# $\frac{\partial F}{\partial y}\neq0\implies$ continuous contour line? (Implicit Function Theorem)

I have a function $$F(x,y)=z$$ and two points $$(x_1,y_1),(x_2,y_2)$$ s.t. $$F(x_1,y_1)=F(x_2,y_2)=c$$, $$x_1. I know that $$\frac{\partial F}{\partial y}<0$$ in $$[x_1,x_2]\times\mathbb{R}$$.

I'd like to prove that there is a continuous contour line between the two points.

I know that there's a rectangle $$V\times W$$ that contains $$(x_1,y_1)$$ s.t. $$F^{-1}(c)\cap V\times W$$ is the graphic of a function, i.e., I have in there a continuous contour line.

I'd like to know if the fact that I have $$\frac{\partial F}{\partial y}<0$$ in the entire interval $$[x_1,x_2]$$ allows that I consider $$V=[x_1,x_2]$$ and $$W=\mathbb{R}$$, so I can prove the statement.

Many thanks!

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Edit for comment on 12/26

Yes, it's correct. Here's a sketch of the argument. Note first of all that by the implicit function theorem, the portion of a level curve of $$F$$ in the given region must be a one-dimensional manifold with no boundary points in the open rectangle $$(x_1,x_2)\times\Bbb R$$.

Let $$\Gamma$$ be the connected component of the level curve $$F(x,y)=c$$ passing through $$(x_1,y_1)$$. The fact that $$\partial F/\partial y\ne 0$$ tells us that $$\Gamma$$ can have no vertical tangent line, and this says that the set of $$x$$-coordinates of points on $$\Gamma$$ cannot have a maximum value $$. That is, $$\Gamma$$ contains a point $$(x_2,y^*)$$. But now the condition $$\partial F/\partial y<0$$ tells us that level curves of $$F$$ can have at most one point for each fixed $$x$$. Therefore, $$y^*=y_2$$ and $$(x_2,y_2)\in\Gamma$$.

• Try an example for yourself to see what goes wrong. In particular, what happens to $\partial F/\partial y$ when $x=x_3$? Dec 26, 2020 at 18:05
• You must have a level curve through every point unless the domain of $F$ has holes in it!! Can you think of a specific function that gives you a picture like yours? Understand that function and then think about my proof. You were never explicit about hypotheses, but we're assuming that $F$ is $C^1$ on $[x_1,x_2]\times\Bbb R$ and that $\partial F/\partial y$ is everywhere negative. Dec 26, 2020 at 18:41
• Yes, that's fine. You're right that this was a gap in my argument; we could have $x_3 = \sup \{x: (x,y)\in\Gamma\}$ with no point $(x_3,y^*)\in\Gamma.$ Since level curves are closed, this can happen only if there is a vertical asymptote at $x=x_3$. If we study $F(x_3,y)$, it must be decreasing everywhere and cannot pass through the value $c$, which, as you've pointed out violates continuity at points for large values of $y$. Dec 26, 2020 at 21:26
• Hmm ... You keep catching me. So for what you've suggested, we'd have to have $\partial F/\partial x$ change sign at $x=x_3$ (positive on the left, negative on the right), as your surface sketch suggests, but I don't see any vertical asymptotes coming from that sketch. At the moment, I don't see a concrete example. Dec 26, 2020 at 23:52
• I am pretty sure a Taylor polynomial approximation of $F$ is good enough to show that what you are suggesting cannot happen. You can't have $F$ smooth, $\partial F/\partial y$ everywhere $<0$, and have an asymptote at $x=0$. Dec 27, 2020 at 0:19

There may not be a contour line between the two points. Indeed, consider the function $$F(x,y) = (x^2+1) \arctan y$$ with $$\frac{\partial F}{\partial y}(x,y) = \frac{x^2+1}{y^2+1}> 0$$ on $$\mathbb{R}^2$$. At the points $$(-1,1)$$ and $$(1,1)$$ the functions takes the value $$c=\frac{\pi}{2}$$, but the level curve $$F(x,y) = c$$, that is $$(x^2+1)\arctan y = \frac{\pi}{2}$$ does not intersect the vertical axis $$x=0$$.

The idea is simple: for every fixed $$x$$ , the function $$F(x,y)$$ is strictly increasing in $$y$$, with image an open interval $$I_x$$. As $$x$$ varies, this interval varies too. It may be that the interval $$I_x$$ contains $$c$$ for $$x=x_1$$, $$x_2$$, but, for some intermediate point $$x_3$$, $$I_{x_3}$$ does not contain $$c$$. The example is chosen so that the $$c$$ will be one end of the interval $$I_{x_3}$$.

Note: The level curve $$F(x,y) = \frac{\pi}{2}$$ is the graph of the function $$y = \tan \frac{\pi}{2(x^2+1)}$$ defined on $$\mathbb{R} \backslash \{0\}$$.

$$\bf{Added:}$$ I changed the sign so that $$\frac{\partial F}{\partial y} > 0$$, with $$F\mapsto -F$$.

• This seems to be an example I was searching for at the end. Two points, though. “For every $y$” should be “for every $x$.” Also you do not have $y_1<y_2$ as the OP required. Dec 27, 2020 at 16:58
• Thank you, indeed it should be "for every $x$", corrected! I have a feeling that the OP meant $x_1 < x_2$, so one can talk about the interval $[x_1, x_2]$. We can also consider the left point to be $(-\sqrt{2}, \frac{1}{\sqrt{3}})$. Dec 27, 2020 at 17:18
• Thank you all, really is $x_1<x_2$ (maybe there's an example with $y_1<y_2$ too, but this does not mean now...). Thank you @TedShifrin and orangeskid Dec 27, 2020 at 17:53
• @Quiet_waters: You are very welcome! A very interesting question. I think if every vertical between $x_1$ and $x_2$ intersects the level set in exatly one point then those points of intersection form a curve connecting the two points. But otherwise they may not be in the same connected components of the level set. The second picture with the asymptote gave the idea, like Ted Shifrin mentioned. Dec 27, 2020 at 18:54
• @Quiet_waters: With the hypothesis given ( $\frac{\partial F}{\partial y}$ constant sign ) : it the level curve joins $(x_i, y_i)$ then this level set in the strip is bounded. Conversely, if the level strip is bounded, we can show it joins the points. In fact, it we can draw two horizontal segments, one above, one below the points, that does not intersect the level curve, then on them the values must be $< c$, respectively $> c$. Then on every vertical between them, there will be a point on the level set. So the issue is: the level set should not escape at infinity.. Dec 27, 2020 at 21:16