EXAMPLE 1 Eigenvalue Problem

Find the eigenvalues and eigenvectors of the matrix

Solution. The characteristic equation is the quadratic equation

It has the solutions λ₁ = −2 and λ₂ = −0.8. These are the eigenvalues of A.

Eigenvectors are obtained from (14*). For λ = λ₁ = −2 we have from (14*)

A solution of the first equation is x₁ = 2, x₂ = 1. This also satisfies the second equation. (Why?) Hence an eigenvector of A corresponding to λ₁ = −2.0 is

is an eigenvector of A corresponding to λ₂ = −0.8, as obtained from (14*) with λ = λ₂. Verify this.

EXAMPLE 1 Mixing Problem Involving Two Tanks

A mixing problem involving a single tank is modeled by a single ODE, and you may first review the corresponding Example 3 in Sec. 1.3 because the principle of modeling will be the same for two tanks. The model will be a system of two first-order ODEs.

Tank T₁ and T₂ in Fig. 78 contain initially 100 gal of water each. In T₁ the water is pure, whereas 150 lb of fertilizer are dissolved in T₂. By circulating liquid at a rate of 2 gal/min and stirring (to keep the mixture uniform) the amounts of fertilizer y₁(t) in T₁ and y₂(t) in T₂ change with time t. How long should we let the liquid circulate so that T₁ will contain at least half as much fertilizer as there will be left in T₂?

Fig. 78. Fertilizer content in Tanks (lower curve) and T₂

Solution. Step 1. Setting up the model. As for a single tank, the time rate of change y′₁(t) of y₁(t) equals inflow minus outflow. Similarly for tank T₂. From Fig. 78 we see that

Hence the mathematical model of our mixture problem is the system of first-order ODEs

As a vector equation with column vector and matrix A this becomes

Step 2. General solution. As for a single equation, we try an exponential function of t,

Dividing the last equation λxe^λt = Axe^λt by e^λt and interchanging the left and right sides, we obtain

We need nontrivial solutions (solutions that are not identically zero). Hence we have to look for eigenvalues and eigenvectors of A. The eigenvalues are the solutions of the characteristic equation

We see that λ₁ = 0 (which can very well happen—don't get mixed up—it is eigen vectors that must not be zero) and λ₂ = −0.04. Eigenvectors are obtained from (14*) in Sec. 4.0 with λ = 0 and λ = −0.04. For our present A this gives [we need only the first equation in (14*)]

respectively. Hence x₁ = x₂ and x₁ = −x₂, respectively, and we can take x₁ = x₂ = 1 and x₁ = −x₂ = 1. This gives two eigenvectors corresponding to λ₁ = 0 and λ₂ = −0.04, respectively, namely,

From (1) and the superposition principle (which continues to hold for systems of homogeneous linear ODEs) we thus obtain a solution

where c₁ are c₂ arbitrary constants. Later we shall call this a general solution.

Step 3. Use of initial conditions. The initial conditions are y₁(0) = 0 (no fertilizer in tank T₁) and y₂(0) = 150. From this and (3) with t = 0 we obtain

In components this is c₁ + c₂ = 0, c₁ − c₂ = 150. The solution is c₁ = 75, c₂ = −75. This gives the answer

In components,

Figure 78 shows the exponential increase of y₁ and the exponential decrease of y₂ to the common limit 75 lb. Did you expect this for physical reasons? Can you physically explain why the curves look “symmetric”? Would the limit change if T₁ initially contained 100 lb of fertilizer and T₂ contained 50 lb?

Step 4. Answer. T₁ contains half the fertilizer amount of T₂ if it contains 1/3 of the total amount, that is, 50 lb. Thus

Hence the fluid should circulate for at least about half an hour.

EXAMPLE 2 Electrical Network

Find the currents I₁(t) and I₂(t) in the network in Fig. 79. Assume all currents and charges to be zero at t = 0, the instant when the switch is closed.

Fig. 79. Electrical network in Example 2

Solution. Step 1. Setting up the mathematical model. The model of this network is obtained from Kirchhoff's Voltage Law, as in Sec. 2.9 (where we considered single circuits). Let I₁(t) and I₂(t) be the currents in the left and right loops, respectively. In the left loop, the voltage drops are LI′₁ = I′₁ [V] over the inductor and R₁(I₁ − I₂) = 4(I₁ − I₂)[V] over the resistor, the difference because I₁ and I₂ flow through the resistor in opposite directions. By Kirchhoff's Voltage Law the sum of these drops equals the voltage of the battery; that is, I′₁ + 4(I₁ − I₂) = 12, hence

In the right loop, the voltage drops are R₂I₂ = 6I₂[V] and R₁(I₂ − I₁) = 4(I₂ − I₁)[V] over the resistors and (I/C)∫ I₂ dt = 4 ∫ I₂ dt[V] over the capacitor, and their sum is zero,

Division by 10 and differentiation gives I′₂ − 0.4I′₁ + 0.4I₂ = 0.

