1.2 Initial-Value Problems

INTRODUCTION

We are often interested in problems in which we seek a solution y(x) of a differential equation so that y(x) satisfies prescribed side conditions—that is, conditions that are imposed on the unknown y(x) or on its derivatives. In this section we examine one such problem called an initial-value problem.

Initial-Value Problem

On some interval I containing x0, the problem

Solve: (1)

Subject to:

where y0, y1, …, yn−1 are arbitrarily specified real constants, is called an initial-value problem (IVP). The values of y(x) and its first n−1 derivatives at a single point x0: y(x0) = y0, y′(x0) = y1, …, y(n−1)(x0) = yn−1, are called initial conditions (IC).

First- and Second-Order IVPs

The problem given in (1) is also called an nth-order initial-value problem. For example,

(2)

and

(3)

are first- and second-order initial-value problems, respectively. These two problems are easy to interpret in geometric terms. For (2) we are seeking a solution of the differential equation on an interval I containing x0 so that a solution curve passes through the prescribed point (x0, y0). See FIGURE 1.2.1. For (3) we want to find a solution of the differential equation whose graph not only passes through (x0, y0) but passes through so that the slope of the curve at this point is y1. See FIGURE 1.2.2. The term initial condition derives from physical systems where the independent variable is time t and where y(t0) = y0 and y′(t0) = y1 represent, respectively, the position and velocity of an object at some beginning, or initial, time t0.

The graph has a shaded region of width I marked on the horizontal axis. Five curves are inside the region. A curve goes up to the right through the marked point (x subscript zero, y subscript zero). Under that curve, there are two curves. The first one goes up to the right and then goes down to the right. The second one starts on the x axis, goes up to the right and then goes down and ends just above the x axis. There are two curves above the first curve. The first one starts on the top left of the shaded area, goes down to the right, reaches a low point and then goes up to the right. The second one starts approximately below the first one, goes up to the right and ends just below the first one.

FIGURE 1.2.1 First-order IVP

The graph has a shaded region of width I marked on the horizontal axis. There are five curves. The first curve goes down to the right and then goes up to the right through the marked point (x subscript zero, y subscript zero), and goes up and right. The slope at (x subscript zero, y subscript zero) is m = y subscript 1. There is one curve below this curve. It starts just above the x axis, goes down, touches the x axis and then goes up and to the right. There are three curves above the first curve. All three curves start on the left, go down to the right and then up to the right.

FIGURE 1.2.2 Second-order IVP

Solving an nth-order initial-value problem frequently entails using an n-parameter family of solutions of the given differential equation to find n specialized constants so that the resulting particular solution of the equation also “fits”—that is, satisfies—the n initial conditions.

EXAMPLE 1 First-Order IVPs

  1. It is readily verified that y = cex is a one-parameter family of solutions of the simple first-order equation y′ = y on the interval (−∞, ∞). If we specify an initial condition, say, y(0) = 3, then substituting x = 0, y = 3 in the family determines the constant 3 = ce0 = c. Thus the function y = 3ex is a solution of the initial-value problem

y′ = y, y(0) = 3.

  1. Now if we demand that a solution of the differential equation pass through the point (1, −2) rather than (0, 3), then y(1) = −2 will yield −2 = ce or c = −2e−1. The function y = −2ex−1 is a solution of the initial-value problem

y′ = y, y(1) = −2.

The graphs of these two solutions are shown in blue in FIGURE 1.2.3. ≡

The graph has four quadrants and consists of six curves. The first curve enters the second quadrant just above the negative x axis, goes up to the right gradually and then sharply through (0, 3) and then goes up into the first quadrant. The second curve has a similar trend and goes up to the left of the first curve, in the second quadrant. The third curve goes up to the right of the first curve, in the first quadrant. The fourth curve enters just below the negative x axis in the third quadrant. It goes down to the right gradually and then sharply. It intersects the negative y axis and goes down though (1, negative 2) and goes down into the fourth quadrant. The fifth curve has a similar trend as the fourth one and it goes down to the left of the fourth curve, passing through (0, negative 3). The sixth curve also as a similar trend to the fourth one and it goes down to the right of the fourth curve, in the fourth quadrant.

