Limit for&nbsp;→∞&nbsp;of a solution to an ODEConsider the Cauchy problem:{y′(t)=t22+sin(t2)cos(y(t)2)y(0)=0&nbsp;

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Limit for&amp;nbsp;→∞&amp;nbsp;of a solution to an ODEConsider the Cauchy problem:{y′(t)=t22+sin(t2)cos(y(t)2)y(0)=0&amp;nbsp;

morpheus1ls1 · Accepted Answer

Step 1First observe that the RHSf:R2⇒R,∼f(t,y)↦t22+sin⁡(t2)cos⁡(y(t)2)is continuously differentiable in y. Therefore, for any pair&amp;nbsp;(s,&amp;nbsp;x)∈R2&amp;nbsp;there is exactly one solution&amp;nbsp;y~&amp;nbsp;such that&amp;nbsp;y~(s)=x.Note that because of this uniqueness property, graphs of solutions do never cross.It is obvious that&amp;nbsp;y(t)=±π2&amp;nbsp;is a solution. So a (unique!) solution with&amp;nbsp;y(0)=0&amp;nbsp;is bounded in between&amp;nbsp;(−π2,π2)&amp;nbsp;(Just graph it and you will see.) Also note that since&amp;nbsp;f&amp;gt;0&amp;nbsp;on(0,R)×(−π2,π2)we have that&amp;nbsp;y′&amp;gt;0&amp;nbsp;which means that y increases strictly. Now, since y is bounded and strictly increasing, it has a limit at&amp;nbsp;t⇒∞.Assume that&amp;nbsp;ell&amp;nbsp;=displaysty≤limt⇒∞y(t)∈(0,π2). Then , on&amp;nbsp;t∈[1,∞)y′(t)=f(t,y(t))&amp;nbsp;has some interesting properties: Since&amp;nbsp;y(0)=0&amp;nbsp;and due to y being strictly increasing, we must have that&amp;nbsp;y(t)∈[y(1),ell]&amp;nbsp;for all&amp;nbsp;t≥1&amp;nbsp;where&amp;nbsp;y(1)∈(0,ℓ]. In fact, one can see:L&amp;nbsp;=∈ft≥1f(t,y(t))&amp;gt;0It follows:limt⇒∞y(t)=limt⇒∞∫0tf(r,y(r))dr≥limt⇒∞∫1tf(r,y(r))dr≥limt⇒∞∫1tLdr=limt⇒∞L(t−1)=∞So somehow&amp;nbsp;limt⇒∞y(t)=∞, which clearly is a contradiction to the assumption. So we must have that&amp;nbsp;ℓ=π2This argument is far from being trivial, but it is one of the most important ones in ODE theory.

Limit for t\rightarrwo\infty of a solution to an ODE Consider the

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