Suppose That F and F Are Continuous and of Exponential Order T to Infinity Use Integration by Parts

Elliptic Equations Involving Measures

Laurent Véron , in Handbook of Differential Equations: Stationary Partial Differential Equations, 2004

Proposition 3.41

Assume g has positive and finite exponential order of growth at infinity, a ₊(g). Let R > 0 and v ∈ ${\mathfrak{M}}^b$ (B_R (0)) with no atom. If c > 4π/a ₊(g) there exists no function u ∈ L ¹(B_R (0)) such that g(u) ∈ L ¹(B_R (0)) and

(3.134) $\begin{array}{l}{\displaystyle {\int }_{B_R(0)}\left(-u\Delta \zeta +g(u)\zeta \right)\text{d}x}\hfill \\ \begin{array}{cc}=c\zeta (0)+{\displaystyle {\int }_{B_R(0)}\zeta \text{d}v}& \forall \zeta \in C_c^{\infty }\left(B_R(0)\right).\end{array}\hfill \end{array}$

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Laplace Transform Methods for Partial Differential Equations

George A. Articolo , in Partial Differential Equations & Boundary Value Problems with Maple (Second Edition), 2009

10.3 Inverse Transform and Convolution Integral

If the function f(t ) satisfies the conditions of exponential order and piecewise continuity, then it can be shown that there exists a uniqueness between the function and its inverse. From complex analysis, it can be shown that we can retrieve the function f(t) from its transform F(s) by performing the following inverse operation:

$f\left(t\right)=\underset{a-i\infty }{\overset{a+i\infty }{\int }}\frac{F\left(s\right){\text{e}}^{st}}{2\pi i}\text{d}s$

This inverse operation is an integration over the variable s in the complex s-plane along the line Re{s} = a, where a is any point greater than σ, the "abscissa of convergence" of f(t). We do not intend to evaluate inverses from integrations in the complex plane here, since this would take us too far from our immediate intentions.

Generally, inverse transforms are evaluated by doing partial fraction expansions on F(s) and then relating the expansion terms to previously evaluated transforms. An alternative procedure is to use a convolution method similar to that developed in Chapter 9.

For Laplace transforms, the convolution integral reads as follows. Let F(s) be the Laplace transform of f(t) and assume that it can be partitioned into a product of two functions F1(s) and F2(s); that is,

$F\left(s\right)=F1\left(s\right)F2\left(s\right)$

where F1(s) is the Laplace transform of f1(t) and F2(s) is the Laplace transform of f2(t):

$F1\left(s\right)=\underset{0}{\overset{\infty }{\int }}f1\left(t\right){\text{e}}^{-st}\text{d}t$

Of course, f1(t) and f2(t) must both be piecewise continuous and of exponential order. It can be shown (see related references) that f(t), the inversion of F(s), can be expressed as the convolution integral

$f\left(t\right)=\underset{0}{\overset{t}{\int }}f1\left(\tau \right)f2\left(t-\tau \right)\text{d}\tau$

We take advantage of the convolution integral in evaluating some inverses in some of the following problems.

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Introduction to the Laplace Transform

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

8.3 Solving Initial-Value Problems with the Laplace Transform

In this section we show how the Laplace transform is used to solve initial value problems. To do this we first need to understand how the Laplace transform of the derivatives of a function relates to the function itself. We begin with the first derivative.

Theorem 8.8 Laplace Transform of the First Derivative

Suppose that $f(t)$ is continuous for all $t⩾0$ and is of exponential order b for $t>T$ . Also, suppose that $f^{\prime }(t)$ is piecewise continuous on any closed subinterval of $[0,\infty )$ . Then, for $s>b$

$\mathcal{L}\{f^{\prime }(t)\}=s\mathcal{L}\{f(t)\}-f(0).$

Proof

Using integration by parts with $u=e^{-st}$ and $dv=f^{\prime }(t)dt$ , we have

$\begin{array}{cc}\hfill & \mathcal{L}\{f^{\prime }(t)\}\hfill \\ \hfill & ={\int }_0^{\infty }{e}^{-st}{f}^{\prime }(t)dt\hfill \\ \hfill & =\underset{M\to \infty }{\lim }(e^{-st}f(t){|}_{t=0}^{t=M}+s{\int }_0^{\infty }{e}^{-st}f(t)dt)\hfill \\ \hfill & =-f(0)+s\mathcal{L}\{f(t)\}\hfill \\ \hfill & =s\mathcal{L}\{f(t)\}-f(0).\hfill \end{array}$

 □

Proof of Theorem 8.8 assumes that $f^{\prime }$ is a continuous function. If we use the assumption that $f^{\prime }$ is continuous on $0<t_1<t_2<\cdots t_n<\infty$ , we complete the proof by using

$\begin{array}{cc}\hfill \mathcal{L}\{f^{\prime }(t)\}& ={\int }_0^{t_1}{e}^{-st}{f}^{\prime }(t)dt+{\int }_{t_1}^{t_2}{e}^{-st}{f}^{\prime }(t)dt\hfill \\ \hfill & +\cdots +{\int }_{t_n}^{\infty }{e}^{-st}{f}^{\prime }(t)dt.\hfill \end{array}$

This is the same integration by parts formula shown in the proof of Theorem 8.8 for each integral. Now we make the same assumptions of $f^{\prime }$ and $f^{″}$ as we did of f and $f^{\prime }$ , respectively, in the statement of Theorem 8.8 and use Theorem 8.8 to develop an expression for $\mathcal{L}\{f^{″}(t)\}$ :

$\begin{array}{cc}\hfill \mathcal{L}\{f^{″}(t)\}& =s\mathcal{L}\{f^{\prime }(t)\}-f^{\prime }(0)\hfill \\ \hfill & =s[s\mathcal{L}\{f(t)\}-f(0)]-f^{\prime }(0)\hfill \\ \hfill & =s^2\mathcal{L}\{f(t)\}-sf(0)-f^{\prime }(0).\hfill \end{array}$

Continuing this process, we can construct similar expressions for the Laplace transform of higher-order derivatives, which leads to the following theorem.

Theorem 8.9 Laplace Transform of Higher Derivatives

More generally, if $f^{(i)}(t)$ is a continuous function of exponential order b on $[0,\infty )$ for $i=0$ , 1, …, $n-1$ and $f^{(n)}(t)$ is piecewise continuous on any closed subinterval of $[0,\infty )$ , then for $s>b$

$\begin{array}{cc}\hfill \mathcal{L}\{f^{(n)}(t)\}& =s^n\mathcal{L}\{f(t)\}-s^{n-1}f(0)\hfill \\ \hfill & -\cdots -s{f}^{(n-2)}(0)-f^{(n-1)}(0).\hfill \end{array}$

We will use this theorem and corollary in solving initial value problems. However, we can also use them to find the Laplace transform of a function when we know the Laplace transform of the derivative of the function.

Example 8.17

Find $\mathcal{L}\{{\sin }^{2}kt\}$ .

Solution: We can use the theorem to find the Laplace transform of $f(t)={\sin }^{2}kt$ . Notice that $f^{\prime }(t)=2k\sin kt\cos kt=k\sin 2kt$ . Then, because $\mathcal{L}\{f^{\prime }(t)\}=s\mathcal{L}\{f(t)\}-f(0)$ and

$\begin{array}{cc}\hfill \mathcal{L}\{f^{\prime }(t)\}& =\mathcal{L}\{k\sin 2kt\}=k\frac{2k}{s^2+{(2k)}^2}\hfill \\ \hfill & =\frac{2k^2}{s^2+4k^2},\hfill \end{array}$

we have $\frac{2k^2}{s^2+4k^2}=s\mathcal{L}\{f(t)\}-0$ . Therefore, $\mathcal{L}\{f(t)\}=\frac{2k^2}{s(s^2+4k^2)}$ .  □

We now show how the Laplace transform can be used to solve initial value problems. Typically, when we solve an initial value problem that involves $y(t)$ , we use the following steps.