FIGURE 1.2.3 Solutions of IVPs in Example 1

The next example illustrates another first-order initial-value problem. In this example, notice how the interval I of definition of the solution y(x) depends on the initial condition y(x0) = y0.

EXAMPLE 2 Interval I of Definition of a Solution

In Problem 6 of Exercises 2.2 you will be asked to show that a one-parameter family of solutions of the first-order differential equation y′ + 2xy2 = 0 is y = 1/(x2 + c). If we impose the initial condition y(0) = −1, then substituting x = 0 and y = −1 into the family of solutions gives −1 = 1/c or c = −1. Thus, y = 1/(x2 − 1). We now emphasize the following three distinctions.

  • Considered as a function, the domain of y = 1/(x2 − 1) is the set of real numbers x for which y(x) is defined; this is the set of all real numbers except x = −1 and x = 1. See FIGURE 1.2.4(a).
  • Considered as a solution of the differential equation y′ + 2xy2 = 0, the interval I of definition of y = 1/(x2 − 1) could be taken to be any interval over which y(x) is defined and differentiable. As can be seen in Figure 1.2.4(a), the largest intervals on which y = 1/(x2 − 1) is a solution are (−∞, −1), (−1, 1), and (1, ∞).
  • Considered as a solution of the initial-value problem y′ + 2xy2 = 0, y(0) = −1, the interval I of definition of y = 1/(x2 − 1) could be taken to be any interval over which y(x) is defined, differentiable, and contains the initial point x = 0; the largest interval for which this is true is (–1, 1). See Figure 1.2.4(b). ≡
Two graphs. The first graph has the function defined for all x except x equals plus-minus 1. It has four quadrants and consists of three curves. Two vertical dashed lines are plotted at x equals 1 and x equals negative 1. The first curve enters the second quadrant, just above the negative x axis. It goes up to the right gradually and then sharply and goes up the dashed vertical line in the second quadrant. The second curve enters the first quadrant, just to the right of the vertical dashed line. It goes down to the right sharply and then gradually, toward the x axis. The third curve enters the third quadrant, just to the right of the vertical dashed line, it goes up to the right, toward the negative y axis. It reaches a high point at (0, negative 1), then it goes down to the right sharply toward the vertical dashed line in the fourth quadrant. The second graph has the solution defined on an interval containing x equals 0. It has four quadrants and consists of one curve. Two vertical dashed lines are plotted on the graph. The first dashed line is at x equals negative one, and the second dashed line is at x equals one. The curve enters the third quadrant, just to the right of the vertical dashed line, it goes up to the right, toward the negative y axis. It reaches a high point at (0, negative 1), then it goes down to the right sharply toward the vertical dashed line in the fourth quadrant.

FIGURE 1.2.4 Graphs of function and solution of IVP in Example 2

See Problems 3–6 in Exercises 1.2 for a continuation of Example 2.

EXAMPLE 3 Second-Order IVP

In Example 10 of Section 1.1 we saw that x = c1 cos 4t + c2 sin 4t is a two-parameter family of solutions of x″ + 16x = 0. Find a solution of the initial-value problem

x″ + 16x = 0, x(π/2) = −2, x′(π/2) = 1.(4)

SOLUTION

We first apply x(π/2) = −2 to the given family of solutions: c1 cos 2π + c2 sin 2π = −2. Since cos 2π = 1 and sin 2π = 0, we find that c1 = −2. We next apply x′(π/2) = 1 to the one-parameter family x(t) = −2 cos 4t + c2 sin 4t. Differentiating and then setting t = π/2 and x′ = 1 gives 8 sin 2π + 4c2 cos 2π = 1, from which we see that c2 = . Hence is a solution of (4). ≡

Existence and Uniqueness

Two fundamental questions arise in considering an initial-value problem:

Does a solution of the problem exist? If a solution exists, is it unique?

For a first-order initial-value problem such as (2), we ask:

Note that in Examples 1 and 3, the phrase “a solution” is used rather than “the solution” of the problem. The indefinite article “a” is used deliberately to suggest the possibility that other solutions may exist. At this point it has not been demonstrated that there is a single solution of each problem. The next example illustrates an initial-value problem with two solutions.

