Differential equations


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Differential equations

  1. 1. Differential Equations by tarungehlots Concepts of Differential Equation Consider a variable k that might denote the per capita capital stock level in an economy.When convenient, we can recognize that the level of capital depends upon time by addition thetime argument. Doing so, we would write the per capita capital stock level as k (t ) , rather thanjust writing k . If we think of time as unfolding continuously, we can also think of capital asbeing a continuous function of time, and we can assume that the function has derivate dk/ dt .This derivative is the instantaneous change in per capita capital. In many presentations, dk/ dt is written as k , but we will here use the more familiar notation dk / dt k (t ) . By writing thederivate as k (t ) , so that we include the time argument, we are emphasizing that the value of thederivative may change with time. Because it is typically cumbersome to repeatedly write downthe argument, we can also write the derivative as just k , while remembering that its value maychange with time. A differential equation is an equation that relates the time derivative of a variable to its level.An example is the equation(1) k sf (k ) gk nk k.The variable k is called a state variable because it gives the state of the system at any givenpoint in time. In our example, k gives the state or level of per capita capital stock.2. A Dynamic System The basic dynamic principle is the idea that “the way things are determines the way thingschange.” A differential equation is one way of modeling the basic dynamic principle. The statevariable k indicates the way things are, while the time derivative k indicates how things change.The differential equation itself is what relates the two. More precisely, a differential equationtypically presents a functional relationship, showing how k depends upon k . When we want toemphasize the functional relationship, we can write the differential equation as k (k ) , wherethe general functional form (k ) is the rule that tells us how the value of the time derivative k is determined from the level of the state variable k . 1
  2. 2. A difference equation can also be used to represent the basic dynamic principle. When adifference equation is used, time unfolds as sequential time periods, rather than unfoldingcontinuously. We will focus on using the differential equation here. A solution to a differential equation is a function k (t ) that satisfies the differential equationfor all points in time t that are of interest. Some differential equations can be solved. However,most differential equations that are of interest in economics cannot be solved. Nonetheless, wecan still learn about the path followed by the state variable over time by using a phase diagramtechnique. Assuming the function f (k ) in equation (1) is nonlinear, we can describe equation (1) asa single state variable system, and equation (1) is a nonlinear and non-autonomous differentialequation. As a dynamic system, equation (1) is a single state variable system because there isonly one time derivative present, which is k . If the system were a two state variable system,you would see two time derivatives and two different state variables. The equation is labelednonlinear because the functional relationship between k and k is not linear. Typically,nonlinear differential equations cannot be solved. However, the presence of the generalfunctional form f (k ) in equation (1) assures us that we cannot solve it. Finally, equation (1) islabeled non-autonomous because it includes variables other than state variables and timederivatives. The variables s , g , n , and in equation (1) are exogenous variables, which arevariables that describe the environment impacting the system. Usually exogenous variables areconstants, which is the case in equation (1). Classifying the variables of a differential equation system helps clarify how you themodel builder believe the system works. Because our system (1) is a single equation system, wecan only have one endogenous variable. The endogenous variables a differential equationsystem are always the time derivatives. So, for our system, k is endogenous. The statevariables of a system are always classified as predetermined, so k is predetermined for oursystem. The level of the variable k is predetermined at point in time t because theinstantaneous change k in the level of k we just determined by the model in the previousinstant of