APrimeronDifferential Equations

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1 APrimeronDifferential Equations Francesco Giavazzi September 2007 Preliminaries: Classification of differential equations Adifferential equation is any equation that contains a differential, or derivative. Differential equations can be ordinary or partial, depending whether they contain ordinary or partial derivatives. They are classified accordingtotheirorder (whether they contain a first derivative, a second derivative or a higher order derivative), whether they are linear or non linear, and whether they are autonomous or non autonomous. 1.1 order The order of a differential equation is determined by the highest order of derivative contained in the equation. An example of a second order differential equation is 3ÿ +2ẏ + y =2 1.2 autonomous, non autonomous Adifferential equation is said to be autonomous if it does not depend on time explicitly; otherwise it is non autonomous. For example, ẏ +5y = t is a non autonomous first-order equation. But ẏ +5y =3 is an autonomous first-order equation. ThesenotesareinspiredbyMaurice Obstfeld s Primer on Differential Equations (unpublished, Spring 1980) which is unfortunately no longer available. IGIER, Università Bocconi 1

2 1.3 linear, non linear Adifferential equation is non linear if it involves any non linear terms in y, ẏ, ÿ, andsoon.itislinearifall the y terms are raised to no power than 1. For example, ẏ + t 2 y =cost is a linear, but non autonomous first-order differential equation, whereas ẏ + y 2 =2 is a non linear, autonomous, first-order differential equation. Before proceeding, it may be useful to recall that a solution to a differential equation is a function that makes the differential equation true. There are usually many solutions to a differential equation. For example, consider the linear, autonomous, first-order differential equation that describes the accumulation of capital in a closed economy without capital depreciation: K = I(t) One solution is Another is The general solution is K(t) =It K(t) =It +1 K(t) =It + C where C is an arbitrary constant. To verify that this is indeed a solution, differentiate it and see that the derivative of K does equal I. Since C can take any value, there is an infinity of solutions to this differential equation. However, if we are given more information, such as the initial value of K, we can determine the value of C. For example, if we are given that K(0) = 3, which means the value of the economy s endowment of capital K at t =0 was 3, then the value of C must be 3. 2

3 PRIMER ON DIFFERENTIAL EQUATION 1. Autonomous systems of differential equations The most general differential equations systems have the form dx/dt = dy/dt = ẋ = Φ (x, y, t) (1) ẏ = Ψ (x, y, t) We shall start by considering the case in which time, t, does not appear explicitly on the right-hand-side. Such differential equations are called autonomous or purely homogeneous Autonomous systems are time-independent, and therefore a stationary state (or steady state, or long run equilibrium) for the variables x and y is defined by the conditions Φ (x, y) = 0 Ψ (x, y) = 0 (2) By shifting the origin, we may take this stationary or long run equilibrium to be the point (0, 0). (2) will in general be a non-linear system. A first pass at studying the properties of such a system consists in linearizing it around the stationary equilibrium and then studying the dynamic properties of the linearized system. What we shall find is an approximation, and will obviously be valid only for small displacements of (x, y) from(0, 0). Linearizing (with a second order Taylor expansion) Φ and Ψ in a neighborhood of the origin we will obtain dx/dt = dy/dt = ẋ = ax + by (3) ẏ = cx + dy where a = Φ x, etc. and all partial derivatives are evaluated at (0, 0). Our goal is to solve the system (3) by expressing x and y as functions of time. Before doing this, it is useful to consider the case of a single differential equation in a single variable. We shall then see how the 2x2 system can be reduced to the simpler case. 3

4 1.1 The single-variable case We are now faced with the simple linear system dx/dt = ẋ = ax (4) Consider a solution of the form x(t) = K exp(at), where K is an arbitrary constant. Differentiating with respect to time we see that dx/dt = ẋ = ak exp(at) =ax(t), so that we have indeed found a solution for (4). Since K = x(0), choosing the arbitrary constant K is equivalent to choosing an initial condition for x. The subsequent evolution of x is then governed by (4). It is important to remember that there are infinitelymanypathscompatible with with a differential equation, and that we can choose a particular path only by choosing an initial condition or starting point for the variables of the system. The analysis of stability is trivial in a single-variable system. If a>0, it is evident that x(t) as t, and so the system fails to attain a stationary state: it is dynamically unstable. When a<0, x(t) 0as t, and so the system is stable with stationary state x = The general case Our strategy is to exploit our knowledge of the solution to (4) in order to solve (3). Writing (3) in matrix form, we have ẋ. y x = A y We notice immediately that if we can find a change in coordinates such that A becomes diagonal, we may reduce the system to one in which our previous result can be used, i.e. to two first-order differential equations such as in (4). We know that we can write A = P ΛP 1,whereΛ is the diagonal matrix whose non-zero elements are the eigenvalues λ 1 and λ 2 of A and P is a matrix whose columns are the eigenvectors of A. Multiplying both sides of (5) by P 1 and defining u v = P 1 x y we obtain.ụ λ1 0 u = v 0 λ 2 v which is the diagonalization we were looking for. (5) (6) 4

5 It is now immediate that the solution to (6) has the form u(t) = K 1 exp(λ 1 t) v(t) = K 2 exp(λ 2 t) where K 1 and K 2 are arbitrary constants. We now need only transform this solution back in terms of x and y, the original state variables. Since P u v 0 = x y 0,wemustlookmorecloselyatP,thematrixofthe eigenvectors of A. To find P,weusethefactthatAP = P Λ. Writing e11 e P = 12 e 21 e 22 we see that ae 11 + be 21 = λ 1 e 11 ce 11 + de 21 = λ 1 e 21 and so the eigenvector 0 e 11 e 21 (which, you should remember, is defined only up to multiplication by a constant, because the rows of the matrix h b of the eigenvectors are not linearly independent) is given by m (λ 1 a) 1 where m 6= 0 is some constant. Observation Note that we could have chosen our eigenvector to be a multiple of [(λ 1 d) /c 1] 0. But as we shall see in a moment, this would make no difference, because any eigenvalue λ must satisfy (λ d) /c = b/ (λ a), which is simply the systems characteristic equation, that is the solution to λ a b det =0 c λ d (end of observation) Returning to the equations that allow us to compute the e ij,wefind with the same steps followed above that 0 e 12 e 22 can be written as h i0 n,wheren is arbitrary. Having found P,wecannowfind b (λ 2 a) 1 u v = P 1 x y 5 i0,

