44 CHAPTER 3. SECOND ORDER LINEAR EQUATIONS
That is, when ω is very close to the natural frequency ω0 we will have maximum oscillation. This phenomenon is called resonance. A graph of 1/ as a function of ω is shown in Fig. 3.1.
3.5Exercises
#1. Euler’s formula
Using Euler’s formula prove the trig identity
cos(4θ) = cos4 θ − 6 cos2 θ sin2 θ + sin4 θ.
Again using Euler’s formula find a formula for cos(2nθ) where n = 1, 2, . . .. In this way one can also get identities for cos(2n + 1)θ as well as sin nθ.
#2. Roots of unity
Show that the n (distinct) solutions to the polynomial equation
xn − 1 = 0
are e2πik/n for k = 1, 2, . . . , n. For n = 6 draw a picture illustrating where these roots lie in the complex plane.
#3. Constant coe cient ODEs
In each case find the unique solution y = y(x) that satisfies the ODE with stated initial conditions:
1.y′′ − 3y′ + 2y = 0, y(0) = 1, y′(0) = 0.
2.y′′ + 9y = 0, y(0) = 1, y′(0) = −1.
3.y′′ − 4y′ + 4y = 0, y(0) = 2, y′(0) = 0.
#4. Higher Order Equations
The third order homogeneous di erential equation with constant coe cients is
a3y′′′ + a2y′′ + a1y′ + a0y = 0 |
(3.21) |
where ai are constants. Assume a solution of the form
y(x) = eλx
and derive an equation that λ must satisfy in order that y is a solution. (You should get a cubic polynomial.) What is the form of the general solution to (3.21)?
3.5. EXERCISES |
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#5. Euler’s equation
A di erential equation of the form
t2y′′ + aty′ + by = 0, t > 0 |
(3.22) |
where a, b are real constants, is called Euler’s equation.8 This equation can be transformed into an equation with constant coe cients by letting x = ln t. Solve
t2y′′ + 4ty′ + 2y = 0 |
(3.23) |
#6 Forced undamped system
Consider a forced undamped system described by
y′′ + y = 3 cos(ωt)
with initial conditions y(0) = 1 and y′(0) = 1. Find the solution for ω 6= 1.
#7. Driven damped oscillator
Let
y′′ + 3y′ + 2y = 0
be the equation of a damped oscillator. If a forcing term is F (t) = 10 cos t and the oscillator is initially at rest at the origin, what is the solution of the equation for this driven damped oscillator? What is the phase angle?
#8. Damped oscillator
A particle is moving according to
y′′ + 10y′ + 16y = 0
with the initial condition y(0) = 1 and y′(0) = 4. Is this oscillatory ? What is the maximum value of y?
#9. Wronskian
Consider (3.3) with p(x) and q(x) continuous on the interval [a, b]. Prove that if two solutions y1 and y2 have a maximum or minimum at the same point in [a, b], they cannot form a basis of V.
8There is perhaps no other mathematician whose name is associated to so many functions, identities, equations, numbers, . . . as Euler.
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CHAPTER 3. SECOND ORDER LINEAR EQUATIONS |
#10. Euler’s equation (revisited) from physics
In Exercise 2.3.9 we obtained a set of three first-order di erential equations for Ω1, Ω2 and Ω3, which are called the Euler equations when there is no torque. Let us assume that I1 = I2 6= I3. (The body with these moments of inertia is called a free symmetric top.) In this case we have
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(I2 − I3)Ω2Ω3 |
(3.24) |
I1Ω1 |
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(I3 − I1)Ω3Ω1 |
(3.25) |
I2Ω2 |
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(3.26) |
I3Ω3 |
Notice that Ω3 is a constant from (3.26). Show that Ω1 and Ω2 have the form of
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A sin(ωt + θ0); |
Ω2(t) |
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A cos(ωt + θ0) |
where A and θ0 are some constants. Here Ω1, Ω2 and Ω3 are three components of the angular
~ ~
velocity vector Ω. Show that it follows that the magnitude (length) of Ω is a constant. Find an explicit expression for ω in terms of Ii and the constant Ω3.
Chapter 4
Di erence Equations
Science is what we understand well enough to explain to a computer. Art is everything else we do.
D.E. Knuth in the preface of A=B by H. Wilf & D. Zeilberger
4.1Introduction
We have learned that the general inhomogeneous second order linear di erential equation is
of the form
a(x) d2y + b(x) dy + c(x)y = f (x). dx2 dx
The independent variable, x, takes values in R. (We say x is a continuous variable.) Many applications lead to problems where the independent variable is discrete; that is, it takes values in the integers. Instead of y(x) we now have yn, n an integer. The discrete version of the above equation, called an inhomogeneous second order linear di erence equation, is
an yn+2 + bn yn+1 + cn yn = fn |
(4.1) |
where we assume the sequences {an}, {bn}, {cn} and {fn} are known. For example,
(n2 + 5)yn+2 + 2yn+1 + |
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is such a di erence equation. Usually we are given y0 and y1 (the initial values), and the problem is to solve the di erence equation for yn.
In this chapter we consider the special case of constant coe cient di erence equations:
a yn+2 + b yn+1 + c yn = fn
where a, b, and c are constants independent of n. If fn = 0 we say the di erence equation is homogeneous. An example of a homogeneous second order constant coe cient di erence equation is
1
6yn+2 + 3 yn+1 + 2yn = 0.
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CHAPTER 4. DIFFERENCE EQUATIONS |
4.2Constant Coe cient Di erence Equations
4.2.1Solution of Constant Coe cient Di erence Equations
In this section we give an algorithm to solve all second order homogeneous constant coe - cient di erence equations
a yn+2 + b yn+1 + c yn = 0. |
(4.2) |
The method is the discrete version of the method we used to solve contant coe cient di erential equations. We first guess a solution of the form
yn = λn, λ 6= 0.
(For di erential equations we guessed y(x) = eλx.) We now substitute this into (4.2) and require the result equal zero,
0= aλn+2 + bλn+1 + cλn = λn aλ2 + bλ + c .
This last equation is satisfied if and only if
aλ2 + bλ + c = 0. |
(4.3) |
Let λ1 and λ2 denote the roots of this quadratic equation. (For the moment we consider only the case when the roots are distinct.) Then
λn1 and λn2
are both solutions to (4.2). Just as in our study of second order ODEs, the linear combination
c1λn1 + c2λn2
is also a solution and every solution of (4.2) is of this form. The constants c1 and c2 are determined once we are given the initial values y0 and y1:
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c1λ1 + c2λ2, |
are two equation that can be solved for c1 and c2.
4.2.2Fibonnaci Numbers
Consider the sequence of numbers
1 1 2 3 5 8 13 21 34 · · ·
that is, each number is the sum of the preceding two numbers starting with
1 1
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as initial values. These integers are called Fibonnaci numbers and are denoted Fn. From their definition, Fn satisfies the di erence equation
Fn+1 = Fn + Fn−1 for n ≥ 1
with
F0 = 0, F1 = 1.
The quadratic equation we must solve is
λ2 = λ + 1,
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Fn = c1λn1 + c2λn2 ,
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Since λ1 > 1 and |λ2| < 1, λn1 grows with increasing n whereas λn2 → 0 as n → ∞. Thus for large n
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1Sometimes the number |
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is called the golden mean.