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16 1. Differential Equations and Their Solutions σp(t) = ektp i.e., in this case the flow map φt is just multiplication by ekt. • Example 1–3. Harmonic Oscillator. If we start from the Harmonic Oscillator Equation, d2x dt2 = −x, and use the trick above to rewrite this second-order equation as a first-order system, we end up with the linear system in R2: dx dt = −y, dy dt = x. In this case the maximal solution curve σ(x 0 ,y0) (t) can again be given explicitly, namely σ(x 0 ,y0) (t) = (x0 cos(t)−y0 sin(t),x0 sin(t)+y0 cos(t)), so that now φt is rotation in the plane through an angle t. It is interesting to observe that this can be considered a special case of (a slightly generalized form of) the preceding example. Namely, if we identify R2 with the complex plane C in the standard way (i.e., a+ib := (a, b)) and write z = (x, y) = x + iy, z0 = (x0,y0) = x0 + iy0, then since iz = i(x + iy) = −y + ix = (−y, x), we can rewrite the above first- order system as dz dt = iz, which has the solution z(t) = eitz0. Of course, multiplication by eit is just rotation through an angle t. It is very useful to have conditions on a vector field V that will guar- antee its completeness. Exercise 1–12. Show that σp(t) − p≤ t 0 V (σp(t)) dt. Use this and the No Bounded Escape Theorem to show that dx dt = V (x) is complete provided that V is bounded (i.e., sup x∈ Rn V (x) ∞). Exercise 1–13. A vector field V may be complete even if it is not bounded, provided that it doesn’t “grow too fast”. Let B(r) = sup x r V (x) . Show that if ∞ 1 dr B(r) = ∞, then V is complete. Hint: How long does it take σp(t) to get outside a ball of radius R? Exercise 1–14. If a vector field is not complete, then given any positive , there exist points p where either α(p) − or ω(p) . 1.2. Euler’s Method Only a few rather special initial value problems can be solved in closed form using standard elementary functions. For the general case it is