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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: $\mathcal{L}(f’(t)) = sF(s)-f(0), \mathcal{L}(f’’(t)) = s^2F(s)-sf(0)-f’(0)$
Convolution is an important operation that combines two functions to produce a third function. The convolution of two functions f(t) and g(t) is denoted by (f*g)(t) and is defined as: (f*g)(t) = $\int_{0}^{t} f(u)g(t-u)du$. It is widely used in various fields such as engineering, physics, signal processing, and applied mathematics to analyze systems and solve differential equations.
The variable u is a dummy variable of integration that runs from 0 to t. f(u) represents the input (original) function evaluated at time u. g(t−u) represents the other function shifted by t−u. The convolution integral sums the product f(u)·g(t−u) over the interval from u = 0 to u = t.
Given two functions f(t) and g(t), their Laplace Transforms are: $F(s) = \mathcal{L}(f(t)) = \int_{0}^{∞} e^{-st}f(t)dt, G(s) = \mathcal{L}(g(t)) = \int_{0}^{∞} e^{-st}g(t)dt$.
A common question arises: Is there a direct formula for the Laplace Transform of the product f(t)·g(t) in terms of their individual Laplace Transforms? The answer is: that there is no direct formula for the Laplace Transform of the product of two functions f(t)g(t) solely in terms of their individual Laplace transforms F(s) and G(s).
However, we can find the Laplace transform of their convolution (f * g)(t) using a special rule.
The Convolution Theorem states that the Laplace Transform of the convolution of two functions f(t) and g(t), denoted (f*g)(t), is equal to the product of their individual Laplace transforms. Mathematically, $\mathcal{L}((f*g)(t)) = F(s)G(s) ↭ F(s)G(s) = \int_{0}^{∞} e^{-st}(f*g)(t)dt$ where F(s) is the Laplace transform of f(t) and G(s) is the Laplace transform of g(t).
The convolution operation in the time domain corresponds to multiplication in the Laplace (frequency) domain. This theorem provides a powerful method for solving differential equations and analyzing systems because it allows us to work with multiplication in the Laplace (frequency) domain instead of convolution in the time domain.
Proof.
We will prove that: $F(s)G(s) = \int_{0}^{∞} e^{-st}(f * g)(t)dt = \mathcal{L}((f * g)(t))$
We begin with the Laplace Transforms of the two functions f(t) and g(t): $F(s) = \int_{0}^{∞} e^{-su}f(u)du$ and $G(s) = \int_{0}^{∞} e^{-sv}g(v)dv$.
Multiplying these two transforms yields:
$F(s)G(s) = [\text{By definition}] \int_{0}^{∞} e^{-su}f(u)du·\int_{0}^{∞} e^{-sv}g(v)dv = \int_{0}^{∞}\int_{0}^{∞}e^{-s(u+v)}f(u)g(v)dudv$
Next, we perform a change of variables to simplify the double integral. Let t = u + v (⇒ v = t -u), and u = u (remains the same).
du·dv = $\frac{∂(u, v)}{∂(u, t)}du·dt = \vert \begin{smallmatrix}\frac{∂u}{∂u} & \frac{∂u}{∂t}\\ \frac{∂v}{∂u} & \frac{∂v}{∂t}\end{smallmatrix} \vert = \vert \begin{smallmatrix}1 & 0\\ -1 & 1\end{smallmatrix} \vert du·dt = 1·du·dt$. Hence, the differentials remain dudt because the Jacobian determinant of the transformation is 1.
$\int_{0}^{∞}\int_{0}^{∞}e^{-s(u+v)}f(u)g(v)dudv = \int_{t=0}^{∞}\int_{u=0}^{t}e^{-st}f(u)g(t-u)dudt = \int_{0}^{∞}e^{-st}\int_{0}^{t} f(u)g(t-u)dudt = \int_{0}^{∞}e^{-st} (f*g)(t)$. Therefore, $F(s)G(s) = \int_{t=0}^{∞} e^{-st}(f * g)(t)dt = \mathcal{L}((f * g)(t))$.
