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The scariest monsters are the ones that lurk within our souls, Edgar Allen Poe.
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., $\frac{dy}{dx} = 3x +5y$):
Definition. A first-order linear ordinary differential equation is an ordinary differential equation (ODE) involving an unknown function y(x), its first derivative y′, and functions of the independent variable x, which can be written in the general form:: a(x)y' + b(x)y = c(x) where:
These equations are termed “linear” because the unknown function y and its derivative y’ appear to the first power and are not multiplied together or composed in any nonlinear way.
If the function c(x)=0 for all x in the interval of interest, the equation simplifies to: a(x)y’ + b(x)y = 0. Such an equation is called a homogeneous linear differential equation.
The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:
Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x0, y0), meaning that it satisfies the initial condition y(x0) = y0.
This theorem ensures that under these conditions, the solution exists and is unique near x = x0.
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) | $\mathcal{L}(u(t)) = \frac{1}{s}, s > 0$ |
| $e^{at}$ | $\frac{1}{s - a}, s > a$ |
| $e^{(a + bi)t}$ | $\mathcal{L}(e^{(a + bi)t}) = \frac{1}{s - (a + bi)}, s > a$ |
| $\cos(\omega t)$ | $\mathcal{L}(\cos(\omega t)) = \frac{s}{s^2 + \omega^2}, s > 0$ |
| $\sin(\omega t)$ | $\mathcal{L}(\sin(\omega t)) = \frac{\omega}{s^2 + \omega^2}, s > 0$ |
| $t^n$ | $\mathcal{L}(t^n) = \frac{n!}{s^{n+1}}, s > 0$ |
| $u(t-a)$ | $\mathcal{L}(u(t-a)) = \frac{e^{-as}}{s}, s > 0$ |
| $\delta(t-a)$ | $\mathcal{L}(\delta(t-a)) = e^{-as}, a \geq 0$ |
| $\frac{1}{t}$ | $\mathcal{L}\left(\frac{1}{t}\right) = \text{not defined}$ |
| $e^{-bt} \cos(\omega t)$ | $\mathcal{L}(e^{-bt} \cos(\omega t)) = \frac{s + b}{(s + b)^2 + \omega^2}, s > -b$ |
| $e^{-bt} \sin(\omega t)$ | $\mathcal{L}(e^{-bt} \sin(\omega t)) = \frac{\omega}{(s + b)^2 + \omega^2}, s > -b$ |
| $e^{at}f(t)$ | $\mathcal{L}(e^{at}f(t)) = F(s-a)$ |
The last entry illustrates the Exponential Shift Theorem, indicating that multiplying a function by an exponential term eat shifts its Laplace Transform.
Additionally, the Laplace Transforms of derivatives are given by: $\mathcal{L}(f’(t)) = sF(s)-f(0), \mathcal{L}(f’’(t)) = s^2F(s)-sf(0)-f’(0)$
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, a collision, or any scenario where a large force is exerted briefly. Understanding how to model and analyze these impulses is crucial for accurately describing the behavior of physical systems under sudden forces.
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. Impulse has units of force multiplied by time (e.g., Newton-seconds).
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. It is simply the area under the force-time graph between t = a and t = b.
To model an impulse applied to a system, we can represent the force as a box function. This function is zero everywhere except over a small interval where it has a constant value.
Consider an undamped mass-spring system with mass m. When an impulse is applied, such as a sudden push or hit, 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 which is the product of the height and the width h) 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 u0h(t), which is 1 in the interval [0, h] and 0 elsewhere.
Impulse = $\int_{a}^{b} f(t)dt = \int_{0}^{h} \frac{1}{h}[u(t) -u(t-h)]dt = \int_{0}^{h} \frac{1}{h}dt = 1$.
The equation of motion without external forces is: my’’ + ky = 0, where y(t) is the displacement of the mass from its equilibrium position; m is the mass; and k is the spring constant. For simplicity, we can set m = 1 and k = 1, leading to: y’’ + y = 0.
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:
To solve the differential equation, we apply the Laplace Transform, which converts differential equations into algebraic equations in the s-domain.
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:
$\frac{1}{h}u_{0h}(t) = \frac{1}{h}[u(t)-u(t-h)] \leadsto \frac{1}{h}[\frac{1}{s}-\frac{e^{-hs}}{s}] = \frac{1-e^{-hs}}{hs}$
$\frac{1}{h}u_{0h}(t) \leadsto \frac{1-e^{-hs}}{hs}, \mathcal{L}(\frac{1}{h}[u(t)-u(t-h)]) = \frac{1-e^{-hs}}{hs}$
This expression represents the Laplace Transform of the force f(t).
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 mathematical construct that models an impulse occurring at a single point in time.
It is 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) = $\int_{-∞}^{∞} δ(t)f(t)dt = f(0)$ (🚀)
It can be thought of as an infinitely high, infinitely narrow spike at t = 0 with area 1.
The most important property of the Dirac delta function is the sifting property, which states that for any function g(t) that is continuos at t = a: $\int_{-∞}^{∞} δ(t-a)g(t)dt = g(a)$. In other words, the delta function centered a t = a “picks out” the value of g(t) at t = a.
The Dirac delta function centered at t = a can be informally described by: $δ_a = δ(t -a) = \begin{cases} ∞, &t = a \\ 0, &t ≠ a \end{cases}$ with the property that its integral over any interval containing t = a is 1.
