Proof of the existence of a solution on IVP with a given interval for $x$

Question

Prove that the solution $y(x)$ exists on the given interval for $x$: $$y'=y^2 + \cos x^2$$ on $0\leq x\leq 1/2$ with $y(0)=0$. I think that I have to use Picard's Theorem. But I don't know why I am given an interval for $x$. And the left bound of $x$ is the initial value. Any help?

Answer 1

The task asks if $[0,\frac12]$ is a subset of the domain of the (or a) maximal solution of the IVP. This is a non-trivial question as the solution will diverge to infinity in finite time as soon as it reaches a value larger than $1$, as $y'>y^2-1$ gives an exploding lower bound.

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First variant, $y'\le y^2+1$ gives that, as long as the solution exists, $y(x)\le \tan(x)$. In the consequence, the solution remains bounded above for $x<\frac\pi2$. As long as $\cos(x^2)\ge 0$, that is, $x\le\sqrt{\frac\pi2}$, it is also clear that $y$ is increasing and thus positive. Both together give that, for instance, $[0,1]$ is a subset of the maximal domain.

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Second variant, if you want to go by the Picard iteration, you first fix a rectangle $R$ via $|x|\le a$, $|y|\le b$. Then the right side function is bounded by $|f(x,y)|\le M=b^2+1$ on $R$. Solutions are guaranteed by this bound to stay inside this rectangle for $|x|\le h$ if $Mh\le b$. Next for the convergence one needs that $q=Lh<1$ for the contraction factor of the Picard iteration, where $L=2b$ is an $y$-Lipschitz constant. Both conditions can be satisfied if $h$ is chosen small enough. As the bounds do not depend on $a$, one can set $a=h$. With $b=1$ these conditions are satisfied for any $h<\frac12$, it follows that the solution exists at least on $[0,1]$, with a limit at $x=\frac12$ due to boundedness.

Answer 2

Observe that the unique solution of the IVP $$ y'=1+y^2, \quad y(0)=0, $$ is $y(x)=\tan x$, and its maximal interval is $(-\pi/2,\pi/2)$.

Next, define the Picard sequence $y_n(x)=\int_0^x \big(\cos^2t+y_{n-1}^2(t)\big)\,dt$ of $$ y'=\cos^2(x)+y^2, \quad y(0)=0, $$ and prove inductively that, $$ 0= y_0(x)\le y_1(x)\le \cdots\le y_n(x)\le y_{n+1}(x)\le \tan x, \quad x\in [0,\pi). $$ Hence, $y_n$ converges to some $y(x)\le \tan x$, for $x\in [0,\pi/2)$.

Also, for $x\in [0,1/2]$, we have $½<\pi/6$, and hence $\tan(1/2)<\tan(\pi/6)=\frac{\sqrt{3}}{3}$. $$ 0\le y_{n+1}(x)-y_n(x)=\int_0^x \big(y_n^2(t)-y_{n-1}^2(t)\big)\,dt =\int_0^x \big(y_n(t)+y_{n-1}(t)\big)\big(y_n(t)-y_{n-1}(t)\big)\,dt \\ \le 2\tan x \int_0^x \big(y_n(t)-y_{n-1}(t)\big)\,dt \le \frac{2\sqrt{3}}{3}\int_0^x \big(y_n(t)-y_{n-1}(t)\big)\,dt $$ In particular, $y_1(x)-y_0(x)=\int_0^x \cos^2t\,dt \le x$, $$ y_2(x)-y_1(x)\le \frac{2\sqrt{3}}{3}\int_0^x \big(y_1(t)-y_0(t)\big)\,dt \le \frac{2\sqrt{3}}{3}\int_0^x t\,dt \le \frac{2\sqrt{3}}{3}\cdot \frac{x^2}{2} $$ $$ y_3(x)-y_2(x)\le \frac{2\sqrt{3}}{3}\int_0^x \big(y_2(t)-y_1(t)\big)\,dt \le \frac{2\sqrt{3}}{3}\int_0^x \frac{2\sqrt{3}}{3}\cdot \frac{t^2}{2}\,dt \le \left(\frac{2\sqrt{3}}{3}\right)^2\cdot \frac{x^3}{3!} $$ and inductively $$ 0 \le y_{n}(x)-y_{n-1}(x)\le \left(\frac{2\sqrt{3}}{3}\right)^{n-1}\cdot \frac{x^{n}}{n!} \le \frac{1}{n!} , \quad x\in [0,1/2], $$ Hence $\{y_n(x)\}$ converges uniformly in $[0,1/2]$ (Weierstrass M-test) and the limit $y(x)$ satisfies $$ y(x)=\int_0^x \big(\cos^2t+y^2(t)\big)\,dt, \quad x\in [0,1/2] $$ and hence $y$ satisfies the IVP.

Note. The interval $[0,1/2]$ could be replaced by $[-a,a]$, for any $0<a<\pi/2$.