Chapter 8 Equilibria in Nonlinear Sstems Recall linearization for Nonlinear dnamical sstems in R n : X 0 = F (X) : if X 0 is an equilibrium, i.e., F (X 0 ) = 0; then its linearization is U 0 = AU; A = DF (X 0 ) Let X (t) is the nonlinear solution satisfing X (0) = X 0 + U 0 if U (t) is the solution to its linearization satisfing U (0) = U 0 then X 0 + U (t) is an approximation to X (t) of order more than one: jx (t) (X 0 + U (t))j ju 0 j! 0 as ju 0 j! 0 But this doesn t mean nonlinear sstem and its linearization are structurall the same near X 0, i.e., the ma be or ma not be conjugate Example 1. x 0 = x + 0 = : The ow is x 0 + 1 (t; x 0 ; 0 ) = 3 0 e t 1 3 0e t 0 e t If initiall, x 0 + 1 3 0 = 0 then the solution approaches to the equilibrium (0; 0) : 1
The nonlinear sstem has an equilibrium solution (x; ) = (0; 0) : The linearization near (0; 0) is x 0 = x 0 = : The ow for the linearization is x0 e t (t; x 0 ; 0 ) = 0 e t If initiall x 0 = 0; all solutions approach to the equilibrium (0; 0) : In this case, both nonlinear sstem and its linearization has a "stable" manifold: the former is a parabola and the latter is axis: And, in this case near (0; 0) ; we can simpl "drop o " the nonlinear term in both sstem and in solution (drop 0): But this is not alwas the case. Example. Consider Solutions are x 0 = x 0 = : x = x 0 1 x 0 t ; = 0e t For x 0 < 0; solution (x; )! (0; 0) from the left, as t! 1 For x 0 > 0; but x 0 ver small, the solution exists up to 0 t < 1=x 0 ; and x! 1 from the left as t! 1=x 0 : However, its linearization x 0 = 0 0 = : has solution x = x 0 ; = 0 e t So x components for these two sstems are structurall different.
Comparing these two examples: the equilibrium X 0 in Example is hperbolic, while in the second example, it is not hperbolic Hperbolic equilibrium X 0 As we discussed before, two linear sstems are conjugate if the have the same numbers of eigenvalues with negative real parts and positive real parts. So in this case the linearization is conjugate to a linear sstem with distinct real eigenvalues. Assuming X 0 = 0 ( otherwise b a translation), and DF (X 0 ) has distinct real eigenvalues, k negative eigenvalues and (n k) positive. Let A be its linearization: X 0 = F (X) = AX + G (X) eigenvalue (A) = f 1 ; :::; k ; 1 ; :::; n k g ; If X 0 is a sink, i.e., k = n In general, when k < n; Write 0 < 1 < < ::: < k ; j > 0 jg (X)j jxj 1! 0 as jxj! 0 S = span fall eigenvectors with negative eigenvaluesg : Then, for an initial value X 0 S and jx 0 j is small, the solution X (t) is moving towards the origin. This is because the tangent direction of the solution X 0 (t) is opposite to X : X 0 X = (AX + G (X)) X = X T AX + G (X) X = X T AX + G (X) X 1 + jg (X)j jxj 1 jxj < 0 if jxj is small So d dt kx (t)k = d dt (X X) = X0 (t) X (t) < 0 Hence kx (t)k is decreasing with a pure negative rate. 3
So when X 0 is sink, all solutions are asmptoticall stable, i.e., X (t)! 0. Stable manifold: In general, if there are k eigenvalues with negative real part, then there is a k dimensional stable manifold (locall) with tangent space S, in the sense that an solution initiating from it (and close to the origin) will sta there and the solution converges to the origin. In D case, if x 0 = x + f 1 (x; ) 0 = + f (x; ) then there is a stable curve x = h () that passes through the origin and is tangent to the axis : h (0) = 0; h 0 (0) = 0: Linearization Theorem: Suppose that A = DF (X 0 ) is hperbolic, i.e., all eigenvalues have non-zero real parts. Then nonlinear ow is conjugate to the linear ow of its linearization sstem near X 0 locall. Example 3. Let us reconsider Example 1 x 0 = x + 0 = The solution for the second equation is Substituting into the rst equation = 0 e t : x 0 = x + 0e t ; we see solution for the nonlinear sstem: x = x 0 + 0 e t 0 3 3 e t = 0 e t 4
If initiall x 0 + 0 3 = 0; then the solution is x = x 0 e t = 0 e t which will go to the origin as t! 1: So the stable curve is x + 3 = 0: On the other hand, its linearization is The solution for the linearization is x 0 = x 0 = x = x 0 e t = 0 e t We now demonstrate these two sstems are globall conjugate. Introduce new variables (u; v) = h (x; ) = (h 1 (x; ) ; h (x; )) u = x + 3 ; v = : Under the new variables, the nonlinear sstem becomes: u 0 = x 0 + 3 0 = x + + 3 3 () = u v 0 = 0 = = v So if (x; ) is a nonlinear solution, then (u; v) solves the linearized sstem: u 0 = u v 0 = v 5