To simplify the solution process, we first get rid of 0.4I′₁, which by (4a) equals 0.4(−4I₁ + 4I₂ + 12). Substitution into the present ODE gives

and by simplification

In matrix form, (4) is (we write J since I is the unit matrix)

Step 2. Solving (5). Because of the vector g this is a nonhomogeneous system, and we try to proceed as for a single ODE, solving first the homogeneous system J′ = AJ (thus J′ − AJ = 0) by substituting J = xe^λt. This gives

Hence, to obtain a nontrivial solution, we again need the eigenvalues and eigenvectors. For the present matrix A they are derived in Example 1 in Sec. 4.0:

Hence a “general solution” of the homogeneous system is

For a particular solution of the nonhomogeneous system (5), since g is constant, we try a constant column vector J_p = a with components a₁, a₂. Then J′_p = 0, and substitution into (5) gives Aa + g = 0; in components,

The solution is a₁ = 3, a₂ = 0; thus . Hence

in components,

The initial conditions give

Hence c₁ = −4 and c₂ = 5. As the solution of our problem we thus obtain

In components (Fig. 80b),

Now comes an important idea, on which we shall elaborate further, beginning in Sec. 4.3. Figure 80a shows I₁(t) and I₂(t) as two separate curves. Figure 80b shows these two currents as a single curve [I₁(t), I₂(t)] in the I₁I₂-plane. This is a parametric representation with time t as the parameter. It is often important to know in which sense such a curve is traced. This can be indicated by an arrow in the sense of increasing t, as is shown. The I₁I₂-plane is called the phase plane of our system (5), and the curve in Fig. 80b is called a trajectory. We shall see that such “phase plane representations” are far more important than graphs as in Fig. 80a because they will give a much better qualitative overall impression of the general behavior of whole families of solutions, not merely of one solution as in the present case.

THEOREM 1 Conversion of an ODE

An nth-order ODE

can be converted to a system of n first-order ODEs by setting

This system is of the form

PROOF

The first n − 1 of these n ODEs follows immediately from (9) by differentiation. Also, y′_n = y⁽ⁿ⁾ by (9), so that the last equation in (10) results from the given ODE (8).

EXAMPLE 3 Mass on a Spring

To gain confidence in the conversion method, let us apply it to an old friend of ours, modeling the free motions of a mass on a spring (see Sec. 2.4)

For this ODE (8) the system (10) is linear and homogeneous,

Setting , we get in matrix form

The characteristic equation is

It agrees with that in Sec. 2.4. For an illustrative computation, let m = 1, c = 2, and k = 0.75. Then

This gives the eigenvalues λ₁ = −0.5 and λ₂ = −1.5. Eigenvectors follow from the first equation in A − λI = 0, which is −λx₁ + x₂ = 0. For λ₁ this gives 0.5x₁ + x₂ = 0, say, x₁ = 2, x₂ = −1. For λ₂ = −1.5 it gives 1.5x₁ + x₂ = 0, say, x₁ = 1, x₂ = −1.5. These eigenvectors

This vector solution has the first component

which is the expected solution. The second component is its derivative

PROBLEM SET 4.1

1–6 MIXING PROBLEMS

1. Find out, without calculation, whether doubling the flow rate in Example 1 has the same effect as halfing the tank sizes. (Give a reason.)
2. What happens in Example 1 if we replace T₁ by a tank containing 200 gal of water and 150 lb of fertilizer dissolved in it?
3. Derive the eigenvectors in Example 1 without consulting this book.
4. In Example 1 find a “general solution” for any ratio a = (flow rate)/(tank size), tank sizes being equal. Comment on the result.
5. If you extend Example 1 by a tank T₃ of the same size as the others and connected to T₂ by two tubes with flow rates as between T₁ and T₂, what system of ODEs will you get?
6. Find a “general solution” of the system in Prob. 5.

7–9 ELECTRICAL NETWORK

In Example 2 find the currents:

• 7. If the initial currents are 0 A and −3 A (minus meaning that I₂(0) flows against the direction of the arrow).
• 8. If the capacitance is changed to C = 5/27 F. (General solution only.)
• 9. If the initial currents in Example 2 are 28 A and 14 A.

10–13 CONVERSION TO SYSTEMS

Find a general solution of the given ODE (a) by first converting it to a system, (b), as given. Show the details of your work.

• 10. y″ + 3y′ + 2y = 0
• 11. 4y″ − 15y′ − 4y = 0
• 12. y″′ + 2y″ − y′ − 2y = 0
• 13. y″ + 2y′ − 24y = 0
• 14. TEAM PROJECT. Two Masses on Springs. (a) Set up the model for the (undamped) system in Fig. 81. (b) Solve the system of ODEs obtained. Hint. Try y = xe^ωt and set ω² = λ. Proceed as in Example 1 or 2. (c) Describe the influence of initial conditions on the possible kind of motions.

  

  Fig. 81. Mechanical system in Team Project

• 15. CAS EXPERIMENT. Electrical Network. (a) In Example 2 choose a sequence of values of C that increases beyond bound, and compare the corresponding sequences of eigenvalues of A. What limits of these sequences do your numeric values (approximately) suggest?

  (b) Find these limits analytically.

  (c) Explain your result physically.

  (d) Below what value (approximately) must you decrease C to get vibrations?

THEOREM 1 Existence and Uniqueness Theorem

Let f₁ …, f_n in (1) be continuous functions having continuous partial derivatives ∂f₁/∂y₁, …, ∂f₁/∂y_n, …, ∂f_n/∂y_n in some domain R of ty₁y₂ … y_n-space containing the point (t₀, K₁, … K_n). Then (1) has a solution on some interval t₀ − α < t < t₀ + α satisfying (2) , and this solution is unique.

THEOREM 2 Existence and Uniqueness in the Linear Case

Let the a_jk's and g_j's in (3) be continuous functions of t on an open interval α < t < β containing the point t = t₀. Then (3) has a solution y(t) on this interval satisfying (2) , and this solution is unique.

THEOREM 3 Superposition Principle or Linearity Principle

If y⁽¹⁾ and y⁽²⁾ are solutions of the homogeneous linear system (4) on some interval, so is any linear combination y = c₁y⁽¹⁾ + c₁y⁽²⁾.

