Econ 204 Differential Equations. 1 Existence and Uniqueness of Solutions

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1 Econ 4 Differential Equations In this supplement, we use the methods we have developed so far to study differential equations. 1 Existence and Uniqueness of Solutions Definition 1 A differential equation is an equation of the form y t Fyt, t where U is an open subset of R n R and F : U R n. An initial value problem is a differential equation combined with an initial condition yt y with y, t U. A solution of the initial value problem is a differentiable function y : a, b R n such that t a, b, yt y and, for all t a, b, dy Fyt, t. The general solution of the differential dt equation is the family of all solutions for all initial values y, t U. Theorem Consider the initial value problem y t Fyt, t, yt y 1 Let U be an open set in R n R containing y, t. Suppose F : U R n is continuous. Then the initial value problem has a solution. If, in addition, F is Lipschitz in y on U i.e. there is a constant K such that for all y, t, ŷ, t U, Fy, t Fŷ, t K y ŷ, then there is an interval a, b containing t such that the solution is unique on a, b. Proof: We will limit our proof to the case in which F is Lipschitz; for the general case, see Coddington and Levinson [1]. Since U is open, we may choose r > such that R {y, t : y y r, t t r} U 1

2 Given the Lipschitz condition, we may assume that Fy, t Fŷ, t K y ŷ for all y, t, ŷ, t R. Let { } 1 δ min K, r M We claim the initial value problem has a unique solution on t δ, t + δ. Let C be the space of continuous functions from [t δ, t + δ] to R n, endowed with the sup norm Let f sup{ ft : t [t δ, t + δ]} S {z C[t δ, t + δ] : zs, s R for all s [t δ, t + δ]} S is a closed subset of the complete metric space C, so S is a complete metric space. Consider the function I : S C defined by Izt y + t Fzs, sds Iz is defined and continuous because F is bounded and continuous on R. Observe that if zs, s R for all s [t δ, t + δ], then Izt y Fzs, sds t t t max{ Fy, s : y, s R} δm r so Izt, t R for all t [t δ, t + δ]. Thus, I : S S. Given two functions x, z S and t [t δ, t + δ], Izt Ixt y + Fzs, sds y Fxs, sds t t Fzs, s Fxs, s ds t t t sup{ Fzs, s Fxs, s : s [t δ, t + δ]} δk sup{ zs xs : s [t δ, t + δ]} z x

3 Therefore, Iz Ix z x, so I is a contraction. Since S is a complete metric space, I has a unique fixed point y S. Therefore, for all t [t δ, t + δ], we have yt t Fys, sds F is continuous, so the Fundamental Theorem of Calculus implies that y t Fyt, t for all t t δ, t + δ. Since we also have yt y + t Fys, sds y y restricted to t δ, t + δ is a solution of the initial value problem. On the other hand, suppose that ŷ is any solution of the initial value problem on t δ, t + δ. It is easy to check that ŷs, s R for all s t δ, t + δ, so we have Fŷs, s M; this implies that ŷ has a extension to a continuous function still denoted ŷ in S. Since ŷ is a solution of the initial value problem, the Fundamental Theorem of Calculus implies that I ŷ ŷ. Since y is the unique fixed point of I, ŷ y. Example Consider the initial value problem y t 1 + y t, y Here, we have Fy, t 1 + y which is Lipschitz in y over sets U V R, provided that V is bounded, but not over all of R R. The theorem tells us that the initial value problem has a unique solution over some interval of times a, b, with a, b. We claim the unique solution is yt tan t. To see this, note that y t d dt tan t d sint dt cos t cos t cost sint sint cos t

4 cos t + sin t cos t 1 + sin t cos t 1 + tan t 1 + yt y tan Notice that yt is defined for t π, π, but lim t π + yt and lim yt t π Thus, the solution of the initial value problem cannot be extended beyond the interval π, π, because the solution blows up at π/ and π/. Example 4 Consider the initial value problem y t y, y The function Fy, t y is not locally Lipschitz in y at y : y y y y y y which is not a bounded multiple of y. Given any α, let y α t { if t α t α if t α We claim that y α is a solution of the initial value problem for every α. For t < α, y αt y α t. For t > α, y αt t α 4

