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Unformatted text preview: 1 CDS 140a: Homework 1 Solutions 1. Consider the planar system ( x,v ) ∈ R 2 given by ˙ x = v (0.1) ˙ v = x 3 (0.2) (a) The equilibrium points for the system can be found by setting equations ( 0.1 ) and ( 0.2 ) equal to 0. from equation ( 0.1 ) we get v = 0 and from equation ( 0.2 ) we get x 3 = 0 and so x = 0. Therefore, there is only one equilibrium points given by ( x,v ) = (0 , 0). (b) From equation ( 0.2 ), we can write ˙ v = v ∂v ∂x = x 3 . We assert that v 2 / 2 + x 4 / 4 = constant = Energy This conservation of energy equation can be proved by d dt E = 0; using equation ( 0.2 ) we get: d dt E = ( v + x 3 )˙ v = 0 (0.3) Hence energy is conserved. (c) the phase portrait for the given problem is : (d) In the energy equation, replacing ( x ( t ) ,v ( t )) by ( x ( t ) , v ( t )) , ( x ( t ) ,v ( t )) , ( x ( t ) , v ( t )) , 2 doesn’t change the energy value. Thus, the trajectory is symmetric about all axis. since, they crosses the axis, this implies that they are a closed trajectory. Also, the energy equation has the topological form of a circle with coordinates ( x 2 ,v ), which says its a closed trajectory. To be more specific, lets check the Jacobian of the given equation. J ( x,v ) = 1 3 x 2 Therefore, J (0 , 0) = 0 1 0 0 The trace of this matrix trace = p = 0, and also det = q = 0, this tell us that both the eigen values are zero. Hence, this is a nonhyperbolic fixed point. therefore it act as a center and hence the orbit surrounding to the fixed point are periodic. 2. Given ¨ x = x 3 ˙ x (0.4) This equation ( 0.4 ) is same as that in problem 1, but with a dissipation term. Since, the dissipation term does not effect the equilibrium points, they are still the same: ( x,v ) = (0 , 0). Since we have a dissipation term, the system will lose its energy, it will move towards stable point. In this case, (0 , 0) is a stable point and hence all the trajectory will move towards (0 , 0) We can check this by writing the equation in the form ˙ x = v (0.5) ˙ v = x 3 v (0.6) by solving this equation, we get the fixed point as ( x,v ) ≡ (0 , 0) The Jacobian of the system of equation is: J ( x,v ) = 1 3 x 2 1 Therefore, J (0 , 0) = 1 1 The trace of this matrix trace = p = 1, and also det = q = 0, this tell us that one of the eigen value is zero. Hence, this is a nonhyperbolic fixed point and therefore it act as a attractor and hence the trajectory will sink on the fixed point. 3 3. Given: planar system ( x,v ) ∈ R 2 : ¨ x = 2 x + x 2 x 3 This can be written in the following form....
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This note was uploaded on 02/28/2011 for the course MATH 140A taught by Professor Marsden during the Fall '09 term at Caltech.
 Fall '09
 Marsden
 Equations

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