Consider the Piecewise Function Find the Value of a That Makes F X Continuous Everywhere

Beyond Wavelets

Tony F. Chan , Hao-Min Zhou , in Studies in Computational Mathematics, 2003

5.4.1 Function Approximation

Constructing approximations to the piecewise continuous functions is a very natural application of the designed ENO-wavelet transform. One simple way is to use the low frequencies f _j (x) to approximate f(x) directly. Here, we use some 1-D numerical examples to illustrate the approximation abilities of the ENO-wavelet transforms. We will demonstrate the error bound 3.2 given in section 5.3. In particular, we show results for the ENO-Haar, ENO-DB4 and ENO-DB6 wavelet transforms.

In all examples, for simplicity, we just consider functions with zero values at the boundary. For non-zero boundary functions, we can easily extend the function by zero and treat the boundaries as discontinuities.

To illustrate the performance of ENO-wavelet transforms, we show graphical comparisons of the standard wavelet approximations and corresponding ENO-wavelet approximations. In addition, we compare the L _∞ and L ₂ errors of the standard wavelet approximations and the ENO-wavelet approximations at different levels by measuring E _∞,j = inf_x ||f(x) − f _j (x||, which is computed by finding the largest difference on the finest grid, and E _2j = ||f(x) − f _j (x)||2. Using them, we compute the orders of accuracy defined by:

$Orde{r}_{\mathrm{\infty }}={log}_{\mathrm{\infty }}\frac{E_{\mathrm{\infty },i}}{E_{\mathrm{\infty },i-1}},$

and

$Orde{r}_2={log}_2\frac{E_{2,i}}{E_{2,i-1}},$

which indicates the order of accuracy of the approximation in the L _∞ norm and L ₂ norm respectively.

Since we consider noise free examples in this part, we use the method for noise free data described in Section 5.2.2 to detect the positions of the discontinuities. And we select a = 2 (as used in the algorithms in section 5.2) for all 1-D examples.

Firstly, we compare the approximations for smooth functions. Table 5.1 shows the results of comparison of DB4 with ENO-DB4 approximations for the function $f\left(x\right)=exp\left[-\left(\frac{1}{x}+\frac{1}{1-x}\right)\right],0<x<1,$

Table 5.1. The Comparison of maximum error of the standard DB4 and the ENO-DB4 approximations for the smooth function $f\left(x\right)=exp\left[-\left(\frac{1}{x}+\frac{1}{1-x}\right)\right],0\&lt;x\&lt;1.$ They have the same error and both achieve second order of accuracy which agrees with the results in Theorem 1 for the smooth functions.

level	DB4 E ∞	ENO-DB4 E ∞	Order ∞
4	3.316e-5	3.316e-5
3	7.650e-6	7.650e-6	2.104
2	1.590e-6	1.590e-6	2.232
1	2.972e-7	2.973e-7	2.406

We see from the table that for smooth functions, the ENO-wavelet transforms have exactly the same approximation error as the standard wavelet transforms. Both of them maintain the approximation order 2, which agree with the results in Theorem 1. In fact, we notice that in this situation, no singularity is detected, the ENO-wavelet algorithms perform the standard transforms for completely smooth functions as we expected.

Next, we apply Haar and ENO-Haar, DB4 and ENO-DB4, and DB6 and ENO-DB6 transforms to a piecewise smooth function and compare the approximation error. Figure 5.4 shows the comparison of the order of accuracy in the L _∞ and L ₂ norm. It is clear that both L _∞ and L ₂ order of accuracy for ENO-wavelet transforms are of the order 1, 2 and 3 for ENO-Haar, ENO-DB4 and ENO-DB6 respectively. And they agree with the results in Theorem 2. In contrast, standard wavelet transforms do not retain the corresponding order of accuracy for piecewise smooth functions.

Figure 5.4. The approximation accuracy comparison of ENO-wavelet and wavelet transforms. Both L _∞ (left) and L ₂ (right) order of accuracy show that ENO-wavelet transforms maintain the order 1, 2 and 3 for ENO-Haar, ENO-DB4 and ENO-DB6 respectively and they agree with the results of Theorem 2. In contrast, standard wavelet transforms do not retain the order of accuracy for piecewise smooth functions.

To see the Gibbs' oscillations, we display the 4-level ENO-DB6 and standard DB6 approximations to a piecewise smooth function in Figure 5.5. In the left picture, we show the original function (dotted line), the standard wavelet linear approximations (dash-dotted) and the ENO-wavelet approximations (solid line). The right pictures is the zoom-in of the left picture near a discontinuity. We clearly see the Gibbs' oscillations in the standard approximations; in contrast, the ENO-wavelet approximations preserve the jump accurately.