5 t α y α t. For t α, y α α + h y α α lim h + h y α α + h y α α lim h h h lim h + h lim h h so y αα y α α. Finally, y α, so y α is a solution of the initial value problem, so we see the solution is decidedly not unique! Remark 5 The initial value problem of Equation 1 has a solution defined on the interval inf {t : s t, t ] ys, s U},sup {t : s [t, t ys, s U} and it is unique on that interval provided that F is locally Lipschitz on U, i.e. for every y, t U, there is an open set V with y, t V U such that F is Lipschitz on V. Autonomous Differential Equations In many situations of interest, the function F in the differential equation depends only on y and not on t. Definition 6 An autonomous differential equation is a differential equation the form y t Fyt where F : R n R depends on t only through the value of yt. A stationary point of an autonomous differential equation is a point y s R n such that Fy s We can study the qualitative properties of solutions of autonomous differential equations by looking for stationary points. Note that the constant function yt y s 5

6 is a solution of the initial value problem y Fy, yt y s If F satisfies a Lipschitz condition, the constant function is the unique solution of the initial value problem. If F is C, then from Taylor s Theorem, we know near a stationary point y s, we have Fy s + h Fy s + DFy s h + O h DFy s h + O h Thus, when we are sufficiently close to the stationary point, the solutions of the differential equation are closely approximated by the solutions of the linear differential equation y y Y s DFy s y y s Complex Exponentials The exponential function e x for x R or x C is given by the Taylor series e x x k k! For x, y C, we have If x C, x a + ib for a, b R, so k e x+y e x e y e x e a+ib e a e ib e a ib k k! k e a k e a k ib k k! + k bk i k k! + i ib k+1 k + 1! k i k b k+1 k + 1! 6

7 e a 1 k bk k e a cosb + i sinb Now suppose that t R, so If a <, then e tx as t k! + i 1 k b k+1 k + 1! k e tx e ta+itb e ta costb + i sintb If a >, then e tx as t If a, then e tx 1 for all t R 4 Homogenous Linear Differential Equations with Constant Coefficients Let M be a constant n n real matrix. Linear differential equations of the type y y y s My y s are called homogeneous linear differential equations with constant coefficients. 1 Differential Equation has a complete solution in closed form. Note that the matrix representing DFy s need not be symmetric, so it need not be diagonalizable. However, if the n n matrix DFy s is diagonalizable over C, the complete solution takes the following particularly simple form: Theorem 7 Consider the linear differential equation y y y s My y s where M is a real n n matrix. Suppose that M can be diagonalized over the complex field C. Let U be the standard basis of R n and V {v 1,..., v n } be a 1 Later, we shall study equations of the type y y y s My y s + Ht, where H : R R n ; such equations are called inhomogeneous; homogenous equations are ones in which H is identically zero. 7

8 basis of complex eigenvectors corresponding to the eigenvectors λ 1,..., λ n C. Then the solution of the initial value problem is given by yt y s + Mtx U,V id e λ 1t t e λ t t.... e λnt t Mtx V,Uidyt y s 4 and the general complex solution is obtained by allowing yt to vary over C n ; it has n complex degrees of freedom. The general real solution is obtained by allowing yt to vary over R n ; it has n real degrees of freedom. Every real solution is a linear combination of the real and imaginary parts of a complex solution. In particular, 1. If the real part of each eigenvalue is less than zero, all solutions converge to y s. If the real part of each eigenvalue is greater than zero, all solutions diverge from y s and tend to infinity. If the real parts of some eigenvalues are less than zero and the real parts of other eigenvalues are greater than zero, solutions follow hyperbolic paths 4. If the real parts of all eigenvalues are zero, all solutions follow closed cycles around y s Remark 8 If one or more of the eigenvalues are complex, each of the three matrices in Equation 4 will contain complex entries, but the product of the three matrices is real. Thus, if the initial condition y is real, Equation 4 gives us a real solution; indeed, it gives us the unique solution of the initial value problem. Remark 9 Given a fixed time t, the general real solution is obtained by varying the initial values of yt over R n, which provides n real degrees of freedom. You might think that varying t provides one additional degree of 8