1.

Compute the Laplace transform of each term in the differential equation;

2.

Solve the resulting equation for $\mathcal{L}\{y(t)\}=Y(s)$ ; and

3.

Determine $y(t)$ by computing the inverse Laplace transform of $Y(s)$ .

The advantage of this method is that through the use of the property

(8.8) $\begin{array}{cc}\hfill \mathcal{L}\{f^{(n)}(t)\}& =s^n\mathcal{L}\{f(t)\}-s^{n-1}f(0)-\cdots \hfill \\ \hfill & -s{f}^{(n-2)}(0)-f^{(n-1)}(0)\hfill \end{array}$

we change the differential equation to an algebraic equation that can be solved for $\mathcal{L}\{f(t)\}$ .

Example 8.18

Solve the initial value problem $y^{\prime }-4y=e^{4t}$ , $y(0)=0$ .

How does the solution change if $y(0)=1$ ?

Solution: We begin by taking the Laplace transform of both sides of the differential equation. Because $\mathcal{L}\{y^{\prime }\}=sY(s)-y(0)=sY(s)$ , we have

$\begin{array}{cc}\hfill \mathcal{L}\{y^{\prime }-4y\}& =\mathcal{L}\{e^{4t}\}\hfill \\ \hfill \mathcal{L}\{y^{\prime }\}-4\mathcal{L}\{y\}& =\frac{1}{s-4}\hfill \\ \hfill \underset{\mathcal{L}\{y^{\prime }\}}{\underbrace{sY(s)-y(0)}}-4\underset{\mathcal{L}\{y\}}{\underbrace{Y(s)}}& =\frac{1}{s-4}\hfill \\ \hfill (s-4)Y(s)& =\frac{1}{s-4}.\hfill \end{array}$

Solving this equation for $Y(s)$ yields $Y(s)=\frac{1}{{(s-4)}^2}$ . By using the shifting property with $\mathcal{L}\{t\}=1/s^2$ , we have $y(t)={\mathcal{L}}^{-1}\left\{\frac{1}{{(s-4)}^2}\right\}=t{e}^{4t}$ .  □

In many cases, we must determine a partial fraction decomposition of $Y(s)$ to obtain terms for which the inverse Laplace transform can be found.

Example 8.19

Solve the initial value problem $y^{″}-4y^{\prime }=0$ , $y(0)=3$ , $y^{\prime }(0)=8$ .

Solution: Let $Y(s)=\mathcal{L}\{y(t)\}$ . Then, applying the Laplace transform to the equation gives us $\mathcal{L}\{y^{″}-4y^{\prime }\}=\mathcal{L}\{0\}$ . Because

$\begin{array}{cc}\hfill & \mathcal{L}\{y^{″}-4y^{\prime }\}\hfill \\ \hfill & =\mathcal{L}\{y^{″}\}-4\mathcal{L}\{y^{\prime }\}\hfill \\ \hfill & =\underset{\mathcal{L}\{y^{″}\}}{\underbrace{s^{2}Y(s)-sy(0)-y^{\prime }(0)}}-4\underset{\mathcal{L}\{y^{\prime }\}}{\underbrace{(sY(s)-y(0))}}\hfill \end{array}$

and $\mathcal{L}\{0\}=0$ (why?), this equation becomes

$s^{2}Y(s)-sy(0)-y^{\prime }(0)-4sY(s)+4Y(0)=0.$

Applying the initial conditions $y(0)=3$ and $y^{\prime }(0)=8$ to this results in

$\begin{array}{cc}\hfill & s^{2}Y(s)-3s-8-4sY(s)+4\cdot 3=0\text{or}\hfill \\ \hfill & s(s-4)Y(s)=3s-4.\hfill \end{array}$

Solving for $Y(s)$ , we find that $Y(s)=\frac{3s-4}{s(s-4)}$ . If we expand the right-hand side of this equation in partial fractions based on the two linear factors in the denominator of $Y(s)$ , we obtain $\frac{3s-4}{s(s-4)}=\frac{A}{s}+\frac{B}{s-4}$ . Multiplying both sides of the equation by the denominator $s(s-4)$ , we have $3s-4=A(s-4)+Bs$ . If we substitute $s=0$ into this equation, we have $3\cdot 0-4=A\cdot (0-4)+B\cdot 0$ or $-4=-4A$ so that $A=1$ . Similarly, if we substitute $s=4$ , we find that $8=4B$ so that $B=2$ . Therefore, $\frac{3s-4}{s(s-4)}=\frac{1}{s}+\frac{2}{s-4}$ so

$y(t)={\mathcal{L}}^{-1}\{\frac{1}{s}+\frac{2}{s-4}\}=1+2e^{4t}.\text{\hspace{1em}□}$

A partial fraction decomposition involving a repeated linear factor is illustrated in the following example.

Example 8.20

Solve the initial value problem $y^{″}+2y^{\prime }+y=6$ , $y(0)=5$ , $y^{\prime }(0)=10$ .

Solution: Let $\mathcal{L}\{y(t)\}=Y(s)$ . Then,

$\begin{array}{cc}\hfill \mathcal{L}\{y^{″}+2y^{\prime }+y\}& =\mathcal{L}\{6\}\hfill \\ \hfill \mathcal{L}\{y^{″}\}+2\mathcal{L}\{y^{\prime }\}+\mathcal{L}\{y\}& =6/s\hfill \\ \hfill \underset{\mathcal{L}\{y^{″}\}}{\underbrace{s^{2}Y(s)-sy(0)-y^{\prime }(0)}}& \hfill \\ \hfill +2\underset{\mathcal{L}\{y^{\prime }\}}{\underbrace{(sY(s)-y(0))}}+Y(s)& =6/s\hfill \\ \hfill (s^2+2s+1)Y(s)& =6/s+5s+20\hfill \\ \hfill (s^2+2s+1)Y(s)& =(6+5s^2+20s)/s.\hfill \end{array}$

Solving for $Y(s)$ and factoring the denominator yields $Y(s)=\frac{5s^2+20s+6}{s{(s+1)}^2}$ . In this case, the denominator contains the nonrepeated linear factor s and the repeated linear factor $(s+1)$ . The partial fraction decomposition of $Y(s)$ is

$\frac{5s^2+20s+6}{s{(s+1)}^2}=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{{(s+1)}^2},$

so multiplication on each side of this equation by $s{(s+1)}^2$ results in the equation

$\begin{array}{cc}\hfill 5s^2+20s+6& =A{(s+1)}^2+Bs(s+1)+Cs\text{ or}\hfill \\ \hfill 5s^2+20s+6& =(A+B)s^2+(2A+B+C)s+A.\hfill \end{array}$

Equating the coefficients, we obtain the system

$\begin{array}{cc}\hfill A+B& =5\hfill \\ \hfill 2A+B+C& =20\hfill \\ \hfill A& =6,\hfill \end{array}$

which has solution $A=6$ , $B=-1$ , and $C=9$ so

$\frac{5s^2+20s+6}{s{(s+1)}^2}=\frac{6}{s}-\frac{1}{s+1}+\frac{9}{{(s+1)}^2}.$

Therefore, $y(t)={\mathcal{L}}^{-1}\{\frac{6}{s}-\frac{1}{s+1}+\frac{9}{{(s+1)}^2}\}=6-e^{-t}+9t{e}^{-t}$ .  □

Use Laplace transforms to solve $y^{\prime }-y=0$ .

In some cases, $F(s)$ involves irreducible quadratic factors as we see in the next example.

