The scariest monsters are the ones that lurk within our souls, Edgar Allen Poe.

An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:

**Dependent and independent variables**. Variables represent unknown quantities. The independent variable is chosen freely, while the dependent variable changes in response to the independent variable.**Constants**. Fixed numerical values that do not change.**Algebraic operations**. Operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction.

Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

It involves (e.g., $\frac{dy}{dx} = 3x +5y$):

**Dependent variables**:*Variables that depend on one or more other variables*(y).**Independent variables**: Variables upon which the dependent variables depend (x).**Derivatives**: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

- The function f(x, y) (the right-hand side of the ODE) in y’ = f(x, y) is continuous in a neighborhood around a point (x
_{0}, y_{0}) and - Its partial derivative with respect to y, $\frac{∂f}{∂y}$, is also continuous near (x
_{0}, y_{0}).

Then the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x_{0}, y_{0}) .

A first-order linear differential equation (ODE) has the general form: a(x)y' + b(x)y = c(x) where y′ is the derivative of y with respect to x, and a(x), b(x), and c(x) are functions of x. If c(x) = 0, the equation is called homogeneous, i.e., a(x)y’ + b(x)y = 0.

The equation can also be written in the standard linear form as: y’ + p(x)y = q(x) where $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$

A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0 where:

- y is the dependent variable (a function of the independent variable t),
- y′ and y′′ are the first and second derivatives of y with respect to t,
- t is the independent variable,
- A and B are constants.

This equation is homogeneous, meaning that there are no external forcing terms (like a function of t) on the right-hand side.

The Laplace Transform of a function f(t), where t ≥ 0, is defined as $\mathcal{L}(f(t)) = \int_{0}^{∞} f(t)e^{-st}dt = F(s)$.

One of the most important properties of the Laplace Transform is linearity, which states: $\mathcal{L}(af(t)+bg(t)) = a\mathcal{L}(f(t))+b\mathcal{L}(g(t))$

Function | Laplace Transform |
---|---|

u(t) | $\mathcal{L}(u(t)) = \frac{1}{s}, s > 0$ |

$e^{at}$ | $\frac{1}{s - a}, s > a$ |

$e^{(a + bi)t}$ | $\mathcal{L}(e^{(a + bi)t}) = \frac{1}{s - (a + bi)}, s > a$ |

$\cos(\omega t)$ | $\mathcal{L}(\cos(\omega t)) = \frac{s}{s^2 + \omega^2}, s > 0$ |

$\sin(\omega t)$ | $\mathcal{L}(\sin(\omega t)) = \frac{\omega}{s^2 + \omega^2}, s > 0$ |

$t^n$ | $\mathcal{L}(t^n) = \frac{n!}{s^{n+1}}, s > 0$ |

$u(t-a)$ | $\mathcal{L}(u(t-a)) = \frac{e^{-as}}{s}, s > 0$ |

$\delta(t-a)$ | $\mathcal{L}(\delta(t-a)) = e^{-as}, a \geq 0$ |

$\frac{1}{t}$ | $\mathcal{L}\left(\frac{1}{t}\right) = \text{not defined}$ |

$e^{-bt} \cos(\omega t)$ | $\mathcal{L}(e^{-bt} \cos(\omega t)) = \frac{s + b}{(s + b)^2 + \omega^2}, s > -b$ |

$e^{-bt} \sin(\omega t)$ | $\mathcal{L}(e^{-bt} \sin(\omega t)) = \frac{\omega}{(s + b)^2 + \omega^2}, s > -b$ |

$e^{at}f(t)$ | $\mathcal{L}(e^{at}f(t)) = F(s-a)$ |

This is the Exponential Shift Theorem, indicating that multiplying a function by an exponential term shifts its Laplace Transform.

Besides, $\mathcal{L}(f’(t)) = sF(s)-f(0), \mathcal{L}(f’’(t)) = s^2F(s)-sf(0)-f’(0)$

In the study of differential equations and physical systems, impulse inputs represent sudden forces applied over a very short period of time. These impulses model events like a hammer strike or a collision, where a large force is exerted briefly.

If f(t) is a force applied over time, the impulse delivered over the time interval [a, b] is defined as: Impulse = $\int_{a}^{b} f(t)dt$.

The impulse represents the total effect of a force applied over a time interval, accounting for both the magnitude and the duration of the force.

