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Consider the initial value problem $$\frac{dy}{dx}=x^2+y^2$$ with $y(0)=1$ where $0≤x≤1$. Then which of the following statements are true?

$(a)$ There exists a unique solution in $\displaystyle\bigg[0,\frac{\pi}{4}\bigg]$.

$(b)$ Every solution is bounded in $\displaystyle\bigg[0,\frac{\pi}{4}\bigg]$.

$(c)$ The solution exhibits a singularity at some point in $[0,1]$.

$(d)$ The solution becomes unbounded in some subinterval of $\displaystyle\bigg[\frac{\pi}{4},1\bigg]$.

For $(a)$; I started out by finding the largest interval of existence by Picard's theorem. Considering a rectangular strip $|x|≤h$ and $|y−1|≤k$, we see that $|x^2+y^2|≤|x|^2+|(y−1)+1|^2≤h^2+k^2+1=M$. Now the maximum interval of existence is $|x|≤h′$ where $\displaystyle h′=\min\bigg\{h,\frac{k}{M}\bigg\}=\min\bigg\{h,\frac{k}{h^2+k^2+1}\bigg\}$. But I'm unable to check that minimum to see whether $\displaystyle h'>\frac{\pi}{4}$ or $\displaystyle h'<\frac{\pi}{4}$. Also how to check the boundedness/singularity of solutions in the above intervals I don't understand. Any help is appreciated.

I know the solution can be found using Bessel function. But that's not what I'm asking here. I want a method without explicitly finding the solution.

am_11235...
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1 Answers1

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For $x\in[0,1]$ you get $$ y^2\le y'\le 1+y^2\implies \frac1{1-x}\le y(x)\le \tan(\tfrac\pi4+x), $$ so that the solution remains bounded for $x<\frac\pi4$.

If b) means for arbitrary initial point, then the left inequality gives $\frac{y(0)}{1-y(0)x}\le y(x)$ which forces a pole before $x=\frac1{y(0)}$.

See also Riccati D.E., vertical asymptotes for similar and other ideas to bound solutions and maximal domains. c) is again for the given initial condition? Use the left inequality to get an estimate for $y(\frac12)$ and then use $\frac14+y^2\le y'$ on $x\in[\frac12,1]$ for as long as the resulting lower bound exists.

d) is a consequence of a) and c).

Lutz Lehmann
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  • solution is bounded for $x<\frac{\pi}{4}$ by $\frac{1}{x-1}\leq y(x)\leq tan(\frac{\pi}{4}+x)$ but in option $2$ it is said bounded in $[0,\frac{\pi}{4}]$ how it is bounded in closed interval $[0,\frac{\pi}{4}]$ ? Thank you . – neelkanth Nov 18 '20 at 12:01
  • In b) the answer depends on what "every solution" actually means. As all solutions are unique, the IVP only has one solution. Thus the interpretation that the initial condition is variable, so that the claim becomes wrong. – Lutz Lehmann Nov 18 '20 at 18:25
  • Hello! Why is the solution bounded for $x<\frac{\pi}{4}$? Do you mean that the solution is bounded on $[0, a]$ for $a<\frac{\pi}{4}$? – psie Sep 29 '23 at 11:26
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    @sunny : In a strict sense your interpretation is correct. However, the bound is not very tight for $x$ close to zero. So one could apply the same strategy to a segmentation of the interval, with corresponding smaller constants in the bounds. This gives an upper bound over a domain that includes $\frac\pi4$. – Lutz Lehmann Sep 29 '23 at 11:37