Lecture Notes 6: Dynamic Equations Part B: Second and Higher-Order Linear Difference Equations in One Variable

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1 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 1 of 56 Lecture Notes 6: Dynamic Equations Part B: Second and Higher-Order Linear Difference Equations in One Variable Peter J. Hammond latest revision 2016 September 11th

2 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 2 of 56 Lecture Outline Solving Second-Order Equations Inhomogeneous Equations Two Special Cases Method of Undetermined Coefficients Higher-Order Linear Equations with Constant Coefficients Stationary States and Stability for Second-Order Equations

3 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 3 of 56 Second-Order Equations A general second-order difference equation specifies the state x t at each time t as a function x t = F t (x t 1, x t 2 ) of the state at two previous times. We focus on linear equations in one variable with constant coefficients. These take the form x t+1 + ax t + bx t 1 = f t where a, b are scalars, and f t is the forcing term. We assume that b 0 because otherwise we have the first-order equation x t+1 + ax t = 0.

4 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 4 of 56 A Coupled Pair of First-Order Equations The equation x t+1 + ax t + bx t 1 = f t can be converted into a coupled pair of first-order equations. To do so, define y t = x t 1, so the equations become x t+1 = ax t by t + f t y t+1 = x t In matrix form, these can be written as ( ) ( ) ( xt+1 a b xt = 1 0 y t+1 This kind of vector difference equation will be considered much more fully in part C. y t ) + ( ) ft 0

5 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 5 of 56 The Homogeneous Case Nevertheless, consider the homogeneous case when the vector equation is ( ) xt+1 y t+1 ( a b 1 0 ) ( ) xt = The solution in matrix form is evidently ( ) ( ) t ( ) xt a b x0 = y t 1 0 y 0 ( ) x0 for an arbitrary initial state. y 0 y t ( ) 0 0 Inspired by our earlier discussion ( ) of matrix powers, x0 consider the case when is an eigenvector, y 0 with corresponding eigenvalue λ that is, suppose ( ) ( ) ( ) a b x0 x0 = λ 1 0 y 0 y 0

6 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 6 of 56 Solving the Homogeneous Case In case ( a b 1 0 ( xt y t ) = ) ( ) x0 = λ y 0 ( a b 1 0 ( x0 y 0 But for this to work, the initial vector ), the solution takes the form ) t ( ) x0 y 0 ( x0 y 0 ) ( ) = λ t x0 y 0 must be a non-zero solution ( ) ( ) a λ b x0 of the matrix equation = 1 λ y 0 ( ) 0. 0 For such a ( non-zero solution) to exist, a λ b the matrix must be singular, implying that 1 λ a λ b 1 λ = λ2 + aλ + b = 0

7 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 7 of 56 The Auxiliary Equation Instead of treating the second-order equation as a coupled pair, consider directly the homogeneous second-order equation x t+1 + ax t + bx t 1 = 0 Inspired by our previous analysis using eigenvalues of a suitable matrix, we look for a solution of the form x t = λ t x 0, for suitable constants λ and x 0. It is a solution provided that λ t+1 x 0 + aλ t x 0 + bλ t 1 x 0 = 0. Ignoring the trivial solutions when x 0 = 0 or λ = 0, cancel λ t 1 x 0 to obtain the auxiliary or characteristic equation λ 2 + aλ + b = 0 This, of course, is the condition for λ to be an eigenvalue.

8 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 8 of 56 The Auxiliary Equation and Its Roots The auxiliary equation λ 2 + aλ + b = 0 is quadratic. It therefore has two roots λ 1, λ 2 satisfying λ 2 + aλ + b = (λ λ 1 )(λ λ 2 ). In particular λ 1 + λ 2 = a and λ 1 λ 2 = b. The assumption that b 0 implies that the two roots λ 1, λ 2 are both non-zero. This leaves three cases: 1. two distinct real roots λ 1, λ 2 R, which is true iff a 2 > 4b; 2. two complex conjugate roots λ 1, λ 2 = re ±iθ C, which is true iff a 2 < 4b; 3. two coincident real roots λ = λ 1 = λ 2 R, which is true iff a 2 = 4b.

