Proof of Nagumo's Theorem of invarianceGiven a continuous system x′(t)=p(x(t)),t≥0 on Rn with initial data x(0)∈Rn and assuming that solutions exist and are unique in Rn, let S⊂Rn be a closed set. Then, S is positively invariant under the flow of the system x(t)∈S for all t≥0) if and only if p(x)∈Kx(S) for x∈∂S (the boundary of S) where Kx(S) is the set of all sub-tangential vectors to S at x, i.e.,Kx(S)={z∈Rn:limh−→0dd(x+hp(x);S)=0h},where d(w;S) is the distance from the vector w to the subset S.

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Proof of Nagumo&#039;s Theorem of invarianceGiven a continuous system x′(t)=p(x(t)),t≥0 on Rn with initial data x(0)∈Rn and assuming that solutions exist and are unique in Rn, let S⊂Rn be a closed set. Then, S is positively invariant under the flow of the system x(t)∈S for all t≥0) if and only if p(x)∈Kx(S) for x∈∂S (the boundary of S) where Kx(S) is the set of all sub-tangential vectors to S at x, i.e.,Kx(S)={z∈Rn:limh−→0dd(x+hp(x);S)=0h},where d(w;S) is the distance from the vector w to the subset S.

Brennan Thompson · Accepted Answer

Step 1Here I have a proof for ** the necessary condition** which is more easier. That is, let x∈S, if we assume that the system x′(t)=p(x(t)),;t≥0 with initial data x(0)=x has a unique solution such thatx(t)∈S  for all  t≥0.Step 2We prove that p(x)∈Kx(S). So, if x(0)∈∫(S), it is clear that Kx(S)=Rn (to prove that) so it is trivial. But, if x(0)∈∂S, then for h&amp;gt;0, we haveh−1d(x(0)+hp(x(0));S)≤h−1∣(x(0)+hp(x(0))−x(h)∣.The right hand side of the last inequality goes to zero as h goes to zero, since x′(0)=p(x(0)) and x(h)∈S by assumption.

Proof of Nagumo's Theorem of invariance Given a continuous

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