Series and Transform Methods

Let $y''+p(x)\,y'+q(x)\,y=0$ be a second-order linear homogeneous equation and $x_0\in\mathbb{R}.$ The point $x_0$ is an ordinary point of the equation if and only if $p$ and $q$ are analytic at $x_0.$ The point $x_0$ is a singular point of the equation if and only if it is not an ordinary point.

Let $y''+p(x)\,y'+q(x)\,y=0$ be a second-order linear homogeneous equation and $x_0$ be an ordinary point. The power series method is the method that finds solutions as power series centred at $x_0$ with unknown coefficients,

$$y(x)=\sum_{n=0}^{\infty}a_n\left(x-x_0\right)^n,$$

by differentiating term by term, substituting into the equation and equating the coefficient of each power of $x-x_0$ to zero. The recurrence relation of the equation is the resulting relation that determines $a_2,a_3,\ldots$ from the free coefficients $a_0=y\left(x_0\right)$ and $a_1=y'\left(x_0\right).$ The power series method has the following properties:

• near an ordinary point every solution is analytic;
• the radius of convergence of the series is at least the distance from $x_0$ to the nearest singular point of the equation in the complex plane.

The Airy equation is the equation $y''-x\,y=0.$ It has only ordinary points and the recurrence relation $\left(n+2\right)\left(n+1\right)a_{n+2}=a_{n-1},$ whose solutions converge on all of $\mathbb{R}.$

Let $y''+p(x)\,y'+q(x)\,y=0$ be a second-order linear homogeneous equation and $x_0$ be a singular point. The point $x_0$ is a regular singular point if and only if $\left(x-x_0\right)p(x)$ and $\left(x-x_0\right)^2q(x)$ are analytic at $x_0.$ The point $x_0$ is an irregular singular point if and only if it is a singular point that is not regular.

Let $y''+p(x)\,y'+q(x)\,y=0$ be a second-order linear homogeneous equation with a regular singular point $x_0,$ and $p_0=\lim_{x\rightarrow x_0}\left(x-x_0\right)p(x)$ and $q_0=\lim_{x\rightarrow x_0}\left(x-x_0\right)^2q(x).$ The indicial equation of the equation at $x_0$ is the quadratic equation

$$r\left(r-1\right)+p_0r+q_0=0.$$

Let $y''+p(x)\,y'+q(x)\,y=0$ be a second-order linear homogeneous equation with a regular singular point $x_0.$ The Frobenius method is the method that seeks solutions of the form

$$y(x)=\left(x-x_0\right)^r\sum_{n=0}^{\infty}a_n\left(x-x_0\right)^n,\qquad a_0\neq 0,$$

where the exponent $r$ is a root of the indicial equation. The Frobenius method has the following properties:

• the larger root always gives a solution of this form;
• if the roots differ by an integer the second solution may contain a term $\log_e\left(x-x_0\right)$ times the first.

Let $\nu\geq 0.$ Bessel’s equation of order $\nu$ is the equation $x^2y''+x\,y'+\left(x^2-\nu^2\right)y=0,$ which has a regular singular point at $0$ with indicial roots $\pm\nu.$

Let $\ell\in\mathbb{R}.$ Legendre’s equation is the equation $\left(1-x^2\right)y''-2x\,y'+\ell\left(\ell+1\right)y=0,$ which has regular singular points at $\pm 1.$ When $\ell\in\mathbb{N},$ Legendre’s equation has a polynomial solution of degree $\ell,$ the Legendre polynomial $P_\ell.$

Let $f:\left[0,\infty\right)\rightarrow\mathbb{R}$ be a function. The function $f$ is of exponential order if and only if there exist $M,a\geq 0$ with $\lvert f(t)\rvert\leq M\,e^{at}$ for all $t\geq 0.$

Let $f:\left[0,\infty\right)\rightarrow\mathbb{R}$ be integrable on every bounded interval and of exponential order with constant $a.$ The Laplace transform of $f$ is the function of $s$ given by the improper integral

$$\mathcal{L}\left\{f\right\}(s)=F(s):=\int_0^{\infty}e^{-st}f(t)\,dt,$$

which converges absolutely for $s>a$ by comparison with $\int_0^{\infty}M\,e^{-\left(s-a\right)t}\,dt.$ The Laplace transform is linear.