In the case where f(t) is a constant force F, then the impulse simplifies to: Impulse = $\int_{a}^{b} f(t)dt = F\int_{a}^{b} dt = F·(b-a)$. The total impulse is proportional to the magnitude of the force and the duration over which it is applied.

Consider an undamped mass-spring system with mass m. When an impulse is applied, we can model the force over a small time interval, say [0, h], e.g., image applying a short burst of force to the mass over this brief period (Refer to Figure i for a visual representation and aid in understanding it), by using a box function.

In mathematical terms, to represent an impulse of area 1 (a unit impulse), the height of the force during the interval must be: ^{1}⁄_{h}. This reflects that as the time interval becomes shorter, the magnitude of the force must increase to maintain the total impulse (area under the curve) as 1.

To represent an impulse of area 1 (a unit impulse), we define the force function f(t) as: $f(t) = \frac{1}{h}[u(t) -u(t-h)]$ where u(t) is the Heaviside Step Function, u(t) -u(t-h) is the unit box function u_{0h}(t), which is 1 in the interval [0, h] and 0 elsewhere.

Impulse = $\int_{a}^{b} f(t)dt = \int_{0}^{h} \frac{1}{h}[u(t) -u(t-h)]dt = \int_{0}^{h} \frac{1}{h}dt = 1$.

- As the time interval h becomes very small, the force magnitude $\frac{1}{h}$ becomes very large to maintain a constant impulse of 1.
- This reflects the idea that an impulse is a very large force applied over an infinitesimally small time period.

The equation of motion for the undamped mass-spring system under the action of the impulse is: $y'' + y = {1}{h}[u(t) -u(t-h)]dt = \frac{1}{h}u_{0h}(t)$ where:

- y(t) is the displacement of the mass from its equilibrium position.
- y′′ +y = 0 is the homogeneous equation describing the natural oscillations of the system.
- u
_{0h}(t) = u(t) - u(t-h) is a box function representing a force that turns on at t = 0 and turns off at t = h. - The right-hand side, $\frac{1}{h}u_{0h}(t)$, represents the force of height
^{1}⁄_{h}applied over the interval [0, h]

Step 1: **Laplace Transform of the Forcing Function**

To solve this equation in the Laplace domain, recall the Laplace Transform of a step function: $u(t) \leadsto \frac{1}{s}, u(t-a) \leadsto e^{-as}\frac{1}{s}$

Thus, applying this to the box function $u_{ab}(t) = u_a(t)-u_b(t) = u(t-a)-u(t-b)$, a = 0, b = h:

$\frac{1}{h}u_{0h}(t) = \frac{1}{h}[u(t)-u(t-h)] \leadsto \frac{1}{h}[\frac{1}{s}-\frac{e^{-hs}}{s}] = \frac{1-e^{-hs}}{hs}$

$\frac{1}{h}u_{0h}(t) \leadsto \frac{1-e^{-hs}}{hs}, \mathcal{L}(\frac{1}{h}[u(t)-u(t-h)]) = \frac{1-e^{-hs}}{hs}$

Step 2: **Taking the Limit as h → 0**

As the width h of the impulse interval approaches zero (h → 0), we get: $\lim_{h \to 0} \frac{1-e^{-hs}}{hs} =[\text{Let u = hs}] \lim_{u \to 0} \frac{1-e^{-u}}{u} =[\text{L’Hospital’s rule }] \lim_{u \to 0} \frac{e^{-u}}{1} = 1$.

On the other side of the Laplace Transform, as h approaches zero, the box function $\frac{1}{h}u_{0h}(t)$ approaches the Dirac delta function, denoted by δ(t).

Conclusion. The Laplace Transform of the Dirac Delta Function is $\mathcal{L}(δ(t)) = 1$.

As h → 0, the interval [0, h] get smaller and smaller and the value of the function at 0 goes to infinity.

The Dirac **delta function** δ(t), also known as the unit impulse, is not a function in the traditional sense but a **a generalized function on the real numbers or distribution, whose value is zero everywhere except at zero t = 0, and whose integral over the entire real line is equal to one**, $\int_{-∞}^{∞} δ(t)dt = 1$ (Refer to Figure ii for a visual representation and aid in understanding it), and for any continuous function f(t) = $\int_{-∞}^{∞} δ(t)f(t)dt = f(0)$ (🚀)

δ(t) is zero everywhere except at t = 0, where it is “infinite” in such a way that its integral over any interval containing t = 0 is 1.