Figure 5.5. The 4-level ENO-DB6 (solid line) and the standard DB6 (dash-dotted line) Approximation. The standard DB6 generates oscillations near discontinuities, but the ENO-DB6 does not.

In Figure 5.8, we also present the standard DB6 wavelet coefficients (dotted line) and the ENO-DB6 wavelet coefficients (solid line) respectively. The left part corresponds to the low frequency coefficients and the right part the high frequency coefficients. We notice that there are some large standard high frequency coefficients near the discontinuities.

Figure 5.8. The 4-level ENO-DB6 coefficients (solid line) and the standard DB6 coefficients (dotted line). There are large high frequency coefficients near the discontinuities in the standard DB6 transform but not in the ENO-DB6 transform.

On the other hand, no large high frequency coefficients are present in the ENO-wavelet coefficients. This illustrates that the ENO-wavelet coefficients have better distribution than standard wavelet coefficients, i.e., no large coefficients in the high frequencies and the energy is concentrated in the low frequency end.

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Diffraction of Acoustic Waves and Boundary Integral Equations

Paul J.T. Filippi , in Acoustics, 1999

3.2.1 Statement of the problem

Let Ω be a domain of ℝn (n = 1, 2 or 3), with a regular boundary σ. It is assumed that a unit normal vector $\overrightarrow{n}$ , pointing out to the exterior of Ω, can be defined almost everywhere on σ. If Ω is bounded, we have an interior problem; if Ω extends up to infinity, we have an exterior problem. Let f be the distribution which represents the energy harmonic sources. The acoustic pressure p is the solution of the non-homogeneous equation:

(3.33) $\left(\mathrm{\Delta }+k^2\right)p\left(M\right)=f\left(M\right),M\in \mathrm{\Omega }$

where k is, as usual, the wavenumber. The boundary σ is assumed to satisfy a local boundary condition:

(3.34) $\alpha \mathrm{Tr}{\partial }_{n}p\left(M\right)+\beta \mathrm{Tr}p\left(M\right)=0,M\in \sigma$

Very often, the symbol 'Tr' can be omitted. But, as has been seen already, the sound field can be represented by surface integrals which are discontinuous or have a discontinuous normal gradient or involve, on σ, a finite part of the integral: in these cases, the symbol 'Tr' is absolutely necessary to indicate that the value on σ of the function is obtained as the limit of a Riemann integral. The possible existence of a solution of this boundary value problem is beyond the scope of this book; among the best textbooks dealing with boundary value problems of classical physics and in which such a proof is given, we can mention Methods of Mathematical Physics by R. Courant and D. Hilbert [7].

Some remarks must be made on the mathematical requirements on which mathematical physics is based. The basic hypothesis which is always made in physics is that any system has a finite energy density and its behaviour is based on the Hamilton principle (energy conservation). In acoustics, this implies that the sound pressure and its gradient are described by functions which are (at least locally) square integrable. Furthermore, an energy flux density across any elementary surface of the propagation domain boundary must be defined. The consequences are: (a) the value on the boundary of the function which represents the sound pressure inside the propagation domain can be different from the value of the sound pressure; (b) the value of the pressure field (or of its gradient) on the boundary of the propagation domain is obtained as a limit of a function defined inside the propagation domain; (c) the limits on the boundary of the functions representing the pressure field and its gradient need to be (at least locally) square integrable functions. These remarks justify the necessity to use the symbol 'Tr' as a reminder of the mathematical requirements which must be respected to get a description of the physical phenomena.

If α = 1 and β = 0, we are left with the Neumann problem (cancellation of the normal component of the particle velocity); α = 0 and β = 1 describe the Dirichlet problem (cancellation of the sound pressure). Finally, let us recall that, in most situations, an absorbing boundary can be characterized by a specific normal impedance ζ which is defined by

$\zeta =-\iota k\frac{\mathrm{Tr}p}{\mathrm{Tr}{\partial }_{n}p}\left(\frac{\beta }{\alpha }=\frac{\iota k}{\zeta }\right)$

This corresponds to α = 1 and β ≠ 0. Such a boundary condition is generally called the Robin condition.

In what follows, it is assumed that α and β can be piecewise continuous functions: this allows the boundary σ to be made of different materials (for example, one part is rather hard, another one being highly absorbent). To ensure the uniqueness of the solution for any real frequency when the domain Ω is unbounded, the principle of energy conservation must be respected: this is done by adding a Sommerfeld condition at infinity (which has already been given) or by using either the principle of limit absorption or the principle of limit amplitude.

Limit absorption principle

Let ε be a positive parameter and p_ε be the unique bounded solution corresponding to a wavenumber k_ε = k(1 + ɩε). For a real wavenumber k, the solution which satisfies the energy conservation principle is the limit, for ε → 0, of p_ε .

