4.2.2 The Browers fixed point theorem One of the classical mathematical results that follow from the very existence of the index function is the famous Brower's fixed point theorem, which states that for every continuous function G: B ={x∈Rn+1:||≤1} The statement is obvious(though still very useful) when n= 1. Let us prove it for n> l, starting with assume the contrary. Then the map g: Bn bn which maps E bn to the point of Sm- which is the(unique) intersection of the open ray starting from G(a)and passing through z with Sn-. Then H: Sn-x0, 1]H Sm- defined by H(, t)=G(tr) is a homotopy between the identity map H(, 1)and the constant map H(, 0). Due to existence of the index function, such a homotopy does not exist, which proves the theorem 4.2.3 Existence of periodic solutions Let a: R"+R be locally Lipschitz and T-periodic with respect to the second a(a, t+T)=a(i,t)va,t where t>0 is a given number. Assume that solutions of the Ode i(t)=a(r(t), t) (4.3) with initial conditions r(O)E Bn remain in Bn for all times. Then(4. 3)has a T-periodic solution a=r(t)=r(t+T) for all tE R Indeed, the map I H.(T, 0, I)is a continuous function G: Bn+B". The solution5 4.2.2 The Brower’s fixed point theorem One of the classical mathematical results that follow from the very existence of the index function is the famous Brower’s fixed point theorem, which states that for every continuous function G : Bn ∞� Bn, where Bn = {x ⊂ Rn+1 : |x| ∀ 1}, equation F(x) = x has at least one solution. The statement is obvious (though still very useful) when n = 1. Let us prove it for n > ˆ 1, starting with assume the contrary. Then the map G : Bn ∞� Bn which maps x ⊂ Bn to the point of Sn−1 which is the (unique) intersection of the open ray starting from G(x) and passing through x with Sn−1. Then H : Sn−1 × [0, 1] ∞� Sn−1 defined by H(x, t) = Gˆ(tx) is a homotopy between the identity map H(·, 1) and the constant map H(·, 0). Due to existence of the index function, such a homotopy does not exist, which proves the theorem. 4.2.3 Existence of periodic solutions Let a : Rn × R ∞� Rn be locally Lipschitz and T-periodic with respect to the second argument, i.e. a(¯x, t + T) = a(¯x, t) � x, t where T > 0 is a given number. Assume that solutions of the ODE x˙ (t) = a(x(t), t) (4.3) with initial conditions x(0) ⊂ Bn remain in Bn for all times. Then (4.3) has a T-periodic solution x = x(t) = x(t + T) for all t ⊂ R. Indeed, the map x¯ ∞� x(T, 0, x¯) is a continuous function G : Bn ∞� Bn. The solution of x¯ = G(¯x) defines the initial conditions for the periodic trajectory