Laplace Transforms and Initial-Value Problems

Definition Let $f$ be defined for $t \ge 0$ and integrable on every interval $[0, T]$. Its Laplace transform $\mathcal{L}\{f\}$ is the function

defined for those real numbers $s$ for which this limit exists as a finite number. Only the values of $f$ for $t \ge 0$ matter, which is why it is called the one-sided Laplace transform.

Definition The region of convergence of $\mathcal{L}\{f\}$ is the set of real $s$ for which the improper integral converges. A function $f$ is of exponential order $a$ if there are constants $M > 0$ and $T_0 \ge 0$ with

Existence theorem. If $f$ is piecewise continuous on every interval $[0, T]$ and of exponential order $a$, then the integral converges absolutely for every $s > a$, and $F(s) \to 0$ as $s \to \infty$. Unless it is empty or the whole real line, the region of convergence is a half-line. For absolute convergence this is easy to see, because if the integral of $e^{-s_0 t} \lvert f(t) \rvert$ is finite, then so is the integral of the smaller function $e^{-st} \lvert f(t) \rvert$ for every $s > s_0$.

Definition Suppose $f$ is continuous on $[0, \infty)$ and of exponential order $a$, and $f'$ is piecewise continuous on every interval $[0, T]$. Then for $s > a$

If also $f'$ is continuous on $[0, \infty)$ and of exponential order $a$, and $f''$ is piecewise continuous on every $[0, T]$, then for $s > a$

The proof is integration by parts on $[0, T]$, where the boundary term $e^{-sT} f(T)$ tends to $0$ because $s > a$.

Definition Uniqueness theorem. Let $f$ and $g$ be piecewise continuous on every $[0, T]$ and of exponential order, with $\mathcal{L}\{f\}(s) = \mathcal{L}\{g\}(s)$ for all $s$ greater than some $s_0$. Then $f(t) = g(t)$ at every $t \ge 0$ where both are continuous. In particular two continuous functions with the same Laplace transform are equal everywhere on $[0, \infty)$. The inverse Laplace transform $\mathcal{L}^{-1}\{F\}$ is the continuous $f$ with $\mathcal{L}\{f\} = F$ when there is one. Otherwise it is a piecewise continuous $f$, determined except at its points of discontinuity, where the values do not matter. The inverse is linear.

Definition For $c \ge 0$ the unit step function is

with $\mathcal{L}\{u_c\} = \frac{e^{-cs}}{s}$ for $s > 0$. Delay rule. If $\mathcal{L}\{f\} = F(s)$ exists for $s > a$ and $c \ge 0$, then for $s > a$

The graph of $u_c(t) f(t - c)$ is the graph of $f$ moved $c$ units to the right, with the value $0$ before $t = c$.

Definition For $c > 0$ and $\varepsilon > 0$, the pulse $d_\varepsilon(t) = \frac{1}{\varepsilon}\bigl(u_c(t) - u_{c + \varepsilon}(t)\bigr)$ has height $\frac{1}{\varepsilon}$ on $[c, c + \varepsilon)$ and total area $1$. For each $s > 0$ its Laplace transform satisfies

The unit impulse at $c$, written $\delta(t - c)$, is the generalized input whose integral against each $g$ continuous on $[0, \infty)$ is the limit of the pulse integrals,

The choice $g(t) = e^{-st}$ gives $\mathcal{L}\{\delta(t - c)\} = e^{-cs}$, the limit above. The impulse is not an ordinary function. The condition $c > 0$ keeps the kick inside the interval of integration, and an impulse at $t = 0$ needs an extra convention about the lower limit. No piecewise continuous function $f$ of exponential order has Laplace transform $e^{-cs}$. Otherwise its running integral $h(t) = \int_0^t f(\tau)\, d\tau$ would have Laplace transform $\frac{e^{-cs}}{s}$ by the derivative rule, so uniqueness would give $h = u_c$ away from $t = c$, which is impossible because $h$ is continuous and $u_c$ jumps at $c$.

Definition For $f$ and $g$ piecewise continuous on every $[0, T]$, their convolution is

for $t \ge 0$. Convolution theorem. If $f$ and $g$ are piecewise continuous on every $[0, T]$ and of exponential order $a$, then $f * g$ is of exponential order $b$ for every $b > a$, and for $s > a$

Definition Consider $a y'' + b y' + c y = g(t)$ with constants $a \ne 0$, $b$, $c$, initial values $y(0) = y_0$ and $y'(0) = y_1$, and $g$ piecewise continuous on every $[0, T]$ and of exponential order. Then the solution $y$ and its derivative $y'$ are continuous and of exponential order, so the rules above apply. Taking Laplace transforms gives

valid for all $s$ large enough. Solve for $Y$, apply the inverse Laplace transform, and the result is the solution for $t \ge 0$. When $g$ has jumps, $y''$ jumps there too and the equation holds except at those points. With an impulse input $k\,\delta(t - t_0)$, where $t_0 > 0$, the solution stays continuous and $y'$ jumps by $\frac{k}{a}$ at $t = t_0$.