why $y'=\frac{2xy}{x^2+y^2}$ is not global Lipschitz

Question

I have this given IVP $y'=\begin{cases}\frac{2xy}{x^2+y^2}&\;\;\;\;(x,y)\neq(0,0)\\0&\;\;\;\;(x,y)=(0,0)\end{cases}$ where $y(-1)=-0.001$. I used matlab obtain this figure

Why the graph blow up after $0$?

$$|y(t)-y(-1)|\leq\left|\int_{-1}^{t}y'(s)\,ds\right|\leq\int_{-1}^{t}|y'(s)|\,ds\leq\int_{-1}^{t}\,ds=t+1$$

$$|y(t)|\geq-0.999,\,\,t\to0$$

From WolfAlpha, it gives http://www.wolframalpha.com/input/?i=y%27%3D2xy%2F(x%5E2%2By%5E2),y(-1)%3D-0.001

Answer 1

Using the same argument as in this post The function $4x^3y/(x^4+y^2)$ fails the Lipschitz condition near the origin we get

$$y'=\frac{2xy}{x^2+y^2}=:f(x,y)$$

$$\partial_x \ f(x,y)=\frac{2y(x^2+y^2)-4x^2y}{(x^2+y^2)^2}, ~\partial_y \ f(x,y)=\frac{2x(x^2+y^2)-4xy^2}{(x^2+y^2)^2}$$

$$\lim_{y \to 0}\partial_y \ f(x,y)=\frac{2x^3}{x^4}=\frac{2}{x}$$

and therefore the partial derivatives of $f$ near the origin are not bounded; and $f$ isn't locally Lipschitz continuous near the origin.

Answer 2

Use polar coordinates to show that it is not continuous at 0. For $x = r\,\text{cos}(\theta)$ and $y = r\,\text{sin}(\theta)$ you get

$$lim_{r\to 0} \frac{2\,r^2\cos(\theta)sin(\theta)}{r^2(\text{cos}^2(\theta) + \text{sin}^2(\theta))} = \lim_{r\to 0} 2 \text{cos}(\theta)\,\text{sin}(\theta)$$

and this limit does not exist.

Answer 3

$$\frac{dy}{dx}=\frac{2xy}{x^2+y^2}\quad\to\quad (x^2+y^2)dy=2xydx$$ $$x^2=X \quad\to\quad (X+y^2)dy=ydX \quad\to\quad y\frac{dX}{dy}=X+y^2$$ $y\frac{dX}{dy}-X=y^2 \quad$is a linear ODE which general solution is easy to find :$\quad X=y^2+2cy$

The general solution of $\quad \frac{dy}{dx}=\frac{2xy}{x^2+y^2}\quad$ is : $\quad y^2+2cy=x^2$ $$(y+c)^2-x^2=c^2$$ $$y(x)=-c\pm\sqrt{c^2+x^2}$$ $y(x)$ is the equation of a conic section of the hyperbola kind, which center is $(0,-c)$ . The asymptotes are $\quad y=-c+x\quad$ and $\quad y=-c-x$.

$y'=\pm\frac{x}{\sqrt{c^2+x^2}}\quad\to\quad y'(0,0)=0\quad$ satisfies the given specification.

The condition $\quad y(-1)=-0.001\quad$ implies $\quad c=\frac{x^2-y^2}{2y}=\frac{1-10^{-6}}{-0.002}\simeq -500$

The "blow-up" is only due to the big value of $|c|$. In reality this is the normal behaviour of hyperbola.

Around $x=0$, the equivalent of the function is : $$y=-c\left(1-\sqrt{1+\frac{x^2}{c^2}}\right) \simeq \frac{x^2}{-2c}\simeq \frac{x^2}{1000}$$ which is close to a parabola as expected.

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