Beedgighref28n

2022-04-28

Limit for $\to \mathrm{\infty}$ of a solution to an ODE

Consider the Cauchy problem:

$\{\begin{array}{l}{y}^{\prime}(t)=\frac{{t}^{2}}{2+\mathrm{sin}({t}^{2})}cos(y(t{)}^{2})\\ y(0)=0\end{array}$

morpheus1ls1

Beginner2022-04-29Added 22 answers

Step 1

First observe that the RHS

$f:{\mathbb{R}}^{2}\Rightarrow \mathbb{R},\sim f(t,y)\mapsto \frac{{t}^{2}}{2+\mathrm{sin}\left({t}^{2}\right)}\mathrm{cos}\left(y{\left(t\right)}^{2}\right)$

is continuously differentiable in y. Therefore, for any pair $(s,\text{}x)\in {\mathbb{R}}^{2}$ there is exactly one solution $\stackrel{~}{y}$ such that $\stackrel{~}{y}\left(s\right)=x$.

Note that because of this uniqueness property, graphs of solutions do never cross.

It is obvious that $y\left(t\right)=\pm \sqrt{\frac{\pi}{2}}$ is a solution. So a (unique!) solution with $y\left(0\right)=0$ is bounded in between $(-\sqrt{\frac{\pi}{2}},\sqrt{\frac{\pi}{2}})$ (Just graph it and you will see.) Also note that since $f>0$ on

$(0,\mathbb{R})\times (-\sqrt{\frac{\pi}{2}},\sqrt{\frac{\pi}{2}})$

we have that ${y}^{\prime}>0$ which means that y increases strictly. Now, since y is bounded and strictly increasing, it has a limit at $t\Rightarrow \mathrm{\infty}$.

Assume that $ell{\textstyle \phantom{\rule{0.222em}{0ex}}}=displaysty\le \underset{t\Rightarrow \mathrm{\infty}}{lim}y\left(t\right)\in (0,\sqrt{\frac{\pi}{2}})$. Then , on $t\in [1,\mathrm{\infty})$

${y}^{\prime}\left(t\right)=f(t,y\left(t\right))$ has some interesting properties: Since $y\left(0\right)=0$ and due to y being strictly increasing, we must have that $y\left(t\right)\in [y\left(1\right),ell]$ for all $t\ge 1$ where $y\left(1\right)\in (0,\ell ]$. In fact, one can see:

$L{\textstyle \phantom{\rule{0.222em}{0ex}}}=\in {f}_{t\ge 1}f(t,y\left(t\right))0$

It follows:

$\underset{t\Rightarrow \mathrm{\infty}}{lim}y\left(t\right)=$

$\underset{t\Rightarrow \infty}{\mathrm{lim}}{\int}_{0}^{t}f(r,y(r\left)\right)\mathrm{d}r$

$\ge \underset{t\Rightarrow \infty}{\mathrm{lim}}{\int}_{1}^{t}f(r,y(r\left)\right)\mathrm{d}r$

$\ge \underset{t\Rightarrow \infty}{\mathrm{lim}}{\int}_{1}^{t}L\mathrm{d}r$

$=\underset{t\Rightarrow \mathrm{\infty}}{lim}L(t-1)=\mathrm{\infty}$

So somehow $\underset{t\Rightarrow \mathrm{\infty}}{lim}y\left(t\right)=\mathrm{\infty}$, which clearly is a contradiction to the assumption. So we must have that $\ell =\sqrt{\frac{\pi}{2}}$

This argument is far from being trivial, but it is one of the most important ones in ODE theory.

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