6.3. Nonlinear Systems of Equations

Transcription

1 G. NAGY ODE November,.. Nonlinear Systems of Equations Section Objective(s): Part One: Two-Dimensional Nonlinear Systems. ritical Points and Linearization. The Hartman-Grobman Theorem. Part Two: ompeting Species: Extinction. ompeting Species: oexistence. Remarks: We know how to solve systems of linear But systems of nonlinear In this section we find qualitative di erential equations. di erential equations are harder to solve. properties of the solutions to nonlinear systems. We first find the critical points We then find the behavior of solutions to nonlinear of the nonlinear system. systems near the critical points. Finally, we glue together the information from all the critical points to get a qualitative (Linearizations.) phase portrait of solutions to the nonlinear system. We focus on two versions of the competing species system: The case when one species goes extinct. The case when both species coexist.

2 G. NAGY ODE november,... Two-Dimensional Nonlinear Systems. Example : (The Nonlinear Pendulum) m(` ) = mg sin( ), that is + sin( ) =. g` Introduce x = and x =, ` m x = x x = g` sin(x ). Example : (Predator-Prey) Let x be the predator and y be the prey. Then, the equation is x = ax + bx x, x = cx x + dx. Example : (ompeting Species) Let x be the rabbit population and x be the sheep population, both competing for the same food resources. The equation is x = r x x K x, x = r x x K x.

3 G. NAGY ODE November,... ritical Points and Linearization. Definition. A critical vector x c solution of point of a system x = f (x) is the end point of a f (x c )=. Remarks: (a) Recall that x =(x,x )isapoint on the x x -plane while x = hx,x i is a vector with origin at (, ) and end point at x =(x,x ). (b) x c is solution of x (t) =f (x), since (x c ) = = f (x c ). (c) In components, the field is f = apple apple f xc, and the vector x f c = x c f (x c,x c )=, is solution of f (x c,x c )=. When there are more than one critical point we write x ci,withi =,,,. Example : Find all the critical points of the two-dimensional (decoupled) system x = x +(x ) x = x. Solution: We need to find all constant vectors x = x 7 solutions of x x +(x ) =, x =. From the second equation we get x =. From the first equation we get x (x ) = ) x =, or x = ±. Therefore, we got three critical points, x c = 7, x c = 7, x c = 7.

4 G. NAGY ODE november, Definition. The linearization critical point given by x c is the linear system of a systemx = f (x) at a u =(Df c ) u, where the Jacobian matrix at x c is, Df x c = f. Remark: : In components, the nonlinear system its linearization are x = f (x,x ), x = f (x,x ),, apple u u = f apple u u. Example : Find the linearization at every critical point of the nonlinear system x = x +(x ) x = x. Solution: We found earlier that this system has three critial points, x = 7, x = 7, x = 7 This means we need to compute three linearizations, one for each critical point. We start computing the derivative matrix at an arbitrary point x, @ = ( x + ( x + x @x ( x ( x ) so we get that Df(x) = +x 7

5 G. NAGY ODE November, We only need to evaluate this matrix Df at the critical points. We start with x, x = 7 Df = 7 u u 7 = 7 u 7 The Jacobian at x and x is the same, so we get the same linearization at these points, x = 7 Df = 7 u u 7 = 7 u u 7 u x = 7 Df = 7 u u 7 = 7 u 7 u

6 G. NAGY ODE november,... The Hartman-Grobman Theorem. Remark: The linearization of a nonlinear system allow us to classify the critical points of nonlinear systems. linearization. Definition. A critical point x c of a systemx = f (x) is: (a) an sink (b) a source (c) a saddle (d) a center i both eigenvalues of Df c have negative real part; i both eigenvalues of Df c have positive real part; i one eigenvalue of Df c is positive and the other is negative; i both eigenvalues of Df c are pure imaginary; A critical point x c is called hyperbolic the real part of all eigenvalues of Df c are nonzero. i it belongs to cases (a-c), that is, Theorem. (Hartman-Grobman) onsider a nonlinear autonomous system, x = f (x), with f continuously di erentiable, and consider its linearization at a hyperbolic critical point given by x c, u =(Df c ) u. Then, there is a neighborhood of x c can be transformed where all the solutions of the linear system into solutions of the nonlinear system by a continuous, invertible, transformation. Remark: The theorem above says that the phase portrait of the linearization at a hyperbolic critical point is enough to determine the qualitative picture of the phase portrait of the nonlinear system near that critical point.