The most important property of the Dirac delta function is the sifting property, which states that for any function g(t) that is continuous at t = a: $\int_{-∞}^{∞} δ(t-a)g(t)dt = g(a)$. In other words, The delta function “picks out” the value of g(t) at t = a.

$δ_a = \begin{cases} ∞, &t = a \\ 0, &t ≠ a \end{cases}$

The Mean Value Theorem for Integrals states that if f(t) is continuous on the closed interval [a, b], then there exists a point c ∈ [a,b] such that: $\int_{a}^{b} f(t)dt = (b-a)f(c)$

The total area under f(t) from a to b equals the area of a rectangle with width b − a and height f(c).

Let δ_{a, ε} be an approximation of the delta function (Rectangular Pulse Approximation) such that is zero outside [a, a + ε] and it has a constant height of $\frac{1}{ε}$

$δ_{a, ε} = \begin{cases} \frac{1}{ε}, &t ∈ [a, a + ε] \\ 0, &otherwise \end{cases}$

Let’s calculate: $\int_{0}^{∞} δ_{a, ε}(t)g(t)dt = \int_{a}^{a+ ε} \frac{1}{ε}g(t)dt =[\text{Th. Mean Value Theorem for Integrals}] ε·\frac{1}{ε}g(c) = g(c)$ where c ∈ [a, a + ε]

Taking the Limit as ε → 0, the interval [a, a+ε] shrinks to the point a, then c → a, hence $\int_{0}^{∞} δ_{a, ε}(t)g(t)dt → g(a)$.

Therefore, $\int_{0}^{∞} δ_a(t)g(t)dt = g(a)$

Special Case: a = 0, the Dirac delta function becomes δ(t), centered at the origin. f(t) = $\int_{-∞}^{∞} δ(t)f(t)dt = f(0)$

Since δ(t) is zero everywhere except at t = 0, the integral picks out the value of f(t) at t = 0.

Another property is: $\frac{d}{dt}u(t-a)=δ_a(t)$

The Laplace Transform of the Dirac Delta Function is: $\mathcal{L}(δ(t)) = \int_{0}^{∞} e^{-st}δ(t)dt = e^{-s·0} = 1$. The delta function picks out the value of f(t) = e^{-st} at t = 0 🚀, which is 1.

This makes the delta function a powerful tool in modeling instantaneous forces or idealized impulses, as it encapsulates the idea of applying a force over an infinitesimally small time period (a force that acts instantaneously at t = 0, e.g., if you strike a mass-spring system with a hammer), while still imparting a finite impulse.

In the context of the Laplace Transform, we typically deal with causal functions —functions that are defined to be zero for t < 0.

Next, we explore how the delta function behaves under convolution. Let f be an arbitrary function. The convolution of f(t) with the delta function δ(t) is defined as: (f * δ)(t) = $\int_{0}^{t} f(τ)δ(t - τ)dτ = f(t)$

Since δ(t - τ) is non-zero only when τ = t. If t ≥ 0, τ = t is withing [0, t]. The integral 🚀 reduces to f(t).

In the Laplace domain, convolution becomes multiplication. The Laplace Transform of δ(t) is 1, so:

$u(t)f(t) * δ(t) \leadsto_{\mathcal{L}} F(s)·1, u(t)f(t)\leadsto_{\mathcal{L}} F(s)$. Therefore, **u(t)f(t) * δ(t) = u(t)f(t)**. In other words, the delta function acts as the identity under convolution,, effectively “sampling” the function at t. In other words, convolving any function with δ(t) simply reproduces the original function. This property reflects the nature of the delta function as a mathematical “impulse” that **isolates the behavior of a system at a specific moment in time.**

Additionally, the derivative of the step function u(t) is the delta function: u’(t) = δ(t).

The step function jumps from 0 to 1 at t = 0, and its derivative is the delta function at that point.

Consider an undamped mass-spring system where the mass m is attached to a spring. The system is subjected to (“kicked with”) an impulse of magnitude A applied at time t = ^{π}⁄_{2}. The initial conditions are given as: y(0) = 1, y’(0) = 0.

The goal is to solve the differential equation that governs the motion of the system using the Laplace Transform.