Example 8.21

Solve the initial value problem $y^{‴}+4y^{\prime }=-10e^t$ , $y(0)=2$ , $y^{\prime }(0)=2$ , $y^{″}(0)=-10$ .

Solution: Let $\mathcal{L}\{y(t)\}=Y(s)$ . Taking the Laplace transform of the equation and solving for $Y(s)$ gives us,

$\begin{array}{cc}\hfill & \mathcal{L}\{y^{‴}\}+4\mathcal{L}\{y^{\prime }\}=\mathcal{L}\{-10e^t\}\hfill \\ \hfill & \underset{\mathcal{L}\{y^{‴}\}}{\underbrace{s^{3}Y(s)-s^{2}y(0)-s{y}^{\prime }(0)-y^{″}(0)}}\hfill \\ \hfill & +4\underset{\mathcal{L}\{y^{\prime }\}}{\underbrace{(sY(s)-y(0))}}=-10/(s-1)\hfill \\ \hfill & (s^3+4s)Y(s)=-10/(s-1)+2s^2+2s-2\hfill \\ \hfill & Y(s)=-\frac{10}{(s-1)(s^3+4s)}+\frac{2s^2+2s-2}{s^3+4s}\hfill \\ \hfill & Y(s)=\frac{2s^3-4s-8}{s(s-1)(s^2+4)}.\hfill \end{array}$

Finding the partial fraction decomposition of the right-hand side of this equation, we obtain $\frac{2s^3-4s-8}{s(s-1)(s^2+4)}=\frac{A}{s}+\frac{B}{s-1}+\frac{Cs+D}{s^2+4}$ . Multiplying each side of this equation by $s(s-1)(s^2+4)$ gives us

$\begin{array}{cc}\hfill 2s^3-4s-8& =A(s-1)(s^2+4)+Bs(s^2+4)\hfill \\ \hfill & +(Cs+D)s(s-1)\hfill \\ \hfill 2s^3-4s-8& =(A+B+C)s^3+(-A-C+D)s^2\hfill \\ \hfill & +(4A+4B-D)s-4A.\hfill \end{array}$

Equating coefficients yields the system of equations

$\begin{array}{cc}\hfill A+B+C& =2\hfill \\ \hfill -A-C+D& =0\hfill \\ \hfill 4A+4B-D& =-4\hfill \\ \hfill -4A& =-8,\hfill \end{array}$

with solution $A=2$ , $B=-2$ , $C=2$ , and $D=4$ so $\frac{2s^3-4s-8}{s(s-1)(s^2+4)}=\frac{A}{s}+\frac{B}{s-1}+\frac{Cs+D}{s^2+4}=\frac{2}{s}-\frac{2}{s-1}+\frac{2s+4}{s^2+4}$ . Therefore,

$\begin{array}{cc}\hfill y(t)& ={\mathcal{L}}^{-1}\{\frac{2}{s}-\frac{2}{s-1}+\frac{2s+4}{s^2+4}\}\hfill \\ \hfill & =2{\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}-2{\mathcal{L}}^{-1}\left\{\frac{1}{s-1}\right\}\hfill \\ \hfill & +2{\mathcal{L}}^{-1}\left\{\frac{s}{s^2+4}\right\}+2{\mathcal{L}}^{-1}\left\{\frac{2}{s^2+4}\right\}\hfill \\ \hfill & =2-2e^t+2\cos 2t+2\sin 2t.\text{\hspace{1em}□}\hfill \end{array}$

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Tropical Radioecology

Keiko Tagami , ... M. Angelica V. Wasserman , in Radioactivity in the Environment, 2012

5.4.3 Biokinetic Models

Assuming constant environmental concentrations of a radionuclide, uptake kinetics in whole individuals are often described by a single-compartment, first-order exponential model:

(5.3) $C{F}_t=C{F}_{\mathit{ss}}(1-e^{-\mathrm{kt}})$

where CF _t and CF _ss represent the CFs at time t (in days, d) and at steady state, respectively, following the introduction of a constant concentration of radioactivity into the environment, and k is the biological loss rate constant (d ^{−   1}) (Whicker and Schultz, 1982). The CF _ss value is equivalent to the equilibrium TF value, defined above (Eq. 5.1).

Loss kinetics can be expressed in terms of percentage of remaining radioactivity, or the radioactivity at time t divided by the initial radioactivity measured in the organisms at the beginning of the depuration (loss) period, assuming that the organism is now growing in a radionuclide-free environment. The simplest loss kinetics are described by a single-component exponential model:

(5.4) $A_t=A_0{e}^{-\mathrm{kt}}$

where A _t and A ₀ are the activities (%) remaining at time t (measured in days, d) and at time zero (i.e., 100%), respectively, and k is the biological loss rate constant (d ^{−   1}). The determination of k allows the calculation of the radionuclide's biological half-life (T _b1/2=   ln2/k). Radioactive decay needs to be considered in these estimates when the isotopic half-life is sufficiently short to have an effect within the time frame of the biological process.

It should always be noted that these models are, at best, simplifications of the real situation. However, they do provide a basis for generic modelling of radionuclide biokinetics within quasi-natural systems and are often included as a component in more complex radiological dose assessment models (see Chapter 7).

A generalized model that accounts for multiple uptake pathways for any individual (e.g., from different prey organisms for a predator), in addition to distribution and subsequent loss from multiple tissue types within any organism, is described more fully in the Chapter 6. However, those more complex biokinetics models are equally pertinent to diverse terrestrial systems and may be amended and applied as required.

It should also be noted that the generalized models assume that the equilibrium concentration of a radionuclide in an organism is linearly proportional to the (bioavailable) concentration of that radionuclide in the supporting environment. That is, for twice the concentration in the environment the amount in the organism will be doubled. There is mounting evidence that this assumption is oversimplified. In many cases the proportionality will be less than a 1:1 relationship and may often be nonlinear (e.g., Yasuda and Uchida, 1995). In those cases, some have suggested the use of a power law function to describe the equilibrium plant concentration (e.g., Blanco Rodrıguez et al., 2006). Thus:

(5.5) $C_p=f{C}_s{}^g$

where C _p and C _s are the equilibrium radionuclide concentrations in plant and soil, respectively, and g is the proportionality constant. For linear TFs, g would be set to 1. Converting Equation 5.5 into TF notation gives the following:

(5.6) $\text{TF}=f{C}_s{{}^{(g}}^{-1)}$

Nonlinearity can, to some degree at least, be explained by changes in the proportion of bioavailable radioactivity decreasing with increasing concentration (e.g., DeForest et al., 2007). Another example is given in some recently published results for radium uptake by the bush passion fruit (Passiflora foetida) in areas affected by uranium mining in tropical Australia (Supervising Scientist, 2009). In these field samples, the TF decreased to a minimum of 0.004 ± 0.0002 based on total soil radium (0.3 ± 0.02 using estimates of bioavailable radium) at activity concentrations in excess of 10³  Bq/kg. This agrees with the general pattern for ²²⁶Ra uptake by plants described by Simon and Ibrahim (1987, 1990). As for K _ds, several other factors discussed in more detail in Section 5.5 may also influence TF linearity.

However, the concept of TF linearity has also been noted to hold true in many cases, across a range of species, radionuclides, and concentrations (e. g., Sheppard, 1980; Tracy et al., 1983; Vandenhove et al., 2001; Blanco Rodrıguez et al., 2002, 2006). Sometimes this is qualified for evaluations made above some critical range, particularly for essential elements or their radionuclide analogues. For example, linearity holds true for uranium soil concentrations above 20 μg U g^{−   1} on an ash weight basis (Sheppard and Sheppard, 1985), but below that it will underestimate the degree of bioaccumulation. Nonetheless, for the purposes of modelling, TF linearity can be reasonably assumed unless specific information to the contrary is available.

