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I have a smooth function $F(x,y)$ which sends an open disc $D\subset \mathbb{R}^2$ in the plane to $\mathbb{R}$. I would like to know how to compute whether the level curves of $F$ are all each (a subset of) a line.

For example, $F$ might be a function like $ax+by+c$ whose the level curves are all parallel lines. Or $F$ might be a function with "radial" contours, e.g. $F(x,y)$ is the slope of the line between $(x,y)$ and the origin, and $D$ is a disc in the first quadrant.

So far, it seems to me that:

  • There are only two basic ways the contours can all be lines --- the radial case, and the parallel lines case. This is because the contours cannot intersect
  • If you have a functional form for $F$, it should be possible to perform a test, possibly involving logarithms of ratios of derivatives, to determine whether the level curves are all straight. I haven't been able to devise such a test yet, however.
  • One way of looking at it is to pick a point $p\in D$ and let $\gamma_p:\mathbb{R}\rightarrow \mathbb{R}^2$ be the line that passes through $p$ parallel to the level curve of $F$. Then the claim is roughly that for all $p$, $F\circ \gamma_p$ is constant, i.e. its derivative vanishes.
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  • $\begingroup$ The parallel case shouldn't be so hard: If the lines are parallel to the $y$-axis, this is the same as saying that $\partial_y f = 0$, and the general, "sloped" case you get by rotation, so it should be enough to check that $\begin{pmatrix} \partial_x f \\ \partial_y f \end{pmatrix}$ has the same slope everywhere, i.e. $\frac{\partial_x f(a)}{\partial_y f(a)} \big/ \bigg|\frac{\partial_x f(a)}{\partial_y f(a)}\bigg|$ is constant when you let $a$ range. $\endgroup$
    – Ben Steffan
    Commented Apr 18 at 21:59

1 Answer 1

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Twice differentiating with respect to $x$, the level curve $F(x,y) = c$, if it's not a vertical line, satisfies $$F_x + F_y \dfrac{dy}{dx} = 0$$ and $$\dfrac{d^2 y}{dx^2} = \frac{- F_{xx} F_y^2 + 2 F_{xy} F_x F_y - F_{yy} F_x^2}{F_y^3}$$ (where subscripts denote partial derivatives) Thus an $F$ whose level curves are all straight lines should satisfy the partial differential equation $$ - F_{xx} F_y^2 + 2 F_{xy} F_x F_y - F_{yy} F_x^2 = 0 $$

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  • $\begingroup$ Thank you! The functional form of the answer makes sense, but I have one confusion about the derivation: when I have a function $\mathbb{R}^m\rightarrow \mathbb{R}$, I can take the derivative with respect to any of the $m$ arguments. But I can't take the derivative of one argument with respect to another, unless there's an implicit composition $F\circ (x,\hat y(x))$. I don't know which function $\widehat{y}(x) : \mathbb{R}\rightarrow \mathbb{R}$ is --- maybe this is the implicit function theorem applied to F(x,y)-c = 0? --- in order to make the derivative $\frac{d\widehat y}{dx}$ make sense. $\endgroup$
    – user326210
    Commented Apr 18 at 22:26
  • $\begingroup$ Follow up: I've confirmed that this works. If (a,b) lies on the level curve with value c, $F(x,y)=c$ implicitly defines the level curve $\hat y_c(x)$ around (a,b). A straight level curve has zero second derivative. By IFT, $\hat{y}^\prime = -F_x(x,\hat{y})/F_y(x,\hat{y})$. We differentiate once more wrt x, and set equal to zero. Because we want this equation to hold for all $c$ (all level curves), we can replace the specific coordinate $(x,\hat{y}_c(x))$ with "all $(x,y)\in D$", eliminating reference to specific curves $\hat{y}_c$. $\endgroup$
    – user326210
    Commented Apr 19 at 20:20

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