By John C. Butcher

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**Example text**

Because the two components of z are small we can approximate f (y + z) by f (y) + (∂f /∂y)z. Hence, the perturbation itself satisﬁes the equation dz2 z2 dx = J(x) dz3 z3 dx and the negative eigenvalues of J(x) guarantee the decay of the components of z. The solution to this problem, together with the value of λ, is shown in Figure 104(i). 105 The Van der Pol equation and limit cycles The simple pendulum, which we considered in Subsection 103, is a non-linear variant of the ‘harmonic oscillator’ problem y = −y.

In particular, if Y (x) is a square matrix, 34 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS y2 y1 Figure 122(i) Solution to problem (122c) with y3 pointing out of the page initially orthogonal, and L(x, Y ) is always skew-symmetric, then Y (x) will remain orthogonal. Denote the elements of Y by yij . An example problem of this type is 0 Y (x) = 1 −µy21 −1 0 µy11 µy21 −µy11 0 Y, Y (0) = I, (122e) with µ a real parameter. The solution to (122e) is cos(x) − sin(x) cos(µx) Y (x) = sin(x) cos(x) cos(µx) 0 sin(µx) 123 sin(x) sin(µx) − cos(x) sin(µx) cos(µx) .

V. 2-C*exp(u+v) evaluates a vector with element number i equal to ui vi2 − C exp(ui + vi ), and that linspace(0,2*pi,n+1) generates a vector with n + 1 components, equally spaced in [0, 2π]. 107 The Euler equations of rigid body rotation For a rigid body on which no moments are acting, the three components of angular velocity, in terms of the principal directions of inertia ﬁxed in the DIFFERENTIAL AND DIFFERENCE EQUATIONS 21 body, satisfy the Euler equations: dw1 = (I2 − I3 )w2 w3 , dt dw2 = (I3 − I1 )w3 w1 , I2 dt dw3 = (I1 − I2 )w1 w2 , I3 dt I1 (107a) where the ‘principal moments of inertia’ I1 , I2 and I3 are positive.