3.9 Linear Models: Boundary-Value Problems

INTRODUCTION

The preceding section was devoted to dynamic physical systems each described by a mathematical model consisting of a linear second-order differential equation accompanied by prescribed initial conditions—that is, side conditions that are specified on the unknown function and its first derivative at a single point. But often the mathematical description of a steady-state phenomenon or a static physical system demands that we solve a linear differential equation subject to boundary conditions—that is, conditions specified on the unknown function, or on one of its derivatives, or even on a linear combination of the unknown function and one of its derivatives, at two different points. By and large, the number of specified boundary conditions matches the order of the linear DE. We begin this section with an application of a relatively simple linear fourth-order differential equation associated with four boundary conditions.

Deflection of a Beam

Many structures are constructed using girders, or beams, and these beams deflect or distort under their own weight or under the influence of some external force. As we shall now see, this deflection y(x) is governed by a relatively simple linear fourth-order differential equation.

To begin, let us assume that a beam of length L is homogeneous and has uniform cross sections along its length. In the absence of any load on the beam (including its weight), a curve joining the centroids of all its cross sections is a straight line called the axis of symmetry. See FIGURE 3.9.1(a). If a load is applied to the beam in a vertical plane containing the axis of symmetry, the beam, as shown in Figure 3.9.1(b), undergoes a distortion, and the curve connecting the centroids of all cross sections is called the deflection curve or elastic curve. The deflection curve approximates the shape of the beam. Now suppose that the x-axis coincides with the axis of symmetry and that the deflection y(x), measured from this axis, is positive if downward. In the theory of elasticity it is shown that the bending moment M(x) at a point x along the beam is related to the load per unit length w(x) by the equation

(1)

The illustration is shown in two parts. Part (a) has a homogeneous beam placed horizontally. A red dashed line passing through the beam represents axis of symmetry. Part (b) has a homogeneous beam deflected downward. A horizontal red line represents the axis of symmetry of the beam when it was at the original position. A red dashed curve passing through the beam represents deflection curve. It takes the shape of the beam.

FIGURE 3.9.1 Deflection of a homogeneous beam

In addition, the bending moment M(x) is proportional to the curvature κ of the elastic curve

M(x) = EIκ, (2)

where E and I are constants; E is Young’s modulus of elasticity of the material of the beam, and I is the moment of inertia of a cross section of the beam (about an axis known as the neutral axis). The product EI is called the flexural rigidity of the beam.

Now, from calculus, curvature is given by κ = y″/[1 + (y′)2]3/2. When the deflection y(x) is small, the slope y′ ≈ 0 and so [1 + (y′)2]3/2 ≈ 1. If we let κ ≈ y″, equation (2) becomes M = EI y″. The second derivative of this last expression is

(3)

Using the given result in (1) to replace d2M/dx2 in (3), we see that the deflection y(x) satisfies the fourth-order differential equation

. (4)

Boundary conditions associated with equation (4) depend on how the ends of the beam are supported. A cantilever beam is embedded or clamped at one end and free at the other. A diving board, an outstretched arm, an airplane wing, and a balcony are common examples of such beams, but even trees, flagpoles, skyscrapers, and the George Washington monument can act as cantilever beams, because they are embedded at one end and are subject to the bending force of the wind. For a cantilever beam, the deflection y(x) must satisfy the following two conditions at the embedded end x = 0:

  • y(0) = 0 since there is no deflection, and
  • y′(0) = 0 since the deflection curve is tangent to the x-axis (in other words, the slope of the deflection curve is zero at this point).

At x = L the free-end conditions are

  • y″(L) = 0 since the bending moment is zero, and
  • y‴(L) = 0 since the shear force is zero.

The function F(x) = dM/dx = EI d3y/dx3 is called the shear force. If an end of a beam is simply supported or hinged (also called pin supported, and fulcrum supported), then we must have y = 0 and y″ = 0 at that end. Table 3.9.1 summarizes the boundary conditions that are associated with (4). See FIGURE 3.9.2.

Three parts show three different types of beams. Part (a) has a beam embedded at both ends. The beam is deflected downward at the center. The left end of the beam is labeled x = 0. The right end of the beam is labeled x = L. Part (b) has a cantilever beam embedded at the left end, free at the right end. The beam is deflected downward at the free end. The left end of the beam is labeled x = 0. The right end of the beam is labeled x = L. Part (c) has a simply supported beam supported at both ends. The beam is slightly deflected downward at the center. The left end of the beam is labeled x = 0. The right end of the beam is labeled x = L.

FIGURE 3.9.2 Beams with various end conditions

TABLE 3.9.1.
Ends of the Beam Boundary Conditions
Embedded y = 0, y′ = 0
Free y″ = 0, y‴ = 0
Simply supported or hinged y = 0, y′ = 0

EXAMPLE 1 An Embedded Beam

A beam of length L is embedded at both ends. Find the deflection of the beam if a constant load w0 is uniformly distributed along its length—that is, w(x) = w0, 0 < x < L.

