# Qualitative study of $\dot x = - y^2$, $\dot y= x^2$

Make the qualitative study of the ODE on the plane $$\begin{cases}\dot x = - y^2\\ \dot y= x^2 \end{cases}$$ and determine how many solutions satisfy $y(0) = x(1) = 0$.

My attempt.

Let $f(x,y):=\begin{pmatrix}-y^2\\x^2 \end{pmatrix}$ be the vectorial field. It is well defined on all plane and $C^1$, then every Cauchy Problem has a unique local solution and every one of these solution can be extended in unique way to a maximal solution (not necessary globally defined). Now I note that the function $$E(x,y):=\frac{x^3}{3}+\frac{y^3}{3}$$ is a constant of motion, i.e. $\dot E(x(t),y(t))=c$, where $(x(t),y(t))$ is a solution. The orbits are the level sets of $E$, that are open curves like this: https://m.wolframalpha.com/input/?i=plot+x%5E3+%2B+y%5E3+%3D+10, or https://m.wolframalpha.com/input/?i=plot+x%5E3+%2B+y%5E3+%3D+-10.

• It is all correct?

• Do you think that it is enough ? The task "Make the qualitative study" is very generic. What would you add?

• I don't know how to determine how many solutions satisfy $y(0) = x(1) = 0$.

• Since $E$ is constant on the trajectories you can determine what $x(0)$ (and $y(1)$) would have to be. This cuts down the number of possibilities significantly. – copper.hat Jun 30 '18 at 20:34

If $y(0)=x(1)=0$ then $x(0)=y(1)=a$ for some nonnegative $a$ and it remains to enmerate the values of $a$ such that it takes a time $1$ to follow the curve $x^3+y^3=a^3$ from $(x,y)=(a,0)$ to $(x,y)=(0,a)$. To do so, note that, along this curve, $$dt=\frac{dy}{x^2}=\frac{dy}{(a^3-y^3)^{2/3}}$$ hence one needs to solve for $a$ the identity $$1=\int_0^a\frac{dy}{(a^3-y^3)^{2/3}}=\frac1a\int_0^1\frac{dy}{(1-y^3)^{2/3}}$$ Finally, for every $a$, consider the curve $$C(a)\ :\ x^3+y^3=a^3$$ Then, each solution stays on a given curve $C(a)$. There is exactly one solution such that $y(0)=x(1)=0$, and it stays on $C(a^*)$, where $$a^*=\int_0^1\frac{dy}{(1-y^3)^{2/3}}\approx1.76639\ldots$$ The curves $C(a)$ for $a>0$ are in the halfplane $x+y>0$, the curve $C(0)$ is the line $x+y=0$ and the curves $C(a)$ for $a<0$ are in the halfplane $x+y<0$. Each solution stays on a given curve $C(a)$ and moves on it from the SE quadrant to the NW quadrant, passing by the NE quadrant if $a>0$ and by the SW quadrant if $a<0$, see below for a sketch of the solutions of the dynamical system.
• Thanks. What we can say about the maximal interval of definition of the solution $(t_{min},\,t_{max})$? I think that it is finite since $t=\int^{y(t)}_{y(0)}\frac{dy}{(a^3-y^3)^{2/3}}$ that is integrable also if $y(t_{max})=+\infty$, Then $t_{max}$ is a real number. Similar for $t_{min}$. Do you agree? – Ef_Ci Jul 4 '18 at 1:12
• Indeed, $$t_{\mathrm{max}}=\int_{y_0}^{+\infty}\frac{dy}{\sqrt[3]{(a^3-y^3)^2}}$$ which is finite, and likewise for $t_{\mathrm{min}}$. – Did Jul 4 '18 at 5:23