DYNAMIC LECTURE 1 UNIVERSITY OF MARYLAND: ECON 600

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1 DYNAMIC LECTURE 1 UNIVERSITY OF MARYLAND: ECON 6 1. differential Equations Basic Concepts for Univariate Equations. We use differential equations to model situations which treat time as a continuous variable. An ordinary differential equation is an expression which describes a relationship between a function of one variable and its derivatives. Definition 1. (Ordinary differential Equation). An ordinary differential equation is an equation of the form: (1.1) x m (t) = F [t, x(t), x 1 (t), x 2 (t),..., x m 1 ; α] where x j (t) = dj x(t) d j t and α Ω R p is a vector of parameters, and F is a function R m+1+p R Note: The solution is a function x(t) that, together with its derivatives, satisfies Equation 1.1 for given values of the parameters. We call 1.1 an ordinary differential equation because x is a function of one argument, t, only. If it was a function of more than one variable, partial derivatives would appear and we would have a partial differential equation. We will only study ordinary differential equations. A differential equation is linear if F is linear in x(t) and its derivatives. A differential equation is autonomous if t does not appear as an independent argument of F, but enters only through x. The order of a differential equation is the order of the highest derivative of x that appears in it. Thus, 1.1 is an m-th order differential equation. Note: Any differential equation can be reduced to a first-order differential equation system by introducing additional variables. For example, given the third-order differential equation Define: x 3 (t) = ax 2 (t) + bx 1 (t) + x(t) y(t) = x 1 (t) z(t) = x 2 (t) Date: Summer Notes compiled on August 25, 213 1

2 DYNAMIC LECTURE 1 2 Then: and we have the first-order system: z 1 (t) y 1 (t) x 1 (t) = y 1 (t) = x 2 (t) = z(t) z 1 (t) = x 3 (t) az(t) + by(t) + cx(t) z(t) y(t) Definition 2. (Particular Solution, General Solution). A particular solution to 1.1 is a differentiable function x(t) that satisfies 1.1 for some subinterval I of the domain of definition of t, I. The set of all solutions is called the general solution, x g (t). To see that the solution to a differential equation is generally not unique, consider the following example: (1.2) x 1 (t) = 2x(t) One solution to Equation 1.2 is x(t) = e 2t. But for any constant c, x(t) = ce 2t is also a solution. The non-uniqueness problem can be overcome by augmenting the differential equation with a boundary condition of the form: x(t ) = x Definition 3. (Boundary Value Problem). A boundary value problem is defined by a differential equation (1.3) and a boundary condition (1.4) x 1 (t) = f[t, x(t)] x(t ) = x, (x, t ) X I The following theorem says that, under certain conditions, every boundary value problem has a unique solution. Theorem 4. (The Fundamental Existence-Uniqueness Theorem). Let F be C 1 in some neighborhood of (x, t ). Then in some subinterval I of I containing t there is a unique solution x(t) to the boundary value problem given by Equations 1.3 and 1.4. Example 5. If F is not C 1 in some neighborhood of (x, t ), the solution may not be unique. Consider the boundary value problem: (1.5) (1.6) It is easily verified that both x 1 (t) = 3x(t) 2/3, x, t R x() = x(t) = t 3 x(t) = are solutions to Equations 1.5 and 1.6. Note that f(x) = 3x(t) 2/3 is not differentiable at x =.

3 DYNAMIC LECTURE 1 3 Figure 1.1. A phase diagram Figure 1.2. No steady state exists Example 6. The solution may not exist globally. Consider the boundary value problem: (1.7) (1.8) x 1 (t) = x(t) 2, x, t R x() = 1 It is easily verified that x(t) = 1 1 t satisfies Equations 1.7 and 1.8, but it is not defined at t = 1. We may be often interested in analyzing the long-run properties of a differential equation, in particular, the properties of its steady state (long-run equilibrium) and whether the solution eventually converges to the steady state (stability). We can do such analysis without having to find an explicit solution of the differential equation Steady States and Stability. Definition 7. (Steady State). Consider the autonomous differential equation x 1 (t) = f[x(t)] f : X R R A steady state is a point x X such that f( x) =. (See DLF page 41) We can use phase diagrams (as in Figure 1.1) to illustrate this definition: A steady state may not exist (Figure 1.2): The steady state need not be unique (Figure 1.3): Consider an equation that is initially at rest at an equilibrium point x, and suppose that some shock causes a deviation from x. We may want to know if the equation will return to the steady state, or at least remain close to it, or if it will get farther and farther away from it over time.