time. Non state variables are exogenous, so s , g , n , and are exogenous. The lastvalue that needs to be specified for the system is the initial condition k0 for the state variable k ,which is the value of the variable k at the initial point in time t 0 . Thus, we can summarizethe classification of variables as follows:Classification of VariablesEndogenous (1): k Exogenous (4): s , g , n ,Predetermined (1): kInitial Conditions (1): k0 It is common that differential equation models contain auxiliary equations, which areequations that endogenized (i.e., determine) the values for other variables by relating them to thestate variable. Using standard techniques for analyzing a differential equation of this type, we 2
  3. 3. can learn how the path for state variable k will evolve over time. Once we know the pathfollowed by the variable k , we can use the following “auxiliary” equations to determine thepaths followed by the other variables of interest. To illustrate, let us add the endogenous variables y , r , w , and c to our system, and assumethat they are determined from the following four equations:(2) y f (k )(3) r f (k )(4) w f (k ) f (k )k(5) c 1 syThe variables y , r , and w are directly related to the variable k , and only related to k . Thus, ifwe know the path followed by k , we can use equations (2)-(4) to determine the paths followedby y , r , and w . Equation (5) indicates that we need to know the path followed by y , alongwith the value of the exogenous variable s , to determine the path followed by c . So we candetermine the path followed by c by first using equation (2) to determine the path followed by y . This discussion indicates that we can classify the variables of the dynamic system (1)-(5) asfollows.Classification of VariablesEndogenous (5): k , y , r , w , cExogenous (4): s , g , n ,Predetermined (1): kInitial Conditions (1): k0 The core of this model is the path followed by the state variable k . Once you understand thecore of a model, you can use that knowledge to then understand the auxiliary aspects of themodel. Notice that adding the auxiliary equations only changes the number of endogenousvariables. We can think of the core of the model, which is equation (1) and the firstclassification of variables as a subsystem, a system upon which the auxiliary aspects of themodel depend.3. Analyzing the Steady StateFinding the Steady StatesThe steady state for the state variable k is the state where the variable is not changing, or wherek 0 . Setting k 0 in (1), we knowSS1 sf (k ) g n k. 3
  4. 4. There are five variables in equation SS1, the model’s four exogenous variables and the statevariable. Equation SS1 determines the steady state values of the state variable as a function ofthe exogenous variables. Assuming f (0) 0 , one value of k for which equation SS1 holds is k 0 . This steady state is not so interesting in that this result is simply telling us that if nocapital initially exists, then none will accumulate over time, (which is because capital is assumedto be essential in the production process). If there is a non zero value for k that satisfies SS1, letit be denoted by k . To illustrate a situation where there is a unique value for the steady state k , consider Figure 2below. The right side of equation SS1 is linear in k , with a zero intercept, as shown in thefigure. The assumption f (0) 0 implies the left side of equation SS1 also has a zero intercept.The conditions f 0 and f 0 would ensure that the production function increases at adecreasing rate as k increases, as shown in the figure. If it were true that f (0) g n , thenthe only steady state would be associated with k 0 , because the two curves sf (k ) and g n k would only intersect at the point (0,0) in Figure 2. The Inada condition lim f (k ) ensures that the sf (k ) curve initial rises above the g n k curve, and thek 0Inada condition lim f (k ) 0 ensures that the sf (k ) curve eventually intersects the k g n k curve. Because f 0 and f 0 for all k , we know that this intersection at acapital level k 0 only occurs once, which implies k is unique, as shown in Figure 2.Figure 2: The steady state for the Solow-Swan growth model y g n k f (k ) sf (k ) k kStability of the Steady State The Inada conditions lim f (k ) and lim f (k ) 0 also ensure that the steady tate k 0 kassociated with k k is stable. A steady state is locally stable if the state variable moves towardthe steady state value for the variable when it is in the “neighborhood” of the steady state.Define the function (k ) so k (k ) . Notice that, if (k ) 0 , then the steady state k islocally stable. This is because (k ) 0 implies k 0 when k k and k 0 when k k . 4