The simplified models are also dependent upon the environmental concentrations of the radionuclides being so low they do not substantially influence biochemical adaptation in the affected organisms. Many organisms have bioaccumulation mechanisms for toxins and nutrients that work to regulate for constant elemental concentrations within their tissues or limit their extent. This is achieved by a variety of mechanisms, including enhanced uptake (of essential elements) and/or exclusion, excretion, and sequestration (of contaminants). When these mechanisms are overwhelmed, regulation no longer functions adequately and linear uptake can result. Further detailed generic discussion of these types of mechanisms (albeit for aquatic examples) can be found in Luoma and Rainbow (2005) and Rainbow (2002), as well as in Chapter 6.

More detailed literature reviews on soil-to-plant radionuclide TFs, although not restricted to tropical environments, can be found in Simon and Ibrahim (1987) and McGee et al. (1996).

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Laplace Transform Methods

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

8.5.1 The Convolution Theorem

In many cases, we are required to determine the inverse Laplace transform of a product of two functions. Just as in integral calculus when the integral of the product of two functions did not produce the product of the integrals, neither does the inverse Laplace transform of the product yield the product of the inverse Laplace transforms. We state the following theorem.

Theorem 32 Convolution Theorem

Suppose that f(t) and g(t) are piecewise continuous on $[0,\infty )$ and both of exponential order b. Further suppose that $\mathcal{L}\left\{f(t)\right\}=F(s)$ and $\mathcal{L}\left\{g(t)\right\}=G(s)$ . Then,

(8.38) $\begin{array}{l}\hfill {\mathcal{L}}^{-1}\left\{F(s)G(s)\right\}={\mathcal{L}}^{-1}\left\{\mathcal{L}\left\{\left(f*g\right)(t)\right\}\right\}=\left(f*g\right)(t)={\int }_0^{t}f(t-\nu )g(\nu )d\nu .\end{array}$

$\text{Clear}[f,g]\text{LaplaceTransform}[{\int }_0^{t}f[t-v]g[v]\mathrm{dv},t,s]$

LaplaceTransform[f[t], t, s]LaplaceTransform[g[t], t, s]

Note that

$\begin{array}{l}\hfill \left(f*g\right)(t)={\int }_0^{t}f(t-\nu )g(\nu )d\nu \end{array}$

is called the convolution integral.

Example 8.5.1

Compute $\left(f*g\right)(t)$ if f(t) = e ^−t and $g(t)=sint$ . Verify the Convolution theorem for these functions.

Solution

We use the definition and integration by parts to obtain

$\begin{array}{l}\hfill \begin{array}{cc}\left(f*g\right)(t)& ={\int }_0^{t}f(t-\nu )g(\nu )d\nu ={\int }_0^t{e}^{-t+\nu }sin\nu d\nu =e^{-t}{\int }_0^t{e}^{\nu }sin\nu d\nu \\ & =e^{-t}{\left[\frac{1}{2}{e}^{\nu }\left(sin\nu -cos\nu \right)\right]}_0^{\nu }=\frac{1}{2}{e}^{-t}\left[e^t(sint-cost)-(sin0-cos0)\right]\\ & =\frac{1}{2}\left(sint-cost\right)+\frac{1}{2}{e}^{-t}.\end{array}\end{array}$

The same results are obtained with Mathematica. After defining convolution, which computes $\left(f*g\right)(t)$ ,

$\text{Clear}[\text{convolution},f,t,g,v];\text{convolution}[\text{f}\_,\text{g}\_]\text{:=}{\int }_0^{t}f[t-v]g[v]\mathrm{dv}$

we define f(t) and g(t)

f[t_]=Exp[−t];g[t_]=Sin[t];

and then use convolution to compute $\left(f*g\right)(t)$ .

convolution[f, g]

$\frac{1}{2}(e^{-t}-\text{Cos}[t]+\text{Sin}[t])$

Note that $\left(f*g\right)(t)=\left(g*f\right)(t)$ .

convolution[g, f]

$\frac{1}{2}(e^{-t}-\text{Cos}[t]+\text{Sin}[t])$

Now, according to the Convolution theorem, $\mathcal{L}\left\{f(t)\right\}\mathcal{L}\left\{g(t)\right\}=\mathcal{L}\left\{\left(f*g\right)(t)\right\}$ . In this example, we have

$\begin{array}{l}\hfill F(s)=\mathcal{L}\left\{f(t)\right\}=\mathcal{L}\left\{e^{-t}\right\}=\frac{1}{s+1}\text{and}G(s)=\mathcal{L}\left\{g(t)\right\}=\mathcal{L}\left\{sint\right\}=\frac{1}{s^2+1}.\end{array}$

Hence, ${\mathcal{L}}^{-1}\left\{F(s)G(s)\right\}={\mathcal{L}}^{-1}\left\{\frac{1}{s+1}\cdot \frac{1}{s^2+1}\right\}$ should equal $\left(f*g\right)(t)$ . We compute ${\mathcal{L}}^{-1}\left\{\frac{1}{s+1}\cdot \frac{1}{s^2+1}\right\}$ with InverseLaplaceTransform.

$\text{InverseLaplaceTransform}[\frac{1}{(s+1)(s^2+1)},s,t]$

$\frac{e^{-t}}{2}+\frac{1}{2}(-\text{Cos}[t]+\text{Sin}[t])$

Hence,

$\begin{array}{l}\hfill {\mathcal{L}}^{-1}\left\{\frac{1}{s+1}\cdot \frac{1}{s^2+1}\right\}=\frac{1}{2}{e}^{-t}-\frac{1}{2}cost+\frac{1}{2}sint,\end{array}$

which is the same result as that obtained for $\left(f*g\right)(t)$ .

Example 8.5.2

Use the Convolution theorem to find the Laplace transform of $h(t)={\int }_0^{t}cos(t-\nu )sin\nu d\nu$ .

Solution

Notice that $h(t)=\left(f*g\right)(t)$ , where $f(t)=cost$ and $g(t)=sint$ . Therefore, by the Convolution theorem, $\mathcal{L}\left\{\left(f*g\right)(t)\right\}=F(s)G(s)$ . Hence,

$\begin{array}{l}\hfill \mathcal{L}\left\{h(t)\right\}=\mathcal{L}\left\{f(t)\right\}\mathcal{L}\left\{g(t)\right\}=\mathcal{L}\left\{cost\right\}\mathcal{L}\left\{sint\right\}=\frac{s}{s^2+1}\cdot \frac{1}{s^2+1}=\frac{s}{{\left(s^2+1\right)}^2}.\end{array}$

The same result is obtained with LaplaceTransform.

$\text{LaplaceTransform}[{\int }_0^t\text{Cos}[t-v]\text{Sin}[v]\mathrm{dv},t,s]\text{//}\text{Simplify}$

$\frac{s}{{(1+s^2)}^2}$

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Laplace transform methods

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

8.5.1 The convolution theorem

In many cases, we are required to determine the inverse Laplace transform of a product of two functions. Just as in integral calculus when the integral of the product of two functions did not produce the product of the integrals, neither does the inverse Laplace transform of the product yield the product of the inverse Laplace transforms. We state the following theorem.

