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Coordinated Differential Equations

Thomas W. Judson, Austin Eide, Adam Larios

Section 3.1 The Laplace Transform

Consider the electrical circuit governed by the differential equation
\begin{equation*} L I'' + RI' + \frac{1}{C} I = E'(t). \end{equation*}
The voltage function, \(E'(t)\text{,}\) might have discontinuities. For example, the voltage in the circuit can be periodically turned on and off. The previous methods that we have used to solve second order linear differential equations may not apply here. However, the Laplace transform, an integral transform, gives a method of solving such equations.
As a second example, let us consider a population of fish that is governed by exponential growth,
\begin{align*} \frac{dP}{dt} & = kP\\ P(0) & = P_0, \end{align*}
and suppose that we wish to determine the effects of seasonal fishing. In other words, harvesting will not be continuous. For example, we might only allow fishing at a constant rate \(r\) during the first half of the year,
\begin{equation*} H(t) = \begin{cases} r & 0 \leq t \leq 1/2, \\ 0 & 1/2 \lt t \lt 1, \\ r & 1 \leq t \leq 3/2, \\ 0 & 3/2 \lt t \lt 2, \\ & \vdots \end{cases} \end{equation*}
Our initial value problem now becomes
\begin{align*} \frac{dP}{dt} & = kP - H\\ P(0) & = P_0, \end{align*}
It should be clear that we need some additional tools to analyze differential equations possessing discontinuous terms.
Given an initial value problem
\begin{align*} ay'' + by' + cy & = g(t)\\ y(0) & = y_0\\ y'(0) & = y_0', \end{align*}
the idea is to use the Laplace transform to change the differential equation into an equation that can be solved algebraically and then transform the algebraic solution back into a solution of the differential equation. Surprisingly, this method will even work when \(g\) is a discontinuous function, provided the discontinuities are not too bad.

Subsection 3.1.1 Definition of the Laplace Transform

We shall define the Laplace transform of a function \(f(t)\) by
\begin{equation*} {\mathcal L}(f)(s)= F(s) = \int_0^\infty e^{-st} f(t) \, dt, \end{equation*}
provided this integral converges. The Laplace transform of a function has many nice properties, especially with respect to the derivative of \(f\text{.}\) However, before we investigate these properties, let us compute several Laplace transforms.

Example 3.1.1.

  1. Let \(f(t) = k\text{,}\) where \(k\) is a constant. Then
    \begin{equation*} {\mathcal L}(f)(s) = \int_0^\infty e^{-st}k \; dt = \frac{k}{s}, \end{equation*}
    for \(s \gt 0\text{.}\)
  2. For \(f(t) = t\text{,}\) the Laplace transform is
    \begin{equation*} {\mathcal L}(f)(s) = \int_0^\infty e^{-st} t \, dt = \frac{1}{s^2}, \end{equation*}
    for \(s \gt 0\text{.}\)
  3. If \(f(t) = e^{at}\) with \(t > 0\text{,}\) then
    \begin{equation*} {\mathcal L}(f)(s) = \int_0^\infty e^{-st} e^{at} \, dt = \int_0^\infty e^{-(s -a)t} \, dt = \left[ - \frac{e^{-(s -a)t}}{s - a} \right]_0^\infty. \end{equation*}
    Noting that \(e^{-(s -a)t} \to 0\) as \(t \to \infty\) for \(s \gt a\text{,}\) we find that
    \begin{equation*} {\mathcal L}(e^{at}) = \frac{1}{s - a} \end{equation*}
    for \(s \gt a\text{.}\)

Example 3.1.2.

The Laplace transform of a function does not always exist, even for functions that are infinitely differentiable. For example, let \(f(t) = e^{t^2}\text{.}\) Then for any real number \(s\text{,}\) we know that
\begin{equation*} e^{t^2} e^{-st} = e^{t(t-s)} \gt 1\text{,} \end{equation*}
for \(t \gt s\) with \(t \geq 0\text{;}\) hence,
\begin{equation*} \int_0^b e^{t^2} e^{-st} \, dt = \int_0^b e^{t(t-s)} \, dt \geq b. \end{equation*}
Thus, the integral
\begin{equation*} \int_0^\infty e^{t^2} e^{-st} \, dt \end{equation*}
does not converge.
The Laplace transform, however, does exist in many cases. In Subsection 3.1.3, we will show that the Laplace transform of a function exists provided the function does not grow too quickly and does not possess bad discontinuities.