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).
One way to understand the delta function is by considering it as the limit of a sequence of functions that become infinitely peaked at a point. For example, consider the rectangular pulse approximation of the delta function centered at t = a:
δa, 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 $\frac{1}{ε}$
$δ_{a, ε} = \begin{cases} \frac{1}{ε}, &t ∈ [a, a + ε] \\ 0, &otherwise \end{cases}$ where ε > 0 is a small number.
Let’s calculate: $\int_{-∞}^{∞} δ_{a, ε}(t)g(t)dt = \int_{a}^{a+ ε} \frac{1}{ε}g(t)dt =$
By the Mean Value Theorem for Integrals, since g(t) is continuous on the interval [a, a+ε], there exists a point c ∈ [a, a+ε] such that: $\int_{a}^{a+ ε} g(t)dt = ε·g(c)$. Therefore, the integral becomes $\int_{∞}^{-∞} δ_{a, ε}(t)g(t)dt = \int_{a}^{a+ ε} \frac{1}{ε}g(t)dt =ε·\frac{1}{ε}g(c) = g(c)$.
Taking the Limit as ε → 0, the interval [a, a+ε] shrinks to the point t = a, and since g(t) is continuous at t = a, g(c) → g(a), hence $\lim_{ε \to 0} \int_{∞}^{-∞} δ_{a, ε}(t)g(t)dt → g(a)$.
Therefore, the sifting property holds: $\int_{∞}^{-∞} δ_a(t)g(t)dt = g(a)$
Special Case: a = 0, the Dirac delta function becomes δ(t), centered at the origin. Thus, for any continuous function f(t): $\int_{-∞}^{∞} δ(t)f(t)dt = f(0)$
Since δ(t) is zero everywhere except at t = 0, the integral effectively picks out or samples the value of f(t) at t = 0.
An important property of the delta function is its relationship with the unit step function u(t). The derivative of the unit step function is the delta function: $\frac{d}{dt}u(t-a)=δ_a(t)$
This reflects the fact that u(t−a) jumps from 0 to 1 at t = a, and the derivative captures this instantaneous change as the Dirac delta function at t = a.
The Laplace Transform of the Dirac Delta Function is: $\mathcal{L}(δ(t)) = \int_{0}^{∞} e^{-st}δ(t)dt = e^{-s·0} = 1, \mathcal{L}(δ(t-a)) = \int_{0}^{∞} e^{-st}δ(t-a)dt = e^{-s·a}$. The delta function picks out the value of f(t) = e-st at t = a and t = 0 respectively 🚀.
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 or a, 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: displacement y(0) = 1, velocity 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
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. $sin(t-\frac{π}{2}) = -cos(t)$
$sin(t-\frac{π}{2}) = sin(t)cos(\frac{π}{2}) -cos(t)sin(\frac{π}{2}) = sin(t)·0 -cos(t)·1 = -cos(t)$
$y = \begin{cases} cos(t), &0 ≤ t ≤ π/2 \\ cost(t) -Acos(t), &t ≥ π/2 \end{cases}$
$y = \begin{cases} cos(t), &0 ≤ t ≤ π/2 \\ (1 - A)cos(t), &t ≥ π/2 \end{cases}$
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:
The system is subjected to impulses (e.g. a hammer) at times t = nπ for n = 0, 1, 2,…. These impulses are 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: $y'' + y = \sum_{n=0}^\infty δ(t-nπ)$ with 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 $\mathcal{L}(y’’)+\mathcal{L}(y) = \mathcal{L}(\sum_{n=0}^\infty δ(t-nπ)) ↭ s^2Y +Y = \sum_{n=0}^\infty e^{-nπs} ↭ (s^2+1)Y = \sum_{n=0}^\infty e^{-nπs}$. 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) = $\frac{1}{s^2+1}, \mathcal{L}^{-1}(\frac{1}{s^2+1}) = f(t) = sin(t)$.
Applying the theorem to each term in Y(s): $Y(t) = \sum_{n=0}^\infty u(t -nπ)sin(t -nπ)$
Recall sin(t -nπ) = (-1)nsin(t) and observe for any arbitrary t: nπ < t < (n+1)π, u(t -nπ) = 0 for t < nπ.
sin(t -nπ) = (-1)nsin(t) because sin(t−π) = −sin(t), sin(t−2π)=sin(t), aºnd so on.
$Y(t) = sin(t)\sum_{n=0}^\infty (-1)^n·u(t -nπ)$
Y(t) = sin(t) -sin(t) + ··· + (-1)nsin(t)
$Y(t) = \begin{cases} sin(t), &n,even \\ 0, &n,odd \end{cases}$
For t in [0, π): The system oscillates naturally with displacement y(t)=sin(t). This is due to the initial impulse at t = 0.
At t=π: An impulse is applied that stops the oscillation.
For t in [π,2π): The displacement is zero (y(t)=0); the system is at rest.
At t=2π: Another impulse is applied that restarts the oscillation.
This pattern repeats every π units of time.
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:
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 characterizes 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).
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: $s^2Y + asY + bY = 1 ⇒[\text{Solving for Y}] Y(s) = \frac{1}{s^2+as+b} = W(s)$, and taking its inverse Laplace transform: $y(t) = δ(t)*W(t) =[\text{The delta function acts as the identity under convolution}] W(t)$. 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).
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