EXAMPLE 4 An IVP Can Have Several Solutions

Each of the functions y = 0 and satisfies the differential equation dy/dx = xy1/2 and the initial condition y(0) = 0, and so the initial-value problem dy/dx = xy1/2, y(0) = 0, has at least two solutions. As illustrated in FIGURE 1.2.5, the graphs of both functions pass through the same point (0, 0). ≡

The graph has four quadrants and consists of a curve and a line. The line goes along the x axis for the y equals 0. The curve represents the equation y equals x to the power of four, over 16. The curve enters the top left of the second quadrant and goes down and right. It goes down to the right sharply and the gradually approaching the negative x axis. It goes to the right passing through the origin. The curve continues into the first quadrant and goes up and right gradually and then sharply.

FIGURE 1.2.5 Two solutions of the same IVP in Example 4

Within the safe confines of a formal course in differential equations one can be fairly confident that most differential equations will have solutions and that solutions of initial-value problems will probably be unique. Real life, however, is not so idyllic. Thus it is desirable to know in advance of trying to solve an initial-value problem whether a solution exists and, when it does, whether it is the only solution of the problem. Since we are going to consider first-order differential equations in the next two chapters, we state here without proof a straightforward theorem that gives conditions that are sufficient to guarantee the existence and uniqueness of a solution of a first-order initial-value problem of the form given in (2). We shall wait until Chapter 3 to address the question of existence and uniqueness of a second-order initial-value problem.

THEOREM 1.2.1 Existence of a Unique Solution

Let R be a rectangular region in the xy-plane defined by a ≤ x ≤ b, c ≤ y ≤ d, that contains the point (x0, y0) in its interior. If f(x, y) and ∂f/∂y are continuous on R, then there exists some interval I0: (x0 − h, x0 + h), h > 0, contained in [a, b], and a unique function y(x) defined on I0 that is a solution of the initial-value problem (2).

The foregoing result is one of the most popular existence and uniqueness theorems for first-order differential equations, because the criteria of continuity of f(x, y) and ∂f/∂y are relatively easy to check. The geometry of Theorem 1.2.1 is illustrated in FIGURE 1.2.6.

The graph consists of a curve. A vertical dashed line labeled a is plotted on the horizontal axis of the graph. Another vertical dashed line labeled b is plotted on the horizontal axis, to the right of the first dashed line. A horizontal line labeled c and another horizontal dashed line labeled d is plotted above the first horizontal line. The common region shared all the four dashed lines is shaded and marked as region R. Two vertical dashed lines are plotted in between the lines a and b, with a distance of I subscript zero between them. Within the two vertical lines I subscript 0 distance apart, a curve starts from the vertical dashed line beside a, and goes up and right, to the vertical dashed line beside b, passing through the point (x subscript zero, y subscript zero).

FIGURE 1.2.6 Rectangular region R

EXAMPLE 5 Example 4 Revisited

We saw in Example 4 that the differential equation dy/dx = xy1/2 possesses at least two solutions whose graphs pass through (0, 0). Inspection of the functions

and

shows that they are continuous in the upper half-plane defined by y > 0. Hence Theorem 1.2.1 enables us to conclude that through any point (x0, y0), y0 > 0, in the upper half-plane there is some interval centered at x0 on which the given differential equation has a unique solution. Thus, for example, even without solving it we know that there exists some interval centered at 2 on which the initial-value problem dy/dx =xy1/2, y(2) = 1, has a unique solution. ≡

In Example 1, Theorem 1.2.1 guarantees that there are no other solutions of the initial-value problems y′ = y, y(0) = 3, and y′ = y, y(1) = −2, other than y = 3ex and y = −2ex−1, respectively. This follows from the fact that f(x, y) = y and ∂f/∂y = 1 are continuous throughout the entire xy-plane. It can be further shown that the interval I on which each solution is defined is (−∞, ∞).