4 DYNAMIC LECTURE 1 4 Figure 1.3. Steady state is not unique Figure 1.4. Stable and unstable steady states Figure 1.5. The steady state is globally asymptotically stable Definition 8. (Stability). Let x be an isolated (or locally unique) steady state of the autonomous differential equation x 1 (t) = f[x(t)], f : X R R We say that x is stable if for any ɛ >, δ (, ɛ] such that x(t ) x < δ x(t) x < ɛ t t That is, any solution x(t) that at some point enters a ball of radius δ around x remains within a ball of (possibly larger) radius ɛ. Definition 9. (Asymptotic Stability). A steady state is asymptotically stable if it is stable (as in the previous definition) and δ can be chosen in such a way that any solution that satisfies x(t ) x < δ for some t, will also satisfy lim t x(t) = x. That is, any solution that gets sufficiently close to x not only remains nearby but also converges to x as t. x 1 (t) > implies that x(t) is increasing (the arrows of motion point to the right), and x 1 (t) < implies that x(t) is decreasing (the arrows of motion point to the left). Then: In Figure 1.4, x 1, x 3 are locally asymptotically stable while x 2 is unstable In Figure 1.5, x 1 : globally asymptotically stable. We can conclude that if for all x in some neighborhood of a steady state x:

5 DYNAMIC LECTURE 1 5 x(t) < x x 1 (t) > and x(t) > x x 1 (t) < then x is asymptotically stable x(t) < x x 1 (t) < and x(t) > x x 1 (t) > then x is unstable. Therefore, we can determine the stability property of a steady state by checking the sign of the derivative of f[x(t)] at x: x is (locally) asymptotically stable if x is unstable if What happens if f [x(t)] x(t)= x =? f [x(t)] x(t)= x < f [x(t)] x(t)= x > If x(t) B ɛ ( x) x 1 (t) = then we have a flat area that is asymptotically stable. If x(t) B ɛ ( x) x 1 (t) (ie on both sides of x, x 1 (t) has the same positive/negative sign) then x is unstable. This suggests that we can study the stability properties of a differential equation from the sign of f [x(t)] x(t)= xi as long as f [x(t)] x(t)= xi. Definition 1. (Hyperbolic Equilibrium). Let x be a steady state of the autonomous differential equation: (1.9) x 1 (t) = f[x(t)], f : X R R, f C 1 We say that x is a hyperbolic equilibrium if f [x(t)] x(t)= x. The previous analysis suggests that we can study the stability properties of a nonlinear differential equation by linearizing it, as long as the equilibrium is hyperbolic. Theorem 11. (Grobman-Hartman Theorem). If x is a hyperbolic equilibrium of the autonomous differential Equation 1.9, then there is a neighborhood U of x such that Equation 1.9 is topologically equivalent to the linear equation (1.1) in U. x 1 (t) = f [x(t)] x(t)= x [x(t) x] Note that 1.1 is a first-order Taylor series approximation of f around x: f[x(t)] = f( x) + f [x(t)] x(t)= x [x(t) x] + R[x(t) x] where (and since f( x) = ). R[x(t) x] lim x(t) x x(t) x The theorem says that near a hyperbolic equilibrium x, the nonlinear differential equation (Equation 1.9) has the same qualitative structure as the linearized differential equation (Equation 1.1). In particular, if x is (locally) asymptotically stable for Equation 1.1, then it is locally asymptotically stable for Equation 1.9, and if it is unstable for Equation 1.1, then it is unstable for Equation 1.9.