6 so that x(t) y(t) = P u(t) v(t) mb/ (λ1 a) nb/ (λ 2 a) m n K1 exp (λ 1 t) K 2 exp (λ 2 t) which allows us to write the general solution of (3) in the form x(t) = [k 1 b/ (λ 1 a)] exp (λ 1 t)+[k 2 b/ (λ 2 a)] exp (λ 2 t) (7) y(t) = k 1 exp (λ 1 t)+k 2 exp (λ 2 t) where now k 1 = mk 1 and k 2 = nk 2 are arbitrary constants to be determined by the initial conditions. 1.3 Stability of the autonomous systems of differential equations Before proceeding with a quantitative analysis of the solutions, it is common in economic applications to conduct a qualitative analysis, often with a phase diagram. In order to do this, we require two types of information. First, we need to know the steady-state values for y and x (i.e. the values for which ẏ, ẋ = 0). Second, we need to know the arrows of motion of the system around the steady state values, to determine whether or not the system would converge to one of these steady state values. Solution (7) makes clear that the stability of the system (3) i.e. its tendency (or lack of) to converge to a stationary position is crucially linked to the signs of its characteristic roots λ 1 and λ 2.Itiseasytofind these roots. If λ is an eigenvalue, we know that there is an eigenvector e such that A e λ e =0andsoifI is the 2x2 identity matrix, A λi must be singular, i.e. its determinant det (A λi) must be zero. The characteristic equation of (3) can thus be written as λ 2 (a + d) λ +(ad bc) =λ 2 tr(a) + det(a) =0. The two roots of this equation are obviously just the eigenvalues of A. From the quadratic formula we obtain λ 1,λ 2 = tr(a) ± tr(a) 2 4det(A) 1/2 2 (8) 6

7 We must now distinguish between real and complex roots: Real Roots: a) λ 1,λ 2 negative; λ 1,λ 2 positive Let us initially assume that the eigenvalues are real. If the two roots of (8) are negative, it is obvious from (7) that, regardless of the initial conditions imposed, x(t) andy(t) will converge to a stationary equilibrium (0, 0) as t, because exp(λt) 0whenλ is negative. In this case the system is stable, i.e. it will converge to a stationary state from any starting point. From (8) it is immediate that a necessary and sufficient condition for both roots to be negative is that tr(a) < 0 and det(a) > 0. (λ 1 + λ 2 = tr(a); λ 1 λ 2 = det(a)) Following the same reasoning, it appears clear that when λ 1, 2 is positive, the system will fail to converge, diverging from the stationary state from any starting point. b) λ 1 positive, λ 2 negative, or vice-versa If, on the contrary, it happens to be det(a) < 0, we see from (8) that one root will be negative and the other positive. This is the case of saddlepath stability that arises in many economic models characterized by: (i) the assumption that agents form their expectations with full knowledge of the future (the assumption of perfect foresight) (ii) the presence, in the system, of non-predetermined variables, that is variable such as the price of a financial asset, but also consumption that canjumpinreactionto news. What does saddlepath stability entail? For concreteness, assume λ 2 < 0. Then we see that if k 1 6= 0, the system s behavior will sooner or later be dominated by its positive root λ 1, and the state variables will exhibit an explosive behavior. However, if we impose the initial condition k 1 = 0, the system s behavior will depend only on its negative root, and a stationary equilibrium will in the long run be attained. The condition k 1 = 0 implies a relationship between x and y µ (λ2 a) y(t) = x(t) (9) b 7

8 The locus described by (9) which is derived from (7) is the stable branch (or saddlepath, or stable manifold) of the system. It is the unique path in the (x, y) plane which converges to (0, 0). A typical saddlepath configuration is illustrated in Figure (1). The condition k 1 = 0 is typically imposed in macroeconomic models. Notice that if one of the two variables in the system is non-predetermined, (9) is necessary to uniquely determine the equilibrium at any instant t. This is because, at any t, one variable (the pre-determined variable, e.g. thestock of capital) will be given by history, but the other variable (e.g. the price of capital, which is non pre-determined, i.e. not given by history) needs an additional condition to be determined. Absent (9), the equilibrium of the system at any t would not be determined. Animportantcaseinwhichwedo not impose k 1 = 0 is when we study the response of the economy to anticipated future changes in some exogenous variables for instance when we ask how does consumption react to the 8

9 announcement of a future increase in government spending. In this case a set of initial conditions naturally emerges from economic reasoning Complex Roots When tr(a) 2 4 det(a) < 0, the characteristic roots are complex numbers. Since the coefficients of the characteristic equation (6) are real, complex roots will always occur as complex conjugate pairs. Hence, if p + iq is one root, the other root must be p iq. For instance, suppose that: [λ 1 ] = [p + iq] [λ 2 ] = [p iq] then, from the fact that x(0) = k 1 b/(λ 1 a)+k 2 b/(λ 2 a) mustbereal, we deduce that k 1 b/(λ 1 a) k 2 b/(λ 2 a) must be complex conjugates, which can be written in polar form as re iθ and re iθ.recalling that x(t) =[k 1 b/ (λ 1 a)] exp (λ 1 t)+[k 2 b/ (λ 2 a)] exp (λ 2 t) and substituting, we have: x(t) =r exp(iθ)exp(p + iq)t + r exp( iθ)exp(p iq)t = = r exp(pt)exp(i(qt + θ)) + r exp(pt)exp( i(qt + θ)) = =2r exp(pt)cos(qt + θ) = =2r exp(pt)(cos(θ)cos(qt) sin(θ)sin(qt)). Rewriting it, for clarity s sake: x(t) = 2r exp(pt)(cos(θ)cos(qt) sin(θ)sin(qt)). (10) On the other hand, substituting λ 1 = p + iq in re iθ = k 1 b/(λ 1 a), and isolating k 1,wehavek 1 = re iθ (p a + iq)/b, andsymmetrically k 2 = re iθ (p a iq)/b. Recalling that: y(t) =k 1 exp (λ 1 t)+k 2 exp (λ 2 t) 9

