3.10 Green’s Functions

INTRODUCTION

We have seen in Section 3.8 that the linear second-order differential equation

(1)

plays an important role in applications. In the mathematical analysis of physical systems it is often desirable to express the response or output y(x) of (1) subject to either initial conditions or boundary conditions directly in terms of the forcing function or input g(x). In this manner the response of the system can quickly be analyzed for different forcing functions.

To see how this is done we start by examining solutions of initial-value problems in which the DE (1) has been put into the standard form

(2)

by dividing the equation by the lead coefficient a₂(x). We also assume throughout this section that the coefficient functions P(x), Q(x), and f(x) are continuous on some common interval I.

3.10.1 Initial-Value Problems

Three Initial-Value Problems

We will see as the discussion unfolds that the solution of the second-order initial-value problem

Here at least one of the numbers y₀ or y₁ is assumed to be nonzero. If both y₀ and y₁ are 0, then the solution of the IVP is y = 0.

(3)

can be expressed as the superposition of two solutions: the solution y_h of the associated homogeneous DE with nonhomogeneous initial conditions

(4)

and the solution y_p of the nonhomogeneous DE with homogeneous (that is, zero) initial conditions

(5)

As we have seen in the preceding sections of this chapter, in the case where P and Q are constants the solution of the IVP (4) presents no difficulties: We use the methods of Section 3.3 to find the general solution of the homogeneous DE and then use the given initial conditions to determine the two constants in that solution. So we will focus on the solution of the IVP (5). Because of the zero initial conditions, the solution of (5) could describe a physical system that is initially at rest and so is sometimes called a rest solution.

Green’s Function

If y₁(x) and y₂(x) form a fundamental set of solutions on the interval I of the associated homogeneous form of (2), then a particular solution of the nonhomogeneous equation (2) on the interval I can be found by variation of parameters. Recall from (3) of Section 3.5, the form of this solution is

Because y₁(x) and y₂(x) are constant with respect to the integration on t, we can move these functions inside the definite integrals.

(6)

The variable coefficients u₁(x) and u₂(x) in (6) are defined by (5) of Section 3.5:

(7)

The linear independence of y₁(x) and y₂(x) on the interval I guarantees that the Wronskian W = W(y₁(x), y₂(x)) ≠ 0 for all x in I. If x and x₀ are numbers in I, then integrating the derivatives in (7) on the interval [x₀, x] and substituting the results in (6) give

(8)

where

From the properties of the definite integral, the two integrals in the second line of (8) can be rewritten as a single integral

. (9)

The function G(x, t) in (9),

, (10)

is called the Green’s function for the differential equation (2). Green’s function is named after the self-educated English mathematician and physicist George Green (1793–1841), who developed this concept around 1830.

Important. Read this paragraph a second time.

Observe that a Green’s function (10) depends only on the fundamental solutions, y₁(x) and y₂(x) of the associated homogeneous differential equation for (2), and not on the forcing function f(x). Therefore all linear second-order differential equations (2) with the same left-hand side but different forcing functions have the same Green’s function. So an alternative title for (10) is the Green’s function for the second-order differential operator L = D² + P(x)D + Q(x).

EXAMPLE 1 Particular Solution

Use (9) and (10) to find a particular solution of y″ − y = f(x).

SOLUTION

The solutions of the associated homogeneous equation y″ − y = 0 are , and It follows from (10) that the Green’s function is

(11)

Thus from (9), a particular solution of the DE is

(12)

≡

EXAMPLE 2 General Solutions

Find the general solution of the following nonhomogeneous differential equations.

(a) (b)

SOLUTION

From Example 1, both DEs possess the same complementary function Moreover, as pointed out in the paragraph preceding Example 1, the Green’s function for both differential equations is (11).

(a) With the identifications and we see from (12) that a particular solution of is Thus the general solution of the given DE on any interval not containing the origin is

(13)

You should compare this solution with that found in Example 3 of Section 3.5.

(b) With in (12), a particular solution of is

The general solution is then

(14) ≡

Now consider the special initial-value problem (5) with homogeneous initial conditions. One way of solving the problem when has already been illustrated in Sections 3.4 and 3.5; that is, apply the initial conditions to the general solution of the nonhomogeneous DE. But there is no actual need to do this because we already have solution of the IVP at hand; it is the function defined in (9).

THEOREM 3.10.1 Solution of the IVP (5)

The function y_p(x) defined in (9) in the solution of the initial-value problem (5).

PROOF:

By construction we know that in (9) satisfies the nonhomogeneous DE. Next, because a definite integral has the property we have

Finally, to show that we utilize the Leibniz formula* for the derivative of an integral:

Hence,

≡

EXAMPLE 3 Example 2 Revisited

Solve the initial-value problems

(a) (b)

SOLUTION

(a) With and it follows from (13) of Example 2 and Theorem 3.10.1 that the solution of the initial-value problem is

where

(b) Identifying and we see from (14) that the solution of the IVP is

(15) ≡

In Part (b) of Example 3, we can carry out the integration in (15), but bear in mind that x is held constant throughout the integration with respect to t:

EXAMPLE 4 Another IVP

Solve the initial-value problem

SOLUTION

We begin by constructing the Green’s function for the given differential equation.