Subsection 3.1.2 Properties of the Laplace Transform

One of the most important properties of the Laplace transform is linearity. That is,
\begin{equation*} {\mathcal L} ( \alpha f + \beta g )(s) = \alpha {\mathcal L}(f)(s) + \beta {\mathcal L} ( g)(s), \end{equation*}
where \(\alpha, \beta \in {\mathbb R}\) provided the Laplace transforms of \(f\) and \(g\) exist. The proof of this statement follows directly from the definition of the Laplace transform and the properties of integration,
\begin{align*} {\mathcal L} ( \alpha f + \beta g )(s) \amp = \int_0^\infty e^{-st} [\alpha f(t) + \beta g(t)] \, dt\\ \amp = \lim_{b \to \infty} \int_0^b e^{-st} [\alpha f(t) + \beta g(t)] \, dt\\ \amp = \alpha \lim_{b \to \infty} \int_0^b e^{-st} f(t) \, dt + \beta \lim_{b \to \infty} \int_0^b e^{-st}\beta g(t) \, dt\\ \amp = \alpha {\mathcal L}(f)(s) + \beta {\mathcal L} ( g)(s) \end{align*}
We state the linearity of the Laplace transform as a theorem.
Transforms of functions having the linearity property are called linear operators.

Example 3.1.4.

Recall that \(\sinh at = (e^{at} - e^{-at})/2\text{.}\) Provided \(a \gt 0\text{,}\) linearity makes it very easy to compute the Laplace transform of \(\sinh at\text{.}\) By Example 3.1.1,
\begin{equation*} {\mathcal L} ( \sinh at) = \frac{1}{2} {\mathcal L} ( e^{at}) - \frac{1}{2} {\mathcal L} ( e^{-at}) = \frac{1}{2} \left( \frac{1}{s - k} - \frac{1}{s + k} \right). \end{equation*}
Thus, for \(s \gt k \gt 0\text{,}\)
\begin{equation*} {\mathcal L} ( \sinh at) = \frac{k}{s^2 - k^2}. \end{equation*}
The Laplace transform of discontinuous functions also exist, provided the disconinuities are not too bad. We say that a function \(f\) is piecewise continuous on an interval \([a, b]\) if \(f\) satisfies the following conditions.
  1. There are a finite number of discontinuities of \(f\) on \([a, b]\text{.}\)
  2. If \(c \in [a, b]\) is a discontinuity, then the one-sided limits
    \begin{equation*} \lim_{t \to c^-} f(t) \qquad \mbox{and} \qquad \lim_{t \to c^+} f(t) \end{equation*}
    both exist. We also assume that \(\lim_{t \to a^+} f(t)\) and \(\lim_{t \to b^-} f(t)\) exist.
We say that a function is function on an interval \([0, \infty)\text{,}\) if it is piecewise continuous on the interval \([0, b]\) for all \(b \gt 0\text{.}\) The types of discontinuities that we are describing are sometimes called jump discontinuities. If a function is piecewise continuous, then the continuities are not too bad.

Example 3.1.5.

One of the simplest piecewise continuous functions is
\begin{equation*} u(t) = \begin{cases} 0, & t \lt 0 \\ 1, & t \geq 0. \end{cases} \end{equation*}
The function \(u(t)\) has a jump discontinuity at \(t = 0\text{.}\) It follows very quickly that \({\mathcal L}(u(t)) = 1/s\) for \(s \gt 0\text{.}\)
If we define the unit step function at \(a\) to be
\begin{equation*} u_a(t) = u(t - a) = \begin{cases} 0, & t \lt a \\ 1, & t \geq a, \end{cases} \end{equation*}
then we have a jump discontinuity at \(t = a\) (Figure 3.1.6). If \(a \gt 0\text{,}\) then
\begin{equation*} {\mathcal L}(u_a(t)) = \int_0^\infty e^{-st} u_a(t) \, dt = \int_a^\infty e^{-st} \, dt = \lim_{b \to \infty} \left[ -\frac{e^{-st}}{s} \right]_a^b = \frac{e^{-sa}}{s}, \end{equation*}
where \(s \gt 0\) and \(a \gt 0\text{.}\) A function of the form \(u_a(t) = u(t - a)\) is called a step function or Heaviside function, named for the British engineer Oliver Heaviside.
described in detail following the image
a function that is equal to one for zero and postive values and equal to zero for negative values
Figure 3.1.6. Unit step function for \(a = 0\)

Example 3.1.7.