Limit amplitude principle

Let us consider the unique solution P_ω (Mt; tt) of the wave equation which corresponds to a harmonic source with angular frequency ω starting at a time t = 0. The solution of the Helmholtz equation which satisfies the energy conservation principle is the limit for t → ∞ of P_ω (M; t).

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Fourier series

Mary Attenborough , in Mathematics for Electrical Engineering and Computing, 2003

16.1 Introduction

Fourier analysis is the theory behind frequency analysis of signals. This chapter is concerned with the Fourier analysis of periodic, piecewise continuous functions. A periodic function can be represented by a Fourier series. A non-periodic function can be represented by its Fourier transform which we shall not be concerned with here. Discrete functions may be represented by a discrete Fourier transform, which also we shall not look at in this book.

Any periodic signal is made up of the sum of single frequency components. These components consist of a fundamental frequency component, multiples of the fundamental frequency, called the harmonics and a bias term, which represents the average off-set from zero. There are three ways of representing this information which are equivalent. We can represent the frequency components as the sum of a sine and cosine terms, or by considering the amplitude and phase of each component, or we can represent them using a complex Fourier series. The use of the complex Fourier series simplifies the calculation.

Having found the Fourier components we can use the system's frequency response function, as found in the previous chapter, to find the steady state response to any periodic signal.

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Complex Analysis

Joseph P.S. Kung , Chung-Chun Yang , in Encyclopedia of Physical Science and Technology (Third Edition), 2003

V.C Harmonic Functions

Harmonic functions, defined in Section II, are real functions u(x, y) satisfying Laplace's equation. We shall use the notation: u(x, y)   = u(x  + iy)   = u(z). Harmonic functions have several important properties.

The mean-value property. Let u(z) be a harmonic function on a region Ω. Then for any disk D with center a and radius r whose closure is contained in Ω,

$\text{u}(\text{a})=\frac{1}{2\pi }{\int }_0^{2\pi }\text{u}(\text{a}+\text{r}{\text{e}}^{\text{i}\theta })\text{d}\theta$

The maximum principle for harmonic functions. Let u(z) be a harmonic function on the domain Ω. If there is a point a in Ω such that u(a) equals the maximum max {u(z)   : z  ∈   Ω}, then u(z) is a constant function.

An important problem in the theory of harmonic function is Dirichlet's problem . Given a simply connected domain Ω and a piecewise continuous function g(z) on the boundary ∂Ω, find a function u(z) in the closure $\overline{\Omega }$ such that u(z) is harmonic in Ω, and the restriction of u(z) to the boundary $\overline{\Omega }\backslash \Omega$ equals g(z). (The boundary ∂Ω is the set $\overline{\Omega }\backslash \Omega$ .)

For general simply connected domains, Dirichlet's problem is difficult. It is equivalent to finding a Green's function. For a disk, the following formula is known.

The Poisson formula.

Let g(z)   = g(e ^iϕ), −π   <   ϕ   ≤   π, be a piecewise continuous function on the boundary {z: ¦z¦   =   1} of the unit disk. Then the function

$\begin{array}{ll}\text{u}(\text{z})\hfill & =\text{u}(\text{r}{\text{e}}^{\text{i}\theta })\hfill \\ \hfill & =\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left(\frac{1-{\text{r}}^2}{1+{\text{r}}^2-2\text{r}cos(\phi -\theta )}\right)\text{g}({\text{e}}^{\text{i}\phi })\text{d}\phi \hfill \end{array}$

is a solution to Dirichlet's problem for the unit disk.

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The Laplace Transform

James R. Kirkwood , in Mathematical Physics with Partial Differential Equations, 2013

Introduction

A second integral transform that plays a prominent role in the solution of differential equations is the Laplace transform. If f(x ) is a piecewise continuous function for which

$f(x)=0\mathrm{if}x<0\text{,}$

then the Laplace transform of f(x), denoted (ℒf)(s), is defined to be

$(ℒf)(s)={\int }_0^{\infty }{e}^{-sx}f(x)dx$

for functions for which the integral converges.

We note a relationship between the Laplace transform and the Fourier transform. We have

$(ℱf)(s)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }f(x)e^{-ixs}dx$

so

$(ℱf)(-is)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }f(x)e^{-ix(-is)}dx=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }f(x)e^{-xs}dx.$

If f(x)=0 for x<0, then

(1) $(ℱf)(-is)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }f(x)e^{-xs}dx=\frac{1}{\sqrt{2\pi }}{\int }_0^{\infty }f(x)e^{-xs}dx=\frac{1}{\sqrt{2\pi }}(ℒf)(s).$

(This is one place where our definition of the Fourier transform makes things a little messier. There is a definition of the Fourier transform for which (ℱf)(−is)=(ℒf)(s).)

