# Is $\frac{\nabla{u(\mathbf{x})}}{(\nabla{u(\mathbf{x})})\cdot\mathbf{x}}$ the gradient of any function?

Suppose

$$\mathbf{p}(\mathbf{x})=\frac{\nabla{u(\mathbf{x})}}{(\nabla{u(\mathbf{x})})\cdot\mathbf{x}},$$

where $\mathbf{x} \in \mathbb{R}^n$, $\mathbf{p}: \mathbb{R}^n \to \mathbb{R}^n$ and $u:\mathbb{R}^n \to \mathbb{R}$. Assume $\mathbf{p}$ and $u$ are convex and defined and differentiable for all $\mathbb{R}^n > \mathbf{0}$.

I would like to show that $\mathbf{p}(\mathbf{x})$ is the gradient of some function $B: \mathbb{R}^n \to \mathbb{R}$, although I don't necessarily need to know the form of the function. What I really need to do is calculate $B(\mathbf{x_1})-B(\mathbf{x_0})$ for arbitrary values of $\mathbf{x_1}$ and $\mathbf{x_0}$.

I can already calculate $\mathbf{p}(\mathbf{x})$ as needed, and if I can show that it is a valid gradient, I think the gradient theorem says that $B(\mathbf{x_1})-B(\mathbf{x_0})$ = $\int^\mathbf{x_1}_\mathbf{x_0}{\mathbf{p}(\mathbf{x}) \cdot d\mathbf{x}}$, which I can calculate numerically. However, if $\mathbf{p}(\mathbf{x})$ is not a valid gradient of any function, then the line integral is path-dependent and $B$ is ill-defined.

So my question is, is it possible to show that $\mathbf{p}(\mathbf{x})$ as defined here is a valid gradient? Or are there conditions that I could apply to guarantee that it is? I've spent a couple of days trying out variations on the chain rule, substituting terms, etc., but haven't made much headway.

• Is $u$, perhaps, homogeneous? Then by Euler's theorem $\nabla u \cdot \mathbf{x} = k u$ ($k$ is the degree of $u$), and then $\mathbf{p} = \nabla\left(\frac{1}{ku}\right)$. – Matthew Leingang Mar 14 '17 at 11:45


In your situation (omitting dependence of the gradient on $\Vec{x}$ out of laziness), $$p_{i} = \frac{1}{(\Del u \cdot \Vec{x})}\, \frac{\dd u}{\dd x_{i}}.$$ Consequently, \begin{align*} \frac{\dd p_{i}}{\dd x_{j}} &= \frac{1}{(\Del u \cdot \Vec{x})}\, \frac{\dd^{2} u}{\dd x_{j}\, \dd x_{i}} - \frac{1}{(\Del u \cdot \Vec{x})^{2}}\, \frac{\dd u}{\dd x_{i}} \, \frac{\dd}{\dd x_{j}}(\Del u \cdot \Vec{x}) \\ % &= \frac{1}{(\Del u \cdot \Vec{x})}\, \frac{\dd^{2} u}{\dd x_{j}\, \dd x_{i}} - \frac{1}{(\Del u \cdot \Vec{x})^{2}}\, \frac{\dd u}{\dd x_{i}} \left[\frac{\dd u}{\dd x_{j}} + \Del\left(\frac{\dd u}{\dd x_{j}}\right) \cdot \Vec{x}\right] \\ % &= \frac{1}{(\Del u \cdot \Vec{x})}\, \frac{\dd^{2} u}{\dd x_{j}\, \dd x_{i}} - \frac{1}{(\Del u \cdot \Vec{x})^{2}}\, \frac{\dd u}{\dd x_{i}} \left[\frac{\dd u}{\dd x_{j}} + \sum_{k} x_{k} \frac{\dd^{2} u}{\dd x_{k}\, \dd x_{j}}\right]. \end{align*} That is, $\Vec{p}$ is a gradient field if and only if $$\frac{\dd u}{\dd x_{i}} \sum_{k} x_{k}\, \frac{\dd^{2} u}{\dd x_{k}\, \dd x_{j}} = \frac{\dd u}{\dd x_{j}} \sum_{k} x_{k}\, \frac{\dd^{2} u}{\dd x_{k}\, \dd x_{i}}\quad\text{for all i, j,}$$ if and only if $$\sum_{k} x_{k} \left[\frac{\dd u}{\dd x_{i}}\, \frac{\dd^{2} u}{\dd x_{k}\, \dd x_{j}} - \frac{\dd u}{\dd x_{j}}\, \frac{\dd^{2} u}{\dd x_{k}\, \dd x_{i}}\right] = 0\quad\text{for all i, j.} \tag{**}$$

Equation (*) does not generally hold for your $\Vec{p}$. For instance, if $u(x, y) = x + y^{2}$, then $\Del u = (1, 2y)$ and $\Del u \cdot \Vec{x} = x + 2y^{2}$, so $$\Vec{p} = \frac{(1, 2y)}{x + 2y^{2}};\qquad p_{1} = \frac{1}{x + 2y^{2}},\quad p_{2} = \frac{2y}{x + 2y^{2}}.$$ A short computation gives $$\frac{\dd p_{1}}{\dd y} = -\frac{4y}{(x + 2y^{2})^{2}} \neq -\frac{2y}{(x + 2y^{2})^{2}} = \frac{\dd p_{2}}{\dd x}.$$

In any event, and (*) or (**) is easier to check than the integral condition you've been considering.

• The algebra might be simplified if we use the identity $\nabla\times(f\vec{G}) = \nabla f \times \vec{G} + f\nabla\times\vec{G}$. Since $\vec{G}$ is a gradient, the second term vanishes. – Amey Joshi Mar 14 '17 at 16:12
• Thanks, this is tremendously helpful! But I don't quite see how the equation before (**) follows from the one before that. Am I missing something obvious? – Matthias Fripp Mar 14 '17 at 18:57
• @mfripp: The display above the equation you ask about gives an expression for $\frac{\dd p_{i}}{\dd x_{j}}$. Swap $i$ and $j$ to get $\frac{\dd p_{j}}{\dd x_{i}}$, equate using (*), and cancel the terms that are obviously invariant under exchanging $i$ and $j$: the second partial, and the product of the two first partials divided by $(\Del u \cdot \Vec{x})^{2}$. – Andrew D. Hwang Mar 14 '17 at 19:12
• @AmeyJoshi: Formally yes, though strictly the curl is only defined when $n = 3$. – Andrew D. Hwang Mar 14 '17 at 19:14
• @AndrewD.Hwang: I think I understand now. At first I assumed that (**) somehow led to (*), but now I see that you are starting with (*) and then deriving (**). So in general, with a simply connected domain, is equation (*) a necessary and sufficient condition for integrability (i.e., it's true for any gradient field)? Does that rule have a name? I didn't come across it anywhere previously. – Matthias Fripp Mar 15 '17 at 19:25