Interval of Existence/Uniqueness

Suppose y(x) represents a solution of the initial-value problem (2). The following three sets on the real x-axis may not be the same: the domain of the function y(x), the interval I over which the solution y(x) is defined or exists, and the interval I0 of existence and uniqueness. In Example 7 of Section 1.1 we illustrated the difference between the domain of a function and the interval I of definition. Now suppose (x0, y0) is a point in the interior of the rectangular region R in Theorem 1.2.1. It turns out that the continuity of the function f(x,y) on R by itself is sufficient to guarantee the existence of at least one solution of dy/dx = f(x, y), y(x0) = y0, defined on some interval I. The interval I of definition for this initial-value problem is usually taken to be the largest interval containing x0 over which the solution y(x) is defined and differentiable. The interval I depends on both f(x, y) and the initial condition y(x0) = y0. See Problems 31–34 in Exercises 1.2. The extra condition of continuity of the first partial derivative ∂f/∂y on R enables us to say that not only does a solution exist on some interval I0 containing x0, but it also is the only solution satisfying y(x0) = y0. However, Theorem 1.2.1 does not give any indication of the sizes of the intervals I and I0; the interval I of definition need not be as wide as the region R, and the interval I0 of existence and uniqueness may not be as large as I. The number h > 0 that defines the interval I0: (x0 − h, x0 + h), could be very small, and so it is best to think that the solution y(x) is unique in a local sense, that is, a solution defined near the point (x0, y0). See Problem 50 in Exercises 1.2.

REMARKS

  1. The conditions in Theorem 1.2.1 are sufficient but not necessary. When f(x, y) and ∂f/∂y are continuous on a rectangular region R, it must always follow that a solution of (2) exists and is unique whenever (x0, y0) is a point interior to R. However, if the conditions stated in the hypotheses of Theorem 1.2.1 do not hold, then anything could happen: Problem (2) may still have a solution and this solution may be unique, or (2) may have several solutions, or it may have no solution at all. A rereading of Example 4 reveals that the hypotheses of Theorem 1.2.1 do not hold on the line y = 0 for the differential equation dy/dx = xy1/2, and so it is not surprising, as we saw in Example 4 of this section, that there are two solutions defined on a common interval (−h, h) satisfying y(0) = 0. On the other hand, the hypotheses of Theorem 1.2.1 do not hold on the line y = 1 for the differential equation dy/dx = |y − 1|. Nevertheless, it can be proved that the solution of the initial-value problem dy/dx = |y − 1|, y(0) = 1, is unique. Can you guess this solution?
  2. You are encouraged to read, think about, work, and then keep in mind Problem 49 in Exercises 1.2.

1.2 Exercises Answers to selected odd-numbered problems begin on page ANS-1.

In Problems 1 and 2, y = 1/(1 + c1e−x) is a one-parameter family of solutions of the first-order DE y′ = y − y2. Find a solution of the first-order IVP consisting of this differential equation and the given initial condition.

  2. y(−1) = 2

In Problems 3–6, y = 1/(x2 + c) is a one-parameter family of solutions of the first-order DE y′ + 2xy2 = 0. Find a solution of the first-order IVP consisting of this differential equation and the given initial condition. Give the largest interval I over which the solution is defined.

  1. y(2) =
  2. y(−2) =
  3. y(0) = 1

In Problems 7–10, x = c1 cos t + c2 sin t is a two-parameter family of solutions of the second-order DE x″ + x = 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions.

  1. x(0) = −1, x′(0) = 8
  2. x(π/2) = 0, x′(π/2) = 1
  3. x(π/6) = , x′(π/6) = 0
  4. x(π/4) = , x′(π/4) =

In Problems 11–14, y = c1ex + c2e−x is a two-parameter family of solutions of the second-order DE y″ − y = 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions.

  1. y(0) = 1, y′(0) = 2
  2. y(1) = 0, y′(1) = e
  3. y(−1) = 5, y′(−1) = −5
  4. y(0) = 0, y′(0) = 0

In Problems 15 and 16, determine by inspection at least two solutions of the given first-order IVP.

  1. y′ = 3y2/3, y(0) = 0
  2. xy′ = 2y, y(0) = 0

In Problems 17–24, determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point (x0, y0) in the region.

  1. y2/3

In Problems 25–28, determine whether Theorem 1.2.1 guarantees that the differential equation possesses a unique solution through the given point.

