MA551 Ordinary Differential Equation, Fall, 214 9/3/214 Linear systems of ordinary differential equations: Consider the first order linear equation for some constant a. The solution is given by such that c = x(). (IVP) Example of uncouple linear system: dx dt = Ax. dx/dt = ax x(t) = ce at dx 1 /dt = x 1 dx 2 /dt = 2x 2. In matrix form, it is ( ) ( ) dx dt = Ax, x = x1 1, A = x 2 2 It is easy to see the solution of the above system is given by ( ) e t x(t) = A = e 2t c with c = x(). Definition: The phase portrait is the solutions curves in the x 1, x 2 plane (phase plane) by using arrows to indicate the direction of the motion along these curves as t increases. The dynamical system defined by the linear system is the mapping: φ : R R 2 R 2 defined by the solution x(t, c). For c 1 =, c 2 =, x 1 (t) = and x 2 (t) = for all t R. The point (, ) is referred to as an equilibrium point. Note that the solution curves coincide with y = k/x 2 with k = c 2 1c 2. The mapping f : R 2 R 2 given by f(x) = Ax defines a vector fields on R 2 ; i.e., to each point x R 2, the mapping f assigns a vector f(x). If we draw each vector f(x) with its initial point at the point x R 2, we can see that the solution curves are tangent to the vectors since at time t = t, the velocity vector v = dx/dt t=t is tangent to x = x(t) at x = x(t ). Now we consider a system in R 3 : The general solution is given by dx 1 /dt = x 1, dx 2 /dt = x 2, dx 3 /dt = x 3. x 1 (t) = c 1 e t, x 2 (t) = c 2 e t, x 3 (t) = c 3 e t. The x 1, x 2 plane is referred to as the unstable subspace of the system. The x 3 is called the stable subspace of the system. 1
9/8/214 Diagonalization Theorem: If the eigenvalues λ 1,..., λ n of A are real and distinct, then any set of corresponding eigenvectors {v 1,..., v n } form a basis for R n, the matrix P = [v 1... v n ] is invertible and Let y = P 1 x. Then x = P y and dy dt P 1 AP = diag[λ 1,..., λ n ]; = P 1 dx dt = P 1 Ax = P 1 AP y = diag[λ 1,..., λ n ]y. This is an uncoupled system and its solution is given by Note that y() = P 1 x(), the solution of dx dt Example 1: Solve the linear system: y(t) = diag[e λ1t,..., e λnt ]. = Ax is then given by x(t) = P E(t)P 1 x(). ẋ 1 = x 1 3x 2 ẋ 2 = 2x 2 λ 1 = 1, λ 2 = 2. P = ( ) 1 1, P 1 = 1 ( ) 1 1, P 1 AP = 1 ( ) 1. 2 Thus y 1 = c 1 e t, y 2 = c 2 e 2t and thus x(t) = P ( ) e t e 2t P 1 c, Phase portraits! Definition: Suppose A has k negative eigenvalues λ 1,..., λ k and n k positive eigenvalues λ k+1,..., λ n. The stable subspace is defined as E s = span{v 1,..., v k }. The unstable subspace is E u = span{v k+1,..., v n }. If A has pure imaginary eigenvalues, then there is also a center subspace E c. Example 2: Let ( ) 3 1 A =. 1 3 Find the eigenvalues of A and solve ẋ = Ax. Then sketch the phase portraits. HW1: Due on Monday, Sep. 15th Page 5: 1.(d), 2.(b), 6 Page 9: 1.(c), 3.(c), 4, 6 2
9/1/124: Exponentials of Operators Let T : R n R n be a linear operator. The operator norm of T is defined as T = max T x x 1 where x denotes the Euclidean norm of x R n. Let L(R n ) be the linear space of linear operators from R n R n. The norm on L(R n ) satisfies T and T = T = ; kt = k T, k R; S + T S + T. Lemma: If T L(R n ) is represented by the matrix A with respect to the standard basis for R n, then A l n where l is the maximum length of the rows of A. Definition: A sequence of linear operators T k L(R n ) is said to converge to a linear operator T L(R n ) as k, i.e., lim k T k = T, if for all ɛ > there exists an N such that for all k > N, T T k < ɛ. Lemma: For S, T L(R n ) and x R n, (1) T x T x ; (2) T S T S ; (3) T k T k, k = 1, 2,.... Theorem: Given T L(R n ) and t >, the series T k t k k= is absolutely and uniformly convergent for all t t. Definition: Let A be an n n matrix. Then for t R, e At = k! A k t k Proposition: If P and T are linear transforms on R n and S = P T P 1, then e S = P e T P 1. Corollary: If P 1 AP = diag(λ j ) then e At = P diag(e λjt )P 1. Proposition: If S and T are linear transforms on R n which commute, then e S+T = e S e T. Corollary: If T is a are linear transforms on R n, the inverse of e T is given by (e T ) 1 = e T. k= k! 3
Corollary: If ( ) a b A = b a then ( ) e A = e a cos b sin b. sin b cos b ( ) ( ) a b Re(λ Proof. Let λ = a + ib and by induction = k ) Im(λ k ) b a Im(λ k ) Re(λ k. Hence ) ( ) ( ) e A Re(λ = ) ( = k /k!) Im(λ k /k!) Im(λ k /k!) Re(λ k = e a cos b sin b. /k!) sin b cos b k= Corollary: If A = ( ) a b a then ( ) e A = e a 1 b. 1 ( ) b Proof. Write A = ai + B where B =. Since B 2 = B 3 =... =, e A = e ai e B = e a e B. e B = I + B 9/12/214 The initial value system: Lemma: Let A be a square matrix, then Proof. d dt eat = lim e At eah I h h ẋ = Ax x() = x. (1) d dt eat = Ae At. ( = e At lim lim A + A2 h +... + Ak h k 1 ) h k 2! k! Theorem: The Fundamental Theorem for Linear Systems. Let A be an n n matrix. Then for a given x R n, the initial value problem (1) has a unique solution given by x(t) = e At x. Proof. Existence and Uniqueness. Example: ( ) 2 1 ẋ = Ax where A = x() = 1 2 ( ) ( x(t) = e At x = e 2t cos t sin t 1 sin t cos t ) ( ) 1. = e 2t ( cos t sin t ). 4
9/15/214 Linear Systems in R 2 x(t) = [ ] [ λ λ 1 B =, B = µ λ [ ] [ ] e λt e µt x, x(t) = e λt 1 t 1 [ ] λ I B =, λ < < µ. Saddle at the origin µ [ ] λ II B =, λ µ <. µ [ ] a b III B =, a <. stable focus b a [ ] b IV B =. center b where Example: Consider Solution: λ = ±2i and Hence one has that P = A = ] [ a b, B = b a ]. [ x, x(t) = e at cos bt sin bt sin bt cos bt ] x. ẋ = Ax (2) [ ] 4 1 [ ] [ ] 2, P 1 1/2 = 1 1 [ ] cos 2t sin 2t x(t) = P sin 2t cos 2t x 1 (t) = c 1 cos 2t 2c 2 sin 2t [, B = P 1 2 AP = 2 [ ] cos 2t 2 sin 2t = c 1/2 sin 2t cos 2t x 2 (t) = 1/2c 1 sin 2t + c 2 cos 2t. It is clear that x 2 1(t) + 4x 2 2(t) = c 2 1 + 4c 2 2. Definition: The linear system (2) is said to have a saddle, a node, a focus or a center at the origin if the matrix A is similar to one of the matrices B in Cases I, II, III, IV, respectively. Theorem: Let δ = deta and τ = tracea and consider the linear system (2). a If δ < then (2) has a saddle at the origin b If δ > and τ 2 4δ then (2) has a node at the origin; it is stable if τ < and unstable if τ >. c If δ >, τ 2 4δ <, and τ, then (2) has a focus at the origin; it is stable if τ < and unstable if τ >. d If δ > and τ = then (2) has a center at the origin. Definition: A stable node or focus of (2) is called a sink of the linear system and an unstable node or focus is called a source. HW2: Due on Monday, Sep. 22th Page 15: 1.(b), 2, 6(b) Page 19: 3(a), 4, 6 Page 27: 3, 5, 7 ], 5
