By V.V. Filippov

Usually, equations with discontinuities in area variables persist with the ideology of the `sliding mode'. This publication comprises the 1st account of the idea which permits the attention of actual options for such equations. the adaptation among the 2 methods is illustrated via scalar equations of the sort *y¿=f(y)* and by way of equations bobbing up less than the synthesis of optimum keep an eye on. an in depth examine of topological results concerning restrict passages in traditional differential equations widens the idea for the case of equations with non-stop right-hand aspects, and makes it attainable to paintings simply with equations with complex discontinuities of their right-hand aspects and with differential inclusions.

*Audience:* This quantity should be of curiosity to graduate scholars and researchers whose paintings contains usual differential equations, useful research and normal topology

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**Extra resources for Basic topological structures of ordinary differential equations**

**Example text**

If p is a periodic point of Ot with period T and {p} is a closed set, then OT (p) = p. Proof. Let A =IT > 0 1 0T(p) = p}. By the definition of period, T = inf A. So, T E cl(A). Because {p} is closed, (T, p) E cl(A) x {p} = cl(A x {p}). 2. Periodic Points 29 Since 0 is continuous and {p} is closed, OT (p) = O(T, p) E cl(q(A x {p})) = cl({p}) = {p}. Therefore, OT (p) = p. 5. Let qt be a flow on a topological space X. If p E X and O' (p) = p for some real number T, then OnT (p) = p for every integer n.

So, each orbit lies within a level set of F. To determine the Poincare recurrent set of Ot we analyze the level sets of F. Case I. F(x, y, z) = 0. In this case the level set of F is the z-axis. Along the z-axis, x = y = 0, and the system of differential equations reduces to the constant differential equation z=1. Thus, the z-axis is an orbit of Ot, and no point on the z-axis is Poincare recurrent. Case II. F(x, y, z) = 2 for r > 0. r 2. Recurrent Points 54 Parametrize the level set F(x, y, z) = 1/r2 by x(a, 8) _ r2 (r -} - 1 cos a) cos 0, = (r + r2 - 1 cos a) sin 9, r2 - 1 sin a, z(a, 9) = y(a, 9) where a e S' and 9 E S1.

The fixed set of a flow on a Hausdorff topological space is closed. Proof. Let Ot be a flow on a Hausdorff topological space X. We shall prove that X \ Fix(gt) is open. Let x E X \ Fix(gt). Thus, there exists a real number T such that 0'(x) L x. Since X is Hausdorff, there exist disjoint open subsets U and V of X such that 0'(x) E U and x E V. Let A = 0--r(U) n V. Since 0' is continuous, A is an open set. We shall show that A C X \ Fix(ot). If y c A, then 0'(y) E U and y E V. Because U and V are disjoint, qT(y) y.