Interval of unique solution of differential equation $y’=x^2+y^2, y(0)=1 .$

Question

Consider the following initial value problem $$y’=x^2+y^2, y(0)=1$$

Which of the following statements is/ are

$1.$ there exist a unique solution in $[0,\frac{\pi}{4}]$.

$2.$ every solution is bounded on $[0,\frac{\pi}{4}]$.

$3.$ The solution has a singularity at some point in $[0,1]$.

$4.$ The solution because unbounded at some sub interval of $[\frac{\pi}{4}, 1]$.

I tried to find interval for unique solution by using existence and uniqueness theorem as

For $$|x|\leq a$$ $$|y-1|\leq b$$ we have

$$|f(x,y)|=|x^2+y^2|\leq a^2+(1+b^2)$$ $$h=min\{a,\frac{b}{M}\}=min\{a,\frac{b}{1+a^2 +b^2+2ab}\}$$ Now to find maximum of $$g(b)= \frac{b}{1+a^2 +b^2+2ab}$$ we have $$g’(b)= \frac{1+a^2-b^2}{(1+a^2 +b^2+2ab)^2}=0\iff b=\pm\sqrt{1+a^2}$$ So $$h=min\{ a,\frac{\sqrt{1+a^2}}{2+2a^2+2a\sqrt{1+a^2}}\}$$ $$a= \frac{\sqrt{1+a^2}}{2+2a^2+2a\sqrt{1+a^2}}$$ gives $$ 2a(1+a^2)=(1-2a^2)\sqrt{1+a^2}$$ $$a=\frac{1}{2\sqrt{2}}$$ Now it seems no conclusions. If it is unique( that I don’t know how ) in $[0,\frac{\pi}{4}]$ then it will be bounded also in this interval as continuous functions are bounded on compact set and if solution can’t be extended to $[0,1]$ then it is because of unboundedness of solution as I guess . I don’t know how to handle last two portions. Please help. Thanks.

My guess is the solution would be roughly to compare the solutions of $y' = y^2$, $y' = x^2 + y^2$, $y' = 1 + y^2$ with $y(0) = 1$ on the interval $x \in [0, 1]$. — Daniel Schepler, Nov 18 '20 at 08:40
@Daniel Schepler is there any theorem or result that relates them? — neelkanth, Nov 18 '20 at 09:14
@ProfessorVector I want location like near $1$ not exactly point . — neelkanth, Nov 18 '20 at 17:24

xpaul · Accepted Answer · 2020-11-18T19:28:39.563

1

Clearly the equation has a unique solution since $f(x,y)$ is locally Lipschitz. Since $y'\ge0$ for all $x\ge0$, we have $y\ge1$. Note $$ y'=x^2+y^2\ge y^2 $$ and hence $$ \frac{y'}{y^2}\ge1. $$ Integrating both sides from $0$ to $x$ gives $$ -\frac1{y(x)}+1>x $$ or $$ y(x)>\frac{1}{1-x}, x\in[0,1). \tag1$$ Similarly for $x\in[0,\frac{\pi}{4}]$, we have $$ y'=x^2+y^2\le (\frac\pi4)^2+y^2 $$ or $$ \frac{y'}{(\frac\pi4)^2+y^2}\le 1. $$ Integrating both sides from $0$ to $x$ gives $$ \frac{4}{\pi}\arctan(\frac{4}{\pi}y(x))-\frac{4}{\pi}\arctan(\frac{4}{\pi}y(0))\le x $$ or $$ y(x)\le\frac{\pi}4\tan(x+\frac{4}{\pi}\arctan(\frac{4}{\pi})). \tag2$$ Note that (2) holds only for $x\in[0,\frac{\pi}{4}]$. We can Combining (1) and (2) gives, for $x\in[0,\frac{\pi}4]$, $$ \frac1{1-x}< y(x)\le\frac{\pi}4\tan(x+\frac{4}{\pi}\arctan(\frac{4}{\pi})). $$ Thus (1), (2), and (3), and (4) are true.

edited Nov 18 '20 at 19:28

answered Nov 18 '20 at 15:30

xpaul

47,821

On interval $[0,1]$ I think $x^2+y^2\leq 1+y^2$? – neelkanth Nov 18 '20 at 15:47
I think I made a mistake. Thanks, I will modify. – xpaul Nov 18 '20 at 15:49
Thank you . how can we say that solution is bounded in $[0, \frac{\pi}{4}]?$? – neelkanth Nov 18 '20 at 16:03
I have doubt only in Second option . – neelkanth Nov 18 '20 at 16:04
Let me think. Thanks. – xpaul Nov 18 '20 at 16:05
You are right. I think (2) is false. – xpaul Nov 18 '20 at 16:07
it seems first two options are wrong but in answer key all of four options are given correct. – neelkanth Nov 18 '20 at 16:08
Yes, I didn't look at carefully. – xpaul Nov 18 '20 at 16:10
Let me think more. – xpaul Nov 18 '20 at 16:14
ok sir thank you . – neelkanth Nov 18 '20 at 16:15
Please see the update. I hope this time it is ok. – xpaul Nov 18 '20 at 16:33
thank you very much .... it is ok now . – neelkanth Nov 18 '20 at 16:38
1

Actually, equation (1) is only valid on the domain of $y$ prior to the singularity, which occurs at some point $x_0 \in (\frac{\pi}{4}, 1)$. So, it doesn't make sense in the following equation to discuss $\lim_{x\to 1^-} y(x)$. – Daniel Schepler Nov 18 '20 at 16:53
@DanielSchepler sir if possible can you give detailed of this . Thank you. – neelkanth Nov 18 '20 at 17:13
If you get $f'(y)y'\le 1$, then you have to conclude $f(y(x))-f(y(0))\le x$. Do not omit the integration constant $f(y(0))=\frac4\pi\arctan(\frac4\pi)$. – Lutz Lehmann Nov 18 '20 at 19:22
@LutzLehmann, yes, you are right. I don't know why I missed this. Thanks. – xpaul Nov 18 '20 at 19:25

score 1 · Answer 2 · answered Nov 18 '20 at 21:37

As you found out, this is a duplicate of checking uniqueness and boundedness of an initial value problem.

As to your solution attempt, the minimum in the determination of $h$ from the "butterfly" condition is just the minimum of two numbers, not of a function. Thus $$ M=a^2+(1+b)^2 $$ gives besides $h\le a$ the condition $h\le \frac{b}{a^2+(1+b)^2}$. You can see that you made some transcription errors up to this point.

Next you want to ensure contractivity of the Picard iteration, to that end determine the Lipschitz constant $L=\max2|y|=2(1+b)$. Contractivity is ensured as long as $Lh<1$.

If you try out values for $a,b$ like $a=b=1$, you will find rather small admissible values for $h$, thus this approach is not very suited for this task.

Interval of unique solution of differential equation $y’=x^2+y^2, y(0)=1 .$

2 Answers2