9/17/214 Complex Eigenvalues Theorem: If the 2n 2n real matrix A has 2n distinct complex eigenvalues λ j = a j + ib j and λ and corresponding complex eigenvectors w j = u j ± iv j, then {u 1, v 1,..., u n, v n } is a basis for R 2n, the matrix is invertible and P = [v 1, u 1,..., v n, u n ] [ ] P 1 aj b AP = diag j. Remark: If one chooses P = [u 1, v 1,..., u n, v n ], then [ ] P 1 aj b AP = diag j. b j a j Corollary: Under the hypotheses of the above theorem, the solution of the initial value problem is given by [ ] x(t) = P diage ajt cos bj t sin b j t P 1 x sin b j t cos b j t. Note that the matrix represents a rotation through bt radians. Example: Let b j a j [ ] cos bt sin bt R = sin bt cos bt A = 3 3 2 1 1 λ 1 = 3, λ 2,3 = 2 ± i. The corresponding eigenvectors are 1 v 1 = w = u 2 + iv 2 = 1 + i 1 Thus P = 1 1 1, P 1 = 1 1 1 P 1 AP = 3 2 1. 1 1 1 2 The solution is given by e 3t e 3t x(t) = P e 2t cos t e 2t sin t P 1 x = e 2t (cos t + sin t) 2e 2t sin t x. e 2t sin t e 2t cos t e 2t sin t e 2t (cos t sin t) Stable subspace E s? Unstable subspace E u? Example: 1 1 A = 1 1 3 2. 1 1 6
9/19/214 Multiple Eigenvalues Example: Let A = [ ] 3 1. 1 1 λ 1 = λ 2 = 2. Eigenvectors(?) Definition: Let λ be an eigenvalue of the n n matrix A of multiplicity m n. Then for k = 1,..., m, any nonzero solution v of (A λi) k v = is called a generalized eigenvector of A. Definition: An n n matrix N is said to be nilpotent of order k if N k 1 and N k =. Theorem: Let A be a real n n matrix with real eigenvalues λ 1,..., λ n repeated according to their multiplicity. Then there exists a basis of generalized eigenvectors for R n. And if {v 1,..., v n } is any basis of generalized eigenvectors for R n, the matrix P = [v 1,..., v n ] is invertible, where A = S + N P 1 SP = diag[λ j ], the matrix N = A S is nilpotent of order k n, and S and N commute, i.e., SN = NS. Corollary: Under the hypotheses of the above theorem, the linear system ẋ = Ax, together with the initial condition x() = x, has the solution [ x(t) = P diag[e λjt ]P 1 I + Nt +... + N k 1 t k 1 ] x. (k 1)! Example: A = 1 1 2 1 1 λ 1 = 1, λ 2 = λ 3 = 2. It is easy to find v 1 = [1 1 2] T and v 2 = [ 1] T. Thus we need to find a generalized eigenvector. (A 2I) 2 v = 1 1 v =. 2 We find that v 3 = (, 1, ) T. Thus 1 1 P = 1 1 P 1 = 2 1 2 1 1 1 Thus S = P diag[1 2 2] 1 1 2 N = A S = 2 2 1 1 Note that N 2 =. e t e t x(t) = P e 2t P 1 [I + Nt]x = e t e 2t e 2t x e 2t 2e t + (2 t)e 2t te 2t e 2t 7
9/22/214 Jordan Forms Theorem: (The Jordan Canonical Form) Let A be a real matrix with real eigenvalues λ j, j = 1,..., k and complex eigenvalues λ j = a j + ib j and λ j = a j ib j, j = k + 1,... n. Then there exists a basis {v 1,... v k, v k+1, u k+1,..., v n, u n } for R 2n k, where v j, j = 1,..., k and w j, j = k + 1,..., n are generalized eigenvectors of A, u j = Re(w j and v j = Im(w j such that the matrix P = [v 1,... v k, v k+1, u k+1,..., v n, u n ] is invertible and P 1 AP = diag[b 1,..., B r where the elementary Jordan blocks B = B j are either of the form λ 1... λ 1... B =............ λ 1......... λ for λ one of the real eigenvalues of A or of the form D I 2... D I 2... B =............ D I 2......... D with [ ] a b D = b a for λ = a + ib one of the complex eigenvalues of A. Thus the solution of the linear system is given by I 2 = [ ] 1 1 x(t) = P diag[e Bjt ]P 1 x. If B j = B is an m m matrix corresponding to a real eigenvalue, then V = λi + N and 1 t t 2 /2!... t m 1 /(m 1)! 1 t... t m 2 /(m 2)! e Bt = e λt e N t = e λt..................... 1 t...... 1 If B j = B is a 2m 2m matrix corresponding to a complex eigenvalue λ = a + ib of A, then R Rt Rt 2 /2!... Rt m 1 /(m 1)! R Rt... Rt m 2 /(m 2)! e Bt = e at..................... R Rt...... R where ( ) cos bt sin bt R = sin bt cos bt 8
Corollary: Each coordinate in the solution x(t) of the initial value problem is a linear combination of functions of the form t k e at cos bt t k e at sin bt where λ = a + ib is an eigenvalue of the matrix A and k n 1. The kernel of a linear operator T : R n R n is defined as Ker(T ) = {x R n T (x) = }. Definition: Let λ be an eigenvalue of A of multiplicity n. The deficiency indices is defined as One has that δ k = dimker(a λi) k. δ 1 δ 2... δ n = n. Let ν k be the number of elementary Jordan blocks of size k k of matrix A. Then δ 1 = ν 1 + ν 2 +... + ν k δ 2 = ν 1 + 2ν 2 +... + 2ν k δ 3 = ν 1 + 2ν 2 + 3ν 3 +... + 3ν n... Example 1: [ λ ] λ [ λ ] 1 λ Example: HW3: Due on Monday, Sep. 29th Page 32: 1, 3 Page 37: 1(a), 2(d) Page 47: 1(a, f), 2(b,d), 8, 1 δ n = = ν 1 + 2ν 2 + 3ν 3 +... + nν n δ 1 = δ 2 = 2. δ 1 = 1, δ 2 = 2. λ λ δ 1 = δ 2 = δ 3 = 3. λ λ 1 λ δ 1 = 2, δ 2 = δ 3 = 3. λ λ 1 λ 1 δ 1 = 1, δ 2 = 2, δ 3 = 3. λ 9
9/26/214 Stability Theory Definition: Let λ j = a j + ib j, w j = u j + v j. Then Example 1: E s = Span{u j, v j a j < } E c = Span{u j, v j a j = } E u = Span{u j, v j a j > } 2 1 A = 1 2 3 λ 1,2 = 2 ± i, w = (, 1, ) T + i(1,, ) T and λ 3 = 3, u = (,, 1) T. The solution of the linear system is x(t) = e At x. The mappings e At : R n R n can be viewed as the description of the motion of x along trajectories of (1), which we call the flow of the linear system (1). Definition: If all eigenvalues of the matrix A have nonzero real part, the the flow is called a hyperbolic flow and (1) is called a hyperbolic system. Definition: A subspace E R n is said to be invariant with respect to the flow e At : R n R n if e At E E for all t R. Lemma: Let E be the generalized eigenspace of A corresponding to an eigenvalue λ, Then AE E. Theorem: Let A be a real n n matrix. Then R n = E s E u E c. Furthermore, E s, E u, E c are invariant with respect to the flow e At. Definition: If all of the eigenvalues of A have negative (positive) real parts, the origin is called a sink (source). Example: 2 1 A = 1 2 3 λ 1,2 = 2 ± i and λ 3 = 3. Theorem: The following statements are equivalent: 1. For all x R n, lim t e At x = and for x, lim t e At x =. 2. All eigenvalues of A have negative real part. 3. There are positive constants a, c, m and M such that for all x R n, e At x Me ct x for all t and e At x me at x for all t. Corollary: If x E s, then e At x E s for all t R and And if x E u, then e At x E u for all tr and lim t eat x =. lim t eat x =. 1
