82Ug 8r8y 22(x- (rtaR)R3 a(z-R)R3 (e+a)2+y2) (e-R)2+y2)8 =22 3Ry B(aR+x】 a(z-BR) (aR+)2+y2)2+ (x-BR)2+y2)572 82Ug 22(y yR ayR3 8y8x (z+aR)2+y2)号 (x-)2+y2)2 =22 3Ry B(aR+E) a(E=BR】 (aR+z)2+y2)572+ (z-8)2+y2)572 82Va n2(y- R ayR 0w2 By (e+aR)2+y2) =22 8 33R3y2 3aRy2 (aR+)2+y2)3+ (aR+x)2+y2572- 2-82+y2+e-97+1 Evaluating these expressions at the symmetrical points LI and L2,we obtain the following results: L1: L2: 82Uua=-922 8r2 82U=922 8x2 82Un=0 Bxoy 82U=0 Bxay 82=0 8yox 82Ug=0 bybx 20=302 82=-322 0y2 Therefore,the eigenvalues of the linearized evolution matrix are: 入1,2=±V1+2W7 Due to the symmetry of LI and L2,it is to be expected that the eigenvalues for each are the same.Since one eigenvalue is real and positive and the other real and negative,we conclude that LI and L2 are saddle points.They are not dynamically stable,meaning that small deviations from the exact equilibrium point will grow over time.Without corrections,an object perturbed slightly from LI or L2 would drift away from the stable equilibrium Next,we analyze the stability at L3.Evaluating these expressions at the symmetrical points L1 and L2, we obtain the following results: Therefore, the eigenvalues of the linearized evolution matrix are: Due to the symmetry of L1 and L2, it is to be expected that the eigenvalues for each are the same. Since one eigenvalue is real and positive and the other real and negative, we conclude that L1 and L2 are saddle points. They are not dynamically stable, meaning that small deviations from the exact equilibrium point will grow over time. Without corrections, an object perturbed slightly from L1 or L2 would drift away from the stable equilibrium. Next, we analyze the stability at L3