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ORDINARY DIFFERENTIAL EQUATIONS Lecture 20 Solution of Linear IVPs by Laplace Transforms

(Revised 13 April, 2009 @ 00:00) Professor Stephen H Saperstone Department of Mathematical Sciences George Mason University Fairfax, VA 22030 email: [email protected] Copyright © 2009 by Stephen H Saperstone All rights reserved

20.1 Solutions with Step Inputs We investigate a sequence of examples of the form where

is a step function (in Example 20.1), and then it is a pulse (in Examples 20.2 and 20.3).

Example 20.1 Calculate the solution to the IVP

Solution:

Apply LT's to both sides of the ODE

and get

where and the right side comes from Property 8 in Table 19.1. Substitute the initial values into the last equation and simplify. We get

so that

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The last equation is in a form suitable for Property 9 in Table 19.1 that is

whose inverse Laplace transform is where

and

Thus we must determine

As

is given by

Write

Now use Entry 5 of Table 18.1 to get

Finally from Property 5 in Table 19.1 we compute

So Hence the solution to the IVP is

Observe that as

since cosine is periodic with period

Thus we can express the solution

The solution is plotted in Figure 20.1.

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Figure 20.1

End of Example 20.1

20.2 Solutions with Pulse Inputs Example 20.2 Calculate the solution to the IVP

Solution:

Note that the input is just the pulse with graph

Figure 20.2

This IVP differs from Example 20.1 in just the additional input term. Apply LT's to both sides of the ODE to get

where and the right side comes from Property 8 in Table 18.1. Substitute the initial values into the last equation and simplify. We get

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so that

From Example 20.1 we readily see that where

was calculated

It follows that the solution is

and with graph plotted in Figure 20.3.

Figure 20.3

Observe how the solution remains zero after This is because both input is turned off at that time, the solution must remain at zero. End of Example 20.2

and

have value zero at

Since the

Example 20.3 Calculate the solution to the IVP

Solution: get

This IVP differs from Example 20.1 in just the additional input term. Apply LT's to both sides of the ODE to

where and the right side comes from Property 8 in Table 19.1. Substitute the initial values into the last equation and simplify. We get

so that

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From Example 20.1 we readily see that where

was calculated

It follows that the solution is This solution doesn't lend itself to an easy trigonometric simplification as did the solutions in Examples 20.1 & 20.2. It'sgraph is more illuminating, though, as Figure 20.4 shows.

Figure 20.4

When the input is turned off at the response is at its peak: and Even though the input is zero from this point on, the response behaves as though simple harmonic motion begins anew from the initial values and End of Example 20.3

20.3 Solutions with Periodic Pulse Inputs Example 20.4 Calculate the solution to the IVP where

is the square wave with graph

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Solution:

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Apply LT's to both sides of the oDE to get

Apply the -periodiodicity property to compute

Note that the period

Then

Thus

so that

Next recognize that

The term

can be expressed as the product

is the sum of the geometric series

Multiply the infinite series termwise by

so that we have

to get

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Combine terms so that

Now use convolution to compute

Finally use the -shift property to obtain

To better visualize

write out the first few terms of the sum:

Observe the "resonance like" response in the plot below of the first 10 terms of the series.

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End of Example 20.4

Example 20.5 (Sawtooth Input) Calculate the solution to the IVP where

is the sawtooth wave

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Solution:

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Apply LT's to both sides of the ODE to get

where and the right side comes from -periodicity: Property 10 in Table 19.1. Substitute the initial values into the last equation and simplify. We get

so that

Then

Set

We invert

by appealing to -integration: Property 5 in Table 18.1. In particular, write

where

In Example 20.1 we showed that Set

and proceed again as in Example 20.1 to get

Next we invert

by creating a geometric series and appealing to the -shift: Property 9 in Table 19.1. First write

Then

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Note that each term in the series has the form specified by the -shift property in Table 19.1: where

Thus

Now

from Eqn. (20.1), so we have

Consequently

Put this together with

to get the solution

We interpret the solution as follows: a. Here for all

The solution on this interval is simply

b.

Here

for all

The solution on this interval is simply

c.

