I was looking into a question about vector functions $\varphi$ satisfying the Laplace equation $\Delta \varphi =0$. I found that this answer to the question was given in most places I looked for, but my problem with it is that it doesn't use the hypothesis that $\Delta \varphi =0$.
My question comes from my understanding that the curl of a gradient is $0$ only if the mixed partials satisfy Clairaut's theorem, which is not always true. So is there a counterexample where $\Delta \varphi =0$ but $\nabla \times \nabla \varphi \neq 0$? Or does the hypothesis of being a solution to the Laplace equation impose that the mixed partials are always equal?
Thank you!
Given $f:\Bbb{R}^3\to \Bbb{R}$ a function of class $C^2$ (twice continuously differentiable), Clairaut's Theorem says the mixed partials commute, i.e. that $$ \frac{\partial^2 f}{\partial x_i\partial x_j}=\frac{\partial^2 f}{\partial x_j\partial x_i}$$ for all $i,j\in \{1,2,3\}$. It is true that whenever $f$ is of class $C^2$, $\nabla \times \nabla f=0$, as a calculation using the symmetry of the partials will show. I think that for this statement you only need the $C^2$ assumption. You should note that usually a harmonic function (satisfying $\Delta f=0$) is assumed to be $C^2$. In particular, harmonicity implies $C^2$ and the proof can be carried out using the weaker statement.