9/29/214 Non-homogeneous Linear Systems ẋ = Ax + b(t) (3) Definition: A fundamental matrix solution of ẋ = Ax is any non-singular n n matrix function Φ(t) such that Φ (t) = AΦ(t) for all t R. Remark: Φ(t) = e At is a fundamental matrix which satisfies Φ() = I. Any fundamental matrix Φ(t) is given by Φ(t) = e At C for some non-singular matrix C. Theorem.1. If Φ(t) is any fundamental matrix solution of ẋ = Ax, then the solution of the nonhomogeneous linear system (3) and the initial condition x() = x is given by Proof. Since Φ (t) = AΦ(t), we have that x(t) = Φ(t)Φ 1 ()x + t x (t) = Φ (t)φ 1 ()x + Φ(t)Φ 1 (t)b(t) + x (t) = A The initial condition is obvious. [ Φ(t)Φ 1 ()x + = Ax(t) + b(t) Remark: With Φ(t) = e At, one has that t Φ(t)Φ 1 (τ)b(τ)dτ. t Φ (t)φ 1 (τ)b(τ)dτ. ] Φ(t)Φ 1 (τ)b(τ)dτ + b(t) t x(t) = e At x + e At e Aτ b(τ)dτ. Example: solve the force harmonic oscillator problem Introducing ẍ + x = f(t) ẋ 1 = x 2 ẋ 2 = x 1 + f(t) Thus one has a linear system with [ ] [ ] 1 A = b = 1 f(t) [ ] [ ] e At cos t sin t = e At cos t sin t = sin t cos t sin t cos t Thus we obtain HW4: Due on Monday, Oct. 6th Page 58: 1(a, d, g, i), 2(b), 3 Page 62: 2, 3 t x(t) = e At x + e At e Aτ b(τ)dτ. x(t) = x() cos t ẋ() sin t + t f(τ) sin(τ t)dτ. 11
1/1/214 2.1 Preliminary Concepts and Definitions We consider the so called autonomous system Example 1: The initial value problem ẋ = f(x). ẋ = 3x 2/3, x() = has two different solutions u(t) = t 3 and v(t) =. f(x) = 3x 2/3 is continuous at but not differentiable there. Example 2: Consider the initial value problem ẋ = x 2, x() = 1. The solution is given by x(t) = 1 1 t. This solution is only defined on (, 1) and lim x(t) =. t 1 The interval (, 1) is called the maximal interval of existence of the solution. Note that the function x(t) = 1 1 t has another branch defined on (1, ). However, it is not part of the solution since the initial value is not satisfied. Definition: The function f : R n R n is differentiable at x R n if there is a linear transformation Df(x ) L(R n ) that satisfies f(x + h) f(x ) Df(x )h lim = h h The linear transformation Df(x ) is called the derivative of f at x. Theorem: If f : R n R n is differentiable at x, then the partial derivatives fi x j, i, j = 1,..., n, all exists at x and for all x R n, n f Df(x )x = (x )x j x j which is the Jacobian matrix. Example: Find the derivative of the function f(x) = j=1 [ ] x1 x 2 2 x 2 + x 1 x 2 and evaluate it at x = (1, 1) T. Definition: Suppose that V 1 and V 2 are two normed linear spaces with respective norms 1 and 2. Then F : V 1 V 2 is continuous at x V 1 if for all ɛ > there exists a δ > such that x V 1 and x x 1 < δ implies that F(x) F(x ) < ɛ. F is said to be continuous on the set E V 1 if it is continuous at each point x E, i.e., F C(E). Definition: Suppose that f : E R n is differentiable on E, writing f C 1 (E) if the derivative Df : E L(R n ) is continuous on E. Theorem: Suppose that E is an open subset of R n and that f : E R n. Then f C 1 (E) iff the partial derivatives fi x j, i, j = 1,..., n exist and are continuous on E. 12
1/3/214 Existence-Uniqueness Definition: Suppose that f C(E) where E is an open subset of R n. Then x(t) is a solution of the differential equation ẋ = f(x) (4) on an interval I if x(t) is differentiable on I and if for all t I, x(t) E and x (t) = f(x(t)) And given x E, x(t) is a solution of the initial value problem ẋ = f(x), x(t ) = x (5) on an interval I if t I x(t ) = x. Definition: Let E be an open subset of R n. A function f : E R n is said to satisfy a Lipschitz condition on E if there is a positive constant K such that for all x, y E f(x f(y) K x y. The function is said to be locally Lipschitz on E if for each point x E there is an ɛ-neighborhood of x, N ɛ (x ) E and a constant K such that for all x, y N ɛ (x ) f(x f(y) K x y. Lemma: Let E be an open subset of R n and let f : E R n. Then if f C 1 (E), f is locally Lipschitz on E. Proof. Since E is open, given x E, there exists an ɛ > such that N ɛ (x ) E. Let K = max Df(x). x x ɛ/2 For x, y N ɛ/2, let u = x y. Then x + su N ɛ/2 for s 1. Define F : [, 1] R n by and therefore Thus we obtain f(y) f(x) F(s) = f(x + su) F (s) = Df(x + su)u f(y) f(x) = F(1) F() = 1 Df(x + su)u ds 1 1 F (s)ds = 1 Df(x + su)uds. Df(x + su) u ds K u = K y x. Picard s method of successive approximation for (5) u (t) = x, u k+1 (t) = x + t f(u k (s))ds, k =, 1, 2,... Example: Solve the initial value problem ẋ = ax, x() = x. 1 ( ) u (t) = x, u 1 (t) = x + ax ds = x (1 + at),..., u k (t) = x 1 + at +... + a k tk k! u C(I), u = sup u(t).cauchysequence I Theorem: For I = [ a, a], C(I) is a complete normed linear space. 13
1/6/214 Theorem.2. Let E be an open set of R n containing x and assume that f C 1 (E). Then there exists an a > such that the initial value problem has a unique solution x(t) on the interval [ a, a]. ẋ = f(x), x() = x Proof. Since f C 1 (E), there is an N ɛ (x ) E and a constant K > such that for all x, y N ɛ (x ), f(y) f(x) K y x. Let b = ɛ/2 and N = {x R n x x b}. Since f is continuous and N is closed, we can set Let the successive approximation u k (t) be defined as u (t) = x, u k+1 (t) = x + M = max x N f(x). t f(u k (s))ds, k =, 1, 2,... Assuming that there exists an a > such that u k (t) is defined and continuous on [ a, a] and satisfies max u k(t) x b, (6) [ a,a] it follows that f(u k (t)) is defined and continuous on [ a, a] and therefore that u k+1 (t) is defined and continuous on [ a, a] and satisfies u k+1 (t) x t f(u k (s)) ds Ma for all t [ a, a]. Thus, choosing < a b/m, it follows by induction that u k (t) is defined and continuous and satisfies (6) for all t [ a, a] and k = 1, 2, 3,... Next since for all t [ a, a] and k =, 1,..., u k (t) N, it follows from the Lipschitz condition of f that for all t [ a, a], And then assuming that u 2 (t) u 1 (t) t t for some integer j 2, it follows that for all t [ a, a] K f(u 1 (s)) f(u (s)) ds u 1 (t) u (t) ds Ka max u 1(t) u (t) [ a,a] Kab. max u j(t) u j 1 (t) (Ka) j 1 b (7) [ a,a] u j+1 (t) u j (t) t t K f(u j (s)) f(u j 1 (s)) ds u j (t) u j 1 (t) ds Ka max [ a,a] u j(t) u j 1 (t) (Ka) j b. 14
