On Which of the Following Intervals is F Continuous the Function F Has a Jump Discontinuity at

Laplace Transform Methods

Martha L. Abell , James P. Braselton , in Differential Equations with Mathematica (Fourth Edition), 2016

8.1.2 Exponential Order, Jump Discontinuities and Piecewise-Continuous Functions

In our first calculus course, we learn that some improper integrals diverge, which indicates that the Laplace transform may not exist for some functions. 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, M > 0, and T > 0 such that

$\begin{array}{l}\hfill \left|f(t)\right|\le M{e}^{bt}\end{array}$

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 ${lim}_{t\to c^{+}}f(t)$ and ${lim}_{t\to c^{-}}f(t)$ are finite, but unequal, values. The function 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). The function 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 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 $[0,\infty )$ if y = f(t) is piecewise continuous on [0, N] for all N.

Theorem 19

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

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, $\mathcal{L}\left\{f(t)\right\}$ exists for s > b.

Example 8.1.6

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

Solution

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

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

Using Cases (\;), we define and graph this piecewise-defined function in Figure 8-1. Because Mathematica uses a "connect the dots" scheme when graphing functions, Mathematica does not automatically detect the discontinuity at t = 4. However, when we use the Exclusions option, we instruct Mathematica to avoid t = 4.

Figure 8-1. Plot of a piecewise-defined function

Clear[f]f [t_]:=−1/;0≤t≤4 f[t_]:=1/;t>4

p1=Plot[f[t], {t, 0, 8}, PlotLabel→"(a)"]

p2=Plot[f[t], {t, 0, 8}, Exclusions→{t==4}, PlotLabel→"(b)"]

Show[GraphicsRow[{p1, p2}]]

However, the LaplaceTransform command is unable to compute the Laplace transform of y = f(t) when f is defined in this manner. To compute the Laplace transform using Mathematica, we take advantage of the UnitStep function, which is defined by

$\begin{array}{l}\hfill \mathrm{UnitStep}[t]=\left\{\begin{array}{l}0,t<0\\ 1,t\ge 0\end{array}\right..\end{array}$

Thus, y = f(t) is given by UnitStep[t-4]-UnitStep[4-t]. After defining y = f(t) in this manner, we see that LaplaceTransform is then able to compute $\mathcal{L}\left\{f(t)\right\}$ .

Note that the commands UnitStep and Heaviside have nearly identical functionality.

Clear[f] f[t_]=UnitStep[t−4]−UnitStep[4−t]; LaplaceTransform[f[t], t, s]

$\frac{e^{-4s}}{s}-\frac{1-e^{-4s}}{s}$

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Laplace transform methods

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

8.1.2 Exponential order, jump discontinuities, and piecewise continuous functions

In the introductory calculus courses, we learn that some improper integrals diverge, which indicates that the Laplace transform may not exist for some functions. 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, $M>0$ , and $T>0$ such that

$|f(t)|⩽M{e}^{bt}$

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. The function $y=f(t)$ has a jump discontinuity at $t=a$ if ${\lim }_{t\to a^{+}}f(t)$ is a finite value different from $f(a)$ . The function $y=f(t)$ 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 $s>b$ .

Example 8.1.6

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:

$\mathcal{L}\{f(t)\}={\int }_0^{\infty }f(t)e^{-st}dt={\int }_0^4-1\cdot e^{-st}dt+{\int }_4^{\infty }{e}^{-st}dt={\left[\frac{1}{s}{e}^{-st}\right]}_{t=0}^{t=4}+\underset{M\to \infty }{\lim }{[-\frac{1}{s}{e}^{-st}]}_{t=4}^{t=M}=\frac{1}{s}(e^{-4s}-1)-\frac{1}{s}\underset{M\to \infty }{\lim }(e^{-Ms}-e^{-4s})=\frac{1}{s}(2e^{-4s}-1).$

Using Cases (\;), we define and graph this piecewise defined function in Fig. 8.1. Because Mathematica uses a "connect the dots" scheme when graphing functions, Mathematica does not automatically detect the discontinuity at $t=4$ . However, when we use the Exclusions option, we instruct Mathematica to avoid $t=4$ .

Figure 8.1. Plot of a piecewise defined function.