  1. (1, 4)
  2. (5, 3)
  3. (2, −3)
  4. (−1, 1)
    1. By inspection, find a one-parameter family of solutions of the differential equation xy′ = y. Verify that each member of the family is a solution of the initial-value problem xy′ = y, y(0) = 0.
    2. Explain part (a) by determining a region R in the xy-plane for which the differential equation xy′ = y would have a unique solution through a point (x0, y0) in R.
    3. Verify that the piecewise-defined function

      satisfies the condition y(0) = 0. Determine whether this function is also a solution of the initial-value problem in part (a).

    1. Verify that y = tan (x + c) is a one-parameter family of solutions of the differential equation y′ = 1 + y2.
    2. Since f(x, y) = 1 + y2 and ∂f/∂y = 2y are continuous everywhere, the region R in Theorem 1.2.1 can be taken to be the entire xy-plane. Use the family of solutions in part (a) to find an explicit solution of the first-order initial-value problem y′ = 1 + y2, y(0) = 0. Even though x0 = 0 is in the interval (−2, 2), explain why the solution is not defined on this interval.
    3. Determine the largest interval I of definition for the solution of the initial-value problem in part (b).
    1. Verify that y = −1/(x + c) is a one-parameter family of solutions of the differential equation y′ = y2.
    2. Since f(x, y) = y2 and ∂f/∂y = 2y are continuous everywhere, the region R in Theorem 1.2.1 can be taken to be the entire xy-plane. Find a solution from the family in part (a) that satisfies y(0) = 1. Find a solution from the family in part (a) that satisfies y(0) = −1. Determine the largest interval I of definition for the solution of each initial-value problem.
    1. Find a solution from the family in part (a) of Problem 31 that satisfies y′ = y2, y(0) = y0, where y0 ≠ 0. Explain why the largest interval I of definition for this solution is either (− ∞, 1/y0) or (1/y0, ∞).
    2. Determine the largest interval I of definition for the solution of the first-order initial-value problem y′ = y2, y(0) = 0.
    1. Verify that 3x2 − y2 = c is a one-parameter family of solutions of the differential equation ydy/dx = 3x.
    2. By hand, sketch the graph of the implicit solution 3x2 − y2 = 3. Find all explicit solutions y = ϕ(x) of the DE in part (a) defined by this relation. Give the interval I of definition of each explicit solution.
    3. The point (−2, 3) is on the graph of 3x2 − y2 = 3, but which of the explicit solutions in part (b) satisfies y(−2) = 3?
    1. Use the family of solutions in part (a) of Problem 33 to find an implicit solution of the initial-value problem ydy/dx = 3x, y(2) = −4. Then, by hand, sketch the graph of the explicit solution of this problem and give its interval I of definition.
    2. Are there any explicit solutions of ydy/dx = 3x that pass through the origin?

    In Problems 35–38, the graph of a member of a family of solutions of a second-order differential equation d2y/dx2 = f(x, y, y′) is given. Match the solution curve with at least one pair of the following initial conditions.

    (a) y(1) = 1, y′(1) = −2

    (b) y(−1) = 0, y′(−1) = −4

    (c) y(1) = 1, y′(1) = 2

    (d) y(0) = −1, y′(0) = 2

    (e) y(0) = −1, y′(0) = 0

    (f) y(0) = −4, y′(0) = −2

  7. A graph has four quadrants and consists of a curve. The curve enters the second quadrant and goes down and right, passing through the point (negative 1, 0). The curve continues into the third quadrant and goes down and right, passing through the point (0, negative 4). The curve continues into the fourth quadrant, reaches a low point at approximately (0.9, negative 4.5), and goes up and right, passing through the point approximately (2.5, 0). The curve continues into the first quadrant and goes up and right.

    FIGURE 1.2.7 Graph for Problem 35

  8. A graph has four quadrants and consists of a curve. The curve enters the second quadrant just above the negative x axis. It goes down and right through approximately (negative 1.5, 0) into the third quadrant. It passes through the point (negative 1, 0) then goes up to the right into the fourth quadrant and intersects the x axis at approximately (1.9, 0). It goes up in the first quadrant, reaches a high point at approximately (5, 1.9) and then goes down to the right.

