# Change of coordinates to make parallel gradients?

Suppose you have two functions $F,G : \mathbb{R}^n \rightarrow \mathbb{R}$ with the property that none of the partial derivatives vanish anywhere.

I think it should be possible to find a function $P:\mathbb{R}^n\rightarrow \mathbb{R}^n$ such that $\nabla F$ is parallel to $\nabla (G\circ P)$ everywhere, but I'm not sure how to begin.

Edit: Here is what I've tried. I want to use the implicit function theorem to find $P$ (or something like it), so I define a function $H:\mathbb{R}^{n+1}\times \mathbb{R}^n \rightarrow \mathbb{R}^n$ by:

$$H(\vec{x}, t; \vec{y}) \equiv \nabla F(\vec{x}) - \frac{1}{t}\nabla G(\vec{y}).$$

If $H$ constitutes an implicit function $\vec{y}(\vec{x}, t)$ and $\nabla F$ is parallel to $\nabla G$ at some point, we will have in the vicinity of that point that $\nabla F (\vec{x}) - \frac{1}{t}\nabla G \circ \vec{y}(\vec{x},t)$, which means that $\nabla F (\vec{x})$ is parallel to $\nabla G[\vec{y}(\vec{x},t)]$. The Jacobian of $H$ is (proportional to) the Hessian matrix of $G$, which is probably invertible due to the strong smoothness conditions on $G$. Beyond this, I don't know.

Another idea would be to try to compute $\vec{y}$ and $t$ as a function of $\vec{x}$. This requires a function $\widehat{H}$ with a higher dimensional range. I'm not sure how to do that without creating a singularity in the Jacobian, but perhaps just adding $t$ as an extra component will work;

$$\widehat{H}(\vec{x}; t, \vec{y}) \equiv \langle \partial_1 G(\vec{y}),\ldots, \partial_n G(\vec{y}), 0\rangle - t\langle \partial_1 F(\vec{x}),\ldots, \partial_n F(\vec{x}), 1\rangle$$

Then the Jacobian is $(n+1)\times (n+1)$. The bulk of it is made up of the Hessian matrix of $G$ as usual. The remaining row looks like a row of an identity matrix, and the remaining column block is $-\nabla F$.

If this matrix is invertible (is it?) and $F$ and $G$ are parallel somewhere, then we can find a function $\hat{P}(\vec{x}) = \langle t, \vec{y}\rangle$ such that $\nabla F$ is parallel to $\nabla G \circ P$.

If this works, then perhaps it solves the original problem because if $\nabla F$ is parallel to $\nabla G \circ P$, then $\nabla(G \circ P) = (\nabla G \circ P)\cdot \mathbf{J}(P) = t \cdot \nabla F \cdot \mathbf{J}(P)$.

• What is $\nabla F$ ? – kelvinn aja Oct 27 '17 at 22:29
• @Sou, the gradient of $F$, i.e. $\langle \partial_1 F, \ldots, \partial_n F \rangle$. – user326210 Oct 27 '17 at 22:38