By induction, (7) holds for j = 2, 3,.... Setting α = Ka and choosing < a < 1/K, for m > k N and t [ a, a] u m (t) u k (t) m 1 j=k u j+1 (t) u j (t) u j+1 (t) u j (t) j=n α j b = αn 1 α b. j=n Therefore, for all ɛ > there exists an N such that m, k > N implies that u m u k = max [ a,a] u m(t) u k (t) < ɛ. Cauchy sequence! Thus u k (t) converges to a continuous function u(t) uniformly on [ a, a] as k. Taking the limit on both sides of successive approximation, we have u(t) = x + t f(u(s))ds t [ a, a]. Since u(t) is continuous, f(u(t)) is continuous and by the fundamental theorem of calculus, u (t) = f(u(t)) on [ a, a] and u() = x. To show uniqueness, let u(t) and v(t) be two solutions. We have that Since Ka < 1, we have that u v = HW5: Due on Monday, Oct. 13th Page 69: 1, 3, 6 Page 77: 1, 2, 3,11 u v = max u(t) v(t) = [ a,a] t1 t1 t1 K f(u(s)) f(v(s))ds f(u(s)) f(v(s)) ds u(s) v(s) ds Ka max u(t) v(t) [ a,a] Ka u v. 15
1/18/214 Stability Lemma.3. (Gronwall) Suppose that g(t) is a continuous real valued function that satisfies g(t) > and g(t) C + K t g(s)ds for all t [, a] where C, K are positive constants. It then follows that for all t [, a], g(t) Ce Kt. Proof. Let G(t) = C + K t g(s)ds for t [, a]. Then G(t) g(t) and G(t) > for all t [, a]. The fundamental theorem of calculus implies G (t) = Kg(t) and thus for all t [, a]. One has Integrating from to t, we obtain i.e., for all t [, a]. Thus g(t) G(t) Ce Kt. G (t) G(t) = Kg(t) G(t) KG(t) = K G(t) d (ln G(t)) K. dt ln G(t) Kt + ln G() G(t) G()e Kt Theorem.4. (Dependence on Initial Conditions) Let E be an open subset of R n containing x and assume that f C 1 (E). Then there exists an a > and a δ > such that for all y N δ (x ) the initial value problem ẋ = f(x), x() = y has a unique solution u(t, y) with u C 1 (G) where G = [ a, a] N δ (x ) R n+1 ; furthermore, for each y N δ (x ), u(t, y) is a twice continuously differentiable function of t for t [ a, a]. Corollary.5. Under the hypothesis of the above theorem, Φ(t, y) = u (t, y) y for t [ a, a] and y N δ (x ) if and only if Φ(t, y) is the fundamental matrix solution of Φ = Df[u(t, y)]φ, Φ(, y) = I for t [ a, a] and y N δ (x ). Theorem.6. (Dependence on Parameters) Let E be an open subset of R n+m containing the point (x, µ ) where x R n and µ R m and assume that C 1 (E). It then follows that there exists an a > and a δ > such that for all y N δ (x ) and µ N δ (µ ), the initial value problem ẋ = f(x, µ), x() = y has a unique solution u(t, y, µ) with u C 1 (G) with G = [ a, a] N δ (x ) N δ (µ ). 16
1/1/214 The Maximal Interval of Existence Example 1: Consider the initial value problem ẋ = x 2 x() = 1 has the solution x(t) = (1 t) 1 defined on its maximal interval of existence (α, β) = (, 1). Furthermore, lim t 1 x(t) =. Example 2: consider the initial value problem ẋ = 1 2x x() = 1 has the solution x(t) = 1 t defined on its maximal interval of existence (α, β) = (, 1). The function f(x) = 1/2x C 1 (E) where E = (, ) and E = {}. Note that Example 3: Consider the initial value problem with x(1/π) = (, 1, 1/pi) T. The solution is lim x(t) = E. t 1 1 ẋ 1 = x 2 x 2, ẋ 2 = x 1 3 x 2, ẋ 3 = 1 3 x(t) = [sin 1/t, cos 1/t, t] T. The maximal interval (α, β) = (, ). At the finite endpoint α =, lim t + x(t) does not exist. Lemma.7. Let E be an open subset of R n containing x and suppose f C 1 (E). Let u 1 (t) and u 2 (t) be solutions of the initial value problem on the intervals I 1 and I 2. Then I 1 I 2 and if I is any open interval containing and contained in I 1 I 2, it follows that u 1 (t) = u 2 (t) for all t I. Proof. Since u 1 (t) and u 2 (t) solves the initial value problem on I 1 and I 2, respectively, it follows that I 1 I 2. If I satisfies the conditions in the lemma, then the fundamental existence-uniqueness theorem implies that u 1 (t) = u 2 (t) on some open interval ( a, a). Let I I be the union of all such interval contained in I on which u 1 (t) = u 2 (t). Clearly, I I and if I is a proper subset of I, then one of the endpoints t of I is contained in I I 1 I 2. It follows from the continuity of u 1 (t) and u 2 (t) on I that lim u 1 (t) = lim u 2 (t) =: u. t t t t It then follows from the uniqueness of solutions that u 1 (t) = u 2 (t) on some interval I = (t a, t + a) I. Thus u 1 (t) = u 2 (t) on the interval I I I and I is a proper subset of I I. But this contradicts the fact that I is the largest open interval contained in I on which u 1 (t) = u 2 (t). Therefore, I = I and we have u 1 (t) = u 2 (t) for all t I. 17
1/13/214 Theorem.8. Let E be an open subset of R n and assume that f C 1 (E). Then for each point x E, there is a maximal interval J on which the initial value problem has a unique solution, x(t); i.e., if the initial value problem has a solution y(t) on a interval I, then I J and y(t) = x(t) for all t I. Furthermore, the maximal interval J is open; i.e., J = (α, β). Proof. The initial value problem (1) has a unique solution on some open interval ( a, a). Let (α, β) be the union of all open intervals I such that (1) has a solution on I. We define a function on (α, β) as follows: Given t (α, β), t belongs to some open interval I such that (1) has a solution u(t) on I, define x(t) = u(t). By the above lemma, x(t) is well-defined. Thus, x(t) is a solution of (1) since each point t (α, β) is contained in some open interval I. The fact that J is open follows from the fact that any solution of (1) on an interval (α, β] can be uniquely continued to a solution on an interval (α, β + a) with a >. Definition: The interval (α, β) is called the maximal interval of existence of the solution x(t) of the initial value problem. Theorem.9. Let E be an open subset of R n containing x, let f C 1 (E), and let (α, β) be the maximal interval of existence of the solution x(t) of the initial value problem (1). Assume that β <. Then given any compact set K E, there exists a t (α, β) such that x(t) / K. Proof. Since K is compact, f(x) < M for all x K. Let x(t) solves (1) on (α, β) and β <. By way of contradiction, assume x(t) K for all t (α, β). If α < t 1 < t 2 < β then Thus x(t 1 ) x(t 2 ) t2 t 1 x(x(s)) ds M t 2 t 1. x(t 1 ) x(t 2 ) t 1, t 2 β. Thus lim t β x(t) exists. Let x 1 = lim t β x(t). Then x 1 K E since E is compact. Now we define a function u(t) on (α, β] by u(t) = { x(t) for t (α, β) x 1 for t = β. (8) Then u(t) solves (1) and is differentiable on (α, β]. Since x 1 E, there is a unique solution x 1 (t) to on some interval (β a, β + a). We define ẋ = f(x) x(β) = x 1 { u(t) for t (α, β)] v(t) = x 1 (t) for t [β, β + a). Then v(t) solve (1) on (α, β + α). It contradicts that (α, β) is the maximal interval of existence. Theorem.1. Let E be an open subset of R n containing x, let f C 1 (E) and let [, β) be the right maximal interval of existence of the solution x(t) of the initial value problem (1). Assume that β <. Then given any compact set K E, there exists a t (, β) such that x(t) / K. Corollary.11. Under the hypothesis of the above theorem, if β < and if lim t β x(t) exists then lim t β x(t) E. 18
