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I've been trying to prove that the initial value problem $$y'=y^2+t^2,\quad y(0)=0$$ has a maximal solution in $[0,2[$. From the uncountable simulations I've run it is clear to me that it is the case. The solution moreover seems to be asymptotic to $\frac1{2-t}$ as one can check by doing the change of variables $u=(2-t)y$ and solving numerically for $u$ (one then sees that $u\to 1$ as $t\to 2^-$).

I've been able to prove the existence of a solution up to $t=1.96$ using a comparison criterion and using as bounds the polynomials that come out from the typical iterated integration step $p_{n+1}(t)=y_0+\int_{t_0}^tf(s,p_n(s))\,ds$. I stopped at $1.96$ because the polynomials started to become pretty big (order $500$+) but I'm confident I could prove existence up to arbitrarily close to $2$.

Using the same comparison criterion I was unable to prove that the solution is controlled by neither $\frac1{2-t}$ nor $\frac1{2-t}+t-2$ despite it being numerically obvious that it is the case.

I've also tried the change of variables $v=\frac1{y+1}$, (the $+1$ so to make the denominator non-zero) which seems to make for a more stable ODE where the goal is to prove that the first positive root is at $v(2)=0$. However I am struggling to get a nice comparison out of that too.

How could one go about proving this fact?

Thanks!

aleph2
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  • The solution to this (classical/archetypical) Riccati equation can be expressed in Bessel functions. The location of the pole is actually at $b=2.003147359426885$, which is around the claimed result, but not exactly so. See https://math.stackexchange.com/questions/2348022/riccati-d-e-vertical-asymptotes, https://math.stackexchange.com/questions/446926/riccati-differential-equation-y-x2y2 and linked topics. – Lutz Lehmann Mar 18 '24 at 14:57
  • Thus for a numerical method, find the first positive root of $u''+x^2u=0$ under $u(0)=1$, $u'(0)=0$. Finding a root of a smooth function is easier than locating a pole singularity. – Lutz Lehmann Mar 18 '24 at 15:16
  • Thank you. The linear ode you mention comes from which change of variables? Does it always work for Riccati equations? – aleph2 Mar 19 '24 at 12:40
  • Details are in my answer in the first link. Yes, a similar approach works for all Riccati equations. A more general approach avoids derivatives of the coefficients and transforms into a linear first-order system, for an example see https://math.stackexchange.com/questions/446926/riccati-differential-equation-y-x2y2, https://math.stackexchange.com/questions/4404625/solve-x2y26-y2xy – Lutz Lehmann Mar 19 '24 at 12:51
  • @LutzLehmann in the linked questions, the bounds on the pole are not tight enough to rule out $t=2$ – Sal Mar 19 '24 at 13:32
  • The number is from a comment to the answer, where the root was refined using a more exact approximation. // When the right side is cubic, not Riccati: https://math.stackexchange.com/questions/2622702/the-ivp-begincases-dotx-x3e-t2-x0-1-endcases-possesses-a-solut – Lutz Lehmann Mar 19 '24 at 14:25

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