ODE-ProjectODE-Project Sinusoidal Forcing

Section 4.3 Sinusoidal Forcing

Objectives

To understand and be able to use Euler’s formula and complexification to solve the equation

$x^{″} + p x^{'} + q x = g (t),$

where the forcing function $g (t)$ is $\sin ω t$ or $\cos ω t .$
To understand and be able to use complex numbers to express solutions in the form

$x (t) = A \cos (ω t - ϕ),$

where $A$ is the amplitude of the solution, $ω$ is the frequency of the solution, and $ϕ$ is the phase angle.

If we consider different forcing functions

g (t)

for the equation

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x^{″} + p x^{'} + q x = g (t),

functions that are periodic are especially important. Recall that a function

g (t)

is periodic if

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g (t + T) = g (t)

for all

t

and some fixed constant

T .

The most familiar periodic functions are

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g (t) = \sin ω t and g (t) = \cos ω t .

The period for each of these two functions is

2 π / ω

and the frequency is

ω / 2 π .

These two functions share the additional property that their average value is zero. That is,

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\frac{1}{T} \int_{0}^{T} g (t) d t = 0.

We say that sinusoidal forcing occurs in the differential equation

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x^{″} + p x^{'} + q x = A \cos ω t + B \sin ω t .

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Subsection 4.3.1 Complexification

Given a second-order linear differential equation

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a x^{″} + b x^{'} + c x = A \cos ω t + B \sin ω t,

we can use Euler’s formula,

e^{i β t} = \cos β t + i \sin β t

to derive a particular solution. That is, we will assume that our particular solution has the form

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x_{c} = x_{Re} + i x_{Im}

and use the properties of complex numbers.

If complex numbers make you uncomfortable, the alternative is to become an expert in trigonometric identities

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Example 4.3.1.

Let us consider the equation

\begin{matrix} (4.3.1) & x^{″} + 6 x^{'} + 5 x = \sin 2 t . \end{matrix}

The solution to the corresponding homogeneous equation,

x^{″} + 6 x^{'} + 5 x = 0,

x_{h} = c_{1} e^{- 5 t} + c_{2} e^{- t} .

To find a particular solution, we can use the method of undetermined coefficients and assume that the solution has the form

x_{p} = A \cos 2 t + B \sin 2 t .

If we carry out the appropriate calculations, we will obtain a particular solution

x_{p} = - \frac{12}{145} \cos 2 t + \frac{1}{145} \sin 2 t .

Thus, the general solution is

x = x_{h} + x_{p} = c_{1} e^{- 5 t} + c_{2} e^{- t} - \frac{12}{145} \cos 2 t + \frac{1}{145} \sin 2 t .

Notice that all solutions of (4.3.1) will approach the particular solution as

t \to \infty .

Example 4.3.2.

Now let us solve (4.3.1) using complex numbers. If we assume that the equation has a complex solution of the form

x_{c} = x_{Re} + i x_{Im},

then

\begin{aligned} \frac{d^{2}}{d t^{2}} x_{c} + 6 \frac{d}{d t} x_{c} + 5 x_{c} & = \frac{d^{2}}{d t^{2}} (x_{Re} + i x_{Im}) + 6 \frac{d}{d t} (x_{Re} + i x_{Im}) + 5 (x_{Re} + i x_{Im}) \\ = e^{2 i t} \\ = \cos 2 t + i \sin 2 t . \end{aligned}

Equating the real and imaginary parts of this equation, we obtain

\begin{aligned} x_{Re}^{″} + 6 x_{Re}^{'} + 5 x_{Re} & = \cos 2 t \\ x_{Im}^{″} + 6 x_{Im}^{'} + 5 x_{Im} & = \sin 2 t . \end{aligned}

Thus, if we can find a complex solution, we can find a solution to

x^{″} + 6 x^{'} + 5 x = \sin 2 t

simply by examining the imaginary part of the solution.

Now let us assume that our solution has the form

x_{c} = A e^{2 i t} .