PROOF

Differentiating and using (4), we obtain

THEOREM 1 General Solution

If the constant matrix A in the system (1) has a linearly independent set of n eigenvectors, then the corresponding solutions y⁽¹⁾, …, y⁽ⁿ⁾ in (4) form a basis of solutions of (1) , and the corresponding general solution is

EXAMPLE 1 Trajectories in the Phase Plane (Phase Portrait)

Find and graph solutions of the system.

In order to see what is going on, let us find and graph solutions of the system

Solution. By substituting y = xe^λt and y′ and y′ = λxe^λt dropping the exponential function we get Ax = λx. The characteristic equation is

This gives the eigenvalues λ₁ = −2 and λ₂ = −4. Eigenvectors are then obtained from

For λ₁ = −2 this is −x₁ + x₂ = 0. Hence we can take x⁽¹⁾ = [1 1]^T. For λ₂ = −4 this becomes x₁ + x₂ = 0, and an eigenvector is x⁽²⁾ = [1 −1]^T. This gives the general solution

Figure 82 shows a phase portrait of some of the trajectories (to which more trajectories could be added if so desired). The two straight trajectories correspond to c₁ = 0 and c₂ = 0 and the others to other choices of c₁, c₂.

EXAMPLE 1 (Continued) Improper Node (Fig. 82)

An improper node is a critical point P₀ at which all the trajectories, except for two of them, have the same limiting direction of the tangent. The two exceptional trajectories also have a limiting direction of the tangent at P₀ which, however, is different.

The system (8) has an improper node at 0, as its phase portrait Fig. 82 shows. The common limiting direction at 0 is that of the eigenvector x⁽¹⁾ = [1 1]^T because e^−4t goes to zero faster than e^−2t as t increases. The two exceptional limiting tangent directions are those of x⁽²⁾ = [1 −1]^T and −x⁽²⁾ = [−1 1]^T.

EXAMPLE 2 Proper Node (Fig. 83)

A proper node is a critical point P₀ at which every trajectory has a definite limiting direction and for any given direction d at P₀ there is a trajectory having d as its limiting direction.

The system

has a proper node at the origin (see Fig. 83). Indeed, the matrix is the unit matrix. Its characteristic equation (1 − λ)² = 0 has the root λ = 1. Any x ≠ 0 is an eigenvector, and we can take [1 0]^T and [0 1]^T. Hence a general solution is

EXAMPLE 3 Saddle Point (Fig. 84)

A saddle point is a critical point P₀ at which there are two incoming trajectories, two outgoing trajectories, and all the other trajectories in a neighborhood of P₀ bypass P₀.

The system

has a saddle point at the origin. Its characteristic equation (1 − λ)(−1 − λ) = 0 has the roots λ₁ = 1 and λ₂ = −1. For λ = 1 an eigenvector [1 0]^T is obtained from the second row of (A − λI)x = 0, that is, 0x₁ + (−1 − 1)x₂ = 0. For λ₂ = −1 the first row gives [0 1]^T. Hence a general solution is

This is a family of hyperbolas (and the coordinate axes); see Fig. 84.

EXAMPLE 4 Center (Fig. 85)

A center is a critical point that is enclosed by infinitely many closed trajectories.

The system

has a center at the origin. The characteristic equation λ² + 4 = 0 gives the eigenvalues 2i and −2i. For 2i an eigenvector follows from the first equation of −2ix₁ + x₂ = 0 of (A − λI)x = 0, say, [1 2i]^T. For λ = −2i that equation is −(−2i)x₁ + x₂ = 0 and gives, say, [1 −2i]^T. Hence a complex general solution is

A real solution is obtained from (12*) by the Euler formula or directly from (12) by a trick. (Remember the trick and call it a method when you apply it again.) Namely, the left side of (a) times the right side of (b) is −4y₁y′₁. This must equal the left side of (b) times the right side of (a). Thus,

This is a family of ellipses (see Fig. 85) enclosing the center at the origin.

EXAMPLE 5 Spiral Point (Fig. 86)

A spiral point is a critical point P₀ about which the trajectories spiral, approaching P₀ as t → ∞ (or tracing these spirals in the opposite sense, away from P₀).

The system

has a spiral point at the origin, as we shall see. The characteristic equation is λ² + 2λ + 2 = 0. It gives the eigenvalues −1 + i and −1 − i. Corresponding eigenvectors are obtained from (−1 − λ)x₁ + x₂ = 0. For λ = −1 + i this becomes −ix₁ + x₂ = 0 and we can take [1 i]^T as an eigenvector. Similarly, an eigenvector corresponding to −1 − i is [1 −i]^T. This gives the complex general solution

The next step would be the transformation of this complex solution to a real general solution by the Euler formula. But, as in the last example, we just wanted to see what eigenvalues to expect in the case of a spiral point. Accordingly, we start again from the beginning and instead of that rather lengthy systematic calculation we use a shortcut. We multiply the first equation in (13) by y₁, the second by y₂, and add, obtaining

We now introduce polar coordinates r, t, where . Differentiating this with respect to t gives 2rr′ = 2y₁y′₁ + 2y₂y′₂. Hence the previous equation can be written

For each real c this is a spiral, as claimed (see Fig. 86).

EXAMPLE 6 No Basis of Eigenvectors Available. Degenerate Node (Fig. 87)

This cannot happen if A in (1) is symmetric (a_kj = a_jk, as in Examples 1–3) or skew-symmetric (a_kj = −a_jk, thus a_jj = 0). And it does not happen in many other cases (see Examples 4 and 5). Hence it suffices to explain the method to be used by an example.