10 and proceeding as above, we can show that: y(t) =2r exp(pt)([p a)/b cos(θ) c/b sin(θ)] cos(qt) [(p a)/b sin(θ)+q/b cos(θ)] sin(qt)). (11) It is evident that the system x(t), y(t) will exhibit damped oscillations if the real part p of its roots is negative, and will be unstable otherwise. Choice of r and θ yields the two initial conditions needed to describe the path of the system. It should be noted that the stability criterion developed above ( ẋ =0,ẏ = 0 ) remains valid. 1.4 Summary The steady-state solutions to an autonomous system of differential equations are said to be stable if the system converges to the steady state solutions and unstable otherwise. We have found that the stability characteristics of differential equations depend on the sign of the characteristic roots. Roots with negative real parts are associated with differential equations that converge to the steady state (stable); roots with positive real parts are associated with differential equations that diverge from the steady state (unstable). The stability of a system of differential equations also depends on the signs of the roots of the characteristic equation. The following theorem recaps the conditions of convergence: THEOREM The steady-state solution of a system of linear, autonomous differential equations is asymptotically stable if and only if the characteristic roots are negative (the real part is negative in the case of complex-valued roots). In economic models, it is common to obtain a system of differential equations in which one of the characteristic roots is positive and the other one is negative. In this case, the steady-state equilibrium is called a saddlepoint equilibrium, and it is unstable. However, y(t) andx(t) convergeto their steady-state solutions if the initial conditions satisfy k 1 =0,which implies y(t) = ³ (λ2 a) b x(t). The locus of points (y, x)defined by this equation is known as the saddle path. 10

11 2. Non autonomous linear systems We shall now consider systems of general form ẋ = [ax + by + v(t)] (12) ẏ = [cx + dy + w(t)] As t is included explicitly, then this is a non-autonomous system. To start, it is again useful to look at the case of a single equation in one variable: ẋ =[ax + v(t)] (13) where the term (ν) is function of time, and t is included explicitly Thesinglevariablecase We refer to ax as the homogenous part of the above equation and to v(t) astheinhomogenous part. Aswehaveseen,ageneralsolutionto the homogenous part is given by x(t) =K exp(at). To solve this form, we guess a solution of form: where f(t) is a function to be determined. To find f(t),we differentiate (14), obtaining x(t) =K exp(at)f(t) (14) ẋ(t) = K exp(at)f 0 (t)+ak exp(at)f(t) = = ax(t)+v(t) (provided that K exp(at)f(t) = x(t) ). Hence v(t) = K exp(at)f 0(t), or f 0(t) =(1/K)exp( at)v(t). As the reader well knows, the second Fundamental Theorem of Calculus holds for any f, continuous on an open interval I, andanya point in I, stating that if F is defined by 11

12 then F (x) = Z x a f(t)dt F 0 (x) =f(x) at each point in I. In our case, being f 0 (t) =(1/K)exp( at)v(t), we may rewrite : Z ½ (1/K) exp( as)v(s)ds + C 1 a > 0 t f(t) = Z t (1/K) exp(as)v(s) ds + C 2 a<0 ¾ where a > 0 is often referred to as the forward solution, and a<0 viceversa. Using this result, we easily obtain the solution for the equation (12): ½ k exp(at) x(t) = k exp(at)+ Z t Z t where k is given by an initial condition. exp(a(t s))v(s)ds a > 0 ¾ exp(a(t s))v(s)ds a < 0 We have made our choice of solution depending on the sign of a: a positive a denotes a forward solution, as we have already noticed. This dependency implies that we are implicitly assuming that x(t) should be finite for all functions v(t) that do not grow too quickly with time. For example, if we were to choose the firsttypeofsolution(the forwardlooking solution), when a<0, the integral term will fail to converge. We illustrate this last remark with an example. 12

13 Example 1: the Cagan hyperinflation model In his classic paper, Cagan (1956) studied seven hyperinflation. Hyperinflation is defined as a period during which the price level of goods in terms of money rises at a rate averaging at least 50% per month. We consider the Cagan hyperinflation model of money demand. Let M denote a country s money supply, and P its price level. Demand for real money balances M P depends entirely on expected future price-level inflation: once expected inflation rises, then M P decreases, by rising the opportunity cost of holding money. If we write the demand for real balances in the log linear form, we have m d p = aπ (a >0) where m log M, p log P and π ṗ. The relation above is a first order differential equation, which explains price level dynamics in terms of money supply (m d = m), taking money supply as an exogenous variable. (E1.1 rational expectations) If we close the model by assuming rational expectations, E(π) = ṗ, we find that the price level is governed by the non-autonomuos d.e.: ṗ =(1/a)p (1/a)m(t) Utilizing the previous findings, a solution for the price level must have the form: p(t) =k exp((1/a)t)+(1/a) Z t exp((1/a)(t s))m(s)ds As in our earlier discussion, it is reasonable to impose the initial condition k = 0, which prevents the behavior of p(t) from being dominated by the positive root 1/a. It is also the only initial condition consistent with the sensible requirement that p(t) = m when m(s) = m for all s > t. The economic reason to set k = 0 is that allows us to rule out speculative bubbles. We thus obtain Z p(t) =(1/a) exp((1/a)(t s))m(s)ds t 13