The two linearly independent solutions of are and From (10), with we find

With the identification a solution of the given initial-value problem is

If we wish to evaluate the integral, we first write

Here we have used the trigonometric identity sin(2x − 2t) = sin 2x cos 2t − cos 2x sin 2t.

and then use integration by parts:

≡

Initial-Value Problems—Continued

We can now make use of Theorem 3.10.1 to find the solution of the initial-value problem posed in (3).

THEOREM 3.10.2 Solution of the IVP (3)

If y_h is the solution of the initial-value problem (4) and y_p is the solution (9) of the initial-value problem (5) on the interval I, then

y = y_h + y_p (16)

is the solution of the initial-value problem (3).

PROOF:

Because is a linear combination of the fundamental solutions, it follows from (10) of Section 3.1 that is a solution of the nonhomogeneous DE. Moreover, since satisfies the initial conditions in (4) and satisfies the initial conditions in (5), we have

≡

Keeping in mind the absence of a forcing function in (4) and the presence of such a term in (5), we see from (16) that the response of a physical system described by the initial-value problem (3) can be separated into two different responses:

(17)

In different symbols, the following initial-value problem represents the pure resonance situation for a vibrating spring/mass system. See pages 165 and 166.

EXAMPLE 5 Using Theorem 3.10.2

Solve the initial-value problem

SOLUTION

We solve two initial-value problems.

First, we solve By applying the initial conditions to the general solution of the homogeneous DE, we find that and Therefore, y_h(x) = cos 2x − sin 2x.

Next we solve Since the left-hand side of the differential equation is the same as the DE in Example 4, the Green’s function is the same; namely, With we see from (9) that the solution of this second problem is .

Finally, in view of (16) in Theorem 3.10.2, the solution of the original IVP is

(18) ≡

If desired, we can integrate the definite integral in (18) by using the trigonometric identity

with and

(19)

Hence, the solution (18) can be rewritten as

(20)

Note that the physical significance indicated in (17) is lost in (20) after combining like terms in the two parts of the solution

The beauty of the solution given in (18) is that we can immediately write down the response of a system if the initial conditions remain the same but the forcing function is changed. For example, if the problem in Example 5 is changed to

we simply replace in the integral in (18) by t and the solution is then

Because the forcing function f is isolated in the particular solution the solution in (16) is useful when f is piecewise defined. The next example illustrates this idea.

EXAMPLE 6 An Initial-Value Problem

Solve the initial-value problem

when the forcing function f is piecewise defined:

SOLUTION

From (18), with sin 2t replaced by f(t), we can write

Because f is defined in three pieces, we consider three cases in the evaluation of the definite integral. For x, 0,

for 0 ≤ x ≤ 2π,

and finally for x > 2π, we can use the integration following Example 5:

Hence is

and so

Putting all the pieces together we get

The graph y(x) is given in FIGURE 3.10.1. ≡

A curve is graphed on an x y coordinate plane. It has three parts. The first part starts at the point (negative pi, 1), goes down and to the right through the negative x axis, and reaches a low point in the third quadrant. Then it goes up and to the right through the negative x axis, reaches a high point in the second quadrant, goes down and to the right, and ends at (0, 1). The second part follows a decaying oscillatory pattern. It starts at (0, 1), goes down and to the right through the positive x axis, and reaches a point in the fourth quadrant. Then it goes up and to the right through the positive x axis, reaches a point in the first quadrant, goes down and to the right, and reaches a point in the fourth quadrant. Again it goes up and to the right through the positive x axis, reaches a point in the first quadrant, goes down and to the right through the positive x axis, and ends at a point in the fourth quadrant corresponding to the x value of 2 pi. The third part starts at the endpoint of the second part, goes down and to the right, then goes up and to the right through the positive x axis, and reaches a point in the first quadrant. Then it goes down and to the right through the positive x axis, and ends at a point in the fourth quadrant corresponding to the x value of 3 pi. — FIGURE 3.10.1 Graph of y(x) in Example 6

We next examine how a boundary-value problem (BVP) can be solved using a different kind of Green’s function.

3.10.2 Boundary-Value Problems

In contrast to a second-order IVP in which y(x) and y′(x) are specified at the same point, a BVP for a second-order DE involves conditions on y(x) and y′(x) that are specified at two different points x = a and x = b. Conditions such as

are just special cases of the more general homogeneous boundary conditions

(21)

and

(22)

where and are constants. Specifically, our goal is to find an integral solution that is analogous to (9) for nonhomogeneous boundary-value problems of the form

(23)

In addition to the usual assumptions that P(x), Q(x), and f(x) are continuous on [a, b], we assume that the homogeneous problem

possesses only the trivial solution y = 0. This latter assumption is sufficient to guarantee that a unique solution of (23) exists and is given by an integral where G(x, t) is a Green’s function.

