Examples 1 and 2 further illustrate how systems of differential equations arise naturally in scientific problems.

Example 1

Dual mass-spring system Consider the system of two masses and two springs shown in Fig. 7.1.1, with a given external force f(t) acting on the right-hand mass $m_2.$ We denote by x(t) the displacement (to the right) of the mass $m_1$ from its static equilibrium position [when the system is motionless and in equilibrium and $f(t)=0$] and by y(t) the displacement of the mass $m_2$ from its static position. Thus the two springs are neither stretched nor compressed when x and y are zero.

In the configuration in Fig. 7.1.1, the first spring is stretched x units and the second by $y-x$ units. We apply Newton’s law of motion to the two “free body diagrams” shown in Fig. 7.1.2; we thereby obtain the system

$$\begin{array}{lll}{m}_{1}x″& =& -k_{1}x+k_2(y-x),\\ m_{2}y″& =& -k_2(y-x)+f(t)\end{array}$$ (3)

of differential equations that the position functions x(t) and y(t) must satisfy. For instance, if $m_1=2,m_2=1,k_1=4,k_2=2,$ and $f(t)=40\sin 3t$ in appropriate physical units, then the system in (3) reduces to

$$\begin{array}{rcl}2x″& =& -6x+2y,\\ y″& =& 2x-2y+40\sin 3t.\end{array}$$ (4)

Example 2

Dual brine tanks Consider two brine tanks connected as shown in Fig. 7.1.3. Tank 1 contains x(t) pounds of salt in 100 gal of brine and tank 2 contains y(t) pounds of salt in 200 gal of brine. The brine in each tank is kept uniform by stirring, and brine is pumped from each tank to the other at the rates indicated in Fig. 7.1.3. In addition, fresh water flows into tank 1 at 20 gal/min, and the brine in tank 2 flows out at 20 gal/min (so the total volume of brine in the two tanks remains constant). The salt concentrations in the two tanks are x/100 pounds per gallon and y/200 pounds per gallon, respectively. When we compute the rates of change of the amount of salt in the two tanks, we therefore get the system of differential equations that x(t) and y(t) must satisfy:

$$\begin{array}{lcl}x\text{′}& =& -30\cdot \frac{x}{100}+10\cdot \frac{y}{200}=-\frac{3}{10}x+\frac{1}{20}y,\\ y\text{′}& =& 30\cdot \frac{x}{100}-10\cdot \frac{y}{200}-20\cdot \frac{y}{200}=\frac{3}{10}x-\frac{30}{20}y\end{array}$$

—that is,

$$\begin{array}{l}20x\text{′}=-6x+y,\\ 20y\text{′}=6x-3y.\end{array}$$ (5)

Consider a system of differential equations that can be solved for the highest-order derivatives of the dependent variables that appear, as explicit functions of t and lower-order derivatives of the dependent variables. For instance, in the case of a system of two second-order equations, our assumption is that it can be written in the form

It is of both practical and theoretical importance that any such higher-order system can be transformed into an equivalent system of first-order equations.

To describe how such a transformation is accomplished, we consider first the “system” consisting of the single nth-order equation

We introduce the dependent variables $x_1,x_2,\dots ,x_n$ defined as follows:

Note that $x_1^{\text{′}}=x\text{′}=x_2,x_2^{\text{′}}=x″=x_3,$ and so on. Hence the substitution of (8) in Eq. (7) yields the system

of n first-order equations. Evidently, this system is equivalent to the original nth-order equation in (7), in the sense that x(t) is a solution of Eq. (7) if and only if the functions $x_1(t),x_2(t),\dots ,x_n(t)$ defined in (8) satisfy the system of equations in (9).

It may appear that the first-order system obtained in Example 3 offers little advantage, because we could use the methods of Chapter 5 to solve the original (linear) third-order equation. But suppose that we were confronted with the nonlinear equation

to which none of our earlier methods can be applied. The corresponding first-order system is

and we will see in Section 7.7 that there exist effective numerical techniques for approximating the solution of essentially any first-order system. So, in this case, the transformation to a first-order system is advantageous. From a practical viewpoint, large systems of higher-order differential equations typically are solved numerically with the aid of the computer, and the first step is to transform such a system into a first-order system for which a standard computer program is available.