Given a function \(f(t)\text{,}\) consider the new function
\begin{equation*} g(t) = u_a(t) f( t - a). \end{equation*}
To obtain the function \(g\) from \(f\text{,}\) we shift the graph of the \(f\) to the right by \(a\) units and let \(g(t) = 0\) for \(0 \leq t \lt a\) (see Figure 3.1.8). Let us compute the Laplace transform of \(g\text{,}\)
\begin{equation*} {\mathcal L}(g) = \int_0^\infty g(t) e^{-st} \, dt =\int_a^\infty f(t - a) e^{-st} \, dt. \end{equation*}
If we make the substitution \(w = t - a\text{,}\) then
\begin{equation*} {\mathcal L}(g) = \int_0^\infty f(w) e^{-s(w+a)} \, dt = e^{-sa} \int_0^\infty f(w) e^{-sw} \, dt = e^{-sa} {\mathcal L}(f). \end{equation*}
In other words, if \({\mathcal L}(f) = F(s)\text{,}\) then \({\mathcal L}(u_a(t) f( t- a)) = e^{-as} F(s)\text{.}\)
described in detail following the image
the graph of two functions with the first function increasing until two and then decreasing and the second function shifted to the right by one and set to zero for values less than one
Figure 3.1.8. \(u_a(t)f(t-a)\text{,}\) where \(a = 1\)

Activity 3.1.1. Finding Laplace Transforms.

Find the Laplace transform for each of the functions below. Recall that \(\cosh t = (e^t + e^{-t})/2\) and \(\sinh t = (e^t - e^{-t})/2\)
(a)
\(f(t) = t^6\)
(b)
\(f(t) = t^7 e^{-t}\)
(c)
\(f(t) = 3\sin^2 \left(\sqrt{2}t\right)\)
(d)
\(f(t) = e^{3t}\sin 2t\)
(e)
\(f(t) = 2 t^2 \cosh t\)

Subsection 3.1.3 Existence and Uniqueness of the Laplace Transform

If we are to use Laplace transforms to study differential equations, we would like to know which functions actually have Laplace transforms. Furthermore, if two functions have the same Laplace transform, we can ask if the functions must be the same. In other words, we wish to know if the Laplace transform of a function exists and is unique. We can answer both of these questions affirmatively if the function \(f(t)\) is piecewise continuous on \([0, \infty)\) and does not grow too quickly as \(t \to \infty\text{.}\) We say a function \(f(t)\) is exponentially bounded on \([0, \infty)\) if there exist constants \(M \geq 0\) and \(a\) such that
\begin{equation*} |f(t)| \leq Me^{at}, \end{equation*}
for all \(t\) in \([0, \infty)\text{.}\) In other words, the graph of \(f\) must lie between the curves \(y = Me^{at}\) and \(y = -Me^{at}\text{.}\) The following two theorems tell us that Laplace transforms exist and are unique for piecewise continuous, exponentially bounded functions on the interval \([0, \infty)\text{.}\) We leave the proofs of these theorem as exercises.
In light of Theorem 3.1.10, it now makes sense to define the inverse Laplace transform of a function \(F(s)\text{,}\) which we will denote by \({\mathcal L}^{-1}(F(s))(t)\text{,}\) as the function \(f(t)\) whose unique Laplace transform is \(F(s)\text{.}\) Furthermore, the inverse Laplace transform is linear,
\begin{equation*} {\mathcal L}^{-1} ( \alpha F + \beta G ) = \alpha {\mathcal L}^{-1}(F) + \beta {\mathcal L}^{-1} ( G). \end{equation*}