$\mathbf{\text{Clear}}\mathbf{[}\mathit{f}\mathbf{]}$

$\mathit{f}\mathbf{[}\mathbf{\text{t}}\mathrm{\_}\mathbf{]}\mathbf{\text{:=}}\mathbf{-}\mathbf{1}\mathbf{\text{/;}}\mathbf{0}\mathbf{⩽}\mathit{t}\mathbf{⩽}\mathbf{4}$

$\mathit{f}\mathbf{[}\mathbf{\text{t}}\mathrm{\_}\mathbf{]}\mathbf{\text{:=}}\mathbf{1}\mathbf{\text{/;}}\mathit{t}\mathbf{>}\mathbf{4}$

$\mathbf{\text{p1}}\mathbf{=}\mathbf{\text{Plot}}\mathbf{[}\mathit{f}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{,}\mathbf{\{}\mathit{t}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{8}\mathbf{\}}\mathbf{,}$

$\mathbf{\text{PlotLabel}}\mathbf{\to }\mathbf{\text{"(a)"}}\mathbf{]}$

$\mathbf{\text{p2}}\mathbf{=}\mathbf{\text{Plot}}\mathbf{[}\mathit{f}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{,}\mathbf{\{}\mathit{t}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{8}\mathbf{\}}\mathbf{,}\mathbf{\text{Exclusions}}\mathbf{\to }\mathbf{\{}\mathit{t}\mathbf{=}\mathbf{=}\mathbf{4}\mathbf{\}}\mathbf{,}$

$\mathbf{\text{PlotLabel}}\mathbf{\to }\mathbf{\text{"(b)"}}\mathbf{]}$

$\mathbf{\text{Show}}\mathbf{[}\mathbf{\text{GraphicsRow}}\mathbf{[}\mathbf{\{}\mathbf{\text{p1}}\mathbf{,}\mathbf{\text{p2}}\mathbf{\}}\mathbf{]}\mathbf{]}$

However, the LaplaceTransform command is unable to compute the Laplace transform of $y=f(t)$ when f is defined in this manner. To compute the Laplace transform using Mathematica, we take advantage of the UnitStep function, which is defined by

$\mathtt{UnitStep}[\mathtt{t}]=\{\begin{array}{c}0,t<0,\hfill \\ 1,t⩾0.\hfill \end{array}$

Thus, $y=f(t)$ is given by UnitStep[t-4]-UnitStep[4-t]. After defining $y=f(t)$ in this manner, we see that LaplaceTransform is then able to compute $\mathcal{L}\{f(t)\}$ .

Note that the commands UnitStep and Heaviside have nearly identical functionality.

$\mathbf{\text{Clear}}\mathbf{[}\mathit{f}\mathbf{]}$

$\mathit{f}\mathbf{[}\mathbf{\text{t}}\mathrm{\_}\mathbf{]}\mathbf{=}\mathbf{\text{UnitStep}}\mathbf{[}\mathit{t}\mathbf{-}\mathbf{4}\mathbf{]}\mathbf{-}\mathbf{\text{UnitStep}}\mathbf{[}\mathbf{4}\mathbf{-}\mathit{t}\mathbf{]}\mathbf{;}$

$\mathbf{\text{LaplaceTransform}}\mathbf{[}\mathit{f}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{,}\mathit{t}\mathbf{,}\mathit{s}\mathbf{]}$

$\frac{e^{-4s}}{s}-\frac{1-e^{-4s}}{s}$   □

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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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Handbook of Numerical Methods for Hyperbolic Problems

P. Bochev , M. Gunzburger , in Handbook of Numerical Analysis, 2016

8.2 A Feedback LSFEM

The feedback least-squares method of Choi (2000) for the model advection–reaction problem (13) is our second example of a least-squares formulation that aims to combine the best properties of Banach and Hilbert space settings. The method itself had been prompted by the adaptively reweighted LSFEM in the last section and so, the two methods are close relatives. The key differences between the methods include (i) the use of a statistical approach by the feedback method, adopted from Carmo and Galeao (1991), to locate the discontinuity, and (ii) a weight function based on the solution gradient rather than on its residual.