Refer to Figure ii for a visual representation and aid in understanding it. t is the outer variable running from 0 to ∞, u is the variable inside the convolution integral and for each t, u ranges from 0 to t.
This property is extremely useful because it allows us to simplify complex operations by working in the frequency domain.
This means that instead of directly computing the convolution integral in the time domain, we can compute the Laplace transforms of the two functions separately, multiply them, and then take the inverse Laplace Transform of the product to obtain to obtain (f*g)(t).
Alternative Method: Using the Laplace Transform.
Convolution with the constant function 1 results in the cumulative integral (accumulation) of f(t) up to time t. This operation is equivalent to finding the accumulated value or the definite integral of f(t) from u = 0 to u = t.
Convolution can be intuitively understood intuitively as a way of combining two processes: the input and the system’s response. It calculates the output of a system based on its input signal and its impulse response.
In this case, the input is the dumping of radioactive waste, and the system’s response is the decay of that waste over time.
Imagine you’re continuously dumping radioactive waste onto a pile over time. The dumping rate at time t, measured in years, is represented by the function f(t).
To determine the total amount of radioactive waste remaining at time t, we need to account for:
Total Waste Remaining at Time t. Now, imagine you are continuously dumping radioactive waste starting from time t = 0. At any later time t, the total amount of radioactive waste left on the pile will depend on two factors = [How much waste was dumped at each earlier time]·[How much has decayed since that earlier time].
For each small time interval [ui, ui+1], the amount of waste dumped is f(ui)Δui, and the amount of radioactive waste left on the pile at time t (after decay) is $f(u_i)Δu_ie^{-k(t-u)}$. The total amount of radioactive waste left at time t can be approximated as: $\sum_{i=1}^n f(u_i)Δu_ie^{-k(t-u)}$.
As Δui → 0 (to do this approximation more accurate), this sum turns into an integral, giving us the exact total amount of radioactive waste at time t: $\int_{0}^{t} f(u)e^{-k(t-u)}du = f(t) * e^{-kt}$.
This integral is the convolution of the dumping rate function f(t) with the decay function $e^{-kt}$. It accounts for the accumulation and decay of the waste over time, combining the input and the system’s response.
Special Cases of Convolution:
The total mass of chickens produced at time t is a function of the production rate and growth of production over time. Each chick produced at time u grows for a duration of t−u until time t. If growth is linear, the weight of each chick at time t is proportional to t− u.
If the growth is linear, the total mass of chickens at time t would be: f(t) * g(t) = (f * g)(t) where f(t) * g(t) represents the convolution of the production rate f(t) (#kg of chickens at time t) and g(t) represents the growth function over time (e.g., t). Total Mass at time t = (f * g)(t) = $\int_{0}^{t} f(u)g(t-u)du = \int_{0}^{t} f(u)(t-u)du$. It accounts for how much each earlier production contributes to the total at the present time. In other words, convolution sums up how past production rates contribute to the total at a given time,accounting for the growth of each chick since its birth.
In many applications, we encounter functions that have sudden changes, known as jump discontinuities. A jump discontinuity occurs when a function abruptly jumps from one value to another at a certain point. A classic example of a function that exhibits a jump discontinuity is the Heaviside step function (also called the unit step function).
The unit step function, denoted as u(t), is a step function, the value of which is zero for negative arguments, one for positive arguments, and u(0) is sometimes defined as 0, 1, or left undefined. The exact value at t = 0 often does not affect practical applications. Refer to Figure i for a visual representation and aid in understanding it. Mathematically, it is defined as: $u(t) = \begin{cases} 0, &t < 0 \\ undefined~ or~ 0, &t = 0 \\ 1, &t > 0 \end{cases}$
The Heaviside step function is a function that “jumps” from 0 to 1 at t = 0. There is a bit of controversy or ambiguity regarding the value of u(0). Some definitions leave u(0) undefined, while others define it as either 0, 1⁄2 or 1. This ambiguity usually doesn’t matter for practical purposes, it is often inconsequential because the Laplace Transform and most physical applications focus on t > 0.