Find and graph a general solution of

Solution. A is not skew-symmetric! Its characteristic equation is

It has a double root λ = 3. Hence eigenvectors are obtained from (4 − λ)x₁ + x₂ = 0, thus from x₁ + x₂ = 0, say, x⁽¹⁾ = [1 −1]^T and nonzero multiples of it (which do not help). The method now is to substitute

with constant u = [u₁ u₂]^T into (14). (The xt-term alone, the analog of what we did in Sec. 2.2 in the case of a double root, would not be enough. Try it.) This gives

On the right, Ax = λx. Hence the terms λxte^λt cancel, and then division by e^λt gives

Here λ = 3 and x = [1 −1]^T, so that

A solution, linearly independent of x = [1 −1]^T, is u = [0 1]^T. This yields the answer (Fig. 87)

The critical point at the origin is often called a degenerate node. c₁y⁽¹⁾ gives the heavy straight line, with c₁ > 0 the lower part and c₁ < 0 the upper part of it. y⁽²⁾ gives the right part of the heavy curve from 0 through the second, first, and—finally—fourth quadrants. −y⁽²⁾ gives the other part of that curve.

PROBLEM SET 4.3

1–9 GENERAL SOLUTION

Find a real general solution of the following systems. Show the details.

1. y′₁ = y₁ + y₂

   y′₂ = 3y₁ − y₂

2. y′₁ = 6y₁ + 9y₂

   y′₂ = y₁ + 6y₂

3. y′₁ = y₁ + 2y₂

   

4. y′₁ = −8y₁ − 2y₂

   y′₂ = 2y₁ − 4y₂

5. y′₁ = 2y₁ + 5y₂

   y′₂ = 5y₁ + 12.5y₂

6. y′₁ = 2y₁ − 2y₂

   y′₂ = 2y₁ + 2y₂

7. y′₁ = y₂

   y′₂ = −y₁ + y₃

   y′₃ = −y₂

8. y′₁ = 8y₁ − y₂

   y′₂ = y₁ + 10y₂

9. y′₁ = 10y₁ − 10y₂ − 4y₃

   y′₂ = −10y₁ + y₂ − 14y₃

   y′₃ = −4y₁ − 14y₂ − 2y₃

10–15 IVPs

Solve the following initial value problems.

• 10. y′₁ = 2y₁ + 2y₂

  y′₂ = 5y₁ − y₂

  y₁(0) = 0, y₂(0) = 7

• 11.
• 12.
• 13. y′₁ = y₂

  y′₂ = y₁

  y₁(0) = 0, y₂(0) = 2

• 14. y′₁ = −y₁ − y₂

  y′₂ = y₁ − y₂

  y₁(0) = 1, y₂(0) = 0

• 15. y′₁ = 3y₁ + 2y₂

  y′₂ = 2y₁ + 3y₂

  y₁(0) = 0.5, y₂(0) = −0.5

16–17 CONVERSION

Find a general solution by conversion to a single ODE.

• 16. The system in Prob. 8.
• 17. The system in Example 5 of the text.
• 18. Mixing problem, Fig. 88. Each of the two tanks contains 200 gal of water, in which initially 100 lb (Tank T₁) and 200 lb (Tank T₂) of fertilizer are dissolved. The inflow, circulation, and outflow are shown in Fig. 88. The mixture is kept uniform by stirring. Find the fertilizer contents y₁(t) in T₁ and y₂(t) in T₂.

  

  Fig. 88. Tanks in Problem 18

• 19. Network. Show that a model for the currents I₁(t) and I₂(t) in Fig. 89 is

  

  Find a general solution, assuming that R = 3 Ω, L = 4 H, C = 1/12 F.

  

  Fig. 89. Network in Problem 19

• 20. CAS PROJECT. Phase Portraits. Graph some of the figures in this section, in particular Fig. 87 on the degenerate node, in which the vector y⁽²⁾ depends on t. In each figure highlight a trajectory that satisfies an initial condition of your choice.

DEFINITIONS Stable, Unstable, Stable and Attractive

A critical point P₀ of (1) is called stable² if, roughly, all trajectories of (1) that at some instant are close to P₀ remain close to P₀ at all future times; precisely: if for every disk D of radius > 0 with center P₀ there is a disk D_δ of radius δ > 0 with center P₀ such that every trajectory of (1) that has a point P₁ (corresponding to t = t₁, say) in D_δ has all its points corresponding to t t₁ in D. See Fig. 90.

P₀ is called unstable if P₀ is not stable.

P₀ is called stable and attractive (or asymptotically stable) if P₀ is stable and every trajectory that has a point in D_δ approaches P₀ as t → ∞. See Fig. 91.

EXAMPLE 1 Application of the Criteria in Tables 4.1 and 4.2

In Example 1, Sec 4.3, we have , p = −6, q = 8, Δ = 4, a node by Table 4.1(a), which is stable and attractive by Table 4.2(a).

EXAMPLE 2 Free Motions of a Mass on a Spring

What kind of critical point does my″ + cy′ + ky = 0 in Sec. 2.4 have?

Solution. Division by m gives y″ = −(k/m)y − (c/m)y′. To get a system, set y₁ = y, y₂ = y′ (see Sec. 4.1). Then y′₂ = y″ = −(k/m)y₁ − (c/m)y₂. Hence

We see that p = −c/m, q = k/m, Δ = (c/m)² − 4k/m. From this and Tables 4.1 and 4.2 we obtain the following results. Note that in the last three cases the discriminant Δ plays an essential role.

No damping. c = 0, p = 0, q > 0, a center.

Underdamping. c² < 4mk, p < 0, q > 0, Δ < 0, a stable and attractive spiral point.

Critical damping. c² = 4mk, p < 0, q > 0, Δ = 0, a stable and attractive node.

Overdamping. c² > 4mk, p < 0, q > 0, Δ > 0, a stable and attractive node.

