^2 (uv) = \nabla \cdot \nabla(uv) = \nabla \cdot (v \nabla u + u \nabla v), \tag{1}$

where we have used what is essentially the Leibniz rule for products

$(fg)^\prime = f^\prime g + fg^\prime \tag{2}$,

applied to the individual components of $\nabla(uv)$. We also have

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

which follows from the identity

$\nabla \cdot (fX) = \nabla f \cdot X + f \nabla \cdot X, \tag{4}$

holding for any (sufficiently differentiable) function$f$ and vector field $X$. (4) is also a well-known formula which again follows from the Leibnitz rule. Combining (1) and (3) yields

$\nabla^2(uv) = u\nabla^2 v + 2\nabla u \cdot \nabla v+ v\nabla^2 u, \tag{5}$

which shows that $uv$ is harnonic if and only if $\nabla u \cdot \nabla v = 0$; that is, if and only if the gradients of $u$ and $v$ are orthogonal, so that their integral curves run normal to one another.

Hope this helps. Cheers.

and as always,

Fiat Lux!!!