14 x e x a the phase paths in the x y plane are given

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1.4. ¨ x + e x = a . The phase paths in the (x , y) plane are given by y d y d x = a e x , (i)
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1 : Second-order differential equations in the phase plane 9 3 2 1 1 x 2 1 1 2 y Figure 1.11 Problem 1.3. 2 1 1 2 x 2 1 1 2 y Figure 1.12 Problem 1.4: a < 0. which has the general solution 1 2 y 2 = ax e x + C . (ii) Case (a) , a < 0. The system has no equilibrium points. From (i), d y/ d x is never zero, negative for y > 0 and positive for y < 0. Some phase paths are shown in Figure 1.12. Case (b) , a = 0. The system has no equilibrium points. As in (a), d y/ d x is never zero. Some phase paths are shown in Figure 1.13. Case (c) , a > 0. This equation has one equilibrium point at x = ln a . The potential V (x) (see Section 1.3) of this conservative system is V (x) = ( a + e x ) d x = − ax + e x , which has the expected stationary value at x = ln a . Since V ( ln a) = e ln a = a > 0, the stationary point is a minimum which implies a centre in the phase diagram. Some phase paths are shown in Figure 1.14 for a = 2.
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10 Nonlinear ordinary differential equations: problems and solutions 2 1 1 2 x 2 1 1 2 y Figure 1.13 Problem 1.4: a = 0. 1 1 2 x 2 1 1 2 y Figure 1.14 Problem 1.4: a > 0. 1.5 Sketch the phase diagrams for the equation ¨ x e x = a , for a < 0, a = 0, and a > 0. 1.5. ¨ x e x = a . The differential equation of the phase paths is given by y d y d x = a + e x , which has the general solution 1 2 y 2 = ax + e x + C . Case (a) , a < 0. There is a single equilibrium point, at x = ln ( a) . The potential V (x) (see Section 1.3 in NODE) of this conservative system is V (x) = ( a e x ) d x = − ax e x ,
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1 : Second-order differential equations in the phase plane 11 which has the expected stationary value at x = ln ( a) . Since V ( ln ( a)) = − e ln ( a) = a < 0, the stationary point is a maximum, indicating a saddle at x = ln ( a) . Some phase paths are shown in Figure 1.15. Case (b) , a > 0. The equation has no equilibrium points. Some typical phase paths are shown in Figure 1.16. Case (c) , a = 0. Again the equation has no equilibrium points, and the phase diagram has the main features indicated in Figure 1.16 for the case a > 0, that is, phase paths have positive slope for y > 0 and negative slope for y < 0. 1 1 2 x 2 1 1 2 y 2 Figure 1.15 Problem 1.5: a < 0. 3 1 1 2 x 2 1 1 2 y Figure 1.16 Problem 1.5: a > 0.
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12 Nonlinear ordinary differential equations: problems and solutions 1.6 The potential energy V (x) of a conservative system is continuous, and is strictly increasing for x < 1, zero for | x | ≤ 1, and strictly decreasing for x > 1. Locate the equilibrium points and sketch the phase diagram for the system. 1.6. From Section 1.3, a system with potential V (x) has the governing equation ¨ x = − d V (x) d x . Equilibrium points occur where ¨ x = 0, or where d V (x)/ d x = 0, which means that all points on the x axis such that | x | ≤ 1 are equilibrium points. Also, the phase paths are given by 1 2 y 2 = V (x) + C .
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