Laplace Transform
4.7 Impulse and Dirac Delta Function
In prior sections, we explored initial value problems for second-order differential equations with constant coefficients, focusing on cases where the forcing function, [asciimath]f(t)[/asciimath], is either continuous or piecewise continuous on the interval [asciimath][0, oo)[/asciimath].
[asciimath]ay''+by'+cy=f(t)\ ;[/asciimath] [asciimath]y(0)=k_0, \ \ y'(0)=k_1[/asciimath]
Now, let’s turn our attention to a different type of forcing function: one that represents an impulsive force. Impulsive forces are characterized by very large magnitudes over extremely short time intervals, effectively appearing as a sudden “jolt” or “spike” in the system. Such impulses occur in various contexts, including electrical circuits during a switch-on event, mechanical systems during a collision, or any scenario where a sudden, significant force is applied for a brief period.
A. Dirac Delta Function
To mathematically model these impulsive forces, we use the Dirac Delta function, denoted as [asciimath]delta(t)[/asciimath]. The Dirac Delta function is not a function in the traditional sense but rather a generalized function or distribution with the following properties.
1. Zero everywhere except at zero:
[asciimath]delta(t)={(0 \ if\ t!=0),(oo \ if \ t=0):}[/asciimath]
2. Integral equals one:
[asciimath]int_-oo^oo delta(t) = 1[/asciimath]
3. Sifting property:
[asciimath]int_-oo^oo f(t) delta(t) = f(0)[/asciimath] for any [asciimath]f(t)[/asciimath] that is continuous on the interval that contains [asciimath]t=0[/asciimath]
By shifting the argument [asciimath]t[/asciimath] in [asciimath]delta(t)[/asciimath], we can model impulses that occur at times other than [asciimath]t=0[/asciimath]. The shifted Dirac Delta function, [asciimath]delta(t-a)[/asciimath], has a spike at [asciimath]t=a[/asciimath] and is defined as
[asciimath]delta(t-a)={(0 \if \ t!=a),(oo \if \ t=a):}[/asciimath]
Thus the sifting property extends to
4. Sifting at [asciimath]t=a[/asciimath]:
[asciimath]int_-oo^oo f(t) delta(t-a) = f(a)[/asciimath] for any [asciimath]f(t)[/asciimath] that is continuous on the interval that contains [asciimath]t=a[/asciimath]
B. Laplace Transform of the Dirac Delta Function
The Laplace Transform provides a convenient way to handle the Dirac Delta function in the context of solving differential equations. The transform of a shifted Dirac Delta function is given by
[asciimath]\mathcal{L}{delta(t-a)}=e^(-as)[/asciimath](4.7.1)
Understanding the Dirac Delta function and its properties is crucial for modeling and analyzing systems subjected to impulsive forces.
Find the solution to the initial value problem
[asciimath]y''+16y=4delta(t-pi),[/asciimath] [asciimath]y(0)=1,[/asciimath] [asciimath]y'(0)=0[/asciimath]
Show/Hide Solution
Taking the Laplace transform of the equation, applying Equation 4.7.1 with [asciimath]a=pi[/asciimath] to the Delta function, yields
[asciimath]s^2 Y(s)-s+16Y(s)=4e^(-pis)[/asciimath]
Solving for [asciimath]Y(s)[/asciimath], we obtain
[asciimath]Y(s)=(4e^(-pis) +s)/(s^2+16)[/asciimath]
[asciimath]=e^(-pis) 4/(s^2+16)+s/(s^2+16)[/asciimath]
Computing the inverse Laplace transform gives
[asciimath]y(t)=u(t-pi) sin(4(t-pi))[/asciimath] [asciimath]+cos(4t)[/asciimath]
Which equivalently is
[asciimath]y(t)= {(cos(4t) if tltpi),(sin(4t)+cos(4t) if t>=pi):}[/asciimath]
[asciimath]={(cos(4t) if tltpi),(sqrt(2)sin(4t +pi/4) if t>=pi):}[/asciimath]
The below figure shows [asciimath]y(t)[/asciimath]. The impulsive force is applied and adds momentum to the system at [asciimath]t=pi[/asciimath]. For comparison, the dotted line represents the undisturbed system.
Try an Example
Section 4.7 Exercises
- Solve the initial value problem
[asciimath]y'' + 25 y = delta(t-4), \quad \ y(0) = y'(0) = 0.[/asciimath]
Show/Hide Answer
[asciimath]y(t)={(0 \ if t<4),(1/5 sin(5(t-4)) \if t>=4):}[/asciimath]
- Solve the initial value problem
[asciimath]y'' + 4 y = 80 e^(4 t) + delta(t-7), \quad \ y(0) = 11,\ y'(0) = 32[/asciimath]
Show/Hide Answer
[asciimath]y(t)={(7cos(2 t) + 8sin(2 t) + 4 e^(4 t) \ if t<7),(7cos(2 t) + 8sin(2 t) + 4 e^(4 t) + 1/2 sin(2(t-7)) \if t>=7):}[/asciimath]