The starting point in the construction of G(x, t) is again the variation of parameters formulas (6) and (7).

Another Green’s Function

Suppose and are linearly independent solutions on of the associated homogeneous form of the DE in (23) and that x is a number in the interval Unlike the construction of (8) where we started by integrating the derivatives in (7) over the same interval, we now integrate the first equation in (7) on and the second equation in (7) on

and (24)

The reason for integrating and over different intervals will become clear shortly. From (24), a particular solution of the DE is

(25)

The right-hand side of (25) can be written compactly as a single integral

, (26)

where the function G(x, t) is

(27)

The piecewise-defined function (27) is called a Green’s function for the boundary-value problem (23). It can be proved that G(x, t) is a continuous function of x on the interval [a, b].

Now if the solutions and used in the construction of (27) are chosen in such a manner that at satisfies and at satisfies then, wondrously, defined in (26) satisfies both homogeneous boundary conditions in (23).

To see this we will need

(28)

and

(29)

The last line in (29) results from the fact that

y₁(x)u′₁(x) + y₂(x)u′₂(x) = 0.

See the discussion in Section 3.5 following (4).

Before proceeding, observe in (24) that u₁(b) = 0 and u₂(a) = 0. In view of the second of these two properties we can show that y_p(x) satisfies (21) whenever y₁(x) satisfies the same boundary condition. From (28) and (29) we have

Likewise, implies that whenever satisfies (22) so does

The next theorem summarizes these results.

THEOREM 3.10.3 Solution of the BVP (23)

Let y₁(x) and y₂(x) be linearly independent solutions of

y″ + P(x)y′ + Q(x)y = 0

on [a, b], and suppose y₁(x) and y₂(x) satisfy boundary conditions (21) and (22), respectively. Then the function y_p(x) defined in (26) is a solution of the boundary-value problem (23).

EXAMPLE 7 Using Theorem 3.10.3

Solve the boundary-value problem

The boundary condition y′(0) = 0 is a special case of (21) with a = 0, A₁ = 0, and B₁ = 1. The boundary condition y(π/2) = 0 is a special case of (22) with b = π/2, A₂ = 1, and B₂ = 0.

SOLUTION

The solutions of the associated homogeneous equation are and and satisfies whereas satisfies The Wronskian is and so from (27) we see that the Green’s function for the boundary-value problem is

It follows from Theorem 3.10.3 that a solution of the BVP is (26) with the identifications and

or after evaluating the definite integrals, . ≡

Don’t infer from the preceding example that the demand that satisfy (21) and satisfy (22) uniquely determines these functions. As we see in the last example, there is a certain arbitrariness in the selection of these functions.

EXAMPLE 8 A Boundary-Value Problem

Solve the boundary-value problem

SOLUTION

The differential equation is recognized as a Cauchy–Euler DE.

From the auxiliary equation the general solution of the associated homogeneous equation is Applying to this solution implies or By choosing we get and On the other hand, applied to the general solution shows or The choice now gives and so The Wronskian of these two functions is

Hence, the Green’s function for the boundary-value problem is

In order to identify the correct forcing function f we must write the DE in standard form:

From this equation we see that and so (26) becomes

Straightforward definite integration and algebraic simplification yields the solution . ≡

3.10 Exercises Answers to selected odd-numbered problems begin on page ANS-8.

3.10.1 Initial-Value Problems

In Problems 1–6, proceed as in Example 1 to find a particular solution of the given differential equation in the integral form (9).

In Problems 7–12, proceed as in Example 2 to find the general solution of the given differential equation. Use the results obtained in Problems 1–6. Do not evaluate the integral that defines

In Problems 13–18, proceed as in Example 3 to find the solution of the given initial-value problem. Evaluate the integral that defines

In Problems 19–30, proceed as in Example 5 to find a solution of the given initial-value problem.

In Problems 31–34, proceed as in Example 6 to find a solution of the initial-value problem with the given piecewise-defined forcing function.

where
where
where
where

3.10.2 Boundary-Value Problems

In Problems 35 and 36, (a) use (25) and (26) to find a solution of the boundary-value problem. (b) Verify that function satisfies the differential equations and both boundary conditions.

In Problem 35 find a solution of the BVP when
In Problem 36 find a solution of the BVP when

In Problems 39–44, proceed as in Examples 7 and 8 to find a solution of the given boundary-value problem.

Discussion Problems

Suppose the solution of the boundary-value problem

is given by where and are solutions of the associated homogeneous differential equation chosen in the construction of so that and Prove that the solution of the boundary-value problem with nonhomogeneous DE and boundary conditions,

is given by

[Hint: In your proof, you will have to show that and Reread the assumptions following (23).]
Use the result in Problem 45 to solve

* See Problems 43 and 44 on pages 31 and 32.