Theorem 8.15 Convolution theorem

Suppose that $f(t)$ and $g(t)$ are piecewise continuous on $[0,\infty )$ and both of exponential order b. Further suppose that $\mathcal{L}\{f(t)\}=F(s)$ and $\mathcal{L}\{g(t)\}=G(s)$ . Then,

(8.18) ${\mathcal{L}}^{-1}\{F(s)G(s)\}={\mathcal{L}}^{-1}\left\{\mathcal{L}\{(f⁎g)(t)\}\right\}=(f⁎g)(t)={\int }_0^{t}f(t-\nu )g(\nu )d\nu .$

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

$\mathbf{\text{LaplaceTransform}}\mathbf{[}{\mathbf{\int }}_{\mathbf{0}}^{\mathit{t}}\mathit{f}\mathbf{[}\mathit{t}\mathbf{-}\mathit{v}\mathbf{]}\mathit{g}\mathbf{[}\mathit{v}\mathbf{]}\mathit{d}\mathit{v}\mathbf{,}\mathit{t}\mathbf{,}\mathit{s}\mathbf{]}$

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

Note that

$(f⁎g)(t)={\int }_0^{t}f(t-\nu )g(\nu )d\nu$

is called the convolution integral.

Example 8.5.1

Compute $(f⁎g)(t)$ if $f(t)=e^{-t}$ and $g(t)=\sin t$ . Verify the convolution theorem for these functions.

Solution

We use the definition and integration by parts to obtain

$(f⁎g)(t)={\int }_0^{t}f(t-\nu )g(\nu )d\nu ={\int }_0^t{e}^{-t+\nu }\sin \nu d\nu =e^{-t}{\int }_0^t{e}^{\nu }\sin \nu d\nu =e^{-t}{\left[\frac{1}{2}{e}^{\nu }(\sin \nu -\cos \nu )\right]}_0^{\nu }=\frac{1}{2}{e}^{-t}[e^t(\sin t-\cos t)-(\sin 0-\cos 0)]=\frac{1}{2}(\sin t-\cos t)+\frac{1}{2}{e}^{-t}.$

The same results are obtained with Mathematica. After defining convolution, which computes $(f⁎g)(t)$ ,

$\mathbf{\text{Clear}}\mathbf{[}\mathbf{\text{convolution}}\mathbf{,}\mathit{f}\mathbf{,}\mathit{t}\mathbf{,}\mathit{g}\mathbf{,}\mathit{v}\mathbf{]}\mathbf{;}$

$\mathbf{\text{convolution}}\mathbf{[}\mathbf{\text{f}}\mathrm{\_}\mathbf{,}\mathbf{\text{g}}\mathrm{\_}\mathbf{]}\mathbf{\text{:=}}\mathbf{\int }{}_{\mathbf{0}}{}^{\mathit{t}}\mathit{f}\mathbf{[}\mathit{t}\mathbf{-}\mathit{v}\mathbf{]}\mathit{g}\mathbf{[}\mathit{v}\mathbf{]}\mathit{d}\mathit{v}$

we define $f(t)$ and $g(t)$

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

$\mathit{g}\mathbf{[}\mathbf{\text{t}}\mathrm{\_}\mathbf{]}\mathbf{=}\mathbf{\text{Sin}}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{;}$

and then use convolution to compute $(f⁎g)(t)$ .

$\mathbf{\text{convolution}}\mathbf{[}\mathit{f}\mathbf{,}\mathit{g}\mathbf{]}$

$\frac{1}{2}(e^{-t}-\text{Cos}[t]+\text{Sin}[t])$

Note that $(f⁎g)(t)=(g⁎f)(t)$ .

$\mathbf{\text{convolution}}\mathbf{[}\mathit{g}\mathbf{,}\mathit{f}\mathbf{]}$

$\frac{1}{2}(e^{-t}-\text{Cos}[t]+\text{Sin}[t])$

Now, according to the convolution theorem, $\mathcal{L}\{f(t)\}\mathcal{L}\{g(t)\}=\mathcal{L}\{(f⁎g)(t)\}$ . In this example, we have

$F(s)=\mathcal{L}\{f(t)\}=\mathcal{L}\left\{e^{-t}\right\}=\frac{1}{s+1}\text{and}G(s)=\mathcal{L}\{g(t)\}=\mathcal{L}\{\sin t\}=\frac{1}{s^2+1}.$

Hence, ${\mathcal{L}}^{-1}\{F(s)G(s)\}={\mathcal{L}}^{-1}\{\frac{1}{s+1}\cdot \frac{1}{s^2+1}\}$ should equal $(f⁎g)(t)$ . We compute ${\mathcal{L}}^{-1}\{\frac{1}{s+1}\cdot \frac{1}{s^2+1}\}$ with InverseLaplaceTransform.

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

$\frac{e^{-t}}{2}+\frac{1}{2}(-\text{Cos}[t]+\text{Sin}[t])$

Hence,

${\mathcal{L}}^{-1}\{\frac{1}{s+1}\cdot \frac{1}{s^2+1}\}=\frac{1}{2}{e}^{-t}-\frac{1}{2}\cos t+\frac{1}{2}\sin t,$

which is the same result as that obtained for $(f⁎g)(t)$ , as expected.   □

Example 8.5.2

Use the convolution theorem to find the Laplace transform of $h(t)={\int }_0^t\cos (t-\nu )\sin \nu d\nu$ .

Solution

Note that $h(t)=(f⁎g)(t)$ , where $f(t)=\cos t$ and $g(t)=\sin t$ . Therefore, by the convolution theorem, $\mathcal{L}\{(f⁎g)(t)\}=F(s)G(s)$ . Hence,

$\mathcal{L}\{h(t)\}=\mathcal{L}\{f(t)\}\mathcal{L}\{g(t)\}=\mathcal{L}\{\cos t\}\mathcal{L}\{\sin t\}=\frac{s}{s^2+1}\cdot \frac{1}{s^2+1}=\frac{s}{{(s^2+1)}^2}.$

The same result is obtained with LaplaceTransform.

$\mathbf{\text{LaplaceTransform}}\mathbf{[}{\mathbf{\int }}_{\mathbf{0}}^{\mathit{t}}\mathbf{\text{Cos}}\mathbf{[}\mathit{t}\mathbf{-}\mathit{v}\mathbf{]}\mathbf{\text{Sin}}\mathbf{[}\mathit{v}\mathbf{]}\mathit{d}\mathit{v}\mathbf{,}\mathit{t}\mathbf{,}\mathit{s}\mathbf{]}\mathbf{\text{//}}\mathbf{\text{Simplify}}$

$\frac{s}{{(1+s^2)}^2}$   □

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Applications related to ordinary and partial differential equations

Martha L. Abell , James P. Braselton , in Mathematica by Example (Sixth Edition), 2022

Application: the Convolution Theorem

Sometimes we are required to determine the inverse Laplace transform of a product of two functions. Just as in differential and integral calculus when the derivative and integral of a product of two functions did not produce the product of the derivatives and integrals, respectively, neither does the inverse Laplace transform of the product yield the product of the inverse Laplace transforms. The Convolution Theorem tells us how to compute the inverse Laplace transform of a product of two functions.

Theorem 6.2 The Convolution Theorem

Suppose that $f(t)$ and $g(t)$ are piecewise continuous on $[0,\infty )$ and both are of exponential order. Furthermore, suppose that the Laplace transform of $f(t)$ is $F(s)$ and that of $g(t)$ is $G(s)$ . Then,

(6.27) ${\mathcal{L}}^{-1}\{F(s)G(s)\}={\mathcal{L}}^{-1}\left\{\mathcal{L}\{(f⁎g)(t)\}\right\}={\int }_0^{t}f(t-\nu )g(\nu )d\nu .$

Note that $(f⁎g)(t)={\int }_0^{t}f(t-\nu )g(\nu )d\nu$ is called the convolution integral.