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Laplace transform methods

Martha L. Abell , James P. Braselton , in Differential Equations with Mathematica (Fifth Edition), 2023

8.2.2 Laplace transform of an integral

We have seen that the Laplace transform of the derivatives of a given function can be found from the Laplace transform of the function. Similarly, the Laplace transform of the integral of a given function can also be obtained from the Laplace transform of the original function.

Theorem 8.8 Laplace transform of an integral

Suppose that $F(s)=\mathcal{L}\{f(t)\}$ , where $y=f(t)$ is a piecewise continuous function on $[0,\infty )$ and of exponential order b. Then, for $s>b$ ,

(8.8) $\mathcal{L}\{{\int }_0^{t}f(\alpha )d\alpha \}=\frac{1}{s}\mathcal{L}\{f(t)\}.$

The theorem tells us that

(8.9) ${\mathcal{L}}^{-1}\left\{\frac{1}{s}\mathcal{L}\{f(t)\}\right\}={\int }_0^{t}f(\alpha )d\alpha .$

Example 8.2.7

Compute ${\mathcal{L}}^{-1}\left\{\frac{1}{s(s+2)}\right\}$ .

Solution

In this case, $\frac{1}{s(s+2)}=\frac{1}{s}\frac{1}{s+2}$ , so $\mathcal{L}\{f(t)\}=\frac{1}{s+2}$ . Therefore, $f(t)={\mathcal{L}}^{-1}\left\{\frac{1}{s+2}\right\}=e^{-2t}$ . Using the theorem, we then have

${\mathcal{L}}^{-1}\left\{\frac{1}{s(s+2)}\right\}={\int }_0^t{e}^{-2\alpha }d\alpha =\frac{1}{2}(1-e^{-2t}).$

Note that the same result is obtained with InverseLaplaceTransform

$\mathbf{\text{InverseLaplaceTransform}}\mathbf{[}\mathbf{1}\mathbf{/}\mathbf{(}\mathit{s}\mathbf{(}\mathit{s}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{)}\mathbf{,}\mathit{s}\mathbf{,}\mathit{t}\mathbf{]}$

$\frac{1}{2}-\frac{e^{-2t}}{2}$

or through a partial fraction expansion of $\frac{1}{s(s+2)}$ : $\frac{1}{s(s+2)}=\frac{1}{2s}-\frac{1}{2(s+2)}$ , ${\mathcal{L}}^{-1}\left\{\frac{1}{s(s+2)}\right\}={\mathcal{L}}^{-1}\{\frac{1}{2s}-\frac{1}{2(s+2)}\}=\frac{1}{2}-\frac{1}{2}{e}^{-2t}$ .   □

The following theorem is useful in determining if the inverse Laplace transform of a function $F(s)$ exists.

Theorem 8.9

Suppose that $y=f(t)$ is a piecewise continuous function on $[0,\infty )$ and of exponential order b. Then,

$\underset{s\to \infty }{\lim }F(s)=\underset{s\to \infty }{\lim }\mathcal{L}\{f(t)\}=0.$

Example 8.2.8

Determine if the inverse Laplace transform of the functions may exist for the functions (a) $F(s)=\frac{2s}{s-6}$ and (b) $F(s)=\frac{s^3}{s^2+16}$ .

Solution

In both cases, we find ${\lim }_{s\to \infty }F(s)$ . If this value is not zero, then ${\mathcal{L}}^{-1}\{F(s)\}$ cannot be found. (a) ${\lim }_{s\to \infty }F(s)={\lim }_{s\to \infty }\frac{2s}{s-6}=2\ne 0$ , so ${\mathcal{L}}^{-1}\left\{\frac{2s}{s-6}\right\}$ does not exist. (b) ${\lim }_{s\to \infty }F(s)={\lim }_{s\to \infty }\frac{s^3}{s^2+16}=\infty \ne 0$ . Thus, ${\mathcal{L}}^{-1}\left\{\frac{s^3}{s^2+16}\right\}$ does not exist.

However, in a more abstract setting, the Laplace transform may exist. We will discuss the DiracDelta function in Section 8.4.