Proof. u(t) defined in (8) is continuous on [, β]. Let K be the image of [, β] of u(t) then K is compact. Assume that x 1 E. Then there exists a t (, β) such that x(t) / K. This is a contradiction. Thus x 1 / E. But since x(t) E for all t [, β), it follows that x 1 = lim t β x(t) E. Therefore, x 1 E. HW6: Due on Monday, Oct. 27th Page 84: 1, 2(a), 5 Page 94: 1(c), 2(b), 4 Page 1: 1, 4, 5 19
Corollary.12. Let E be an open subset of R n containing x, let f C 1 (E) and let [, β) be the right maximal interval of existence of the solution x(t) of (1). Assume that there exists a compact set K E such that {y R n y = x(t) for some t [, β)} K It then follows that β =. Theorem.13. Let E be an open subset of R n containing x, let f C 1 (E). Suppose that (1) has a solution x(t, x ) defined on a closed interval [a, b]. then there exists a δ > and a positive constant K such that for all y N + δ(x ) the initial value problem ẋ = f(x) x() = y has a unique solution x(t, y) defined on [a, b] which satisfies x(t, y) x(t, x ) y x e K t and uniformly for all t [a, b]. lim x(t, y) = x(t, x ) y x 1/15/214 The Flow Defined by a DE We define the flow, e At : R n R n of the linear system ẋ = Ax. The mapping φ t = e At satisfies (i) φ (x) = x (ii) φ s (φ t (x)) = φ s+t (x) for all s, t R (iii) φ t (φ t (x)) = φ t (φ t (x)) for all t R Definition.14. Let E be an open subset of R n containing x, let f C 1 (E). For x E let φ(t, x ) be the solution of the initial value problem ẋ = f(x) x() = x defined on this maximal interval of existence I(x ). Then for t I(x ), the set of mappings φ t defined by φ t (x ) = φ(t, x ) is called the flow of the differential equation; φ t is also referred to as the flow of the vector field f(x). Ω = {(t, x ) R E t I(x )}. Example: Consider the differential equation ẋ = 1/x with f(x) = 1/x C 1 (E) and E = {x R x > }. The solution for x() = x is given by φ(t, x ) = 2t + x 2 on its maximal interval of existence I(x ) = ( x 2 /2, ). Find Ω. Theorem.15. Let E be an open subset of R n containing x, let f C 1 (E). Then Ω is an open subset of R E and φ C 1 (Ω). Proof. If (t, x ) Ω and t >, the solution x(t) = φ(t, x ) is defined on [, t ]. Thus the solution x(t) can be extended to an interval [, t +ɛ] for some ɛ > ; i.e., φ ( t, x ) is define d on the closed interval [t ɛ, t +ɛ]. Then there exists a neighborhood of x, N δ (x ), such that φ(t, y) is defined on [t ɛ, t + ɛ] N δ (x ); i.e., [t ɛ, t + ɛ] N δ (x ) Ω. Similar proof holds for t <. Therefore, Ω is open. Since φ C 1 (G) where G = [t ɛ, t + ɛ] N δ (x ) by a theorem in previous section, and (t, x ) is arbitrary, φ C 1 (Ω). 2
1/2/214 Theorem.16. Let E be an open subset of R n containing x, let f C 1 (E). Then for all x E, if t I(x ) and s I(φ t (x )), it follows that s + t I(x ) and φ s+t (x ) = φ s (φ t (x )). Proof. Suppose that s >, t I(x ) and s I(φ t (x )). Let I(x ) = (α, β) and define the function x : (α, s + t] E by { φ(r, x ) α < r t x(r) = φ(r t, φ t (x )) t r s + t. Then x(r) is a solution of the initial value problem ẋ = f(x). on (α, s + t]. Hence s + t I(x ) and by uniqueness of solutions φ s+t (x ) = x(s + t) = φ(s, φ t (x )) = φ s (φ t (x )). The case of s = is trivial. If s <, we define the function x : [s + t, β)e by { φ(r, x ) t r β x(r) = φ(r t, φ t (x )) s + t r t. Then x(r) is a solution of the initial value problem on [s + t, β) and similar argument to s > holds. Theorem.17. Let E be an open subset of R n containing x, let f C 1 (E). If (t, x ) Ω then there exists a neighborhood U of x such that {t} U Ω. It follows that the set V = φ t (U) is open in E and that φ t (φ t (x)) = x x U and φ t (φ t (y)) = y y V. Definition.18. Let E be an open subset of R n, let f C 1 (E), and let φ t : E E be the flow of the differential equation ẋ = f(x) defined for all t R. Then a set S E is called invariant with respect to the flow φ t if φ t (S) S for all t R and S is called positively (or negatively) invariant with respect to the flow φ t if φ t (S) S for all t (or t ). Example: Consider the nonlinear system with The solution for x() is given by The set is invariant under the flow φ t. If c S then c 2 c 2 1/3 and it follows that f(x) = [ x 1, x 2 + x 2 1] T. φ t (c = φ(t, c) = [c 1 e t, c 2 e t + c2 1 3 (et e 2t )] T. S = {x R 2 x 2 = x 2 1/3} Thus φ t (S) S for all t R. φ t (c) = [c 1 e t, c2 1 3 e 2t ] T S. 21
1/27/214 Linearization The linear function Ax = Df(x )x is called the linear part of f at x. ẋ = f(x) (9) Definition.19. A point x R n is called an equilibrium point or critical point of (9) if f(x ) =. An equilibrium point x is called a hyperbolic equilibrium point of (9) none of the eigenvalues of the matrix Df(x ) has zero real part. The linear system ẋ = Ax with A = Df(x ) is called the linearization of (9) at x. If x = is an equilibrium point of (9), we have f() = and f() = Df()x + 1 2 D2 f()(x, x) +... If x is an equilibrium point of (9) and φ t : E R n is the flow of (9), then φ t (x ) = x for all t R. Thus x is called a fix point of the flow φ t (zero, critical point, singular point of the vector field f). Definition.2. An equilibrium point x of (9) is called a sink if all of the eigenvalues of matrix Df(x ) have negative real part; it is called a source if all of the eigenvalues of matrix Df(x ) have positive real part; and it is called a saddle if it is a hyperbolic equilibrium point point and at least one eigenvalue with a positive real part and at least one with a negative real part. Example: Classify all the equilibrium points of (9) with [ ] x 2 f(x) = 1 x 2 2 1. 2x 2 Df(x) = f(x) = x = [ ] 2x1 2x 2 Df(1, ) = 2 ( 1 ), x = ( ) 1 [ ] 2, Df( 1, ) = 2 [ ] 2. 2 Thus, (1, ) is a source and ( 1, ) is a saddle. If x is a hyperbolic equilibrium point of (9) then the local behavior of the nonlinear system (9) is topologically equivalent to the local behavior of the linear system ẋ = Ax (1) with A = Df(x ), i.e., there is a continuous 1-1 map of a neighborhood of x onto an open set U containing the origin, H : N ɛ (x ), which transform (9) into (1), maps trajectories of (9) in N ɛ (x ) onto trajectories of (1) in the open set U and preserves the orientation of the trajectories by time, i.e., H preserves the direction of the flow along the trajectories. Example: Consider H : R 2 R 2 y = H(x) = H transforms the nonlinear system (9) into the linear system (1) with [ x1 x 2 + x2 1 3 ] [, H 1 (y) = [ ] x1 f(x) = x 2 + x 2 1 A = Df() = [ ] 1 1 in the sense that if y = H(x) then [ ] [ ] ẋ ẏ = 1 x ẋ 2 + 2 3 x = 1 1ẋ 1 x 2 + x 2 1 + 2 3 x = 1( x 1 ) y 1 y 2 y1 3 ]. [ ] y1, i.e., ẏ = y 2 [ ] 1 y. 1 22