Subsection 3.1.4 Finding Laplace Transforms and Inverse Transforms

To use the method of Laplace transforms effectively, we need either tables or computer software so that we can easily find Laplace transforms and inverse Laplace transforms.
Table 3.1.11. Table of Laplace Transforms
\(f(t) = {\mathcal L}^{-1}(F(s))\) \(F(s) = {\mathcal L}(f(t))\)
1 \(1/s\text{,}\) \(s \gt 0\)
\(e^{at}\) \(1/(s - a)\text{,}\) \(s \gt a\)
\(t^n\text{,}\) \(n \in {\mathbb N}\) \(n!/s^{n+1}\text{,}\) \(s \gt 0\)
\(t^a\text{,}\) \(a > -1\) \(\Gamma(a + 1)/s^{a + 1}\text{,}\) \(s \gt 0\)
\(\sin at\) \(a/(s^2 + a^2)\text{,}\) \(s \gt 0\)
\(\cos at\) \(s/(s^2 + a^2)\text{,}\) \(s \gt 0\)
\(\sinh at\) \(a/(s^2 - a^2)\text{,}\) \(s \gt |a|\)
\(\cosh at\) \(s/(s^2 - a^2)\text{,}\) \(s \gt |a|\)
\(e^{at} \sin bt\) \(b/[(s-a)^2 + b^2]\text{,}\) \(s \gt a\)
\(e^{at} \cos bt\) \((s-a)/[(s-a)^2 + b^2]\text{,}\) \(s \gt a\)
\(t^n e^{at}\text{,}\) \(n \in {\mathbb N}\) \(n!/(s - a)^{n + 1}\text{,}\) \(s \gt a\)
\(u_c(t)\) \(e^{-cs}/ s\text{,}\) \(s \gt 0\)
\(u_c(t) f(t - c)\) \(e^{-cs} F(s)\)
\(e^{ct} f(t)\) \(F(s - c)\)
\(f(ct)\) \((1/c) F(s/c)\text{,}\) \(c \gt 0\)
\(\delta(t - c) = \delta_c(t)\) \(e^{-cs}\)
\(f'(t)\) \(s F(s) - f(0)\)
\(f''(t)\) \(s^2 F(s) - sf(0) - f'(0)\)

Example 3.1.12.

Find the Laplace transformation of the function
\begin{equation*} f(t) = 7 + 2e^t\sin(6t) - \frac{e^{3t}t^4}{4} \end{equation*}
Table 3.1.11 tells us that the Laplace transform of \(1\) is \(\frac1s\text{,}\) so the first term becomes \(\frac7s\text{.}\) We also see the Laplace transformation of \(e^{at}\sin(bt)\) is \(\frac{b}{(s-a)^2+b^2}\text{,}\) so the second term gives us \(\frac{12}{(s-1)^2+36}\text{.}\) The Laplace transformation of \(t^ne^{at}\) is \(\frac{n!}{(s-a)^{n+1}}\text{,}\) so the third term gives us \(\frac{6}{(s-3)^5}\text{.}\) Putting this together gives
\begin{equation*} F(s) = \frac7s + \frac{12}{(s-1)^2+36} - \frac{6}{(s-3)^5} \end{equation*}

Example 3.1.13.

Find the inverse Laplace transformation of the function
\begin{equation*} F(s) = \frac{14}{s^2+s-2} \end{equation*}
Factoring the denominator gives \(s^2+s-2 = (s - 1)(s + 2)\text{,}\) and we can split the expression via partial fractions (as if we are integrating) to written \(\frac{14}{s^2+s-2} = \frac{14}{s-1} + \frac{14}{s+2}\text{.}\) Taking the inverse Laplace of this gives \(14e^t + 14e^{-2t}\text{.}\)

Example 3.1.14.

Find the inverse Laplace transformation of the function
\begin{equation*} \frac{2s + 1}{s^2 + 2s + 5} \end{equation*}
The denominator does not factor (even if we allow irrational coefficients), but we can complete the square and obtain \(s^2+2s+5 = (s + 1)^2 + 4\text{.}\) We then want the numerator in the form \(A(s+1)+B\text{,}\) for some constants \(A\) and \(B\text{.}\) Setting \(A(s+1)+B = 2s + 1\) and equating coefficients (as in partial fractions) gives us \(A = 2, B = -1\text{,}\) and we are really trying to find
\begin{equation*} \mathcal{L}^{-1}(\frac{2(s+1)}{(s+1)^2+4}) - \mathcal{L}^{-1}(\frac{1}{(s+1)^2+4}) \end{equation*}
Recall from Table 3.1.11 that \(\mathcal{L}(e^{at}\sin(bt)) = \frac{b}{(s-a)^2+b^2}\) and \(\mathcal{L}(e^{at}\cos(bt)) = \frac{s}{(s-a)^2+b^2}\text{,}\) so we end up with
\begin{equation*} 2e^{-t}\cos(2t) + \frac12 e^{-t}\sin(2t) \end{equation*}

Example 3.1.15.