  5. 5. For our model given by SS1, (k ) sf (k ) g n k . So, (k ) sf (k ) g n .Because (k ) sf (k ) g n , our steady state k is locally stable if sf (k ) g n 0. Notice, in Figure 2, that the slope of the sf (k ) curve, at the steady state k k , is less than theslope of the g n k curve. That is, sf (k ) g n k . So, we find (k ) 0 , meaningour steady state k is locally stable. Examining Figure 2, we can also see that the steady state k 0 is locally unstable. In theneighborhood of k 0 , with k positive, notice in Figure 2 that the slope of the sf (k ) curve isgreater than the slope of the g n k curve. This implies sf (k ) g n 0 , whichimplies ( k ) 0 . If ( k ) 0 in the neighborhood of k 0 , then k is increasing in thatneighborhood, meaning it increasing away from the steady state value of zero. Figure 3 is a phase diagram. The traditional phase diagram for a single state variabledifferential equation is constructed by plotting the state variable along the horizontal axis andplotting the time derivative of the state variable along the vertical axis. So, for our model k isplotted horizontally and dk / dt k is plotted vertically. The function (k ) shows how k depends upon k . The Inada conditions tell us that the function (k ) is positive and divergingto infinity as k approaches zero, and they tell us (k ) decreases as k increases, eventuallybecoming negative, as shown in Figure 3. The steady state value for k is where (k ) equalszero. The arrows along the horizontal axis are presented to emphasize k is increasing when k k but decreasing when k k .Figure 3: Phase diagram for the Solow-Swan growth model (k ) k k (k )To summarize, we know that the steady state k 0 is locally unstable and the steady state k kis locally stable. The Inada conditions indicate that the path taken over time by the statevariable k is as shown in Figure 3. If the initial capital stock level k0 is between zero and k ,then k increases over time and approaches k . Alternatively, if k0 is greater than k , then kdecreases over time and approaches k . Figure 4 shows these two potential time paths for k byplotting k as it depends upon time. 5
  6. 6. Figure 4: Two possible time paths for the state variable k in the Solow-Swan growth model y k0 k k k0 k t4. A System of Two Differential EquationsSuppose a dynamic system is represented by the following system of equations and thefollowing classification of variables(A) k (t ) f (k (t )) c(t ) [ g n]k (t ) U (c(t ))(B) c (t ) g n f (k (t )) U (c(t ))Endogenous (2): k (t ) , c (t )Exogenous (2): , g, nPredetermined (2): k (t ) , c (t )Initial Conditions (2): k (0) , c(0)How can we characterize the paths followed by the two state variables?5. Using a Phase Diagram to Characterize the Two Dimensional Differential EquationSystem We now characterize the path followed by c (t ) and k (t ) diagrammatically in a “phasediagram.” The first step in constructing the phase diagram is to plot the “nullclines” associatedwith the steady state conditions c (t ) 0 and k (t ) 0 . From (B), we can see that c (t ) 0implies [ g n] f . The usual restrictions on the production function f imply there is aunique value k such that [ g n] f (k ) . (Notice that we need the Inada conditions lim f (k ) and lim f (k ) 0 to conclude this. The restrictions f ( k ) 0 and f (k ) 0k 0 k 6
  7. 7. alone are not enough.) When k k , the assumption f 0 implies f (k ) f (k ) , so f (k ) [ g n] which implies the right side of (B) is negative. Because U 0 and U 0are typically assumed for the utility function, we know c (t ) must be positive for the left side of(B) to be negative. Thus, when k k , c (t ) 0 . Using analogous reasoning, when k k ,c (t ) 0 . This is shown by the arrows in Figure 1. We obtain the k (t ) 0 nullcline by setting k (t ) 0 in condition (A), which impliesc(t ) f (k (t )) [ g n]k (t ) . Because f (0) 0 , we know c(t ) 0 when k (t ) 0 . The Inadaconditions lim f (k ) and lim f (k ) 0 allow us to draw conclusions about the shape of k 0 kthis nullcline consumption level c (t ) . The assumption lim f (k ) ensures that k 0f (k (t )) [ g n]k (t ) when k (t ) is near zero, which implies c(t ) 0 when k (t ) is near zero.Moreover, lim f (k ) ensures f (k (t )) increases faster than [ g n]k (t ) as k (t ) increases k 0from a number near zero, which implies c (t ) is increasing as k (t ) increases from zero. Theassumption lim f (k ) 0 ensures that, when k (t ) is large