Assume that g has a jump discontinuity at x ∈ Γ₋ that is propagated by the solution ϕ along the characteristic $\chi \subset \mathrm{\Omega }$ . Similarly to the iteratively reweighted method from Section 8.1, the feedback LSFEM relies on the residual of the finite element solution ϕ ^h to locate χ. However, in the feedback method, the unweighting of the LSF is confined to a discontinuity set M _χ containing the elements that are near χ. To define this set, each element $\kappa \in {\mathcal{T}}_h$ is ranked using the following quantities.

1.

The mean residual for the element κ:

$\ll R{\gg }_{\kappa }={\int }_{\kappa }|R_{\kappa }({\varphi }^h)|/\mu (\kappa )d\mathrm{\Omega }\forall \kappa \in {\mathcal{T}}_h,$

where μ(κ) is the element measure and $R_{\kappa }(\varphi )=\left(\nabla \cdot (\mathit{b}{\varphi }^h)+c{\varphi }^h-f\right){|}_{\kappa }$ is the element residual.

2.

The mean residual for the finite element partition ${\mathcal{T}}_h$ :

$\ll R{\gg }_h=\frac{1}{N_e}\sum _{\kappa \in {\mathcal{T}}_h}\ll R{\gg }_{\kappa },$

where N _e is the number of elements in ${\mathcal{T}}_h.$

3.

The mean deviation for ${\mathcal{T}}_h$ :

$\ll \delta {\gg }_h={\left(\frac{1}{N_e}\sum _{\kappa \in {\mathcal{T}}_h}{\left(\ll R{\gg }_h-\ll R{\gg }_{\kappa }\right)}^2\right)}^{1/2}.$

An initial set M ₀ is defined by including all elements κ whose mean element residual exceeds the mean residual for ${\mathcal{T}}_h$ , plus a term proportional to the mean deviation:

(51) $\begin{array}{l}\hfill M_0=\{\kappa \in {\mathcal{T}}_h|\ll R{\gg }_{\kappa }\ge \ll R{\gg }_h+ϵ\ll \delta {\gg }_h\}.\end{array}$

The positive parameter ϵ may be used to adjust the sensitivity of this detection criterion. For triangular partitions ${\mathcal{T}}_h$ , the set M _χ is constructed from M ₀ using the following recursive process (see Fig. 1).

Fig. 1. The shaded triangles in the leftmost plot are in the set M ₀. The centre plot shows an intermediate set obtained at step 2. The rightmost plot shows the set M _χ .

1.

Tag all elements in M ₀ by 1 and all elements in ${\mathcal{T}}_h\setminus M_0$ by 0.

2.

Update element tags according to the following rules: initialize m _Ω = 3;

–

if κ has tag 0 and m _Ω adjacent elements have tags 1, set the tag of κ to 1;

–

if κ has tag 1 and all adjacent elements have tag 0, set the tag of κ to 0;

–

set m _Ω = 2 and repeat until no tags change.

3.

M _χ is the set of all elements whose tag equals 1.

This procedure is used to define the feedback LSFEM as follows.

1.

Set ω _κ = 1 for all $\kappa \in {\mathcal{T}}_h$ and compute a minimizer ${\varphi }_0^h$ of (50).

2.

Use ${\varphi }_0^h$ to construct the set M _χ .

3.

Set

${\omega }_{\kappa }=\left\{\begin{array}{ll}1& \text{for}\kappa \notin M_{\chi }\\ \frac{1}{1+\alpha \parallel \nabla {\varphi }_0^h{\parallel }_{\kappa }}& \text{for}\kappa \in M_{\chi },\end{array}\right.$

where α is an amplifying constant (see Jiang, 1993) and solve (50).

If necessary, step 3 or steps 2–3 can be repeated until the relative error between two consecutive solutions becomes smaller than some prescribed tolerance.

Usually, the construction of M _χ from M ₀ takes just few steps which are cheaper than solving the LSFEM problem itself. As a result, the feedback method tends to be more efficient than the iteratively reweighted approach for which the first couple of iterations are essentially used to determine the same information as provided by the discontinuity set M _χ .

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COMPUTATION FOR TRANSIENT AND IMPACT DYNAMICS

D.J. Benson , J. Hallquist , in Encyclopedia of Vibration, 2001

Shock viscosity

Shocks propagate via a thermodynamically irreversible process and appear as jump discontinuities in the solution variables. Their accurate resolution and propagation are critical for solving high-speed impact problems. Oscillations will occur behind the jump in the stress unless some form of damping, which is called the shock viscosity, is included in the calculation. The physical thickness of a shock is typically on the order of microns, but in computational practice, they are smeared over three to six elements regardless of the element size. Although this introduces a large error in the shock width itself, the critical aspects of the shock, namely its speed and the stress states on either side of it, are accurately calculated.