14 The price level consists simply in an esponentially-weighted average of all future money stocks, a solution consistent with the hypotesized forwardlooking behavior of agents in rational expectations models. Looking closer, the remaining integral term shows that the price level depends on a discounted value of future money supply m with weight summing to one. This implies that money is fully neutral: changing the level of the money supply or the nominal unit of account by the same proportion on all dates leads to an immediate equal proportional change in the price level. (This feature is characteristic of all models that lack nominal rigidities or money illusion). (E1.2 adaptive expectations) We might alternatively close the Cagan model by allowing expectations to evolve according to the adaptive formula π = b(ṗ π) = bπ + bṗ (b >0) Now the system s root ( b) is negative, and so the backward solution is the convergent one. Under adaptive expectations, as we shall see in a moment, agents take the past alone into account in setting today s price level. The solution for the expected inflation rate is then: Z t π(t) =m(t)+ab exp( b(t s))p(s)ds Our convergence assumption allows today s price level to depend only on today s money stock and the past history of inflation. This is no surprise. The reader should be aware, however, that there exist rational expectations models in which only the backward-looking solution converges, so that it is rational to ignore the future in setting prices. The implication is that the convergence assumption should not be taken lightly. Example 2: Capital growth in an economy with technological change (from Hoy, Livernois, McKenna, Rees, Stengos, 2001) Consider a simple model of an economy that produces output using only capital. The amount produced is given by the production function which, we assume, shifts over time because of technological change. Letting y denote output, and k denote capital, we assume that the technical relationship is 14

15 y =(a + ak)t 1/2 This says that output is a linear function of the amount of capital in the economy k, but that, over time, the entire function increases. We further assume that a constant share s of output is saved, where 0 <s<1. Capital accumulation in the economy is equal to savings. Therefore Rearranging, it gives k = s(a + αk)t 1/2 k sαt 1/2 k = sαt 1/2 We wish to solve this to obtain an expression showing k as a function of t. Thecoefficient is sαt 1/2.Therefore A(t) = which can be integrated to obtain Z 0 sαt 1/2 dt A(t) = 2 3 sαt3/2 An equivalent expression for the differential equation then is d h i e 2sαt3/2 /3 k = sαt 1/2 e 2sαt3/2 /3 dt Integrating both sides gives e 2sαt3/2 /3 k = sa Z 0 t 1/2 e 2sαt3/2 /3 dt + C The integration on the right-hand side can be taken further to get " # e 2sαt3/2 e 2sαt3/2 /3 /3 k = sa sα Solving for k(t) gives k(t) = a α + Ce 2sαt3/2 /3 Assuming an initial condition of k(0) = k 0,wemustset 15

16 C = a α + k 0 After substituting, we get k(t) = a ³ a α + α + k 0 e 2sαt3/2 /3 as the explicit solution for capital in this model. 2.2 The general case To deal with the general case ẋ = [ax + by + v(t)] ẏ = [cx + dy + w(t)] we may once again change coordinates so that we have a diagonal system and then invoke the solution just derived. At this point, the reader now has at hand the tools needed to find a general solution Concluding remarks: higher dimension systems We have restricted our attention to systems of dimension 2 or lower. Our techniques obviously extend to higher-dimensional systems, but the algebra becomes tedious. Our methods also handle higher order equations like ẍ = aẋ + bx To solve, we merely define y = ẋ, and then solve the two-dimensional first-order system ẋ = y ẏ = ay + bx Finally, all our results have exact analogues in the discrete-time, difference equation framework favored by many. The interested reader should consult the texts by Sargent. 3. Introducing disturbances in a perfect foresight model 16

17 This paragraph discusses the response to an anticipated future monetary disturbance in Dornbusch s model of exchange-rate dynamics, which is the perfect-foresight extension of the Mundell Fleming model. In the first section we discuss an important continuity condition implicitly imposed whenever we describe the paths available to economies in which agents have perfect foresights. In the following section, we apply this condition to obtain an analytical solution for the path chosen by the economy in response to an announced future increase in the money supply. In the final section, we present an alternative derivation using the original method of Sargent and Wallace. This alternative approach has the advantage of making explicit the effects of the change in expectations that occurs when the announcement is made THE CONTINUITY CONDITION.* Empirical evidence has shown that, during the 70s and the 80s, the industrialized economies in the world saw years of economic instability. If volatile money supplies had amplified effects on exchange rates, Dornbusch reasoned, they might be substantially responsible for the sharp exchange rate fluctuations observed. Buthowdothepricelevelandtheexchangerateinaneconomy respond to a change in the money supply? The Dornbusch overshooting model provides a framework for conducting an interesting analysis of this and related questions. In addition, it provides a classic example of a system of two linear, first-order differential equations for which the steady state is a saddle point equilibrium. We can introduce a simplified version of the Dornbusch exchange-rate dynamics model considering a small open economy, that faces an exogenous world rate. The money demand is described by m D p = λr + ψy where m is the demand of money, p the price level, r the domestic interest rate, y the domestic level of output, and λ and ψ are constant positive coefficients of the money demand function. All lower case are logarithms in the Dornbusch model. As a result, the real supply of money is given by m S = m p, wherem is the exogenous nominal supply of money. Equilibrium in the asset market requires that supply equals demand : 17

18 m p = λr + ψy If this economy were not open to trade with other countries, this equation would determine the equilibrium interest rate, r. However, because the economy is assumed to be open to trade, international capital movements will force the interest rate to equal the world interest rate,r,inthe absence of exchange rate fluctuations. However, if we allow variations in the exchange rate, an interest rate differential equal to the expected rate of depreciation of the home currency can persist. Symbolically this means that r = r + E(ė) where E(ė) is the expected rate of depreciation of the exchange rate. The exchange rate, e, is defined as the domestic price of foreign currency. Note that it is correct to interpret ė as a rate of change or depreciation because e is the logarithm of the actual exchange rate. For example, if the U.S. dollar is expected to depreciate by 1% against other currencies, then the U.S. interest rate will be 1% higher than the world rate in equilibrium. Investors are not willing to exploit the interest differential any further because their earnings are denominated in dollars, which are expected to depreciate by 1%, thereby canceling the higher interest rate. Dornbusch assumes that economic agents have perfect foresight when forecasting exchange rate movements. This means that expected and actual rates of depreciation of the exchange rate are assumed to be equal. This implies that E(ė) =ė and thus the interest parity implies r = r + ė Substituting the interest parity condition in the money market equation, and taking y = r = 0 for simplicity, we get, after some rearranging: ė = m λ + p λ This is one of the two linear differential equations in the model. The second one, which we derive next, describes the dynamics of the domestic 18

