# Consider the differential system { <mtable columnalign="left left" rowspacing=".2em" co

Rachel Villa 2022-05-22 Answered
Consider the differential system
$\left\{\begin{array}{ll}& {y}^{\prime }\left(t\right)=ay\left(t{\right)}^{3}+bz\left(t\right)\\ & {z}^{\prime }\left(t\right)=cz\left(t{\right)}^{5}-by\left(t\right)\end{array}$
with $t>0$
$y\left(0\right)={y}_{0},z\left(0\right)={z}_{0},\phantom{\rule{1em}{0ex}}a<0,\phantom{\rule{1em}{0ex}}c<0,\phantom{\rule{1em}{0ex}}b\in \mathbb{R}$
the question is to prouve that this system admits a unique solution on $\left[0,+\mathrm{\infty }\right]$?
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Erick Blake
By standard ODE theorems, we know that if a solution doesn't exist, it will only not exist because the solution "blows up," that is, if there exists a $T>0$ such that $x\left(t\right)$ or $y\left(t\right)$ is unbounded as$t\to {T}^{-}$.
Multiply the first equation by $y\left(t\right)$, and the second solution by$z\left(t\right)$, and then add them together, you obtain
$\frac{1}{2}\frac{d}{dt}\left(\left(y\left(t\right){\right)}^{2}+\left(z\left(t\right){\right)}^{2}\right)=a\left(y\left(t\right){\right)}^{4}+c\left(z\left(t\right){\right)}^{6}.$
Since $a$ and $c$ are negative, it follows that $\left(y\left(t\right){\right)}^{2}+\left(z\left(t\right){\right)}^{2}$ is bounded. Hence a blow up can never happen.