9 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 9 of 56 Case 1: Two Distinct Real Roots In this case λ 2 + aλ + b = (λ λ 1 )(λ λ 2 ), where λ 1, λ 2 = 1 2 a ± 1 2 a 2 4b. Note that a = λ 1 + λ 2 and b = λ 1 λ 2 with a 2 4b = (λ 1 + λ 2 ) 2 4λ 1 λ 2 = (λ 1 λ 2 ) 2 > 0. There are two degrees of freedom in the difference equation, so we look for two linearly independent solutions x (1) (2) H (t) and x H (t) of the homogeneous difference equation x t+1 + ax t + bx t 1 = 0. that is two solutions for which Ax (1) (2) H (t) + Bx H (t) 0 implies that the two scalars A and B satisfy A = B = 0.

10 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 10 of 56 Two Linearly Independent Solutions Note that Aλ t 1 + Bλt 2 = 0 for both t = 0 and t = 1 if and only if ( ) ( ) ( ) 1 1 A 0 = λ 1 λ 2 B 0 This has a non-trivial solution in the two constants A and B iff 0 = 1 1 λ 1 λ 2, or if and only if 0 = λ 2 λ 1. So when λ 1 λ 2, the only solution is trivial, with A = B = 0. Hence, the two functions x 1 t = x 0 λ t 1 and x 2 t = x 0 λ t 2 with x 0 0 are linearly independent solutions of x t+1 + ax t + bx t 1 = 0. There are two degrees of freedom in the difference equation. Therefore, its general solution with these two degrees of freedom is x t = Aλ t 1 + Bλt 2 for arbitrary real constants A and B.

11 Example: The Fibonacci Sequence The Fibonacci sequence is (x t ) t=0 = (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233,...) It is the unique solution with x 0 = 0 and x 1 = 1 of the Fibonacci difference equation x t+1 x t x t 1 = 0. The characteristic equation is λ 2 λ 1 = 0, with characteristic roots λ 1,2 = 1 2 ( 1 ± 5). Its two roots are: the golden ratio ϕ := λ 1 = 1 2 (1 + 5) ; and λ 2 = 1 λ 1 = 1 2 (1 5) The general solution of the Fibonacci difference equation is x t = Aλ t 1 + Bλt 2 for arbitrary constants A and B. To obtain the Fibonacci sequence with x 0 = 0 and x 1 = 1 requires B = A and 1 = A(λ 1 λ 2 ) = A 5, so B = A = Hence x t = t [ (1 + 5) t (1 5) t], so x t N. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 11 of 56

12 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 12 of 56 Case 2: Two Complex Conjugate Roots In this case λ 2 + aλ + b = (λ re iθ )(λ re iθ ) where a = re iθ + re iθ = r(cos θ + i sin θ) + r(cos θ i sin θ) = 2r cos θ and b = (re iθ )(re iθ ) = r 2 with sin θ 0. It follows that a 2 4b = 4r 2 cos 2 θ 4r 2 = 4r 2 sin 2 θ < 0. Note that r = ( a ) ( ) b and θ = arccos = arccos 1 2r 2 a b 1 2.

13 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 13 of 56 Case 2: Oscillating Solutions In the complex plane C, two possible solutions of the difference equation x t+1 + ax t + bx t 1 = 0 with x 0 0 are x (1) t = x 0 (re iθ ) t = x 0 r t e iθt = x 0 r t (cos θt + i sin θt) and x (2) t = x 0 (re iθ ) t = x 0 r t e iθt = x 0 r t (cos θt i sin θt) In the real line R, two possible solutions are x (1) t = r t cos θt and x (2) t = r t sin θt These are linearly independent because x (1) 0 x (2) 0 = 1 0 r cos θ r sin θ = r sin θ 0 x (1) 1 x (2) 1 The general solution is therefore x t = r t (A cos θt + B sin θt) for arbitrary real constants A and B, where A = x 0.

14 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 14 of 56 Case 3: Two Coincident Roots In this case λ 2 + aλ + b = (λ λ) 2, where b = λ 2 with a 2 4b = 0. Consider the perturbed equation x t+1 + ax t + bx t 1 = 0 where ã = 2 λ and b = λ 2 ɛ 2 with ɛ a small positive number. We consider the behaviour of its general solution as ɛ 0. The auxiliary equation λ 2 + aλ + b = 0 can be written as λ λλ + λ 2 ɛ 2 = 0. Note that λ λλ + λ 2 ɛ 2 = (λ + λ + ɛ)(λ + λ ɛ). So the perturbed auxiliary equation has the two real roots λ = λ ± ɛ.