Example 6.23 L–R–C circuits

The initial-value problem used to determine the charge $q(t)$ on the capacitor in an L–R–C circuit is

$L\frac{d^{2}Q}{d{t}^2}+R\frac{dQ}{dt}+\frac{1}{C}Q=f(t),Q(0)=0,\frac{dQ}{dt}(0)=0,$

where L denotes inductance, $dQ/dt=I$ , $I(t)$ current, R resistance, C capacitance, and $E(t)$ voltage supply. Because $dQ/dt=I$ , this differential equation can be represented as

$L\frac{dI}{dt}+RI+\frac{1}{C}{\int }_0^{t}I(u)du=E(t).$

Note also that the initial condition $Q(0)=0$ is satisfied, because $Q(0)=\frac{1}{C}{\int }_0^{0}I(u)du=0$ . The condition $dQ/dt(0)=0$ is replaced by $I(0)=0$ . (a) Solve this integrodifferential equation, an equation that involves a derivative as well as an integral of the unknown function, by using the Convolution theorem. (b) Consider this example with constant values $L=C=R=1$ and $E(t)=\{\begin{array}{c}\sin t,0⩽t<\pi /2\hfill \\ 0,t⩾\pi /2\hfill \end{array}$ . Determine $I(t)$ and graph the solution.

Solution

We proceed as in the case of a differential equation by taking the Laplace transform of both sides of the equation. The Convolution theorem, Eq. (6.27), is used in determining the Laplace transform of the integral with

$\mathcal{L}\{{\int }_0^{t}I(u)du\}=\mathcal{L}\{1⁎I(t)\}=\mathcal{L}\left\{1\right\}\mathcal{L}\{I(t)\}=\frac{1}{s}\mathcal{L}\{I(t)\}.$

Therefore application of the Laplace transform yields

$Ls\mathcal{L}\{I(t)\}-LI(0)+R\mathcal{L}\{I(t)\}+\frac{1}{C}\frac{1}{s}\mathcal{L}\{I(t)\}=\mathcal{L}\{E(t)\}.$

Because $I(0)=0$ , we have $Ls\mathcal{L}\{I(t)\}+R\mathcal{L}\{I(t)\}+\frac{1}{C}\frac{1}{s}\mathcal{L}\{I(t)\}=\mathcal{L}\{E(t)\}$ . Simplifying and solving for $\mathcal{L}\{I(t)\}$ results in $\mathcal{L}\{I(t)\}=\frac{Cs\mathcal{L}\{E(t)\}}{LC{s}^2+RCs+1}$

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

$\mathbf{\text{LaplaceTransform}}\mathbf{[}\mathit{l}{\mathit{i}}^{\mathbf{\prime }}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{+}\mathit{r}\mathit{i}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{,}\mathit{t}\mathbf{,}\mathit{s}\mathbf{]}$

$r\text{LaplaceTransform}[i[t],t,s]+l(-i[0]+s\text{LaplaceTransform}[i[t],t,s])$

$\mathbf{\text{Solve}}\mathbf{[}\mathit{l}\mathit{s}\mathbf{\text{lapi}}\mathbf{+}\mathit{r}\mathbf{\text{lapi}}\mathbf{+}\frac{\mathbf{\text{lapi}}}{\mathit{c}\mathit{s}}\mathbf{\text{==}}\mathbf{\text{lape}}\mathbf{,}\mathbf{\text{lapi}}\mathbf{]}$

$\left\{\{\text{lapi}\to \frac{c\text{lape}s}{1+crs+cl{s}^2}\}\right\}$

so that $I(t)={\mathcal{L}}^{-1}\left\{\frac{Cs\mathcal{L}\{E(t)\}}{LC{s}^2+RCs+1}\right\}$ . In the Solve command, we use lapi to denote $\mathcal{L}\{I(t)\}$ and lape to denote $\mathcal{L}\{E(t)\}$ . For (b), we note that $E(t)=\{\begin{array}{c}\sin t,0⩽t<\pi /2\hfill \\ 0,t⩾\pi /2\hfill \end{array}$ can be written as $E(t)=\sin t(\mathcal{U}(t)-\mathcal{U}(t-\pi /2))$ . We define and plot the forcing function $E(t)$ on the interval $[0,\pi ]$ in Fig. 6.23 (a).

Figure 6.23. (a) Plot of $E(t)=\sin t(\mathcal{U}(t)-\mathcal{U}(t-\pi /2))$ ; (b) I(t) (University of Missouri colors).

We use lowercase letters to avoid any possible ambiguity with built-in Mathematica functions, like E and I.

$\mathit{e}\mathbf{[}\mathbf{\text{t}}\mathrm{\_}\mathbf{]}\mathbf{\text{:=}}\mathbf{\text{Sin}}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{(}\mathbf{\text{UnitStep}}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{-}\mathbf{\text{UnitStep}}\mathbf{[}\mathit{t}\mathbf{-}\frac{\mathit{\pi }}{\mathbf{2}}\mathbf{]}\mathbf{)}$

$\mathbf{\text{p1}}\mathbf{=}\mathbf{\text{Plot}}\mathbf{[}\mathit{e}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{,}\mathbf{\{}\mathit{t}\mathbf{,}\mathbf{0}\mathbf{,}\mathit{\pi }\mathbf{\}}\mathbf{,}\mathbf{\text{PlotStyle}}\mathbf{\to }$

$\mathbf{\{}\mathbf{\{}\mathbf{\text{Thickness}}\mathbf{[}\mathbf{.}\mathbf{01}\mathbf{]}\mathbf{,}\mathbf{\text{CMYKColor}}\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{.}\mathbf{25}\mathbf{,}\mathbf{.}\mathbf{9}\mathbf{,}\mathbf{.}\mathbf{05}\mathbf{]}\mathbf{\}}\mathbf{\}}\mathbf{]}$

Next, we compute the Laplace transform of $\mathcal{L}\{E(t)\}$ with LaplaceTransform. We call this result lcape.

$\mathbf{\text{lcape}}\mathbf{=}\mathbf{\text{LaplaceTransform}}\mathbf{[}\mathit{e}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{,}\mathit{t}\mathbf{,}\mathit{s}\mathbf{]}$

$\frac{1}{1+s^2}-\frac{e^{-\frac{\pi s}{2}}s}{1+s^2}$

Using the general formula obtained for the Laplace transform of $I(t)$ , we note that the denominator of this expression is given by $s^2+s+1$ , which is entered as denom. Hence the Laplace transform of $I(t)$ , called lcapi, is given by the ratio slcape/denom.

$\mathbf{\text{denom}}\mathbf{=}{\mathit{s}}^{\mathbf{2}}\mathbf{+}\mathit{s}\mathbf{+}\mathbf{1}\mathbf{;}$

$\mathbf{\text{lcapi}}\mathbf{=}\mathit{s}\mathbf{\text{lcape}}\mathbf{/}\mathbf{\text{denom}}\mathbf{;}$

$\mathbf{\text{lcapi}}\mathbf{=}\mathbf{\text{Simplify}}\mathbf{[}\mathbf{\text{lcapi}}\mathbf{]}$

$\frac{s-e^{-\frac{\pi s}{2}}{s}^2}{1+s+2s^2+s^3+s^4}$

We determine $I(t)$ with InverseLaplaceTransform. Note that HeavisideTheta[x] is defined by $\theta (x)=\{\begin{array}{c}0,\text{ if}x<0\hfill \\ 1,\text{ if}x>0\hfill \end{array}$ .

$\mathit{i}\mathbf{[}\mathbf{\text{t}}\mathrm{\_}\mathbf{]}\mathbf{=}\mathbf{\text{InverseLaplaceTransform}}\mathbf{[}\mathbf{\text{lcapi}}\mathbf{,}\mathit{s}\mathbf{,}\mathit{t}\mathbf{]}$

$\text{Sin}[t]-\text{HeavisideTheta}[-\frac{\pi }{2}+t](-\frac{e^{\frac{1}{4}(\pi -2t)}(\sqrt{3}\text{Cos}[\frac{1}{4}\sqrt{3}(\pi -2t)]+\text{Sin}[\frac{1}{4}\sqrt{3}(\pi -2t)])}{\sqrt{3}}+\text{Sin}[t])-\frac{2e^{-t/2}\text{Sin}[\frac{\sqrt{3}t}{2}]}{\sqrt{3}}$

This solution is plotted in p2 (in black) and displayed with the forcing function (in gray) in Fig. 6.23 (b). Notice the effect that the forcing function has on the solution to the differential equation.