By shifting the Heaviside Step Function, we can control the point at which the jump occurs.
Let’s shift the unit step by a constant a. This creates a new function, ua(t), which “activates” at t = a. The translated or shifted Heaviside step function is defined as: ua(t) = $u(t -a) = \begin{cases} 0, &t < a \\ 1, &t ≥ a \end{cases}$
This shift allows us to control when the jump occurs. Instead of jumping from 0 to 1 at t = 0, the function activates at t = a. This shift allows us to model inputs or forces that start at a specific time t = a.
We can use two translated Heaviside functions to create a box-shaped function that is 1 between two points a and b, and 0 outside this interval. This is called the unit box function and is denoted by uab(t). It “turns on” 🔦 at t = a (u(t -a)) and “turns off” at t = b (u(t -b)). The difference creates a “window” where the function is active between t = a and t = b.
It is defined as:
$u_{ab}(t) = \begin{cases} 0, &t < a \\ 1, &a ≤ t < b \\ 0, &t ≥ b \end{cases}$
When we multiply a function f(t) by ua(t), we effectively “turn on” the function at t = a and “turn it off” at t = b.
$f(t) = \begin{cases} 0, &t < a \\ f(t), &a ≤ t < b\\ 0, &t ≥ b \end{cases}$
This is useful for modeling systems where an input or force is applied starting at time t = a.
When using the Unit Box Function uab(t), we “window” the function f(t) between t = a and t = b. In other words, we can express uab(t) using the difference of two Heaviside step functions. $u_{ab}(t) = u_a(t)-u_b(t) = u(t-a)-u(t-b)$. The main idea is that ua(t) turns on the function at t = a and ub(t) turns it off at t = b, creating a window between a and b where the function is 1.
Consider multiplying the unit box function uab(t) by another function f(t), i.e., uab(t)f(t). This product has the effect of “windowing” the function f(t), meaning that:
$f(t)u_{ab}(t) = \begin{cases} 0, &t < a \\ f(t), &a ≤ t < b \\ 0, &t ≥ b \end{cases}$
Let’s calculate the Laplace transform of the unit step function u(t): $\mathcal{L}(u(t)) = \int_{0}^{∞} e^{-st}u(t)dt =[\text{Since u(t) = 1 for positive values of t}] = \int_{0}^{∞} e^{-st}dt = \frac{1}{s}$, s > 0
Thus, the Laplace transform of the Heaviside step function is 1⁄s,which is consistent with the Laplace Transform of the constant function 1 for s > 0.
If we know that $\mathcal{L}(1) = \mathcal{L}(u(t)) = \frac{1}{s}$, s > 0, the question becomes: What is the inverse Laplace transform of 1⁄s, $\mathcal{L}^{-1}(\frac{1}{s})$?
Suppose that the Laplace Transform of f is F, $f(t)\leadsto_F F(s)$, the inverse Laplace transform of F(s) is f(t), but any function that behaves like f(t) for positive arguments (and possibly has different behavior for negative arguments) will also work. This is because the Laplace transform is only concerned with what happens for t ≥ 0.
To ensure uniqueness, we usually agree that f(t) is 0 for all t < 0. This is equivalent to multiplying f(t) by the Heaviside step function u(t), which forces f(t) to be 0 for negative arguments: $F(s) \leadsto_{L^{-1}} u(t)f(t)$ ↭ $\mathcal{L}^{-1}(F(s)) = f(t)u(t)$. By applying the Heaviside step function, we effectively cut off any behavior of f(t) for t < 0 (its tail) so this convention ensures that f(t) is defined only for non-negative time and the Laplace Transform and its inverse are well-defined.