9 freedom, but it doesn t. Given any solution satisfying the initial condition yt y, the solution is defined on some interval t δ, t + δ; given t 1 t δ, t + δ, let y 1 yt 1 ; then the solution with initial condition yt 1 y 1 is the same as the solution with initial condition yt y. The same holds true for the general complex solution. Proof: We rewrite the differential equation in terms of a new variable z Mtx V,U id y which is the representation of the solution with respect to the basis V of eigenvectors. Let z s Mtx V,U id y s. Then we have where z z s Mtx V,U id y y s z z s z Mtx V,U id y Mtx V,U id M y y s Mtx V,U M Mtx U,V z z s Bz z s B λ 1 λ..... λ n Thus, the i th component of zt z s satisfies the differential equation zt z s i λ i zt z s i so so zt z s i e λ it t zt z s i zt z s e λ 1t t e λ t t.... e λnt t zt z s 9

10 yt y s Mtx U,V id zt z s e λ 1t t e λ t t Mtx U,V id.... e λnt t e λ 1t t e λ t t Mtx U,V id.... e λnt t zt z s Mtx V,Uid yt y s which shows that the solution to the initial value problem is as claimed; the general complex solution is found by varying yt over C n. Given a differentiable function y : [t, C n, write yt wt + ixt as the sum of its real and imaginary parts. Then y t w t + ix t. Since the matrix M is real, y is a complex solution if and only if w and x are real solutions. A solution is real if and only if yt R n, so the general real solution has n real degrees of freedom; the solution of the initial value problem is determined from the n real initial conditions yt. Statements 1-4 about the asymptotic properties of the solutions follow from the properties of complex exponentials in Section. 5 The Form of Real Solutions When the eigenvalues of a real matrix M are complex, the corresponding eigenvectors must be complex also. Consequently, even for a real solution y of the undiagonalized equation, the solution z of the diagonalized equation, which is the representation of y in terms of the basis V of eigenvectors, will be complex. However, as noted in the previous section, the product of the three matrices in Equation 4 is real, and hence the equation gives us a real solution, indeed the unique solution of the initial value problem. The general solution can be obtained by varying the initial condition y. We can determine the form of the real solutions once we know the eigenvalues, and in one important case, this provides us with an easier way to find 1

11 the general solution of the differential equation, or the unique solution of the initial value problem. Theorem 1 Consider the differential equation y y y s My y s Suppose that the matrix M can be diagonalized over C. Let the eigenvalues of M be a 1 + ib 1, a 1 ib 1,..., a m + ib m, a m ib m, a m+1,..., a n m Then for each fixed i 1,..., n, every real solution is of the form yt y s i m e a jt t C ij cos b j t t + D ij sin b j t t j1 + n m jm+1 C ij e a jt t The n parameters {C ij : i 1,..., n; j 1,..., n m} {D ij : i 1,..., n; j 1,..., m} have n real degrees of freedom. The parameters are uniquely determined from the n real initial conditions of an Initial Value Problem. Proof: Rewrite the expression for the solution y as n yt y s i γ ij e λ jt t j1 Recall that the non-real eigenvalues occur in conjugate pairs, so suppose that λ j a + ib, λ k a ib so the expression for yt y s contains the pair of terms γ ij e λ jt t + γ ik e λ kt t γ ij e at t cos bt t + i sin bt t +γ ik e at t cos bt t i sinbt t e at t γ ij + γ ik cos bt t + i γ ij γ ik sinbt t e at t C ij cos bt t + D ij sin bt t 11

12 Since this must be real for all t, we must have C ij γ ij + γ ik R and D ij i γ ij γ ik R so γ ij and γ ik are complex conjugates; this can also be shown directly from the matrix formula for y in terms of z. Thus, if the eigenvalues λ 1,..., λ n are a 1 + ib 1, a 1 ib 1, a + ib, a ib,..., every real solution will be of the form yt y s i a m + ib m, a m ib m, a m+1,..., a n m m e a jt t C ij cos b j t t + D ij sin b j t t j1 + n m jm+1 C ij e a jt t Since the differential equation satisfies a Lipschitz condition, the Initial Value Problem has a unique solution determined by the n real initial conditions. Thus, the general solution has exactly n real degrees of freedom in the n coefficients. Remark 11 The constraints among the coefficients C ij, D ij can be complicated. One cannot just solve for the coefficients of y 1 from the initial conditions, then derive the coefficients for y,..., y n. For example, consider the differential equation y1 y The eigenvalues are and 1. If we set we get 1 y1 y y 1 t C 11 e t t + C 1 e t t y t C 1 e t t + C e t t y 1 t C 11 + C 1 y t C 1 + C 1