19 price level. Whereas we have assumed that the asset market is always in equilibrium (exchange rate and interest rate adjust continuously), we assume that prices adjust sluggishly in response to excess demand. Specifically, we assume that ṗ = δ(y D y S ) δ>0 where y D and y S are the (logarithms of) aggregate quantity demanded and supplied respectively, and δ is the speed-of-adjustment coefficient. Under the assumption that the (logarithm of) supply is fixed at 0, and the demand is given by y D =(e p), we get ṗ = δ(e p) which is the second linear differential equation in the model, and describes the price-level adjustment. The model can be expressed in the dynamic system ṗ δ δ p p = ė 1/λ 0 e ē (15) δ δ Labelling A = 1/λ 0, the determinant of this system is negative: det(a) = δ λ < 0 therefore, its real roots are of opposite sign one another. This means that the economy is governed by a saddle path of equilibrium. The system steady state,as is easily verified, is described by p =ē = m. To draw the isoclines for p and e, and to describe their motion in response to changes of the fundamentals of the economy, we proceed as follows. isoclines ṗ =0 = e = p ė =0 = p = m = p arrows of motion 19

20 p: δṗ/δp = δ <0 (the derivative is negative above the line ṗ =0) δṗ/δe =+δ>0 (the derivative is postive below the line ṗ =0) e: δė/δp =+1/λ > 0 (the derivative is positive above ė =0) δė/δe = 1/λ < 0 (the derivative is negative below ė =0) We can now draw our economy: Letting μ 1 and μ 2 denote the eigenvalues of the matrix, μ 1 > 0 >μ 2, we find the associated eigenvectors are multiples of [λμ 1 1] and [λμ 2 1 ], and we may thus write the general (continuous) solution to(15)as: p(t) p = k 1 λμ 1 exp(μ 1 t)+k 2 λμ 2 exp(μ 2 t) (16) e(t) ē = k 1 exp(μ 1 t)+k 2 exp(μ 2 t) 20

21 ,where k 1 and k 2 are given by choice of initial conditions, while p is predetermined. By restricting ourselves to the above continuous solutions, we have ruled out paths of the type shown in Fig.3, which satisfy the differential equations in the dymamic system (15), if we interpret the derivatives appearing there as right-hand side derivatives. In other words, we rule out anticipated discrete jumps, or, what come to the same thing, anticipated shifts in the system s initial condition. There is a good economic reason for imposing this CONTINUITY CON- DITION. Suppose agents expect a discrete jump in the exchange rate at some point in the future. Then they also must expect that the interest-parity condition will be violated, for the jump in e will cause an instantaneously infinite capital gain or loss. Thus, the instant before the jump is expected to occur, agents will have an incentive to bring about a sharp change in e. But this would violate the assumption of perfect foresight.(for an reference example of this, see the Krugman model of balance-sheet financial 21

22 crises). It follows that we can rule out paths of the discontinuous type on economic grounds. When a disturbance occurs, asset-holders will bid the exchange-rate to a level such that no further jumps occur; and in this way they will ensure that foreign exchange holdings earn exactly the equilibrium rate of return, r, at every point of the economy s anticipated subsequent path ANTICIPATORY DYNAMICS Permanent not foreseen change in m Suppose now that an exogenous increase in the nominal supply of money m were to occur at time t 1.Before the instant t 1, the economy is stuck in its initial long run equilibrium, point 1, and the relevant isoclines are labelled with subscript 1. The exogenous increase in the nominal supply of money m makes the isoclines e to shift up to the upper path, labelled with subscript 2, while the path of p would remain unaffected. As a result, p and ē rise by the same amount, which is clear because the slope of the p isocline is unity. We know that the economy must end in the long run equilibrium point 2, but it cannot jump to the new steady state instantly, however, because domestic price, p, changes sluggishly. On the other hand, e can adjust instantly. Given the strong assumption of perfect foresight on the part of economic agents in the model, the exchange rate will jump to the point e 2, to reach the new saddle path. (consider how only this jump is consistent with the economy reaching the new long-run equilibrium at the new steady state). From that point, as shown in Figure 4, the economy follows the new saddle path toward the new steady state 2. Even from this simple dynamic exercise, we see how in this model the short-run increase in the exchange rate therefore overshoots the long run increase, by changing more than proportionately to the money stock, and this 22

23 is the reason why we speak about an overshooting exchange rate Permanent and foreseen change in m Let now the economy be in its long-run equilibrium at t 0, when the central bank announces that at time t0 (> t 0 ) in the future, the money supply will be increased from m to m 0 = m+ m. What happens? We have two clues. First, since p is a predetermined variable, we know that p cannot change from its original path p(0) = p = m when the announcement is made. Second, we know from the continuity condition that when m is actually increased, the exchange rate cannot be expected to take a discrete jump (such a jump would be arbitraged away the moment it was foreseen). Figure 5 shows the path of the economy consistent with these two requirements. Initially, e depreciates to e0, withp remaining at p = m = p(0).the economy must then move along one of its unstable paths, for the money supply has yet to increase and so (14) still describes the motion of p and e over 23

