Section 1.4: Second-Order and Higher-Order Equations. Consider a second-order, linear, homogeneous equation with constant coefficients

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1 Section 1.4: Second-Order and Higher-Order Equations Consider a second-order, linear, homogeneous equation with constant coefficients x t+2 + ax t+1 + bx t = 0. (1) To solve this difference equation, we consider solutions of the form x t = λ t, where λ 0. Definition: The equation λ 2 +aλ+b = 0 is called the characteristic equation and λ 2 +aλ+b is called the characteristic polynomial of the difference equation (1). The two solutions of the characteristic equation are called the eigenvalues. The form of the general solution depends on the eigenvalues and can be separated into three cases. Case 1: Real Distinct Eigenvalues If the eigenvalues are real and distinct (λ 1 λ 2 ), then the general solution is x t = c 1 λ t 1 + c 2 λ t 2. Case 2: Real Repeated Eigenvalues If the eigenvalues are real and repeated (λ 1 = λ 2 ), then the general solution is x t = c 1 λ t 1 + c 2 tλ t 2. Case 3: Complex Conjugate Eigenvalues If the eigenvalues are complex conjugates (λ = a ± ib), then the general solution is where r = a 2 + b 2 and φ = arctan(b/a). x t = c 1 r t cos(φt) + c 2 r t sin(φt), In each case, the general solution is an arbitrary linear combination of two linearly-independent solutions x 1 (t) and x 2 (t). This is known as the superposition principle. 1

2 A check for linear independence involves a quantity called the Casoratian, which is very similar to the Wronskian in differential equations. Definition: The Casoratian of x 1 (t) and x 2 (t) is C(t) = C(x 1 (t), x 2 (t)) = x 1 (t) x 2 (t) x 1 (t + 1) x 2 (t + 1) = x 1(t)x 2 (t + 1) x 2 (t)x 1 (t + 1). The solutions x 1 (t) and x 2 (t) are linearly independent if C(t) 0 for some t = 0, 1, 2,.... In this case, x 1 (t) and x 2 (t) are said to form a fundamental set of solutions. Example: Find the Casoratian of each set of functions. (a) x 1 (t) = λ t 1, x 2 (t) = λ t 2 (b) x 1 (t) = λ t, x 2 (t) = tλ t 2

3 Example: Find the general solution of each difference equation. (a) x t+2 + 3x t+1 + 2x t = 0 (b) x t+2 10x t x t = 0 (c) x t+2 + 4x t+1 + 8x t = 0 3

4 Example: Solve the initial value problem x t+2 + 4x t+1 + 3x t = 0, x 0 = 0, x 1 = 1. Example: Solve the initial value problem x t+2 + 4x t+1 + 4x t = 0, x 0 = 2, x 1 = 4. 4

5 Higher-Order Linear Equations Consider a kth-order, linear, homogeneous difference equation with constant coefficients x t+k + a 1 x t+k a k x t = 0. (2) Let x t = λ t, where λ 0. The characteristic equation for this difference equation is λ k + a 1 λ k a k = 0. The characteristic equation has k eigenvalues λ 1,..., λ k. Case 1: Real Distinct Eigenvalues If the eigenvalues are real and distinct, then the general solution is x t = c 1 λ t c k λ t k. Case 2: Real Repeated Eigenvalues If there is a real eigenvalue of multiplicity m, then m linearly independent solutions can be formed as λ t 1, tλ t 1,..., t m 1 λ t 1. Case 3: Complex Conjugate Eigenvalues If there are complex conjugate eigenvalues λ 1,2 = r (cos φ ± i sin φ) of multiplicity m, then there are 2m linearly independent solutions r t cos(tφ), r t sin(tφ), tr t cos(tφ), tr t sin(tφ),..., t m 1 r t cos(tφ), t m 1 r t sin(tφ). In this case the Casoratian is the determinant of a k k matrix and the solutions can be shown to be linearly independent. Example: Find the general solution of x t+3 + x t+2 + x t+1 + x t = 0. 5

6 Nonhomogeneous Linear Equations Consider a kth-order, nonhomogeneous, linear difference equation where b t is a nonzero function of t = 0, 1, 2,.... x t+k + a 1 x t+k a k x t = b t, (3) The general solution of the nonhomogeneous equation is given by x(t) = x h (t) + x p (t), where x h (t) is a solution of the corresponding homogeneous equation and x p (t) is a particular solution of the nonhomogeneous equation. To find a particular solution, we use the Method of Undetermined Coefficients. That is, we assume the particular solution has the same form as the forcing term b t. Example: Find the general solution of the nonhomogeneous equation x t+2 5x t+1 + 6x t = t

7 Example: Find the general solution of the nonhomogeneous equation x t+2 + x t+1 12x t = t2 t. 7

8 Limiting Behavior of Solutions The long-term behavior of solutions of linear difference equations is determined by the eigenvalues of the characteristic equation. The magnitude of the eigenvalues determine whether solutions are bounded or unbounded. The types of eigenvalues (real or complex) determine whether solutions oscillate or whether solutions converge or diverge monotonically. Of particular interest is whether the magnitude of all eigenvalues is less than one. Recall that the magnitude of a real eigenvalue λ = a is its absolute value λ = a. The magnitude of a complex eigenvlaue λ = a + bi is λ = a + bi = a 2 + b 2. Definition: Suppose that the k eigenvalues of a characteristic equation are λ 1,..., λ k. An eigenvalue λ i such that λ i λ j for all j i is called a dominant eigenvalue. If the inequality is strict, λ i > λ j for all j i, then λ i is called a strictly dominant eigenvalue. In particular, if there exists a dominant eigenvalue λ 1 such that λ 1 < 1, then solutions to the difference equation converge to zero. Example: Consider the difference equation 4x t+2 + x t = 0. (a) Find the eigenvalues of this difference equation. Identify the dominant eigenvalue. Is it strictly dominant? (b) Determine the limiting behavior of x t without solving the equation. (c) Find the general solution of the difference equation. limiting behavior you expected? Does the solution exhibit the 8