I need help in finding maximal solution for the problem: $$ \cases {{\dot{x} = x^2+t}\\{x(0)=0}}$$ I know that because $x(1) \geq \frac{1}{2}$ and that every solution $x(t)$ of the problem is greater than or equal to $\tan (t+x(1) - 1)$ gives that $$\displaystyle \lim_{t \rightarrow \frac{\pi +1}{2}}x(t) = \infty$$ Is this enough or can I tighten this?
What do I do with negative $t$'s ?

(I also tried to bound the absolute value of the integral equation and use Gronwell inequality i.e: $$\lvert x \left( t\right)\rvert = \rvert \int_{0}^{t}x^2+t\ dt\lvert < something$$ as well as $$|\int_0 ^t \frac{dx}{x^2}| = |\int_0 ^ t 1+\frac{t}{x^2}dt|< anything$$ but no success.)

An other attempt with Pickard iterations:
$\varphi_{0} = 0$ and $\varphi_{n} = \int_0 ^t \varphi^2_{n-1}(s) + s\ ds$, after some calculations, (and I need help in verifying correctness) I got $$\varphi_n(t) = \sum _{i=0}^n \frac{t^{a_i}}{a_i \cdot {a_{i-1}^2 \cdot... \cdot {a_{0}^{\log_2{i}}}}} = \frac{t^{a_n}}{a_n \cdot a_{n-1}^2 \cdot ... \cdot a_{0}^{\log n}} + O( \frac{t^{a_{n-1}}}{{a_{n-1}}^{n-1}} )$$

When $a_n = 3\cdot 2^{n-1}-1$

By ratio test - the series diverges when t>1, but $ \frac{1}{2} \leq x(1) \leq \tan(1)$ (previous lines) gives that the solution can be extended for $t$'s larger than 1. What is going on Here?


1 Answer 1


The analytic solution given below involves the Bessel functions. The powers of $t$ are non-integers. So, the formula for $x(t)$ is valid only in case of $t\geq 0$.

$x(t)$ is finite except in particular values $t=\left(\frac{3}{2}j_{-1/3}(n)\right)^{2/3}$, where $j_{-1/3}(n)$ is th n-th root of the Bessel function of order $-1/3$.

For example, starting from $t=0$ and $t$ increassing, $x(t)$ blow-up at $t=\left(\frac{3}{2}j_{-1/3}(1)\right)^{2/3}\simeq 1.86635086$

A similar study was done in case of $t<0$ (in addition, at the end of the post). The formula is similar, but with "modified Bessel functions" instead of "Bessel functions". For any $t<0$ the fonction $x(t)$ is continuous, without blow-up.

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