SOLUTION

From (4), we see that the deflection y(x) satisfies

Because the beam is embedded at both its left end (x = 0) and right end (x = L), there is no vertical deflection and the line of deflection is horizontal at these points. Thus the boundary conditions are

y(0) = 0, y′(0) = 0, y(L) = 0, y′(L) = 0.

We can solve the nonhomogeneous differential equation in the usual manner (find yc by observing that m = 0 is a root of multiplicity four of the auxiliary equation m4 = 0, and then find a particular solution yp by undetermined coefficients), or we can simply integrate the equation d4y/dx4 = w0/EI four times in succession. Either way, we find the general solution of the equation y = yc + yp to be

Now the conditions y(0) = 0 and y′(0) = 0 give, in turn, c1 = 0 and c2 = 0, whereas the remaining conditions y(L) = 0 and y′(L) = 0 applied to y(x) = c3x2 + c4x3 + x4 yield the simultaneous equations

Solving this system gives c3 = w0L2/24EI and c4 = − w0L/12EI. Thus the deflection is

or . By choosing w0 = 24EI, and L = 1, we obtain the graph of the deflection curve in FIGURE 3.9.3. ≡

A deflection curve of a beam embedded at both ends is graphed on an x y coordinate plane. The curve starts at the origin (0, 0), goes down and to the right, and reaches a low point in the fourth quadrant. The low point corresponds to the x value of 0.5. Then, it goes up and to the right and ends at the point (1, 0). The curve is slightly flat at both ends.

FIGURE 3.9.3 Deflection curve for Example 1

The discussion of the beam notwithstanding, a physical system that is described by a two-point boundary-value problem usually involves a second-order differential equation. Hence, for the remainder of the discussion in this section we are concerned with boundary-value problems of the type

Solve: (5)

Subject to: (6)

In (5) we assume that the coefficients a0(x), a1(x), a2(x), and g(x) are continuous on the interval [a, b] and that a2(x) ≠ 0 for all x in the interval. In (6) we assume that A1 and B1 are not both zero and A2 and B2 are not both zero. When g(x) = 0 for all x in [a, b] and C1 and C2 are 0, we say that the boundary-value problem is homogeneous. Otherwise, we say that the boundary-value problem is nonhomogeneous. For example, the BVP y″ + y = 0, y(0) = 0, y(π) = 0 is homogeneous, whereas the BVP y″ + y = 1, y(0) = 0, y(2π) = 0 is nonhomogeneous.

Eigenvalues and Eigenfunctions

In applications involving homogeneous boundary-value problems, one or more of the coefficients in the differential equation (5) may depend on a constant parameter λ. As a consequence the solutions y1(x) and y2(x) of the homogeneous DE (5) also depend on λ. We often wish to determine those values of the parameter for which the boundary-value problem has nontrivial solutions. The next example illustrates this idea.

EXAMPLE 2 Nontrivial Solutions of a BVP

Solve the homogeneous boundary-value problem

y″ + λy = 0, y(0) = 0, y(L) = 0.

SOLUTION

We consider three cases: λ = 0, λ < 0, and λ > 0.

Case I: For λ = 0, the solution of the DE y″ = 0 is y = c1x + c2. Applying the boundary conditions y(0) = 0 and y(L) = 0 to this solution yields, in turn, c2 = 0 and c1 = 0. Hence for λ = 0, the only solution of the boundary-value problem is the trivial solution y = 0.

Case II: For λ < 0, it is convenient to write λ = −α2, where α > 0. With this new notation the auxiliary equation is m2 −α2 = 0 and has roots m1 = α and m2 = −α. Because the interval on which we are working is finite, we choose to write the general solution of y″ − α2y = 0 in the hyperbolic form y = c1 cosh α x + c2 sinh α x. From y(0) = 0 we see

y(0) = c1 cosh 0 + c2 sinh 0 = c1 · 1 + c2 · 0 = c1

implies c1 = 0. Hence y = c2 sinh αx. The second boundary condition y(L) = 0 then requires c2 sinh αL = 0. When α ≠ 0, sinh αL ≠ 0, and so we are forced to choose c2 = 0. Once again the only solution of the BVP is the trivial solution y = 0.

Case III: For λ > 0 we write λ = α2, where α > 0. The auxiliary equation m2 + α2 = 0 now has complex roots m1 = iα and m2 = −iα, and so the general solution of the DE y″ + α2y = 0 is y = c1 cos αx + c2 sin αx. As before, y(0) = 0 yields c1 = 0 and so y = c2 sin αx. Then y(L) = 0 implies

c2 sin αL = 0.

If c2 = 0, then necessarily y = 0. But this time we can require c2 ≠ 0 since sin αL = 0 is satisfied whenever α L is an integer multiple of π:

.

