3. Second-Order Differential Equations
3.2 Linear Equations with Constant Coefficients
We recall two models from Chapter 1. For a spring-mass system with damping the displacementx(t) satisfies
mx+cx+kx= 0.
The currentI(t) in an RCL circuit with no emf satisfies LI+RI+ 1
CI= 0.
The similarity between these two models is called themechanical-electrical analogy. The spring constantk is analogous to the inverse capacitance 1/C;
both a spring and a capacitor store energy. The damping constantcis analogous to the resistance R; both friction in a mechanical system and a resistor in an electrical system dissipate energy. The massmis analogous to the inductance
L; both represent “inertia” in the system. All of the equations we examine in the next few sections can be regarded as either circuit equations or mechanical problems.
After dividing by the leading coefficient, both equations above have the form
u+pu+qu= 0, (3.4)
wherepandqare constants. An equation of the form (3.4) is called asecond- order linear equation with constant coefficients. Because zero is on the right side (physically, there is no external force or emf), the equation ishomo- geneous. Often the equation is accompanied by initial data of the form
u(0) =A, u(0) =B. (3.5)
The problem of solving (3.4) subject to (3.5) is called theinitial value prob- lem (IVP). Here the initial conditions are given at t = 0, but they could be given at any time t = t0. Fundamental to our discussion is the following existence-uniqueness theorem, which we assume to be true. It is proved in ad- vanced texts.
Theorem 3.2
The initial value problem (3.4)–(3.5) has a unique solution that exists on−∞<
t <∞.
The plan is this. We first note that the DE (3.4) always has two independent solutions u1(t) and u2(t) (byindependent we mean one is not a multiple of the other). We prove this fact by actually exhibiting the solutions explicitly. If we multiply each by an arbitrary constant and form the combination
u(t) =c1u1(t) +c2u1(t),
wherec1andc2are the arbitrary constants, then we can easily check thatu(t) is also a solution to (3.4). This combination is called the general solution to (3.4). We prove at the end of this section that all solutions to (3.4) are contained in this combination. Finally, to solve the initial value problem we use the initial conditions (3.5) to uniquely determine the constantsc1 andc2.
We try a solution to (3.4) of the formu=eλt,whereλis to be determined.
We suspect something like this might work because every term in (3.4) has to be the same type of function in order for cancellation to occur; thusu,u, and u must be the same form, which suggests an exponential foru. Substitution ofu=eλt into (3.4) instantly leads to
λ2+pλ+q= 0, (3.6)
which is a quadratic equation for the unknownλ. Equation (3.6) is called the characteristic equation. Solving, we obtain roots
λ= 1
2(−p±
p2−4q).
These roots of the characteristic equation are called the characteristic values (or roots) corresponding to the differential equation (3.4). There are three cases, depending upon whether the discriminant p2−4q is positive, zero, or negative. The reader should memorize these three cases and the forms of the solution.
Case 1.Ifp2−4q >0 then there are two real unequal characteristic values λ1andλ2.Hence, there are two independent, exponential-type solutions
u1(t) =eλ1t, u2(t) =eλ2t, and the general solution to (3.4) is
u(t) =c1eλ1t+c2eλ2t. (3.7) Case 2. If p2−4q = 0 then there is a double root λ =−p/2. Then one solution is u1 =eλt. A second independent solution in this case isu2 =teλt. Therefore the general solution to (3.4) in this case is
u(t) =c1eλt+c2teλt. (3.8) Case 3. If p2−4q < 0 then the roots of the characteristic equation are complex conjugates having the form
λ=α±iβ.
Therefore twocomplex solutions of (3.4) are e(α+iβ)t, e(α−iβ)t.
To manufacture real solutions we use a fundamental result that holds for all linear, homogeneous equations.
Theorem 3.3
Ifu=g(t) +ih(t) is a complex solution to the differential equation (3.4), then its real and imaginary parts,g(t) andh(t),are real solutions.
The simple proof is requested in the Exercises.
Let us take the first of the complex solutions given above and expand it into its real and imaginary parts using Euler’s formula: eiβt = cosβt+isinβt.
We have
e(α+iβ)t=eαteiβt=eαt(cosβt+isinβt) =eαtcosβt+ieαtsinβt.