$\mathbf{\text{p2}}\mathbf{=}\mathbf{\text{Plot}}\mathbf{[}\mathit{i}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{,}\mathbf{\{}\mathit{t}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{10}\mathbf{\}}\mathbf{,}\mathbf{\text{PlotStyle}}\mathbf{\to }$

$\mathbf{\{}\mathbf{\{}\mathbf{\text{Thickness}}\mathbf{[}\mathbf{.}\mathbf{01}\mathbf{]}\mathbf{,}\mathbf{\text{CMYKColor}}\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{.}\mathbf{25}\mathbf{,}\mathbf{.}\mathbf{9}\mathbf{,}\mathbf{.}\mathbf{05}\mathbf{]}\mathbf{\}}\mathbf{\}}\mathbf{]}\mathbf{;}$

$\mathbf{\text{Show}}\mathbf{[}\mathbf{\text{p1}}\mathbf{,}\mathbf{\text{p2}}\mathbf{,}\mathbf{\text{PlotRange}}\mathbf{\to }\mathbf{\text{All}}\mathbf{]}$

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

In this case, we see that we can use DSolve to solve the initial-value problem

$Q^{″}+Q^{\prime }+Q=E(t),Q(0)=0,Q^{\prime }(0)=0$

as well. However, the unsimplified result is very lengthy, so we use FullSimplify to attempt to simplify the result as much as possible.

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

$\mathbf{\text{sol}}\mathbf{=}\mathbf{\text{DSolve}}\mathbf{[}\mathbf{\{}{\mathit{q}}^{\mathbf{″}}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{+}{\mathit{q}}^{\mathbf{\prime }}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{+}\mathit{q}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{\text{==}}\mathit{e}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{,}\mathit{q}\mathbf{[}\mathbf{0}\mathbf{]}\mathbf{\text{==}}\mathbf{0}\mathbf{,}{\mathit{q}}^{\mathbf{\prime }}\mathbf{[}\mathbf{0}\mathbf{]}\mathbf{\text{==}}\mathbf{0}\mathbf{\}}\mathbf{,}\mathbf{.}$

$\mathit{q}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{,}\mathit{t}\mathbf{]}\mathbf{\text{//}}\mathbf{\text{FullSimplify}}$

$\{\{q[t]\to \begin{array}{cc}\{\hfill & \begin{array}{c}-\text{Cos}[t]+\frac{1}{3}{e}^{-t/2}(3\text{Cos}[\frac{\sqrt{3}t}{2}]+\sqrt{3}\text{Sin}[\frac{\sqrt{3}t}{2}])\hfill \\ t⩾0\mathrm{\&}\mathrm{\&}2t⩽\pi \hfill \\ \frac{1}{3}{e}^{-t/2}(3\text{Cos}[\frac{\sqrt{3}t}{2}]+\sqrt{3}(-2e^{\pi /4}\text{Sin}[\frac{1}{4}\sqrt{3}(\pi -2t)]+\text{Sin}[\frac{\sqrt{3}t}{2}]))\hfill & 2t>\pi \hfill \\ 0\hfill & \text{True}\hfill \end{array}\hfill \end{array}\}\}$

We see that this result is a real-valued function using ComplexExpand, followed by Simplify.

$\mathit{q}\mathbf{[}\mathbf{\text{t}}\mathrm{\_}\mathbf{]}\mathbf{=}\mathbf{\text{ComplexExpand}}\mathbf{[}\mathbf{\text{sol}}\mathbf{[}\mathbf{[}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{]}\mathbf{]}\mathbf{]}\mathbf{\text{//}}\mathbf{\text{Simplify}}$

$\begin{array}{cc}\{\hfill & \begin{array}{cc}-\text{Cos}[t]+\frac{3\text{Cos}[\frac{\sqrt{3}t}{2}]+\sqrt{3}\text{Sin}[\frac{\sqrt{3}t}{2}]}{3\sqrt{e^t}}\hfill & t⩾0\mathrm{\&}\mathrm{\&}2t⩽\pi \hfill \\ \frac{3\text{Cos}[\frac{\sqrt{3}t}{2}]+\sqrt{3}(-2e^{\pi /4}\text{Sin}[\frac{1}{4}\sqrt{3}(\pi -2t)]+\text{Sin}[\frac{\sqrt{3}t}{2}])}{3\sqrt{e^t}}\hfill & 2t>\pi \hfill \\ 0\hfill & \text{True}\hfill \end{array}\hfill \end{array}$

We use this result to graph $Q(t)$ and $I(t)=Q^{\prime }(t)$ in Fig. 6.24.

Figure 6.24. Q(t) and I(t)=Q′(t).

$\mathbf{\text{Plot}}\mathbf{[}\mathbf{\{}\mathit{q}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{,}{\mathit{q}}^{\mathbf{\prime }}\mathbf{[}\mathit{t}\mathbf{]}\mathbf{\}}\mathbf{,}\mathbf{\{}\mathit{t}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{10}\mathbf{\}}\mathbf{,}$

$\mathbf{\text{PlotStyle}}\mathbf{\to }\mathbf{\{}\mathbf{\{}\mathbf{\text{CMYKColor}}\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{.}\mathbf{25}\mathbf{,}\mathbf{.}\mathbf{9}\mathbf{,}\mathbf{.}\mathbf{05}\mathbf{]}\mathbf{\}}\mathbf{,}$

$\mathbf{\{}\mathbf{\text{CMYKColor}}\mathbf{[}\mathbf{.}\mathbf{6}\mathbf{,}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{.}\mathbf{4}\mathbf{,}\mathbf{1}\mathbf{]}\mathbf{\}}\mathbf{\}}\mathbf{]}$   □

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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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Fractional Differential Equations

In Mathematics in Science and Engineering, 1999

2.8.1 Basic Facts on the Laplace Transform

Let us recall Home basic facts about the Laplace transform.

The function F(s) of the complex variable s defined by

(2.235) $F\left(s\right)=L\left\{f\left(t\right);s\right\}={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt}$

is called the Laplace transform of the function f(t), which is called the original. For the existence of the integral (2.235) the function f(t ) must be of exponential order α, which means that there exist positive constants M and T such that

$e^{-\alpha t}\left|f\left(t\right)\right|\le M\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}for\hspace{0.17em}all}t>T.$

In other words, the function f(t) must not grow faster then a certain exponential function when t → ∞.

We will denote the Laplace transforms by uppercase letters and the originals by lowercase letters.

The original f(t) can be restored from the Laplace tranform F(s) with the help of the inverse Laplace transform

(2.236) $f\left(t\right)=L^{-1}\left\{F\left(s\right);t\right\}={\displaystyle \underset{c-i\infty }{\overset{c+i\infty }{\int }}{e}^{st}F\left(s\right)ds,c=Re\left(s\right)>c_0,}$

where c ₀ lies in the right half plane of the absolute convergence of the Laplace integral (2.235).

The direct evaluation of the inverse Laplace transform using the formula (2.236) is often complicated; however, sometime it gives useful information on the behaviour of the unknown original f(t) which we look for.