Here

for all

The solution on this interval is

End of Example 20.5

20.4 Solutions with Impulsive Inputs What happens when the input pulse approaches that of a delta function?

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Example 20.6 Calculate the solution to the IVP

Solution:

Apply LT's to both sides of the ODE to get

where and the right side comes from Entry 13 of Table 18.1 and Property 9 in Table 19.1. Substitute the initial values into the last equation and simplify. We get

so that

Since

we must have that

It follows that

since sine is periodic with period is just the sine wave

Thus a plot of the solution

starting at

The input is turned on at End of Example 20.6

and the ensuing motion is simple harmonic.

Application Example 20.7 [Pendulum Clock]

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The motion will die out as a result of friction. An escape mechanism is frequently employed to overcome the frictional forces as illustrated in the graphic of the pendulum above. An escapement is a toothed wheel that drives the hands of the clock through a sequence of gears. The hanging weight, the bob, is supported by a cable that winds around the spindle. The resulting torque causes the spindle to turn. The motion of the escape wheel is stopped by a toothed anchor that rocks back and forth with the bob. The teeth are designed so that when the pendulum swings through in one direction, the escape wheel exerts a small impulse (force) on the anchor, thereby giving the bob an extra push in the same direction. The challenge is to model the motion of the bob by an appropriate ODE. In the absence of any forcing to keep the pendulum clock in motion, an appropriate IVP for the motion of the pendulum bob is given by Here, represents an initial angular velocity imparted to the pendulum bob to get it moving at time verify that that the solution to this IVP when (underdamped motion) is

Note that when

the solution

represents simple harmonic motion with angular frequency

You should

The introduction

of friction causes the (natural) angular frequency to decrease to In order to better visualize the motion of the bob in the presence of friction, we plot the motion for the assigned values and

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To create the effect of an impulse that acts on the pendulum when the bob is at the bottom of its swing, we introduce the restoring impulsive force

The right side of this equation for

represents a sequence of impulsive forces of magnitude

The alternating sign determined by model the motion of the bob is

applied at times

represents the alternating directions of the impulses. Thus the ODE we use to

Note that we time the impulses to coincide with the quasi-frequency of the unforced (but damped) motion. An analysis of this ODE with initial values and the values and yields the solution

It's graph is

It is instructive to examine the graph of

so that we can see the role the impulsive forces play in the solution.

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A phase portrait of the motion is given by

End of Example 20.7

20.5 GREEN'S KERNEL SOLUTION (AGAIN!) Consider the general IVP with causal input Apply Laplace transforms to both sides to obtain

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where

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and

Then

Collect terms and transpose so that

or

where we have set

Then upon taking inverse LTs we get

In view of the zero-state solution to Eqn. (20.2) for causal inputs

we must have by the uniqueness of Laplace transforms that

Thus,

If we set

then we have

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which is precisely what the convolution theorem above tells us. In particular, when solution to Eqn. (20.2) is given by the inverse Laplace transform

and

the

Definition 20.8 [Transfer Function] The function

defined by

is called the transfer function of the Eqn. (21.2). Observe that the expression is precisely the characteristic polynomial of the associated homogeneous ODE. Thus the impulse response is the inverse LT of the transfer function, i.e.,

Also note that

must be the Laplace transforms of the ideal BSS for

20.6 INTEGRAL EQUATIONS Some linear equations other than ODEs can be solved with the aid of Laplace transforms. We illustrate the usefulness of Laplace transforms in the next few examples. The first example is a linear integral equation.

Example 20.9 Use Laplace transforms and the convolution theorem to solve the following integral equation for

Solution: equation:

Observe that the integral is a convolution of

and

Take the Laplace transorm of both sides of the

to get

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Let

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. Then

Solve for

to obtain

It follows that

End of Example 20.9

Volterra Integral Equations An important class of integral equations is Volterra's integral equation, which has the form

where and are known functions with Laplace transforms that the integral is a convolution so that Eqn. (20.4) can be rewritten as Apply the Laplace transform operator where

and

We recognize

to Eqn. (20.5) to obtain

We can readily solve for

to get

Finally, the solution to Volterra's integral equation can be represented as

Example 20.10 Use Laplace transforms and the convolution theorem to solve the following Volterra integral equation for

Solution:

Observe that the integral is the convolution

Then we have Now apply the Laplace transform operator

to this equation and get

Apply the convolution theorem and use Entries 2 and 4 of Table 18.1 to obtain

where

Next, solve for

:

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Now we invert each of the preceding terms. The first term is easy:

The second term requires the -integration property. Set

Then

so

Putting it all together yields the solution

End of Example 20.10

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