Therefore for any real nonzero c2, yn = c2 sin(nπx/L) is a solution of the problem for each positive integer n. Since the differential equation is homogeneous, any constant multiple of a solution is also a solution. Thus we may, if desired, simply take c2 = 1. In other words, for each number in the sequence

,

the corresponding function in the sequence

,

is a nontrivial solution of the original problem. ≡

The numbers λn = n2π2/L2, n = 1, 2, 3, ... for which the boundary-value problem in Example 2 possesses nontrivial solutions are known as eigenvalues. The nontrivial solutions that depend on these values of λn, yn = c2 sin(nπx/L) or simply yn = sin(nπx/L) are called eigenfunctions. The graphs of the eigenfunctions for n = 1, 2, 3, 4, 5 are shown in FIGURE 3.9.4. Note that each of the five graphs passes through the two points (0, 0) and (0, L).

Five curves are graphed on an x y coordinate plane. The curve for n = 1 starts at the point (0, 0), goes up and to the right, reaches a high point (L over 2, 1), then goes down and to the right, and ends at the point (L, 0). The curve for n = 2 starts at the point (0, 0), goes up and to the right, reaches a high point (L over 4, 1), then goes down and to the right through the point (L over 2, 0), and reaches a low point (3L over 4, negative 1). Again, it goes up and to the right and ends at the point (L, 0). The curve for n = 3 starts at the point (0, 0), goes up and to the right, reaches a high point in the first quadrant corresponding to y = 1, then goes down and to the right through the positive x axis, and reaches a low point (L over 2, negative 1). Again, it goes up and to the right through the positive x axis, reaches a high point in the first quadrant corresponding to y = 1, then goes down and to the right and ends at the point (L, 0). The curve for n = 4 starts at the point (0, 0), goes up and to the right, reaches a high point in the first quadrant corresponding to y = 1, then goes down and to the right through the point (L over 4, 0), and reaches a low point in the fourth quadrant corresponding to y = negative 1. Again, it goes up and to the right through the point (L over 2, 0), and reaches a high point in the first quadrant corresponding to y = 1. Then, it goes down and to the right through the point (3L over 4, 0), reaches a low point in the fourth quadrant corresponding to y = negative 1, again goes up and to the right, and ends at the point (L, 0). The curve for n = 5 starts at the point (0, 0), goes up and to the right, reaches a high point in the first quadrant corresponding to y = 1, then goes down and to the right through the positive x axis, and reaches a low point in the fourth quadrant corresponding to y equals negative 1. Again, it goes up and to the right through the positive x axis, and reaches a high point (L over 2, 1). Then, it goes down and to the right through the positive x axis, reaches a low point in the fourth quadrant corresponding to y = negative 1, again goes up and to the right, reaches a high point in the first quadrant corresponding to y = 1, then goes down and to the right, and ends at the point (L, 0).

FIGURE 3.9.4 Graphs of eigenfunctions yn = sin(nπx/L), for n = 1, 2, 3, 4, 5

EXAMPLE 3 Example 2 Revisited

If we choose L = π in Example 2, then the eigenvalues of the problem

y″ + λy = 0, y(0) = 0, y(π) = 0

are λn = n2, n = 1, 2, 3, .... It follows then that the boundary-value problem

y″ + 10y = 0, y(0) = 0, y(π) = 0

possesses only the trivial solution y = 0 because 10 is not an eigenvalue; that is, 10 is not the square of a positive integer. ≡

Buckling of a Thin Vertical Column

In the eighteenth century Leonhard Euler was one of the first mathematicians to study an eigenvalue problem in analyzing how a thin elastic column buckles under a compressive axial force.

Consider a long slender vertical column of uniform cross section and length L. Let y(x) denote the deflection of the column when a constant vertical compressive force, or load, P is applied to its top, as shown in FIGURE 3.9.5. By comparing bending moments at any point along the column we obtain

(7)

where E is Young’s modulus of elasticity and I is the moment of inertia of a cross section about a vertical line through its centroid.

The illustration has two parts. Part (a) has a long vertical elastic column of uniform cross section fixed at both ends. The length of the vertical column is labeled L. Part (b) has a deflected vertical column fixed at both ends. The top end of the column is labeled x = 0. The bottom end of the column is labeled x = L. A force P is applied on the top end and it acts downward. A dashed vertical line connects the two ends of the columns. A dashed curve passes through the center of the column taking the shape of the column. A downward arrow and a right arrow are placed at the top end where x = 0.

FIGURE 3.9.5 Elastic column buckling under a compressive force

EXAMPLE 4 The Euler Load

Find the deflection of a thin vertical homogeneous column of length L subjected to a constant axial load P if the column is simply supported or hinged at both ends.

SOLUTION

The boundary-value problem to be solved is

First note that y = 0 is a perfectly good solution of this problem. This solution has a simple intuitive interpretation: If the load P is not great enough, there is no deflection. The question then is this: For what values of P will the column bend? In mathematical terms: For what values of P does the given boundary-value problem possess nontrivial solutions?

