The Solow model - analytical solution, linearization, shooting, relaxation (complete) (Quantitative Dynamic Macroeconomics, Lecture Notes, Thomas Steger, University of Leipzig). Solow model: analytical solution. Model setup Consider the Solow model with exponential population growth L = L 0 e nt (n 0) and without technical change (A = const.). The equation of motion for capital per capita k := K L reads k = sy- d +n k, where y = Ak a. The reduced form of the Solow is hence given by k = k a - d +n k with k 0 = k 0 () From k = 0 the steady state level of capital is as follows k è = d+n -a Assume that the economy is in a steady state initially and the saving rate increases once and for all. How does the transition from the initial steady state to the new steady state look like? (2) 2. Finding an analytical solution We start with the non-linear differential equation (DE) k = k a - d +n k, which will be transformed into a linear DE. Define z = k -a such that z = -a k -a k fl k = z -a - k a (3) Plugging this into k = k a - d +n k gives z -a - k a = k a - d +n k z = -a - -a d +n k -a z = -a - -a d +n z (4) (5) (6) This is a linear DE which can be solved analytically. Here we employ Mathematica to find the analytical solution.
2 Solow_complete.nb DSolve z' t s A z t, z 0 z0, z t, t FullSimplify z t As t n A s z0 n n The solution can be written as z t = A s n +d + z 0 - A s n +d a-+d t (7) This solution is in line with the following principle: The solution of the DE y + ay= b is given by y = b a + Be-at, where B is a constant of intergration determined by y 0 = y 0 (e.g., Chiang and Wainwright, 2005, Chapter 5). Noting z = k -a we get k t = As n +d + k -a 0 - As n +d a-+d t -a (8) This result can be verified by simply by forming k and plugging this into k = sk a - d +n k timepathk As n k0 n As n n t ; D timepathk, t s A timepathk n timepathk PowerExpand FullSimplify True 3. Plotting time paths We can now describe the evolution of an economy that experiences a once and for all increase in s. The following set of parameters is employed setparameter A, s 0.2, 0.3, 0.02, n 0.0 ; setparameter2 A, s 0.3, 0.3, 0.02, n 0.0 ; The initial steady state level of k is given by k0. setparameter; The time paths k t, y t, c t can then be written in Mathematica syntax as follows
Solow_complete.nb 3 As timepathkexact n k0 As n timepathy A timepathkexact ; timepathc s timepathy; n t ; We finally plot the transition from the initial to the new steady state (notice that the shock is assumed to occur at t = 0) tende 50; p Plot If t 0, k0, timepathkexact. setparameter2, t, 0, tende, AxesLabel t, "k t ", PlotStyle Thickness 0.008 ; p2 Plot If t 0, A k0. setparameter, timepathy. setparameter2, t, 0, tende, AxesLabel t, "y t ", PlotStyle Thickness 0.008 ; p3 Plot If t 0, s A k0. setparameter, timepathc. setparameter2, t, 0, tende, AxesLabel t, "c t ", PlotStyle Thickness 0.008 ; Show Graphicrray p, p2, p3 k t 26 24 22 20 8 c t.85.80.75.70.65 y t 2.6 2.5 2.4
4 Solow_complete.nb 2. Solow model: linearization. Some conceptual remarks Consider the following (non-linear) differential equation (think of the reduced-form Solow model) k = f k with k 0 = k 0 Linearizing this DE around the steady state k = k è by means of a first-order Taylor approximation gives (border conditions omitted) k = è f k è + f k k k - k è where f k is evaluated at k = k è. Notice f k è = 0 by definition recall the steady state condition k = 0. Moreover, k since k = k - k è we can write k - k è = è f k k - k è k () (2) (3) This is a linear and homogenous DE in the variable k - k è. We know that the solution may be expressed as (see, for instance, Chiang, 2004) k - k è = è Be lt (4) where B is an "arbitrary constant of integration" (determined by initial conditions, i.e. k 0 = k 0 ) and l represents an eigenvalue given by l = f k. k 2. Solution of the linearized equation Again, we start with the reduced form of the Solow model given by k = k a - d +n k with k 0 = k 0 Linearizing the above DE around the steady state k è by means of a first-order Taylor approximation gives k - k è = è ak è a- è - d +n k - k This is a linear DE with constant coefficients of the form k - k è k - k è = è Be at > a k - k è, which has the following solution (5) (6) (7) Noting a = a k è a- - d +n and k è = d+n -a we get
