Product $u v$ of harmonic functions is harmonic if $\langle \nabla u, \nabla v \rangle$ is $\equiv 0$ everywhere

Question

Let $\Omega \subseteq \mathbb R^d$ be open, and let $u, v: \Omega \to \mathbb R$ be harmonic functions. I now want to prove: the product function $u v: \Omega \to \mathbb R$ is harmonic on $\Omega$, if and only if $\langle \nabla u, \nabla v \rangle \equiv 0$ on $\Omega$ (where $\langle \cdot,\cdot \rangle$ is the standard scalar product).

Now to show that, I tried to write out

$$\langle \nabla u, \nabla v \rangle = \sum_{i=1}^n \frac{\partial}{\partial x_i}u \cdot \frac{\partial}{\partial x_i} v \\ = \frac{\partial}{\partial x_1} (u v) + \dots + \frac{\partial}{\partial x_n} (u v)$$

but from there on, I'm not really sure how to continue. I would need to show that this sum over the first-order derivatives is equal to $0$ if and only if the sum over the second-order derivatives

$$\Delta u v = \sum_{i=1}^n \frac{\partial^2}{\partial x_n^2} (u v)$$

is also equal to $0$ everywhere. But I'm not sure how I can show that and how the harmonicity of $u, v$ themselves comes into play? Am I missing something obvious here?

Answer 1

Direct calculation gives you

$$\Delta uv=\sum_{i=1}^{n}\partial^{2}_{i}\left(uv\right)=\sum_{i=1}^{n}\partial_{i}\left(v\partial_{i}u+u\partial_{i}v\right)=$$

$$=\sum_{i=1}^{n}\left(v\partial_{i}^{2}u+2\left(\partial_{i}v\right)\left(\partial_{i}u\right)+u\partial_{i}^{2}v\right)=v\Delta u+u\Delta v+2\left<\nabla u,\nabla v\right>$$

and its pretty clear how to finish the proof.

Answer 2

$$\partial^2_i (u v)=\partial_i^2(u)v+2\partial_i(u)\partial_i(v)+u\partial^2_i(v)$$ Thus $$\Delta (u v) = \displaystyle \sum_{i=1}^n \partial^2_i (u v)= \sum_{i=1}^n(\partial_i^2(u)v+2\partial_i(u)\partial_i(v)+u\partial^2_i(v))=v\displaystyle \sum_{i=1}^n \partial^2_i (u)+u\displaystyle \sum_{i=1}^n \partial^2_i (v)+2\displaystyle \sum_{i=1}^n \partial_i (u)\partial_i(v)=v\Delta u+ 2\langle\nabla u,\nabla v\rangle+u\Delta v=2\langle\nabla u,\nabla v\rangle$$

Answer 3

hint: show the following: $$\Delta (u v) = u \Delta v + v\Delta u + 2 \langle \nabla u,\nabla v \rangle $$

by using the product rule for the second derivative.