The linear second-order differential equation

(with constant coefficients and independent variable t) transforms via the substitutions $x\text{′}=y,x″=y\text{′}$ into the two-dimensional linear system

Conversely, we can solve this system in (13) by solving the familiar single equation in (12).

Example 5

To solve the two-dimensional system

$$\begin{array}{lcl}x\text{′}& =& -2y,\\ y\text{′}& =& \frac{1}{2}x,\\ & & \end{array}$$ (14)

we begin with the observation that

$$x″=-2y\text{′}=-2\left(\frac{1}{2}x\right)=-x.$$

This gives the single second-order equation $x″+x=0$ having general solution

$$x(t)=A\cos t+B\sin t=C\cos (t-\alpha ),$$

where $A=C\cos \alpha$ and $B=C\sin \alpha .$ Then

$$\begin{array}{lcl}y(t)& =& -\frac{1}{2}x\text{′}(t)=-\frac{1}{2}(-A\sin t+B\cos t)\\ & =& \frac{1}{2}C\sin (t-\alpha ).\end{array}$$

The identity ${\cos }^2\theta +{\sin }^2\theta =1$ therefore implies that, for each value of t, the point (x(t), y(t)) lies on the ellipse

$${\displaystyle \frac{x^2}{C^2}}+{\displaystyle \frac{y^2}{{(C/2)}^2}}=1$$

with semiaxes C and C/2. Figure 7.1.4 shows several such ellipses in the xy-plane.

A solution (x(t), y(t)) of a two-dimensional system

can be regarded as a parametrization of a solution curve or trajectory of the system in the xy-plane. Thus the trajectories of the system in (14) are the ellipses of Fig. 7.1.4. The choice of an initial point (x(0), y(0)) determines which one of these trajectories a particular solution parametrizes.

The picture showing a system’s trajectories in the xy-plane—its so-called phase plane portrait—fails to reveal precisely how the point (x(t), y(t)) moves along its trajectory. If the functions f and g do not involve the independent variable t, then a direction field—showing typical arrows representing vectors with components (proportional to) the derivatives $x\text{′}=f(x,y)$ and $y\text{′}=g(x,y)$—can be plotted. Because the moving point (x(t), y(t)) has velocity vector $(x\text{′}(t),y\text{′}(t)),$ this direction field indicates the point’s direction of motion along its trajectory. For instance, the direction field plotted in Fig. 7.1.4 indicates that each such point moves counterclockwise around its elliptical trajectory. Additional information can be shown in the separate graphs of x(t) and y(t) as functions of t.

Example 6

To find a general solution of the system

$$\begin{array}{rcl}x\text{′}& =& y,\\ y\text{′}& =& 2x+y,\end{array}$$ (15)

we begin with the observation that

$$x″=y\text{′}=2x+y=x\text{′}+2x.$$

This gives the single linear second-order equation

$$x″-x\text{′}-2x=0$$

having the characteristic equation

$$r^2-r-2=(r+1)(r-2)=0$$

and the general solution

$$x(t)=A{e}^{-t}+B{e}^{2t}.$$ (16)

Therefore,

$$y(t)=x\text{′}(t)=-A{e}^{-t}+2B{e}^{2t}.$$ (17)

Typical phase plane trajectories of the system in (15) parametrized by Eqs. (16) and (17) are shown in Fig. 7.1.6. These trajectories may resemble hyperbolas sharing common asymptotes, but Problem 23 shows that their actual form is somewhat more complicated.