Solow_complete.nb 5 a = a d+n a- -a - d +n =a d +n - d +n = a - d +n (8) such that k - k è = è a- d+n t Be The constant of integration B results from k 0 = k 0 at t = 0 which gives k 0 - k è = è Be a- d+n 0 fl B = k 0 - k è (9) (0) The complete solution to the linear DE under study is hence given by k = è k è + k 0 - k è a- d+n t e () 3. Plotting time paths We can now describe the evolution of an economy that experiences a once and for all increase in s. The following set of parameters is employed setparameter A, s 0.2, 0.3, 0.02, n 0.0 ; setparameter2 A, s 0.3, 0.3, 0.02, n 0.0 ; The initial steady state level of k is given by k0. setparameter; The time paths k t, y t, c t can then be written in Mathematica syntax as follows timepathk k0 E t ; timepathy A timepathk ; timepathc s timepathy; Next we plot the time paths of k t, y t, c t
6 Solow_complete.nb tende 50; p2 Plot If t 0, k0, timepathk. setparameter2, t, 0, tende, AxesLabel t, "k t ", PlotStyle Thickness 0.008 ; p22 Plot If t 0, A k0. setparameter, timepathy. setparameter2, t, 0, tende, AxesLabel t, "y t ", PlotStyle Thickness 0.008 ; p32 Plot If t 0, s A k0. setparameter, timepathc. setparameter2, t, 0, tende, AxesLabel t, "c t ", PlotStyle Thickness 0.008 ; Show Graphicrray p2, p22, p32 k t 26 24 22 20 8.85.80.75.70.65 c t y t 2.6 2.5 2.4 Here we compare the exact solution and the linearized solution for k t
Solow_complete.nb 7 pcompare Plot If t 0, k0, timepathk timepathkexact. setparameter2, t, 0, tende, AxesLabel t, "k ex t k lin t ", PlotStyle Thickness 0.008 ; pcompare2 Show p, p2 ; Show Graphicrray pcompare2, pcompare k t k ex t k lin t 26 0.5 24 0.4 22 0.3 20 0.2 8 0. 3. Solow model: "shooting". Preliminaries The DE under study reads k = k a - d +n k with k 0 = k 0 (9) The set of parameters must be specified numerically setparameter A, s 0.2, 0.3, 0.02, n 0.0 ; setparameter2 A, s 0.3, 0.3, 0.02, n 0.0 ; For future uses, let us label the initial stationary solution (steady state) as ksol ksol. setparameter; 2. Applying NDSolve We define the DE to be solved numerically in Mathematica syntax
8 Solow_complete.nb Clear dek ; dek k' t k t k t ; Now the command NDSolve can be applied Clear ns ; ns NDSolve dek. setparameter2, k 0 ksol, k t, t, 0, 255 k t InterpolatingFunction 0., 255., t 3. Plotting time paths In the next step we have to extract and evaluate these interpolating functions to be able to visualize the result. kn t : Evaluate ns,, 2 Here we define y t and c t as functions of k t Clear y, c ; y t : Akn t. setparameter2; c t : s y t. setparameter2; Next we plot the time paths of k t, y t, c t
Solow_complete.nb 9 tende 50; p3 Plot If t 0, ksol, kn t, t, 0, tende, AxesLabel t, "k t ", PlotStyle Thickness 0.008 ; p23 Plot If t 0, A ksol. setparameter, y t, t, 0, tende, AxesLabel t, "y t ", PlotStyle Thickness 0.008 ; p33 Plot If t 0, s A ksol. setparameter, c t, t, 0, tende, AxesLabel t, "c t ", PlotStyle Thickness 0.008 ; Show Graphicrray p3, p23, p33 k t 26 24 22 20 8.85.80.75.70.65 c t y t 2.6 2.5 2.4 4. Solow model: relaxation. Relaxation: a sketch à The Relaxation Algorithm: a sketch "The relaxation method determines the solution by starting with a guess and improving it, iteratively. As the iteration improves the solution, the result is said to relax to the true solution." (Press et al., p. 765) à Starting point: a general ODE Consider the following autonomous, ordinary differential equation system (ODE), expressed in vector notation x t = f x t with f : N Ø N (0)
