3.3.1 Lyapunov First MethodTheorem 3.6 For a nonlinear system described byX(t) = f[X(t),t](1)If all eigenvalues of the linearized differential equationafir=_of.^X = J. AXJaxTX=XeaxAX = X-XoxnX=Xhave negative real part, then the equilibrium point X。 isasymptotically stable in a small neighborhood of it
3.3.1 Lyapunov First Method
3.3.1 Lyapunov First MethodTheorem 3.6 For a nonlinear system described byX(t) = f[X(t),t)(2) If there is at least one eigenvalue of the linearizeddifferential equation4=.0f. ax =J.AXIaxtIX-X.Ox,X=Xhas positive real part, then the equilibrium point X。 isunstable in a small neighborhood of it
3.3.1 Lyapunov First Method
3.3.1 Lyapunov First MethodTheorem 3.6 For a nonlinear system described byX(t) = f[X(t),t)(3) If some eigenvalues of the linearized differentialX=fequation .ax =J.axaxtIX=Xehave zero real part and others have negative real part, thestability of the equilibrium point X。 is related to thesummation of the higher-order terms in the Taylor seriesg(X). In the case, the equilibrium point X。 may be stable.asymptotically stable or unstable
3.3.1 Lyapunov First Method
Example 3.7Consider the system described byX=X-XX2X2 =-X2 + XiX2Determine the equilibrium points and the stability of thesystem on them..Obviously, the system is a non-linear system.SolutionBy letting X = [xi x2」, the system can be described by[fi(X)]_[xi -xix2X = f(X) :fi(X)]L-x2 + xix2Letting X = O, two equilibrium points of the system can be[1][o]obtained as.X.Xel =2[1][0
Example 3.7Consider the system described byX=X-XX2X2=-X2+XiX2SolutionThe Jacobi matrix can be calculated asafiafiarOx,1-x,-Xi1af2af.-1+xiX2axOx.The linearized differential equation of the system on Xcan be obtained as[o]0aXX.:aX00