By writing λ = P/EI we see that

y″ + λy = 0, y(0) = 0, y(L) = 0

is identical to the problem in Example 2. From Case III of that discussion we see that the deflection curves are yn(x) = c2 sin(nπx/L), corresponding to the eigenvalues λn = Pn/EI = n2π2/L2, n = 1, 2, 3, .... Physically this means that the column will buckle or deflect only when the compressive force is one of the values Pn = n2π2EI/L2, n = 1, 2, 3, .... These different forces are called critical loads. The deflection curve corresponding to the smallest critical load P1 = π2EI/L2, called the Euler load, is y1(x) = c2 sin(πx/L) and is known as the first buckling mode. ≡

The deflection curves in Example 4 corresponding to n = 1, n = 2, and n = 3 are shown in FIGURE 3.9.6. Note that if the original column has some sort of physical restraint put on it at x = L/2, then the smallest critical load will be P2 = 4π2EI/L2 and the deflection curve will be as shown in Figure 3.9.6(b). If restraints are put on the column at x = L/3 and at x = 2L/3, then the column will not buckle until the critical load P3 = 9π2EI/L2 is applied and the deflection curve will be as shown in Figure 3.9.6(c). See Problem 25 in Exercises 3.9.

Three graphs show three deflection curves of a thin vertical homogeneous column of length L. The curves are graphed on an x y coordinate plane. However, the x axis is now vertical and directed downward and the y axis is horizontal and directed to the right. The curve shown in part (a) starts at the point (0, 0), reaches a point to the right of the x axis corresponding to x = L over 2, and ends at the point (L, 0). The curve shown in part (b) starts at the point (0, 0), reaches a point to the right of the x axis corresponding to x = L over 4, and goes to the point (L over 2, 0). Then, it goes to a point to the left of the x axis corresponding to x = 3L over 4, and ends at the point (L, 0). The curve shown in part (c) starts at the point (0, 0), reaches a point to the right of the x axis, goes through the x axis, and reaches a point to the left of the x axis corresponding to x = L over 2. Then, it goes through the x axis, reaches a point to the right of the x axis, and ends at the point (L, 0).

FIGURE 3.9.6 Deflection curves for compressive forces P1, P2, P3

Rotating String

The simple linear second-order differential equation

y″ + λy = 0 (8)

occurs again and again as a mathematical model. In Section 3.8 we saw (8) in the forms d2x/dt2 + (k/m)x = 0 and d2q/dt2 + (1/LC)q = 0 as models for, respectively, the simple harmonic motion of a spring/mass system and the simple harmonic response of a series circuit. It is apparent when the model for the deflection of a thin column in (7) is written as d2y/dx2 + (P/EI)y = 0 that it is the same as (8). We encounter the basic equation (8) one more time in this section: as a model that defines the deflection curve or the shape y(x) assumed by a rotating string. The physical situation is analogous to when two persons hold a jump rope and twirl it in a synchronous manner. See FIGURE 3.9.7 parts (a) and (b).

Three illustrations are shown in three parts. Part (a) has two persons holding a jump rope at both ends and rotating it. Two positions of the rope are shown. When the rope is at top, it is concave downward. When the rope is at bottom, it is concave upward. The direction of rotation of the rope is marked with a downward rotating arrow. Part (b) has the deflection curve of the rope. A horizontal line represents the rope at normal position. The curve starts at x = 0, reaches a high point corresponding to x = L over 2, and ends at the point (L, 0). An upward arrow starts at (L over 2, 0) and reaches the high point. It is labeled: y(x). The direction of rotation of the rope is marked with a downward rotating arrow labeled lowercase omega. Part (c) has the forces acting on the portion of the rope between x and x + delta x. At x, an arrow along the rope points left and is labeled T subscript 1. A horizontal line is drawn to the left from the end of the rope that corresponds to x. The angle between T subscript 1 and the horizontal line is labeled theta subscript 1. At x + delta x, an arrow along the rope points right and is labeled T subscript 2. A horizontal line is drawn to the right from the end of the rope that corresponds to x + delta x. The angle between T subscript 2 and the horizontal line is labeled theta subscript 2.

FIGURE 3.9.7 Rotating rope and forces acting on it

Suppose a string of length L with constant linear density ρ (mass per unit length) is stretched along the x-axis and fixed at x = 0 and x = L. Suppose the string is then rotated about that axis at a constant angular speed ω. Consider a portion of the string on the interval [x, x + Δx], where Δx is small. If the magnitude T of the tension T, acting tangential to the string, is constant along the string, then the desired differential equation can be obtained by equating two different formulations of the net force acting on the string on the interval [x, x + Δx]. First, we see from Figure 3.9.7(c) that the net vertical force is

F = T sin θ2 − T sin θ1. (9)

When angles θ1 and θ2 (measured in radians) are small, we have sin θ2 ≈ tan θ2 and sin θ1 ≈ tan θ1. Moreover, since tan θ2 and tan θ1 are, in turn, slopes of the lines containing the vectors T2 and T1, we can also write

tan θ2 = y′(x + Δx) and tan θ1 = y′(x).