13 which doesn t have a unique solution. However, from the original differential equation, we have so y 1 t y 1 t e t t, y t y t e t t C 11 y 1 t C 1 C 1 C y t One can find the solution to the Initial Value Problem by plugging the n real initial conditions into Equation 4, and the general solution by varying the initial conditions. However, in the important special case the coefficients ȳ y y. y n 1 C 11,..., C 1n m, D 11,..., D 1m in the general solution are arbitrary real numbers; once they are set, the other coefficients are determined. Write m yt y s e a jt t C j cos b j t t + D j sin b j t t j1 + n m jm+1 C j e a jt t For the Initial Value Problem, compute the first n 1 derivatives of y at t and set them equal to the initial conditions. This yield n linear equations in the n coefficients, which have a unique solution. Note also that C j e a jt t C j e a jt e a j t cos b j t t cosb j t b j t cosb j t cos b j t + sinb j t sinb j t sinb j t t sinb j t b j t cosb j t sinb j t + sinb j t cosb j t 1

14 so we can also write yt y s m e ajt C j cosb j t + D j sinb j t j1 + n m jm+1 C j e a jt 6 Second Order Linear Differential Equations Consider the second order differential equation y cy + by. Let ȳt yt y t Then we have ȳ t y t y t 1 yt c b y t 1 ȳ c b The eigenvalues are b± b +4c, the roots of the characteristic polynomial λ bλ c. We have the following cases: 1. c b: the eigenvalues are equal to b ; the matrix may not be diagonalizable, so the solutions may have a more complicated 4 form.. c > b 4 : b + 4c >, so the eigenvalues are real and distinct, hence the matrix is diagonalizable. a c > : the eigenvalues are real and of opposite sign, so the solutions follow roughly hyperbolic trajectories. The asymptotes are the two eigenvectors; along one eigenvector, the solutions converge to zero, while along the other, they tend to infinity. b c < the eigenvalues are real and of the same sign as b 14

15 i. b > : the solutions tend to infinity. ii. b < : the solutions converge to zero. iii. b : impossible since > c > b 4 >. c c : one eigenvalue is, while the other equals b. i. b > : the solutions tend to infinity along one eigenvector, and are constant along the other eigenvector. ii. b < : the solutions tend to zero along one eigenvector, and are constant along the other eigenvector. iii. b : impossible since c > b 4.. c < b. In this case, the eigenvalues are complex conjugates whose 4 real parts have the same sign as b. a b < : the solutions spiral in toward b b > : the solutions. spiral outward and tend to infinity. c b : the solutions are closed cycles around zero. Example 1 Consider the second order linear differential equation We let so the equation becomes y y + y ȳ ȳ y y 1 1 The eigenvalues are roots of the characteristic polynomial λ λ so the eigenvalues and corresponding eigenvectors are given by λ 1 v 1 1, λ 1 v 1, 1 ȳ 15

16 From this information alone, we know the qualitative properties of the solutions are as given in the phase plane diagram: The solutions are roughly hyperbolic in shape with asymptotes along the eigenvectors. Along the eigenvector v 1, the solutions flow off to infinity; along the eigenvector v, the solutions flow to zero. The solutions flow in directions consistent with the flows along the asymptotes. On the y-axis, we have y, which means that everywhere on the y- axis except at the stationary point, the solution must have a vertical tangent. On the y -axis, we have y, so we have y y + y y Thus, above the y-axis, y y >, so y is increasing along the direction of the solution; below the y-axis, y y <, so y is decreasing along the direction of the solution. Along the line y y, y y y, so y achieves a minimum or maximum where it crosses that line. We seek to find the general solution. We have described two methods. y This method does not depend on the special structure ȳ y. From Equation 4, the general solution is given by yt e t t yt y Mtx t U,V id e t t Mtx V,U id y t 1 1 e t t 1/ 1/ yt 1 e t t / 1/ y t 1 1 e t t e t t yt 1 e t t e t t y t e t t +e t t e t t e t t yt y t e t t e t t 16 e t t +e t t