The standard form of shock viscosity, q, a function of the volume strain rate, ${\stackrel{.}{\varepsilon }}_{\upsilon }$ , is:

(14) $q=-\rho l{\stackrel{.}{\varepsilon }}_{\upsilon }(c_{1}C+c_2|l{\stackrel{.}{\varepsilon }}_{\upsilon }|)$

and it differs little from the one originally introduced by von Neumann and Richtmeyer to solve shock problems in the design of the atom bomb (they did not include the linear term). The shock viscosity is treated like a contribution to the pressure for the calculation of the nodal forces.

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Handbook of Numerical Methods for Hyperbolic Problems

U.S. Fjordholm , in Handbook of Numerical Analysis, 2016

4.5.1 Uniform kth-Order Accuracy up to Discontinuities

If v is a piecewise $C^{\infty }$ function with finitely many jump discontinuities ("shocks"), then for sufficiently small Δx, the ENO reconstruction is a kth-order approximation of v in all cells not containing a shock (Harten et al., 1986). Indeed, if Δx is sufficiently small, then there are at least k cells in-between the shocked cells. Moreover, the ℓth divided difference $[{\overline{v}}_s,\dots ,{\overline{v}}_{s+\ell }]$ over any stencil containing a shocked cell behaves as O(Δx ^−ℓ ). Thus, if Δx is small enough then in every nonshocked cell, the ENO stencil selection procedure can, and will, select an ENO stencil {s _i ,…, s _i + k − 1}not containing a shock. The property of uniform kth-order accuracy then follows as in Section 4.1.3.

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Inference for nonsmooth regression curves and surfaces using kernel-based methods

I. Gijbels , in Recent Advances and Trends in Nonparametric Statistics, 2003

4.1 Indirect methods

As explained in the introduction, once estimates for the locations of the jump discontinuity point are obtained, say ${\widehat{\mathit{\tau }}}_1,\cdots ,{\widehat{\mathit{\tau }}}_{\mathrm{k}}$ , the interval of estimation is divided into k +   1 intervals $[0,{\widehat{\mathit{\tau }}}_1[,[{\widehat{\mathit{\tau }}}_1,{\widehat{\mathit{\tau }}}_2[,\cdots ,\left[{\widehat{\mathit{\tau }}}_k,1\right]$ and on each segment the regression function is continuous and estimated using standard smoothing techniques (such as for example a local linear smoother with cross-validation bandwidth).

An alternative indirect method has been proposed by Kang, Koo and Park (2000) in the fixed design regression case. They suggest to estimate the location τ and γ via (3) and (4) respectively, to obtain suitably adjusted data

$Z_i=Y_i-\widehat{\mathrm{\gamma }}{1}_{\left[\widehat{\mathit{\tau }},1\right]}\left(x_i\right),i=1,\cdots ,n$

and then to use standard kernel regression methods to estimate the function g ₀ in (2).

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Symmetric Hyperbolic Systems and Shock Waves

S. Kichenassamy , in Encyclopedia of Mathematical Physics, 2006

Jump Discontinuities: Shock Waves

A "shock wave" is a weak solution of a system of conservation laws admitting a jump discontinuity. By definition, weak solutions satisfy, for any smooth function ${\varphi }_A(x)$ with compact support,

$\int \int \{f^{A\alpha }{\partial }_{\alpha }{\varphi }_A+N^A\varphi A\}\text{d}t\text{d}\mathit{x}=0$

The theory of shock waves is an attempt to understand solutions of conservation laws which are limits of solutions of diffusion equations; the hope is that the influence of second-derivative terms is appreciable only near shocks, and that, for given initial data, there is a unique weak solution of the conservation law which may be obtained as such a limit, if modeling has been done correctly. This problem may be difficult already for a single shock ("shock structure").