$\mathbf{\text{InverseLaplaceTransform}}\mathbf{[}\mathbf{2}\mathit{s}\mathbf{/}\mathbf{(}\mathit{s}\mathbf{-}\mathbf{6}\mathbf{)}\mathbf{,}\mathit{s}\mathbf{,}\mathit{t}\mathbf{]}$

$2(6e^{6t}+\text{DiracDelta}[t])$

$\mathbf{\text{InverseLaplaceTransform}}\mathbf{[}{\mathit{s}}^{\mathbf{\wedge }}\mathbf{3}\mathbf{/}\mathbf{(}{\mathit{s}}^{\mathbf{\wedge }}\mathbf{2}\mathbf{+}\mathbf{16}\mathbf{)}\mathbf{,}\mathit{s}\mathbf{,}\mathit{t}\mathbf{]}$

$-16\text{Cos}[4t]+{\text{DiracDelta}}^{\prime }[t]$   □

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Introduction to differential equations

Henry J. Ricardo , in A Modern Introduction to Differential Equations (Third Edition), 2021

1.3.1 An integral form of an IVP solution

If a first-order equation can be written in the form $y^{\prime }=f(x)$ —that is, if the right-hand side is a continuous (or piecewise continuous) function of the independent variable alone—then we can always express the solution to the IVP $y^{\prime }=f(x)$ , $y(x_0)=y_0$ on an interval $(a,b)$ as

(1.3.1) $y(x)={\int }_{x_0}^{x}f(t)dt+y_0$

for x in $(a,b)$ . Note that we use the x value of the initial condition as the lower limit of integration and the y value of the initial condition as a particular constant of integration. We use t as a dummy variable. Given Eq. (1.3.1), the Fundamental Theorem of Calculus (FTC) (Section A.4) implies that $y^{\prime }=f(x)$ , and we see that $y(x_0)={\int }_{x_0}^{x_0}f(t)dt+y_0=0+y_0=y_0$ , which is what we want. This way of handling certain types of IVPs is common in physics and engineering texts. In Example 1.2.6 the solution of the equation with $y(-1)=2$ , for example, is

$y(x)={\int }_{-1}^x{t}^2-2t+7dt+2={(\frac{t^3}{3}-t^2+7t)|}_{t=x}{-(\frac{t^3}{3}-t^2+7t)|}_{t=-1}+2=(\frac{x^3}{3}-x^2+7x)-\left(\frac{-25}{3}\right)+2=\frac{x^3}{3}-x^2+7x+\frac{31}{3}.$

You could also solve this problem the way we did in Example 1.3.1—that is, without using a definite integral formula.

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Introduction to the Laplace Transform

Martha L. Abell , James P. Braselton , in Introductory Differential Equations (Fourth Edition), 2014

Exponential Order, Jump Discontinuities, and Piecewise-Continuous Functions

In calculus, we learn that some improper integrals diverge, which indicates that the Laplace transform may not exist for some functions. For example, f (t) =   t ^{−  1} grows too rapidly near t   =  0 for the improper integral ∫ ₀ ^∞ e ^{− st} f(t)   dt to exist and f (t) =  e^t2 grows toot rapidly as t  →   ∞ for the improper integral ∫ ₀ ^∞ e ^{− st} f(t)   dt to exist. Therefore, we present the following definitions and theorems so that we can better understand the types of functions for which the Laplace transform exists.

Definition 31 (Exponential Order). A function y   =   f (t) is of exponential order b if there are numbers b,C  > 0, and T  > 0 such that

$\left|f\right.\left.\left(t\right)\right|\le C{\mathrm{e}}^{\mathit{bt}}$

for t  >   T.

In the following sections, we will see that the Laplace transform is particularly useful in solving equations involving piecewise or recursively defined functions.

Definition 32 (Jump Discontinuity). A function y   =   f (t) has a jump discontinuity at t   =   c on the closed interval [a, b] if the one-sided limits lim _{t  → c   +} f (t) and lim _{t  → c   −} f (t) are finite, but unequal, values. y   =   f (t) has a jump discontinuity at t   =   a if lim _{t  → a   +} f (t) is a finite value different from f (a). y   =   f (t) has a jump discontinuity at t   =   b if lim _{t  → b   −} f (t) is a finite value different from f (b).

Definition 33 (Piecewise Continuous). A function y   =   f (t) is piecewise continuous on the finite interval [a, b] if y   =   f (t) is continuous at every point in [a, b] except at finitely many points at which y   =   f (t) has a jump discontinuity.

A function y   =   f (t) is piecewise continuous on [0, ∞) if y   =   f (t) is piecewise continuous on [0, N] for all N.

Theorem 39 (Sufficient Condition for Existence of ℒ{f (t)}). Suppose that y  = f (t) is a piecewise-continuous function on the interval [0, ∞) and that it is of exponential order b for t  > T. Then, ℒ{f (t)} exists for s  > b.

Proof. We need to show that the integral ∫ ₀ ^∞ e ^{− st} f(t)   dt converges for s  > b, assuming that f (t) is a piecewise-continuous function on the interval [0, ∞) and that it is of exponential order b for t  > T. First, we write the integral as

${\displaystyle {\int }_0^{\infty }{e}^{-\mathit{st}}f\left(t\right)\mathrm{d}t}={\displaystyle {\int }_0^T{e}^{-\mathit{st}}f\left(t\right)\mathrm{d}t}+{\displaystyle {\int }_T^{\infty }{e}^{-\mathit{st}}f\left(t\right)\mathrm{d}t}\text{,}$

where T is selected so that |f(t)|   ≤ Ce ^bt for the constants b and C, C  >   0.

