# Find a curve $\gamma$ satisfying $\int_\gamma y^3 \sin^2(x) \, dx - x^5 \cos^2(y) \, dy = 0$

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?

• 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 $$\oint_\gamma P \,dx = -\iint_D \frac{\partial P}{\partial y} \,dx\,dy$$. – Travis Apr 26 at 16:21
• @Travis I think I made a mistake when copying down the problem. It should be $$\int_\gamma y^3 \sin^2(x) dx - x^5 \cos^2(y) dy = 0$$ – mathdoge Apr 26 at 16:27
• Nevertheless, you need $3y^2$, not $2y^3$! Note that the integrand is everywhere $\ge 0$. – Ted Shifrin Apr 26 at 16:39
• @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 $D$? – mathdoge Apr 27 at 0:01
• Well, of course, you cannot find such a region! – 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.