17 yt +y t e t t + yt y t e t t yt +y t e t t + yt +y t e t t The general solution has two real degrees of freedom; a specific solution is determined by specifying the initial conditions yt and y t. The second method depends on the special structure ȳ y y In this setting, it is easier to find the general solution by setting yt C 1 e t t + C e t t and solving for the coefficients C 1 and C : yt C 1 + C y t C 1 e t t C e t t y t C 1 C C 1 yt + y t C yt y t yt yt + y t e t t + yt y t e t t Typically, we are not interested in the solution for y ; however, we can always obtain it by differentiating the solution for y. 7 Inhomogeneous Linear Differential Equations, with Nonconstant Coefficients Consider the inhomogeneous linear differential equation y Mty + Ht 5 17

18 Here, M is a continuous function from t to the set of n n matrices and H is a continuous function from t to R n. We first show that there is a close relationship between solutions of the inhomogeneous linear differential equation 5 and the associated homogeneous linear differential equation y Mty 6 Theorem 1 The general solution of the inhomogeneous linear differential equation 5 is y h + y p where y h is the general solution of the homogeneous linear differential equation 6 and y p is any particular solution of the inhomogeneous linear differential equation 5. Proof: Fix any particular solution y p of the inhomogeneous differential equation 5. Suppose that y h is any solution of the corresponding homogeneous differential equation 6. Let y i t y h t + y p t. Then y it y ht + y pt Mty h t + Mty p t + Ht Mty h t + y p t + Ht Mty i t + Ht so y i is a solution of the inhomogeneous differential equation 5. Conversely, suppose that y i is any solutions of the inhomogenous differential equation 5. Let y h t y i t y p t. Then y ht y pt y it Mty i t + Ht Mty p t Ht Mty i t y p t Mty h t so y h is a solution of the homogeneous differential equation 6 and y i y + h + y p. Thus, in order to find the general solution of the inhomogeneous equation, one needs to find the general solution of the homogeneous equation and a particular solution of the inhomogeneous equation. 18

19 Here for an n n matrix M, we define e M k M k k! I + M + M + and e tm t k M k k A particular solution of the inhomogeneous equation in the constant coefficient case has the following simple form: Theorem 14 Consider the inhomogeneous linear differential equation 5, and suppose that Mt is a constant matrix M, independent of t. A particular solution of the inhomogeneous linear differential equation 5, satisfying the initial condition y p t y, is given by y p t e t t M y + Proof: We verify that y p solves 7: y p t e t t M y + k! t e t sm Hsds 7 t e t sm Hsds e t tm y + e t tm e s tm Hsds t e t t M y + e s tm Hsds t y pt Me t t M y + e s tm Hsds t +e t t M e t tm Ht My p t + Ht y p t e t t M y + y t e s t M Hsds 19

20 Example 15 Consider the inhomogeneous linear differential equation y1 y 1 1 y1 By Theorem 14, a particular solution is given by y + sint cost y p t e t tm y + e t sm Hsds t e t 1 t e t s e t + 1 e t s e t t e e t + t s sin s e s t ds coss et 1 + e s sinsds e t 1 + e s cos sds e s sinsds e s sins t e s cos sds e t sint + e sin + e t sint + e s cos s e s cos sds t e t sint + e t cos t + e cos e t sin t + cos t + 1 e s sinsds e t sin t + cos t + 1 e s sinsds e t sin t + cos t + 1 e s cossds e s cos s t e t cos t e cos + e s sinsds e s sinsds e s sinsds e s sinsds e t cos t 1 + e s sin s t e s cossds sins coss e s sinsds ds