Then

\begin{aligned} x_{c}^{″} + 6 x_{c}^{'} + 5 x_{c} & = - 4 A e^{2 i t} + 12 A i e^{2 i t} + 5 A e^{2 i t} \\ = (1 + 12 i) A e^{2 i t} \\ = (1 + 12 i) A (\cos 2 t + i \sin 2 t) . \end{aligned}

Equating the real and imaginary parts of this equation, we obtain

\begin{aligned} x_{Re}^{″} + 6 x_{Re}^{'} + 5 x_{Re} & = \cos 2 t \\ x_{Im}^{″} + 6 x_{Im}^{'} + 5 x_{Im} & = \sin 2 t \end{aligned}

and we immediately see that

A = \frac{1}{1 + 12 i} = \frac{1}{145} - \frac{12}{145} i .

Therefore, the complex solution to

x^{″} + 6 x^{'} + 5 x = e^{2 i t}

\begin{aligned} x_{c} & = A e^{2 i t} \\ = (\frac{1}{145} - \frac{12}{145} i) \cdot (\cos 2 t + i \sin 2 t) \\ = (\frac{1}{145} \cos 2 t + \frac{12}{145} \sin 2 t) + i (- \frac{12}{145} \cos 2 t + \frac{1}{145} \sin 2 t) . \end{aligned}

The imaginary part of this function is

x_{Im} = - \frac{12}{145} \cos 2 t + \frac{1}{145} \sin 2 t,

which is the particular solution that we have been seeking. Thus, our general solution agrees with what we found in Example 4.3.1.

Activity 4.3.1. Second-Order Linear Differential Equations and Complexification.

Find (1) a particular solution and (2) a general solution for each of the following differential equations.

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(a)

x^{″} + 4 x^{'} - 21 x = 2 e^{4 t}

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(b)

x^{″} + 4 x^{'} - 21 x = 3 e^{3 t}

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(c)

x^{″} - 4 x^{'} + 20 x = 3 \sin 3 t

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(d)

x^{″} - 4 x^{'} + 20 x = 2 e^{2 t} \cos 4 t

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(e)

x^{″} - 14 x^{'} + 49 x = \sin 3 t

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Subsection 4.3.2 Qualitative Analysis

We can use the complex solution of

a x^{″} + b x^{'} + c x = A \cos ω t + B \sin ω t

to analyze the qualitative behavior of solutions.

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Example 4.3.3.

We discovered that the complex solution of

x^{″} + 6 x^{'} + 5 x = e^{2 i t}

to be

x_{c} = A e^{2 i t},

where

A = (1 - 12 i) / 145 .

Let us rewrite

A

in polar form. Since

| A | = \frac{1}{\sqrt{145}},

we know that

A = \frac{1}{\sqrt{145}} e^{i θ},

where

θ = \arctan (- 12) \approx - 1.4877 .

Therefore,

x_{c} = A e^{2 i t} = \frac{1}{\sqrt{145}} e^{i θ} e^{2 i t} = \frac{1}{\sqrt{145}} e^{i (2 t + θ)} .

Our particular solution is the imaginary part of

x_{c},

x_{p} (t) = \frac{1}{\sqrt{145}} \sin (2 t + θ) = \frac{1}{\sqrt{145}} \cos (2 t + θ - \frac{π}{2}) = \frac{1}{\sqrt{145}} \cos (2 t - ϕ),

where

ϕ \approx 3.058451 .

We say that

ϕ

is the phase angle of our solution. The amplitude of our solution is

1 / \sqrt{145}

and the period is

π

(Figure 4.3.4).

Activity 4.3.2. Finding Particular Solutions of the Form $y_{p} = A \cos (ω t - ϕ)$ .

Consider the differential equation

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\begin{matrix} (4.3.2) & y^{″} + 10 y^{'} + 34 y = \sin 2 t . \end{matrix}

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(a)

Find the general solution to the homogeneous equation

y^{″} + 10 y^{'} + 34 y = 0 .