Let $F$ be the Laplace transform of a continuous function $f.$ Lerch’s theorem states that two continuous functions with the same Laplace transform are equal. The inverse Laplace transform of $F,$ written $\mathcal{L}^{-1}\left\{F\right\},$ is the function $f.$ The following common transforms are valid for $s$ larger than the stated constant:

$$\mathcal{L}\left\{1\right\}=\frac{1}{s},\qquad\mathcal{L}\left\{t^n\right\}=\frac{n!}{s^{n+1}},\qquad\mathcal{L}\left\{e^{at}\right\}=\frac{1}{s-a}$$ $$\mathcal{L}\left\{\cos\omega t\right\}=\frac{s}{s^2+\omega^2},\qquad\mathcal{L}\left\{\sin\omega t\right\}=\frac{\omega}{s^2+\omega^2}$$

Let $f$ be $C^1$ with $f$ and $f'$ of exponential order. The derivative rule states that the Laplace transform turns differentiation into multiplication by $s$ and builds in the initial values.

$$\mathcal{L}\left\{f'\right\}(s)=s\,F(s)-f(0),\qquad\mathcal{L}\left\{f''\right\}(s)=s^2F(s)-s\,f(0)-f'(0)$$

In general $\mathcal{L}\left\{f^{(n)}\right\}(s)=s^nF(s)-\sum_{k=0}^{n-1}s^{n-1-k}f^{(k)}(0).$ The derivative rule follows from integration by parts.

The Heaviside step function $H$ is the function that is $0$ for $t<0$ and $1$ for $t\geq 0.$

Let $f$ have Laplace transform $F$ and $a\geq 0.$ The shift theorems state that multiplying $f$ by an exponential shifts its transform and that delaying $f$ multiplies its transform by an exponential.

$$\mathcal{L}\left\{e^{at}f(t)\right\}(s)=F\left(s-a\right),\qquad\mathcal{L}\left\{H\left(t-a\right)f\left(t-a\right)\right\}(s)=e^{-as}F(s)$$

The Laplace transform also satisfies $\mathcal{L}\left\{t\,f(t)\right\}=-F'(s)$ and $\mathcal{L}\left\{\int_0^tf(\tau)\,d\tau\right\}=\frac{F(s)}{s}.$

Let $f$ and $g$ be functions on $\left[0,\infty\right).$ The convolution of $f$ and $g$ is the function

$$\left(f*g\right)(t):=\int_0^tf(\tau)\,g\left(t-\tau\right)d\tau.$$

The convolution is commutative, associative and bilinear.

Let $f$ and $g$ have Laplace transforms $F$ and $G.$ The convolution theorem states that the Laplace transform of a convolution is the product of the Laplace transforms.

$$\mathcal{L}\left\{f*g\right\}=F\,G$$

An initial value problem for a linear equation with constant coefficients is solved by transforming both sides, which by the derivative rule gives an algebraic equation for $Y=\mathcal{L}\left\{y\right\},$ solving for $Y$ and inverting, usually by partial fractions and the table of transforms.

Let $a\,y''+b\,y'+c\,y=g(t)$ be a linear equation with constant coefficients and $y(0)=y'(0)=0.$ The Laplace transform of the solution is

$$Y(s)=\frac{G(s)}{a\,s^2+b\,s+c}.$$

The transfer function of the equation is $\frac{1}{a\,s^2+b\,s+c}.$ The impulse response of the equation is the inverse Laplace transform $h$ of the transfer function. By the convolution theorem the solution is $y=h*g,$ which is variation of parameters for constant coefficients.

The Dirac delta $\delta$ is the limit of unit-area pulses of shrinking width, defined by $\int_{-\infty}^{\infty}\delta\left(t-a\right)f(t)\,dt=f(a)$ for continuous $f.$ The Dirac delta is not a function. Its Laplace transform is $\mathcal{L}\left\{\delta\left(t-a\right)\right\}=e^{-as},$ and the response of the equation to $\delta$ is the impulse response $h.$ The Laplace transform handles discontinuous forcing through the Heaviside step function and impulsive forcing through the Dirac delta.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be periodic with period $2\pi$ and Riemann integrable on $\left[-\pi,\pi\right].$ The functions $1,\cos nx,\sin nx$ for $n\in\mathbb{N}$ are orthogonal with respect to the inner product $\langle f,g\rangle:=\int_{-\pi}^{\pi}f(x)\,g(x)\,dx,$ with $\langle\cos nx,\cos nx\rangle=\langle\sin nx,\sin nx\rangle=\pi$ for $n\geq 1$ and $\langle 1,1\rangle=2\pi.$ The Fourier coefficients of $f$ are its normalised inner products with these functions.

$$a_n:=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx\,dx,\qquad b_n:=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin nx\,dx$$

Let $f$ be periodic with period $2\pi$ and Fourier coefficients $a_n,b_n.$ The Fourier series of $f$ is the series of functions with these coefficients, whose partial sums $S_N$ are trigonometric polynomials.

