ODE-ProjectODE-Project Delta Functions and Forcing

Section 6.3 Delta Functions and Forcing

Objectives

To understand Impulse forcing, a term used to describe a very quick push or pull on a system, such as the blow of a hammer or the force of an explosion, and that an impulse function can be described by Dirac delta function, $δ (t),$ which has the properties

$\begin{matrix} δ (t) = 0, t \neq 0; \\ \int_{- \infty}^{\infty} δ (t) d t = 1. \end{matrix}$
To understand that we can use the Dirac delta function to solve initial value problems such as

$\begin{aligned} \frac{d^{2} y}{d t^{2}} + 2 \frac{d y}{d t} + 26 y & = δ_{4} (t) \\ y (0) & = 1 \\ y^{'} (0) & = 0, \end{aligned}$

or

$\frac{d^{2} y}{d t^{2}} + p \frac{d y}{d t} + q y = g (t),$

where $g (t)$ is a function that is very large in a very short time interval.

Subsection 6.3.1 Impulse Forcing

Impulse forcing is the term used to describe a very quick push or pull on a system, such as the blow of a hammer or the force of an explosion. For example, consider the equation for a damped harmonic oscillator

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\frac{d^{2} y}{d t^{2}} + 2 \frac{d y}{d t} + 26 y = g (t),

where

g (t)

is a function that is very large in a very short time interval, say

| t - t_{0} | < τ

and zero otherwise. The integral

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I (τ) = \int_{t_{0} - τ}^{t_{0} + τ} g (t) d t

or since

g (t)

is zero outside of the interval

| t - t_{0} | < τ

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I (τ) = \int_{- \infty}^{\infty} g (t) d t

measures the strength or impulse of the forcing function

g (t) .

In particular, assume that

t_{0} = 0

and

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g (t) = d_{τ} (t) = {\begin{cases} 1 / 2 τ, & - τ < t < τ \\ 0, & otherwise. \end{cases}

It is easy to see that

I (τ) = 1

in this case.

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Examining this forcing function over shorter and shorter time intervals with

τ

getting closer and closer to zero, we find that

I (τ) = 1

in all cases. Thus,

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lim_{τ \to 0} d_{τ} (t) = 0

for

t \neq 0;

however,

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lim_{τ \to 0} I (τ) = 1.

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We can use this information to define the unit impulse function,

δ (t),

to be the ``function’’ that imparts an impulse of magnitude one at

t = 0,

but is zero for all values of

t

other than zero. In other words,

δ (t)

has the properties

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\begin{matrix} δ (t) = 0, t \neq 0; \\ \int_{- \infty}^{\infty} δ (t) d t = 1. \end{matrix}

Of course, we study no such function in calculus. The ``function’’

δ

is an example of what is known as a generalized function. We call

δ,

the Dirac delta function.

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We can define a unit impulse at a point

t_{0}

by considering the function

δ (t - t_{0}) .

In this case,

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\begin{matrix} δ (t - t_{0}) = 0, t \neq t_{0}; \\ \int_{- \infty}^{\infty} δ (t - t_{0}) d t = 1. \end{matrix}

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Subsection 6.3.2 The Laplace Transform of the Dirac Delta Function

Even though the Dirac delta function is not a piecewise continuous, exponentially bounded function, we can define its Laplace transform as the limit of the Laplace transform of

d_{τ} (t)

τ \to 0 .

More specifically, assume that

t_{0} > 0

and

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L (δ (t - t_{0})) = lim_{τ \to 0} L (d_{τ} (t - t_{0})) .

Assuming that

t_{0} - τ > 0,

the Laplace transform of

d_{τ} (t - t_{0})

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\begin{aligned} L (d_{τ} (t - t_{0})) & = \int_{0}^{\infty} e^{- s t} d_{τ} (t - t_{0}) d t \\ = \int_{t_{0} - τ}^{t_{0} + τ} e^{- s t} d_{τ} (t - t_{0}) d t \\ = \frac{1}{2 τ} \int_{t_{0} - τ}^{t_{0} + τ} e^{- s t} d t \\ = \frac{1}{2 s τ} e^{- s t} |_{t = t_{0} - τ}^{t = t_{0} + τ} \\ = \frac{1}{2 s τ} e^{- s t_{0}} (e^{s τ} - e^{- s τ}) \\ = \frac{\sinh s τ}{s τ} e^{- s t_{0}} . \end{aligned}

We can use l’Hospital’s rule to evaluate

(\sinh s τ) / s τ

τ \to 0,

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lim_{τ \to 0} \frac{\sinh s τ}{s τ} = lim_{τ \to 0} \frac{s \cosh s τ}{s} = 1.