24 time. Speculators choose e 0 so that at time t 0,whenm increases, (p(t 0 ),e(t 0 )) lies on the stable arm or saddlepath of the dynamic system associated with money stock m + m. They thus prevent a foreseen jump in the exchange rate that might have otherwise been necessary to place the economy on the new saddlepath at t 0. After t 0, the economy travels toward the stationary state ( p 1, ē 1 ). We now turn to the derivation of an analytical representation of the implied path. Our task is to determine two sets of initial conditions (k 1,k 2 ), and (k 1,1,k 1,2 ). The first set is associated with the economy s path between t =0andt = t 0 ; the second gives its path between t = t0 and t =. For clarity s reasons, we label (k 1,1,k 1,2 )as(k0 1,k0 2 ) From the assumption of price stickiness, (15) tells us that p(0) p =0=k 1 λμ 1 + k 2 λμ 2 24

25 (16), and this is our first constraint on λ 1 and k 2. The saddlepath of the system associated with the higher money stock m + m is described by the equation p(t) p0 = λμ 2 (e(t) ē0), or, (4), p(t) p = λμ 2 (e(t) ē)+(1 λμ 2 ) m (17) The unstable path chosen by the economy for the period between t =0 and t = t 0 must, by the continuity condition, be such that (p(t 0 ),e(t 0 )) lies on the locus given by (17). The price level and exchange rate at time t are p(t 0 ) p = k 1 λμ 1 exp(μ 1 t 0 ) k 1 λμ 1 exp(μ 2 t 0 ) e(t 0 ) ē = k 1 exp(μ 1 t 0 ) (μ 1 /μ 2 )k 1 exp(μ 2 t 0 ) where we have used (15) and (16), and substituting into (17) yields: or k 1 λμ 1 exp(μ 1 t 0 )=k 1 λμ 2 exp(μ 1 t 0 )+(1 λμ 2 ) m k 1 = m(1 λμ 2 )exp( μ 1 t 0 )/λ(μ 1 μ 2 ) (18) k 2 = (μ 1 /μ 2 ) m(1 λμ 2 )exp( μ 1 t 0 )/λ(μ 1 μ 2 ). The initial conditions imposed on k 1 and k 2 determine the economy s path between the time of the announcement and the time of the monetary expansion. The initial conditions are associated with the economy s path after m rises, since the economy must now be on the saddlepath. We must have k 1 =0, so that at t = t 0 : e(t 0 ) ē 0 = k 0 2 exp(μ 2 t 0 ) (19) But from (18a, b), we have seen that e(t 0 ) ē 0 = e(t 0 ) ē m (20) = m(1 λμ 2) λ(μ 1 μ 2 ) [1 (μ 1/μ 2 )exp((μ 1 μ 2 )t 0 )] m 25

26 and this, together with (19), implies k 0 2 = m λ(μ 1 μ 2 ) [(1 λμ 1)exp( μ 2 t 0 ) (μ 1 /μ 2 )(1 λμ 2 )exp( μ 1 t 0 ) To summarize, the path of the economy is given by: p(t) = p + μ 1 m(1 λμ 2 ) (μ 1 μ 2 ) exp( μ 1 t 0 )[exp(μ 1 t) exp(μ 2 t)] (0 t t 0 ) p 0 + μ 2 m (μ 1 μ 2 ) exp(μ 2t)[(1 λμ 1 )exp( μ 2 t 0 ) (μ 1 /μ 2 )(1 λμ 2 )exp( μ 1 t 0 )] (t t 0 ) e(t) = ē 0 + ē + m(1 λμ 2 ) λ(μ 1 μ 2 ) exp( μ 1 t 0 )[exp(μ 1 t) (μ 1 /μ 2 )exp(μ 2 t)] (0 t t 0 ) m λ(μ 1 μ 2 ) exp(μ 2t)[(1 λμ 1 )exp( μ 2 t 0 ) (μ 1 /μ 2 )(1 λμ 2 )exp( μ 1 t 0 )] (t t 0 ) These expressions are readily derived using (15) and the initial conditions we have calculated. Some implications of these results are now briefly explored. Setting t = 0, we can examine the jump taken by the exchange rate when the announcement occurs. The spot depreciation is just e(0) ē = m(1 1/λμ 2 )exp( m 1 t 0 ) which implies that as t 0, e(0) ē. The farther in the future is the policy action, the smaller is the impact on today s exchange rate. When t 0 =0,wehavetheeffect of an unanticipated change in m, and see immediately that since μ 2 < 0,theexchangerateovershootē0. What is the value of the exchange rate when the increase in money takes place? at t = t 0,wehave 26

27 e(t 0 ) ē 0 = m λ(μ 1 μ 2 ) [(1 λμ 1) (μ 1 /μ 2 )(1 λμ 2 )exp((μ 1 μ 2 )t)] and since λμ 1 < 1,theexchangeratemustbeaboveitslong-runlevel, m actually rises. An interesting point to note is that lim t e(t0 ) ē 0 = m (1 λμ 1 ) /λ(μ 1 /μ 2 ) > 0 Even when the disturbance is anticipated infinitely far in advance, economy will have not adjusted completely when it occurs. This is a peculiarity of the Dornbusch model, due to the fact that the price adjustment rule does not look forward into time (as would, say, a Phillips curve augmented by inflationary expectations), but instead reflects only current excess demand. 4. An alternative approach An alternative approach to solving the problem of the previous section serves as a check on the solution we have derived. The alternative approach has the advantage that it makes explicit the role of expectations regarding the future level of the money supply. In addition, it shows how we can still interpret continuity condition as a once-for-all choice of initial condition. To implement this second method, we write the path of the money stock anticipated after the announcement is made as m(t) = m (0 <t<t 0 ) (21) m(t) = m + m (t t 0 ) (22) Our system of equations is now non-autnomous, given by ṗ = δ(e p) ė = (1/λ)p [1/λm(t)] 27

