496 Ordinary Differential Equations
The critical points of system (2) can be found by setting the right-hand-sides
of equations (2.1)- (2.4) equal to zero. Doing so , we find from (2.1) that Y2 = 0
and from (2.3) that y 4 = 0. Substituting y2 = 0 into the right-hand-side of
equation (2.4), we see either (i) y3 = 0 or (ii) the quantity in square brackets
in (2.4) is zero. Assuming (i) y3 = 0 and substituting into the right-hand-side
of (2.2), we find y 1 must satisfy
(3) f(z) = z - (1 - μ)(z + μ) - μ(z + μ - 1) =^0
lz + μ1^3 lz + μ - 113
for (y 1 , 0, 0, 0) to be a critical point of system (2). The function f (z) is defined
and continuous on (-00,00) for z -=J -μ and z -=J 1 - μ. At z =-μand
z = 1 - μ , the function f(z ) has a vertical asymptote. Figure 10.28 is a
graph of f(z ) on the interval [-2.5, 2.5] forμ= .012129. Notice that f(z ) has
three real zeros- one in the interval (-oo, -μ), since limt->-oo f(z ) = -oo;
one in the interval ( -μ , 1 - μ); and one in the interval ( 1 - μ , oo), since
limt_,oo f(z) = oo.
f(z)4-2 -1 0 2 z-20-40Figure 10.28 A Graph of Equation (3) forμ= .012129For z < -μ, lz+μI = -(z+μ) and lz+μ-11 = -(z+μ-1). Substituting
these expressions into equation (3), multiplying by (z + μ)^2 (z + μ - 1)^2 and
simplifying, we find z must satisfy the quintic equation
(4) z^5 + 2(2μ - l)z^4 + (6μ^2 - 6μ + l)z^3 + (4μ^3 - 6μ^2 + 2μ + l)z^2+(μ^4 - 2μ^3 + μ^2 + 4μ - 2)z + (3μ^2 - 3μ + 1) = 0.