Coordinated Differential Equations

Subsection2.1.1RLC Circuits

Recall the RC circuits that we studied earlier (see Section1.3). Such circuits contained a voltage source, a capacitor, and a resistor. A battery or generator is an example of a voltage source, and a toaster or an electric stove is an example of something that might provide a resistance in a circuit. Capacitors store an electrical charge and are used in electronic flashes for cameras. We will now add an inductor such as a solenoid, a coil that generates a magnetic field. Inductor applications include transformers, power supplies, televisions, and radios. Our new circuit is called an RLC circuit (Figure2.1.1).

Current, \(I(t)\text{,}\) is the rate at which a charge flows through this circuit and is measured in amperes or amps. We assign a direction to the current, and a current flowing in the opposite direction will be given negative values. The impressed voltage, \(E(t)\text{,}\) is measured in volts, the resistance \(R\) is measured in ohms, and the capacitance \(C\) is measured in farads. The charge on the capacitor \(Q(t)\) at time \(t\) is measured in coulombs. Inductance on the coil, \(L\text{,}\) is measured in henrys.

The following laws from physics govern how our circuit behaves.

• \(I = \dfrac{dQ}{dt}\text{.}\)

• The voltage drop across a resistor is \(IR\) (Ohm's Law).

• The voltage drop across a capacitor is \(Q/C\text{.}\)

• The voltage drop across an inductor is \(L (dI/dt)\text{.}\)

• In a closed circuit the impressed voltage is equal to the sum of the voltage drops in the rest of the circuit (Kirchhoff's Second Law).

Applying Kirchhoff's Second Law to our circuit, we have the differential equation

or

Differentiating both sides of (2.1.1), we have

For example, we might consider an RLC circuit with \(R = 1\text{,}\) \(L = 1\text{,}\) and \(C = 1\text{.}\) At \(t = 0\) when both \(I(0) =0\) and \(I'(0) = Q(0) = 0\text{,}\) the impressed voltage on the circuit is given by \(E(t) = \sin(t)\text{.}\) Our equation becomes

This is an example of a second-order linear differential equation.

Subsection2.1.2Solving Second-Order Linear Equations

Suppose that we have a homogeneous second-order linear differential equation with constant coefficients,

The goal of this section is to be able to solve all such equations. To build some intuition, we start by considering the homogeneous first-order linear differential equation

Our work in Chapter1 suggests that this equation has a solution of the form \(x(t) = e^{\lambda t}\) for some constant \(\lambda\text{.}\) We can identify \(\lambda\) by substuting our guess for the solution into (2.1.3) and solving algebraically. Since \(\frac{d}{dt}e^{\lambda t} = \lambda e^{\lambda t}\text{,}\) this substitution yields the equation

The term \(e^{\lambda t}\) is always non-zero, and therefore we can safely divide both sides of the equation above by \(e^{\lambda t}\) to reach

which has solution \(\lambda = -\frac{b}{a}\text{.}\) Thus (2.1.3) has \(x(t) = e^{-\frac{b}{a}t}\) as a solution.

A similar strategy works for homogeneous second-order linear differential equations. We guess that (2.1.2) has a solution of the form \(x=e^{\lambda t}\) and substitute our guess into the left-hand side of the equation. In detail,

Dividing out the non-zero term \(e^{\lambda t}\) as before, we see that \(\lambda\) must satisfy

We call (2.1.4) the characteristic equation for \(ax'' + bx' + cx = 0\text{.}\) Its roots are

If \(b^2 - 4ac \gt 0\text{,}\) we get distinct real roots

which correspond to a pair of solutions

to (2.1.2). Each solution \(x_{i}(t)\) exhibits exponential growth or decay depending on the sign of \(\lambda_{i}\text{.}\)

The characteristic equation (2.1.4) also has distinct roots when \(b^{2} - 4ac \lt 0\text{,}\) although now the roots are complex numbers. Even so, we can make sense of the corresponding solutions using tools from Section1.8. In this case, it is customary to write \(\lambda = \alpha \pm i\beta\text{,}\) where

We then have a pair of complex conjugate roots

corresponding, as in the real case, to a pair of solutions

We can rewrite \(x_{1}(t)\) as follows:

From Euler's formula (1.8.1), we have \(e^{i\beta t}=\cos(\beta t) + i\sin(\beta)\text{,}\) and so

Using the facts that \(\cos(-\beta t)=\cos(\beta t)\) and \(\sin(-\beta t) = -\sin(\beta t)\text{,}\) a similar computation on \(x_{2}(t)\) results in the expression

To get real solutions, we can take the real part of \(x_{1}(t)\) and (negative) the imaginary part of \(x_{2}(t)\text{.}\) Going forward, we will use the resulting real functions as the canonical solutions to (2.1.2) when \(b^{2} - 4ac \lt 0\text{,}\) re-defining \(x_{1}(t)\) and \(x_{2}(t)\) as

Note that complex roots of the characteristic equation induce oscillatory behavior in the solutions of the differential equation due to the presence of the \(\sin\) and \(\cos\) terms.

