The Laplace Transform, differential equations lesson

The signal $e^{3t}$ encloses infinite area on $[0, \infty)$ , so its plain integral says nothing useful. Multiply it first by a decaying exponential $e^{-st}$ , though, and the area becomes finite as soon as the decay is fast enough. The Laplace transform records that weighted area as a function of the decay parameter $s$ . This lesson defines it, gives conditions under which it exists, finds the values of $s$ where the integral converges and builds the first entries of a table by linearity. Throughout, $t \ge 0$ is time and $s$ is a complex number.

Key terms

One-sided Laplace transform

Definition Let $f$ be defined for $t \ge 0$ and integrable on every interval $[0, T]$ . Its one-sided Laplace transform is the function

\begin{aligned} F(s) &= \mathcal{L}\{f\}(s) \\[0.4em] &= \int_0^\infty e^{-st} f(t)\,dt, \end{aligned}

where the improper integral means $\lim_{T \to \infty} \int_0^T e^{-st} f(t)\,dt$ . It is defined at each complex number $s$ for which this limit exists as a finite number. Writing $s = \sigma + i\omega$ with real $\sigma$ and $\omega$ , the weight has modulus $\lvert e^{-st} \rvert = e^{-\sigma t}$ , so the real part of $s$ sets the rate of decay and the imaginary part adds oscillation. Only the values of $f$ for $t \ge 0$ enter the integral, which is why this Laplace transform is called one-sided. A lowercase letter names the function of $t$ , and the matching capital names its Laplace transform, a function of $s$ .

In plain words Multiply the signal by the decaying weight $e^{-st}$ and take the total area from $0$ to $\infty$ . Each admissible $s$ produces one number. The time variable is integrated away, so the result depends on $s$ alone.

Example For the constant signal $f(t) = 1$ and any $s \ne 0$ ,

\int_0^T e^{-st}\,dt = \frac{1 - e^{-sT}}{s}.

If $\operatorname{Re} s > 0$ , then $\lvert e^{-sT} \rvert = e^{-(\operatorname{Re} s)T} \to 0$ , so $\mathcal{L}\{1\}(s) = 1/s$ . If $\operatorname{Re} s < 0$ , the term $e^{-sT}$ grows without bound. If $\operatorname{Re} s = 0$ and $s \ne 0$ , it keeps moving around the unit circle, and at $s = 0$ the integral equals $T$ . None of these cases has a finite limit, so the pair $\mathcal{L}\{1\} = 1/s$ holds exactly on the half-plane $\operatorname{Re} s > 0$ .

Piecewise continuity

Definition A function $f$ is piecewise continuous on $[0, \infty)$ if on every bounded interval $[0, T]$ it is continuous except at finitely many points, and at each of those points its one-sided limits exist and are finite. At $t = 0$ only the right-hand limit is required. Such a function is bounded and integrable on every $[0, T]$ . Changing its value at finitely many points of a bounded interval does not change any integral, because single points carry no area. For the same reason a Laplace transform cannot see the value at a jump. Two piecewise continuous functions of exponential order with the same Laplace transform agree at every point where both are continuous, while at a jump their values may differ.

In plain words The graph may jump, but only finitely often in any finite stretch of time, and each jump goes between two finite heights. The function never runs off to infinity at a finite time.

Example The switched signal equal to $1$ for $0 \le t < 2$ and to $0$ for $t \ge 2$ is piecewise continuous, with one jump at $t = 2$ . Redefining its value at $t = 2$ as $1$ , or as $5$ , leaves its Laplace transform unchanged. The function $1/(t - 1)$ is not piecewise continuous on $[0, \infty)$ , because it has no finite one-sided limit at $t = 1$ . Neither is $t^{-1/2}$ , which is unbounded near $t = 0$ . Even so, the integral of $e^{-st}t^{-1/2}$ converges for $\operatorname{Re} s > 0$ and equals $\sqrt{\pi/s}$ , with the principal square root, so piecewise continuity is a convenient sufficient condition and not a necessary one.

Exponential order

Definition Let $a$ be a real number. A function $f$ on $[0, \infty)$ is of exponential order $a$ if there are constants $M > 0$ and $T_0 \ge 0$ with

\begin{gathered} \lvert f(t) \rvert \le M e^{at} \\[0.4em] \text{for all } t \ge T_0. \end{gathered}

A function of exponential order $a$ is also of exponential order $b$ for every $b > a$ , so the order is a growth bound and not a single characteristic number.

In plain words After some time the size of the signal stays under a fixed exponential curve. What the signal does early on does not matter, only how fast it can grow in the long run.