28 The inverse of the matrix of eigenvectors of the system s homogeneous part is easily calculated to be p 1 = 1 1 λμ2 λ(μ 1 μ 2 ) 1 λμ 1 and so we may write the system in the diagonal form as μ 2 u = μ 1 u + λ(μ 1 μ 2 ) m(t) μ v = μ 2 v 1 λ(μ 1 μ 2 ) m(t) where [u v] 0 = p 1 [pe] 0.The solution of this system is of general form u(t) = k 1 exp(μ 1 t) v(t) = k 2 exp(μ 2 t) μ 2 λ(μ 1 μ 2 ) μ 1 λ(μ 1 μ 2 ) Z t Z t exp(μ 1 (t s))m(s)ds exp(μ 2 (t s))m(s)ds Multiplying by P and so changing back to the original coordinates p and e yields Z p(t) = λμ 1 k 1 exp(μ 1 t)+λμ 2 k 2 exp(μ 2 t) μ 1μ 2 exp(μ (μ 1 μ 2 ) 1 (t s)) t μ Z t 1μ 2 exp(μ (μ 1 μ 2 ) 2 (t s)m(s)ds Z μ e(t) = k 1 exp(μ 1 t)+λexp(μ 2 t) 2 exp(μ λ(μ 1 μ 2 ) 1 (t s)m(s)ds μ 1 λ(μ 1 μ 2 ) Z t exp(μ 2 (t s)m(s)ds t We may use this general solution to calculate the path of the economy in the wake of the announcement of monetary expansion at time t. But first we need to determine the initial conditions k 1 and k 2.Asusual,weset 28

29 k 1 = 0: this places the economy on a generalized saddle path having the property that(p(t),e(t)) will approach a steady state when m(t)converges to a constant value t.thisleavesk 2 to be determined, and we do so by recalling that, by price-level stickiness,p(0) = p = m. Now, for 0 >t>t 0 : = Z t Z t Z t exp((μ 1 (t s))m(s)ds + exp((μ 1 (t s))mds + Z = m/μ 1 m/μ 2 +exp(μ 1 (t t )) m/μ 1 t exp(μ 2 (t s))m(s)ds exp((μ 1 (t s)) mds + Z t exp(μ 2 (t s))mds (7). For t t 0 Z t Z t + Z t exp((μ 1 (t s))m(s)ds + exp(μ 2 (t s))mds + Z t t exp(μ 2 (t s))m(s)ds(t s))(m + m)ds exp(μ 2 (t s))mds = (m + m)μ 1 m/μ 2 [1 exp(μ 2 (t t )) m/μ 2 (8). Thus, at time t =0,wehave: m = λμ 2 k 2 μ 1μ 2 (μ 1 μ 2 ) [m/μ 1 m/μ 2 +exp( μ 1 t )) m/μ 1 ] = λμ 2 k 2 + m μ 2 m (μ 1 μ 2 ) exp( μ 1t ) With this second initial condition on hand, we may solve for the path; we start with the price level.. When 0 >t>t, we find, using (7), that: 29

30 p(t) = λμ 2 m exp(μ 2 t μ 1 t ) μ 1μ 2 λ(μ 1 μ 2 ) (μ 1 μ 2 ) [m/μ 1 m/μ 2 +exp(t t )) m = p + μ 2 m (μ 1 μ 2 ) exp( μ 1t )[exp(μ 2 t) exp(μ 1 t)] (for p <m). Now we know that μ 1 μ 2 = δλ and μ 1 + μ 2 = δ. Soμ 2 = μ 2 + δ δ = μ 1 + λμ 1 μ 2 = μ 1 (1 λμ 2 ),andwemaywritetheaboveas: p(t) = p + μ 1 m(1 λμ 2 ) (μ 1 μ 2 ) exp( μ 1 t )[exp(μ 1 t) exp(μ 2 t)] This is the expression given on page 6. When t t 0,(8)implies p(t) = λμ 2 m λ(μ 1 μ 2 ) exp(μ 2t μ 1 t 0 ) μ 1μ 2 (m + m)/μ1 (m + m)μ (μ 1 μ 2 ) 2 +exp(μ 2 (t t 0 )) m/μ 2 and, since p0 = m + m, p(t) = p 0 + m (μ 1 μ 2 ) exp(μ 2t) μ 2 exp( μ 1 t) μ 1 exp( μ 2 t 0 ) But μ 2 = μ 1 (1 λμ 2 )= μ 2 (μ 1 /μ 2 )(1 λμ 2 )and μ 1 = μ 2 + δ = μ 2 λμ 1 μ 2 = μ 2 (1 λμ 1 ), and so p(t) = p 0 + μ 2 m (μ 1 μ 2 ) exp(μ 2t) (1 λμ 1 )exp( μ 2 t 0 ) (μ 1 /μ 2 )(1 λμ 2 )exp( μ 1 t 0 ) Again, this is identical to the path previously calculated. The path of the exchange rate is derived in similar fashion, and can be shown, using the above relations, to be identical to the path derived using the method of the previous section. This is left as an exercise. 30

31 Where have we imposed the continuity condition? We have implicitly done so by imposing, once and for all, an initial condition: for once this is done the solution of a non-autonomous is continuous. REFERENCES. Birkoff, G. andg.-c. Rota. Ordinary Differential Equations. Lexington: Xerox College Publishing,1969 Blanchard, O.J., Backward and Forward Solutions for Economies with Rational Expectations. American Economic Review 69 (May 1979): Hirsh, M.W., and S. Smale. Differential Equations, Dynamical Systems and Linear Algebra. NewYork:AcademicPress;1974 Hoy, M., Livernois, J., McKenna, C., Rees, R., Stengos, T., Mathematics for Economies, TheMITPress,2001 Obstfeld, M., Rogoff, K., Foundation of International Macroeconomics, The MIT Press, Sargent, T.J. Macroeconomic Theory. NewYork: AcademicPress, 1979 Calvo, G. The Stability of Models of Money and Perfect Foresight: A Comment. Econometrica 45 (Oct. 1977): Rogoff, Essays on Expectations and Exchange-Rate Volatility. Doctoral dissertation, MIT, ( ) ANALYTICAL DERIVATION MUNDELL-FLEMING-DORNBUSCH MODEL The model is described by m p = λr + ψy r = r + ė ṗ = δ(e p) provided that for simplicity reason we have set y = r = 0 we can rewrite m p = λr r = ė ṗ = δ(e p) In order to get the system of differential equations we isolate 31