15 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 15 of 56 The Solution with Fixed Initial Conditions Fix x 0 and x 1. The general solution satisfying x 0 = x 0 and x 1 = x 1 is x t = A( λ + ɛ) t + B( λ ɛ) t where x 0 = A + B and x 1 = A( λ + ɛ) + B( λ ɛ) = (A + B) λ + (A B)ɛ. Hence A + B = x 0 and A B = (1/ɛ)( x 1 x 0 λ), implying that A = 1 2 [ x0 + (1/ɛ)( x 1 x ] 0 λ) and B = 1 2 [ x0 (1/ɛ)( x 1 x ] 0 λ). The solution for fixed ɛ is therefore x ɛ t = 1 2 [ x0 + (1/ɛ)( x 1 x 0 λ) ] ( λ + ɛ) t [ x0 (1/ɛ)( x 1 x 0 λ) ] ( λ ɛ) t which can be rewritten as xt ɛ = 1 2 x [ 0 ( λ + ɛ) t + ( λ ɛ) t] ( x 1 x [ 0 λ) ( λ + ɛ) t ( λ ɛ) t] /ɛ

16 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 16 of 56 The Limiting Solution as ɛ 0 The limit of x ɛ t as ɛ 0 takes the form x 0 λ t ( x 1 x 0 λ) lim ɛ 0 [ ( λ + ɛ) t ( λ ɛ) t] /ɛ To evaluate the last limit, apply l Hôpital s rule to obtain lim ɛ 0 [ ( λ + ɛ) t ( λ ɛ) t] /ɛ = lim ɛ 0 [ t( λ + ɛ) t 1 + t( λ ɛ) t 1] /1 = 2t λ t Two linearly independent possible solutions of the difference equation x t+1 + ax t + bx t 1 = 0 with x 0 0 are x (1) t = x 0 λ t and x (2) t = x 0 tλ t. There are two degrees of freedom in the difference equation. Its general solution is x t = (C + Dt)λ t for arbitrary real constants C and D.

17 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 17 of 56 A Simpler Approach, I We are trying to solve the homogeneous second-order difference equation with a repeated root λ, taking the form x t+1 2λx t + λ 2 x t 1 = 0 We know that one solution is x t = x 0 λ t for arbitrary x 0. To find a second linearly independent solution that we know must exist, try putting x t = λ t y t. Substituting into the original equation gives λ t+1 y t+1 2λ t+1 y t + λ t+1 y t 1 = 0 Disregarding the trivial case when λ = 0, one has y t+1 2y t + y t 1 = 0.

18 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 18 of 56 A Simpler Approach, II To solve y t+1 2y t + y t 1 = 0, try introducing yet another new variable z t = y t+1 y t. This leads to the new difference equation z t z t 1 = 0 whose solution is obviously z t = z 0 for all t = 1, 2,.... Then y t+1 y t = z 0 for all t, implying that y t = y 0 + z 0 t. It follows that x t = λ t y t = (y 0 + z 0 t)λ t. To conclude, two solutions are x (1) t = λ t and x (2) t = tλ t. These are linearly independent because x (1) 0 x (2) 0 = 1 0 λ λ = λ 0 x (1) 1 x (2) 1 The general solution is therefore x t = (A + Bt)λ t for arbitrary real constants A and B, where A = x 0.

19 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 19 of 56 Lecture Outline Solving Second-Order Equations Inhomogeneous Equations Two Special Cases Method of Undetermined Coefficients Higher-Order Linear Equations with Constant Coefficients Stationary States and Stability for Second-Order Equations

20 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 20 of 56 From Particular to General Solutions The homogeneous equation with constant coefficients takes the form x t+1 + ax t + bx t 1 = 0 An associated inhomogeneous equation takes the form x t+1 + ax t + bx t 1 = f t for a general forcing term f t on the RHS. Let xt P denote a particular solution, and xt G any alternative general solution, of the inhomogeneous equation.

21 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 21 of 56 Characterizing the General Solution Our assumptions imply that, for each t = 1, 2,..., one has x P t+1 + ax P t + bx P t 1 = f t x G t+1 + ax G t + bx G t 1 = f t Subtracting the first equation from the second implies that x G t+1 x P t+1 + a(x G t x P t ) + b(x G t 1 x P t 1) = 0 This shows that x H t := x G t x P t solves the homogeneous equation. So the general solution x G t of the inhomogeneous equation x t+1 + ax t + bx t 1 = f t with forcing term f t is the sum x P t + x H t of any particular solution x P t of the inhomogeneous equation; the general solution x H t of the homogeneous equation x t+1 + ax t + bx t 1 = 0.