The assumption “kicked with impulse A at t = ^{π}⁄_{2} can be expressed mathematically using the Dirac delta function $Aδ(t-\frac{π}{2})$, which models the instantaneous application of force at t = ^{π}⁄_{2}.

Therefore, the model that control our motion is $y'' + y = Aδ(t-\frac{π}{2})$ where

- y is the displacement of the mass from its equilibrium position.
- The term y′′ +y = 0 describes the natural behavior of an undamped mass-spring system, which would oscillate without any external force.
- The term Aδ(t -
^{π}⁄_{2}) introduces the external force, modeled as an impulse of magnitude A applied at t =^{π}⁄_{2}.

Using the standard Laplace Transform rules: $\mathcal{L}(y’’) = s^2Y(s) −sy(0) −y’(0), \mathcal{L}(y) = Y(s), \mathcal{L}(Aδ(t-\frac{π}{2})) = Ae^{\frac{-π}{2}s}$, the model’s equation Laplace Transform is $s^2Y -s +Y = Ae^{\frac{-π}{2}s} ↭ (s^2+1)Y - s = Ae^{\frac{-π}{2}s}$. Then, we solve by $Y = \frac{s}{s^2+1} + \frac{Ae^{\frac{-π}{2}s}}{s^2+1}$.

We now take the Inverse Laplace Transform to find y(t) in the time domain. The expression for Y(s) consists of two terms that we can invert separately:

The first term is a standard Laplace Transform, and its inverse is: $\frac{s}{s^2+1} \leadsto_{\mathcal{L^{-1}}} cos(t)$

For the second term, we use the T-axis translation formula: $u(t-a)f(t-a) \leadsto e^{-as}F(s)$ where $F(s) = \frac{A}{s^2+1}$, and $\mathcal{L}^{-1}(\frac{A}{s^2+1}) = A·sin(t)$. Therefore, $\mathcal{L}^{-1}(\frac{Ae^{\frac{-π}{2}s}}{s^2+1}) = u(t-\frac{π}{2})Asin(t-\frac{π}{2})$

Therefore y(t) = $cos(t) + Au(t-\frac{π}{2})sin(t-\frac{π}{2})$

**Simplify the expression**. $sin(t-\frac{π}{2}) = -cos(t)$

$sin(t-\frac{π}{2}) = sin(t)cos(\frac{π}{2}) -cos(t)sin(\frac{π}{2}) = sin(t)·0 -cos(t)·1 = -cos(t)$

$y = \begin{cases} cos(t), &0 ≤ t ≤ π/2 \\ cost(t) -Acos(t), &t ≥ π \end{cases}$

$y = \begin{cases} cos(t), &0 ≤ t ≤ π/2 \\ (1 - A)cos(t), &t ≥ π/2 \end{cases}$

Refer to Figure iii for a visual representation and aid in understanding it.

Interpretation. 0 ≤ t ≤ π/2, The system oscillates naturally with displacement y(t) = cos(t). No external force has been applied yet. At t = π/2, an impulse of magnitude A is applied. This causes an instantaneous change in the amplitude of the oscillations. t ≥ π/2, The amplitude of the oscillations is scaled by (1 −A). The system continues to oscillate, but the amplitude is altered due to the impulse.

Consider an **undamped mass-spring system** where:

**Mass m**: Attached to a spring with stiffness k k (for simplicity, set k = 1 for normalization).**Displacement y(t)**: Represents the displacement of the mass from its equilibrium position at time t.**Impulse Force f(t)**: External forces applied as impulses at specific times.

The system is subjected to impulses (e.g. a hammer) at times t = nπ for n = 0, 1, 2,…. These impulses can be modeled using the Dirac delta function δ(t− nπ), which represents an instantaneous force applied at t = nπ.

Therefore, the differential equation governing the motion of the system is: $y'' + y = \sum_{n=0}^\infty δ(t-nπ)$. Initial conditions y(0) = y’(0) = 0. These conditions indicate that the system starts from rest with no initial displacement or velocity.

The goal is to solve the differential equation that governs the motion of the system using the Laplace Transform.