Note: In this textbook, typically we work with functions that are piecewise continuous and of exponential order. However, in the exercises, we explore functions that may or may not have these properties.

Notice that because f (t) is a piecewisecontinuous function, so is e^–st f (t). The first of these integrals, ∫ ₀ ^T e^{− st} f(t)   dt, exists because it can be written as the sum of integrals over which e ^–st f (t) is continuous. The fact that e ^–st f (t) is piecewise continuous on [T, ∞) is also used to show that the second integral, ∫ _T ^∞e^{− st} f(t)   dt, converges. Because there are constants C and b such that |f(t)|   ≤ Ce ^bt , we have

$\begin{array}{l}\left|{\displaystyle {\int }_T^{\infty }{\mathrm{e}}^{–\mathit{st}}f\left(t\right)\mathrm{d}t}\right|\le {\displaystyle {\int }_T^{\infty }\left|{\mathrm{e}}^{–\mathit{st}}f\left(t\right)\right|\mathrm{d}t}\\ \le C{\displaystyle {\int }_T^{\infty }{\mathrm{e}}^{–\mathit{st}}{\mathrm{e}}^{\mathit{bt}}\mathrm{d}t}=C{\displaystyle {\int }_T^{\infty }{\mathrm{e}}^{–\left(s-b\right)t}\mathrm{d}t}\\ =C\underset{M\to \infty }{lim}{\displaystyle {\int }_T^M{\mathrm{e}}^{–\left(s-b\right)t}\mathrm{d}t}\\ =C\underset{M\to \infty }{lim}{\left[-\frac{1}{s-b}{\mathrm{e}}^{–\left(s-b\right)t}\right]}_{t=T}^{t=M}\\ =-\frac{C}{s-b}\underset{M\to \infty }{lim}\left({\mathrm{e}}^{–\left(s-b\right)M}-{\mathrm{e}}^{–\left(s-b\right)T}\right)\text{.}\end{array}$

Then, if s   −   b   >   0, lim _{M→   ∞} e ^{–   (s–b)M}   =   0, so

$\left|{\displaystyle {\int }_T^{\infty }{\mathrm{e}}^{-\mathit{st}}f\left(t\right)\mathrm{d}t}\right|\le \frac{C}{s-b}{\mathrm{e}}^{–\left(s-b\right)T},s>b\text{.}$

Because both of the integrals ∫ ₀ ^T e^{− st} f(t)   dt and ∫ _T ^∞e^{− st} f(t)   dt exist, ∫ ₀ ^∞e^{− st} f(t)   dt also exists for s  > b.

Example 8.1.7

Find the Laplace transform of $f\left(t\right)=\left\{\begin{array}{l}\begin{array}{cc}\hfill -1,\hfill & \hfill \hfill \end{array}0\le t<4\\ 1,\begin{array}{cc}\hfill \hfill & \hfill \hfill \end{array}t\ge 4\end{array}\right.\text{.}$

Solution

Because y  = f (t) is a piecewise-continuous function on [0, ∞) and of exponential order, ℒ {f (t)} exists. We use the definition and evaluate the integral using the sum of two integrals. We assume that s  >   0:

$\begin{array}{ll}\mathrm{ℒ}\left\{f\left(t\right)\right\}& ={\displaystyle {\int }_0^{\infty }f\left(t\right){\mathrm{e}}^{–\mathit{st}}\mathrm{d}t}={\displaystyle {\int }_0^4-1\times {\mathrm{e}}^{–\mathit{st}}\mathrm{d}t}+{\displaystyle {\int }_4^{\infty }{\mathrm{e}}^{–\mathit{st}}\mathrm{d}t}\hfill \\ \hfill & ={\left[\frac{1}{s}{\mathrm{e}}^{–\mathit{st}}\right]}_{t=0}^{t=4}+\underset{M\to \infty }{lim}{\left[-\frac{1}{s}{\mathrm{e}}^{–\mathit{st}}\right]}_{t=4}^{t=M}\hfill \\ \hfill & =\frac{1}{s}\left(e^{–4s}-1\right)-\frac{1}{s}\underset{M\to \infty }{lim}\left(e^{–\mathit{Ms}}-e^{–4s}\right)\hfill \\ \hfill & =\frac{1}{s}\left(2e^{–4s}-1\right)\text{.}\hfill \end{array}$

Theorem 39 gives a sufficient condition and not a necessary condition for the existence of the Laplace transform. In other words, there are functions such as f (t) =   t ^{−   1/2} that do not satisfy the hypotheses of the theorem for which the Laplace transform exists (see Exercises 62 and 63).

