## Nonlinear First Order Differential Equations

In chapter 21 we saw that we could apply a single solution technique to solve any first-order differential equation that is linear. When the differential equation is nonlinear, however, no single solution technique will work in all cases. In fact only a few special classes of nonlinear, first-order differential equations can be solved at all. We will examine two of the more common classes in section 22.2. Even though solutions are known to exist for any nonlinear differential equation of the first order that satisfies certain continuity restrictions, it is simply not possible to find that solution in many cases. An alternative commonly used in economics is to do a qualitative analysis with the aid of phase diagrams. This technique is examined in section 22.1

### 22.1 Autonomous Equations and Qualitative Analysis

Before attempting a qualitative analysis of an autonomous, nonlinear differential equation, it is essential to know the conditions under which a solution exists. We begin by defining Lhe initial-value problem and then state the conditions for the existence of a solution.

The initial-value problem for an autonomous, nonlinear, first-order differential equation is expressed as y = £(y) (22.1)

The function g(y) must satisfy the properties stated in theorem 22.1 if we are to be guaranteed a solution to the initial-value problem.

Theorem 22.1 If the function g and its partial derivative dg(i)y are continuous in some closed rectangle containing the point Un.yu). then in a neighborhood around /q contained in the rectangle, there is a unique solution y = £(') satisfying equations (22.1 > and (22.2).

Theorem 22.1 assures us that a solution to the initial-value problem for nonlinear. first-order differential equations does indeed exist if g{y) satisfies the continuity conditions staled. However, the knowledge that a soluLion exists by no means ensures that we will be able to find it. In fact, it is rare in economic applications to be able to find explicit solutions to nonlinear differential equations. Instead, it is common to conduct a qualitativ e analysis, often with the aid of a phase diagram.

Consider the nonlinear, first-order differential equation y = v - y2 (22.3)

It turns out that equation (22.3) is a member of one of the classes of nonlinear differential equations that we can solve using some specialized techniques, as we will later see. However, for now, let us suppose diat we do not know these specialized techniques. It quickly becomes clear that our knowledge of solving linear differential equations will not help us find a solution to equation (22.3). Instead, we will attempt a qualitative analysis of the solution with the aid of a phase diagram. To this end. rearrange equation (22.3) to get y = y( 1 — y) (22.4)

For an autonomous, first-order differential equation, we can find the steady-state values of y by setting v =0. Doing this in equation (22.4) reveals that the system will be at rest (y = 0) at

This information is used to construct the phase diagram for equation (22.4) in figure 22.1.

The phase diagram of a single differential equation shows y as a function of v. We are particularly interested in finding the range of y values over which y is increasing over time (y > 0) and the range over which y is decreasing over lime (y < 0). We can construct llie diagram by graphing equation (22.4) with y as the variable on the vertical axis and y as the variable on the horizontal axis. For example, we have just determined that the value of v is zero when v = 0 and y = 1. This gives us two points on the graph. We now want to use equation (22.4) to plol the rest of ihe relationship We can use standard qualitative graphing techniques

to do this. First, determine the value of y at which the curve reaches an extreme value by taking the derivative of y with respect to y and setting it equal to zero:

dy dy

This gives us y = I /2. Determine whether the curve reaches a maximum or a minimum at y =1/2 by taking the second derivative of y with respect to y:

The negative value of the second derivative tells us that die curve reaches a maximum at y = 1 /2. Accordingly we know the curve must have the shape shown in figure 22.1.

A phase diagram helps us to determine the qualitative properties of a solution to a differential equation without actually having to solve it. In figure 22.1 we see that if the value of y ever reaches zero or one, it will remain at that value forever because y = 0 at each of those values. What happens if y has a value between zero and one? The phase diagram shows that y > 0 for 0 < v < I. This means that whenever the value of y is between zero and one. v will tend to increase over time. Of course, we already know that if it ever reaches one, the value will remain there.

What happens when y > I? We see from figure 22.1 that v <0 for y > I. If ever y exceeds one. its value will tend to decrease toward one. Finally, what happens if v < 0? We see that v < 0 for y < 0. which means that if ever the value of y is less than zero, its value will continue to decrease.

All this information is placed in the diagram in the form of arrows, to remind us of the direction of motion of the variable y in the different regions. Our analysis, summarized in the arrows of motion in figure 22.1, indicates that the tendency is for V to converge to the value one if it starts at a value exceeding zero, diverge to -so if it starts at a value less than zero, and to remain at zero if it starts at zero.

Thus, without solving the differential equation, we have obtained a great deal of insight into its solution by relying on a qualitative analysis with the aid of a phase diagram. Once we are given an initial condition for y, we can say with confidence how y will behave over time (increase or decrease) and to what value it will converge. In particular, provided that y<) > 0, we are able to conclude thai y(f) will converge to the value v = I.

### Stability Analysis

We required two types of information in the above qualitative analysis. First, we needed to know the steady-state values for y (i.e.. the values at which y=0)

Second, we needed to know the arrows of motion, or the 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.

