Econ Lecture 14. Outline

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1 Econ Lecture 14 Outline 1. Differential Equations and Solutions 2. Existence and Uniqueness of Solutions 3. Autonomous Differential Equations 4. Complex Exponentials 5. Linear Differential Equations with Constant Coefficients 6. The Form of Real Solutions 1

2 Differential Equations Definition 1. A differential equation is an equation of the form y (t) = F(y(t), t) where F : U R n and U is an open subset of R n R. An initial value problem is a differential equation combined with an initial condition with (y 0, t 0 ) U. y(t 0 ) = y 0 A solution of the initial value problem is a differentiable function y : (a, b) R n such that t 0 (a,b), y(t 0 ) = y 0 and, for all t (a, b), dy dt = F(y(t), t). 2

3 The general solution of the differential equation is the family of all solutions for all initial values (y 0, t 0 ) U.

4 Existence and Uniqueness of Solutions Theorem 1. Consider the initial value problem y (t) = F(y(t), t), y(t 0 ) = y 0 (1) Let U be an open set in R n R containing (y 0, t 0 ). 1. Suppose F : U R n is continuous. Then the initial value problem has a solution. 2. 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, F(y, t) F(ŷ, t) K y ŷ then there is an interval (a,b) containing t 0 such that the solution is unique on (a, b). 3

5 Proof. We consider only the case in which F is Lipschitz. Since U is open, we may choose r > 0 such that R = {(y, t) : y y 0 r, t t 0 r} U Since F is continuous, we may find M R such that F(y, t) M for all (y, t) R. Given the Lipschitz condition, we may assume that Let F(y, t) F(ŷ, t) K y ŷ for all (y, t),(ŷ, t) R { 1 δ = min 2K, r M We claim the initial value problem has a unique solution on (t 0 δ, t 0 + δ). } 4

6 Let C be the space of continuous functions from [t 0 δ, t 0 + δ] to R n, endowed with the sup norm Let f = sup{ f(t) : t [t 0 δ, t 0 + δ]} S = {z C : (z(s), s) R for all s [t 0 δ, t 0 + 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 I(z)(t) = y 0 + t t 0 F(z(s), s) ds I(z) is defined and continuous because F is bounded and continuous on R. Observe that if (z(s), s) R for all s [t 0 δ, t 0 + δ],

7 then t I(z)(t) y 0 = F(z(s), s) ds t 0 t t 0 max{ F(y, s) : (y, s) R} δm r so (I(z)(t), t) R for all t [t 0 δ, t 0 + δ]. Thus, I : S S.

8 Given two functions x, z S and t [t 0 δ, t 0 + δ], I(z)(t) I(x)(t) = = y 0 + t t t 0 F(z(s), s) ds y 0 t t 0 (F(z(s), s) F(x(s), s)) ds t 0 F(x(s), s) ds t t 0 sup F(z(s), s) F(x(s), s) s [t 0 δ,t 0 +δ] δk sup{ z(s) x(s) : s [t 0 δ, t 0 + δ]} 1 2 z x Therefore, I(z) I(x) 1 2 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 0 δ, t 0 + δ], we have y(t) = y 0 + t t 0 F(y(s), s) ds

9 F is continuous, so the Fundamental Theorem of Calculus implies that y (t) = F(y(t), t) for all t (t 0 δ, t 0 + δ). Since we also have y(t 0 ) = y 0 + t0 t 0 F(y(s), s) ds = y 0 y (restricted to (t 0 δ, t 0 + δ)) is a solution of the initial value problem (1). On the other hand, suppose that ŷ is any solution of the initial value problem (1) on (t 0 δ, t 0 + δ). It is easy to check that (ŷ(s), s) R for all s (t 0 δ, t 0 +δ), so we have F(ŷ(s), 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,

10 the Fundamental Theorem of Calculus implies that I (ŷ) = ŷ. Since y is the unique fixed point of I, ŷ = y.

11 Example: Consider the initial value problem y (t) = 1 + y 2 (t), y(0) = 0 Here, we have F(y, t) = 1 + y 2 which is Lipschitz in y over 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 0 (a, b). We claim the unique solution is y(t) = tan t. To see this, note 5

12 that y (t) = d dt tan t = d sin t dtcos t cos tcos t sin t( sin t) = cos 2 t = cos2 t + sin 2 t cos 2 t = 1 + sin2 t cos 2 t = 1 + tan 2 t = 1 + (y(t)) 2 y(0) = tan0 = 0

13 Notice that y(t) is defined for t ( π 2, π ) 2, but lim y(t) = and lim y(t) = t π + t π 2 2 Thus, the solution of the ( initial value problem cannot be extended beyond the interval π 2, π ) 2, because the solution blows up at π/2 and π/2.