The theory of shock waves follows the one-dimensional theory closely. We therefore describe the main facts for a conservation law in one space dimension $\left(u=u\left(t,x\right)\right)$ :

${\partial }_{t}u+{\partial }_{x}f\left(u\right)=0$

If a shock travels at speed c, the weak formulation of the equations gives the Rankine–Hugoniot relation $c\left[u\right]=\left[f\left(u\right)\right]$ , where square brackets denote jumps. There may be several weak solutions having the same initial condition. One restricts solutions by making two further requirements: (1) the system admits an entropy pair $\left(U,F\right)$ with a convex entropy and (2) to be admissible, weak solutions must be limits of "viscous approximations"

${\partial }_{t}u+{\partial }_{x}f\left(u\right)=\epsilon {\partial }_x^{2}u$

as $\epsilon \to 0$ . One then finds easily that the entropy equality $\left({\partial }_{t}U+{\partial }_{x}F=0\right)$ must be replaced, for such weak solutions, by the entropy condition: ${\partial }_{t}U+{\partial }_{x}F\le 0$ in the weak sense. This condition admits a concrete interpretation if the gradient of each characteristic speed is never orthogonal to the corresponding right eigenvector ("genuine nonlinearity"); in that case, characteristics must impinge on the shock ("shock inequalities").

For the equations of gas dynamics with polytropic law $\left(p{\upsilon }^{\text{γ}}=\text{const}.\right)$ , there is a unique solution with initial condition $u=u_l$ for $x<0,u=u_r$ for $x>0$ , where $u_l$ and $u_r$ are constant ("Riemann problem") which satisfies the entropy condition, provided $|u_l-u_r|$ is small. More generally, if the equation of state $p=p\left(v,s\right)>0$ satisfies $\partial p/\partial v<0$ and ${\partial }^{2}p/\partial v^2>0$ , the shock inequalities are equivalent to the fact that the entropy increases after the passage of a shock with $|u_l-u_r|$ small.

On the numerical side, one should mention: (1) the widely used idea of upstream differencing; (2) the Lax–Wendroff scheme, the complete analysis of which requires tools from soliton theory; and (3) the availability of general results for dissipative schemes for SH systems.

Recent trends include: (1) admissibility conditions when genuine nonlinearity does not hold and (2) other approximations of shock wave problems, most notably kinetic formulations.

Some of the ideas of shock wave theory have been applied to Hamilton–Jacobi equations and to motion by mean curvature, with applications to front propagation problems and "computer vision."

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The Dirac Delta Function

R.F. Hoskins Research Professor , in Delta Functions (Second Edition), 2011

2.2.1 Naive definition of delta function

Suppose that we try to extend the definition of differentiation in such a way that it applies to functions with jump discontinuities. In particular we would need to define a 'derivative' for the unit step function, u. For all t ≠ 0 this is of course well-defined in the classical sense as:

$u^{\prime }\left(t\right)=0\mathit{for}\mathit{all}t\ne 0\text{,}$

corresponding to the obvious fact that the graph of y = u(t) has zero slope for all non-zero values of t. At t = 0, however, there is a jump discontinuity, and the definition of derivative accordingly fails. A glance at the graph suggests that it would not be unreasonable to describe the slope as 'infinite' at this point. Moreover, if we take any specific representation, u_c, of the unit step, then the ratio $\frac{u_c\left(h\right)-u_c\left(0\right)}{h}$ becomes arbitrarily large as h approaches 0, or using the familiar convention,

$\underset{h\to 0}{lim}\frac{u_c\left(h\right)-u_c\left(0\right)}{h}=+\infty \text{.}$

Thus, from a descriptive point of view at least, the derivative of u would appear to be a function which (for the moment) we shall denote by δ(t) and which has the following pointwise specification:

(2.7a) $\delta \left(t\right)\equiv u^{\prime }\left(t\right)=0,\mathit{for}\mathit{all}t\ne 0\text{,}$

and, once again adopting the conventional use of the ∞ sign,

(2.7b) $\delta \left(0\right)=+\infty \text{.}$

However, as will shortly become clear, we shall need eventually to consider more carefully the significance of the symbol δ(t). Does it really denote a function, in the proper sense of the word, and in what sense does it represent the derivative of the function u(t)?

