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Do not worry about your difficulties in mathematics. I can assure you mine are still greater, Albert Einstein
All those who seem stupid, they are, and also so are half of those who do not, Quevedo
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.
A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0, where A and B are constants.
To solve this ODE, we seek two linearly independent solutions y1(t) and y2(t). The general solution is then a linear combination of these solutions: $c_1y_1 + c_2y_2$ where c1 and c2 are two arbitrary constants determined by initial conditions. The key to solving the ODE is the characteristic equation, whose roots determine the behavior of the solutions.
In this section, we introduce the concept of Fourier series, a powerful mathematical tool used to represent periodic functions as sums of sine and cosine functions. This is particularly useful when dealing with inhomogeneous linear differential equations of the form: y’’ + ay’ + by = f(t) where
Up to this point, we have often encountered input functions f(t) that are exponential, sine, or cosine functions. Although these functions might seem quite limited and special, they are in fact fundamental building blocks because they allow us to build more complex periodic functions through superposition.
Due to the linear nature of the differential equation of the form y’’ + ay’ + by = f(t), we can decompose more complicated inputs into sums of sines and cosines, solve for their responses individually, and then combine the results to obtain the overall solution. This principle is known as superposition and is a cornerstone in solving linear differential equations.
A Fourier series allows us to expand any periodic function f(t) (with period 2π) as an infinite sum of sine and cosine functions. The general form of a Fourier series is $f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty [a_ncos(nt)+b_nsin(nt)]$ where $a_n=\frac{1}{π}\int_{-π}^{π} f(t)cos(nt)dt, b_n = \frac{1}{π}\int_{-π}^{π} f(t)sin(nt)dt$
Understanding the convergence of Fourier series is essential when representing functions using trigonometric series. By expressing periodic functions as infinite sums of sines and cosines, we can analyze and solve differential equations more effectively.
Definition. A function f(t) is called even if it satisfies the condition: f(-t) = f(t) for all t. Examples: x2, cos(x), ∣x∣, x2 + 4. The graph of an even function is symmetric with respect to the y-axis.
A function f(t) is called odd if it satisfies the condition f(-t) = -f(t) for all t. Examples: x3, sin(x), 2x3 −3x. The graph of an odd function has rotational symmetry around the origin.
The Fourier series for a general periodic function f(t), with period 2π, is given by: $f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_ncos(nt)+b_nsin(nt))$
Since cosine is an even function and sine is an odd function, the behavior of the Fourier coefficients an and bn depends on whether the function f(t) is even, odd, or neither.
If a function f(t) is even, we know that f(-t) = f(t). Let’s evaluate the Fourier series at −t:
$f(-t) = \frac{a_0}{2} + \sum_{n=1}^\infty [a_ncos(n(-t))+b_nsin(n(-t))]$ =[cos(n(−t)) = cos(−nt) = cos(nt) (since cosine is even), sin(n(−t)) = sin(−nt) = −sin(nt) (since sine is odd).] $\frac{a_0}{2} + \sum_{n=1}^\infty [a_ncos(nt)-b_nsin(nt)]$
Comparing this with the original Fourier series:
$f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty [a_ncos(nt)+b_nsin(nt)]$
Since f(−t) = f(t) (because f is even) and because of the Uniqueness of Fourier Series, we conclude that all the sine terms must be zero, i.e., bn = 0. Therefore, the Fourier series for an even function contains only cosine terms:
bn = 0 ∀n, $f(t) = \frac{a_0}{2} + \sum_{n=0}^\infty a_ncos(nt)$ where the coefficients an are given by: $a_n=\frac{1}{π}\int_{-π}^{π} f(t)cos(nt)dt$
Because f(t)cos(nt) is also an even function (product of two even functions), the integral over [−π, π] simplifies to twice the integral over [0, π]:
$a_n=\frac{2}{π}\int_{0}^{π} f(t)cos(nt)dt$
This simplification reduces computation and is valid because the integrand is symmetric about the y-axis.
Similarly, if f(t) is an odd function, we know that f(-t) = -f(t).
