Infer boundedness from differential inequalitydxdt≤x(t)2+y(t)?

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eldgamliru9x · Accepted Answer

Step 1This is not a complete answer as it doesn&#039;t consider the case that your function y is merely integrable, but it illustrates the idea. Let&#039;s suppose that y is sufficiently nice such that the IVPg˙(t)=g(t)2+y(t), g(0)=x0has a solution g∈C1(0T). Furthermore, consider a function x∈C1(0T) that satisfiesx˙(t)&amp;lt;x(t)2+y(t),x(0)=x0for all t∈[0,T]Claim: The function h(t)=g(t)−x(t)&amp;gt;0 for all t∈(0,T]Proof: Since h˙(0)&amp;gt;0, there exists a small interval where h is strictly positive. Consider the minimal t~∈(0,T] that satisfies h(t~)=0. Since h(t)&amp;gt;0 for t∈(0,t~), we conclude that h˙(t~)≤0. On the other hand, we haveh˙(t~)=g˙(t~)−x˙(t~)&amp;gt;g(t~)2−x(t~)2=0,which is obviously a contradiction. Therefore no such point t~ exists and h(t)&amp;gt;0 for t∈(0,T]If you don&#039;t have a strict inequality, then the argument is similar but a bit more tedious to write down.

Infer boundedness from differential inequality \frac{dx}{dt} \leq x(t)^2 + y(t)?

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