6 DYNAMIC LECTURE Application: The Solow Growth Model. Output (Y ) is produced using capital (K) and labor (L) according. The production function takes the form: (1.11) Y (t) = F [K(t), L(t)] F is C 1, it has constant returns to scale and positive and diminishing marginal products with respect to each input. Using the constant returns to scale property, we can rewrite output as: Y (t) = F [K(t), L(t)] = L(t)F [K(t)/L(t), 1] = L(t)f[k(t)] where k(t) = K(t)/L(t). then (1.12) y(t) = Y (t) L(t) = f[k(t)] That is, we can write output per unit of labor as a function of capital per unit of labor. We also assume that the so-called Inada conditions hold: (1.13) f() =, lim f[k(t)] =, f () =, k(t) lim f [k(t)] = k(t) The economy is closed, thus savings equals investment, and a constant fraction s of output is saved. Assume that there is no depreciation of capital. Then: (1.14) K 1 (t) = I(t) = sf [K(t), L(t)], s [, 1] where I denotes investment. Assume that labor grows at a constant, exogenous rate n: (1.15) Then: (1.16) and thus (1.17) Using Equations 1.17 and 1.14 we get: (1.18) L(t) = L()e nt = L e nt K(t) = L(t) L(t) K(t) = k(t)l e nt K 1 (t) = k 1 (t)l e nt + k(t)nl e nt k 1 (t) = sf[k(t)] nk(t) We can interpret Equation 1.18using the graph in Figure 1.6: The steady state: (1.19) k 1 (t) = sf( k) = n k k f( k) = s n The capital-output ratio: K/ Ȳ = k/f( k) is constant, and the capital stock and output in steady state grow at the rate of growth of the population, n. Stability: k 1 (t) > for k(t) < k (since sf[k(t)] > nk(t)) k 1 (t) < for k(t) > k (since sf[k(t)] < nk(t)) Therefore, the steady state is globally asymptotically stable.

7 DYNAMIC LECTURE 1 7 Figure 1.6. The Solow growth model 1.4. Solving Autonomous differential Equations. Consider the following firstorder autonomous linear differential equation: (1.2) x 1 (t) = ax(t) + b All solutions to Equation 1.2 can be written as: (1.21) x g (t) }{{} = x c (t) }{{} + x p (t) }{{} general solution complementary f unction particular solution The complementary function x c (t) is the general solution to the homogenous equation associated with Equation 1.2: (1.22) x 1 (t) = ax(t) and x p (t) is any particular solution to the full non-homogeneous equation 1.2. We can use the method of separation of variables to solve Equation Rewrite Equation 1.22 as: (1.23) dx(t) = ax(t) dx(t) = ax(t) dx(t) x(t) = a

8 DYNAMIC LECTURE 1 8 and integrating both sides of Equation 1.23 we obtain: ˆ ˆ 1 x(t) dx(t) = a (1.24) ln x(t) = at + c 1 e ln x(t) = e at e c1 x c (t) = ce at, c = e c1 Now, to verify that Equation 1.21 holds, we want to show that if x p (t) is any solution to 1.2, then ce at + x p (t) also satisfies Equation 1.2: d [ceat + x p (t)] = ace at + dxp (t) = ace at + ax p (t) + b = a[ce at + x p (t)] + b Thus, if we add ce at to any solution, we get another solution. Moreover, we get all solutions in that way: Let x p (t) and x g (t) be two arbitrary solutions to Equation 1.2. Then dx g (t) (1.25) = ax g (t) + b dx p (t) (1.26) = ax p (t) + b Subtracting Equation 1.26 from Equation 1.25 and defining y(t) = x g (t) x p (t), we obtain: dy(t) = ay(t) with general solution: y(t) = ce at Therefore, the difference between any two arbitrary solutions has the form ce at, and thus we can get all solutions by adding ce at to x p (t). To solve Equation 1.2, we still need to find x p (t). The simplest alternative is usually a steady state solution: Therefore: x 1 (t) = ax(t) + b = x p (t) = b a, if a (1.27) x g (t) = ce at b a, if a and (1.28) x g (t) = bt + c, if a = (Show the last equality) We can pin down the arbitrary constant by means of a boundary condition. Suppose we have: (1.29) x() = x

