MA 30300: Differential Equations

7.6. Impulses and Delta Functions

Goal: Understand the Dirac delta function and how to apply it in solving ODEs using Laplace transforms.

1. Delta Functions: Introduction

Delta Function as a Limit: The Dirac delta function can be denoted as \(\delta(t-a)\) or \(\delta_a(t)\), representing an instantaneous impulse at \(t=a\). It can be interpreted as the limit of a rectangular pulse of height \(1/\epsilon\) and width \(\epsilon\) starting at \(t=a\):

\[ \delta_a(t) = \delta(t-a) = \lim_{\epsilon \to 0} d_\epsilon(t), \quad d_\epsilon(t) = \begin{cases} \frac{1}{\epsilon}, & a \le t < a+\epsilon \\[1mm] 0, & \text{otherwise} \end{cases} \]

Also, the integral of the delta function itself is \(1\):

\[ \int_0^\infty \delta_a(t) \, dt = 1 \]

Remark: No actual function can satisfy both being zero everywhere except at a single point and having an integral equal to \(1\). Such a “function” is not a classical function but is interpreted as a generalized function called the Dirac delta function.

The delta function is particularly useful in solving ODEs using Laplace transforms:

\[ \mathcal{L}\{\delta_a(t)\} = e^{-as}, \quad a \ge 0 \]

This allows us to handle instantaneous inputs directly in the Laplace domain.

Applications in ODEs:

  1. Modeling sudden forces or impulses in mechanical systems (e.g., a hammer strike on a spring-mass system)
  2. Handling instantaneous changes in electrical circuits or other dynamical systems

Example Problems

Solve the following initial value problems:

  1. \[ x'' + 6x' + 9x = 1+\delta(t-7), \quad x(0) = 0, \quad x'(0) = 0 \]
  2. \[ x'' + 10x' + 26x = \delta(t-\pi) + \delta(t-2\pi), \quad x(0) = 0, \quad x'(0) = 2 \]
  3. \[ x'' + 4x = \delta(t) + \delta(t-\pi), \quad x(0) = x'(0) = 0 \]
  4. \[ x'' + 9x = \delta(t-3\pi) +\cos(3t), \quad x(0) = x'(0) = 0 \]
  5. \[ x'' + 2x' + x = \delta(t) - \delta(t-2), \quad x(0) = x'(0) = 2 \]

2. Systems Analysis and Duhamel’s Principle

Consider a second-order linear system with zero initial conditions:

\[ x'' + a x' + b x = f(t), \quad x(0)=0, \quad x'(0)=0 \]

Taking the Laplace transform of the system, we obtain:

\[ (s^2 + a s + b) X(s) = F(s) \] \[ X(s) = W(s) F(s), \quad \text{where } W(s) = \frac{1}{s^2 + a s + b} \]

The function \( W(s) \) is called the transfer function of the system. The weight function \( w(t) \) is defined as the inverse Laplace transform of the transfer function:

\[ w(t) = \mathcal{L}^{-1}\{W(s)\} \]

It represents the response of the system to an impulse input.

\[ X(s) = W(s) F(s) \]

Taking the inverse Laplace transform gives the time-domain solution:

\[ x(t) = \mathcal{L}^{-1}\{ X(s) \} = \mathcal{L}^{-1}\{ W(s) F(s) \} = (w * f)(t) \]

(the convolution of the weight function \(w(t)\) and the input \(f(t)\)).

Duhamel’s Principle: The solution of the system can be expressed in the Laplace domain as:

\[ x(t) = (w * f)(t) = \int_0^t w(t-\tau)\, f(\tau)\, d\tau = \int_0^t f(t-\tau)\, w(\tau)\, d\tau \]

Here, \(w(t) = \mathcal{L}^{-1}\{ W(s) \}\) is the weight function, and \(*\) denotes convolution. Both integral forms are equivalent because convolution is commutative.

Example 2.1

Apply Duhamel's principle to provide an integral formula for the solution of the following initial value problem:

\[ x'' + 6x' + 8x = f(t); \quad x(0)=0, \quad x'(0)=0 \]

Example 2.2

Use Duhamel’s principle to express the solution of the initial value problem as an integral:

\[ x'' + 4x = f(t); \quad x(0)=0, \quad x'(0)=0 \]

Example 2.3

Apply Duhamel's principle to provide an integral formula for the solution of the following initial value problem:

\[ x'' + 5x' + 6x = f(t); \quad x(0)=0, \quad x'(0)=0 \]

Example 2.4

Find the integral form of the solution using Duhamel’s principle for:

\[ x'' + 2x' + x = f(t); \quad x(0)=0, \quad x'(0)=0 \]

Example 2.5

Apply Duhamel's principle to provide an integral formula for the solution of the following initial value problem:

\[ x'' + 9x = f(t); \quad x(0)=0, \quad x'(0)=0 \]