The Stable Manifold Theorem Near a hyperbolic equilibrium point x, the nonlinear system ẋ = f(x) has stable and unstable manifolds S and U tangent at x to the stable and unstable subspace subspaces E s and E u of the linearized system ẋ = Ax with A = Df(x ). Furthermore, S and U are of the same dimensions as E s and E u, and if φ t is the flow of the nonlinear system, then S and U are positively and negatively invariant under φ t respectively and satisfy lim φ t(c) = x c S t and Example: Consider the linear system The solution of the system is given by lim φ t(c) = x c U. t ẋ 1 = x 1 ẋ 2 = x 2 + x 2 1 ẋ 3 = x 3 + x 2 1. x 1 (t) = c 1 e t x 2 (t) = c 2 e t + c 2 1(e t e 2t ), x 3 (t) = c 3 e t + c2 1 3 (et e 2t ) It is easy to see that is the only equilibrium point and A = Df() = 1 1 1 The stable and unstable subspaces E s and E u of ẋ = Ax are the x 1, x 2 plane and the x 3 -axis, respectively. lim t φ t (c) = iff c 3 + c 2 1/3 =. Thus S = {c R 3 c 3 = c 2 1/3}. Similarly, lim t φ t (c) = iff c 1 = c 2 = and therefore U = {c R 3 c 1 = c 2 = }. Definition.21. Let X be a metric space and let A and B be subsets of X. A homeomorphism of A onto B is a continuous 1-1 map of A onto B, h : A B, such that h 1 : B A is continuous. The sets A and B are called homeomorphic or topologically equivalent if there is a homeomorphism of A onto B. Definition.22. An n-dimensional differentiable manifold, M, is a connected metric space with an open covering {U α }, i.e., M = α U α, such that (1) for all α, U α is homeomorphic to the open unit ball in R n, B = {x R n x < 1}, i.e., for all α there exists a homeomorphism of U α onto B, h α B, and (2) if U α U β and h α : U α B, h β B are homeomorphism, then h α (U α U β ) and h β (U α U β ) are subsets of R n and the map h = h α h 1 β : h β(u α U β ) h α (U α U β ) is differentiable and for all x h β (U α U β ), the Jacobian determinant detdh(x). The manifold M is said to be analytic if the maps h = h α h 1 β are analytic. 23
The cylindrical surface S in the above example is a 2D differentiable manifold. The project of the x 1, x 2 plane onto S maps the unit disks centered at the points (m, n) in the x 1, x 2 plane onto homeomorphic images of the unit disk B = {x R 2 x 2 1 + x 2 2 < 1}. These sets U mn S form a countable open cover of S. The pair (U α, h α ) is called a chart for the manifold M and the set of all charts is called an atlas for M. The differentiable manifold M is called orientable if there is an atlas with detdh α h 1 β for all α, β and x h β (U α U β ). Theorem.23. (The Stable Manifold Theorem) Let E be an open subset of R n containing the origin, let f C 1 (E), and let φ t be the flow of the nonlinear system (1). Suppose that f() = and that Df() has k eigenvalues with negative real part and n k eigenvalues with positive real part. Then there exists a k-dimensional differentiable manifold S tangent to the stable subspace E s of the linear system (2) at such that for all t, φ t (S) S and for all x S lim φ t(x ) = ; t and there exists an n k dimensional differentiable manifold U tangent to the unstable subspace E u of (2) such that for all t, φ t (U) U and for all x U, lim φ t(x ) =. t 24
1/31/214 Stability and Liapunov Functions The stability of (1) is determined by the signs of the real parts of the eigenvalues λ j of Df(x ). A hyperbolic equilibrium point x is asymptotically stable iff Re(λ j ) < for 1 j n, i.e., iff x is a sink; is unstable iff it is a source or saddle. Definition.24. Let φ t denote the flow of (1) for t R. An equilibrium point x is stable if for all ɛ > there exists a δ > such that for all x N δ (x ) and t >, φ t (x) N ɛ (x ). The equilibrium point x is unstable if it is not stable. x is asymptotically stable if it is stable and if there exists a δ > such that for all x N δ (x ) lim φ t(x ) = x. t Remark.25. Any hyperbolic equilibrium point of (1) is either asymptotically stable or unstable. Theorem.26. If x is a sink of the nonlinear system (1) and Re(λ j ) < α < for all the eigenvalues λ j of the matrix Df(x ), then given ɛ > there exists a δ > such that for all x N δ (x ), the flow φ t (x) of (1) satisfies φ t (x) x ɛ αt for all t. Theorem.27. If x is a stable equilibrium point of (1), no eigenvalue of Df(x ) has positive real part. Definition.28. If f C 1 (E), V C 1 (E) and φ t is the flow of the differential equation (1), then for x E the derivative of the function V (x) along the solution φ t (x) V (x) = d dt V (φ t(x)) = DV (x)f(x) t= Theorem.29. Let E be an open subset of R n containing x. Suppose that f C 1 (E) and that f(x ) =. Suppose further that there exists a real valued function V C 1 (E) satisfying V (x ) = and V (x) > if x x. Then (a) if V (x) for all x E, x is stable; (b) if V (x) < for all x E \ {x }, x is asymptotically stable; (c) if V > for all x E \ {x }, x is unstable. A function V : R n R satisfying the hypotheses in the above theorem is called a Liapunov function. Remark.3. If V (x) = for all x E then the trajectories of (1) line on the surfaces in R n defined by Example 1: Consider the system V (x) = c. ẋ 1 = x 3 2, ẋ 2 = x 3 1. The origin is a nonhyperbolic equilibrium point. Consider the function Hence the solution curves lie on the closed curves V (x 1, x 2 ) = x 4 1 + x 4 2. V (x1, x 2 ) = 4x 3 1ẋ 1 + 4x 3 2ẋ 2 =. x 4 1 + x 4 2 = c which encircle the origin. The origin is thus stable, but not asymptotically stable. Example 2: Consider the system ẋ 1 = 2x 2 + x 2 x 3 ẋ 2 = x 1 x 1 x 3 ẋ 3 = x 1 x 2. 25
It is easy to find 2 Df() = 1 x = is nonhyperbolic equilibrium point. Try function Then V (x) = c 1 x 2 1 + c 2 x 2 2 + c 3 x 2 3. 1 2 V (x) = (c 1 c 2 + c 3 )x 1 x 2 x 3 + ( 2c 1 + c 2 )x 1 x 2. Hence if c 2 = 2c 1 and c 3 = c 1 >, we have V (x) > for x and V (x) = for all x R 3. Thus x = is stable and the trajectories of this system line on the ellipsoids x 2 1 + 2x 2 2 + x 2 3 = c 2. 26