9 DYNAMIC LECTURE 1 9 Then, if a, x() = c b a c = x + b a and the solution to the boundary value problem given by Equations 1.2 and 1.29 is: ( x g (t) = x + b ) (1.3) e at b a a, if a and (1.31) x g (t) = bt + x, if a = What are the stability properties of Equation 1.3? If a >, then e at as t and x(t) explodes unless x = b a If a <, then e at as t and x(t) is asymptotically stable (it converges to the steady state x = b a, independently of the initial position) Solving Non-Autonomous differential Equations. Consider the firstorder non-autonomous linear differential equation: (1.32) x 1 (t) = a(t)x(t) + b(t) where the coefficients a and b are functions of time. Rewrite Equation 1.32 as: (1.33) x 1 (t) a(t)x(t) = b(t) Consider the function e α(t), with α(t) = t a(s)ds. Multiplying both sides of Equation 1.33 by e α(t), we have: (1.34) x 1 (t)e α(t) a(t)x(t)e α(t) = b(t)e α(t) Note that the left hand side of Equation 1.34 is the derivative of x(t)e α(t). Therefore, Equation 1.34 can be rewritten as: (1.35) d ( x(t)e α(t)) = b(t)e α(t) The expression e α(t) that made the left hand side of Equation 1.34 the exact derivative of a function is called an integrating factor. We can use Equation 1.35 to derive two (equivalent) forms of the general solution of Equation 1.32: the backward solution and the forward solution.

10 DYNAMIC LECTURE 1 1 (1.36) (1.37) The backward solution is obtained by integrating Equation 1.35 backward between and s: ˆs ( ) ˆs d x(t)e α(t) = b(t)e α(t) x(t)e α(t) s = x(s)e α(s) x() = ˆs ˆs b(t)e α(t) b(t)e α(t) ˆs x(s) = x()e α(s) + b(t)e α(s) α(t) This expression gives us the value of x(s) as a function of its initial value x() and a weighted sum of past values of the forcing term, b(t). This form of the solution is convenient when there is a natural initial condition for x. Otherwise, the second form of the solution may be more convenient: The forward solution is obtained by integrating Equation 1.35 forward between s and : s ( d x(t)e α(t) ) = s b(t)e α(t) lim t x(t)e α(t) x(s)e α(s) = b(t)e α(t) provided that the limit exists. s x(s) = e α(s) lim x(t)e α(t) b(t)e α(s) α(t) t 1.6. Application: Households Intertemporal Budget Constraint. Suppose that an (inter-generationally) altruistic household has the following budget constraint: (1.38) a 1 (t) = ra(t) + y(t) c(t) where a(t) denotes assets (or wealth) held at t, y(t) is labor income received at t, c(t) denotes consumption expenditure incurred t, and r is the interest rate. We can find the forward solution. Multiplying both sides of Equation 1.38 by e rt and rearranging: (1.39) [a 1 (t) ra(t)]e rt = [y(t) c(t)]e rt Note that the left hand side of Equation 1.39 is the derivative of a(t)e rt. Therefore, Equation 1.34 can be rewritten as: d ( a(t)e rt (1.4) ) = [y(t) c(t)]e rt s

11 DYNAMIC LECTURE 1 11 Suppose that a() is given,. Let s integrate Equation 1.4 forward starting from : Then (1.41) (1.42) a(t)e rt = [y(t) c(t)]e rt a() = [y(t) c(t)]e rt c(t)e rt = a() + y(t)e rt Equation 1.41 is the households lifetime budget constraint. Note: If the interest rate changes over time, we need to replace r by r(t) and the integrating factor becomes exp t r(s)ds, yielding: ˆt (1.43) c(t) exp r(s)ds = a() + y(t) exp as the intertemporal budget constraint (instead of Equation 1.41). ˆt r(s)ds