PROBLEM SET 4.4

1–10 TYPE AND STABILITY OF CRITICAL POINT

Determine the type and stability of the critical point. Then find a real general solution and sketch or graph some of the trajectories in the phase plane. Show the details of your work.

1. y′₁ = y₁

   y′₂ = 2y₂

2. y′₁ = −4y₁

   y′₂ = −3y₂

3. y′₁ = y₂

   y′₂ = −9y₁

4. y′₁ = 2y₁ + y₂

   y′₂ = 5y₁ − 2y₂

5. y′₁ = −2y₁ + 2y₂

   y′₂ = −2y₁ − 2y₂

6. y′₁ = −6y₁ − y₂

   y′₂ = −9y₁ − 6y₂

7. y′₁ = y₁ + 2y₂

   y′₂ = 2y₁ + y₂

8. y′₁ = −y₁ + 4y₂

   y′₂ = 3y₁ − 2y₂

9. y′₁ = 4y₁ + y₂

   y′₂ = 4y₁ + 4y₂

10. y′₁ = y₂

    y′₂ = −5y₁ − 2y₂

11–18 TRAJECTORIES OF SYSTEMS AND SECOND-ORDER ODEs. CRITICAL POINTS

• 11. Damped oscillations. Solve y″ + 2y′ + 2y = 0. What kind of curves are the trajectories?
• 12. Harmonic oscillations. Solve . Find the trajectories. Sketch or graph some of them.
• 13. Types of critical points. Discuss the critical points in (10)–(13) of Sec. 4.3 by using Tables 4.1 and 4.2.
• 14. Transformation of parameter. What happens to the critical point in Example 1 if you introduce τ = −t as a new independent variable?
• 15. Perturbation of center. What happens in Example 4 of Sec. 4.3 if you change A to A + 0.1I, where I is the unit matrix?
• 16. Perturbation of center. If a system has a center as its critical point, what happens if you replace the matrix A by with any real number k ≠ 0 (representing measurement errors in the diagonal entries)?
• 17. Perturbation. The system in Example 4 in Sec. 4.3 has a center as its critical point. Replace each a_jk in Example 4, Sec. 4.3, by a_jk + b. Find values of b such that you get (a) a saddle point, (b) a stable and attractive node, (c) a stable and attractive spiral, (d) an unstable spiral, (e) an unstable node.
• 18. CAS EXPERIMENT. Phase Portraits. Graph phase portraits for the systems in Prob. 17 with the values of b suggested in the answer. Try to illustrate how the phase portrait changes “continuously” under a continuous change of b.
• 19. WRITING PROBLEM. Stability. Stability concepts are basic in physics and engineering. Write a two-part report of 3 pages each (A) on general applications in which stability plays a role (be as precise as you can), and (B) on material related to stability in this section. Use your own formulations and examples; do not copy.
• 20. Stability chart. Locate the critical points of the systems (10)–(14) in Sec. 4.3 and of Probs. 1, 3, 5 in this problem set on the stability chart.

THEOREM 1 Linearization

If f₁ and f₂ in (1) are continuous and have continuous partial derivatives in a neighborhood of the critical point P₀: (0, 0), and if det A ≠ 0 in (2), then the kind and stability of the critical point of (1) are the same as those of the linearized system

Exceptions occur if A has equal or pure imaginary eigenvalues; then (1) may have the same kind of critical point as (3) or a spiral point.

EXAMPLE 1 Free Undamped Pendulum. Linearization

Figure 93a shows a pendulum consisting of a body of mass m (the bob) and a rod of length L. Determine the locations and types of the critical points. Assume that the mass of the rod and air resistance are negligible.

Solution. Step 1. Setting up the mathematical model. Let θ denote the angular displacement, measured counterclockwise from the equilibrium position. The weight of the bob is mg (g the acceleration of gravity). It causes a restoring force mg sin θ tangent to the curve of motion (circular arc) of the bob. By Newton's second law, at each instant this force is balanced by the force of acceleration mLθ″, where Lθ″ is the acceleration; hence the resultant of these two forces is zero, and we obtain as the mathematical model

Dividing this by mL, we have

When θ is very small, we can approximate sin θ rather accurately by θ and obtain as an approximate solution , but the exact solution for any θ is not an elementary function.

Step 2. Critical points (0, 0), (±2π, 0), (±4π, 0), …, Linearization. To obtain a system of ODEs, we set θ = y₁, θ′ = y₂. Then from (4) we obtain a nonlinear system (1) of the form

The right sides are both zero when y₂ = 0 and sin y₁ = 0. This gives infinitely many critical points (nπ, 0), where n = 0, ±1, ±2, …. We consider (0, 0). Since the Maclaurin series is

the linearized system at (0, 0) is

To apply our criteria in Sec. 4.4 we calculate p = a₁₁ + a₂₂ = 0, q = det A = k = g/L (>0), and Δ = p² − 4q = −4k. From this and Table 4.1(c) in Sec. 4.4 we conclude that is a center, which is always stable. Since sin θ = sin y₁ is periodic with period 2π, the critical points (nπ, 0), n = ±2, ±4, …, are all centers.

Step 3. Critical points (±π, 0), (±3π, 0), (±5π, 0), …, Linearization. We now consider the critical point (π, 0), setting θ − π = y₁ and (θ − π)′ = θ′ = y₂. Then in (4),

and the linearized system at (π, 0) is now

We see that p = 0, q = −k (<0), and Δ = −4q = 4k. Hence, by Table 4.1(b), this gives a saddle point, which is always unstable. Because of periodicity, the critical points (nπ, 0), n= ±1, ±3, …, are all saddle points. These results agree with the impression we get from Fig. 93b.

