同济大学：《常微分方程》课程教学资源（讲义）非线性微分方程及现象 Nonlinear Systems and Phenomena

Nonlinear Systems and Phenomena In previous chapters we have often used explicit solutions of differential equations to answer specific numerical questions. But even when a given ode is difficult or impossible to solve explicitly,it often is possible to extract qualitative information about general properties of its solutions. 口1①y元200

Nonlinear Systems and Phenomena In previous chapters we have often used explicit solutions of differential equations to answer specific numerical questions. But even when a given ode is difficult or impossible to solve explicitly, it often is possible to extract qualitative information about general properties of its solutions

Example Let x(t)denote the temperature of a body with initial tempera- turex(0)=x0.Att=0 this body is immersed in a medium with constant temperature A.Assuming Newton's law of cooling, 合=--A)k>0 conctant) 4日10y至，无2000

Example Let x(t) denote the temperature of a body with initial tempera￾ture x(0) = x0. At t = 0 this body is immersed in a medium with constant temperature A. Assuming Newton’s law of cooling, dx dt = −k(x−A) (k > 0 constant) x(t) = A+ (x0 −A)e −kt It follows immediately that lim t→∞ x(t) = A. So the temperature of the body approaches that of the surrounding medium

Example Let x(t)denote the temperature of a body with initial tempera- turex(0)=x0.At t=0 this body is immersed in a medium with constant temperature A.Assuming Newton's law of cooling, 告=-k-A刻>0 x(1)=A+(xO-A)e-k It follows immediately that limx(t)=A.So the temperature of the body approaches that of the surrounding medium. 4日10y至，无2000

Example Let x(t) denote the temperature of a body with initial tempera￾ture x(0) = x0. At t = 0 this body is immersed in a medium with constant temperature A. Assuming Newton’s law of cooling, dx dt = −k(x−A) (k > 0 constant) x(t) = A+ (x0 −A)e −kt It follows immediately that lim t→∞ x(t) = A. So the temperature of the body approaches that of the surrounding medium

r'>0 r'0 =A Phase iogram (a) (6) 口0y至，1无12000

t x x=A 0 Typicalsolution curves (a) x0 x>A x '<0 Phase diagram (b)

Example d西 d =Ay-By2,AB>0,0)=0 Solution A y- B+(分-B)eA 生0 vftl-0 0 -0 牛0 (e).t-0,B-0 (64<0,B<0 0a0

Example dy dt = Ay−By2 ,AB > 0, y(0) = y0 Solution y = A B+ ( A y0 −B)e−At t y 0 A B y2 (t)= A B y1(t)=0 dy dt 0 (a ) A>0, B>0 (c) t y y2 (t)= A B y1(t)=0 dy dt >0 dy dt 0 (b) A<0, B<0 0 A B (d)

Definition f网 dx (1) is an autonomous first-order differential equation-one in which the independent variable t does not appear explicitly.The solu- tions of the equation f(x)=0 are called critical points of(1). Ifx=c is a critical point of(1),the constant solution x=c of a differential equation is called an equilibrium solution. 4口0y至无2000

Definition dx dt = f(x) (1) is an autonomous first-order differential equation-one in which the independent variable t does not appear explicitly. The solu￾tions of the equation f(x) = 0 are called critical points of (1). If x = c is a critical point of (1), the constant solution x = c of a differential equation is called an equilibrium solution

Stability of Critical Points Definition A critical point x=c of an autonomous first-order equation is said to be stable provided that,if the initial value xo is suffi- ciently close to c,thenx(t)remains close to c for all t>0.More precisely,the critical point c is stable if,for each e>0,there existsδ>0 such that lxo-cl0.The critical point x=c is unstable if it is not stable. 4口0y至无2000

Stability of Critical Points Definition A critical point x = c of an autonomous first-order equation is said to be stable provided that, if the initial value x0 is suffi- ciently close to c, then x(t) remains close to c for all t > 0. More precisely, the critical point c is stable if, for each ε > 0, there exists δ > 0 such that |x0 −c| 0. The critical point x = c is unstable if it is not stable

Example Logistic differential equation dx =kx(M-x),k>O,M>0 dr It has two critical pointsx=0 and x=M.The solution Mxo x()= x和+(M-xo)e-M satisfying the initial condition x(0)=xo. 4日10y至，无2000

Example Logistic differential equation dx dt = kx(M −x), k > 0, M > 0 It has two critical points x = 0 and x = M. The solution x(t) = Mx0 x0 + (M −x0)e−kMt satisfying the initial condition x(0) = x0. Remark The importance of stable critical point

Example Logistic differential equation dx =kx(M-x),k>0,M>0 dt It has two critical pointsx=0 and x=M.The solution Mxo x()= x和+(M-xo)e-M satisfying the initial condition x(0)=xo. Remark The importance of stable critical point. 4口0y至，无2000

Example Logistic differential equation dx dt = kx(M −x), k > 0, M > 0 It has two critical points x = 0 and x = M. The solution x(t) = Mx0 x0 + (M −x0)e−kMt satisfying the initial condition x(0) = x0. Remark The importance of stable critical point

'c01 r'>0 I=0 I-M tsat业 S业 Phuse diogan (e) () 4日1①y至，无2000

t x x=M 0 Typicalsolution curves (e) x=0 x=M x '0 x '<0 Unstable Stable Phase diagram (f)