11/3/214 Proof. WLOG, assume x =. (a) Choose ɛ > such that N ɛ () E and let m ɛ be the minimum of V (x) on S ɛ = {x R n x = ɛ}. Since V (x) > for x, it follows that m ɛ >. Since V (x) is continuous and V () =, there exists a δ > such that x < δ implies that V (x) < m ɛ. Since V (x) for x E, V (x) is decreasing along trajectories of (1). Thus if φ t is the flow of (1), for all x N δ () and t > we have that V (φ t (x )) V (x ) < m ɛ. Now suppose that for x < δ, there is a t 1 > such that φ t1 (x ) = ɛ, i.e., φ t1 (x ) S ɛ. Then V (φ t1 (x )) m ɛ which leads to a contradiction. Thus for x δ and t, φ t (x ) < ɛ. (b) Suppose that V (x) < for all x E. Then V (x) is strictly decreasing along the trajectories of (1). Let φ t be the flow of (1) and let x N δ (). If x < δ, φ t (x ) N ɛ () for all t. Let {t k } be any sequence with t k. Since N ɛ () is compact, there is a subsequence {φ tn (x )} of {φ tk (x )} that converges to a point y in N ɛ (). Since V (x) is strictly decreasing along trajectories of (1) and V (φ tn (x )) V (y ), then V (φ t (x )) > V (y ) for all t >. But if y, then for s > we have V (φ s (x )) < V (y ) and by continuity it follows that for all y sufficiently close to y we have V (φ s (y)) < V (y ) for s >. But then for y = φ tn (x ) and n sufficiently large, we have V (φ s+tn (x )) < V (y ) which contradicts the above inequality. Thus y =. Thus φ tk (x ) for an sequence t k and φ t (x ) as t. (c) Let M be the maximum of the continuous function V (x) on the compact set N ɛ (). Since V (x) >, V (x) is strictly increasing along trajectories of (1). If φ t is the flow of (1), then for any δ > and x N δ () \ {} we have V (φ t (x )) > V (x ) > for all t >. And since V (x) is positive, we have Thus, for all t >. Therefore inf V (φ t (x )) = m >. t V (φ t (x )) V (x ) mt V (φ t (x )) > mt > M for t sufficiently large, i.e., φ t (x ) lies outside the closed set N ɛ (). Hence is unstable. Example 3: Consider the system ẋ 1 = 2x 2 + x 2 x 3 x 3 1 ẋ 2 = x 1 x 1 x 3 x 2 2 ẋ 3 = x 1 x 2 x 3 3. 27
The Liapunov function satisfies V (x) > and V (x) = x 2 1 + 2x 2 2 + x 2 3 V (x) = 2(x 4 1 + 2x 4 2 + x 4 3) < for x. Thus the origin is asymptotically stable, but it is not a sink since the eigenvalues λ 1 = and λ 2,3 = ±2i. Example 4: Consider ẍ + q(x) = where q(x) is continuous and xq(x) > for x. Write it into a system ẋ 1 = x 2 ẋ 2 = q(x 1 ). The total energy of the system V (x) = x2 x1 2 2 + q(s)ds serves as a Liapunov function of this system V (x) = q(x 1 )x 2 + x 2 [ q(x 1 )] =. The solution curves are given by V (x) = c; i.e., the energy is constant on the solution curves. The origin is a stable equilibrium point. HW 7: Due Mon. Nov. 1th, 214 P134: 2, 3, 5, 7 28
11/5/214 Saddle, Nodes, Foci and Centers The nonlinear system have a saddle, a sink or a source at a hyperbolic equilibrium point x if the linear part of f at x had eigenvalues with both positive and negative real parts, only had eigenvalues with negative real parts, or only had eigenvalues with positive real parts, respectively. Consider the nonlinear system written in the following form ẋ = P (x, y) ẏ = Q(x, y). In polar coordinates, we have Thus we have rṙ = xẋ + yẏ, r 2 θ = xẏ yẋ. Or as Example 1: In polar coordinates, we have ṙ = P (r cos θ, r sin θ) cos θ + Q(r cos θ, r sin θ) sin θ, r θ = Q(r cos θ, r sin θ) cos θ P (r cos θ, r sin θ) sin θ. dr r[p (r cos θ, r sin θ) cos θ + Q(r cos θ, r sin θ) sin θ] = F (r, θ) = dθ Q(r cos θ, r sin θ) cos θ P (r cos θ, r sin θ) sin θ. ṙ = ẋ = y xy ẏ = x + x 2. xẋ + yẏ r = xy x2 y + xy + x 2 y r and xẏ yẋ θ = r 2 = x2 + x 3 + y 2 + xy 2 r 2 = 1 + x > for x > 1. Thus along any trajectory of this system in the half plan x > 1, r(t) is constant and θ(t) increases as t. Example 2: Consider the system In polar coordinates, we have Thus Thus r(t) and θ(t) as t. Example 3: Consider the system In polar coordinates, we have Thus = ẋ = y x 3 xy 2, ẏ = x y 3 x 2 y. ṙ = r 3, θ = 1. θ(t) = θ + t, r(t) = r (1 + 2rt) 2 1/2 for t > 1 2r 2. ẋ = y + x 3 + xy 2, ẏ = x + y 3 + x 2 y. ṙ = r 3, θ = 1. θ(t) = θ + t, r(t) = r (1 2rt) 2 1/2 for t < 1 2r 2. Then we have that r(t) and θ(t) as t. 29
Definition.31. The original is called a center if there exists a δ > such that every solution curve in the deleted neighborhood N δ () \ {()} is a closed curve with in its interior. Definition.32. The origin is called a center-focus if there exists a sequence of closed solution curves Γ n with Γ n+1 in the interior of Γ n such that Γ n as n and such that every trajectory between Γ n and Γ n+1 spirals toward Γ n or Γ n+1 as t ±. Definition.33. The origin is called a stable focus if there exists a δ > such that for < r < δ and θ R, r(t, r, θ ) and θ(t, r, θ) as t. It is called an unstable focus if r(t, r, θ ) and θ(t, r, θ) as t. Any trajectory which satisfies r(t) and θ(t) as t ± is said to spiral toward the origin as t ±. Definition.34. The origin is called a stable node if there exists a δ > such that for < r < δ and θ R, r(t, r, θ ) as t and lim t θ(t, r, θ ) exists; i.e., each trajectory in a deleted neighborhood of the origin approaches the origin along a well-defined tangent line as t. the origin is called an unstable node if r(t, r, θ ) as t and lim t θ(t, r, θ ) exists for all r (, δ) and θ R. The origin is called a proper node if it is a node and if every ray through the origin is tangent to some trajectory. Definition.35. The origin is a a (topological) saddle if there exist two trajectories Γ 1 and Γ 2 which approach as t and two trajectories Γ 3 and Γ 4 which approach as t and if there exists a δ > such that all other trajectories which start in the deleted neighborhood of the origin N δ () \ {} leave N δ () as t ±. The special trajectories Γ 1,..., Γ 4 are called separatrices. Theorem.36. (Bendixson) Let E be an open subset of R 2 containing the origin and let f C 1 (E). If the origin is an isolated critical points, then either every neighborhood of the origin contains a closed solution curve with in its interior or there exists a trajectory approaching as t ±. Theorem.37. Suppose that P (x, y) and Q(x, y) are analytic functions of x and y in some open subset E of R 2 containing the origin and suppose that the Taylor expansions of P and Q about (, ) begin with mth-degree terms P m (x, y) and Q m (x, y) with m 1. Then any trajectory which approaches the origin as t either spirals toward the origin as t or it tends toward the origin in a definite direction θ = θ as t. If xq m (x, y) yp m (x, y) is not identically zero, then all directions of approach, θ satisfy the equation cos θ Q m (cos θ, sin θ ) sin θ P m (cos θ, sin θ ) =. Furthermore, if one trajectory spirals toward the origin as t then all trajectories in a deleted neighborhood of the origin spiral toward as t. If P and Q begin with first-degree terms, i.e., if P 1 (x, y) = ax + by Q 1 (x, y) = cx + dy with a, b, c, d not all zero, then the only possible directions in which trajectories can approach the origin are given by directions θ which satisfy For cos θ, if b, the equation is equivalent to b sin 2 θ + (a d) sin θ cos θ c cos 2 θ =. b tan 2 θ + (a d) tan θ c =. This quadratic has at most two solutions θ ( π/2, π/2] and if θ = θ 1 is a solution, then θ = θ 1 + π are solutions. It is equivalent to find the eigenvectors of [ ] a b A =. c d 3