There is no a direct formula for $\mathcal{L}(f(t-a))$ in term of $\mathcal{L}(f(t))$ without additional considerations. The problem is basically that the Laplace of f does not care about the function’s behaviour for negative t because it is only defined for t ≥ 0 (it loses all information about f in (-a, 0)), and shifting the function f(t) to f(t-a) involves values of f(t) for t < a and this information is needed to calculate $\mathcal{L}(f(t-a))$ (Refer to Figure ii for a visual representation and aid in understanding it).
We can address this issue by wiping out this area. When we shift the function, we effectively change when the function starts. If the original function starts at t = 0, the shifted function starts at t = a. The Heaviside function u(t−a) takes care of this, making the function zero for t < a and nonzero for t ≥ a. In other words, we can get a formula from the function u(t-a)·f(t-a).
In the context of the Laplace Transform, the T-axis translation formula deals with shifting a function in time. When you shift a function f(t) by a constant a (i.e., replace t with t −a), the Laplace Transform of the shifted function is related to the Laplace Transform of the original function through multiplication by an exponential factor. More formally, $u(t-a)f(t-a) \leadsto e^{-as}F(s)↭ \mathcal{L}(u(t-a)f(t-a)) = e^{-as}F(s)$ or equivalently $u(t-a)f(t) [\text{Observe t = (t +a) -a}] \leadsto e^{-as} \mathcal{L}(f(t+a)) = e^{-as}F(s+a)$ where:
Proof
We aim to prove that $u(t-a)f(t-a) \leadsto e^{-as}F(s)$.
The Laplace Transform of u(t-a)f(t-a) is defined as:
$\int_{0}^{∞} e^{-st}u(t-a)f(t-a)dt =[\text{Since u(t−a) = 0 for t < a, the integral from 0 to a is zero. Therefore, we can adjust the limits of integration:}] \int_{a}^{∞} e^{-st}f(t-a)dt$
We make a substitution (change of variables) to simplify the integral. Let τ = t - a. This implies that when t = a, τ = 0. When t → ∞, τ → ∞, and the differential remain the same, i.e., dt = dτ.
$\int_{a}^{∞} e^{-st}f(t-a)dt = \int_{0}^{∞} e^{-s(τ+a)}f(τ)dτ =[\text{Simplify the Exponential Term:}] e^{-sa}\int_{0}^{∞} e^{-sτ}f(τ)dτ =[\text{The remaining integral is exactly the definition of the Laplace Transform of f(t)}] e^{-sa}F(s)$
$u(t-a)f(t-a) \leadsto e^{-as}\mathcal{L}(f(t)) = e^{-as}F(s) ↭ u(t-a)f(t-a) \leadsto e^{-as}\mathcal{L}(f(t))$
Similarly, $u(t-a)f(t+a-a) \leadsto e^{-as}\mathcal{L}(f(t+a))↭ u(t-a)f(t) \leadsto e^{-as}\mathcal{L}(f(t+a))$
The unit box function uab(t) is a piecewise function, which is 1 in the interval [a, b] and 0 otherwise. uab(t) = u(t -a) - u(t -b) where u(t) is the Heaviside Step Function, and a and b are constants with a < b.
$u_{a, b}(t) = \begin{cases} 0, &t < a \\ 1, &a ≤ t < b \\ 0, &t ≥ b \end{cases}$
Recall that $u(t) \leadsto \frac{1}{s}, u(t-a)f(t-a) \leadsto e^{-as}F(s), u(t -a)·u(t -a) = u(t -a)\leadsto e^{-as}F(s) = \frac{e^{-as}}{s}$
Using the Linearity of the Laplace Transform: $u_{a, b}(t) = u(t -a) - u(t -b) \leadsto \frac{e^{-as}}{s}-\frac{e^{-bs}}{s} = \frac{e^{-as}-e^{-bs}}{s}$
The Laplace Transform captures the effect of the unit box function being active between t = a and t = b. The exponential terms correspond to the activation (e-as) and deactivation (e-bs) times.