There is one more case to consider: when \(b^{2}-4ac=0\text{.}\) In this case, the charactersitic equation has a repeated real root at \(\lambda = -\frac{b}{2a}\text{,}\) which appears to give only one solution \(x_{1}(t) = e^{-\frac{b}{2a}t}.\) However, as demonstrated in the next example, it is possible to generate a second solution using \(x_{1}(t)\text{.}\)

The technique that we have used in Example2.1.4 is called reduction of order. In general, given a second-order linear differential equation

such that \(b^2 - 4ac = 0\text{,}\) we have a pair of solutions

where \(\lambda = -b/2a\) is the repeated real root of the characteristic equation.

Subsection2.1.3General Solutions to Second-Order Linear Equations

In Subsection2.1.2 we learned how to construct some solutions to the differential equation (2.1.2) given by

using the roots of the characteristic equation \(a\lambda^{2} + b \lambda + c = 0.\) In this section, we will show that these solutions can in fact be used to describe all solutions to the differential equation.

A key idea in this direction is the principle of superposition, which states that scalar multiples and sums of solutions to (2.1.2) are again solutions to the same equation.

This fact can be established using basic properties of the derivative. For instance, suppose that \(x(t)\) is a solution to \(ax'' + bx' + cx = 0\text{,}\) and let \(A\) be an arbitrary real constant. Recalling that for any differentiable function \(f\text{,}\) we have \(\frac{d}{dt}Af(t)=A\frac{d}{dt}f(t)\text{,}\) we may compute

where in the second-to-last line we use the fact that \(x(t)\) is a solution to (2.1.2), that is, \(ax'' + bx' + cx = 0\text{.}\) So, \(Ax(t)\) is also a solution.

Similarly, since \(\frac{d}{dt}(f(t) + g(t)) = \frac{d}{dt}f(t) + \frac{d}{dt}g(t)\) for any differentiable \(f\) and \(g\text{,}\) if \(x_{1}(t)\) and \(x_{2}(t)\) are solutions to (2.1.2), then

So \(x_{1}(t) + x_{2}(t)\) is also a solution to (2.1.2). In conjunction, the preceding computations yield the following theorem.

Before continuing, it is worth noting that Theorem2.1.5 is still true if the constant coefficients \(a\text{,}\) \(b\text{,}\) and \(c\) are replaced with variable coefficients \(a(t)\text{,}\) \(b(t)\text{,}\) and \(c(t)\) which depend on \(t\text{.}\) For our present purposes, however, we continue to focus solely on the constant coefficient case.

The expression \(c_{1}x_{1}(t) + c_{2}x_{2}(t)\) is called a linear combination of the functions \(x_{1}\) and \(x_{2}\text{.}\) Provided that \(x_{1}\) and \(x_{2}\) solve (2.1.2) and \(x_{2}\) is not simply a constant multiple of \(x_{1}\text{,}\) we can think of the linear combination \(c_{1}x_{1}(t) + c_{2}x_{2}(t)\) as a two-parameter family of solutions to (2.1.2).

Does such a family encompass all solutions? To make this question more tractable, we first state an existence and uniqueness result for second-order linear equations which is analogous to Theorem1.6.1.

In light of Theorem2.1.6, to show that a two-parameter family of solutions \(c_{1}x_{1}(t) + c_{2}x_{2}(t)\) contains all solutions to the differential equation, it is enough to demonstrate that the constants \(c_{1}\) and \(c_{2}\) can be chosen to satisfy arbitrary initial conditions. That is, for any \(t_{0}\text{,}\) \(x_{0}\text{,}\) and \(v_{0}\text{,}\) there exists an assignment of \(c_{1}\) and \(c_{2}\) satisfying the system of equations

When this holds, we say that \(x(t) = c_{1}x_{1}(t) + c_{2}x_{2}(t)\) is a general solution to the differential equation.

We begin with an example before proceding with the general theory.

As we saw in Example2.1.7, one can demonstrate that \(x(t) = c_{1}x_{1}(t) + c_{2}x_{2}(t)\) is a general solution by explicitly solving the system (2.1.6)-(2.1.7).

