Let a closed curve, $\gamma$, be parameterized by a function $f : [0, 1] → \mathbb{R}^2$ with a continuous derivative and f(0) = f(1). Suppose that $$ \int_\gamma y^3 \sin^2(x) \, dx - x^5 \cos^2(y) \, dy = 0 $$ Show that there exists a pair $\{x, y\} \neq \{0, 1\}$ with $x \neq y$ and $f(x) = f(y)$. Give an example of a curve satisfying above requirement and the only pairs $\{x, y\}$ with $x \neq y$ and $f(x) = f(y)$ are subsets of $\{0, 1/2, 1\}$.

My attempts: I know how to prove that there are other pairs of $\{x,y\}$ such that $f(x) = f(y)$. But I have no idea how to construct such curve $\gamma$. I think that by using Green's Formula, I will have

$$\iint_D (5x^4 \cos^2 (y) + 3y^2 \sin^2(x))\, dx\, dy = 0. $$

What else information could I derive from above?

  • $\begingroup$ I think you're on the right track, but the application of Green's Theorem doesn't look correct to me. Specializing to the case that the coefficient of $dy$ in the integral over $\gamma$ is $0$ gives $$\oint_\gamma P \,dx = -\iint_D \frac{\partial P}{\partial y} \,dx\,dy$$. $\endgroup$ – Travis Apr 26 at 16:21
  • $\begingroup$ @Travis I think I made a mistake when copying down the problem. It should be $$\int_\gamma y^3 \sin^2(x) dx - x^5 \cos^2(y) dy = 0$$ $\endgroup$ – mathdoge Apr 26 at 16:27
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    $\begingroup$ Nevertheless, you need $3y^2$, not $2y^3$! Note that the integrand is everywhere $\ge 0$. $\endgroup$ – Ted Shifrin Apr 26 at 16:39
  • $\begingroup$ @TedShifrin Thanks for pointing it out. My fault. Yes, the integrand is non-negative everywhere, so if the integral is $0$, I should find a region $D$ such that $5x^4 \cos^2(y) + 3y^2 \sin^2(x) \equiv 0$ on $D$? $\endgroup$ – mathdoge Apr 27 at 0:01
  • $\begingroup$ Well, of course, you cannot find such a region! $\endgroup$ – Ted Shifrin Apr 27 at 0:05

Here I would like to answer my own question. Here I would like to thank @TedShifrin for helping me with this.

First, we show that there exists another pair of $\{x,y\} \neq \{0,1\}$ such that $f(x) = f(y)$. This indeed requires us to show that $\gamma$ is not simple, i.e., it is self-crossed. We could prove it by contradiction: suppose $\gamma$ is simple. Then by Green's Formula, we have that

$$\int\int_D \left(5x^4 \cos^2(y) + 3y^2 \sin^2(x) \right)dxdy = 0.$$

Define the integrand is $F(x,y) := 5x^4 \cos^2(y) + 3y^2 \sin^2(x).$ Note that $ F(x,y) \geq 0$ for all $(x,y) \in \mathbb{R}^2$, so $\int\int_D \left(5x^4 \cos^2(y) + 3y^2 \sin^2(x) \right)dxd y > 0$ for any region $D$ enclosed by simple closed curve $\gamma$. (We could argue this by considering a small neighborhood inside $D$ with $(x,y) \in D$ such that $F(x,y) > 0$, and using the continuity) This is a contradiction, and we proved that $\gamma$ is not simple.

To construct an example, notice that $F(x,y) = F(-x,-y)$. Therefore, we could let $\Gamma: [0, \frac{1}{2}] \to \mathbb{R}^2$ be a closed, simple curve with continuous derivatives such that $\Gamma(0) = \Gamma(\frac{1}{2})$. Define $f$, the parametrization of $\gamma$ as follows: $f(x) = \Gamma(x)$ for $x \in [0,\frac{1}{2}]$ and $f(x) = -\Gamma(1-x)$ for $x \in (\frac{1}{2},1]$. By using Chain Rule, we know that $f$ is continuously differentiable; $f([0,\frac{1}{2}])$ and $f([\frac{1}{2},1])$ are simple and closed. Thus,

$$\int\int_\gamma F = \int\int_{f([0,\frac{1}{2}])} F + \int\int_{f([\frac{1}{2},1])} F = \int\int_{f([0,\frac{1}{2}])} F + \int\int_{-f([0,\frac{1}{2}])} F = 0.$$

For a concrete example, we could consider a unit circle centered at $(1,0)$: $\Gamma(t) = (\cos(4\pi t + \pi)+1, \sin(4\pi t + \pi))$, and extend it oddly.


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