14 Example Consider the initial value problem The function F(y, t) = 2 y (t) = 2 y, y(0) = 0 (2) y is not locally Lipschitz in y at y = 0: 2 y 2 0 = 2 y = y 2 y = y 2 y 0 which is not a bounded multiple of y 0. Given any α 0, let y α (t) = { 0 if t α (t α) 2 if t α We claim that y α is a solution of the initial value problem (2) for 6

15 every α 0. For t < α, y α 0 (t) = 0 = = 2 y α (t). For t > α, y α(t) = 2(t α) = 2 (t α) 2 = 2 y α (α + h) y α (α) lim h 0 + h y α (α + h) y α (α) lim h 0 h y α (t). For t = α, h 2 = lim h 0 + h = 0 0 = lim h 0 h = 0 so y α(α) = 0 = 2 y α (α). Finally, y α (0) = 0, so y α is a solution of the initial value problem (2), so we see the solution is decidedly not unique!

16 Remark: The initial value problem of Equation (1) has a solution defined on the interval (inf {t : s (t, t 0 ] (y(s), s) U},sup {t : s [t 0, t) (y(s), s) U}) and it is unique on this 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. 7

17 Autonomous Differential Equations In many situations of interest, the function F in the differential equation does not depend on t. Definition 2. An autonomous differential equation is a differential equation of the form y (t) = F(y(t)) where F : R n R n depends on t only through the value of y(t). A stationary point of an autonomous differential equation is a point y s R n such that F(y s ) = 0. 8

18 Autonomous Differential Equations We study the qualitative properties of autonomous differential equations by looking for stationary points. The constant function y(t) = y s is a solution (and the unique solution when F is Lipschitz) of the initial value problem y = F(y), y(t 0 ) = y s If F is C 2, then Taylor s Theorem implies that near a stationary point y s, F(y s + h) = F(y s ) + DF(y s )h + O ( h 2) = DF(y s )h + O ( h 2) 9

19 Thus, when we are sufficiently close to the stationary point, the solutions of the autonomous differential equation are closely approximated by the solutions of the linear differential equation y = (y y s ) = DF(y s )(y y s ) Thus, we study solutions of linear differential equations, using linear algebra.

20 Complex Exponentials The exponential function e x (for x R or x C) is given by the Taylor series For x, y C, we have e x = k=0 x k k! e x+y = e x e y 10

21 If x C, x = a + ib for a,b R, so e x = e a+ib = e a e ib = e a = e a = e a = e a k=0 k=0 (ib) k k! (ib) 2k (2k)! + k=0 ( 1) k b2k k=0 = e a (cos b + isin b) k=0 (ib) 2k+1 (2k + 1)! i 2k b2k (2k)! + i i 2k b 2k+1 (2k + 1)! k=0 (2k)! + i k=0 ( 1) k b 2k+1 (2k + 1)!

22 Now suppose that t R, so e tx = e ta+itb = e ta (costb + isintb) If a < 0, then e tx 0 as t If a > 0, then e tx as t If a = 0, then e tx = 1 for all t R

23 Linear Differential Equations with Constant Coefficients Let M R n n. The linear differential equation y = (y y s ) = M(y y s ) has a complete solution in closed form. The matrix representation M = DF(y s ) need not be symmetric, hence may not be diagonalizable. If M is diagonalizable over C, the complete solution takes the following simple form: 11

24 Theorem 2. Consider the linear differential equation y = (y y s ) = M(y 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 basis of (complex) eigenvectors corresponding to the eigenvalues λ 1,..., λ n C. Then the solution of the initial value problem is given by y(t) = y s +P 1 e λ 1(t t 0 ) e λ 2(t t 0 ) e λ n(t t 0 ) P(y(t 0) y s ) (3) where P = (Mtx) V,U (id), and the general complex solution is obtained by allowing y(t 0 ) to vary over C n ; it has n complex

25 degrees of freedom. The general real solution is obtained by allowing y(t 0 ) 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. 2. If the real part of each eigenvalue is greater than zero, all solutions diverge from y s and tend to infinity. 3. If the real parts of some eigenvalues are less than zero and the real parts of other eigenvalues are greater than zero, solutions follow roughly hyperbolic paths.

26 4. If the real parts of all eigenvalues are zero, all solutions follow closed cycles around y s.

27 Remark: If one or more of the eigenvalues are complex, each of the three matrices in Equation (3) will contain complex entries, but the product of the three matrices is real. Thus, if the initial condition y 0 is real, Equation (3) gives us a real solution; indeed, it gives us the unique solution of the initial value problem. Remark: Given a fixed time t 0, the general real solution is obtained by varying the initial values of y(t 0 ) over R n, which provides n real degrees of freedom. You might think that varying t 0 provides one additional degree of freedom, but it doesn t. Given any solution satisfying the initial condition y(t 0 ) = y 0, the solution is defined on some interval (t 0 δ, t 0 + δ); given t 1 (t 0 δ, t 0 + δ), let y 1 = y(t 1 ); then the solution with initial condition y(t 1 ) = y 1 is the same as the solution with initial condition y(t 0 ) = y 0. The same holds true for the general complex solution. 12