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Properties of the Delta Function

R.F. Hoskins Research Professor , in Delta Functions (Second Edition), 2011

3.3.1 Differentiation at jump discontinuities

We can now attach meaning to the term 'derivative' in the case of a function which has one or more jump discontinuities. Let ϕ ₁ and ϕ ₂ be continuously differentiable functions and let f be defined as the function which is equal to ϕ ₁ for all t  < a and equal to ϕ ₂ for all t  > a:

$f\left(t\right)={\varphi }_1\left(t\right)u\left(a-t\right)+{\varphi }_2\left(t\right)u\left(t-a\right)\text{.}$

The function f(t) may not be defined at the point t  = a itself, but we surely have f(a−)   = ϕ ₁(a) and f(a+)   = ϕ ₂(a).

Then.

$\begin{array}{l}\frac{d}{\mathit{dt}}f\left(t\right)=u\left(a-t\right)\frac{d}{\mathit{dt}}{\varphi }_1\left(t\right)+{\varphi }_1\left(t\right)\frac{d}{\mathit{dt}}u\left(a-t\right)\\ +u\left(t-a\right)\frac{d}{\mathit{dt}}{\varphi }_2\left(t\right)+{\varphi }_2\left(t\right)\frac{d}{\mathit{dt}}u\left(t-a\right)\text{.}\end{array}$

Now,

$\frac{d}{\mathit{dt}}u\left(t-a\right)=\frac{\mathit{du}\left(t-a\right)}{d\left(t-a\right)}\frac{d\left(t-a\right)}{\mathit{dt}}=\delta \left(t-a\right)\equiv {\delta }_a\left(t\right)$

and

$\frac{d}{\mathit{dt}}u\left(a-t\right)=\frac{\mathit{du}\left(a-t\right)}{d\left(a-t\right)}\frac{d\left(a-t\right)}{\mathit{dt}}=-\delta \left(a-t\right)\equiv -{\delta }_a\left(t\right)\text{.}$

Hence,

(3.16) $\begin{array}{l}\frac{d}{\mathit{dt}}f\left(t\right)={\varphi }_1^{\prime }\left(t\right)u\left(a-t\right)+{\varphi }_2^{\prime }\left(t\right)u\left(t-a\right)-{\varphi }_1\left(t\right){\delta }_a\left(t\right)+{\varphi }_2\left(t\right){\delta }_a\left(t\right)\\ ={\varphi }_1^{\prime }\left(t\right)u\left(a-t\right)+{\varphi }_2^{\prime }\left(t\right)u\left(t-a\right)+\left[{\varphi }_2\left(a\right)={\varphi }_1\left(a\right)\right]{\delta }_a\left(t\right)\text{.}\end{array}$

The discontinuity at t  = a thus gives rise to a delta function at t  = a multiplied by the saltus of the function f at that point, that is to say, by the number k  = f(a+)   − f(a−). In other words we have a delta function of strength k at t  = a. This means that integration of (3.16) will reintroduce the jump discontinuity in the primitive, f.

In particular, suppose that ϕ ₁(t)   = ϕ(t) and ϕ ₂(t)   = ϕ(t)   + k. Then (3.16) becomes

$\begin{array}{c}\frac{d}{\mathit{dt}}f\left(t\right)={\varphi }^{\prime }\left(t\right)u\left(a-t\right){\varphi }^{\prime }\left(t\right)u\left(t-a\right)+\left[\left(\varphi \left(a\right)+k\right)-\varphi \left(a\right)\right]{\delta }_a\left(t\right)\\ ={\varphi }^{\prime }\left(t\right)+k{\delta }_a\left(t\right)\text{.}\end{array}$

Further, if x and t are any two numbers such that x  < a  < t then we get

$\begin{array}{c}{\displaystyle {\int }_x^t\frac{d}{\mathit{d\tau }}f\left(\tau \right)\mathit{d\tau }={\displaystyle {\int }_x^a{\varphi }^{\prime }}\left(\tau \right)\mathit{d\tau }+{\displaystyle {\int }_a^t{\varphi }^{\prime }\left(\tau \right)\mathit{d\tau }+k{\displaystyle {\int }_x^t\delta \left(\tau -a\right)\mathit{d\tau }}}}\\ =\varphi \left(a\right)-\varphi \left(x\right)+\varphi \left(t\right)-\varphi \left(a\right)+k=f\left(t\right)-f\left(x\right)\text{,}\end{array}$

the integral involving the delta function at t  = a being well-defined precisely because we have x  < a  < t. (See Sec.3.6.2 below.)

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