Compute the Laplace Transform of the function: $u(t-1)t^2$
Relevant Laplace Transform Properties:
Identify a = 1, f(t) = t2, $F(s+a)=\mathcal{L}(f(t+a)) = \mathcal{L}((t+1)^2)$
$u(t-1)t^2 \leadsto e^{-s}\mathcal{L}(t+1)^2 = e^{-s}\mathcal{L}(t^2+2t+1) = e^{-s}(\frac{2}{s^3}+ \frac{2}{s^2}+\frac{1}{s})$
Recall: $\mathcal{L}^{-1}(F(s)·G(s)) = f(t)*g(t), \mathcal{L}(sin(t)) = \frac{1}{s^2+1}$
$\mathcal{L^{-1}}(\frac{1}{s^2+1}·\frac{1}{s^2+1}) = sin(t)*sin(t) = \int_{0}^{t} sin(τ)sin(t -τ)dτ =$
Using the property: $sin(a)sin(b) = \frac{1}{2}[cos(a-b)-cos(a+b)]$
$= \frac{1}{2}\int_{0}^{t} cos(2τ-t)-cos(t)dτ$
This can be separated into two integrals:
$\frac{1}{2}\int_{0}^{t} cos(2τ-t)dτ =[\text{Using the substitution u=2τ-t, du = 2dτ. When τ = 0, u = -t, When τ = t, u = t}] \frac{1}{2}\frac{1}{2}\int_{-t}^{t}cos(u)du = \frac{1}{4}sin(t)\bigg|_{-t}^{t} =$
$= \frac{1}{4}(sin(t)-sin(-t)) = \frac{1}{4}(sin(t)+sin(t)) = \frac{1}{2}sin(t)$
$ \frac{1}{2}\int_{0}^{t} -cos(t)dτ = -\frac{1}{2}cos(t)\int_{0}^{t}dτ = -\frac{1}{2}cos(t)t$
$\mathcal{L^{-1}}(\frac{1}{s^2+1}·\frac{1}{s^2+1}) = \frac{1}{2}sin(t) -\frac{1}{2}tcos(t)$
Relevant Laplace Transform Properties:
Step 1: Decompose F(s) We can break this into two simpler terms: $\frac{1}{s^2+1}+\frac{e^{-πs}}{s^2+1}$. Then, we apply the inverse Laplace transform of each term.
Step 2: Compute the Inverse Laplace Transform of Each Term
$\frac{1}{s^2+1} \leadsto_{\mathcal{L^{-1}}} u(t)sin(t)$, t ≥ 0 The reason that we need to add the step function is because of the presence of exponential in the original formula.
$\frac{e^{-πs}}{s^2+1} \leadsto_{\mathcal{L^{-1}}} u(t-π)sin(t-π)$, t ≥ π and we are using the First Shifting Theorem where a = π, f(t) = sin(t), and F(s) = $\frac{1}{s^2+1}$.
Step 3: Combine the Results
$\mathcal{L^{-1}}(\frac{1+e^{-πs}}{s^2+1}) = u(t)sin(t)+u(t-π)sin(t-π)$
Step 4: Simplify
$f(t) = \begin{cases} sin(t), &0 ≤ t < π \\ sin(t)+sin(t-π), &t ≥ π \end{cases}$
Since sin(t -π) = -sin(t), then sin(t) + sin(t-π) =[It simplifies to] sin(t) - sin(t) = 0.
$f(t) = \begin{cases} sin(t), &0 ≤ t < π \\ 0, &t ≥ π \end{cases}$
This is a piecewise function where the sine function is “turned off” at t = π due to the time shift. This reflects the effect of the exponential term e-πs.
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