I have yet to see any problem, however complicated, which, when you looked at it in the right way, did not become still more complicated, Poul Anderson.
An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:
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., $\frac{dy}{dx} = 3x +5y$):
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) .
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:
This equation is homogeneous, meaning that there are no external forcing terms (like a function of t) on the right-hand side.
A power series is a type of infinite series where each term is a power of a variable, typically written in the form: $\sum_{n=0}^\infty a_nx^n = A(x)$.
In this notation, the an’s represent the coefficients of the power series, and A(x) is the resulting function that the series converges to.
We can rewrite the series by treating the coefficients an as the values of a discrete function, say a(n). This gives us a new way to represent the power series as a discrete sum: $\sum_{n=0}^\infty a(n)x^n = A(x)$.
Here, a(n) is a function defined on the discrete set of non-negative integers. Now, the key idea is that we can relate this discrete function a(n) to a real, continuous function A(x), through known series expansions, for example:
These examples illustrate that discrete functions a(n) can correspond to well-known continuous functions A(x).
The Laplace Transform is based on a similar concept but extends this relationship to integrals instead of sums. The continuous analog of the power series is an integral involving a continuous variable t, rather than a discrete sum over n. The goal is to construct an integral of the form: $\int_{0}^{∞} a(t)x^tdt = A(x)$ where a(t) is a continuous function of t.
However, integrals involving terms like xt can be challenging to compute directly because they are not always convenient to work with. To simplify this expression, we make a transformation that will result in a more manageable form. Let’s proceed step-by-step.
We begin by rewriting xt in terms of the exponential function: x = eln(x) ⇒ $x^t = (e^{ln(x)})^t= [\text{Exponents Laws}] e^{t·ln(x)}$.
To make the resulting expression easier to handle, we now make a substitution for x. Assume that 0 < x <1, which implies that ln(x) < 0. Let: s = -ln(x). Since ln(x) is negative for 0 < x <1, this substitution guarantees that s > 0. Now, we have: $x^t = e^{t·ln(x)} = e^{-st}$ where the new variable s replaces the logarithm.
For clarity (aka cosmetic purposes or consistency with standard notation), we rename the function a(t) as f(t), a more common notation when discussing transforms. Thus, our integral now becomes: $\int_{0}^{∞} f(t)e^{-st}dt = F(s)$. This is the Laplace Transform. In this transformation, the input is a function of t (i.e., f(t)), and the output is a function of s (i.e., F(s)).
We denote the Laplace Transform of f(t) as: $\mathcal{L}(f(t)) = F(s)$ or in an alternative notation, simply: $f(t) \leadsto F(s)$
It’s important to recognize that a transform takes a function of one variable (in this case, t) and transforms it into a function of a different variable (in this case, s). This is different from an operator, which typically maps a function of a given variable into another function of the same variable. For example, a differentiation operator acts on f(t) and produces another function of t, such as f’(t).
In contrast, the Laplace Transform maps a function of t into a new function of a different variable s.
One of the most useful properties of the Laplace Transform is its linearity. If we have two functions f(t) and g(t), then: $\mathcal{L}(f+g) = \mathcal{L}(f)+\mathcal{L}(g), \mathcal{L}(af+bg) = a\mathcal{L}(f)+b\mathcal{L}(g)$ for any constants a and b. In other words, the Laplace Transform of a sum is the sum of the individual Laplace Transforms.
$\mathcal{L}(af+bg)(t) = \mathcal{L}(af(t) +bg(t)) = \int_{0}^{∞} e^{-st}[af(t) + bg(t)]dt = a\int_{0}^{∞} e^{-st}f(t)dt + b\int_{0}^{∞} e^{-st}g(t)dt = a\mathcal{L}(f(t))+b\mathcal{L}(g(t))$
The Laplace Transform is a powerful tool in mathematics and engineering, particularly for solving differential equations.
By definition, the Laplace Transform is given by: $\mathcal{L}(1) = \int_{0}^{∞} 1e^{-st}dt = \int_{0}^{∞} e^{-st}dt =[\text{To evaluate this improper integral, we consider the limit as the upper bound approaches infinity:}] \lim_{R \to ∞}\int_{0}^{R} e^{-st}dt$
Compute the definite integral over the finite interval [0, R]: $\int_{0}^{R} e^{-st}dt = \frac{e^{-st}}{-s}\bigg|_{0}^{R} = \frac{e^{-sR}-1}{-s}$
$\int_{0}^{∞} e^{-st}dt = \lim_{R \to ∞}\int_{0}^{R} e^{-st}dt = \lim_{R \to ∞} \frac{e^{-sR}-1}{-s} =[\text{Since s > 0, as R → ∞, } e^{-sR}→ 0] \frac{1}{s}$, this is true only for s > 0.