EXAMPLE 2 Linearization of the Damped Pendulum Equation

To gain further experience in investigating critical points, as another practically important case, let us see how Example 1 changes when we add a damping term cθ′ (damping proportional to the angular velocity) to equation (4), so that it becomes

where k > 0 and c 0 (which includes our previous case of no damping, c = 0). Setting θ = y₁, θ′ = y₂, as before, we obtain the nonlinear system (use θ″ = y′₂)

We see that the critical points have the same locations as before, namely, (0, 0), (±π, 0), (±2π, 0), …. We consider (0, 0). Linearizing sin y₁ ≈ y₂ as in Example 1, we get the linearized system at (0, 0)

This is identical with the system in Example 2 of Sec. 4.4, except for the (positive!) factor m (and except for the physical meaning of y₁). Hence for c = 0 (no damping) we have a center (see Fig. 93b), for small damping we have a spiral point (see Fig. 94), and so on.

We now consider the critical point (π, 0). We set θ − π = y₁, (θ − π)′ = θ′ = y₂ and linearize

This gives the new linearized system at (π, 0)

For our criteria in Sec. 4.4 we calculate p = a₁₁ + a₂₂ = −c, q = det A = −k, and Δ = p² − 4q = c² + 4k. This gives the following results for the critical point at (π, 0).

No damping. c = 0, p = 0, q < 0, Δ > 0, a saddle point. See Fig. 93b.

Damping. c > 0, p < 0, q < 0, Δ > 0, a saddle point. See Fig. 94.

Since is periodic with period, 2π, the critical points (±2π, 0), (±4π, 0), … are of the same type as (0, 0), and the critical points (−π, 0), (±3π, 0), … are of the same type as (π, 0), so that our task is finished.

Figure 94 shows the trajectories in the case of damping. What we see agrees with our physical intuition. Indeed, damping means loss of energy. Hence instead of the closed trajectories of periodic solutions in Fig. 93b we now have trajectories spiraling around one of the critical points (0, 0), (±, 0), …. Even the wavy trajectories corresponding to whirly motions eventually spiral around one of these points. Furthermore, there are no more trajectories that connect critical points (as there were in the undamped case for the saddle points).

EXAMPLE 3 Predator–Prey Population Model³

This model concerns two species, say, rabbits and foxes, and the foxes prey on the rabbits.

Step 1. Setting up the model. We assume the following.

1. Rabbits have unlimited food supply. Hence, if there were no foxes, their number y₁(t) would grow exponentially, y′₁ = ay₁.
2. Actually, y₁ is decreased because of the kill by foxes, say, at a rate proportional to y₁, y₂, where y₂(t) is the number of foxes. Hence y′₁ = ay₁ − by₁y₂, where a > 0 and b > 0.
3. If there were no rabbits, then y₂(t) would exponentially decrease to zero, y′₂ = −ly₂. However, y₂ is increased by a rate proportional to the number of encounters between predator and prey; together we have y′₂ = −ly₂ + ky₁y₂, where k > 0 and l > 0.

This gives the (nonlinear!) Lotka–Volterra system

Step 2. Critical point (0, 0), Linearization. We see from (7) that the critical points are the solutions of

The solutions are (y₁, y₂) = (0, 0) and . We consider (0, 0). Dropping −by₁y₂ and ky₁y₂ from (7) gives the linearized system

Its eigenvalues are λ₁ = a > 0 and λ₂ = −l < 0. They have opposite signs, so that we get a saddle point.

Step. 3. Critical point (l/k, a/b), Linearization. We set . Then the critical point (l/k, a/b) corresponds to . Since , we obtain from (7) [factorized as in (7*)]

Dropping the two nonlinear terms and , we have the linearized system

The left side of (a) times the right side of (b) must equal the right side of (a) times the left side of (b),

This is a family of ellipses, so that the critical point (l/k, a/b) of the linearized system (7**) is a center (Fig. 95). It can be shown, by a complicated analysis, that the nonlinear system (7) also has a center (rather than a spiral point) at (l/k, a/b) surrounded by closed trajectories (not ellipses).

We see that the predators and prey have a cyclic variation about the critical point. Let us move counterclockwise around the ellipse, beginning at the right vertex, where the rabbits have a maximum number. Foxes are sharply increasing in number until they reach a maximum at the upper vertex, and the number of rabbits is then sharply decreasing until it reaches a minimum at the left vertex, and so on. Cyclic variations of this kind have been observed in nature, for example, for lynx and snowshoe hare near the Hudson Bay, with a cycle of about 10 years.

For models of more complicated situations and a systematic discussion, see C. W. Clark, Mathematical Bioeconomics: The Mathematics of Conservation,3rd ed. Hoboken, NJ, Wiley, 2010.

EXAMPLE 4 An ODE (8) for the Free Undamped Pendulum

If in (4)θ″ + k sin θ = 0 we set θ = y₁, θ′ = y₂ (the angular velocity) and use

Separation of variables gives y₂dy₂ = −k sin y₁dy₁. By integration,

Multiplying this by mL², we get

We see that these three terms are energies. Indeed, y₂ is the angular velocity, so that Ly₂ is the velocity and the first term is the kinetic energy. The second term (including the minus sign) is the potential energy of the pendulum, and mL²C is its total energy, which is constant, as expected from the law of conservation of energy, because there is no damping (no loss of energy). The type of motion depends on the total energy, hence on C, as follows.