$$f(x)\sim\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n\cos nx+b_n\sin nx\right)$$

Let $f$ be periodic with period $2\pi.$ The complex Fourier coefficients of $f$ are $c_n:=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\,e^{-inx}\,dx$ for $n\in\mathbb{Z}.$ By Euler’s formula the Fourier series is equally $\sum_{n=-\infty}^{\infty}c_ne^{inx},$ with $c_{\pm n}=\frac{1}{2}\left(a_n\mp ib_n\right).$ Fourier series have the following properties:

• for a function of period $2L$ the same formulae hold with $x$ replaced by $\frac{\pi x}{L}$ and integrals over $\left[-L,L\right]$ divided by $L$ in place of $\pi;$
• if $f$ is even then $b_n=0$ and the Fourier series is a cosine series;
• if $f$ is odd then $a_n=0$ and the Fourier series is a sine series;
• a function on $\left[0,L\right]$ is expanded in a cosine or sine series by extending it to an even or odd function of period $2L.$

Let $f$ be periodic with period $2\pi$ and Riemann integrable on $\left[-\pi,\pi\right],$ and $\lVert f\rVert_2:=\sqrt{\langle f,f\rangle}.$ By orthogonality, the partial sum $S_N$ is the trigonometric polynomial of degree $N$ closest to $f$ in the norm $\lVert\cdot\rVert_2,$ since $\lVert f-T\rVert_2^2=\lVert f-S_N\rVert_2^2+\lVert S_N-T\rVert_2^2$ for every trigonometric polynomial $T$ of degree $N.$ Bessel’s inequality states that the sum of the squared Fourier coefficients is bounded by the squared norm of $f.$

$$\frac{a_0^2}{2}+\sum_{n=1}^{N}\left(a_n^2+b_n^2\right)\leq\frac{1}{\pi}\lVert f\rVert_2^2$$

Let $f$ be periodic with period $2\pi$ and Riemann integrable on $\left[-\pi,\pi\right]$ with partial sums $S_N.$ The Fourier series converges in mean square to $f$ if and only if $\lVert f-S_N\rVert_2\rightarrow 0.$ The Fourier series of every such $f$ converges in mean square, so Bessel’s inequality becomes Parseval’s identity.

$$\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)^2\,dx=\frac{a_0^2}{2}+\sum_{n=1}^{\infty}\left(a_n^2+b_n^2\right)$$

The Riemann–Lebesgue lemma states that the Fourier coefficients tend to zero, $a_n,b_n\rightarrow 0$ as $n\rightarrow\infty,$ which follows from Parseval’s identity. Mean-square convergence is the density of the trigonometric polynomials in the continuous periodic functions, from the Stone–Weierstrass theorem, extended to integrable functions.

Let $f$ be a function and $x\in\mathbb{R}.$ The one-sided limits of $f$ at $x$ are the limits $\lim_{t\rightarrow x^-}f(t)$ and $\lim_{t\rightarrow x^+}f(t)$ of $f$ restricted to $t<x$ and to $t>x$ respectively.

Let $f$ be a function on $\left[-\pi,\pi\right].$ The function $f$ is piecewise $C^1$ on $\left[-\pi,\pi\right]$ if and only if $f$ and $f'$ are continuous except at finitely many points, at each of which both have one-sided limits.

Let $f$ be periodic with period $2\pi$ and piecewise $C^1$ on $\left[-\pi,\pi\right]$ with partial sums $S_N.$ The Dirichlet convergence theorem states that the Fourier series converges at every $x$ to the average of the one-sided limits of $f,$

$$\lim_{N\rightarrow\infty}S_N(x)=\frac{1}{2}\left(\lim_{t\rightarrow x^-}f(t)+\lim_{t\rightarrow x^+}f(t)\right),$$

which is $f(x)$ wherever $f$ is continuous, and that the convergence is uniform on every closed interval on which $f$ is continuous. Pointwise convergence of Fourier series has the following properties:

• if $f$ is continuous and piecewise $C^1$ the convergence is uniform on $\mathbb{R};$
• the series may be differentiated term by term wherever $f'$ is itself piecewise $C^1;$
• the series of any piecewise $C^1$ function may be integrated term by term;
• if $f\in C^k$ is periodic then $a_n,b_n$ decay faster than $n^{-k}.$

The Gibbs phenomenon is the overshoot of the partial sums near a jump of $f$ by about $9\%$ of the jump, however large $N$ is, which is why the convergence is not uniform at a jump. For example, the odd function equal to $x$ on $\left(-\pi,\pi\right)$ has $b_n=\frac{2\left(-1\right)^{n+1}}{n},$ and Parseval’s identity gives $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}.$