The Laplace transform of the convolution

(2.237) $f\left(t\right)*g\left(t\right)={\displaystyle \underset{0}{\overset{t}{\int }}f\left(t-\tau \right)g\left(\tau \right)d\tau }={\displaystyle \underset{0}{\overset{t}{\int }}f\left(\tau \right)g\left(t-\tau \right)d\tau }$

of the two functions f(t) and g(t), which are equal to zero for t < 0, is equal to the product of the Laplace transform of those function:

(2.238) $L\left\{f\left(t\right)*g\left(t\right);s\right\}=F\left(s\right)G\left(s\right)$

under the assumption that both F(s) and G(s) exist. We will use the property (2.238) for the evaluation of the Laplace transform of the Riemann–Liouville fractional integral.

Another useful property which we need is the formula for the Laplace transform of the derivative of an integer order n of the function f(t):

(2.239) $L\left\{f^n\left(t\right);s\right\}=s^{n}F\left(s\right)-{\displaystyle \sum _{k=0}^{n-1}{s}^{n-k-1}{f}^{\left(k\right)}\left(0\right)=s^{n}F\left(s\right)-}{\displaystyle \sum _{k=0}^{n-1}{s}^k{f}^{\left(n-k-1\right)}\left(0\right),}$

which can be obtained from the definition (2.235) by integrating by parts under the assumption that the corresponding integrals exist.

In the following sections on the Laplace transforms of fractional derivatives we consider the lower terminal a = 0.

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Flow past a shrinking sheet

Kuppalapalle Vajravelu , Swati Mukhopadhyay , in Fluid Flow, Heat and Mass Transfer At Bodies of Different Shapes, 2016

3.3.1 Mathematical formulation

Consider the flow of an incompressible viscous electrically conducting fluid past a flat sheet coinciding with the plane y  =   0 in the presence of a magnetic field. The flow is confined to y   >  0. Two equal forces are applied along the x -axis toward the origin so that the surface shrinks keeping the origin fixed. Here the velocity of shrinking is of exponential order. A variable magnetic field $B\left(x\right)=B_0{e}^{\frac{x}{2L}}$ is applied normal to the sheet, B ₀ being a constant. These radiative effects have important applications in space technology and high-temperature processes. The continuity, momentum, and energy equations in the presence of thermal radiation are written as

(3.29) $\frac{\partial u}{\partial x}+\frac{\partial \upsilon }{\partial y}=0\text{,}$

(3.30) $u\frac{\partial u}{\partial x}+\upsilon \frac{\partial u}{\partial y}=\nu \frac{{\partial }^{2}u}{\partial y^2}-\frac{\sigma B^2}{\rho }u\text{,}$

(3.31) $u\frac{\partial T}{\partial x}+\upsilon \frac{\partial T}{\partial y}=\frac{\kappa }{\rho c_p}\frac{{\partial }^{2}T}{\partial y^2}-\frac{1}{\rho c_p}\frac{\partial q_r}{\partial y}$

where u and υ are the components of velocity respectively in the x and y directions, $\nu =\frac{\mu }{\rho }$ is the kinematic viscosity, ρ is the fluid density, μ is the coefficient of fluid viscosity, q _r is the radiative heat flux, c _p is the specific heat at constant pressure, and κ is the thermal conductivity of the fluid.

Using Rosseland approximation for radiation (Brewster [25]), we can write

(3.31a) $q_r=-\frac{4\sigma }{3k^{*}}\frac{\partial T^4}{\partial y}$

where σ is the Stefan-Boltzman constant, k* is the absorption coefficient.

Assuming that the temperature difference within the flow is such that T ⁴ may be expanded in a Taylor series and expanding T ⁴ about $T_{\infty }$ , and neglecting higher orders, we get $T^4\text{≡}4T_{\infty }^{3}T-3T_{\infty }^4\text{.}$ Therefore, equation (3) becomes

(3.32) $u\frac{\partial T}{\partial x}+\upsilon \frac{\partial T}{\partial y}=\frac{\kappa }{\rho c_p}\frac{{\partial }^{2}T}{\partial y^2}+\frac{16\sigma T_{\infty }^3}{3\rho c_p{k}^{*}}\frac{{\partial }^{2}T}{\partial y^2}\text{.}$

The flow field can be changed significantly by the application of suction or injection (blowing) of a fluid through the bounding surface. In the design of thrust bearing and radial diffusers and thermal oil recovery, the process of suction and blowing is very important.

The appropriate boundary conditions for the problem are given by

(3.33a) $u=U+N\nu \frac{\partial u}{\partial y},\upsilon =-V\left(x\right),T=T_w+D\frac{\partial T}{\partial y}\mathrm{at}y=0\text{,}$

(3.33b) $u\to 0,T\to 0\mathrm{a}\mathrm{s}y\to \infty \text{.}$

Here $U=-U_0{e}^{\frac{x}{L}}$ is the shrinking velocity; $T_w=T_0{e}^{\frac{x}{2L}}$ is the temperature at the sheet; U ₀ and T ₀ are the reference velocity and temperature, respectively; N $=N_1{e}^{-\frac{x}{2L}}$ is the velocity slip factor, which changes with x; N ₁ is the initial value of velocity slip factor; $D=D_1{e}^{-\frac{x}{2L}}$ is the thermal slip factor, which also changes with x; and D ₁ is the initial value of thermal slip factor. The no-slip case is recovered for $N=0$ = D. V(x) > 0 is the velocity of suction, and V(x)< 0 is the velocity of blowing. V(x) = $V_0{e}^{\frac{x}{2L}}$ , a special type of velocity at the wall, is considered, where V ₀ is the initial strength of suction.

Introducing the similarity variable and similarity transformations as

(3.34) $\begin{array}{l}\eta =\sqrt{\frac{U_0}{2\nu L}}{e}^{\frac{x}{2L}}y,u=U_0{e}^{\frac{x}{L}}{f}^{/}\left(\eta \right),\\ \upsilon =-\sqrt{\frac{\nu U_0}{2L}}{e}^{\frac{x}{2L}}\left\{f\left(\eta \right)+\eta f^{/}\left(\eta \right)\right\},T=T_0{e}^{\frac{x}{2L}}\theta \left(\eta \right)\text{,}\end{array}$

upon substitution of (3.34) in equations (3.30) and (3.32), the governing equations reduce to

(3.35) $f^{///}+f{f}^{//}-2f^{/2}-{\mathit{M}}^2{f}^{/}=0\text{,}$

(3.36) $\left(1+\frac{4}{3}\mathit{R}\right){\theta }^{//}+\mathit{Pr}\left(f{\theta }^{/}-f^{/}\theta \right)=0$

and the boundary conditions take the following form:

(3.37a) $f^{/}\left(\eta \right)=-1+\lambda f^{//}\left(\eta \right),f\left(\eta \right)=\mathit{S},\theta \left(\eta \right)=1+\delta {\theta }^{/}\left(\eta \right)\mathrm{at}\eta =0$

and

(3.37b) $f^{/}\left(\eta \right)\to 0,\theta \left(\eta \right)\to 0\mathrm{a}\mathrm{s}\eta \to \infty \text{.}$

The prime denotes differentiation with respect to η, $\mathit{M}=\sqrt{\frac{2\sigma B_0^{2}L}{\rho U_0}}$ is the magnetic parameter, $\lambda =$ $N_1\sqrt{\frac{U_0\nu }{2L}}$ is the velocity slip parameter, $\delta ={\mathit{D}}_1\sqrt{\frac{U_0}{2\nu L}}$ is the thermal slip parameter, $\mathit{S}=$ $\frac{V_0}{\sqrt{\frac{U_0\nu }{2L}}}$ > 0 (or   <   0) is the suction (or blowing) parameter, $\mathit{R}=$ $\frac{4\sigma T_{\infty }^3}{\kappa k^{*}}$ is the radiation parameter, and Pr = $\frac{\mu c_p}{\kappa }$ is the Prandtl number.

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