Example The function $e^{3t}$ is of exponential order $3$ , with $M = 1$ and $T_0 = 0$ , and $\sin(5t)$ is of exponential order $0$ . A power $t^n$ with $n \ge 1$ is of exponential order $a$ for every $a > 0$ , because $t^n e^{-at}$ is continuous and tends to $0$ , so it is bounded on $[0, \infty)$ . It is not of exponential order $0$ , since $t^n$ is unbounded. The function $e^{t^2}$ is of no exponential order at all, because $e^{t^2}/e^{at} = e^{t(t - a)}$ tends to infinity for every real $a$ .

Existence theorem for the Laplace transform

Definition Let $f$ be piecewise continuous on $[0, \infty)$ and of exponential order $a$ , with constants $M$ and $T_0$ as above. Then the Laplace integral of $f$ converges absolutely for every complex $s$ with $\operatorname{Re} s > a$ , so $F(s) = \mathcal{L}\{f\}(s)$ exists on that half-plane. The proof is a comparison. Write $\sigma = \operatorname{Re} s$ . The integral over $[0, T_0]$ is finite because $f$ is bounded there, and for $t \ge T_0$

\lvert e^{-st} f(t) \rvert \le M e^{-(\sigma - a)t},

a bound whose integral over $[T_0, \infty)$ is finite because $\sigma > a$ . The theorem gives sufficient conditions. A function that fails them, such as $t^{-1/2}$ , may still have a Laplace transform.

In plain words If the signal grows no faster than $e^{at}$ , then every weight $e^{-st}$ with $\operatorname{Re} s > a$ decays faster than the signal grows, and the weighted area is finite.

Example Let $f(t) = e^{2t}$ for $0 \le t < 1$ and $f(t) = 3$ for $t \ge 1$ . This signal is piecewise continuous, with one jump at $t = 1$ , and it is of exponential order $0$ with $M = 3$ and $T_0 = 1$ . The theorem guarantees $F(s)$ for $\operatorname{Re} s > 0$ , even though the early piece $e^{2t}$ grows faster. Computing the two pieces separately gives, for $\operatorname{Re} s > 0$ and $s \ne 2$ ,

F(s) = \frac{e^{2 - s} - 1}{2 - s} + \frac{3e^{-s}}{s}.

Region of convergence

Definition The region of convergence of $\mathcal{L}\{f\}$ is the set of complex $s$ at which the Laplace integral converges. It has a simple shape. If the integral converges at $s_0$ , it converges at every $s$ with $\operatorname{Re} s > \operatorname{Re} s_0$ . So the region is empty, the whole plane, or an open half-plane $\operatorname{Re} s > \sigma_0$ together with possibly some points of the boundary line $\operatorname{Re} s = \sigma_0$ . The existence theorem places the half-plane $\operatorname{Re} s > a$ inside the region, but the true boundary $\sigma_0$ can lie further to the left. A formula for $F(s)$ gives values of the Laplace transform only inside the region of convergence.

In plain words Every Laplace transform comes with the set of $s$ where its integral makes sense. Moving $s$ to the right only strengthens the decay of the weight, so once the integral converges it keeps converging further right.

Example For $s \ne 3$ ,

\begin{aligned} &\int_0^T e^{-st}e^{3t}\,dt \\[0.4em] &\quad = \int_0^T e^{-(s - 3)t}\,dt \\[0.4em] &\quad = \frac{1 - e^{-(s - 3)T}}{s - 3}, \end{aligned}

which tends to $1/(s - 3)$ exactly when $\operatorname{Re} s > 3$ . So $\mathcal{L}\{e^{3t}\} = 1/(s - 3)$ with region $\operatorname{Re} s > 3$ , although the fraction itself is defined for every $s \ne 3$ . Since $e^{3t} \le e^{5t}$ for $t \ge 0$ , the function is also of exponential order $5$ , and the theorem with $a = 5$ only promises $\operatorname{Re} s > 5$ . That statement is true but weaker, and it does not locate the boundary. At the other extreme, $e^{t^2}$ has an empty region. For real $s$ the integrand $e^{t^2 - st}$ is positive and grows without bound, so the integral diverges for every real $s$ , and by the half-plane property it converges for no complex $s$ either.

