The problem I'm considering is $$|Y(x)|^2Y(x)+\left<x,Y(x)\right>^2x=0,$$ where $Y:\mathbb{R}^2\to\mathbb{R}^2$ and $x\in\mathbb{R}^2$. We're also told that $Y(0,1)=(0,-1)$. We're asked to show that $-|x|^2x$ is the unique continuous solution. Now, I got into a discussion with my professor, because he doesn't believe my proof works. I think it does, and I'd like to get verification if possible. Thank you very much.
Now first off, me and the Professor agree that for all $x\not=0$ there exists a ball $B(x,r)$ such that there exists a unique continuous $Y':B(x,r)\to\mathbb{R}^2$ satisfying the equation $$|Y'(y)|^2Y'(y)+\left<y,Y'(y)\right>^2y=0,$$ for all $y\in B(x,r)$ and $Y'(x)=-|x|^2x.$ So we'll just assume that for the proof.
$\textbf{Proof:}$
By computation $-|x|^2x$ is a continuous solution, so we're guaranteed existence. We need only show uniqueness.
Let $Y:\mathbb{R}^2\to\mathbb{R}^2$ be such that $$Y(0,1)=Z(0,1)=(0,-1),$$ and for all $x\in\mathbb{R}^2$ we have $$|Y(x)|^2Y(x)+\left<x,Y(x)\right>^2x=0,$$ Let $x\in\mathbb{R}^2\backslash\{0\}.$ Let $\gamma:[0,1]\to\mathbb{R}^2\backslash\{0\}$ be a path from $x_0$ to $x.$ Let $$B:=\{y\in[0,1]:Y(\gamma(y))=-|\gamma(y)|^2\gamma(y)\}.$$ Let $z=\sup B$ and take $z_n\in B$ with $z_n\to z.$ Then by continuity $$Y(\gamma(z))=\lim_{n\to\infty}Y(\gamma(z_n))=\lim_{n\to\infty}-|\gamma(z_n)|^2\gamma(z_n)=-|\gamma(z)|^2\gamma(z),$$ so $z\in B.$
Suppose that $z\not=1.$ Choose $r$ such that there exists a unique continuous $Y':B(\gamma(z),r)\to\mathbb{R}^2$ satisfying the equation $$|Y'(y)|^2Y'(y)+\left<y,Y'(y)\right>^2y=0,$$ and $$Y'(\gamma(z))=-|\gamma(z)|^2\gamma(z).$$ Well then as $Y$ is a continuous local solution satisfying $$Y(\gamma(z))=-|\gamma(z)|^2\gamma(z)$$ it follows that $$Y(y)=-|y|^2y$$ for all $y\in B(\gamma(z),r).$ Now by continuity since $z<1$ we can and do choose $z<z'\leq 1$ such that $\gamma(z')\in B(\gamma(z),r).$ Well then $$Y(\gamma(z'))=-|\gamma(z')|^2\gamma(z'),$$ but then $z'\in B$ contradicting the maximality of $z.$ It then follows that $z=1,$ so $$Y(x)=Y(\gamma(1))=-|\gamma(1)|^2\gamma(1)=-|x|^2x.$$ Since $x$ was arbitrary it follows that $Y(x)=-|x|^2x$ for all $x\in\mathbb{R}^2\backslash\{0\}.$
Finally we note that if $x=0,$ then the equation reduces to $$|Y|^2Y=0,$$ so $$Y(0)=0=-|0|^20.$$ Then $Y(x)=-|x|^2x$ for all $x\in\mathbb{R}^2,$ hence $-|x|^2x$ is the unique continuous solution satisfying the initial conditions (the conditions imposed at $(0,1)$). This completes the proof.$\blacksquare$
As a remark, this problem can be solved rather simply by noting that $Y(x)$ must be a scalar multiple of $x,$ then setting $Y(x)=cx,$ and solving for $c.$ The goal was to prove the claim without appealing to that.