$\mathcal{L}(1) = \frac{1}{s}$ or $1 \leadsto \frac{1}{s}$. This means that the function f(t) = 1 maps to F(s) = 1⁄s under the Laplace Transform.
First, we are going to calculate the Laplace Transform of eatf(t), where f(t) is a function whose Laplace Transform is F(s).
By definition, the Laplace Transform of eatf(t) is: $\mathcal{L}(e^{at}f(t)) = \int_{0}^{∞} e^{at}f(t)e^{-st}dt =[\text{Simplify the exponentials:}] \int_{0}^{∞} e^{-(s-a)t}f(t)dt = F(s-a)$ provided that s-a > 0, which implies s > a.
$e^{at}f(t) \leadsto F(s-a)$ for s > a, assuming $\int_{0}^{∞} f(t)e^{-st}dt = F(s)$. This result is known as the exponential shift formula. This formula shows that multiplying f(t) by an exponential term eat shift the Laplace Transform F(s) to F(s -a).
Consider the specific case where f(t) = 1. Then, $e^{at}·1 = e^{at}\leadsto F(s−a) = \frac{1}{s-a}$ for s > a, since $1 \leadsto \frac{1}{s}$.
The unit step function u(t) is defined as: u(t) = $\begin{cases} 0, t < 0 \\ 1, t ≥ 0 \end{cases}$
The shifted unit step function u(t−a) is defined as: u(t-a) = $\begin{cases} 0, t < a \\ 1, t ≥ a \end{cases}$
$\mathcal{L}(u(t-a)) = \int_{0}^{∞} e^{-st}u(t-a)dt =[\text{Since u(t−a) = 0 for t < a, we can change the limits of integration}] \int_{a}^{∞} e^{-st}·1dt = \frac{e^{-st}}{-s}\bigg|_{a}^{∞} = 0-\frac{e^{-sa}}{-s} = \frac{e^{-sa}}{s}$
The window function $\Pi_{a, b}(t)$, also known as the rectangular function or boxcar function is defined as: $\Pi_{a, b}(t) = \begin{cases} 1, &a < t < b \\ 0, &otherwise \end{cases} =[\text{It can be expressed as the difference of two unit step functions}] u(t-a)-u(t-b)$ where u(t) is the unit step function (also known as the Heaviside step function).
$\mathcal{L}(\Pi_{a, b}(t)) = \mathcal{L}(u(t-a)-u(t-b)) =[\text{Using the linearity property of the Laplace transform}] \mathcal{L}(u(t-a)) -\mathcal{L}(u(t-b)) = \frac{e^{-sa}}{s}-\frac{e^{-sb}}{s} = \frac{e^{-sa}-e^{-sb}}{s}$. This result is valid for $s > 0$, which ensures the convergence of the Laplace transform.
The Dirac delta function δ(t) is a distribution that is defined as: δ(t) = $\begin{cases} 0, t ≠ a \\ ∞, t = 0 \end{cases}$
However, it is not a function in the traditional sense but rather a distribution that satisfies the following properties:
To find the Laplace transform of δ(t−a)δ(t−a), we compute: $\mathcal{L}(δ(t-a)) = \int_{0}^{∞} e^{-st}δ(t-a)dt =[\text{Since the delta function is zero everywhere except at t=a, we can also express this integral over the full range:}] \int_{-∞}^{∞} e^{-st}δ(t-a)dt =[\text{Applying the Sifting Property}] e^{-st}\bigg|_{a} = e^{-sa}$
$\mathcal{L}(f(t-a)u(t-a)) = \int_{0}^{∞} e^{-st}f(t-a)u(t-a)dt =[\text{The integral starts from a because u(t-a) = 0 for t < a.}] \int_{a}^{∞} e^{-st}f(t-a)dt = $
Substitution of variables v = t -a, dv = dt, t = a ⇒v = 0, t = ∞ ⇒v = ∞
$= \int_{0}^{∞} e^{-s(v+a)}f(v)dv = \int_{0}^{∞} e^{-sv}e^{-sa}f(v)dv = e^{-sa}\int_{0}^{∞} e^{-sv}f(v)dv = e^{-sa}\mathcal{L}(f(t)) = e^{-sa}F(s)$ where F(s) is the Laplace transform of f(t).
$\mathcal{L}(f(t)u(t-a)) =$[The trick here is to let g(t-a) = f(t), which allows us to use the result from the first part.] =$\mathcal{L}(g(t-a)u(t-a)) = e^{-sa}\mathcal{L}(g(t))$= [g(t) = g((t+a)-a) = f(t+a)] $e^{-sa}\mathcal{L}(f(t+a))$
Recall the previous result $\mathcal{L}(f(t-a)u(t-a)) = e^{-sa}F(s)$ where F(s) is the Laplace transform of f(t).
