Applications of Systems of Equations 497
For μ = .012129 t his equation is
(5) z^5 - l.951484z^4 + .9281087z^3 + l.023382z^2 - l.951340z + .9640543 = 0.
We used the computer program POLYRTS to solve this equation. There
is one real root, a = -1.00505, and two pairs of complex conjugate roots.
Thus, system (2) has a critical point at (y 1 , y 2 ,y3, y 4 ) = (-1.00505,0, 0,0).
This critical point is represented in the xy-plane (the y 1 y3-plane) shown in
Figure 10 .27 by L 1 : (a, 0).
For-μ< z <I-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 see z must satisfy
(6) z^5 + 2(2μ - l)z^4 + (6μ^2 - 6μ + l)z^3 + (4μ^3 - 6μ^2 + 4μ - l)z^2
+(μ^4 - 2μ^3 + 5μ^2 - 4μ + 2)z + (2μ^3 - 3μ^2 + 3μ - 1) = 0.
For μ = .012129, this equation is
(7) z^5 - l.951484z^4 + .9281087z^3 - .9523595z^2 + l.952216z - .9640508 = 0.
We used POLYRTS to solve equation (7). We found a single real root of
b = .837022 and two pairs of complex conjugate roots. Hence, there is a
critical point of system (2) at (b, 0, 0, 0). This critical point is represented in
the y1y3-plane shown in Figure 10.27 by L2 : (b, 0).
For z > 1 - μ , lz +μI = z + μ and lz + μ - II = z + μ - 1. Substituting
into equation (3) and simplifying, we find z must satisfy
(8) 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.
For μ = .012129 this equation is
(9) z^5 - l.951484z^4 + .9281087z^3 - .9766175z^2 + l.951628z - .9640543 = 0.
Using POLYRTS to solve equation (9), we found the single real root c =
1.15560 and two pairs of complex conjugate roots. The critical point (c, 0, 0 , 0)
of system (2) is represented in Figure 10.27 by L 3 : (c, 0).
The three critical points of system (2) corresponding to the points L 1 , L 2 ,
and L 3 of Figure 10.27 were first discovered by Leonhard Euler (1707-1783).
They are all unstable critical points. In 1772 , Lagrange discovered the stable
critical points L 4 and L 5 shown in Figure 10 .27. The three points E, 111 , and
L 4 form an equilateral triangle as do the three points E, NI, and L 5. Hence,
L4 is located at ((1 - 2μ)/2, ../3/2) and L5 is located at ((1 - 2μ)/2, -../3/2)
in Y1 y3-space.