If \(f(t)\) is a function such that \(F(s) = \mathcal{L}(f)(s) = \frac{1}{s^8+1}\text{,}\) find \(\mathcal{L}(3e^{4t}f(t))(s)\)
To solve this, we look at the part of Table 3.1.11 that says the Laplace transformation of \(e^{ct}f(t)\) is \(F(s-c)\text{.}\) In our case, with \(c = 4\text{,}\) we get
\begin{equation*} \mathcal{L}(e^{4t}f(t))(s) = \frac{3}{(s-4)^8+1} \end{equation*}

Example 3.1.16.

Find the inverse Laplace transformation of the function
\begin{equation*} F(s) = \frac{7s^2+2s+20}{s^3+4s} \end{equation*}
The denominator factors as \(s^3+4s = s(s^2+4)\text{,}\) and we can rewrite this as a partial fraction \(\frac{A}{s}+\frac{Bs+C}{s^2+4}\text{.}\) Clearing denominators, we end up at the equation \(7s^2+2s+20 = A(s^2+4) + s(Bs+C)\text{;}\) solving gives \(A = 5, B = 2, C = 2\text{,}\) so we are to find the inverse Laplace of
\begin{equation*} \frac{5}{s} + \frac{2s}{s^2+4} + \frac{2}{s^2+4} \end{equation*}
The inverse Laplace of the first term is \(5\text{,}\) and using the facts that \(\mathcal{L}(\sin(at)) = \frac{a}{s^2+a^2}\) and \(\mathcal{L}(\cos(at)) = \frac{s}{s^2+a^2}\text{,}\) we end up at
\begin{equation*} 5+2\cos(2t)+\sin(2t) \end{equation*}

Activity 3.1.2. Finding Inverse Transforms.

Find the inverse Laplace transform for each of the following functions.
(a)
\(F(s) = \dfrac{2}{s^2}\)
(b)
\(F(s) = \dfrac{1}{s^4}\)
(c)
\(F(s) = \dfrac{2}{(s - 3)^2 + 4}\)
(d)
\(F(s) = \dfrac{2s^2}{(s^2 + 1)(s - 1)}\)
(e)
\(F(s) = \dfrac{2s - 13}{s(s^2 - 4s + 13)}\)

Subsection 3.1.5 Important Lessons

  • Many initial value problems have discontinuous forcing terms.
  • The Laplace transform of a function \(f(t)\) by
    \begin{equation*} {\mathcal L}(f)(s)= F(s) = \int_0^\infty e^{-st} f(t) \, dt, \end{equation*}
    provided the integral converges.
  • If \(f\) is a piecewise continuous, exponentially bounded function defined \([0, \infty)\text{,}\) then the Laplace transform of \(f\text{,}\)
    \begin{equation*} {\mathcal L}(f)(s)= F(s) = \int_0^\infty e^{-st} f(t) \, dt \end{equation*}
    exists.
  • The Laplace transform is a linear operator; i.e.,
    \begin{equation*} {\mathcal L} ( \alpha f + \beta g )(s) = \alpha {\mathcal L}(f)(s) + \beta {\mathcal L} ( g)(s), \end{equation*}
    where \(\alpha, \beta \in {\mathbb R}\) provided the Laplace transforms of \(f\) and \(g\) exist.
  • If we know the Laplace transform \(F(s)\) of a function, we can recover the original function using the inverse Laplace transform of a function \({\mathcal L}^{-1}(F(s))(t)\text{.}\)
  • The inverse Laplace transform is linear,
    \begin{equation*} {\mathcal L}^{-1} ( \alpha F + \beta G ) = \alpha {\mathcal L}^{-1}(F) + \beta {\mathcal L}^{-1} ( G). \end{equation*}
  • If \({\mathcal L}(f) = F(s)\text{,}\) then \({\mathcal L}(u_c(t) f( t- c)) = e^{-sc} F(s)\text{.}\)
  • If \(c \geq 0\text{,}\) we define the Heaviside function to be
    \begin{equation*} u_c(t) = \begin{cases} 0 & t \lt c \\ 1 & t \geq c. \end{cases} \end{equation*}
    The Laplace transform of \(u_c\) is
    \begin{equation*} {\mathcal L}(u_c(t))(s) = \frac{e^{-cs}}{s}. \end{equation*}

Reading Questions 3.1.6 Reading Questions

1.