Example 7

To solve the initial value problem

$$\begin{array}{l}x\text{′}=-y,\\ y\text{′}=(1.01)x-(0.2)y,\\ x(0)=0,y(0)=1,\end{array}$$ (18)

we begin with the observation that

$$x″=-y\text{′}=-[(1.01)x-(0.2)y]=(-1.01)x-(0.2)x\text{′}.$$

This gives the single linear second-order equation

$$x″+(0.2)x\text{′}+(1.01)x=0$$

having the characteristic equation

$$r^2+(0.2)r+1.01={(r+0.1)}^2+1=0,$$

characteristic roots $-0.1\pm i,$ and the general solution

$$x(t)=e^{-t/10}(A\cos t+B\sin t).$$

Then $x(0)=A=0,$ so

$$\begin{array}{rcl}x(t)& =& B{e}^{-t/10}\sin t,\\ y(t)& =& -x\text{′}(t)=\frac{1}{10}B{e}^{-t/10}\sin t-B{e}^{-t/10}\cos t.\end{array}$$

Finally, $y(0)=-B=1,$ so the desired solution of the system in (18) is

$$\begin{array}{rcl}x(t)& =& e^{-t/10}\sin t,\\ y(t)& =& \frac{1}{10}{e}^{-t/10}(\sin t+10\cos t).\end{array}$$ (19)

These equations parametrize the spiral trajectory in Fig.7.1.7; the trajectory approaches the origin as $t\to +\infty .$ Figure 7.1.8 shows the x- and y-solution curves given in (19).

When we study linear systems in subsequent sections, we will learn why the superficially similar systems in Examples 5 through 7 have the markedly different trajectories shown in Figs. 7.1.4, 7.1.6, and 7.1.7.

In addition to practical advantages for numerical computation, the general theory of systems and systematic solution techniques are more easily and more concisely described for first-order systems than for higher-order systems. For instance, consider a linear first-order system of the form

We say that this system is homogeneous if the functions $f_1,f_2,\dots ,f_n$ are all identically zero; otherwise, it is nonhomogeneous. Thus the linear system in (5) is homogeneous, whereas the linear system in (11) is nonhomogeneous. The system in (10) is nonlinear because the right-hand side of the second equation is not a linear function of the dependent variables $x_1$ and $x_2$.

A solution of the system in (20) is an n-tuple of functions $x_1(t),x_2(t),\dots ,x_n(t)$ that (on some interval) identically satisfy each of the equations in (20). We will see that the general theory of a system of n linear first-order equations shares many similarities with the general theory of a single nth-order linear differential equation. Theorem 1 (proved in Appendix A) is analogous to Theorem 2 of Section 5.2. It tells us that if the coefficient functions $p_{ij}$ and $f_j$ in (20) are continuous, then the system has a unique solution satisfying given initial conditions.

Thus n initial conditions are needed to determine a solution of a system of n linear first-order equations, and we therefore expect a general solution of such a system to involve n arbitrary constants. For instance, we saw in Example 4 that the second-order linear system

which describes the position functions x(t) and y(t) of Example 1, is equivalent to the system of four first-order linear equations in (11). Hence four initial conditions would be needed to determine the subsequent motions of the two masses in Example 1. Typical initial values would be the initial positions x(0) and y(0) and the initial velocities $x\text{′}(0)$ and $y\text{′}(0).$ On the other hand, we found that the amounts x(t) and y(t) of salt in the two tanks of Example 2 are described by the system

of two first-order linear equations. Hence the two initial values x(0) and y(0) should suffice to determine the solution. Given a higher-order system, we often must transform it into an equivalent first-order system to discover how many initial conditions are needed to determine a unique solution. Theorem 1 tells us that the number of such conditions is precisely the same as the number of equations in the equivalent first-order system.

Around the turn of the seventeenth century, Johannes Kepler analyzed a lifetime of planetary observations by the astronomer Tycho Brahe. Kepler concluded that the motion of the planets around the sun is described by the following three propositions, now known as Kepler’s laws of planetary motion:

1. The orbit of each planet is an ellipse with the sun at one focus.

2. The radius vector from the sun to each planet sweeps out area at a constant rate.

3. The square of the planet’s period of revolution is proportional to the cube of the major semiaxis of its elliptical orbit.

In his Principia Mathematica (1687) Isaac Newton deduced the inverse square law of gravitation from Kepler’s laws. In this application we lead you (in the opposite direction) through a derivation of Kepler’s first two laws from Newton’s law of gravitation. Newton’s explanation of planetary motion in terms of gravitation is a landmark of scientific and intellectual history.