Therefore, by Theorem 3.3, u1 =eαtcosβt and u2 =eαtsinβtare two real, independent solutions to equation (3.4). If we take the second of the complex solutions,e(α−iβ)tinstead ofe(α+iβ)t, then we get the same two real solutions.
Consequently, in the case that the characteristic values are complexλ=α±iβ, the general solution to DE (3.4) is
u(t) =c1eαtcosβt+c2eαtsinβt. (3.9) In the case of complex eigenvalues, we recall from trigonometry that (3.9) can be written differently as
u(t) =eαt(c1cosβt+c2sinβt) =eαtAcos(βt−ϕ),
whereAis called theamplitudeandϕis the phase. This latter form is called thephase–amplitude formof the general solution. Written in this form,A and ϕplay the role of the two arbitrary constants, instead ofc1 and c2. One can show that that all these constants are related by
A=
c21+c22, ϕ= arctanc2 c1. This is because the cosine of difference expands to
Acos(βt−ϕ) =Acos(βt) cosϕ+Asin(βt) sinϕ.
Comparing this expression toc1cosβt+c2sinβt, gives Acosϕ=c1, Asinϕ=c2.
Squaring and adding this last set of equations determinesA, and dividing the set of equations determinesϕ.
Observe that the solution in the complex case is oscillatory in nature with eαt multiplying the amplitudeA. Ifα <0 then the solution will be a decaying oscillation and ifα >0 the solution will be a growing oscillation. Ifα= 0 then the solution is
u(t) =c1cosβt+c2sinβt=Acos(βt−ϕ),
and it oscillates with constant amplitudeAand period 2π/β. The frequencyβ is called thenatural frequencyof the system.
There is some useful terminology used in engineering to describe the motion of a spring-mass system with damping, governed by the equation
mx+cx+kx= 0.
The characteristic equation is
mλ2+cλ+k= 0, with roots
λ= −c±√
c2−4mk
2m .
If the roots are complex (c2<4mk) then the system isunder-damped(rep- resenting a decaying oscillation); if the roots are real and equal (c2 = 4mk) then the system iscritically damped(decay, no oscillations, and at most one pass through equilibriumx= 0); if the roots are real and distinct (c2>4mk) then the system is over-damped (a strong decay toward x= 0). The same terminology can be applied to an RCL circuit.
Example 3.4
The differential equation u−u−12u= 0 has characteristic equation λ2− λ−12 = 0 with rootsλ=−3,4.These are real and distinct and so the general solution to the DE is u = c1e−3t+c2e4t. Over a long time the contribution e−3t decays and the solution is dominated by the e4t term. Thus, eventually the solution grows exponentially.
Example 3.5
The differential equation u+ 4u+ 4u= 0 has characteristic equation λ2+ 4λ+ 4 = 0, with roots λ=−2,−2. Thus the eigenvalues are real and equal, and the general solution isu=c1e−2t+c2te−2t. This solution decays as time gets large (recall that a decaying exponential dominates the linear growth term tso thatte−2tgoes to zero).
Example 3.6
The differential equationu+2u+2u= 0 models a damped spring-mass system withm= 1,c= 2, and k= 2. It has characteristic equationλ2+ 2λ+ 1 = 0.
The quadratic formula gives complex rootsλ=−1±2i.Therefore the general solution is
u=c1e−tcos 2t+c2e−tsin 2t,
representing a decaying oscillation. Here, the natural frequency of the un- damped oscillation is 2. In phase–amplitude form we can write
u=Ae−tcos(2t−ϕ).
Let us assume that the mass is given an initial velocity of 3 from an initial position of 1. Then the initial conditions are u(0) = 1, u(0) = 3.We can use these conditions directly to determine eitherc1 and c2 in the first form of the solution, or Aand ϕin the phase-amplitude form. Going the latter route, we apply the first condition to get
u(0) =Ae−0cos(2(0)−ϕ) =Acosϕ= 1.
To apply the other initial condition we need the derivative. We get u=−2Ae−tsin(2t−ϕ).
Then
u(0) =−2Ae−0sin(2(0)−ϕ) = 2Asinϕ= 3.
Therefore we have
Acosϕ= 1, Asinϕ= 3 2.
Squaring both equations and summing gives A2 = 13/4, so the amplitude is
A =
13/4. Note that the cosine is positive and the sine is positive, so the phase angle lies in the first quadrant. The phase is
ϕ= arctan(3 2) .
