Please consider $$\left\{\begin{matrix}Y'&=&f(Y)\\Y(0)&=&y_0\end{matrix}\right.\tag{1}$$ with $$f(u,v):=\begin{pmatrix}+2u-v-u^2v\\-2u+v+u^2v\end{pmatrix}\;\;\;\text{and}\;\;\;y_0:=\begin{pmatrix}u_0\\v_0\end{pmatrix}\in\mathbb{R}_{\ge 0}^2$$ I want to show that
- For all $u_0$, there is a $v_0$ such that the solution $Y$ of $(1)$ globally exists
- For all $u_0$, there is a $v_0$ such that the solution $Y$ of $(1)$ explodes in finite time
I know that all that is somehow related to the zeros of $f$ and the eigenvalues of $-f'$. We've got $$A:=-f'(u,v)=\begin{pmatrix}-2+2uv&+1+u^2\\+2-2uv&-1-u^2\end{pmatrix}$$ So, the eigenvalues of $A$ are given by $$\sigma(A)=\left\{0,2uv-u^2-3\right\}$$ Moreover, $$f^{-1}(\left\{0\right\})=\left\{(u,v) : v=\frac{2u}{1+u^2}\right\}$$ However, all theorems I know so far require that all eigenvalues are positive, but here they are all non-positive.
Is there some clever way, to show (1) and (2)?
The sum of the two entries of $f(u,v)$ is zero for every $(u,v)$ hence $u(t)+v(t)=u_0+v_0$ for every $t$ and $u'=A_{u_0+v_0}(u)$ where, for every $c$, $$A_c(x)=3x-c(1+x^2)+x^3.$$ Fix some $u_0$.
Let $v_0=\dfrac{2u_0}{1+u_0^2}$, then $A_{u_0+v_0}(u_0)=0$ hence $(u(t),v(t))$ is defined globally (and constant).
Let $c_0$ such that $A_{c_0}(x)\gt0$ for every $x\geqslant u_0$ and let $v_0=c_0-u_0$. Then $t\mapsto u(t)$ is increasing to $+\infty$ and, after some time $t_0$, $u(t)$ is so large that $u(t)\geqslant1$ and $u'(t)\geqslant\frac12u^3(t)$. Thus, $$\frac1{u(t_0)^2}-\frac1{u(t)^2}\geqslant t-t_0,$$ for every $t\geqslant t_0$ such that $u(t)$ is well defined. This implies that $u$ explodes at most at time $$t_0+\frac1{u(t_0)^2}.$$
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