Thus (9) becomes

F ≈ T [y′(x + Δx) − y′(x)]. (10)

Second, we can obtain a different form of this same net force using Newton’s second law, F = ma. Here the mass of string on the interval is m = ρΔx; the centripetal acceleration of a body rotating with angular speed ω in a circle of radius r is a = rω2. With Δx small we take r = y. Thus the net vertical force is also approximated by

F ≈ −(ρΔx)yω2, (11)

where the minus sign comes from the fact that the acceleration points in the direction opposite to the positive y-direction. Now by equating (10) and (11) we have

(12)

For Δx close to zero the difference quotient in (12) is approximately the second derivative d2y/dx2. Finally we arrive at the model

(13)

Since the string is anchored at its ends x = 0 and x = L, we expect that the solution y(x) of equation (13) should also satisfy the boundary conditions y(0) = 0 and y(L) = 0.

REMARKS

(i) We will pursue the subject of eigenvalues and eigenfunctions for linear second-order differential equations in greater detail in Section 12.5.

(ii) Eigenvalues are not always easily found as they were in Example 2; you may have to approximate roots of equations such as tan x = −x or cos x cosh x = 1. See Problems 32 and 38–42 in Exercises 3.9.

(iii) Boundary conditions can lead to a homogeneous algebraic system of linear equations where the unknowns are the coefficients ci in the general solution of the DE. Such a system is always consistent, but in order to possess a nontrivial solution (in the case when the number of equations equals the number of unknowns) we must have the determinant of the coefficients equal to zero. See Problems 21 and 22 in Exercises 3.9.

3.9 Exercises Answers to selected odd-numbered problems begin on page ANS-7.

Deflection of a Beam

In Problems 1−5, solve equation (4) subject to the appropriate boundary conditions. The beam is of length L, and w0 is a constant.

    1. The beam is embedded at its left end and free at its right end, and w(x) = w0, 0 < x < L.
    2. Use a graphing utility to graph the deflection curve when w0 = 24EI and L = 1.
    1. The beam is simply supported at both ends, and w(x) = w0, 0 < x < L.
    2. Use a graphing utility to graph the deflection curve when w0 = 24EI and L = 1.
    1. The beam is embedded at its left end and simply supported at its right end, and w(x) = w0, 0 < x < L.
    2. Use a graphing utility to graph the deflection curve when w0 = 48EI and L = 1.
    1. The beam is embedded at its left end and simply supported at its right end, and w(x) = w0 sin(πx/L), 0 < x < L.
    2. Use a graphing utility to graph the deflection curve when w0 = 2π3EI and L = 1.
    3. Use a root-finding application of a CAS (or a graphic calculator) to approximate the point in the graph in part (b) at which the maximum deflection occurs. What is the maximum deflection?
    1. The beam is simply supported at both ends, and w(x) = w0x, 0 < x < L.
    2. Use a graphing utility to graph the deflection curve when w0 = 36EI and L = 1.
    3. Use a root-finding application of a CAS (or a graphic calculator) to approximate the point in the graph in part (b) at which the maximum deflection occurs. What is the maximum deflection?
    1. Find the maximum deflection of the cantilever beam in Problem 1.
    2. How does the maximum deflection of a beam that is half as long compare with the value in part (a)?
    3. Find the maximum deflection of the simply supported beam in Problem 2.
    4. How does the maximum deflection of the simply supported beam in part (c) compare with the value of maximum deflection of the embedded beam in Example 1?
  7. A cantilever beam of length L is embedded at its right end, and a horizontal tensile force of P pounds is applied to its free left end. When the origin is taken at its free end, as shown in FIGURE 3.9.8, the deflection y(x) of the beam can be shown to satisfy the differential equation

    Find the deflection of the cantilever beam if w(x) = w0x, 0 < x < L, and y(0) = 0, y′(L) = 0.

    A cantilever beam of length L is embedded at its right end. A horizontal force P acts to the left at the free end. The beam is deflected downward at its free end. A deflection curve is graphed on an x y coordinate plane. The curve starts at the origin labeled O which is at the free end, goes up and to the right, and ends at the embedded right end of the beam. A downward force (w subscript 0)x acts on the curve at a distance x from the origin.

    FIGURE 3.9.8 Deflection of cantilever beam in Problem 7

  8. When a compressive instead of a tensile force is applied at the free end of the beam in Problem 7, the differential equation of the deflection is

    Solve this equation if w(x) = w0x, 0 < x < L, and y(0) = 0, y′(L) = 0.