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Introduction to the Laplace Transform

Martha L. Abell , James P. Braselton , in Introductory Differential Equations (Fifth Edition), 2018

Exponential Order, Jump Discontinuities and Piecewise-Continuous Functions

In calculus, we learn that some improper integrals diverge, which indicates that the Laplace transform may not exist for some functions. For example, $f(t)=t^{-1}$ grows too rapidly near $t=0$ for the improper integral ${\int }_0^{\infty }{e}^{-st}f(t)dt$ to exist and $f(t)=e^{t^2}$ grows too rapidly as $t\to \infty$ for the improper integral ${\int }_0^{\infty }{e}^{-st}f(t)dt$ to exist. Therefore, we present the following definitions and theorems so that we can better understand the types of functions for which the Laplace transform exists.

Definition 8.2 Exponential Order

A function $y=f(t)$ is of exponential order b if there are numbers b, $C>0$ , and $T>0$ such that

$|f(t)|⩽C{e}^{bt}$

for $t>T$ .

In the following sections, we will see that the Laplace transform is particularly useful in solving equations involving piecewise or recursively defined functions.

Definition 8.3 Jump Discontinuity

A function $y=f(t)$ has a jump discontinuity at $t=c$ on the closed interval $[a,b]$ if the one-sided limits ${\lim }_{t\to c^{+}}f(t)$ and ${\lim }_{t\to c^{-}}f(t)$ are finite, but unequal, values. $y=f(t)$ has a jump discontinuity at $t=a$ if ${\lim }_{t\to a^{+}}f(t)$ is a finite value different from $f(a)$ . $y=f(t)$ has a jump discontinuity at $t=b$ if ${\lim }_{t\to b^{-}}f(t)$ is a finite value different from $f(b)$ .

Definition 8.4 Piecewise Continuous

A function $y=f(t)$ is piecewise continuous on the finite interval $[a,b]$ if $y=f(t)$ is continuous at every point in $[a,b]$ except at finitely many points at which $y=f(t)$ has a jump discontinuity.

A function $y=f(t)$ is piecewise continuous on $[0,\infty )$ if $y=f(t)$ is piecewise continuous on $[0,N]$ for all N.

Theorem 8.2

Sufficient Condition for Existence of $\mathcal{L}\{f(t)\}$

Suppose that $y=f(t)$ is a piecewise continuous function on the interval $[0,\infty )$ and that it is of exponential order b for $t>T$ . Then, $\mathcal{L}\{f(t)\}$ exists for $s>b$ .

Proof

We need to show that the integral ${\int }_0^{\infty }{e}^{-st}f(t)dt$ converges for $s>b$ , assuming that $f(t)$ is a piecewise continuous function on the interval $[0,\infty )$ and that it is of exponential order b for $t>T$ . First, we write the integral as

$\begin{array}{cc}\hfill & {\int }_0^{\infty }{e}^{-st}f(t)dt\hfill \\ \hfill & ={\int }_0^T{e}^{-st}f(t)dt+{\int }_T^{\infty }{e}^{-st}f(t)dt,\hfill \end{array}$

where T is selected so that $|f(t)|⩽C{e}^{bt}$ for the constants b and C, $C>0$ .

In this textbook, typically we work with functions that are piecewise continuous and of exponential order. However, in the exercises we explore functions that may or may not have these properties.

Notice that because $f(t)$ is a piecewise continuous function, so is $e^{-st}f(t)$ . The first of these integrals, ${\int }_0^T{e}^{-st}f(t)dt$ , exists because it can be written as the sum of integrals over which $e^{-st}f(t)$ is continuous. The fact that $e^{-st}f(t)$ is piecewise continuous on $[T,\infty )$ is also used to show that the second integral, ${\int }_T^{\infty }{e}^{-st}f(t)dt$ , converges. Because there are constants C and b such that $|f(t)|⩽C{e}^{bt}$ , we have

$\begin{array}{cc}\hfill & |{\int }_T^{\infty }{e}^{-st}f(t)dt|\hfill \\ \hfill & ⩽{\int }_T^{\infty }|e^{-st}f(t)|dt\hfill \\ \hfill & ⩽C{\int }_T^{\infty }{e}^{-st}{e}^{bt}dt=C{\int }_T^{\infty }{e}^{-(s-b)t}dt\hfill \\ \hfill & =C\underset{M\to \infty }{\lim }{\int }_T^M{e}^{-(s-b)t}dt\hfill \\ \hfill & =C\underset{M\to \infty }{\lim }{[-\frac{1}{s-b}{e}^{-(s-b)t}]}_{t=T}^{t=M}\hfill \\ \hfill & =-\frac{C}{s-b}\underset{M\to \infty }{\lim }(e^{-(s-b)M}-e^{-(s-b)T}).\hfill \end{array}$