Solving the system (2.1.6)-(2.1.7) symbolically yields the following expressions for \(c_{1}\) and \(c_{2}\text{:}\)

This represents a valid (and unique) solution to the system if and only if the denominator which appears in the expressions for both \(c_{1}\) and \(c_{2}\) is not equal to \(0\text{.}\) We call this quantity the Wronskian of \(x_{1}\) and \(x_{2}\) and denote it by

Notice that \(W(x_{1},x_{2})\) is a function of \(t\text{.}\) Provided that \(W(x_{1},x_{2}) \neq 0\) for all values of \(t\text{,}\) by our computations above we may solve the system (2.1.6)-(2.1.7) for any values of \(t_{0}\text{,}\) \(x_{0}\text{,}\) and \(v_{0}\text{.}\) Thus we have the following criterion.

In the example below, we revisit the differential equation from Example2.1.7 in light of Theorem2.1.8.

We can now give an explicit general solution to the equation \(ax'' + bx' + cx = 0\text{.}\) This task combines our efforts from Subsection2.1.2, where we constructed pairs of solutions using the characteristic equation, with the theory developed in this section.

With general solutions in hand, we can solve an initial value problem involving \(ax'' + bx' + cx = 0\) with initial conditions of the form \(x(0) = x_{0}\text{,}\) \(x'(0) = v_{0}\text{.}\) We demonstrate in the next example.

Subsection2.1.4Classifying Harmonic Oscillators

Recall from Subsection1.1.3 that we can model harmonic motion using the equation

In the case of a spring-mass system, \(m\) is the oscillating mass, \(b\) is the damping coefficient, and \(k\) is the spring constant. It is important to remember that both \(m\) and \(k\) are positive constants and \(b \geq 0\text{.}\)

If \(b = 0\text{,}\) then the oscillator is undamped In this case,

The characteristic equation is

and we roots at \(\lambda = \pm i \sqrt{k/m}\text{.}\) Writing \(\omega = \sqrt{k/m}\) for simplicity, Theorem2.1.10 gives the general solution

which represents the position of the mass at time \(t\text{.}\) Now, of course, it is easy to determine the velocity of the mass at time \(t\) to be

Suppose that an undamped harmonic oscillator is modeled by the initial value problem

\begin{align*} \frac{d^2x}{dt^2} + 9 x \amp = 0\\ x(0) \amp = 2\\ x'(0) \amp = 1. \end{align*}

We can quickly determine the solution of this initial value problem to be

\begin{align*} x(t) \amp = 2 \cos 3t +\frac{1}{3} \sin 3t\\ v(t) \amp = \cos 3t - 6 \sin 3t, \end{align*}

where \(v(t) = x'(t)\) is the velocity of the oscillator (Figure2.1.13). Examining the phase plane of the undamped oscillator, we find that the period of the oscillations is given by \(2 \pi / \omega = 2 \pi / 3 \approx 2.094\) (Figure2.1.14).

If we add damping to the oscillator, the equation becomes

where \(b \gt 0\text{.}\) The charactersitic equation of (2.1.9) is

which has roots

There are three possible types of types of motion for the oscillator depending on the nature of the roots of (2.1.10).

• If the damping value of \(b\) is small when compared to \(4mk\text{,}\) then \(b^2 - 4mk \lt 0\) and the roots of (2.1.10) will be complex. Furthermore, the real part of each root, \(-b/ 2m\text{,}\) is always negative. In such a situation, we say that the oscillator is under-damped.

• If the damping value of \(b\) is large , then \(b^2 - 4mk \gt 0\text{,}\) and we obtain distinct real negative roots for (2.1.10). The oscillator is over-damped.

• Finally, we say that the oscillator is critically-damped if \(b^2 - 4mk = 0\text{.}\)

Suppose that an oscillator is modeled by the initial value problem

\begin{align*} \frac{d^2 x}{dt^2} + 0.4 \frac{dx}{dt} + 1.04 x \amp = 0\\ x(0) \amp = 0\\ x'(0) \amp = 1. \end{align*}

Notice that the damping \(b = 0.4\) is very small compared with the spring constant \(k = 1.04\text{.}\) The characteristic equation of the differential equation is \(\lambda^2 + 0.4 \lambda + 1.04 = 0\text{,}\) which has roots \(\lambda = -0.2 \pm i\text{.}\) Therefore, the general solution is

\begin{equation*} x(t) = c_1 e^{-0.2t} \cos t + c_2 e^{-0.2t} \sin t. \end{equation*}

Applying the initial conditions, our solution becomes

\begin{align*} x(t) \amp = e^{-0.2t} \sin t\\ v(t) \amp = x'(t) = e^{-0.2t} (\cos t - 0.2 \sin t). \end{align*}

Notice that the period of the oscillations, \(2 \pi / \omega = 2 \pi \approx 6.283\text{,}\) does not change; however, the amplitude slowly decreases (Figure2.1.16 and Figure2.1.17).