Contributed Problem

Rick Wicklin, PhD
Senior Researcher in Computational Statistics, SAS Institute Inc.

  1. Blowing in the Wind In September 1989, Hurricane Hugo hammered the coast of South Carolina with winds estimated at times to be as high as 60.4 m/s (135 mi/h). Of the billions of dollars in damage, approximately $420 million of this was due to the market value of loblolly pine (Pinus taeda) lumber in the Francis Marion National Forest. One image from that storm remains hauntingly bizarre: All through the forest and surrounding region, thousands upon thousands of pine trees lay pointing exactly in the same direction, and all the trees were broken 5–8 meters from their base. In September 1996, Hurricane Fran destroyed over 8.2 million acres of timber forest in eastern North Carolina. As happened seven years earlier, the planted loblolly trees all broke at approximately the same height. This seems to be a reproducible phenomenon, brought on by the fact that the trees in these planted forests are approximately the same age and size.
    A photo of broken loblolly pines in the Francis Marion National Forest shows damaged caused by a hurricane.

    Courtesy of Texas Forest Service

    Wind damage to loblolly pines

    In this problem, we are going to examine a mathematical model for the bending of loblolly pines in strong winds, and then use the model to predict the height at which a tree will break in hurricane-force winds.*

    Wind hitting the branches of a tree transmits a force to the trunk of the tree. The trunk is approximately a big cylindrical beam of length L, and so we will model the deflection y(x) of the tree with the static beam equation EIy(4) = w(x) (equation (4) in this section), where x is distance measured in meters from ground level. Since the tree is rooted into the ground, the accompanying boundary conditions are those of a cantilevered beam: y(0) = 0, y′(0) = 0 at the rooted end, and at the free end, which is the top of the tree.

    1. Loblolly pines in the forest have the majority of their crown (that is, branches and needles) in the upper 50% of their length, so let’s ignore the force of the wind on the lower portion of the tree. Furthermore, let’s assume that the wind hitting the tree’s crown results in a uniform load per unit length w0. In other words, the load on the tree is modeled by

      We can determine y(x) by integrating both sides of Integrate w(x) on [0, L/2] and then on [L/2, L] to find an expression for on each of these intervals. Let c1 be the constant of integration on [0, L/2] and c2 be the constant of integration on [L/2, L]. Apply the boundary condition and solve for c2. Then find the value of c1 that ensures continuity of the third derivative at the point x = L/2.

    2. Following the same procedure as in part (a) show that

      Integrate EIy′ to obtain the deflection y(x).

    3. Note that in our model y(L) describes the maximum amount by which the loblolly will bend. Compute this quantity in terms of the problem’s parameters.
  2. Blowing in the Wind—Continued By making some assumptions about the density of crown’s foliage, the total force F on the tree can be calculated using a formula from physics: F = ρAv2/6 where ρ ≈ 1.225 kg/m3 is the density of air, v is the wind speed in m/s, and A is the cross-sectional area of the tree’s crown. If we assume that the crown is roughly cylindrical, then its cross section is a rectangle of area A = (2R)(L/2) = RL, where R is the average radius of the cylinder. Then the total force per unit length is then w0 = F/(L/2) = 0.408Rv2.

    The cross-sectional moment of inertia for a uniform cylindrical beam is where r is the radius of the cylinder (tree trunk).

    By your answer to part (c) in Problem 9 and the explanations above, the amount that a loblolly will bend depends on each of the parameters in the following table.

    Symbol Description Typical values
    r radius of trunk 0.15–0.25 m
    R radius of crown 3–4 m
    L height of pine 15–20 m
    E modulus of elasticity 11–14 N/m2
    v Hugo wind speeds 40–60 m/s (90–135 mi/h)
    1. Mathematically show how each parameter affects the bending, and explain in physical terms why this makes sense. (For example, a large value of E results in less bending since E appears in the denominator of y(L). This means that hard wood like oak bends less than a soft wood such as palm.)
    2. Graph y(x) for 40 m/s winds for an “average” tree by choosing average values of each parameter (for example, r = 0.2, R = 3.5, and so on). Graph y(x) for 60 m/s winds for a “tall” (but otherwise average) pine.
    3. Recall that in the derivation of the beam equation on pages 173–174 it was assumed that the deflection of the beam was small. What is the largest possible value of y(L) that is predicted by the model if all parameters are chosen from the given table? Is this prediction realistic, or is the mathematical model no longer valid for parameters in this range?
    4. The beam equation always predicts that a beam will bend, even if the load and flexural rigidity reflect an elephant standing on a toothpick! Different methods are used by engineers to predict when and where a beam will break. In particular, a beam subject to a load will break at the location where the stress function y″(x)/I(x) reaches a maximum. Differentiate the function in part (b) of Problem 9 to obtain EI y″, and use this to obtain the stress y″(x)/I(x).
    5. Real pine tree trunks are not uniformly wide; they taper as they approach the top of the tree. Substitute r(x) = 0.2 − x/(15L) into the equation for I and then use a graphing utility to graph the resulting stress as a function of height for an average loblolly. Where does the maximum stress occur? Does this location depend on the speed of the wind? On the radius of the crown? On the height of the pine tree? Compare the model to observed data from Hurricane Hugo.
    6. A mathematical model is sensitive to an assumption if small changes in the assumption lead to widely different predictions for the model. Repeat part (e) using r(x) = 0.2 − x/(20L) and r(x) = 0.2 − x/(10L) as formulas that describe the radius of a pine tree trunk that tapers. Is our model sensitive to our choice for these formulas?