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(b)

Find the complex solution particular solution,

y_{c}

y^{″} + 10 y^{'} + 34 y = e^{2 i t} .

That is, find

a

for

y_{p} = a e^{2 i t} .

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(c)

Determine

A

and

B,

so that

y_{p} = A \cos 2 t + B \sin 2 t

is a particular solution to (4.3.2)

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(d)

Write

a

from Task 4.3.2.b in polar form,

| a | e^{i θ}

to obtain the solution

y_{c} = | a | e^{2 t + θ} .

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(e)

Find a real particular solution in the form

y_{p} = A \cos (ω t - ϕ) .

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(f)

Plot the solution you found in Task 4.3.2.e, labeling the amplitude, period, and frequency of your solution.

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The corresponding first order system for the differential equation

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x^{″} + p x^{'} + q x = g (t),

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\begin{aligned} x^{'} & = y \\ y^{'} & = - q x - p y + g (t) . \end{aligned}

This is a nonautonomous system, and the tangent vector of a solution curve in the phase plane depends not only on the position

(x, y),

but also on the time

t .

In other words, the direction field changes with time. Since the direction field changes with time, two solutions with the same

(x, y)

value at different times can follow different paths. Consequently, solutions can cross each other in the

x y

-plane without violating the Existence and Uniqueness Theorem.

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Example 4.3.5.

Consider the harmonic oscillator that is modeled by the differential equation

\begin{matrix} (4.3.3) & x^{″} + 2 x^{'} + 17 x = - 2 \sin 3 t . \end{matrix}

The solution to the homogeneous equation

x^{″} + 2 x^{'} + 17 x = 0

x_{h} = c_{1} e^{- t} \cos 4 t + c_{2} e^{- t} \sin 4 t

The complex version of this equation is

x^{″} + 2 x^{'} + 17 x = - 2 e^{3 i t},

and we will use the Method of Undetermined Coefficients and assume that we can find a particular solution of the form

x_{c} = A e^{3 i t} .

Substituting

x_{c}

into equation (4.3.3), we find that

(8 + 6 i) A e^{3 i t} = - 2 e^{3 i t} .

Thus,

x_{c}

is a solution if

A = \frac{- 2}{8 + 6 i} = - \frac{4}{25} + \frac{3}{25} i

We have

\begin{aligned} x_{c} (t) & = (- \frac{4}{25} + \frac{3}{25} i) (\cos 3 t + i \sin 3 t) \\ = (- \frac{4}{25} \cos 3 t - \frac{3}{25} \sin 3 t) + i (\frac{3}{25} \cos 3 t - \frac{4}{25} \sin 3 t) . \end{aligned}

The imaginary part of this function is the solution that we seek,

x_{p} = \frac{3}{25} \cos 3 t - \frac{4}{25} \sin 3 t .

Thus, the general solution to (4.3.3) is

x (t) = c_{1} e^{- t} \cos 4 t + c_{2} e^{- t} \sin 4 t + \frac{3}{25} \cos 3 t - \frac{4}{25} \sin 3 t .

Now suppose that

x (0) = 0

and

x^{'} (0) = 0 .

We can quickly determine that

x^{'} (t) = c_{1} e^{- t} (- \cos 4 t - 4 \sin 4 t) + c_{2} e^{- t} (4 \cos 4 t - \sin 4 t) - \frac{12}{25} \cos 3 t - \frac{9}{25} \sin 3 t

To solve this initial value problem, we must solve the linear system

\begin{aligned} c_{1} + \frac{3}{25} & = 0 \\ - c_{1} + 4 c_{2} - \frac{12}{25} & = 0. \end{aligned}

We obtain

c_{1} = - 3 / 25

and

c_{2} = 9 / 100,

and the solution to our initial value problem is

\begin{aligned} x (t) & = - \frac{3}{25} e^{- t} (- \cos 4 t - 4 \sin 4 t) + \frac{3}{20} e^{- t} (4 \cos 4 t - \sin 4 t) - \frac{12}{25} \cos 3 t - \frac{9}{25} \sin 3 t \\ = - \frac{3}{100} (4 \cos 4 t - 3 \sin 4 t)) e^{- t} - \frac{12}{25} \cos 3 t - \frac{9}{25} \sin 3 t \end{aligned}

The graph of our solution is given in Figure 4.3.6.

