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Consider the differential equation in $\mathbb R$: $$x' = x^2-\lambda x^3; \,\,\,\,\,\, x(0) = x_0; \,\,\,\,\, t\geq 0$$ where $\lambda $ is a parameter. For which initial conditions is the solution bounded for $t\geq 0$? For which initial conditions does the solution blowup in finite time?

I notice that $x=0$ and $x=1/{\lambda}$ are boundaries for solutions. I think as long a $x_0$ is in between the $0$ and $1/{\lambda}$ for ${\lambda} \ne 0$ then the solution is bounded in between there and it cannot escape.

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Please consider editing your question with some work you have done and explain where exactly you are stuck –  Jean-Sébastien Nov 7 '12 at 14:59
    
@ Jean Thank you for your comment. I will try to think this over as I go –  Klara Nov 7 '12 at 15:57
    
Hint: Rewrite your equation as $$x' = x^2(1-\lambda x)$$ and see what happens in the phase line. –  Pragabhava Nov 8 '12 at 3:30
    
@ pragahava I actually had that in my mind, that's how I came with the lines x=0 and x=1\{\lambda}. –  Klara Nov 8 '12 at 3:54
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@Klara What can you say about the cubic $x^2(1-\lambda x)$ as $\lambda$ varies? How will the sign of $x'$, and hence the nature of the critical points will change? –  Pragabhava Nov 8 '12 at 4:38

1 Answer 1

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If $\lambda =0$, $x'=x^2>0$ then IVP's solutions blow up for $x_0>0$ and are bounded for $x_0<0$

If $\lambda <0 $, for $x_0 \in [1/\lambda, 0]$ solutions are bounded. For $x_0 >0, x'>0 $ and $x_0 <1/\lambda, x'<0 $ all the solutions blow up.

If $\lambda >0 $, for $x_0 \in [0,1/\lambda]$ solutions are bounded. For $x_0 <0, x'>0 $ and $x_0 >1/\lambda, x'<0 $ all the solutions are bounded.

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