Eigenvalues and Eigenfunctions

In Problems 11–20, find the eigenvalues and eigenfunctions for the given boundary-value problem.

  1. y″ + λy = 0, y(0) = 0, y(π) = 0
  2. y″ + λy = 0, y′(0) = 0, y(π/4) = 0
  3. y″ + λy = 0, y′(0) = 0, y(L) = 0
  4. y″ + λy = 0, y(0) = 0, y′(π/2) = 0
  5. y″ + λy = 0, y′(0) = 0, y(π) = 0
  6. y″ + λy = 0, y(−π) = 0, y(π) = 0
  7. y″ + 2y′ + (λ + 1)y = 0, y(0) = 0, y(5) = 0
  8. y″ + (λ + 1)y = 0, y′(0) = 0, y′(1) = 0
  9. x2y″ + xy′ + λy = 0, y(1) = 0, y(eπ) = 0
  10. x2y″ + xy′ + λy = 0, y′(e−1) = 0, y(1) = 0

In Problems 21 and 22, find the eigenvalues and eigenfunctions for the given boundary-value problem. Consider only the case λ = α4, α > 0. [Hint: Read (iii) in the Remarks.]



Buckling of a Thin Column

  1. Consider Figure 3.9.6. Where should physical restraints be placed on the column if we want the critical load to be P4? Sketch the deflection curve corresponding to this load.
  2. The critical loads of thin columns depend on the end conditions of the column. The value of the Euler load P1 in Example 4 was derived under the assumption that the column was hinged at both ends. Suppose that a thin vertical homogeneous column is embedded at its base (x = 0) and free at its top (x = L) and that a constant axial load P is applied to its free end. This load either causes a small deflection δ as shown in FIGURE 3.9.9 or does not cause such a deflection. In either case the differential equation for the deflection y(x) is

    1. What is the predicted deflection when δ = 0?
    2. When δ ≠ 0, show that the Euler load for this column is one-fourth of the Euler load for the hinged column in Example 4.
      A thin vertical homogeneous column is embedded at its base and free at its top. The base is labeled x = 0. The free end at the top is labeled x = L. The x axis is along the axis of symmetry of the column before deflection and the y axis is along the horizontal base. An axial load P acts downward at the free end. The column is deflected at its free end to the right of the x axis. The horizontal distance between the x axis and the free end of the deflected column is labeled lowercase delta.

      FIGURE 3.9.9 Deflection of vertical column in Problem 24

  3. As was mentioned in Problem 24, the differential equation (7) that governs the deflection y(x) of a thin elastic column subject to a constant compressive axial force P is valid only when the ends of the column are hinged. In general, the differential equation governing the deflection of the column is given by

    Assume that the column is uniform (EI is a constant) and that the ends of the column are hinged. Show that the solution of this fourth-order differential equation subject to the boundary conditions y(0) = 0, y″(0) = 0, y(L) = 0, y″(L) = 0 is equivalent to the analysis in Example 4.

  4. Suppose that a uniform thin elastic column is hinged at the end x = 0 and embedded at the end x = L.
    1. Use the fourth-order differential equation given in Problem 25 to find the eigenvalues λn, the critical loads Pn, the Euler load P1, and the deflections yn(x).
    2. Use a graphing utility to graph the first buckling mode.

Rotating String

  1. Consider the boundary-value problem introduced in the construction of the mathematical model for the shape of a rotating string:

    For constant T and ρ, define the critical speeds of angular rotation ωn as the values of ω for which the boundary-value problem has nontrivial solutions. Find the critical speeds ωn and the corresponding deflections yn(x).

  2. When the magnitude of tension T is not constant, then a model for the deflection curve or shape y(x) assumed by a rotating string is given by

    Suppose that 1 < x < e and that T(x) = x2.

    1. If y(1) = 0, y(e) = 0, and ρω2 > 0.25, show that the critical speeds of angular rotation are

      ωn =

      and the corresponding deflections are

    2. Use a graphing utility to graph the deflection curves on the interval [1, e] for n = 1, 2, 3. Choose c2 = 1.

Additional Boundary-Value Problems

  1. Temperature in a Sphere Consider two concentric spheres of radius r = a and r = b, a < b. See FIGURE 3.9.10. The temperature u(r) in the region between the spheres is determined from the boundary-value problem

    where u0 and u1 are constants. Solve for u(r).

