t1sol_6243_2003

t1sol_6243_2003 - Massachusetts Institute of Technology...

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�� �± Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.243j (Fall 2003): DYNAMICS OF NONLINEAR SYSTEMS by A. Megretski Take-Home Test 1 Solutions 1 Problem T1.1 Find all values of µ R for which the function V : R 2 ∞≈ R , defined by ¯ x 1 V ¯ = max {| ¯ x 1 , ¯ x 2 | | x 2 |} is monotonically non-increasing along solutions of the ODE x ˙ 1 ( t ) = µx 1 ( t ) + sin( x 2 ( t )) , x ˙ 2 ( t ) = µx 2 ( t ) sin( x 1 ( t )) . Answer: µ 1. Proof For µ 1, x 1 = 0 we have 1 d 2 2 x 1 = µx 1 + x 1 sin( x 2 ) 2 dt < ( x 1 x 2 ) , −| x 1 | | | − | | and hence x 1 is strictly monotonically decreasing when x = 0 and x 1 x 2 . Similarly, | | | | √ | | x 2 is strictly monotonically decreasing when x = 0 and x 2 x 1 . Hence, when µ 1, | | | | √ | | V ( x ) is strictly monotonically decreasing along non-equilibrium trajectories of the system. For µ > 1, x 1 (0) = r , x 2 (0) = r , where r > 0 is sufficiently small we have x ˙ 1 (0) = µr sin( r ) > 0 , hence V ( x ( t )) x 1 ( t ) > r = V ( x (0)) when t > 0 is small enough, which proves that V is not monotonically decreasing. 1 Version of October 20, 2003
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2 Problem T1.2 Find all values of r R for which differential inclusion of the form x ˙( t ) ( x ( t )) , x (0) = ¯ x 0 , where : R 2 is defined by ∞≈ 2 R 2 x/ | x | ) } for ¯ x ) = { f ¯ x = 0 , (0) = { f ( y ) : y = [ y 1 ; y 2 ] R 2 , y 1 + y 2 r } , | | | | has a solution x : [0 , ) ∞≈ R 2 for every continuous function f : R 2 ∞≈ R 2 and for every initial condition ¯ Answer: r 2. x 0 R 2 . Proof First, let us show that existence of solutions is not guaranteed when r < 2. Let τ > 0 be such that 2 τ < 2 r . Define �� �± x 1 0 . 5 2(1 τ ) x 1 f = . x 2 0 . 5 2(1 τ ) x 2 Let us show that, for this f , the differential inclusion ˙ x ( t ) = ( x ( t )) will have no solutions x : [0 , ) ∞≈ R 2 with x (0) = 0. Indeed, since x f ( x/ | x | ) < 0 x = 0 , x R 2 , x ( t ) is strictly monotonically decreasing when x ( t ) = 0. Therefore x (0) = 0
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