Now, for nonlinear solution (x; ) : x = x 0 + 0 e t 0 3 3 e t = 0 e t we see that under new variables u = h 1 (x; ) = x + 3 = x 0 + 0 e t 0 3 3 e t + 1 3 0e t = x 0 + 0 e t 3 = h 1 (x 0 ; 0 ) e t v = h (x; ) = = 0 e t = h (x 0 ; 0 ) e t So if (t; X 0 ) is the nonlinear ow, and (t; U 0 ) is the linear ow, then h ( (t; X 0 )) = (t; h (X 0 )) hence, h (x; ) is a homomorphism that maps solutions of nonlinear sstem to its linearization, i.e., the are conjugate. The stable manifold for the linearization is u = 0 The stable manifold for the nonlinear sstem is h 1 (x; ) = x + =3 = 0: In general, it is not alwas possible to nd such a global conjugac. Example 4: Nonlinear sstem (page 163) x 0 = 1 x 1 x x + 0 = x + 1 1 + x : For an functions (x (t) ; (t)) satisfing x + = 1; we see x 0 = 1 x 1 x = 0 = x + 1 1 = x 6
and thus a periodic solution x = cos t = sin t: Introducing polar coordinates (r; ) ; under the polar coordinates, the sstem becomes r 0 = 1 r 1 0 = 1: r So if initiall r 0 = p x 0 + 0 > 1; then r 0 < 0; so r (t) is decreasing to r = 1: Otherwise if initiall r < 1; r 0 > 0 and r increases to r = 1: In other words, the unit circle is an attractor: an solution tends to the unit circle. On the other hand, its linearization x 0 = 1 x 0 = x + 1 has eigenvalue 1 i; so all solutions spiral awa from the origin. In this case, it is impossible to nd a global conjugac. It can onl be de ned near the origin. Example 5. Consider x 0 = 0 = x z 0 = z + x + has eigenvalues 1; 1; 1: The change of variables u = x; v = ; w = z + 1 3 x + converts the nonlinear sstem to its linearized sstem u 0 = v 0 = w 0 = w u v 7
So the stable manifold is w = z+ 1 3 (x + ) = 0; which is paraboloid opening downward. Stabilit An equilibrium X 0 is called stable if for ever neighborhood G of X 0 ; there is a neighborhood G 1 G of X 0 such that an solution initiated from inside G 1 will remain in G for all t: For linear sstem, sinks, spiral sinks, and centers are stable. An equilibrium X 0 is called asmptoticall stable if it is stable and it converges to X 0 as t! 1: For linear sstem, sinks and spiral sinks are asmptoticall stable. The centers are stable, but not asmptoticall stable. Stabilit Theorem: An equilibrium that is a sink or spiral sink for its linearization is asmptoticall stable. What we discussed above shows that Non-hperbolic equilibrium solutions are onl "partiall stable",.i.e., there is a stable manifold. Bifurcations: For non-hperbolic equilibrium solutions, there will be structural change. This is when bifurcation happens. x 0 = f a (x) (one dimension) Saddle-Node bifurcation in 1d: a bifurcation point a = a 0 is called saddle-node tpe if a = a 0 ; it has onl one equilibrium point (often saddle) On one side of a 0 (for instance, a < a 0 ); it has two or more equilibria On the other side of a 0 (for instance, a > a 0 ); no equilibrium Example: x 0 = x + a 8
x 1.5 1.0 0.5 1.0 0.8 0.6 0.4 0. 0. 0.4 0.5 a 1.0 1.5 Example: x 0 = x + a Theorem: x 0 = f a (x) undergoes a saddle-node bifurcation if f a0 (x 0 ) = f 0 a 0 (x 0 ) = 0 f 00 a 0 (x 0 ) 6= 0; Pitchfork Bifurcation: x 0 = x 3 ax @ @a f a (x 0 ) 6= 0: x 1 1 1 3 4 5 1 a Bifurcatio diagram x 3 ax = 0 looks like a pitchfork for a 0; x = 0 is the onl equilibrium (source, since f 0 a (0) = a >) 9
for a > 0; there are three equilibria x = 0 (sink), x = p a (source) Saddle-Node bifurcation in d: Consider x 0 = x 0 = a For a = 0; it has one equilibrium (0; 0) : For a < 0; no equilibrium. For a > 0; there are two ( p a; 0) : The Jacobian is x 0 0 1 So at ( p a; 0) ; the linearization is p a 0 0 1 So for ( p a; 0) ; one saddle and on sink. For ( p a; 0) ; both are sink. Example: Sstem in polar coordinate sstem Find bifurcation point r 0 = r r 3 0 = sin + a Sol: r r 3 = 0 has roots r = 0; r = 1: Now, if a > 0; there is no equilibrium solution. If a = 0;equilibrium solutions are sin = 0: If 1 < a < 0; there are equilibrium solution sin = a: (see gure 8.9 in page 181 for local behavior) Hopf Bifurcation: as a passes through a 0 ;there is no change of the number of equilibria. But there is a structural change. For instance, periodic solutions occur at a = a 0 ;while there is no periodic solution for a 6= a 0 Example: Consider x 0 = ax x x + 0 = x + a x + 10
Its linearization at (0; 0) is x 0 = ax 0 = x + a The eigenvalues are a i: a > 0 (spiral source) a < 0 (spiral sink) a = 0 is a bifurcation (center) As we discussed in a previous example, when a = 0; the unit circle is a periodic solution that is an attractor. All solution initiated outside the unit circle will converges to the unit circle Homework: #1ac(i, iii, iv), (Onl do the following: nd the stable manifold at (0; 0; 0) ), 5, 9 11