Perpendicular Gradients

Question

Suppose $f:\mathbb{R}^2\to \mathbb{R}$ is smooth. Further suppose $\nabla f$ vanishes no where. When is it possible to find a smooth non-singular $g:\mathbb{R}^2\to \mathbb{R}$ satisfying $\nabla f\cdot \nabla g = 0$?

Answer 1

If $\nabla f \cdot \nabla g=0$ then we know $\nabla g = \phi J \nabla f$ for some function $\phi$ and $J$ the counterclockwise rotation by $\pi/2$. Applying the two-dimensional "curl" $-\text{div}(J \cdot)$ we get

$$ -\text{div} (J\nabla g) = \text{div}(-\phi J^2 \nabla f) = \phi \Delta f + \nabla \phi \cdot \nabla f.$$

So by the Poincaré Lemma, the necessary and sufficient condition for such a $g$ to exist is that $f$ satisfies this equation for some smooth $\phi$. (Side note: when $\phi=1$ the condition is that $f$ is harmonic, and the resulting $g$ will be the harmonic conjugate of $f$.)

Interpreting this as a first-order PDE for $\phi$, I think the fact that $\nabla f$ is conservative and non-vanishing is enough to get a non-zero global solution $\phi$ - the characteristic curves (integral curves of $\nabla f$) all have topology $\mathbb R$.