Does the Laplace transform of a function always exist? If so, why? If not, given an example?

2.

What is the Heaviside function? Explain.

Exercises 3.1.7 Exercises

Finding Laplace transforms.

Find the Laplace transform for each of the functions in Exercise Group 3.1.7.1–8. Recall that \(\cosh t = (e^t + e^{-t})/2\) and \(\sinh t = (e^t - e^{-t})/2\text{.}\)
1.
\(f(t) = \sin at\)
2.
\(f(t) = \cos at\)
3.
\(f(t) = e^{at} \sin bt\)
4.
\(f(t) = \cosh at\)
5.
\(f(t) = 2 \cos at - 3\sinh at\)
6.
\(f(t) = t e^{at}\)
7.
\(f(t) = t \sin at\)
8.
\(f(t) = t^2 \cos at\)

Finding inverse Laplace transforms.

Find the inverse Laplace transform for each of the functions in Exercise Group 3.1.7.9–16. You will find partial fraction decomposition very useful. A couple exercises have solutions added.
9.
\(F(s) = \dfrac{1}{s(s + 1)}\)
10.
\(F(s) = s^{-3/2}\)
11.
\(F(s) = \dfrac{1}{s + 7}\)
12.
\(F(s) = \dfrac{2}{s^2 + 9}\)
13.
\(F(s) = \dfrac{5s - 4}{2s^2 + s - 1}\)
14.
\(F(s) = \dfrac{2s^2 - s + 4}{s^3 + 4s}\)
15.
\(F(s) = \dfrac{1}{(s - 2)^4}\)
16.
\(F(s) = \dfrac{1}{(s + 1)^2(s^2 - 4)}\)

17.

Prove Theorem 3.1.9: If \(f\) is a piecewise continuous, exponentially bounded function defined \([0, \infty)\text{,}\) then the Laplace transform of \(f\text{,}\)
\begin{equation*} {\mathcal L}(f)(s)= F(s) = \int_0^\infty e^{-st} f(t) \, dt \end{equation*}
exists.

18.

Let \(f\) be a function whose Laplace transform \({\mathcal L}(f)(s)= F(s)\) exists for \(s \gt a\text{.}\) Prove that
\begin{equation*} {\mathcal L}(e^{ct} f(t))(s)= F(s - c) \end{equation*}
for \(s \gt a + c\text{.}\)

19.

Prove Theorem 3.1.10: Suppose that \(f\) and \(g\) are piecewise continuous, exponentially bounded functions defined \([0, \infty)\text{,}\) with Laplace transforms of \(F(s)\) and \(G(s)\text{,}\) respectively. If \(F(s) = G(s)\text{,}\) then \(f(t) = g(t)\text{.}\)

20.

If \(f\) is a piecewise continuous, exponentially bounded function defined \([0, \infty)\text{,}\) prove that the Laplace transform of \(f\text{,}\)
\begin{equation*} {\mathcal L}(f)(s)= F(s) = \int_0^\infty e^{-st} f(t) \, dt \end{equation*}
exists.

21.

Define the gamma function to be
\begin{equation*} \Gamma(x) = \int_0^\infty e^{-t} t^{x-1} \; dt \end{equation*}
for \(x \gt 0\text{.}\) The gamma function is very useful for expressing the Laplace transform of the function \(t^\alpha\text{.}\)
  1. Show that \(\Gamma(1) = 1\text{.}\)
  2. Prove that \(\Gamma(x + 1) = x\Gamma(x)\) and deduce that \(\Gamma(n+1) = n!\) for \(n \in {\mathbb N}\text{.}\)
  3. Show that the Laplace transform of \(f(t) = t^\alpha\) is
    \begin{equation*} {\mathcal L}(f)(s) = \frac{\Gamma(\alpha + 1)}{s^{\alpha + 1}}. \end{equation*}
    If \(n\) is a positive integer, then \({\mathcal L}(t^n)(s) = n!/s^{n + 1}\text{.}\)
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