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Let us consider the following Cauchy problem on $\mathbb{R}_+$: $$ g'(t)=-g^2(t)+g^3(t)+g^4(t)h(g(t)),\quad g(0)=\alpha $$ where $\alpha$ is real and positive and $h$ is analytic in a neighbourhood of $0$. I would like to prove, for $\alpha$ small enough, the existence of a bounded function $y$ such that the unique solution of this Cauchy problem writes: $$ \frac{\alpha}{g(t)}=1+\alpha t-\alpha\log(1+\alpha t)-\alpha y(t),\quad y(0)=0. $$ In the simple case $h\equiv 0$, I can prove the existence of a bounded $y$ by posing $u=\alpha/g$ and plugging the ansatz into the ODE for $u$: $$ u'=\alpha\big(1-\frac{\alpha}u\big). $$ Then I integrate this last equation, get an implicit equation on $u$ and plug the ansatz into it. I can conclude by some standard arguments.

In the general case, I cannot use the same strategy and thus would like to prove boundedness of $y$ directly from its ODE. Already in the case $h\equiv 0$, I do not know how to proceed. Here is the ODE satisfied by $y$ in this simple case: $$ y'(t)=\frac{\alpha^2}{1+\alpha t}\frac{\log(1+\alpha t)+y(t)}{1+\alpha t-\alpha\log(1+\alpha t)-\alpha y(t)}. $$ Does anyone know how to prove boundedness of the solution of this last ODE with initial condition $y(0)=0$?

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Such a bounded function $y$ does not exist in general. Indeed, let $\alpha=1$ and $h=0$. Then the unique solution of your Cauchy problem is given by $$g(t)=\frac1{1+t}$$ for real $t\ge0$, and hence from your condition on $y$ we find $y(t)=-\ln(1+t)$, which is unbounded.

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  • $\begingroup$ Alright, you have a point. The existence of a bounded y is not granted in full generality but I think that if I add the condition $h(0)\neq 0$ it should. $\endgroup$
    – Fabien
    Jan 27, 2021 at 14:54
  • $\begingroup$ @Fabien : What does $h(0)$ have to do with this problem? Did you mean $h(g(0))=h(\alpha)$ instead? Anyhow, your question has been fully answered. If you have any further questions, with any further conditions imposed (or otherwise), please post them separately. $\endgroup$ Jan 27, 2021 at 15:57
  • $\begingroup$ My question was not precise enough, I am sorry. The cubic term is indeed required. Moreover I very probably need $g(0)$ small enough. I have had those precisions in the question. $\endgroup$
    – Fabien
    Jan 28, 2021 at 8:32
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    $\begingroup$ @Fabien : (i) You still did not answer my questions: What does $h(0)$ have to do with this problem? Did you mean $h(g(0))=h(\alpha)$ instead? (ii) In your initial post, you did not have the requirement for $g(0)$ to be small. Thus, your question was indeed fully answered. So, I suggest that you revert your later edits, invalidating the valid answer, and maybe post separately any further questions, with any further conditions imposed (or otherwise). Indeed, if your "question was not precise enough", who do you think should assume the responsibility for that? $\endgroup$ Jan 28, 2021 at 15:29

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