22 Linearity in the Forcing Term Theorem Suppose that x P t and y P t are particular solutions of the two respective difference equations x t+1 + ax t + bx t 1 = d t and y t+1 + ay t + by t 1 = e t Then, for any scalars α and β, the linear combination z P t := αx P t + βy P t is a particular solution of the equation z t+1 + az t + bz t 1 = αd t + βe t. Proof. Routine algebra. Consider any equation of the form x t+1 + ax t + bx t 1 = f t where f t is a linear combination n k=1 α kft k of n forcing terms. The theorem implies that a particular solution is the corresponding linear combination n k=1 α kxt Pk of particular solutions to the equations x t+1 + ax t + bx t 1 = f k University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 22 of 56 t.

23 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 23 of 56 Deriving an Explicit Particular Solution, I In part A we were able to derive an explicit solution to the general first-order linear equation x t a t x t 1 = f t. Here, for the special case of constant coefficients, we derive an explicit particular solution satisfying x 0 = x 1 = 0 to the general second-order linear equation x t+1 + ax t + bx t 1 = f t. Indeed, suppose that λ 2 + aλ + b = (λ λ 1 )(λ λ 2 ) because λ 1 and λ 2 are the roots (possibly coincident, or possibly complex conjugates) of the auxiliary equation λ 2 + aλ + b = 0. Introduce the new variable y t = x t λ 1 x t 1, implying that y t+1 λ 2 y t = x t+1 λ 1 x t λ 2 x t + λ 1 λ 2 x t 1 = x t+1 (λ 1 + λ 2 )x t + λ 1 λ 2 x t 1 = x t+1 + ax t + bx t 1 = f t

24 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 24 of 56 Deriving an Explicit Particular Solution, II Instead of the second-order equation x t+1 + ax t + bx t 1 = f t, we have the recursive pair of first-order equations x t λ 1 x t 1 = y t and y t+1 λ 2 y t = f t (for t = 1, 2,...) where λ 1 and λ 2 are the roots of λ 2 + aλ + b = 0. Given the initial conditions x 0 = x 1 = 0 and so y 1 = 0, the explicit solutions like those derived in Part A are the sums y t = t 1 2 f k and x t = t k=1 λt k 1 s=2 λt s 1 y s for t = 1, 2,... Substituting the first equation in the second yields the double sum x t = t s 1 s=2 λt s 1 k=1 λs k 1 2 f k which we would like to reduce to x t = t 1 k=1 ξ t k 1f k i.e., a linear combination of the forcing terms (f 1, f 2,..., f t 1 ).

25 Deriving an Explicit Particular Solution, III We begin by introducing the mapping N N (k, s) 1 ks {k < s} {0, 1} defined by { 1 if k < s 1 ks {k < s} := 0 if k s Then we can rewrite x t = t as the double sum x t = t s=2 s=2 λt s 1 Interchanging the order of summation gives x t = t 1 t k=1 s=2 1 ks{k < s}λ t s = ( t 1 t ) s=k+1 λt s k=1 = t 1 k=1 s 1 k=1 λs k 1 2 f k t 1 k=1 1 ks{k < s}λ t s 1 λ s k 1 2 f k. 1 λ s k λ s k 1 2 f k f k ( λ1 t k 1 + λ t k 2 1 λ λ 1 λ2 t k 2 + λ t k 1 2 This reduces to x t = t 1 k=1 ξ t k 1f k where ξ m := m j=0 λm j 1 λ j 2. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 25 of 56 ) f k

26 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 26 of 56 Deriving an Explicit Particular Solution: IV The value of the sum ξ m = m j=0 λm j 1 λ j 2 depends on whether: λ 1 λ 2 in the general case; λ 1 = λ 2 = λ in the degenerate case. In the general case one has (λ 1 λ 2 )ξ m = m ( ) λ m+1 j 1 λ j 2 λm j 1 λ j+1 2 j=0 implying the particular solution x P t = 1 t 1 ( ) λ t k λ 1 λ 2 k=1 1 λ t k 2 f k In the degenerate case one has ξ m = (m + 1)λ m, implying the particular solution = λ m+1 1 λ m+1 2 x P t = t 1 k=1 (t k)λt k f k

27 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 27 of 56 Lecture Outline Solving Second-Order Equations Inhomogeneous Equations Two Special Cases Method of Undetermined Coefficients Higher-Order Linear Equations with Constant Coefficients Stationary States and Stability for Second-Order Equations