Evaluate f at −t: f(-t) = $\frac{a_0}{2} + \sum_{n=0}^\infty (a_ncos(n(-t))+b_nsin(n(-t)))$ =[cos(n(−t)) = cos(−nt) = cos(nt) (since cosine is even), sin(n(−t)) = sin(−nt) = −sin(nt) (since sine is odd).] $\frac{a_0}{2} + \sum_{n=0}^\infty (a_ncos(nt))-b_nsin(nt)$
Comparing with Original Fourier Series: $f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_ncos(nt)+b_nsin(nt))$
Multiply both sides of the original equation by −1:-f(t) = $-\frac{a_0}{2} + \sum_{n=0}^\infty (-a_ncos(nt)-b_nsin(nt))$
If a function f is odd f(-t) = -f(t), then by the Uniqueness of the Fourier Series we conclude that all the cosine terms must be zero, i.e., an = 0. Therefore, the Fourier series for an odd function contains only sine terms:
$a_n = 0, b_n=\frac{1}{π}\int_{-π}^{π} f(t)sin(nt)dt$
If f is odd, then f(n)sin(nt) is an even function (product of two odd functions), the integral over [−π, π] simplifies to twice the integral over [0, π] and the coefficients bn are given by: $b_n = \frac{2}{π}\int_{0}^{π} f(t)sin(nt)dt.$
This simplification reduces computation and is valid because the integrand is symmetric about the origin.
Examples:
Theorem. If a function f(t) is periodic with period 2π and is continuous at a point t0, then its Fourier series converges to the value of the function at that point: $f(t_0) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_ncos(nt_0)+b_nsin(nt_0))$.
When f(t) has a jump discontinuity at a point t0, the Fourier series does not converge to the value of the function at that point. Instead, it converges to the midpoint of the jump (the average of the left-hand and right-hand limits) at that location. This is known as the Gibbs phenomenon. It refers to the oscillatory behavior of the Fourier series and overshoots that occur near a jump discontinuity, but still converges to the average value at the point of discontinuity.
Increasing the number of terms in the Fourier series reduces the width of the overshoot but not their height.
If the function f(t) has a jump from f(t0-) to f(t0+) at t0, then the Fourier series converges to: $\lim_{N \to ∞}[\frac{a_0}{2} + \sum_{n=1}^N (a_ncos(nt)+b_nsin(nt))] = \frac{f(t_0^+)+f(t_0^-)}{2}$
This behavior is one of the key reasons Fourier series are widely used to represent functions with jump discontinuities. Applications: modeling electrical signals (square waves, sawtooth waves); analyzing mechanical vibrations; solving partial differential equations (heat equation, wave equation).
When dealing with functions defined on a finite interval (0 , L), we often need to extend them periodically to apply Fourier series techniques. This section explains how to extend such functions to the entire real line in a way that makes it periodic with period 2L.
Let’s consider a function f(t) that is periodic with period 2L. The goal is to express the Fourier series in terms of a normalized variable. Specifically, we want to map the interval [0, L] to [0, π] for easier computation and interpretation.
To do this, we perform a change of variables: $t = \frac{L}{π}u ↭ u = \frac{π}{L}t$. This maps t ∈ [0, L] to u ∈ [0, π].
Thus, the Fourier series expansion for a function with period 2L becomes: $f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_ncos(\frac{nπ}{L}t)+b_nsin(\frac{nπ}{L}t)$ where the Fourier coefficients are defined as: $a_n=\frac{1}{L}\int_{-L}^{L} f(t)cos(\frac{nπ}{L}t)dt, b_n = \frac{1}{L}\int_{-L}^{L} f(t)sin(\frac{nπ}{L}t)dt$
If f is even (i.e., f(-t) = f(t)) and periodic with period 2L, then the Fourier series will contain only cosine terms. In this case: $a_n=\frac{2}{L}\int_{0}^{L} f(t)cos(\frac{nπ}{L}t)dt, b_n = 0$, and the Fourier series simplifies to $f(t) = \frac{a_0}{2}+\sum_{n=1}^\infty a_ncos(\frac{nπ}{L}t)$.
If f is odd (i.e., f(-t) = -f(t)) and periodic with period 2L, then the Fourier series will contain only sine terms. In this case: $b_n=\frac{2}{L}\int_{0}^{L} f(t)sin(\frac{nπ}{L}t)dt, a_n = 0$, and the Fourier series simplifies to $f(t) = \sum_{n=1}^\infty b_nsin(\frac{nπ}{L}t)$.
We know that every periodic function has a Fourier series expansion. It is fairly common for functions arising from certain applications to be defined only on a finite interval 0 < x < L. In such cases, it’s common to extend the function periodically over the entire real line to make use of Fourier series. This can be done in two distinct ways:
We define F(t) to be a periodic extension of f is:
There are two periodic extensions for f(t) that are very useful (Refer to Figure iii for a visual representation and aid in understanding it).
Even Periodic extension. Given any f(t), which is defined only for 0 < t < L, we define its even periodic extension, Fe(t), by the conditions that
Odd Periodic extension. We define its odd periodic extension, Fo(t), by the conditions that
Since Fe is even, it has a Fourier cosine series. Similarity, since F0 is odd, it has a Fourier sine series.