21 e t cos t 1 + e t sint + e sin e t sint + cos t 1 e s cossds e t sint + cos t 1 e s cossds et sint + cos t 1 y p t e s cos sds et 1 + e s sinsds e t 1 + es cos sds et 1 + e t sint+cos t+1 e t 1 + et sint+cos t 1 et e t sint+cos t e t 1+e t sint+cost e t sin t cost e t +sin t+cost e s cossds Thus, the general solution of the original inhomogeneous equation is given by y1 C1 e t e t sin t cost y C e t + e t +sin t+cost D1 e t sint+cos t D e t + sin t+cost where D 1 and D are arbitrary real constants. 8 Nonlinear Differential Equations: Linearization Nonlinear differential equations may be very difficult to solve in closed form. There are a number of specific techniques that solve special classes of equations, and there are numerical methods to compute numerical solutions of 1

22 any ordinary differential equation. Unfortunately, time does not permit us to discuss either these special classes of equations, or numerical methods. Linearization allows us to obtain qualitative information about the solutions of nonlinear autonomous equations. As we noted in Section, the idea is to find stationary points of the equation, then study the solutions of the linearized equation near the stationary points. This gives us a reasonably, but not completely, reliable to the behavior of the solutions of the original equation. Example 16 Consider the second-order nonlinear autonomous differential equation y α siny, α > This is the equation of motion of a pendulum, but it has much in common with all cyclical processes, including processes such as business cycles. The equation is very difficult to solve because of the nonlinearity of the function sin y. We define yt ȳt y t so that the differential equation becomes Let ȳ t Fȳ y t α siny 1 t y t α siny 1 t We solve for the points ȳ such that Fȳ : Fȳ so the set of stationary points is y t α sin y 1 t siny 1 and y y 1 nπ and y {nπ, : n Z}

23 We linearize the equation around each of the stationary points. In other words, we take the first order Taylor polynomial for F: For n even, the eigenvalues are Fnπ + h, + k + o h + k F 1 F 1 y 1 y h Fnπ, + F F y 1 y k 1 h + α cosnπ k 1 h 1 n+1 α k λ + α λ 1 iα, λ iα Close to nπ, for n even, the solutions spiral around the stationary point. For y y 1 >, y 1 is increasing, so the solutions move in a clockwise direction. For n odd, the eigenvalues and eigenvectors are λ α λ 1 v 1 α, λ α 1, α, v 1, α Close to nπ, for n odd, the solutions are roughly hyperbolic; along v, they converge to the stationary point, while along v 1, they diverge from the stationary point. From this information alone, we know the qualitative properties of the solutions are as given in the phase plane diagram on the next page: The solutions flow in directions consistent with the flows along the asymptotes. On the y-axis, we have y, which means that everywhere on the y- axis except at the stationary points, the solution must have a vertical tangent.

24 For y nπ, we have y α siny so the derivative of y is zero, so the tangent to the curve is horizontal. If the initial value of y is sufficiently large, the solutions no follow longer closed curves; this corresponds to the pendulum going over the top rather than oscillating back and forth. 9 Nonlinear Differential Equations Stability As we saw in the last section, linearization provides information about the qualitative properties of solutions of nonlinear differential equations near the stationary points. Suppose that y s is a stationary point, and that the eigenvalues of the linearized equation at y s all have strictly negative real parts. Then there exists ε > such that, if y y s < ε, then lim t yt y s ; all solutions of the original nonlinear equation which start sufficiently close to the stationary point y s converge to y s. On the other hand, if the egivenvalues of the linearized equation all have strictly positive real parts, then no solutions of the original nonlinear differential equation converge to y s. If the eigenvalues of the linearized equation at the stationary point y s all have real part zero, then the solutions of the linearized equation are closed curves around y s. In this case, the linearized equation tells us little about the solutions of the nonlinear equation. They may follow closed curves around y s, converge to y s, converge to a limit closed curve around y s, diverge from y s, or converge to y s along certain directions and diverge from y s along other directions. In this section, we illustrate techniques to analyze the asymptotic behavior of solutions of the linearized equation when the eigenvalues all have real part zero. Example 17 Consider the initial value problem y 1 t y t 9y t + 4y 1 t + 4y 1ty t 4y 1 t + 9y 1ty t + 9y t, y 1, y 8 4