28 First Special Case with Distinct Real Roots, I Consider the equation x t+1 + ax t + bx t 1 = f t in the first special case when f t is the power function µ t with µ 0. In the general case when the two roots λ 1 and λ 2 of the auxiliary equation λ 2 + aλ + b = 0 are distinct, the particular solution with x0 P = x 1 P = 0 is x P t = 1 λ 1 λ 2 t 1 k=1 ( λ1 t k λ t k 2 ) µ k But (λ µ) t 1 k=1 λt k µ k = t 1 ( k=1 λ t k+1 µ k λ t k µ k+1), so ( xt P 1 λ t = 1 µ λ 1 µ t λt 2 µ λ 2µ t ) λ 1 λ 2 λ 1 µ λ 2 µ in case µ {λ 1, λ 2 }. Disregarding the terms in λ t 1 and λt 2 that solve the corresponding homogeneous equation, the solution reduces to x P t = αµ t for a suitable constant α. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 28 of 56

29 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 29 of 56 First Special Case with Distinct Real Roots, II The degenerate case when µ {λ 1, λ 2 } is more complicated. In case λ 1 λ 2 = µ, the particular solution with x0 P = x 1 P = 0 is still xt P 1 t 1 ( ) = λ t k λ 1 λ 2 k=1 1 λ t k 2 µ k Because λ 2 = µ, this reduces to x P t = = 1 t 1 ( λ t k λ 1 µ k=1 1 µ k µ t) [ 1 λ t 1 µ λ 1 µ t ] (t 1)µ t λ 1 µ λ 1 µ Disregarding the terms in λ t 1 and in λt 2 = µt that solve the corresponding homogeneous equation, the solution reduces to x P t = αtµ t for a suitable constant α.

30 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 30 of 56 First Special Case with Coincident Real Roots Consider now the degenerate case with coincident real roots λ 1 = λ 2 = λ. So the inhomogeneous equation is x t+1 2λx t + λ 2 x t 1 = µ t. As before, put y t = x t λx t 1 so that y t+1 λy t = x t+1 λx t λx t + λ 2 x t 1 = µ t We consider again the particular solution with x 0 = x 1 = 0 and so y 1 = 0.

31 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 31 of 56 First Special Case with Coincident Real Roots: λ µ Provided that λ µ, for t = 2, 3,... one has yt P = t k=2 λt k µ k 1 = µ(λt 1 µ t 1 ) λ µ and then xt P = t k=2 λt k yk P = t k=2 λt k µ λk 1 µ k 1 = t µλ t 1 λ t k µ k k=2 λ µ = µ(t 1)λt 1 λ µ λt 1 µ 2 µ t+1 (λ µ) 2 λ µ Hence xt P = (α + βt)λ t + γµ t for suitable constants α, β and γ that depend on λ and µ, but not on t. Because (α + βt)λ t is a complementary solution of the homogeneous equation, the particular solution can be reduced to x P t = γµ t.

32 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 32 of 56 First Special Case with Coincident Real Roots: λ = µ In case λ = µ, however, for t = 2, 3,... one has yt P = t k=2 λt k µ k 1 = (t 1)λ t 1 and then xt P = t k=2 λt k yk P = t k=2 λt k (k 1)λ k 1 = t k=2 (k 1)λt 1 = 1 2t(t 1)λt 1 Hence xt P = (αt + βt 2 )λ t for suitable constants α and β that depend on λ = µ, but not on t. Because αtλ t is a complementary solution of the homogeneous equation, the particular solution can be reduced to x P t = βt 2 µ t.

33 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 33 of 56 Second Special Case: Theorem Consider next the equation x t+1 + ax t + bx t 1 = f t in the second special case when f t = t r µ t with µ 0 and r N. As before, let λ 1 and λ 2 denote the roots of the auxiliary equation λ 2 + aλ + b = 0. Theorem The difference equation x t+1 + ax t + bx t 1 = t r µ t has a particular solution of the form xt P = ξ P (t)µ t where ξ P (t) = d j=0 ξ rjt j is a polynomial in t which has degree: d = r in case µ {λ 1, λ 2 }; d = r + 2 in case µ = λ 1 = λ 2 ; d = r + 1 otherwise. We begin the proof by introducing, as before, the new variable y t := x t λ 1 x t 1, implying that y t+1 λ 2 y t = x t+1 λ 1 x t λ 2 x t + λ 1 λ 2 x t 1 = x t+1 + ax t + bx t 1 = t r µ t