We can expect a different type of behavior in the case of an over-damped oscillator. For example,

\begin{align*} \frac{d^2 x}{dt^2} + 4 \frac{dx}{dt} + 3 x \amp = 0\\ x(0) \amp = 0\\ x'(0) \amp = 1. \end{align*}

The characteristic equation of the differential equation is

\begin{equation*} \lambda^2 + 4 \lambda + 3 = (\lambda + 3)(\lambda + 1) = 0; \end{equation*}

hence, we have the general solution

\begin{equation*} x(t) = c_1 e^{-t} + c_2 e^{-3t}. \end{equation*}

Applying the initial conditions, our solution is

\begin{align*} x(t) \amp = x(t) = \frac{1}{2}e^{-t} - \frac{1}{2} e^{-3t}\\ v(t) \amp = x'(t) = -\frac{1}{2}e^{-t} + \frac{3}{2} e^{-3t}. \end{align*}

Notice that the damping is too strong for any oscillations to occur (Figure2.1.19 and Figure2.1.20).

As we increase the damping, the oscillations will cease to occur for some value of \(b\text{.}\) This will happen when \(b^2 - 4mk = 0\text{.}\) the At this point we have critical damping. Consider the system

\begin{align*} \frac{d^2 x}{dt^2} + 4 \frac{dx}{dt} + 4 x \amp = 0\\ x(0) \amp = 0\\ x'(0) \amp = 1. \end{align*}

The general solution to this initial value problem is

\begin{equation*} x(t) = c_1 e^{-2t} + c_2 t e^{-2t}. \end{equation*}

The solution to the initial value problem is

\begin{align*} x(t) \amp = x(t) = t e^{-2t}\\ v(t) \amp = x'(t) = (1 - 2t)e^{-2t}. \end{align*}

Although we see that no oscillations for this oscillator (Figure2.1.22 and Figure2.1.23), oscillations will commence as soon as we start to reduce the damping constant \(b = 4\text{.}\)

Subsection2.1.5Important Lessons

• A second-order linear differential equation with constant coefficients is an equation of the form

  \begin{equation*} a x'' + bx' + cx = 0. \end{equation*}

• The characteristic equation of \(a x'' + bx' + cx = 0\) is

  \begin{equation*} a\lambda^{2} + b\lambda + c = 0. \end{equation*}

  The roots of this equation are

  \begin{equation*} \lambda = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}. \end{equation*}

  These roots govern the behavior of solutions to the differential equation.

• If \(b^2 - 4ac \gt 0\text{,}\) the characteristic equation has distinct real roots

  \begin{equation*} \lambda_{1} = \frac{-b + \sqrt{b^{2}-4ac}}{2a} \quad \text{and} \quad \lambda_{2} = \frac{-b - \sqrt{b^{2}-4ac}}{2a} \end{equation*}

  and a general solution to \(a x'' + bx' + cx = 0\) is given by

  \begin{equation*} x(t) = c_{1}e^{\lambda_{1}t} + c_{2}e^{\lambda_{2}t}. \end{equation*}

• If \(b^2 - 4ac = 0\text{,}\) the characteristic equation has a repeated real root at

  \begin{equation*} \lambda = -\frac{b}{2a} \end{equation*}

  and a general solution to \(a x'' + bx' + cx = 0\) is given by

  \begin{equation*} x(t) = c_{1}e^{\lambda t} + c_{2}t e^{\lambda t}. \end{equation*}

• If \(b^2 - 4ac \lt 0\text{,}\) characteristic equation has complex conjugate roots

  \begin{equation*} \lambda = \alpha \pm \beta i \end{equation*}

  where \(\alpha = -\frac{b}{2a}\) and \(\beta = \frac{\sqrt{4ac - b^{2}}}{2a}\text{.}\) A general solution to \(a x'' + bx' + cx = 0\) is given by

  \begin{equation*} x(t) = c_{1}e^{\alpha t}\cos(\beta t) + c_{2}e^{\alpha t}\sin(\beta t). \end{equation*}

• A simple harmonic oscillator can be modeled by the equation

  \begin{equation*} m \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + k x = 0, \end{equation*}

  where \(m \gt 0\text{,}\) \(k \gt 0\text{,}\) and \(b \geq 0\text{.}\) There are three possible types of motion for the oscillator depending on the sign of \(b^2 - 4mk\text{.}\)

  • If \(b^2 - 4mk \lt 0\text{,}\) the oscillator is under-damped.

  • If \(b^2 - 4mk \gt 0\text{,}\) the oscillator is over-damped.

  • If \(b^2 - 4mk = 0\text{,}\) the oscillator is critically damped.