    FIGURE 1.2.8 Graph for Problem 36

  9. The graph has four quadrants and consists of a curve. The curve enters the second quadrant, just above the negative x axis, and goes down and right, through approximately (negative 2.1, 0). The curve continues into the third quadrant and goes down to the right and reaches a low point at approximately (negative 1, negative 2), then it goes up through (0, negative 1). The curve continues into the fourth quadrant and goes up and right, passing through approximately (0.5, 0). The curve continues into the first quadrant and goes up and right and reaches a high point at approximately (3, 3) and goes down to the right. It passes through approximately (6.5, 0) and goes down to the right in the fourth quadrant.

    FIGURE 1.2.9 Graph for Problem 37

  10. The graph has four quadrants and consists of a curve. The curve enters the third quadrant just below the negative x axis, goes up and right, passing through approximately (negative 1.8, 0). The curve continues into the second quadrant and reaches a high point at approximately at (negative 0.5, 2.5) and then goes down to the right through approximately (0, 2.2) and goes into the first quadrant. It goes down to the right through approximately (1.5, 0). It goes into the fourth quadrant and reaches a low point at approximately (4.6, 3.8). After this point, the curve goes up and right.

    FIGURE 1.2.10 Graph for Problem 38

In Problems 39–44, y = c1 cos 3x + c2 sin 3x is a two-parameter family of solutions of the second-order DE y″ + 9y = 0. If possible, find a solution of the differential equation that satisfies the given side conditions. The conditions specified at two different points are called boundary conditions.

  1. y(0) = 0, y(π/6) = −1
  2. y(0) = 0, y(π) = 0
  3. y′(0) = 0, y′(π/4) = 0
  4. y(0) = 1, y′(π) = 5
  5. y(0) = 0, y(π) = 4
  6. y′(π/3) = 1, y′(π) = 0

Discussion Problems

In Problems 45 and 46, use Problem 63 in Exercises 1.1 and (2) and (3) of this section.

  1. Find a function y = f(x) whose graph at each point (x, y) has the slope given by 8e2x + 6x and has the y-intercept (0, 9).
  2. Find a function y = f(x) whose second derivative is y″ = 12x − 2 at each point (x, y) on its graph and y = −x + 5 is tangent to the graph at the point corresponding to x = 1.
  3. Consider the initial-value problem y′ = x − 2y, y(0) = . Determine which of the two curves shown in FIGURE 1.2.11 is the only plausible solution curve. Explain your reasoning.
    The graph consists of two curves. The first curve starts from the second quadrant and goes up and right, passing through the point (0, 1 over 2). The curve continues into the first quadrant and goes up and right. The second curve starts from the second quadrant, above the first curve, and goes down and right, passing through the same point (0, 1 over 2). The curve continues into the first quadrant and goes down and right.

    FIGURE 1.2.11 Graph for Problem 47

  4. Without attempting to solve the initial-value problem y′ = x2 + y2, y(0) = 1, find the values of y′(0) and y″(0).
  5. Suppose that the first-order differential equation dy/dx = f(x, y) possesses a one-parameter family of solutions and that f(x, y) satisfies the hypotheses of Theorem 1.2.1 in some rectangular region R of the xy-plane. Explain why two different solution curves cannot intersect or be tangent to each other at a point (x0, y0) in R.
  6. The functions

    and

    have the same domain but are clearly different. See FIGURES 1.2.12(a) and 1.2.12(b), respectively. Show that both functions are solutions of the initial-value problem dy/dx = xy1/2, y(2) = 1 on the interval (− ∞, ∞). Resolve the apparent contradiction between this fact and the last sentence in Example 5.

    Two graphs. The first graph has four quadrants and consists of a curve. The curve enters the top left in the second quadrant and goes down and right sharply and then gradually, passes through the origin. The curve continues into the first quadrant and goes up and right, passing through the point (2, 1). The second graph has four quadrants and consists of a curve. The curve enters the second quadrant, along the negative x axis. It passes through the origin. The curve continues into the first quadrant and goes up and to the right, passing through the point (2, 1).

    FIGURE 1.2.12 Two solutions of the IVP in Problem 50

  1. Show that

    is an implicit solution of the initial-value problem

    Assume that [Hint: The integral is nonelementary. See (ii) in the Remarks at the end of Section 1.1.]