Sorry for my bad english.

Let the differential equation defined by $\begin{cases} y'=y\ln(1+y) \\ y(0) = a > -1\end{cases}$.

We want to find the maximal solution defined on the maximal interval $J$. We can notice that $u_1=0$ is a maximal solution.

Let an other solution $(u_2,J)$ distinct of $u_1$. Then $\forall x \in J, u_2(x)>0$ OR $-1<u_2(x)<0$. In the second case, I found that $J=\mathbb{R}$ because $u_2$ is bounded.

In the first case, $\inf(J) = -\infty$ because $u_2$ is incrasing and bounded from below.

But how to study $\sup(J)$ ? Someone could help me ? Thank you in advance !

  • $\begingroup$ Note that $u_1\equiv 0$ is a solution if and only if $a=0$. $\endgroup$ – Chee Han Apr 24 '17 at 23:34

If $y(x)>0$ for all $x\in J$ then $y(x)\ln(1+y(x))\le (1+y(x))\ln(1+y(x))$ and so $$\frac{1}{(1+y(x))\ln(1+y(x))}y'(x)\le 1.$$ Since $a>0$, integrating both sides you get
$$ \ln(\ln(1+y(x)))-\ln(\ln(1+a))\le x,$$ which gives $$1+y(x)\le e^{\ln(1+a)e^{x}}.$$In particular, if $J$ is bounded from above, then so is $y$ but then, since it is increasing, there exists $$\lim_{x\to\sup J}y(x)=\ell<\infty$$ which is a contradiction since we could extend $y$ by considering the initial value problem $y(\sup J)=\ell$. Thus $J=\mathbb {R}.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.