    Two concentric spheres are centered at the origin of a three-dimensional coordinate system. The inner sphere is labeled u = u subscript 0. The outer sphere is labeled u = u subscript 1.

    FIGURE 3.9.10 Concentric spheres in Problem 29

  2. Temperature in a Ring The temperature u(r) in the circular ring or annulus shown in FIGURE 3.9.11 is determined from the boundary-value problem

    ,

    where u0 and u1 are constants. Show that

    Two concentric circles of the circular ring are centered at the origin of a coordinate plane. The radius of the inner circle is labeled a, and the radius of the outer circle is labeled b. The inner circle is labeled u = u subscript 0. The outer circle is labeled u = u subscript 1.

    FIGURE 3.9.11 Circular ring in Problem 30

  3. Rotation of a Shaft Suppose the x-axis on the interval is the geometric center of a long straight shaft, such as the propeller shaft of a ship. When the shaft is rotating at a constant angular speed about this axis the deflection of the shaft satisfies the differential equation

    ,

    where is density per unit length. If the shaft is simply supported, or hinged, at both ends the boundary conditions are then

    .

    1. If , then find the eigenvalues and eigenfunctions for this boundary-value problem.
    2. Use the eigenvalues in part (a) to find corresponding angular speeds . The values are called critical speeds. The value is called the fundamental critical speed and, analogous to Example 4, at this speed the shaft changes shape from to a deflection given by .
      A photo of Propeller shaft of the battleship U S S Missouri.

      Courtesy of the National Archives and Records Administration

      Propeller shaft of the battleship USS Missouri

  4. In Problem 31, suppose . If the shaft is fixed at both ends then the boundary conditions are

    .

    1. Show that the eigenvalues are defined by the positive roots of . [Hint: See the instructions to Problems 21 and 22.]
    2. Show that the eigenfunctions are

                     .

Discussion Problems

  1. Simple Harmonic Motion The model mx″ + kx = 0 for simple harmonic motion, discussed in Section 3.8, can be related to Example 2 of this section.

    Consider a free undamped spring/mass system for which the spring constant is, say, k = 10 lb/ft. Determine those masses mn that can be attached to the spring so that when each mass is released at the equilibrium position at t = 0 with a nonzero velocity v0, it will then pass through the equilibrium position at t = 1 second. How many times will each mass mn pass through the equilibrium position in the time interval 0 < t < 1?

  2. Damped Motion Assume that the model for the spring/mass system in Problem 33 is replaced by mx″ + 2x′ + kx = 0. In other words, the system is free but is subjected to damping numerically equal to two times the instantaneous velocity. With the same initial conditions and spring constant as in Problem 33, investigate whether a mass m can be found that will pass through the equilibrium position at t = 1 second.

In Problems 35 and 36, determine whether it is possible to find values y0 and y1 (Problem 35) and values of L > 0 (Problem 36) so that the given boundary-value problem has (a) precisely one nontrivial solution, (b) more than one solution, (c) no solution, and (d) the trivial solution.

  1. y″ + 16y = 0, y(0) = y0, y(π/2) = y1
  2. y″ + 16y = 0, y(0) = 1, y(L) = 1
  3. Consider the boundary-value problem

    y″ + λy = 0, y(− π) = y(π), y′(− π) = y′(π).

    1. The type of boundary conditions specified are called periodic boundary conditions. Give a geometric interpretation of these conditions.
    2. Find the eigenvalues and eigenfunctions of the problem.
    3. Use a graphing utility to graph some of the eigenfunctions. Verify your geometric interpretation of the boundary conditions given in part (a).
  4. Show that the eigenvalues and eigenfunctions of the boundary-value problem

    y″ + λy = 0, y(0) = 0, y(1) + y′(1) = 0

    are λn = α2n and yn = sin αnx, respectively, where αn, n = 1, 2, 3, ... are the consecutive positive roots of the equation tan α = −α.

Computer Lab Assignments

  1. Use a CAS to plot graphs to convince yourself that the equation tan α = −α in Problem 38 has an infinite number of roots. Explain why the negative roots of the equation can be ignored. Explain why λ = 0 is not an eigenvalue even though α = 0 is an obvious solution of the equation tan α = −α.
  2. Use a root-finding application of a CAS to approximate the first four eigenvalues λ1, λ2, λ3, and λ4for the BVP in Problem 38.
  3. Use a CAS to approximate the eigenvalues of the boundary-value problem:

    .

    Give the corresponding approximate eigenfunctions .

  4. Use a CAS to approximate the eigenvalues defined by the equation in part (a) of Problem 32.

 

* For further information about the bending of trees in high winds, see the articles by W. Kubinec (Phys. Teacher, March, 1990), or F. Mergen (J. Forest.) 52(2), 1954.