Using the standard Laplace Transform rules: $\mathcal{L}(y’’) = s^2Y(s) −sy(0) −y’(0), \mathcal{L}(y) = Y(s), \mathcal{L}(δ(t-a)) = e^{-as}$, the model’s equation Laplace Transform is $\mathcal{L}(y’’)+\mathcal{L}(y) = \mathcal{L}(y)(\sum_{n=0}^\infty δ(t-nπ)) ↭ s^2Y +Y = \sum_{n=0}^\infty e^{-nπs} ↭ (s^2+1)Y = \sum_{n=0}^\infty e^{-nπs}$. Then, we solve by $Y = \sum_{n=0}^\infty \frac{e^{-nπs}}{s^2+1}$.

Recall $u(t-a)f(t-a) \leadsto e^{-as}F(s)$, where F(s) = $\frac{1}{s^2+1}, \mathcal{L}^{-1}(\frac{1}{s^2+1}) = sin(t)$.

$Y(t) = \sum_{n=0}^\infty u(t -nπ)sin(t -nπ)$

Recall sin(t -nπ) = (-1)^{n}sin(t) and observe for any arbitrary t: nπ < t < (n+1)π, u(t -nπ) = 0 for t < nπ.

$Y(t) = sin(t)\sum_{n=0}^\infty (-1)^n·u(t -nπ)$

Y(t) = sin(t) -sin(t) + ··· + (-1)^{n}sin(t)

$Y(t) = \begin{cases} sin(t), &n,even \\ 0, &n,odd \end{cases}$

The system oscillates naturally with displacement y(t) = sin(t). An impulse is applied, instantaneously altering the system’s state. The system alternates between oscillating and being at rest due to the periodic impulses.

In this explanation, we will explore how to solve a general second-order linear differential equation using Laplace Transforms.

We begin by considering the following general second-order linear differential equation: y’’ + ay’ + by = f(t) where:

- y(t) is the output or response of the system (e.g., the displacement of a mass, the voltage in a circuit, etc.).
- f(t) is the input or external forcing function (e.g., an applied force or voltage).
- a and b are constants related to the physical parameters of the system, such as damping coefficient or stiffness in mechanical systems.
- The initial conditions are given as y(0) = 0 and y′(0) = 0, meaning the system starts from rest.

Recall the following properties of the Laplace Transform: $\mathcal{L}(y’’(t))=s^2Y(s)-sy(0)-y’(0) =[\text{Applying initial conditions}] s^2Y(s), \mathcal{L}(y’(t)) = sY(s)-y(0) = [\text{Applying initial conditions}] = sY(s), \mathcal{L}(y(t)) = Y(s), \mathcal{L}(f(t)) = F(s)$

The Laplace transform of the equation y’’ + ay’ + by = f(t) is $s^2Y +aSY + bY = F(s)$. Combining like terms: $(s^2 +aS + b)Y = F(s)$.

Solve for Y: $Y = F(s)·\frac{1}{s^2+as+b}$

The term $\frac{1}{s^2+as+b}$ is referred to as the transfer function of the system, denoted by W(s). The transfer function describes how the system responds to inputs in the Laplace (frequency) domain.

The inverse Laplace transform of the transfer function, $\mathcal{L}^{-1}(\frac{1}{s^2+as+b}) = W(t)$ is called the weight function (or impulse response function). It represents the system’s response over time to a unit impulse applied at t = 0.

The solution y(t) in the time domain can be found by taking the inverse Laplace Transform of Y(s):

Taking the inverse Laplace transformation: $y(t) = f(t)*W(t)$. Thus, the solution in the time domain can be written as a convolution between the input function f(t) and the weight function W(t).

By the definition of convolution, this is written as: $y(t)= \int_{0}^{t} f(u)W(t-u)du$

To understand what W(t) represents, consider the following scenario: we subject the system to a unit impulse δ(t), which is a sharp “kick” (unit impulse) applied at t = 0. This is equivalent to solving the differential equation with f(t) = δ(t)

So, the equation becomes: y’’ + ay’ + by = δ(t) with initial conditions y(0) = 0 = y’(0).

Taking the Laplace Transform yields: $s^2Y’’ + asY + bY = 1 ⇒[\text{Solving for Y}] Y = \frac{1}{s^2+as+b}$, and taking its inverse Laplace transform: $y(t) = δ(t)*W(t) = W(t)$. Therefore, the weight function W(t) is the solution to to the system when it is subjected to an impulse at t = 0 and starts from rest. It characterizes how the system responds over time to a sudden or sharp impulse (kick).

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].

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