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Prove that the solution to Cauchy's problem $$ \begin{cases} y'=y^2+x^2\\ y(0)=0 \end{cases} $$ is an odd function, and that this solution can't be extended on $\mathbb{R}$.

I tried to substitute $y(x)$ with $-y(-x)$ and check if cancels out: $$ \begin{align} (-y(-x))'=(-y(-x))^2+x^2\\ y'(-x)=y^2(-x)+(-x)^2 && \text{substiute } t=-x\\ y'(t)=y^2(t)+t^2 \end{align} $$ I think that proves the first part.

As for the second part of the question:

let $y$ be a solution to the Cauchy's problem, then: $$ \begin{align} y'=y^2+x^2\\ y'\leq y^2\\ \int \frac{dy}{y^2}\leq \int dx\\ \frac{-1}{y}\leq x+c\\ \frac{1}{y}\geq -x+c\\ y\geq \frac{1}{-x+c} \stackrel{x\rightarrow c}{\longrightarrow}\infty \end{align} $$

therefore $y$ does not extend on $\mathbb{R}$. Is that correct?

MartinYakuza
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  • Actually the numerical solution gives a result only in $(-2,2)$ – Raffaele Feb 10 '21 at 21:43
  • See also https://math.stackexchange.com/questions/2348022/riccati-d-e-vertical-asymptote, https://math.stackexchange.com/questions/1353727/y-x2-y2-asymptote, https://math.stackexchange.com/questions/446926/riccati-differential-equation-y-x2y2 and articles linked from there. – Lutz Lehmann Feb 11 '21 at 06:47

1 Answers1

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You almost got it correctly:

  1. You have only proven that $-y(-x)$ is also a solution. You need to patch this result together with a uniqueness result to achieve that $-y(-x)$ is in fact equal to $y(x)$.

  2. You got two mistakes with the inequality signs but they cancel each other out... $$ \begin{align} y'=y^2+x^2\\ y'\color{red}{\geq} y^2\\ \int \frac{dy}{y^2}\geq \int dx\\ \frac{-1}{y}\geq x+c\\ \frac{1}{y}\leq -x+c\\ y\color{\red}{\geq} \frac{1}{-x+c} \stackrel{x\rightarrow c}{\longrightarrow}\infty \end{align} $$ Additionally, to achieve the last line, you should give an explanation why $y>0$, which is straightforward for $x>0, y(0)=0$ and $y'(x)>0$.

maxmilgram
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