= 0.983 radians.
Therefore the solution to the initial value problem is u=
13
4 e−tcos(2t−0.983).
This solution represents a decaying oscillation. The oscillatory part has natural frequency 2 and the period is π. See figure 3.1. The phase has the effect of translating the cos 2t term by 0.983/2, which is called the phase shift.
To summarize, we have observed that the differential equation (3.4) always has two independent solutionsu1(t) andu2(t), and that the combination
u(t) =c1u1(t) +c2u1(t)
is also a solution, called the general solution. Now, as promised, we show that the general solution contains all possible solutions to (3.4). To see this letu1(t) andu2(t) be special solutions that satisfy the initial conditions
u1(0) = 1, u1(0) = 0,
0 1 2 3 4 5 6 7 8 9 10
−0.4
−0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4
t
u
Figure 3.1 Plot of the solution.
and
u2(0) = 0, u2(0) = 1,
respectively. Theorem 3.2 implies these two solutions exist. Now letv(t) be any solution of (3.4). It will satisfy some conditions at t = 0, say, v(0) = a and v(0) =b.But the function
u(t) =au1(t) +bu1(t)
satisfies those same initial conditions, u(0) = a and u(0) = b. Must u(t) therefore equalv(t)? Yes, by the uniqueness theorem, Theorem 3.2. Therefore v(t) =au1(t)+bu1(t),and the solutionv(t) is contained in the general solution.
Two equations occur so frequently that it is worthwhile to memorize them along with their solutions. The pure oscillatory equation
u+k2u= 0
has characteristic rootsλ=±ki,and the general solution is u=c1coskt+c2sinkt.
On the other hand, the equation
u−k2u= 0
has characteristic rootsλ=±k, and thus the general solution is u=c1ekt+c2e−kt.
This latter equation can also be written in terms of the the hyperbolic functions cosh and sinh as
u=C1coshkt+C2sinhkt, where
coshkt= ekt+e−kt
2 , sinhkt=ekt−e−kt
2 .
Sometimes the hyperbolic form of the general solution is easier to work with.
EXERCISES
1. Find the general solution of the following equations:
a) u−4u+ 4u= 0.
b) u+u+ 4u= 0.
c) u−5u+ 6u= 0.
d) u+ 9u= 0.
e) u−2u= 0.
f) u−12u= 0.
2. Find the solution to the initial value problem u+u +u = 0, u(0) = u(0) = 1,and write it in phase-amplitude form.
3. A damped spring-mass system is modeled by the initial value problem u+ 0.125u+u= 0, u(0) = 2, u(0) = 0.
Find the solution and sketch its graph over the time interval 0≤t≤50.If the solution is written in the form u(t) =Ae−t/16cos(ωt−ϕ), find A, ω, and ϕ.
4. For which values of the parameters a and b (if any) will the solutions to u−2au+bu= 0 oscillate with no decay (i.e., be periodic)? Oscillate with decay? Decay without oscillations?
5. An RCL circuit has equation LI+I +I = 0. Characterize the types of current responses that are possible, depending upon the value of the inductanceL.
6. An oscillator with damping is governed by the equationx+ 3ax+bx= 0, whereaandbare positive parameters. Plot the set of points in theabplane (i.e., abparameter space) where the system is critically damped.
7. Find a DE that has general solutionu(t) =c1e4t+c2e−6t.
8. Find a DE that has solution u(t) = e−3t+ 2te−3t. What are the initial conditions?
9. Find a DE that has solution u(t) = sin 4t+ 3 cos 4t.
10. Find a DE that has general solutionu(t) =Acosh 5t+Bsinh 5t,whereA andBare arbitrary constants. Find the arbitrary constants whenu(0) = 2 and u(0) = 0.
11. Find a DE that has solution u(t) = e−2t(sin 4t+ 3 cos 4t). What are the initial conditions?
12. Describe the current responseI(t) of a LC circuit withL= 5 henrys,C= 2 farads, withI(0) = 0,I(0) = 1.
13. Prove Theorem 3.3 by substitutinguinto the equation and separating real and imaginary parts, using linearity. Then use the fact that a complex quantity is zero if, and only if, its real and imaginary parts are zero.