9780199212033

# Substitution of these derivatives into i results in z

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Substitution of these derivatives into (i) results in ¨ z f (x) f (x) 2 ˙ z 2 + g(x) f (x) ˙ z 2 + f (x)h(x) = 0.

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50 Nonlinear ordinary differential equations: problems and solutions The ˙ z 2 can be eliminated by choosing f (x) so that f (x) f (x) = g(x) . Aside from a constant we can put f (x) = exp x g(u) d u , and a further integration leads to f (x) = exp x g(u) d u d x . In terms of z the equation becomes ¨ z + p(z) = 0, where p(z) = f (f 1 (z))h(f 1 (z)) . Obviously this equation is conservative of the form (1.23). 1.36 Sketch the phase diagrams for the following. ( i ) ˙ x = y , ˙ y = 0, ( ii ) ˙ x = y , ˙ y = 1, ( iii ) ˙ x = y , ˙ y = y . 1.36. (i) ˙ x = y , ˙ y = 0. All points on the x axis are equilibrium points. The solutions are x = t + A and y = B . The phase paths are lines parallel to the x axis (see Figure 1.48). (ii) ˙ x = y , ˙ y = 1. There are no equilibrium points. The equation for phase paths is d y d x = 1 y , whose general solution is given by y 2 = 2 x + C . The phase paths are congruent parabolas with the x axis as the common axis (see Figure 1.48). (iii) ˙ x = y , ˙ y = y . All points on the x axis are equilibrium points. The phase paths are given by d y d x = 1 y = x + C , which are parallel inclined straight lines (see Figure 1.49). All equilibrium points are unstable.
1 : Second-order differential equations in the phase plane 51 x y 1 2 3 x 1 2 y 3 2 1 2 1 Figure 1.48 Problem 1.36: The phase diagrams for (i) and (ii). x y Figure 1.49 Problem 1.36: The phase diagram (iii). 1.37 Show that the phase plane for the equation ¨ x εx ˙ x + x = 0, ε > 0 has a centre at the origin, by finding the equation of the phase paths. 1.37. The differential equation for the phase paths of ¨ x εx ˙ x + x = 0, is y d y d x εxy + x = 0. This is a separable equation having the general solution ε x d x = y d y y ε 1 + C = 1 + ε 1 y ε 1 d y + C , or 1 2 εx 2 = y + ε 1 ln | y ε 1 | + C , (i) where C is a constant. Note that there is a singular solution y = ε 1 . The system has a single equilibrium point, at the origin.

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52 Nonlinear ordinary differential equations: problems and solutions To establish a centre it is sufficient to show that all paths in some neighbourhood of the point are closed, so we may restrict consideration to the region y < ε 1 . On this range put F(y) = − y ε 1 ln | y ε 1 | = − y ε 1 ln 1 y) . ( ii ) Then from (i) and (ii) we can express the paths as the union of two families of curves: x = 2 ε ( 1 / 2 ) { C F(y) } 1 / 2 0, ( iii ) and x = − 2 ε ( 1 / 2 ) { C F(y) } 1 / 2 , ( iv ) (wherever C F(y) is non-negative). The curves in (iv) are the reﬂections in the y axis of those in (iii), and the families join up smoothly across this axis. Evidently, for y < ε 1 , F( 0 ) = − ε 1 ln 1 ) ( v ) and F (y) = y ε 1 y = < 0 if y < 0 zero if y = 0 > 0 if y > 0 . ( vi ) Therefore F(y) has a minimum at y = 0. Also F(y) is strictly increasing in both directions away from y = 0 and (from (ii)) F(y) → +∞ as y → −∞ and as y ε 1 from below.
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