34 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 34 of 56 Continuing the Proof By the result in part A, the first-order equation y t+1 λ 2 y t = t r µ t has a particular solution of the form y t = Q(t)µ t, where Q(t) = d j=0 q rjt j is a polynomial in t which has degree: (i) d = r in case µ λ 2 ; (ii) d = r + 1 in case µ = λ 2. By the linearity property of particular solutions, the equation x t λ 1 x t 1 = y t = Q(t)µ t = d j=0 q rjt j µ t has a particular solution x P t = ξ P (t)µ t where x P t = ξ P (t)µ t = d j=0 q rjp j (t)µ t is the appropriate linear combination of the particular solutions x t = P j (t)µ t (j = 0, 1, 2,..., d) of the d + 1 first-order equations x t λ 1 x t 1 = t j µ t.

35 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 35 of 56 Ending the Proof Again, using the result in part A, for each j = 0, 1, 2,..., r, the solution x t = P j (t)µ t of the first-order difference equation x t λ 1 x t 1 = t j µ t involves a polynomial P j (t) in t which has degree: (i) j in case µ λ 1 ; (ii) j + 1 in case µ = λ 1. So the degree of the highest order polynomial P d (t) is (i) d in case µ λ 1 ; (ii) d + 1 in case µ = λ 1. Combined with our previous result on whether d = r or d = r + 1, the degree of ξ P (t) is now easily seen to be r in case µ {λ 1, λ 2 }; r + 2 in case µ = λ 1 = λ 2 ; r + 1 otherwise.

36 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 36 of 56 Lecture Outline Solving Second-Order Equations Inhomogeneous Equations Two Special Cases Method of Undetermined Coefficients Higher-Order Linear Equations with Constant Coefficients Stationary States and Stability for Second-Order Equations

37 First Special Case: A Simpler Approach We have proved that the second-order difference equation x t+1 + ax t + bx t 1 = µ t has a particular solution of the form x P t = αµ t. But there is a much easier way to find x P t, treating the parameter α as an undetermined coefficient. Indeed, for x t = αµ t to be a solution, one needs αµ t+1 + aαµ t + bαµ t 1 = µ t. Dividing each side by µ t 1 yields the equation α(µ 2 + aµ + b) = µ. In the non-degenerate case when µ 2 + aµ + b 0 because µ is not a root of the characteristic equation λ 2 + aλ + b = 0, one has α = µ(µ 2 + aµ + b) 1. Hence, a particular solution is x P t = (µ 2 + aµ + b) 1 µ t+1. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 37 of 56

38 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 38 of 56 Degenerate Case When µ is a Characteristic Root The simple degenerate case occurs when µ 2 + aµ + b = 0 because µ equals one of the two distinct roots λ 1 and λ 2 of the characteristic equation λ 2 + aλ + b = 0. Then we have proved that the second-order difference equation x t+1 + ax t + bx t 1 = µ t has a particular solution of the form x P t = αtµ t. To determine the undetermined coefficient α, we must solve α(t + 1)µ t+1 + aαtµ t + bα(t 1)µ t 1 = µ t Dividing each side by µ t 1 and gathering terms yields the equation αt(µ 2 + aµ + b) + α(µ 2 b) = µ. Provided that µ 2 b, this reduces to α = (µ 2 b) 1 µ.

39 Doubly Degenerate Case When µ 2 = b, however, the degenerate case is more complicated. Indeed, the equation µ 2 + aµ + b = 0 implies that aµ + 2b = 0. Hence µ = 2b/a, so µ 2 = b = 4b 2 /a 2 implying that a 2 = 4b. Then the characteristic equation λ 2 + aλ + b = 0 reduces to (λ µ) 2 = 0, with µ as its repeated root. Inspired by the earlier theorem, we look for a particular solution of the form x P t = αt 2 µ t. To determine the undetermined coefficient α, we must solve α(t + 1) 2 µ t+1 + aαt 2 µ t + bα(t 1) 2 µ t 1 = µ t Dividing each side by µ t 1 and gathering terms yields αt 2 (µ 2 + aµ + b) + α(2t + 1)µ 2 + αb( 2t + 1) = µ Because µ 2 + aµ + b = 0 and b = µ 2, one has α = 1/2µ. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 39 of 56

40 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 40 of 56 Second Special Case Again, inspired by earlier theorems, we can apply the method of undetermined coefficients to the equation x t+1 + ax t + bx t 1 = m k=1 rk j=1 α kjt j µ t k where we naturally assume that the constants µ k (k = 1, 2,..., m) are all different. A particular solution takes the form x P t = m k=1 dk j=1 β kjt j µ t k where the degree d k of each polynomial d k j=1 β kjt j with undetermined coefficients β kj d k j=1 m k=1 is r k in case µ k {λ 1, λ 2 }; r k + 2 in case µ k = λ 1 = λ 2 ; r k + 1 otherwise.