7 G. NAGY ODE November, 7 Example : Use the Hartman-Grobman theorem to sketch the phase portrait of x = x +(x ) x = x. Solution: We already know that this system has three critical points, x = 7, x = 7, x = 7 We have already computed the linearizations at these critical points too. Df = 7, Df = Df = 7 We now need to compute the eigenvalues of the Jacobian matrices above. For the critical point x we have + =, - =, so x is an attractor. For the critical points x and x we have + =, - =, so x and x are saddle points. x x

8 G. NAGY ODE november,... ompeting Species: Extinction. Example 7: Find the linearization at every critical point of the competing species system r = r ( r s), s = s ( s r), Remark: We call this model a rabbits-sheep model, where r(t) is the rabbit population and s(t) is the sheep population at the time t. Solution: We start finding all the critical points of the rabbit-sheep system. r ( r s) =, s ( s r) =. There are four solutions to the equations above: () r = and s = ; () r = and s r = ; () r s = and s = ; () r s = and s r =. From these equations we get () (r =,s= ); () (r =,s= ); () (r =,s= ); () the intersection of the lines s =( r)/ and s =( r) which is given by r = r ) r = r ) r =, ) (r =,s= ). Summarizing, we got the four critical points x =(, ), x =(, ), x =(, ), x =(, ). we can always think the points as the end points of the vectors apple apple x =, x =, x = apple, x = apple. x x

9 G. NAGY ODE November, 9 Now we find the linearization of the rabbit-sheep system. If x = r 7, thesystemisx = s F(x), F F(x) = F 7 r ( r s) 7 = s ( s r) The derivative of F at an arbitrary point x is 7 = ( r s) r s ( s r) We now evaluate the matrix DF(x) at each of the critical points we found. 7 () At x = 7 we get (DF )= 7 + = - =. The critical point x is a source node. To sketch the phase portrait we will need the corresponding eigenvectors, v + = 7 and v - = 7. () At x = 7 we get (Df )= 7 + = - =. The critical point x is an sink node. One can check that the corresponding eigenvectors are v + = 7 and v - = 7. () At x = 7 we get (Df )= 7 + = - =. The critical point x is an source node. One can check that the corresponding eigenvectors are v + = 7 and v - = 7. () At x = 7 we get (Df )= 7 + = + p - = p.

10 G. NAGY ODE november, The critical point x isa saddle node. One can check that the corresponding eigenvectors are v + 7 p p = and v - 7 =. x basin for sheep basin boundary basin for rabbits x

11 G. NAGY ODE November,... ompeting Species: oexistence. Example 7: Find the linearization at every critical point of the competing species system r = r ( r s), s = s ( s r), Remark: This is also a rabbits-sheep model, where r(t) is the rabbit population and s(t) is the sheep population at the time t. Solution: Th equation for the critical points are r ( r s) =, s ( s r) =. heck that the critical points for this system are x =(, ), x =,, x =(, ), x =,. The fector field of this system is F = r ( r s) 7 s ( s r) The derivative of F is 7 = ( r s) r s ( s 7 r) Then, one can check that the critical points above satisfy the following: () x = 7, (DF )= 7 + =, v + = 7, - =, v+ = 7

12 G. NAGY ODE november, () x = 7, (DF )= 7 + =, v+ = 7, - =, v+ = 7 () x = 7, (DF )= 7 7 x =, (DF )= 7 + =, v+ = 7, - =, v + = 7 p + = ( +p ), v + 7 =, p - = ( p ), v + 7 = x x