0 Solow_complete.nb B x 0 = 0 with B : N Ø n F x = 0 with F : N Ø n 2 ; n + n 2 = N () (2) A solution is a function x = x t which maps each tœ into a unique x N such that (0), (), and (2) are satisfied. à Discretization: corresponding algebraic problem The ODE x = f x can then be approximated by a "finite difference equation" (FDE) x v+ - x v = t v+ - t v f x v for all v œ, 2,..., M (3) where x n denotes the value of x at t v for v œ, 2,..., M, t v œ T = t, t 2,..., t M (the time mesh, equal to set of natural numbers in the simplest case), and x v = x v + x v+ 2. A solution of this equation (plus boundary conditions) consists of a set of points x v with v œ, 2,..., M in the state space of dimension N (approximating the solution trajectory). Notice: The initial problem of solving an ODE has been transformed into a problem of solving a set of N M - + N = NM algebraic equations (the above set of equations plus the Ò of finite difference equation Ò of boundary conditions boundary conditions)! à An illustrative example Consider the Solow model (N = ) with three mesh points (M = 3) x ~k; k = f k ; v œ, 2, 3 (4) Between the two mesh points v = 2 and v = we have k 2 - k = 2 - f k + k 2 2 (5) Between the two mesh points v = 3 and v = 2 we have k 3 - k 2 = 3-2 f k 2 + k 3 2 (6) Boundary conditions can be stated as: k = k initial. This represents an algebraic system comprising 3 equations in 3 unknowns. In what follows, we solve such a system for {k, k 2, k 3 } using the FindRoot command. 2. Preliminaries Some settings nx umber of dynamic variables ; The DE is stated as follows
Solow_complete.nb dynequ X, j X, j ; The set of parameters is (shock specification) paraminitial A, s 0.2, 0.3, 0.02, n 0.0 ; paramfinal A, s 0.3, 0.3, 0.02, n 0.0 ; The initial and the final steady state are given by solinitial, solfinal. paraminitial,. paramfinal ; 3. Discretization: algebraic system nn 50;umber of mesh points X, j X, j subslist X, j ; 2 equmain Table X, j X, j dynequ. subslist, j,, nn ; discretization equborder X, 0 solinitial; completesys Join equmain, equborder ; 4. Rootfinding start X, j, X, nn. X, nn solfinal; startvalues Table start, j, 0, nn ; sol FindRoot completesys. paramfinal, startvalues ; Max equations. paramfinal. sol 3.2965 0 5
2 Solow_complete.nb 5. Plots timepathk Table sol i, 2, i,, nxn, nx ; timepathy A timepathk. paramfinal; timepathc s timepathy. paramfinal; p4 ListPlot Join Table solinitial, z,, 0. nn, timepathk, AxesLabel t, "k t " ; p24 ListPlot Join Table A solinitial. paraminitial, z,, 0. nn, timepathy, AxesLabel t, "y t " ; p34 ListPlot Join Table s A solinitial. paraminitial, z,, 0. nn, timepathc, AxesLabel t, "c t " ; Show Graphicrray p4, p24, p34 k t 26 24 22 20 8 50 00 50 t y t 2.6 2.5 2.4 50 00 50 t c t.85.80.75.70.65 50 00 50 t Center manifold for the Solow model solk Solve s ak L x 0, k ; Plot solk,, 2. 0.33, x 0.0, 0.0, s 0.2, L, a, 0, 0, AxesLabel a, k
Solow_complete.nb 3 k 400 300 200 00 2 4 6 8 0 a This is the center manifold, i.e. the continuum of stationary solutions of the dynamic system in terms of (scaleadjusted) k and a. Starting with some initial combination (a 0, k 0 ), the economy converges to a point on this center manifold. Since A = A 0 e x t is exogenous (assumed to grow at constant exponential rate), the economy does not move along the a-dimension, i.e. a always remains constant at a t = a 0. This is implies that knowledge about (a 0, k 0 ) suffices to determine the point on the center manifold that gets ultimately reached.