41 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 41 of 56 Determining the Coefficients The coefficients β kj d k j=1 m k=1 of the particular solution x P t = m k=1 dk j=1 β kjt j µ t k can be found (in principle!) by solving, for k = 1, 2,..., m, the m independent systems of linear equations that result from equating coefficients of powers of t in the expansions dk j=1 β kj[(t + 1) j µ 2 k + atj µ t k + b(t 1)j ] = r k j=1 α kjt j µ k

42 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 42 of 56 Lecture Outline Solving Second-Order Equations Inhomogeneous Equations Two Special Cases Method of Undetermined Coefficients Higher-Order Linear Equations with Constant Coefficients Stationary States and Stability for Second-Order Equations

43 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 43 of 56 Higher-Order Linear Equations with Constant Coefficients An nth order linear equation with constant coefficients takes the form x t + n r=1 a r x t r = f t in the inhomogeneous case, and in the homogeneous case. x t + n r=1 a r x t r = 0 The corresponding auxiliary equation is λ n + n r=1 a r λ n r = 0.

44 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 44 of 56 Roots of the Auxiliary Equation The auxiliary equation can be written as p n (λ) = 0 whose LHS is the polynomial λ n + n r=1 a r λ t r of degree n. By the fundamental theorem of algebra, this equation has at least one root λ 1, which may be complex. Then p n (λ) can be factored as p n (λ) (λ λ 1 )p n 1 (λ). But now the equation p n 1 (λ) = 0 also has at least one root λ 2, which may also be complex. Repeating this argument n times, the auxiliary equation p n (λ) = 0 has n roots λ 1, λ 2,..., λ n, some of which may be repeated. In particular, p n (λ) n i=1 (λ λ i).

45 Solving the Homogeneous Equation Theorem Consider the homogeneous equation x t + n r=1 a r x t r = 0, and suppose that the auxiliary equation can be written as 0 = λ n + n r=1 a r λ t r = k j=1 (λ ρ j) m j with k distinct roots ρ j (j = 1, 2,..., k) whose respective multiplicities m j satisfy k j=1 m j = n. Then the general solution of the homogeneous equation takes the form x t = k j=1 mj h=1 α jht h 1 ρ t j for n arbitrary constants α jh (h = 1, 2,..., m j and j = 1, 2,..., k). That is, the general solution is an arbitrary linear combination of the functions t h 1 ρ t j (h = 1, 2,..., m j and j = 1, 2,..., k). University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 45 of 56

46 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 46 of 56 Solving the Inhomogeneous Equation Theorem The general solution of the inhomogeneous equation x t + n r=1 a r x t r = ih=1 qh j=1 α hjt j µ t h is the sum of the general complementary solution of the corresponding homogeneous equation x t + n r=1 a r x t r = 0 and any particular solution. One particular solution takes the form xt P = i dh h=1 j=1 β hjt j µ t h where the degree d h of each polynomial d h j=1 β hjt j with undetermined coefficients β hj d h j=1 i h=1 is q h in case µ h {ρ 1, ρ 2,..., ρ k }; q h + m j in case µ h = ρ j, a root of multiplicity m j.

47 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 47 of 56 Lecture Outline Solving Second-Order Equations Inhomogeneous Equations Two Special Cases Method of Undetermined Coefficients Higher-Order Linear Equations with Constant Coefficients Stationary States and Stability for Second-Order Equations

48 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 48 of 56 Stationary States of a Linear Equation Consider the second-order equation x t+1 + ax t + bx t 1 = f for a constant forcing term f R. Here a stationary state x R has the defining property that x t 1 = x t = x = x t+1 = x. This is satisfied if and only if x + ax + bx = f, or equivalently, if and only if (1 + a + b)x = f. In case a + b = 1, there is: no stationary state unless f = 0; the whole real line R of stationary states if f = 0. Otherwise, if a + b 1, the only stationary state is x = (1 + a + b) 1 f.