25 y s is a stationary point. The linearization around this stationary point is 9 y t y 4 The characteristic equation is λ +6, so the matrix has distinct eigenvalues λ 1 6i and λ 6i; since both of these have real part zero, we know the solutions of the linearized differential equation follows closed curves around zero. The associated eigenvectors are v 1 i/ 1 so the change of basis matrices are i/ i/ Mtx U,V id and Mtx 1 1 V,U id since and v i/ 1 i/ 1/ i/ 1/ Then the solution of the linearized initial value problem is i/ i/ e 6ti i/ 1/ y 1 1 e 6ti i/ 1/ i/ i/ ie 6ti / e 6ti / 1 1 ie 6ti / e 6ti / e 6ti + e 6ti / e 6ti e 6ti i/4 e 6ti e 6ti i/ e 6ti + e 6ti / cos 6t sin 6t/ sin 6t/ cos 6t cos 6t sin 6t, e 6ti + e 6ti e 6ti e 6ti cos 6t + i sin6t + cos 6t + i sin 6t cos 6t + i sin6t + cos 6t i sin 6t cos 6t cos 6t + i sin6t cos 6t i sin 6t cos 6t + i sin6t cos6t + i sin 6t i sin 6t 5

26 Notice that y 1 t 9 + y t 4 9cos t + 4sin t 9 4 cos t + sin t 1 so the solution of the linearized initial value problem is a closed curve running counterclockwise around the ellipse with principal axes along the y 1 and y axes, of length and respectively. Let Gy y y 4 so that the solution runs around a level set of G. We can get more information about the asymptotic behavior of the solution of the orginal, unlinearized, initial value problem 8 by computing dgyt dgyt dt G y 1 y 1 t 9 G y 1 t y y t y t : dt 9y t + 4y 1 t + 4y 1ty t 4y 1 t + 9y 1ty t + 9y t y 1 ty t + 8y 4 1 t/9 + 8y 1 ty t/9 +y 1 ty t + 9y 1ty t/ + 9y 4 t/ 8y1t/ y1ty t/18 + 9yt/ 4 > unless yt y t is tangent to the solution at every t, and y t always points outside the level curve of G through yt, as in the green arrows in the diagram. Thus, the solution of the initial value problem 8 spirals outward, always moving to higher level curves of G. Indeed, for Gy 1 i.e., outside the ellipse which the solution of the linearized initial value problem falls, it is easy to see that 8y1 4 t/9 + 97y 1 ty t/18 + 9y4 t/ > 8 y 9 1 t + y t is uniformly bounded away from zero, so Gyt dgys ds ds as t. The linear terms become dwarfed by the higher order terms, so dgyt dt 6

27 which will determine whether the solution continues to spiral as it heads off into the distance. Suppose now we consider instead the initial value problem y 1 t y t 9y t 4y 1t 4y 1 ty t 4y 1 t 9y 1 ty t 9y t, y 1, y 9 The linearized initial value problem has not changed. As before, we compute dgyt dt G y 1 y 1 t 9 G y 1 t y y t y t 9y t 4y 1t 4y 1 ty t 4y 1 t 9y 1ty t 9y t y 1 ty t 8y 4 1t/9 8y 1ty t/9 +y 1 ty t 9y 1 ty t/ 9y4 t/ 8y1t/9 4 97y1ty t/18 9yt/ 4 < unless yt As before, y t is tangent to the solution at every t, and y t always points inside the level curve of G through yt, as in the blue arrows in the diagram. Since dgyt < except at the origin, then for all C >, dt { } dgyt α inf : C Gyt Gy < dt since the region {y : C Gy Gy} is compact. If Gyt C for all t, Gyt Gy + Gy + αt as t dgys ds which is a contradiction. Thus, Gyt and the solution of the initial value problem 9 converges to the stationary point as t. ds 7