28 Proof. Let P = (Mtx) V,U (id). Rewrite the differential equation in terms of a new variable z = Py the representation of the solution with respect to the basis V of eigenvectors. Let z s = Py s. Then we have z z s = P(y y s ) (z z s ) = z = Py = PM(y y s ) = PMP 1 (z z s ) = B(z z s ) 13

29 where B = λ λ λ n Thus, the i th component of (z(t) z s ) satisfies the differential equation so (z(t) z s ) i = λ i(z(t) z s ) i (z(t) z s ) i = e λ i(t t 0 ) (z(t0 ) z s ) i

30 so z(t) z s = e λ 1(t t 0 ) e λ 2(t t 0 ) e λ n(t t 0 ) (z(t 0) z s ) y(t) y s = P 1 (z(t) z s ) = P 1 = P 1 e λ 1(t t 0 ) e λ 2(t t 0 ) e λ n(t t 0 ) e λ 1(t t 0 ) e λ 2(t t 0 ) e λ n(t t 0 ) (z(t 0) z s ) P(y(t 0) y s )

31

32 The Form of Real Solutions We can determine the form of the real solutions once we know the eigenvalues. In an important special case, we can solve for the solution of the initial value problem without calculating the diagonalization, as in Equation (3). Theorem 3. Consider the differential equation y = (y y s ) = M(y y s ) Suppose that the matrix M can be diagonalized over C. Let the eigenvalues of M with the correct multiplicity be a 1 + ib 1, a 1 ib 1,..., a m + ib m, a m ib m, a m+1,..., a n m 14

33 Then for each fixed i = 1,..., n, every real solution is of the form (y(t) y s ) i = m j=1 + The n 2 parameters e a j(t t 0 ) ( C ij cos b j (t t 0 ) + D ij sin b j (t t 0 ) ) n m j=m+1 C ij e a j(t t 0 ) {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.

34 Proof. Rewrite the expression for the solution y as (y(t) y s ) i = n j=1 γ ij e λ j(t t 0 ) 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 (y(t) y s ) contains the pair of terms γ ij e λ j(t t 0 ) + γ ik e λ k(t t 0 ) = = γ ij e a(t t 0) (cos b(t t0 ) + isin b(t t 0 )) +γ ik e a(t t 0) (cos b(t t0 ) isin b(t t 0 )) = e a(t t 0) (( γ ij + γ ik ) cos b(t t0 ) + i ( γ ij γ ik ) sin b(t t0 ) ) = e a(t t 0) ( C ij cos b(t t 0 ) + D ij sin b(t t 0 ) )

35 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 2 +ib 2, a 2 ib 2,..., a m +ib m, a m ib m, a m+1,..., a n m every real solution will be of the form (y(t) y s ) i = m j=1 + e a j(t t 0 ) ( C ij cos b j (t t 0 ) + D ij sin b j (t t 0 ) ) n m j=m+1 C ij e a j(t t 0 )

36 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 2 coefficients.

37 Remark: 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 2,..., y n. For example, consider the differential equation ( ) ( ) ( ) y1 2 0 y1 = 0 1 y 2 The eigenvalues are 2 and 1. If we set we get y 2 y 1 (t) = C 11 e 2(t t 0) + C12 e t t 0 y 2 (t) = C 21 e 2(t t 0) + C 22 e t t 0 y 1 (t 0 ) = C 11 + C 12 y 2 (t 0 ) = C 21 + C 22 15

38 which doesn t have a unique solution. However, from the original differential equation, we have so y 1 (t) = y 1 (t 0 )e 2(t t 0), y 2 (t) = y 2 (t 0 )e t t 0 C 11 = y 1 (t 0 ) C 12 = 0 C 21 = 0 C 22 = y 2 (t 0 ) One can find the solution to the initial value problem by plugging the n real initial conditions into Equation (3) in Theorem 2 above, and the general solution by varying the initial conditions.

39 However, in the special case ȳ = the coefficients y y. y (n 1) C 11,..., C 1(n m), D 11,..., D 1m in the general solution are arbitrary real numbers; once they are set, the other coefficients are determined. Write y(t) y s = m j=1 + e a j(t t 0 ) ( ) C j cos b j (t t 0 ) + D j sin b j (t t 0 ) n m j=m+1 C j e a j(t t 0 ) 16

40 For the initial value problem, compute the first n 1 derivatives of y at t 0 and set them equal to the initial conditions. This yields n linear equations in the n coefficients, which have a unique solution. Note also that C j e a j(t t 0 ) = ( C j e a jt 0 ) e a j t cosb j (t t 0 ) = cos ( ) b j t b j t 0 = cos b j tcos b j t 0 + sin b j tsin b j t 0 sin b j (t t 0 ) = sin ( ) b j t b j t 0 so we can also write y(t) y s = m j=1 = cos b j tsin b j t 0 + sin b j tcosb j t 0 e a jt ( C j cos b j t + D j sin b j t ) + n m j=m+1 C j e a jt 17

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