49 Stability of a Linear Equation If a + b 1, let us denote by y t := x t x the deviation of state x t from the stationary state x = (1 + a + b) 1 f. Then y t+1 = x t+1 x = ax t bx t 1 f x = a(y t + x ) b(y t 1 + x ) + f x = ay t by t 1 Thus y t solves the homogenous equation x t+1 + ax t + bx t 1 = 0. As already seen, the solution to this homogeneous equation depends on the roots λ 1,2 = 1 2 a ± 1 2 a 2 4b of the quadratic characteristic equation f (λ) λ 2 + aλ + b (λ λ 1 )(λ λ 2 ) = 0 There are three cases to consider: 1. two distinct real roots because a 2 4b > 0; 2. two complex conjugate roots because a 2 4b < 0; 3. two coincident real roots because a 2 4b = 0. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 49 of 56

50 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 50 of 56 Stability Condition With two distinct roots λ 1 and λ 2, real or complex, the general solution of the homogeneous equation is x t = Aλ t 1 + Bλt 2. Stability is satisfied if and only if for all A, B R one has x t 0 as t. This is true if and only if the absolute values of both roots satisfy λ 1 < 1 and λ 2 < 1. With two coincident roots λ 1 = λ 2 = 1 2 a = b, the general solution of the homogeneous equation is x t = (A + Bt)λ t. Again, stability is satisfied if and only if for all A, B R one has x t 0 as t. This is true if and only if the absolute value of the double root satisfies λ < 1.

51 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 51 of 56 Two Distinct Real Roots Here λ 2 + aλ + b = (λ λ 1 )(λ λ 2 ) where λ 1 and λ 2 are both real. Note that the quadratic function f (λ) λ 2 + aλ + b is convex and satisfies f (λ) + as λ ±. So the real roots of f (λ) = 0 satisfy λ 1 < 1 and λ 2 < 1 iff f ( 1) > 0 and f (1) > 0 with f ( 1) < 0 and f (1) > 0 These conditions are equivalent to 1 a + b > 0 and 1 + a + b > 0 with 2 + a < 0 and 2 + a > 0 or to a < 2 and a < 1 + b. Together with the condition a 2 > 4b for the equation f (λ) = 0 to have two distinct real roots, these inequalities are equivalent to a 1 < b < 1.

52 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 52 of 56 Two Complex Conjugate Roots With two complex conjugate roots because a 2 4b < 0, one has λ 1,2 = 1 2 a ± 1 2 i 4b a 2 = r e±iθ = r(cos θ ± i sin θ) where r = b and θ = arccos(a/2 b) Then the general solution of the homogeneous equation can be written as x t = r t (A cos θt + B sin θt). Stability is satisfied if and only if for all A, B R one has x t 0 as t. This is true if and only if b < 1, as well as a 2 4b < 0.

53 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 53 of 56 A Repeated Real Root With two coincident real roots both equal to 1 2 a, the general solution of the homogeneous equation is x t = (A + Bt)( 1 2 a)t. Stability is satisfied if and only if for all A, B R one has x t 0 as t. This is true if and only if the modulus of the repeated root λ = 1 2a satisfies λ < 1, and so if and only if a < 2.

54 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 54 of 56 A Simpler Test Theorem The linear autonomous equation x t+1 + ax t + bx t 1 = f is stable, both locally and globally, if and only if a < 1 + b < 2. Proof. Stability requires one of the following three to hold: 1. distinct real roots because a 2 > 4b, with a 1 < b < 1; 2. complex conjugate roots because a 2 < 4b, with b < 1; 3. a repeated real root because a 2 = 4b, with a < 2. A diagram in the (a, b)-plane shows that one of these three holds if and only if a < 1 + b < 2.

55 Diagram of Stable Region The stable region occurs where a 1 < b < 1, in the interior of an isosceles right-angled triangle with corners at (a, b) = (0, 1) and (a, b) = (±2, 1). University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 55 of 56

56 A Variable Forcing Term Consider now the second-order equation x t+1 + ax t + bx t 1 = f t for a variable forcing term f t. The general solution takes the form x G t = x H t + x P t where: x P t is one particular solution of x t+1 + ax t + bx t 1 = f t ; is any one of a two-dimension continuum of solutions of the homogeneous equation x t+1 + ax t + bx t 1 = 0. x H t The stability condition a < 1 + b < 2 is necessary and sufficient for any solution of the homogeneous equation to satisfy x H t 0 as t. It is therefore also necessary and sufficient for the difference between any two solutions x (1) t and x (2) t of the inhomogeneous equation x t+1 + ax t + bx t 1 = f t to satisfy x (1) t x (2) t 0 as t. In the long run, this means that there is an asymptotically unique solution to x t+1 + ax t + bx t 1 = f t. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 56 of 56