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Let $f:[0,\alpha]\to\mathbb{R}$ be a solution of the Cauchy problem:

$\begin{cases} f'(t)=(f(t))^2+t \\ f(0)=0 \end{cases} $

The question is: prove that $\alpha<3$.

It is clear that the problem admits a unique solution locally on some $[-\epsilon,\epsilon]$, since $(f)^2+t$ is $C^1$ in $(f,t)$, and we have $f(t)>0$ for all $t>0$, looking at the sign of $f'$. But now I don't know how to proceed. Thank you all!

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  • $\begingroup$ Seems like a Gronwall's inequality to me! $\endgroup$ – Behnam Esmayli Aug 8 '17 at 17:40
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One has $$y(x)=\int_0^x \bigl(y^2(t)+t\bigr)\>dt\geq {x^2\over2}\qquad(x>0)\tag{1}$$ with equality only if $y(t)\equiv0$, which is not the case, by $(1)$. It follows that $y(1)>{1\over2}$.

From $y'(t)\geq y^2(t)$ we infer $${y'(t)\over y^2(t)}\geq1\qquad(t\geq1)\ ,$$ and integrating this from $1$ to $x$ gives $${1\over y(1)}-{1\over y(x)}\geq x-1\ .$$ This implies ${1\over y(x)}\leq1+{1\over y(1)}-x$, hence $$y(x)\geq{1\over\beta -x}\qquad(x\geq1)$$ for some $\beta<3$.

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I try to give an answer to my question.

Let $\delta>0$ be such that $f:(-\delta,\delta)\to\mathbb{R}$ is the maximal extension of the solution $f$; we want to show that $\delta<3$. We already know that $f(0)=0$, $f(t)>0$ for all $t>0$ and $f$ is strictly increasing on $[0,\delta)$.

Let us consider $a\in(0,\delta)$.

(1) For all $t\in[0,\delta)$ we have $f'(t)\ge t$ and then $f(t)\ge \frac{t^2}{2}$. In particular $\frac{1}{f(a)}\le\frac{2}{a^2}$.

(2) For all $t\in[a,\delta)$ we have $f'(t)\ge (f(t))^2$ and then $f(t)\ge y(t)$ with $y(t)$ solution of the problem:

$\begin{cases}y'(t)=(y(t))^2 \\ y(a)=f(a) \end{cases}$

Solving by separation of variables we get that $f(t)\ge y(t)=\frac{1}{a+\frac{1}{f(a)}-t}$ for all $t\in[a,\delta)$.

Now if $\delta\le \frac{3}{2}$ we already have the thesis. Otherwise we can take $a=\frac{3}{2}$ in the above cases, and then we obtain:

$a+\frac{1}{f(a)}\le a+\frac{2}{a^2}=\frac{43}{18}<3$

and for all $t\in[3/2,\delta)$ we have:

$f(t)\ge\frac{1}{\frac{43}{18}-t}$

that implies $\delta\le\frac{43}{18}<3$ as desired.

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    $\begingroup$ You could also use $f'(t)>f(t)^2+1$ for $t\ge1$ with $f(1)\ge \frac12$ so that $f(t)>\tan(t-1+\arctan(\frac12))$ thus $δ<1+\frac\pi2-\arctan(\frac12)=2.10714871779409…$ $\endgroup$ – Lutz Lehmann Aug 10 '17 at 13:32
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We have $f'(t)\geq t$ for all $t>0$. So, by the FTC, $$f(t) = \int_0^tf'(s)ds \geq \int _0 ^t s \ ds = \frac{1}{2}t^2, \ \ \ \ \ \ \forall t>0 \ .$$ Now, use this new piece of information about $f$ to gain a better estimate: $$f'(t) = (f(t))^2+t \geq (\frac{1}{2}t^2)^2+t = \frac{1}{4}t^4+t , \ \ \ \ \forall t>0.$$

Integrating again yields, $$f(t) = \int_0^t f'(s)ds \geq \int_0^t (\frac{1}{4}s^4+s ) \ ds = \frac{1}{5.4}t^5+\frac{1}{2}t^2, \ \ \ \ \ \ \forall t>0 \ .$$ Continuing the process, we will have that $f(t)$ is bounded below by a series in powers of $t$. Now, we may be able to easily show that that series diverges to infinity unless $t$ is less than $3$. I am optimistic that by ignoring certain terms in squaring the expression in each step (for instance by only considering squares of terms and ignoring two times this times that terms) we may have a simpler lower bound that diverges for $t>3$. I give up as this ending is not as elegant as the start!

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  • $\begingroup$ Thank you, you gave me some ideas. I'll try to post an answer later. $\endgroup$ – Pozz Aug 10 '17 at 9:59
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Similarly to another Riccati equation, set $f=-\frac{u'}{u}$ to get $$ -\frac{u''}{u}+\frac{u'^2}{u^2}=\frac{u'^2}{u^2}+t \implies u''+tu=0 $$ with initial conditions $u(0)=1$, $u'(0)=0$. This is an Airy differential equation.

The roots of $u$ are the poles of $f$. Now one can find a power series solution for $u$ $$ \sum_{k=2}^\infty k(k-1)c_kt^{k-2}+\sum_{k=0}^\infty c_kt^{k+1}=0 \iff 2c_2+\sum_{k=1}^\infty \Bigl[(k+2)(k+1)c_{k+2}+c_{k-1}\Bigr]\,t^k=0\\ \implies \begin{aligned}[t] c_0&=1\\ c_1&=0\\ c_2&=0\\ c_{3k}&=-\frac{c_{3k-3}}{3k(3k-1)}\\ c_{3k+1}&=0\\ c_{3k+2}&=0 \end{aligned} $$ so that the solution starts as $$ u(t)=1-\frac{t^3}{3·2}+\frac{t^6}{6·5·3·2}\pm… $$ and for the function $f$ $$ f(t)=\frac{\frac{t^2}{2}-\frac{t^5}{5·3·2}\pm…}{1-\frac{t^3}{3·2}+\frac{t^6}{6·5·3·2}\pm…} $$

To determine the root location of the denominator consider the roots of the partial series polynomials and what they tell of the sign changes of the series. By the Leibniz test for alternating series, as long as $0\le t^3<8·9=72$, one gets the bounds $$ 1-\frac{t^3}{3·2}\le u(t)\le1-\frac{t^3}{3·2}+\frac{t^6}{6·5·3·2}=\frac{(t^3-15)^2-45}{180} $$ The first positive root $t_*$ of $u$ lies between the first positive roots of the bounding polynomials, which gives $$ 6\le t_*^3\le 15-\sqrt{45}=8+\frac{4}{7+\sqrt{45}} $$ or numerically $$ 1.81712059283213\le t_* \le 2.0240265456829 $$ with the numerical root at $t_*=1.9863527074304723$.

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