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Prove that the equation $$y'=1+y^4 \:, y(0)=0$$ has a maximal solution with a bounded domain.

The existence and unicity of such a maximal solution is granted by Cauchy-Lipschitz theorem.

I tried a proof by contradiction for the boundedness. One promising way would be to use the fundamental theorem of calculus and taking limits at infinity...

Thanks for your suggestions.

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2 Answers 2

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Here is an indirect solution. If $y$ does not blow up in finite time, then the maximal solution is defined on all of $[0, \infty)$. We make the following two easy observations:

  1. Sine $y' \geq 1$, we get $y(t) \geq t$.
  2. Since $y' = 1 + y^{4} \geq 2y^{2}$, for any $0 < t_{0} < t_{1}$ we get $$ \frac{y'}{y^{2}} \geq 2 \quad \Longrightarrow \quad \frac{1}{y(t_{0})} - \frac{1}{y(t_{1})} \geq 2(t_{1} - t_{0}). $$

Combining two observations, for any $0 < t_{0} < t_{1}$ we get

$$ \frac{1}{t_{0}} \geq \frac{1}{y(t_{0})} - \frac{1}{y(t_{1})} \geq 2 (t_{1} - t_{0}). $$

Then we get a contradiction by choosing large $t_{0}$ and $t_{1} = t_{0} + 1$.

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  • $\begingroup$ This is exactly the abstraction I was looking for! $\endgroup$ Mar 13, 2014 at 7:46
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Let $y(t)$ be the solution, so that we know that $y(0)=0$ and $y'(t)=1+y(t)^4$. Clearly the function is strictly increasing in its domain, so it is positive on the positive part of its domain.

We have $\frac{y'(t)}{1+y(t)^4}=1$. Integrating this from $0$ to $\tau$, we have $\int_0^\tau\frac{y'(t)}{1+y(t)^4}dt=\tau$, and the integral can be rewritten as $\int_0^{y(\tau)}\frac{d\xi}{1+\xi^4}$, and this can be computed, so we can conclude that $$ \frac{-\tan ^{-1}\left(1-\sqrt{2} y(t)\right)+\tan ^{-1}\left(\sqrt{2} y(t)+1\right)+\tanh ^{-1}\left(\frac{\sqrt{2} y(t)}{y(t)^2+1}\right)}{2 \sqrt{2}}=\tau$$ for all $\tau$ in the positive part of the domain of $y$. Now the function $$\frac{\tanh ^{-1}\left(\frac{\sqrt{2} t}{t^2+1}\right)-\tan ^{-1}\left(1-\sqrt{2} t\right)+\tan ^{-1}\left(\sqrt{2} t+1\right)}{2 \sqrt{2}}$$ is bounded. This means that the domain has to be bounded above. In fact, the upper limit of the domain has to be the limit value of this function, which is $\frac{\pi }{2 \sqrt{2}}$.

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  • $\begingroup$ This is a nice "numerical" solution which yields the limit value of the interval (that's scarce!). I didn't choose it as an answer because it resort to a heavy computation that I couldn't have done by hand. $\endgroup$ Mar 13, 2014 at 7:50
  • $\begingroup$ Well, integrating rational functions is a basic calculus skill, really, and the limit is very easy to compute! $\endgroup$ Mar 13, 2014 at 8:03
  • $\begingroup$ Notice that you can bound from above $(1+\xi^4)^{-1}$ by $(1+\xi^2)^{-1}$ in the integral, and that integral and the corresponding limit you can surely do by memory :-) $\endgroup$ Mar 13, 2014 at 8:04
  • $\begingroup$ I love this answer! (+1 upvote, too!) In fact, the only important point is that $(1 + \xi^{4})^{-1}$ is integrable on $[0, \infty)$. So if $\tau$ is the maximal time of existence, then $y(\tau^{-}) = \infty$ and hence $$ \tau = \int_{0}^{\infty} \frac{d\xi}{1 + \xi^{4}} < \infty. $$ $\endgroup$ Mar 13, 2014 at 8:13
  • $\begingroup$ +1: Nice. While you are right that it is basic calculus, I would still struggle to get the integral correct in reasonable time without my Maxima... $\endgroup$
    – copper.hat
    Mar 13, 2014 at 16:12

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