Figure 93b shows trajectories for various values of C. These graphs continue periodically with period 2π to the left and to the right. We see that some of them are ellipse-like and closed, others are wavy, and there are two trajectories (passing through the saddle points (nπ, 0), n = ±1, ±3, …) that separate those two types of trajectories. From (9) we see that the smallest possible C is C = −k; then y₂ = 0, and cos y₁ = 1, so that the pendulum is at rest. The pendulum will change its direction of motion if there are points at which y₂ = θ′ = 0. Then k cos y₁ + C = 0 by (9). If y₁ = π, then cos y₁ = −1 and C = k. Hence if −k < C < k, then the pendulum reverses its direction for a |y₁| = |θ| < π, and for these values of C with |C| < k the pendulum oscillates. This corresponds to the closed trajectories in the figure. However, if C > k, then y₂ = 0 is impossible and the pendulum makes a whirly motion that appears as a wavy trajectory in the y₁y₂-plane. Finally, the value C = k corresponds to the two “separating trajectories” in Fig. 93b connecting the saddle points.

EXAMPLE 5 Self-Sustained Oscillations. Van der Pol Equation

There are physical systems such that for small oscillations, energy is fed into the system, whereas for large oscillations, energy is taken from the system. In other words, large oscillations will be damped, whereas for small oscillations there is “negative damping” (feeding of energy into the system). For physical reasons we expect such a system to approach a periodic behavior, which will thus appear as a closed trajectory in the phase plane, called a limit cycle. A differential equation describing such vibrations is the famous van der Pol equation⁴

It first occurred in the study of electrical circuits containing vacuum tubes. For μ = 0 this equation becomes y″ + y = 0 and we obtain harmonic oscillations. Let μ > 0. The damping term has the factor −μ(1 − y²). This is negative for small oscillations, when y² < 1, so that we have “negative damping,” is zero for y² = 1 (no damping), and is positive if y² > 1 (positive damping, loss of energy). If μ is small, we expect a limit cycle that is almost a circle because then our equation differs but little from y″ + y = 0. If ν is large, the limit cycle will probably look different.

Setting y = y₁, y′ = y₂ and using y″ = (dy₂/dy₁)y₂ as in (8), we have from (10)

The isoclines in the y₁y₂-plane (the phase plane) are the curves dy₂/dy₁ = K = const, that is,

Solving algebraically for y₂, we see that the isoclines are given by

Fig. 96. Direction field for the van der Pol equation with μ = 0.1 in the phase plane, showing also the limit cycle and two trajectories. See also Fig. 8 in Sec. 1.2

Figure 96 shows some isoclines when μ is small, μ = 0.1, the limit cycle (almost a circle), and two (blue) trajectories approaching it, one from the outside and the other from the inside, of which only the initial portion, a small spiral, is shown. Due to this approach by trajectories, a limit cycle differs conceptually from a closed curve (a trajectory) surrounding a center, which is not approached by trajectories. For larger μ the limit cycle no longer resembles a circle, and the trajectories approach it more rapidly than for smaller μ. Figure 97 illustrates this for μ = 1.

PROBLEM SET 4.5

1. Pendulum. To what state (position, speed, direction of motion) do the four points of intersection of a closed trajectory with the axes in Fig. 93b correspond? The point of intersection of a wavy curve with the y₂-axis?
2. Limit cycle. What is the essential difference between a limit cycle and a closed trajectory surrounding a center?
3. CAS EXPERIMENT. Deformation of Limit Cycle. Convert the van der Pol equation to a system. Graph the limit cycle and some approaching trajectories for μ = 0.2, 0.4, 0.6, 0.8, 1.0, 1.5, 2.0. Try to observe how the limit cycle changes its form continuously if you vary μ continuously. Describe in words how the limit cycle is deformed with growing μ.

4–8 CRITICAL POINTS. LINEARIZATION

Find the location and type of all critical points by linearization. Show the details of your work.

• 4.
• 5.
• 6.
• 7.
• 8.

9–13 CRITICAL POINTS OF ODEs

Find the location and type of all critical points by first converting the ODE to a system and then linearizing it.

• 9. y″ − 9y + y³ = 0
• 10. y″ + y − y³ = 0
• 11. y″ + cos y = 0
• 12. y″ + 9y + y² = 0
• 13. y″ + sin y = 0
• 14. TEAM PROJECT. Self-sustained oscillations. (a) Van der Pol equation. Determine the type of the critical point at (0, 0) when μ > 0, μ = 0, μ = < 0.

  (b) Rayleigh equation. Show that the Rayleigh equation⁵

  

  also describes self-sustained oscillations and that by differentiating it and setting y = Y′ one obtains the van der Pol equation.

  (c) Duffing equation. The Duffing equation is

  

  where usually |β| is small, thus characterizing a small deviation of the restoring force from linearity. β > 0 and β < 0 are called the cases of a hard spring and a soft spring, respectively. Find the equation of the trajectories in the phase plane. (Note that for β > 0 all these curves are closed.)

• 15. Trajectories. Write the ODE y″ − 4y + y³ = 0 as a system, solve it for y₂ as a function of y₁, and sketch or graph some of the trajectories in the phase plane.

  

  Fig. 98. Trajectories in Problem 15

EXAMPLE 1 Method of Undetermined Coefficients. Modification Rule

Find a general solution of

Solution. A general equation of the homogeneous system is (see Example 1 in Sec. 4.3)

Since λ = −2 is an eigenvalue of A, the function e^−2t on the right side also appears in y^(h), and we must apply the Modification Rule by setting

Note that the first of these two terms is the analog of the modification in Sec. 2.7, but it would not be sufficient here. (Try it.) By substitution,

Equating the te^−2t-terms on both sides, we have −2u = Au. Hence u is an eigenvector of A corresponding to λ = −2; thus [see (5)] u = a[1 1]^T with any a ≠ 0. Equating the other terms gives

Collecting terms and reshuffling gives

By addition, 0 = −2a − 4, a = −2, and then ν₂ = ν₁ + 4, say, ν₁ = k, ν₂ = k + 4, thus v = [k k + 4]^T. We can simply choose k = 0. This gives the answer

For other k we get other v; for instance, k = −2 gives v = [−2 2]^T, so that the answer becomes

EXAMPLE 2 Solution by the Method of Variation of Parameters

Solve (3) in Example 1.