28 Example 18 This example was assigned as a problem in 7, much to the regret of the students and the professor; the latter did penance by writing out the solution. The principal axes of elliptical solutions of the linearized equation align with the eigenvectors. In Example 17, the eigenvectors aligned with the standard basis. When the eigenvectors do not align with the standard basis, the algebraic computations become quite challenging and tedious to do by hand. This example shows that they can in fact be done by hand; with an appropriate symbolic manipulation computer program, they would be straightforward. Consider the initial value problem y 1 t y t 5y1 t 1y t y 1t y 1 ty t 1y 1 t 5y t y 1 ty t y t The linearization around the stationary point y t y, y 1, y is 1 The characteristic equation is 5 λ 5 λ+169, λ +144, so there are distinct eigenvalues λ 1 1i and λ 1i; since both of these have real part zero, we know the solution of the linearized initial value problem follows a closed curve around zero. The associated eigenvectors are v 1 1 and v 5+1i, so the change of basis matrices are 1 y Mtx U,V id i 1 5+1i 1 and Mtx V,U id 1 5i 4 Then the solution of the linearized initial value problem is i 1 5+1i i 1 5+1i 1 e 1ti 1 5i 1i e 1ti i 1i i 1 1i 4 1+5i 1i ie 1ti /4 1ie 1ti / ie 1ti /4 1ie 1ti /4 8

29 e 1ti + e 1ti / 5ie 1ti e 1ti /4 1ie 1ti e 1ti /4 1ie 1ti e 1ti /4 e 1ti + e 1ti / + 5ie 1ti e 1ti /4 cos 1t + 5sin 1t/1 1sin1t/1 1sin 1t/1 cos 1t 5sin 1t/1 cos 1t sin 1t cos 1t + sin 1t since e 1ti + e 1ti e 1ti e 1ti cos 1t + i sin1t + cos 1t + i sin 1t cos 1t + i sin1t + cos1t i sin1t cos 1t cos 1t + i sin1t cos 1t i sin 1t cos 1t + i sin1t cos 1t + i sin1t i sin 1t Let z1 t z t y1 t + y t y t y 1 t 6cos 1t 4sin 1t so z1 t 6 + z t 16 cos 1t + sin 1t 1 This shows that the solution of the linearized initial value problem follows an ellipse with principal axes of length and along the vectors 1, 1 and 1, 1 counterclockwise. In order to determine the asymptotic behavior of solutions of the nonlinear equation, we need to find a function G whose level sets are the solutions of the linearized equation. Let z Gy z 16 y y 1 y + y 6 1y 1 1y 1 y + 1y + y 1 y 1 y + y 16 9

30 dgyt dt 6y 1 t 1y t 6y t 1y 1 t 5y 1 t 1y t y 1t y 1 ty t 1y 1 t 5y t y 1ty t y t 1y 1t 8y 1 ty t 6y 4 1t 6y 1ty t 5y 1 ty t + 1y t + 1y ty 1 t + 1y 1ty t +8y 1 ty t 1y t 6y 1ty t 6y 4 t 1y 1t + 5y 1 ty t + 1y 1ty t + 1y 1 ty t 6y 4 1 t + y 1 ty t + y4 t +y 1 ty ty 1t + y t 6y 1t + y t + y 1 ty ty 1t + y t 16y 1 t + y t 1y 1 t + y ty 1 t y 1ty t + y t 16y1t + yt 1y1t + yty 1 t y t < unless y Therefore, the tangent to the solution of the nonlinear equation always points inward to the elliptical level sets of the linearized equation and, as in Example 17, the solution of the initial value problem 1 converges to the stationary point as t. In the initial value problems 8, 9 and 1, we were lucky to some extent. We took G to be the function whose level sets are the solutions of the linearized differential equation, and found that the tangent to the solution always pointed outside the level curve in 8 and always pointed inside the level curve in 9. It is not hard to construct examples in which the tangent points outward at some points and inward at others, so the value Gyt is not monotonic. One may be able to show by calculation that Gyt, so the solution disappears off into the distance, or that Gyt, so the solution converges to the stationary point. An alternative method is to choose a different function G, whose level sets are not the solutions of the linearized differential equation, but for which one can prove that dgyt is dt always positive or always negative; this is called Liapunov s Second Method see Ritger and Rose [].

31 References [1] E. A. Coddington and N. Levinson, Theory of Ordinary Differential Equations, McGraw-Hill, New York, [] Paul D. Ritger and Nicholas J. Rose, Differential Equations with Applications, McGraw-Hill, New York,