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# 18 c the fixed points for the system are found by

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(c) The fixed points for the system are found by solving ˙ x = 0, ˙ y = 0. Their values are (1 , 1), ( - 1 , - 1), (1 , - 1) and ( - 1 , 1). The linearized system about these fixed points is obtained by evaluating the Jacobian at those points, and the eigenvalues and eigenvectors for the each of the fixed points in which we are interested are found. The fixed points of interest are ( - 1 , 1) and ( - 1 , - 1). The Jacobian of the system is bracketleftbigg 2 xy 1 - x 2 0 - 2 y bracketrightbigg . The Jacobian at the fixed point ( - 1 , 1) is ( - 1 , 1) : bracketleftbigg 2 0 2 - 2 bracketrightbigg . Eigenvalues are λ 1 = - 2 and λ 2 = 2, and the eigenvectors are e 1 = [0 , 1] and e 2 = [1 , 0]. The Jacobian at the fixed point ( - 1 , - 1) is ( - 1 , - 1) : bracketleftbigg - 2 0 2 2 bracketrightbigg . Eigenvalues are λ 1 = - 2 and λ 2 = 2, and the eigenvectors are e 1 = [1 , 0] and e 2 = [0 , 1]. (d) At the fixed point ( - 1 , - 1) the signs of ˙ x are such that the solutions are kept in towards the nullcline x = - 1 and since the signs of ˙ y > 0 the solutions move upwards, for all - 1 < y < 0. The nullcline of ˙ x , y = 0 make the solutions move vertically so solutions continue vertically across the axis. Therefore the manifold intersects the x -axis. To show that the system is reversible, consider ˙ x = y (1 - x 2 ) = f ( x, y ) , ˙ y = 1 - y 2 = g ( x, y ). Now f ( x, - y ) = y - x 2 y = - y + x 2 y = - y (1 - x 2 ) = - f ( x, y ) g ( x, - y ) = 1 - ( - y ) 2 = 1 - y 2 = g ( x, y ) which illustrates that f is odd in y and g is even in y . Thus there is existence of a heteroclinic trajectory connecting ( - 1 , - 1) to ( - 1 , 1). (e) Similarly, at the fixed point (1 , - 1), the signs of ˙ x and ˙ y are such that the solutions move vertically from the fixed point toward the fixed point (1 , 1). As already shown above, f is odd in y and g is even in y so there is the existence of a heteroclinic trajectory connecting (1 , - 1) to (1 , 1). The phase portrait of the system can be seen in Figure 20. –2 –1 0 1 2 y –2 –1 1 2 x Figure 20: Phase portrait of the system in Exercise 4. 19
Exercises for Numerics Module 1. Purpose: Determining whether an IVP is well-posed. Exercise: For each of the following IVPs,determine whether or not it is well-posed, and explain why. (a) y = - 5 y y (0) = 1 and 0 < t < 5 . (b) y = 2 y 1 / 2 y (0) = 0 and 0 < t < 2 . Solution: (a) Well-posed. The general solution is y ( t ) = Ce 5 t , and here C = 1. f ( y ) = - 5 y , which is Lipschitz continuous with constant L = 5. (b) Not well-posed, solution is not unique. Two solutions: y ( t ) = 0 or y ( t ) = t 2 . Furthermore, a small perturbation in the initial condition, say y (0) = 1 . 01 results in a significantly different solution y ( t ) = t 2 ± 0 . 2 t + 0 . 1. 2. Purpose: Understanding stability of solutions, stability of a numerical method, computing solutions for Forward and Backward Euler methods. Note to Instructor: If you have chosen not to present the slides on stability of an ODE solution, you may wish to skip part (a) of this exercise. From [Hea02, p.417,#9.4] . Exercise : Consider the ODE with y = - 5 y with initial condition y (0) = 1 . We will solve this ODE numerically using a step-size of h = 0 . 5 .

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