11/1/214 Theorem.38. Suppose that E is an open subset of R 2 containing the origin and that f C 1 (E). If the origin is a hyperbolic equilibrium point the non-linear system, then the origin is a (topological) saddle if and only if the origin is a saddle for the linear system with A = Df(). Example: Consider the system ẋ = x + x + 2y + x 2 y 2 ẏ = 3x + 4y 2xy. The origin is a (topological) saddle since the determinant of the linear part δ = 2. directions which the separatrices approach the origin as t ± are given by Furthermore, the 2 tan 2 θ 3 tan θ 3 = i.e., θ = tan 1 ( 3 ± ) 33. 4 Example: Consider the nonlinear system ẋ = x x ln x 2 + y 2 ẏ = x y + ln x 2 + y 2 for x 2 + y 2 and define f() =. In polar coordinates we have ṙ = r, θ = 1 ln r. Thus r(t) = r e t and θ(t) = θ ln(1 t/ ln r ). We set that for r < 1, r(t) and θ(t) as t and therefore the origin is a stable focus. However, it is a stable proper node for the linearized system. One has that xy Df(x, y) = 1 y 2 1 (x 2 +y 2 ) ln x 2 +y 2 (x 2 +y 2 )(ln x 2 +y 2 ) 2 ln x 2 +y 2 1 x 2 xy 1 ln x 2 +y 2 (x 2 +y 2 )(ln x 2 +y 2 ) 2 (x 2 +y 2 ) ln x 2 +y 2 Df(r, θ) = r 1 2 cos 2 θ r 2 1 2 ln r r 2 1 2 ln r r 2 sin 2 θ xy 1 r 2 1 2 ln r (x 2 +y 2 ) ln x 2 +y 2 r 2 cos θ sin θ Theorem.39. Let E be an open subset of R 2 containing the origin and let f C 2 (E). Suppose that the origin is a hyperbolic critical point of (2). Then the origin is a stable (or unstable) node for the nonlinear system (2) if and only if it is a stable (or unstable) node for the linear system (1) with A = Df(). And the origin is a stable (or unstable) focus for the nonlinear system (2) if and only if it is a stable (or unstable) focus for the linear system (1) with A = Df(). Example: Consider the nonlinear system ẋ = y + x x 2 + y 2 sin(1/ x 2 + y 2 ) ẏ = x + y x 2 + y 2 sin(1/ x 2 + y 2 ) 31
for x 2 + y 2 where we define f() =. In polar coordinates, we have ṙ = r 2 sin(1/r), θ = 1 for r > with ṙ = at r =. Clearly, ṙ = for r = 1/(nπ), i.e., each of the circles r = 1/(nπ) is a trajectory. Furthermore, for nπ < 1/r < (n + 1)π, ṙ < if n is odd and ṙ > if n is even. Thus the trajectories between the circles spiral inward or outward to one of these circles. Thus the origin is a center-focus. Theorem.4. Let E be an open subset of R 2 containing the origin and let f C 1 (E) with f() =. Suppose that eh origin is a center for the linear system (1) with A = Df(). Then the origin is either a center, a center-focus or a focus for the nonlinear system (2). Corollary.41. Let E be an open subset of R 2 containing the origin and let f is analytic in E with f() =. Suppose that the origin is a center for the linear system (1) with A = Df(). Then the origin is either a center or a focus for the nonlinear system (2). Definition.42. The system ẋ = P (x, y) ẏ = Q(x, y) is said to be symmetric with respect to the x-axis if it is invariant under the transformation (t, y) = to( t, y); it is said to be symmetric with respect to the y-axis if it is invariant under the transformation (t, x) = to( t, x). Theorem.43. Let E be an open subset of R 2 containing the origin and let f C 1 (E) with f() =. If the nonlinear system (2) is symmetric with respect to the x-axis or the y-axis, and if the origin is a center for the linear system (1) with A = Df(), then the origin is a center for the nonlinear system (2). HW 8: Due Mon. Nov. 17th, 214 P145: 1(a,c), 2, 3, 4(b,e). 32
11/12/214 2.11 Nonhyperbolic Critical Points in R 2 Consider the system ẋ = P (x, y) ẏ = Q(x, y) where P and Q are analytic in some neighborhood of the origin. Note that if P and Q begin with mth degree terms P m and Q m, and if the function g(θ) = cos θq m (cos θ, sin θ) sin θp m (cos θ, sin θ) is not identically zero, then there are at most 2(m + 1) directions θ = θ along which a trajectory may approach the origin. There directions are given by solutions of g(θ) =. Suppose that g(θ) is not identically zero, then the solutions curves which approach the origin along these tangent lines divide a neighborhood of the origin into a finite number of open regions called sectors. Definition.44. Three types of sections: a hyperbolic sector, a parabolic sector, and an elliptic sector. For planar analytic systems, in addition to nodes, foci, topological saddles and centers, the only other critical points are a critical point with an elliptic domain: a deleted neighborhood of the origin consists of one elliptic sector, one hyperbolic sector, two parabolic sectors, and four separatrices saddle-node: two hyperbolic sectors and one parabolic sector cusp: two hyperbolic sectors and two separatrices We consider the case when A has one zero eigenvalue, i.e., deta=, but tra. Then the system can be written as ẋ = p 2 (x, y), ẏ = y + q 2 (x, y) where p 2 and q 2 are analytic in a neighborhood of the origin and have expansion that begin with seconddegree term in x and y. Theorem.45. Let the origin be an isolated critical point for the above analytic system. Let y = φ(x) be the solution of the equation y + q 2 (x, y) = in a neighborhood of the origin and let the expansion of function ψ(x) = p 2 (x, φ(x)) in a neighborhood of x = have the form ψ(x) = a m x m +... where m 2 and a m. Then (1) for m odd and a m >, the origin is an unstable node, (2) for m odd and a m <, the origin is a topological saddle, (3) for m even, the origin is a saddle-node. Next we consider the case when A has two zero eigenvalues, i.e., deta =, tra =, but A. The system can be put in the normal form ẋ = y ẏ = a k x k [1 + h(x)] + b n x n y[1 + g(x)] + y 2 R(x, y) where h(x), g(x) and R(x, y) are analytic in a neighborhood of the origin, h() = h() =, k 2, a k and n 1. 33