Linearity of the Laplace transform

Definition Suppose $\mathcal{L}\{f\}(s) = F(s)$ exists for $\operatorname{Re} s > a$ and $\mathcal{L}\{g\}(s) = G(s)$ exists for $\operatorname{Re} s > b$ . Then for all constants $\alpha$ and $\beta$ and every $s$ with $\operatorname{Re} s > \max(a, b)$ ,

\begin{aligned} &\mathcal{L}\{\alpha f + \beta g\}(s) \\[0.4em] &\quad = \alpha F(s) + \beta G(s). \end{aligned}

The rule follows from the linearity of integrals over $[0, T]$ and of limits. A combination can converge on a larger set than this common half-plane, for example when terms cancel. But if one term converges at $s$ and the other diverges there, their sum diverges at $s$ , so linearity cannot rescue a divergent term.

In plain words The Laplace transform of a sum is the sum of the Laplace transforms, and constants come out in front, wherever all the pieces converge.

Example Integration by parts on $[0, T]$ , differentiating $t$ and integrating $e^{-st}$ , gives for $s \ne 0$

\begin{aligned} \int_0^T t e^{-st}\,dt &= -\frac{T e^{-sT}}{s} \\[0.4em] &\quad + \frac{1 - e^{-sT}}{s^2}. \end{aligned}

For $\operatorname{Re} s > 0$ the terms containing $e^{-sT}$ tend to $0$ , because exponential decay beats the factor $T$ , so $\mathcal{L}\{t\} = 1/s^2$ there. Together with $\mathcal{L}\{1\} = 1/s$ and $\mathcal{L}\{e^{3t}\} = 1/(s - 3)$ , linearity gives for the signal $f(t) = 2 + t + e^{3t}$

\begin{gathered} F(s) = \frac{2}{s} + \frac{1}{s^2} + \frac{1}{s - 3}, \\[0.4em] \operatorname{Re} s > 3. \end{gathered}

The constant and the ramp converge for $\operatorname{Re} s > 0$ , while the exponential needs $\operatorname{Re} s > 3$ , and that half-plane is exactly the region of convergence of $F$ . Reading the three terms backward returns $2$ , $t$ and $e^{3t}$ , which is a quick check of the bookkeeping. Because $f$ is continuous, no other continuous signal has this Laplace transform.

Check yourself

Find the Laplace transform of $f(t) = 5 - 4e^{-2t}$ and its region of convergence.

Show the answer

The exponential calculation with $-2$ in place of $3$ gives $\mathcal{L}\{e^{-2t}\} = 1/(s + 2)$ for $\operatorname{Re} s > -2$ , and $\mathcal{L}\{1\} = 1/s$ for $\operatorname{Re} s > 0$ . By linearity

\begin{aligned} F(s) &= \frac{5}{s} - \frac{4}{s + 2} \\[0.4em] &= \frac{s + 10}{s(s + 2)}. \end{aligned}

The region of convergence is $\operatorname{Re} s > 0$ , where both pieces converge. For $-2 < \operatorname{Re} s \le 0$ the exponential piece converges but the constant does not, so the sum diverges, and further left it diverges by the half-plane property.

A classmate uses $\mathcal{L}\{e^{5t}\} = 1/(s - 5)$ to conclude that the Laplace transform of $e^{5t}$ equals $-1$ at $s = 4$ . What is wrong?

Show the answer

The pair holds only in the region of convergence $\operatorname{Re} s > 5$ . At $s = 4$ the integrand is $e^{-4t}e^{5t} = e^{t}$ , and its integral over $[0, T]$ is $e^{T} - 1$ , which grows without bound. So the Laplace transform of $e^{5t}$ does not exist at $s = 4$ . The number $-1$ is a value of the fraction $1/(s - 5)$ , not of the integral.

Let $f(t) = 1$ for $0 \le t < 2$ and $f(t) = 0$ for $t \ge 2$ . Find $\mathcal{L}\{f\}$ from the definition, and give its region of convergence.

Show the answer

The integrand is zero for $t \ge 2$ , so for every $T \ge 2$ the integral over $[0, T]$ is the integral over $[0, 2]$ . For $s \ne 0$

\begin{aligned} F(s) &= \int_0^2 e^{-st}\,dt \\[0.4em] &= \frac{1 - e^{-2s}}{s}, \end{aligned}

and $F(0) = 2$ , which is also the limit of the fraction as $s \to 0$ . No decay is needed, so the region of convergence is the whole complex plane. This agrees with the existence theorem, because $f$ is of exponential order $a$ for every real $a$ , with $M = 1$ and $T_0 = 2$ . The value of $f$ at the jump $t = 2$ plays no part.

The Laplace Transform

Key terms

One-sided Laplace transform

Piecewise continuity

Exponential order

Existence theorem for the Laplace transform

Region of convergence

Linearity of the Laplace transform

Check yourself

How this lesson connects

Builds on

Used later in