enough, [ g n]k (t ) increases faster kthan f (k (t )) as k (t ) increases. This implies that, as k (t ) increases, c (t ) must reach a peak andthen decline thereafter. This k (t ) 0 nullcline separates the region where k (t ) is increasingfrom where it is decreasing. Condition (A) indicates that, when c (t ) is greater than its nullclinevalue for a given level k (t ) , then k (t ) 0 . Conversely, when c (t ) is less than its nullclinevalue for a given level k (t ) , then k (t ) 0 . This is shown by the arrows in Figure 1. Figure 1: Dynamics implied by the Euler Equation and Capital Accumulation Constraint for the Neoclassical Model of Optimal Growth c (t ) c (t ) 0 c(t ) f (k (t )) [ g n]k (t ) k (t ) k k (t ) 0 7
  8. 8. Starting at any point in the ( k , c ) of Figure 1, the arrows tell us the direction ofmovement. This allows us to trace the path from any starting point (k0 , c0 ) . The starting pointfor k 0 is given to us in the problem. The starting point for c0 is a choice. Without additionalrestrictions, we cannot determine which value of c0 will occur. However, proceeding with someanalysis can provide an indication of what those restrictions might be. Figure 2 is drawn under the assumption that k 0 k. (After understanding this analysis,it should be apparent how to present the case for k 0 k .) Path 1 is associated with a relativelyhigh initial consumption choice c0 . Because (k0 , c0 ) is in the upper left quadrant of the phasediagram, the direction of movement must be up and to the left ( c increasing and k decreasing).k 0 k . Path 2 is still associated with a relatively high initial consumption choice c0 , but onewhere (k0 , c0 ) falls in the lower left quadrant. Initially, the direction of movement must be upand to the right ( c increasing and k increasing). However, the k (t ) 0 nullcline is eventuallyreached, and the direction of movement of k is reversed. Thus, the paths of type 2 eventuallybecome like those of type 1. For all paths of type 1 and type 2, k (t ) 0 must eventually occur.To rule out such paths as equilibrium paths, we need only add the reasonable condition k (t ) 0to the problem. Figure 2: Phase Diagram for the Neoclassical Model of Optimal Growth c (t ) c (t ) 0 1 2 c 4 * c0 3 k (t ) k0 k k (t ) 0 Path 3 is associated with a relatively low initial consumption choice c0 . Because (k0 , c0 )is in the lower left quadrant of the phase diagram, the direction of movement must initially be upand to the right ( c increasing and k increasing). However, on this path, the c (t ) 0 nullclineis reached, and the direction of movement of c is reversed. Eventually, such a path mustinvolve a negative consumption level. One way to rule out such a path is to recognize that, atany given point in time t , when the capital-consumption combination is ( k (t ), c(t )) , the 8
  9. 9. consumer can discontinuously increase or decrease the consumption level, for such an action isequivalent to the choosing c0 when t 0 . On path 3, by choosing to jump paths bydiscontinuously increase c (t ) , (by choosing to reduce the saving level k (t ) ), the consumer canmove to a new path that will provide a higher consumption level in every period, for all time,which is clearly preferable. Thus, it is not optimal to choose a consumption level c0 thatgenerates a path of type 3. Having ruled out paths of types 1 and 2 because they are not feasible, and paths of type 3because they are clearly not optimal, we can conclude that the unique optimal choice for c0 , *given the initial capital level k 0 is the level for c0 that generates path 4, or the level c 0 shown inFigure 2. On this path, consumption and capital each increase en route to the steady state (k , c ) ,never crossing a nullcline. It is worth noting that, if the consumer believes that the economy would end at aparticular point in time T , rather than assuming the economy will never end, then it would berational to choose a consumption path such that k (T ) 0 . That is, the consumer should choosethe consumption path where capital is just used up when the economy terminates. (The Eulerequation and capital accumulation equation would be the same; only the “boundary condition”condition would change to k (T ) 0 .) This implies, rather than choosing path 4 in Figure 2, theconsumer would choose a path either of Type 1 or Type 2, so that vertical axis where k (t ) 0 iseventually reached. Which path is chosen would depend upon the length of time T. The optimalinitial consumption level becomes lower as the length of time T becomes longer. 9