Then, if $s-b>0$ , ${\lim }_{M\to \infty }{e}^{-(s-b)M}=0$ , so

$|{\int }_T^{\infty }{e}^{-st}f(t)dt|⩽\frac{C}{s-b}{e}^{-(s-b)T},s>b.$

Because both of the integrals ${\int }_0^T{e}^{-st}f(t)dt$ and ${\int }_T^{\infty }{e}^{-st}f(t)dt$ exist, ${\int }_0^{\infty }{e}^{-st}f(t)dt$ also exists for $s>b$ . □

Example 8.7

Find the Laplace transform of $f(t)=\{\begin{array}{c}-1,0⩽t<4\hfill \\ 1,t⩾4\hfill \end{array}$ .

Solution: Because $y=f(t)$ is a piecewise continuous function on $[0,\infty )$ and of exponential order, $\mathcal{L}\{f(t)\}$ exists. We use the definition and evaluate the integral using the sum of two integrals. We assume that $s>0$ :

$\begin{array}{cc}\hfill \mathcal{L}\{f(t)\}& ={\int }_0^{\infty }f(t)e^{-st}dt\hfill \\ \hfill & ={\int }_0^4-1\cdot e^{-st}dt+{\int }_4^{\infty }{e}^{-st}dt\hfill \\ \hfill & ={\left[\frac{1}{s}{e}^{-st}\right]}_{t=0}^{t=4}+\underset{M\to \infty }{\lim }{[-\frac{1}{s}{e}^{-st}]}_{t=4}^{t=M}\hfill \\ \hfill & =\frac{1}{s}(e^{-4s}-1)-\frac{1}{s}\underset{M\to \infty }{\lim }(e^{-Ms}-e^{-4s})\hfill \\ \hfill & =\frac{1}{s}(2e^{-4s}-1).\text{\hspace{1em}□}\hfill \end{array}$

Theorem 8.2 gives a sufficient condition and not a necessary condition for the existence of the Laplace transform. In other words, there are functions such as $f(t)=t^{-1/2}$ that do not satisfy the hypotheses of the theorem for which the Laplace transform exists. (See Exercises 62 and 63.)

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Analytical, Approximate-Analytical and Numerical Methods in the Design of Energy Analyzers

Victor S. Gurov , ... Andrey A. Trubitsyn , in Advances in Imaging and Electron Physics, 2015

1 Methods for Numerical Simulation of Electrostatic Fields

Calculating the potential distribution in an electron/ion-optical system consists of solving the Dirichlet problem for the Laplace equation (or the Poisson equation if the space charge is taken into account) in a certain domain, with the known values of potentials defined on the boundary electrodes. The general mathematical formulation of the problem is as described next.

It is necessary to solve the boundary-value problem for the Poisson equation in the domain Ω with the boundary Γ:

(1) $\mathrm{\Delta }u\left(\xi \right)=\rho ,\xi \in \Omega \mathit{\cup }\mathit{\Gamma }$

with the boundary condition

(2) $u\left(\xi \right)=\overline{u}\left(\chi \right),\chi \in \mathit{\Gamma },$

where ū(χ ) is a piecewise continuous function and ρ is the space charge density.

Three methods are currently popular as basic approaches to the numerical solution of this problem: the finite difference method (FDM), the finite element method (FEM), and the boundary element method (BEM).

FDM (see, for example, Morton & Mayers, 2005) can be most efficiently employed to calculate the fields in the systems with comparatively simple configurations of electrodes, when the nodes of partition mesh fit well the boundary of the calculation domain. The external problem requires introducing a closing surface with artificially imposed boundary conditions—for example, the condition of vanishing of the normal derivative of potential (Manura & Dahl, 2006). Among the finite-difference methods, the estimations of the convergence rate give preference to iterative methods, such as the longitudinal-transverse sweep method with optimal sequence of parameters and the alternating triangular method with Chebyshev's acceleration (Samarskiy, 1983).

FEM (Zienkiewicz, Taylor, & Zhu, 2005) can be successfully used to study systems with a complicated geometry of electrodes. However, irregular arrangement of the mesh nodes for potential calculation complicates the trajectory analysis, which limits the FEM application mainly due to internal problems.

BEM (see, for example, Wrobel & Aliabadi, 2002) represents a genuine numerical realization of the well-known method of integral equations (Cheng & Cheng, 2005) and may be considered the most advanced of the numerical methods of field calculation that have appeared in the computer age. The main advantages of BEM is its simplicity of algorithmization, high accuracy, ability to solve both external and internal problems of potential theory with virtually arbitrary boundary geometry and the presence of different-scale geometry elements.

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