We found thai the system converged to one of the steady-state values (.)■' - I) but not the other (y =0). We could say that y = 1 is a stable equilibrium and y = 0 is an unstable equilibrium. What makes one stable and the other unstable? In terms of ligure 22.1. the arrows of motion point toward the stable equilibrium but away from the unstable equilibrium. But what makes the arrows of motion do this? It turns out that t!nb slope of the y line in figure 22.1 .as it cuts through the equilibrium points, determines whether diose equilibrium points will be stable. The slope is negative at the point y = 1 bui it is positive at the point y = 0. Because the slope of the y line as it cuts through the equilibrium points is just the derivative of y with respect to y evaluated at the equilibrium points, we can state this new result as the following:

Theorem 22.2 A steady-state equilibrium point of a nonlinear, first-order differential equation is stable if the derivative dv/dy is negative at that point and unstable if the derivative is positive at that point.

Let us apply theorem 22.2 to the differential equation (22.4).

We conclude that y = 0 is an unstable equilibrium and y = 1 is a stable equilibrium.

The stability condition (dy/dy < 0 at equilibrium values) is intuitively sensible. It says that if the system is at equilibrium and is somehow pushed away from the equilibrium point, the motion of v will be in the opposite direction (negative derivative) of the push. This means that the system will tend to return toequilibrium. If the motion of y is in the same direction (positive derivative) a;- the push (as it isai \ =0 in ligure 22.1). the system will lend lo move further away from equilibrium,

Finally it is worth noting that this stability condition does not apply only to nonlinear, first-order differential equations. It also applies to the linear, first-order differential equations with constant coefficients: y 4- ay = b. Here the derivative dy/dy — —a is a constant. Thus the value of y will converge to the equilibrium (which is b/a) only if« > 0, which is the same convergence condition we derived in chapter 21.

Example 22.1 Find the steady-state points and determine their stability properties for the following:

Solution

Hie steady-state points occur where y - 0. This gives v(3y - 2) = 0

Therefore y=0 und v = 2/3 are the steady-state points. Applying theorem 22.2 gives il - h _ -> _ I ~ 2 at v = 0

Therefore y = 0 is stable, but y = 2/3 is unstable. ■

A Fishery Model with a Constant Harvest Rate

Suppose that a fish population grows according to the function g(y) = 2y( I - 0

where y is the stock of fish. The fish population is subjected to a constant level of harvesting by a lishing industry. If the harvest is a constant amount equal to 3/4. will the fish population reach a steady-state (positive) size, in which case the harvest is a sustainable activity, or will the fish population decline and become extinct?

To answer these questions, we conduct a qualitative analysis or the dynamics of the population. The growth of the fish population is reduced by 3/4 at each point in time, owing to the lishing industry. The change in the stock of lish is

The steady-state value of the population occurs where y = 0 which, after multiplying through the brackets, implies that for which the solutions arc y -

Solving gives y — 1/2 and y —3/2 as the two steady-state values of the fish population. Are they stable? We can determine the answer by applying theorem 22.2. First, take the derivative of y widi respect to y. This gives us

Evaluated at y - 1 /2, the derivative equals I. a positive value, indicating instability. Evaluated at y = 3/2, the derivative equals -1, a negative value, indicating stability. We conclude that y = 1 /2 is an unstable steady-state stock size and y = 3/2 is a stable one.

How can we determine which equilibrium is likely to arise and whether die fish population is likely to become extinct? A qualitative analysis using the phase diagram in figure 22.2 will answer these questions.

Rather than graphing the curve for y, we have graphed the curve for g(y) and the line 3/4 separately. We need only remember that when g(y) > 3/4, the fish population grows: when g(y) <3/4. the fish population declines; and wher g(y) = 3/4. the fish population has reached a steady state.

The two steady-state values for y are shown in figure 22.2. What is the rnotior, of the dynamic system? That is, in which direction does y move when it is nol at one of the steady-state values? The phase diagram makes it clear that when v is between 0 and 1/2, then g(y) < 3/4, so the population declines, as indicated by the arrow of motion. When y is between 1/2 and 3/2. then g(y) > 3/4. so the population grows. Finally, when y is larger than 3/2. then g(y) < 3/4, so the population declines. We conclude that the fish population will reach the stable steady-state value of 3/2 and remain there forever, provided that y() > 1 ¡2. On the other hand, if the initial population were less than 1 /2, the population would decline to zero.

### The Neoclassical Model of Economic Growth

What determines the long-run growth rate of an economy? The theoiy of economic growth was developed to answer this question and is still evolving. One of the important building blocks for modern refinements of the theory is the model that Nobel laureate Roben Solow developed in the 195t).s.

In this model we assume that output per person in an economy can be expressed as a concave function of the capital-labor ratio v = f(k)

where y is output per person and k = K/ L is the capital-labor ratio. Here K is the aggregate capital stock and L is total labor, which is equal to the total number of persons, assuming, for simplicity, that everyone works. Concavity implies that /'<*)> 0 and f"(k)<0.

The economy's output can be consumed or saved. The economy's capital stock, K, increases by the amount of investment, which is by definition equal to the amount of output saved. We assume a constant savings rate, ,s. Hence is the change in the capital stock, where Y is aggregate output. Since k = K/L. then

Rearrange this to obtain

The labor force is assumed to grow at the constant rate «.Making this substitution

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