Restrict to 0 < t < L, $f(t)=F^e(t) = \frac{a_0}{2}+\sum_{n=1}^\infty a_ncos(\frac{nπt}{L}) = F^0(t) = +\sum_{n=1}^\infty b_nsin(\frac{nπt}{L})$ where $a_n = \frac{2}{L}\int_{0}^{L} F^e(t)cos(\frac{nπt}{L}) = \frac{2}{L}\int_{0}^{L} f(t)cos(\frac{nπt}{L})$ and $b_n = \frac{2}{L}\int_{0}^{L} F^0(t)sin(\frac{nπt}{L})dt = \frac{2}{L}\int_{0}^{L} f(t)sin(\frac{nπt}{L})dt$
This shows that f(t) can be represented either by its cosine series (even extension) or sine series (odd extension) within (0, L).
For example consider the function f(x) = 1 defined on 0 < x < π. The even and odd periodic extensions Fe(x) and F0(x) take the value 1 for all 0 < x < π and have period 2π (Refer to Figure ii for a visual representation and aid in understanding it).
$F^0(t) = \sum_{n=0}^\infty b_nsin(\frac{nπt}{L}) =[\text{2L=2π}] \sum_{n=0}^\infty b_nsin(nt)$ with $b_n = \frac{2}{L}\int_{0}^{L} F^0(t)sin(\frac{nπt}{L}) = \frac{2}{π}\int_{0}^{π} 1·sin(nt)dt = \frac{2}{π}\frac{-cos(nt)}{n}\bigg|_{0}^{π} = \frac{2}{πn}(-cos(nπ)+cos(0))$
Since cos(nπ)=(-1)n and cos(0) = 1, we have:
$= \begin{cases} 0, &n,even \\ \frac{2}{nπ}(-(-1)+1) = \frac{4}{nπ}, &n,odd \end{cases}$
Using the calculated coefficients bn, the odd periodic extension F0(t) becomes:
$F^0(t) = \sum_{n=1, n~odd}^\infty \frac{4}{nπ}sin(nt)$. This gives us the Fourier sine series representation of the odd periodic extension.
$F^e(t) = \frac{a_0}{2}+\sum_{n=1}^\infty a_ncos(\frac{nπt}{L}) = \frac{a_0}{2}+\sum_{n=1}^\infty a_ncos(nt)$ where an are the Fourier cosine coefficients.
$a_n = \frac{2}{L}\int_{0}^{L} F^e(t)cos(\frac{nπt}{L}) = \frac{2}{π}\int_{0}^{π} 1·cos(nt)dt = \frac{2}{π}\frac{sin(nt)}{n}\bigg|_{0}^{π}=0$ because sin(nπ) = 0 for all n. Therefore, an = 0 for all n ≥ 1.
However, the constant term (the average value of the function over the interval) is: $a_0 = \frac{2}{π}\int_{0}^{π} 1dt = \frac{2}{π}·π = 2$
Thus, the Fourier cosine series for the even periodic extension Fe(t) is:
$F^e(t) = \frac{a_0}{2}+\sum_{n=0}^\infty a_ncos(nt) = \frac{a_0}{2} + 0 + 0 + ··· + 0 = 1$. This shows that the even periodic extension of f(x) = 1 is just the constant function Fe(t) = 1. There are no higher cosine terms since all an = 0 ∀ n ≥ 1.
For a function f(t) with period 2π, the Fourier series is generally expressed as: $f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_ncos(nt)+b_nsin(nt)$ where $a_n=\frac{1}{π}\int_{-π}^{π} f(t)cos(nt)dt, b_n = \frac{1}{π}\int_{-π}^{π} f(t)sin(nt)dt$
However, since f(t) is an odd function, we have:
f(t) is periodic with a period 2π and odd. Therefore, we only need to compute the sine terms of the Fourier series. Specifically, $a_n = 0, b_n=\frac{2}{π}\int_{0}^{π} f(t)sin(nt)dt = \frac{2}{π}\int_{0}^{π} tsin(nt)dt$
We integrate from 0 to π because f(t) is odd, and integrating over [−π, π] would be redundant.
To do this, we apply integration by parts. Recall that the integration by parts formula is: ∫udv = uv −∫vdu. u = t ⇒ du = dt. dv = sin(nt)dv, so v = -1⁄ncos(nt)
$b_n = \frac{2}{π}[-\frac{tcos(nt)}{n}]\bigg|_{0}^{π}$
$-\int_{0}^{π} \frac{-cos(nt)}{n}dt = \frac{sin(nt)}{n^2}\bigg|_{0}^{π} = 0$. This is because sin(nπ) = 0 and sin(0) = 0, so the entire integral vanishes.