Assume that the sun is located at the origin in the plane of motion of a planet, and write the position vector of the planet in the form

where $\mathbf{\text{i}}=(1,0)$ and $\mathbf{\text{j}}=(0,1)$ denote the unit vectors in the positive x- and y-directions. Then the inverse-square law of gravitation implies (Problem 27) that the acceleration vector $\mathbf{\text{r}}″(t)$ of the planet is given by

where $r=\sqrt{x^2+y^2}$ is the distance from the sun to the planet. If the polar coordinates of the planet at time t are $(r(t),\theta (t)),$ then the radial and transverse unit vectors shown in Fig. 7.1.13 are given by

The radial unit vector ${\mathbf{\text{u}}}_r$ (when located at the planet’s position) always points directly away from the origin, so ${\mathbf{\text{u}}}_r=\mathbf{\text{r}}/r,$ and the transverse unit vector ${\mathbf{\text{u}}}_{\theta }$ is obtained from ${\mathbf{\text{u}}}_r$ by a $90°$ counterclockwise rotation.

Step 1: Differentiate the equations in (3) componentwise to show that

Step 2: Use the equations in (4) to differentiate the planet’s position vector $\mathbf{\text{r}}=r{\mathbf{\text{u}}}_r$ and thereby show that its velocity vector is given by

Step 3: Differentiate again to show that the planet’s acceleration vector $\mathbf{\text{a}}=d\mathbf{\text{v}}/dt$ is given by

Step 4: The radial and transverse components on the right-hand sides in Eqs. (2) and (6) must agree. Equating the transverse components—that is, the coefficients of ${\mathbf{\text{u}}}_{\theta }$—we get

so it follows that

where h is a constant. Because the polar-coordinate area element—for computation of the area A(t) in Fig. 7.1.14—is given by $dA=\frac{1}{2}{r}^{2}d\theta ,$ Eq. (8) implies that the derivative $A\text{′}(t)$ is constant, which is a statement of Kepler’s second law.

Step 5: Equate radial components in (2) and (6), and then use the result in (8) to show that the planet’s radial coordinate function r(t) satisfies the second-order differential equation

Step 6: Although the differential equation in (9) is nonlinear, it can be transformed to a linear equation by means of a simple substitution. For this purpose, assume that the orbit can be written in the polar-coordinate form $r=r(\theta ),$ and first use the chain rule and Eq. (8) to show that if $r=1/z,$ then

Differentiate again, to deduce from Eq. (9) that the function $z(\theta )=1/r(\theta )$ satisfies the second-order equation

Step 7: Show that the general solution of Eq. (10) is

Step 8: Finally, deduce from Eq. (11) that $r(\theta )=1/z(\theta )$ is given by

with $e=C{h}^2/k,C\cos \alpha =A,C\sin \alpha =B,$ and $L=h^2/k.$ The polar-coordinate graph of Eq. (12) is a conic section of eccentricity e—an ellipse if $0\leqq e<1,$ a parabola if $e=1,$ and a hyperbola if $e>1$—with focus at the origin. Planetary with perihelion distance $r_1=L/(1+e)$ and aphelion distance $r_2=L/(1-e).$ orbits are bounded and therefore are ellipses, with eccentricity $e<1.$ As indicated in Fig. 7.1.15, the major axis of the ellipse lies along the radial line $\theta =\alpha$.

Step 9: Plot some typical elliptical orbits—as described by (12)—having different eccentricities, sizes, and orientations. In rectangular coordinates you can write

to plot an elliptical orbit with eccentricity e, semilatus rectum L (Fig. 7.1.15), and rotation angle $\alpha .$ The eccentricity of the earth’s orbit is $e\approx 0.0167,$ so close to zero that the orbit looks nearly circular (though with the sun off center), and the eccentricities of the other planetary orbits range from 0.0068 for Venus and 0.0933 for Mars to 0.2056 for Mercury (and 0.2486 for Pluto, no longer classified a planet). But many comets have highly eccentric orbits—Halley’s comet has $e\approx 0.97$ (Fig.7.1.16).