$\mathcal{L}(f(t-2)^2u(t-2)) = e^{-s·2}\mathcal{L}(t^2) =[\mathcal{L}(t^n) = \frac{n!}{s^{n+1}}] e^{-2s}\frac{2}{s^3} = \frac{2e^{-2s}}{s^3}$
$\mathcal{L}(t^2u(t-2))$ = [Recall the previous result $\mathcal{L}(f(t)u(t-a)) = e^{-sa}\mathcal{L}(f(t+a))$] $e^{-s·2}\mathcal{L}((t+2)^2) = e^{-2s}\mathcal{L}(t^2+4t+4) =[\text{Using linearity and } \mathcal{L}(t^n) = \frac{n!}{s^{n+1}}] e^{-2s}(\frac{2}{s^3}+\frac{4}{s^2}+\frac{4}{s})$
We can extend the exponential shift formula to complex exponents.
Let b ∈ ℝ, consider $e^{(a+bi)t}, \mathcal{L}(e^{(a+bi)t}) = \int_{0}^{∞} e^{(a+bi)t}e^{-st}dt =[\text{Combine the exponents:}] \int_{0}^{∞} e^{-(s-a-ib)t}dt$
The integrand is now in the form e-ct where c = s -a -ib. Evaluate the integral:
$\int_{0}^{∞} e^{-ct}dt = -\frac{1}{c}e^{-ct}\bigg|_{0}^{∞}$ =
$-\frac{1}{s -a -ib}e^{-(s -a -ib)t}\bigg|_{0}^{∞}$
Apply the limits: As t → ∞, $e^{-(s -a -ib)t} = e^{-(s-a)t} \cdot e^{ibt} → 0$ 💡 assuming s > a. t = 0, $e^{-(s -a -ib)·0} = 1.$
💡 The $e^{ibt}$ part is just oscillating between -1 and 1 (it’s cos(bt) + i·sin(bt)), so it doesn’t affect whether the overall expression approaches zero or not. The key part is $e^{-(s-a)t}$. For this to approach zero as t approaches infinity, we need the exponent to be negative, i.e., -(s-a) < 0 ↭ s > a
$\mathcal{L}(e^{(a+bi)t}) = -\frac{1}{s -a -ib}·0 +\frac{1}{s -a -ib}·1 = \frac{1}{s -a -ib}$
$\mathcal{L}(e^{(a+bi)t}) = \frac{1}{s-(a+ib)}$ valid for s > a, or alternatively, $e^{(a+bi)t} \leadsto \frac{1}{s-(a+bi)}$.
Thus, the complex exponential function e(a + bi)t transform into $\frac{1}{s-(a+bi)}$, demonstrating a direct application of the exponential shift formula with a complex shift.
Euler’s formula states that: $e^{iat} = cos(at) + isin(at), e^{-iat} = cos(at) - isin(at)$
From these, we can express the trigonometric functions in terms of exponentials: $cos(at) = \frac{e^{iat}+e^{-iat}}{2}, sin(at) = \frac{e^{iat}-e^{-iat}}{2i}$
$cos(at) =[\text{Euler’s formula}] \frac{e^{iat}+e^{-iat}}{2}, \mathcal{L}(cos(at)) =[\text{By linearity of the Laplace Transform}, e^{(a+bi)t} \leadsto \frac{1}{s-(a+bi)}] \frac{1}{2}(\frac{1}{s-ia}+\frac{1}{s+ia}) =$
If we have a complex expression and replacing i with -i doesn’t change the result, it means the expression is real, that’s the case with the previous formula.
$\frac{1}{2}·\frac{(s+ia)+(s-ia)}{(s-ia)(s+ia)} = \frac{1}{2}\frac{s+ia+s-ia}{s^2+a^2} = \frac{1}{2}\frac{2s}{s^2+a^2} = \frac{s}{s^2+a^2}, cos(at) \leadsto \frac{s}{s^2+a^2}$ where s > 0.
$sin(at) =[\text{Euler’s formula}] \frac{e^{iat}-e^{-iat}}{2i}, \mathcal{L}(sin(at)) =[\text{By linearity of the Laplace Transform}, e^{(a+bi)t} \leadsto \frac{1}{s-(a+bi)}] \frac{1}{2i}(\frac{1}{s-ia}-\frac{1}{s+ia}) = \frac{1}{2i}\frac{(s+ia)-(s-ia)}{s^2+a^2} = \frac{1}{2i}\frac{2ia}{s^2+a^2} = \frac{a}{s^2+a^2}$. Thus, the Laplace Transform of sin(at) is: $sin(at) \leadsto \frac{a}{s^2+a^2}$. This result is also valid for s > 0.