32 ṗ ė = δ δ (p p) 1 λ 0 (e ē) Let s now denote μ 1 > 0 >μ 2 the eigenvalues of the matrix: the associated eigenvectors have form [λμ 1 1] 0 [λμ 2 1] 0 The general solution, as previous shown in sect. 1.2, is x(t) = [k 1 b/ (λ 1 a)] exp (λ 1 t)+[k 2 b/ (λ 2 a)] exp (λ 2 t) y(t) = k 1 exp (λ 1 t)+k 2 exp (λ 2 t) that in our case translates in p(t) p = k 1 λμ 1 exp(μ 1 t)+k 2 λμ 2 exp(μ 2 t) e(t) ē = k 1 exp(μ 1 t)+k 2 exp(μ 2 t) Recalling that, for the assumption of price stickiness, at the time t = t 0 of the announcement we have p(0) p = 0 = k 1 λμ 1 + k 2 λμ 2 k 1 λμ 1 = (k 2 λμ 2 ) which represents the first constraint on λ and k 2. We are now ready to derive the saddlepath of the system associated with the higher money stock m + m p(t) p 0 e(t) ē 0 = k 1λμ 1 exp(μ 1 t) k 2 λμ 2 exp(μ 2 t) k 1 exp(μ 1 t) k 2 exp(μ 2 t) (k 1 =0) 32

33 which simplifies in p(t) p 0 = λμ 2 e(t) ē 0 p(t) p m = λμ 2 (e(t) ē) λμ 2 m p(t) p = λμ 2 (e(t) ē)+(1 λμ 2 ) m So that our saddle path is p(t) p = λμ 2 (e(t) ē)+(1 λμ 2 ) m t > t0 Consider now the t between the initial and the final equilibrium, t 0 < t<t 0. In that period, the economy follows an unstable path, which is required to bring, at t = t0, the economy in (p(t 0 ),e(t 0 )), along the higher saddle path, allowing the economy to converge to its new long run equilibrium. At the announcement, this equality must hold: and substituting it in k 1 λμ 1 = (k 2 λμ 2 ) μ k 2 = k 1 1 μ 2 we have p(t 0 ) p = k 1 λμ 1 exp(μ 1 t) k 1 λμ 1 exp(μ 2 t) e(t 0 ) ē = k 1 exp(μ 1 t) k 2 exp(μ 2 t) p(t 0 ) p = k 1 λμ 1 exp(μ 1 t 0 ) k 1 λμ 1 exp(μ 2 t 0 ) e(t 0 ) ē = k 1 exp(μ 1 t 0 ) (μ 1 /μ 2 )k 1 exp(μ 2 t 0 ) that denotes the price level and the exchange rate at time t0. Now, if we substitute these values in the equation for the higher saddlepath, we will get the initial conditions to be imposed on k 1 and k 2,to determine the economy s path between the time of the announcement and the time of monetary expansion: p(t) p 0 = λμ 2 (e(t) ē 0 ) 33

34 p(t) p 0 = k 1 λμ 1 exp(μ 1 t 0 ) k 1 λμ 1 exp(μ 2 t 0 ) = λμ 2 e(t) ē 0 +(1 λμ 2 ) m λμ 2 k1 exp(μ 1 t 0 ) (μ 1 /μ 2 )k 1 exp(μ 2 t 0 ) +(1 λμ 2 ) m k 1 λμ 1 exp(μ 1 t 0 ) = k 1 λμ 2 exp(μ 1 t 0 )+(1 λμ 2 ) m and, isolating k 1 and k 2, k 1 = m(1 λμ 2 )exp( μ 1 t 0 )/λ(μ 1 μ 2 ) k 2 = (μ 1 /μ 2 ) m(1 λμ 2 )exp( μ 1 t 0 )/λ(μ 1 μ 2 ) To derive now the saddle path after the m rise, we can recall that k 0 1 =0 (we are on a saddle path), and thus: e(t 0 ) ē 0 = k1(= 0 0) exp(μ 1 t 0 )+k2 0 exp(μ 2 t 0 ) e(t 0 ) ē 0 = k2 0 exp(μ 2 t 0 ) = e(t 0 ) ē m = m(1 λμ 2) λ(μ 1 μ 2 ) [1 (μ 1/μ 2 )exp((μ 1 μ 2 )t 0 )] m and setting e(t 0 ) ē 0 m = k2 0 exp(μ 2 t 0 ) m(1 λμ 2 ) λ(μ 1 μ 2 ) [1 (μ 1/μ 2 )exp((μ 1 μ 2 )t 0 )] m = k2 0 exp(μ 2 t 0 ) we have k 0 2 = m λ(μ 1 μ 2 ) [(1 λμ 2)exp( μ 2 t 0 ) μ 1 μ 2 (1 λμ 2 )exp( μ 1 t 0 )] Summarizing, and substituting the k 1,2 that we have found, the path of the economy is described by p(t) = p + μ 1 m(1 λμ 2 ) (μ 1 μ 2 ) exp( μ 1 t 0 )[exp(μ 1 t) exp(μ 2 t)] (t 0 t t 0 ) p 0 + μ 2 m (μ 1 μ 2 ) exp(μ 2t)[(1 λμ 1 )exp( μ 2 t 0 ) (μ 1 /μ 2 )(1 λμ 2 )exp( μ 1 t 0 )] (t t 0 ) 34

35 e(t) = ē + m(1 λμ 2 ) λ(μ 1 μ 2 ) ē 0 + exp( μ 1 t 0 )[exp(μ 1 t) (μ 1 /μ 2 )exp(μ 2 t)] (t 0 t t 0 ) m λ(μ 1 μ 2 ) exp(μ 2t)[(1 λμ 1 )exp( μ 2 t 0 ) (μ 1 /μ 2 )(1 λμ 2 )exp( μ 1 t 0 )] (t t 0 ) 35