Solution. A basis of solutions of the homogeneous system is [e^−2t e^−2t]^T and [e^−4t −e^−4t]^T. Hence the general solution (4) of the homogeneous system may be written

Here, Y(t) = [y⁽¹⁾ y⁽²⁾] is the fundamental matrix (see Sec. 4.2). As in Sec. 2.10 we replace the constant vector c by a variable vector u(t) to obtain a particular solution

Substitution into (3) y′ = Ay + g gives

Now since y⁽¹⁾ and y⁽²⁾ are solutions of the homogeneous system, we have

Hence Y′u = AYu, so that (8) reduces to

here we use that the inverse Y⁻¹ of Y (Sec. 4.0) exists because the determinant of Y is the Wronskian W, which is not zero for a basis. Equation (9) in Sec. 4.0 gives the form of Y⁻¹,

We multiply this by g, obtaining

Integration is done componentwise (just as differentiation) and gives

(where + 2 comes from the lower limit of integration). From this and Y in (7) we obtain

The last term on the right is a solution of the homogeneous system. Hence we can absorb it into y^(h). We thus obtain as a general solution of the system (3), in agreement with (5*).

PROBLEM SET 4.6

1. Prove that (2) includes every solution of (1).

2–7 GENERAL SOLUTION

Find a general solution. Show the details of your work.

• 2. y′₁ = y₁ + y₂ + 10 cos t

  y′₂ = 3y₁ − y₂ − 10 sin t

• 3. y′₁ = y₂ + e^3t

  y′₂ = y₁ − 3e^3t

• 4. y′₁ = 4y₁ − 8y₂ + 2 cosh t

  y′₂ = 2y₁ − 6y₂ + cosh t + 2 sinh t

• 5. y′₁ = 4y₁ + y₂ + 0.6t

  y′₂ = 2y₁ + 3y₂ − 2.5t

• 6. y′₁ = 4y₂

  y′₂ = 4y₁ − 16t₂ + 2

• 7. y′₁ = −3y₁ − 4y₂ + 11t + 15

  y′₂ = 5y₁ + 6y₂ + 3e^−t − 15t − 20

• 8. CAS EXPERIMENT. Undetermined Coefficients. Find out experimentally how general you must choose y^(p), in particular when the components of g have a different form (e.g., as in Prob. 7). Write a short report, covering also the situation in the case of the modification rule.
• 9. Undetermined Coefficients. Explain why, in Example 1 of the text, we have some freedom in choosing the vector v.

10–15 INITIAL VALUE PROBLEM

Solve, showing details:

• 10.
• 11.
• 12.
• 13.
• 14.
• 15.
• 16. WRITING PROJECT. Undetermined Coefficients. Write a short report in which you compare the application of the method of undetermined coefficients to a single ODE and to a system of ODEs, using ODEs and systems of your choice.

17–20 NETWORK

Find the currents in Fig. 99 (Probs. 17–19) and Fig. 100 (Prob. 20) for the following data, showing the details of your work.

• 17. R₁ = 2 Ω, R₂ = 8 Ω, L = 1 H, C = 0.5 F, E = 200 V
• 18. Solve Prob. 17 with E = 440 sin t V and the other data as before.
• 19. In Prob. 17 find the particular solution when currents and charge at t = 0 are zero.

  

  Fig. 99. Problems 17–19

• 20. R₁ = 1 Ω, R₂ = 1.4 Ω, L₁ = 0.8 H, L₂ = 1 H, E = 100 V, I₁(0) = I₂(0) = 0

  

  Fig. 100. Problem 20

Whereas single electric circuits or single mass–spring systems are modeled by single ODEs (Chap. 2), networks of several circuits, systems of several masses and springs, and other engineering problems lead to systems of ODEs, involving several unknown functions y₁(t), …, y_n(t). Of central interest are first-order systems (Sec. 4.2):

to which higher order ODEs and systems of ODEs can be reduced (Sec. 4.1). In this summary we let n = 2, so that

Then we can represent solution curves as trajectories in the phase plane (the y₁y₂-plane), investigate their totality [the “phase portrait” of (1)], and study the kind and stability of the critical points (points at which both f₁ and f₂ are zero), and classify them as nodes, saddle points, centers, orspiral points (Secs. 4.3, 4.4). These phase plane methods are qualitative; with their use we can discover various general properties of solutions without actually solving the system. They are primarily used for autonomous systems, that is, systems in which t does not occur explicitly.

A linear system is of the form

If g = 0, the system is called homogeneous and is of the form

If a₁₁, …, a₂₂ are constants, it has solutions y = xe^λt, where λ is a solution of the quadratic equation

and x ≠ 0 has components x₁, x₂ determined up to a multiplicative constant by

(These λ's are called the eigenvalues and these vectors x eigenvectors of the matrix A. Further explanation is given in Sec. 4.0.)

A system (2) with g ≠ 0 is called nonhomogeneous. Its general solution is of the form y = y_h + y_p, where y_h is a general solution of (3) and y_p a particular solution of (2). Methods of determining the latter are discussed in Sec. 4.6.

The discussion of critical points of linear systems based on eigenvalues is summarized in Tables 4.1 and 4.2 in Sec. 4.4. It also applies to nonlinear systems if the latter are first linearized. The key theorem for this is Theorem 1 in Sec. 4.5, which also includes three famous applications, namely the pendulum and van der Pol equations and the Lotka–Volterra predator–prey population model.