Thus,

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L (δ (t - t_{0})) = e^{- s t_{0}} .

We can extend this result to allow

t_{0} = 0,

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lim_{t_{0} \to 0} L (δ (t - t_{0})) = lim_{t_{0} \to 0} e^{- s t_{0}} = 1.

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Example 6.3.1.

Let us now solve the initial value problem

\begin{aligned} \frac{d^{2} y}{d t^{2}} + 2 \frac{d y}{d t} + 26 y & = δ_{4} (t) \\ y (0) & = 1 \\ y^{'} (0) & = 0. \end{aligned}

We can think of this as a damped harmonic oscillator that is struck by a hammer at time

t = 4 .

Let

Y (s) = L (y) (s)

and take the Laplace transform of both sides of the differential equation to obtain

s^{2} Y (s) - s y (0) - y^{'} (0) + 2 (s Y (s) - y (0)) + 26 Y (s) = L (δ_{4}) (s)

s^{2} Y (s) - s + 2 s Y (s) - 2 + 26 Y (s) = e^{- 4 s} .

Solving for

Y (s),

we have

Y (s) = \frac{s + 2}{s^{2} + 2 s + 26} + \frac{e^{- 4 s}}{s^{2} + 2 s + 26} .

The inverse Laplace transform of

Y (s)

\begin{aligned} y & = L (\frac{s + 2}{s^{2} + 2 s + 26}) + L (\frac{e^{- 4 s}}{s^{2} + 2 s + 26}) \\ = L (\frac{s + 2}{(s + 1)^{2} + 25}) + \frac{1}{5} L (\frac{5 e^{- 4 s}}{(s + 1)^{2} + 25}) \\ = e^{- t} \cos 5 t + \frac{1}{5} e^{- t} \sin 5 t + \frac{1}{5} u_{4} (t) e^{- (t - 4)} \sin (5 (t - 4)) . \end{aligned}

Figure 6.3.2. Solution to $y^{″} + 2 y^{'} + 26 y - δ_{4} (t)$

It is important to notice that we are using the Dirac delta function like an ordinary function. This requires some rigorous mathematics to justify that we can actually do this.

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Subsection 6.3.3 Important Lessons

Impulse forcing is the term used to describe a very quick push or pull on a system, such as the blow of a hammer or the force of an explosion. For example, consider the equation for a damped harmonic oscillator
$\frac{d^{2} y}{d t^{2}} + p \frac{d y}{d t} + q y = g (t),$
where $g (t)$ is a function that is very large in a very short time interval, say $| t - t_{0} | < τ$ and zero otherwise. The integral
$I (τ) = \int_{t_{0} - τ}^{t_{0} + τ} g (t) d t$
or since $g (t)$ is zero outside of the interval $| t - t_{0} | < τ$
$I (τ) = \int_{- \infty}^{\infty} g (t) d t$
measures the strength or impulse of the forcing function $g (t) .$
We define the unit impulse function, $δ (t),$ to be the ``function’’ that imparts an impulse of magnitude one at $t = 0,$ but is zero for all values of $t$ other than zero. In other words, $δ (t)$ has the properties
$\begin{matrix} δ (t) = 0, t \neq 0; \\ \int_{- \infty}^{\infty} δ (t) d t = 1. \end{matrix}$
The “function” $δ$ is an example of what is known as a generalized function. We call $δ,$ the Dirac delta function.
Similarly, we can define a unit impulse at a point $t_{0}$ by considering the function $δ (t - t_{0}) .$ In this case,
$\begin{matrix} δ (t - t_{0}) = 0, t \neq 0; \\ \int_{- \infty}^{\infty} δ (t - t_{0}) d t = 1. \end{matrix}$
The Laplace transform of the Dirac delta function is
$L (δ (t - t_{0})) = e^{- s t_{0}} .$
We can extend this result to allow $t_{0} = 0,$ by
$lim_{t_{0} \to 0} L (δ (t - t_{0})) = lim_{t_{0} \to 0} e^{- s t_{0}} = 1.$
We can use the Dirac delta function to solve initial value problems such as
$\begin{aligned} \frac{d^{2} y}{d t^{2}} + 2 \frac{d y}{d t} + 26 y & = δ_{4} (t) \\ y (0) & = 1 \\ y^{'} (0) & = 0, \end{aligned}$
or
$\frac{d^{2} y}{d t^{2}} + p \frac{d y}{d t} + q y = g (t),$
where $g (t)$ is a function that is very large in a very short time interval.