Now that we have the Fourier coefficients, we can write the Fourier series for f(t): f(t) = $2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}sin(nt) = 2(sin(t)-\frac{1}{2}sin(2t)+\frac{1}{3}sin(3t)-\frac{1}{4}sin(4t)+···)$
The Fourier series tries to approximate the function f(t) not just at individual points, but over the entire interval [-π, π]. At points of discontinuity, such as t = −π, t = π, and t = (2n +1)π, the Fourier series converges to the midpoint of the jump (average of the left-hand and right-hand limits of f(t)).
For example, at t = π, there is a jump discontinuity, and the Fourier series will converge to the average value at the jump, which is: $\frac{f(π^+)+f(π^-)}{2} = \frac{(-π)+π}{2}=0$
In the discontinuity points, the Fourier series converges to zero (the mid point of the jump). This behavior is known as the Gibbs phenomenon, where the Fourier series oscillates near discontinuities but still converges to the average value of the jump.
Gibbs Phenomenon: The oscillatory behavior and overshoot of the Fourier series near points of discontinuity.
The function f(t) resembles a rectangular pulse centered at t = 0, with a height of 4 and width π. It repeats every 2π. Since f(−t) mirrors f(t) across the vertical axis, f(t) is even.
For a function f(t) with period 2π, the Fourier series is generally expressed as: $f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_ncos(nt)+b_nsin(nt)$ where $a_n=\frac{1}{π}\int_{-π}^{π} f(t)cos(nt)dt, b_n = \frac{1}{π}\int_{-π}^{π} f(t)sin(nt)dt$
f(t) is periodic with a period 2π and even. Therefore, we only need to compute the cosine terms of the Fourier series. Specifically, $b_n = 0, a_n=\frac{2}{π}\int_{0}^{π} f(t)cos(nt)dt$
$a_0 = \frac{2}{π}\int_{0}^{π} f(t)cos(0·t)dt = \frac{2}{π}[\int_{0}^{\frac{π}{2}} 4cos(0·t)dt+\int_{\frac{π}{2}}^{π} 0·cos(0·t)dt] = \frac{2}{π}4t\bigg|_{0}^{\frac{π}{2}} = \frac{2}{π}·4·\frac{π}{2} = 4.$
$a_n=\frac{1}{π}\int_{-π}^{π} f(t)cos(nt)dt = \frac{2}{π}\int_{0}^{π} f(t)cos(nt)dt = \frac{2}{π}\int_{0}^{\frac{π}{2}} 4cos(nt)dt = \frac{8}{π}\frac{sin(nt)}{n}\bigg|_{0}^{\frac{π}{2}} = \frac{8}{π}\frac{sin(\frac{nπ}{2})}{n}$
$k = 1, sin(\frac{(2k-1)π}{2}) = sin(\frac{π}{2}) = 1, k = 2, sin(\frac{(2k-1)π}{2}) = sin(\frac{3π}{2}) = -1, k = 3, sin(\frac{(2k-1)π}{2}) = sin(\frac{5π}{2}) = 1$ ···
f(t) = $\frac{4}{2} + \sum_{n=1, n~ odd}^\infty (\frac{8}{π}\frac{(-1)^{\frac{n-1}{2}}}{n})cos(nt) = 2 + \sum_{n=1, n~ odd}^\infty (\frac{8}{π}\frac{(-1)^{\frac{n-1}{2}}}{n})cos(nt)$
Expressing an in terms of k: $a_n = \frac{8}{π}\frac{(-1)^{k-1}}{2k-1}$. The Fourier Series becomes: $2 + \sum_{k=1}^\infty (\frac{8}{π}\frac{(-1)^{k-1}}{2k-1})cos((2k-1)t) = 2 + \frac{8}{π}\sum_{k=1}^\infty (\frac{(-1)^{k-1}}{2k-1})cos((2k-1)t)$
Discontinuities occur at t = ±π⁄2 and every t = ±nπ⁄2, where n is an integer. At these points, the Fourier series converges to the average of the left-hand and right-hand limits of f(t).
Average value = $\frac{f(\frac{π}{2}^-) + f(\frac{π}{2}^+)}{2} = \frac{4+0}{2} = 2$. Therefore, the Fourier series converges to 2 at t = π⁄2. Near the discontinuities at t = ±π⁄2, the Fourier series overshoots the function value but converges to the average value.
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