Figure 4.3.6. Solution to $x^{″} + 2 x^{'} + 17 x = - 2 \sin 3 t,$ $x (0) = 0,$ $x^{'} (0) = 0$

Since

y = x^{'} (t),

we can now graph the solution curve in the phase plane (Figure 4.3.7). Notice how the solution curve can intersect itself. The restoring force and damping are proportional to

x

and

y = x^{'},

respectively. When

x

and

y

are close to the origin, the external force is as large or larger than the restoring and damping forces. In this part of the

x y

-plane, the external force overcomes the damping and pushes the solution away from the origin.

Figure 4.3.7. Phase Plane for $x^{″} + 2 x^{'} + 17 x = - 2 \sin 3 t,$ $x (0) = 0,$ $x^{'} (0) = 0$

On the other hand, suppose we have initial conditions

x (0) = 2

and

x^{'} (0) = 2,

we can solve the linear system

\begin{aligned} c_{1} + \frac{3}{25} & = 2 \\ - 4 c_{1} - c_{2} - \frac{9}{25} & = 2. \end{aligned}

to obtain

c_{1} = 47 / 25

and

c_{2} = 109 / 100 .

Thus, solution to our initial value problem is

x (t) = \frac{47}{25} e^{- t} \cos 4 t + \frac{109}{100} e^{- t} \sin 4 t + \frac{3}{40} \cos 2 t - \frac{3}{20} \sin 2 t .

The graph of our solution is given in Figure 4.3.8.

Figure 4.3.8. Solution to $x^{″} + 2 x^{'} + 17 x = - 2 \sin 3 t,$ $x (0) = 2,$ $x^{'} (0) = 2$

If we examine the phase plane for this solution (Figure 4.3.9), we see that the initial damping and restoring forces are much larger than the external force. Thus, if we are far from the origin, the solutions in the

x y

-plane tend to spiral towards the origin and are similar to the solutions of the unforced equation.

Figure 4.3.9. Solution to $x^{″} + 2 x^{'} + 17 x = - 2 \sin 3 t,$ $x (0) = 2,$ $x^{'} (0) = 2$

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Subsection 4.3.3 Important Lessons

The functions $\sin ω t$ and $\cos ω t$ are periodic with period $2 π / ω$ and frequency $ω / 2 π .$ The average value of each of these functions is zero.
We can use Euler’s formula and complexification to solve the equation
$x^{″} + p x^{'} + q x = g (t),$
where the forcing function $g (t)$ is $\sin ω t$ or $\cos ω t .$ Furthermore, we can use complex numbers to express our solution in the form
$x (t) = A \cos (ω t - ϕ),$
where $A$ is the amplitude of the solution, $ω / 2 π$ is the frequency of the solution, and $ϕ$ is the phase angle.
If we write the equation
$x^{″} + p x^{'} + q x = g (t),$
as a first-order system,
$\begin{aligned} x^{'} & = y \\ y^{'} & = - q x - p y + g (t), \end{aligned}$
we obtain a nonautonomous system. In this case the direction field changes with time, and two solutions with the same $(x, y)$ value at different times can follow different paths. Therefore, solutions can cross each other without violating the Existence and Uniqueness Theorem.
If we are far from the origin, the solutions in the $x y$ -plane tend to spiral towards the origin and are similar to the solutions of the unforced equation. When $x$ and $y$ are close to the origin, the external force is as large or larger than the restoring and damping forces. In this part of the $x y$ -plane, the external force overcomes the damping and pushes the solution away from the origin.