Laplace Transforms of Derivatives and Initial-Value Problems

Definition Let $y$ be continuous on $[0, T]$ and differentiable except at finitely many points, with $y'$ piecewise continuous. For every complex $s$,

This is integration by parts with $u = e^{-st}$ and $dv = y'(t)\,dt$, carried out on each piece between the points where $y'$ fails to exist. The boundary terms at those interior points cancel because $y$ is continuous. If instead $y$ has a jump at a point $t_0$ in $(0, T)$, the terms there do not cancel, and the right side gains the extra term $-e^{-st_0}\bigl(y(t_0^+) - y(t_0^-)\bigr)$.

Definition Let $y$ be continuous on $[0, \infty)$ and of exponential order $\gamma$, so $\lvert y(t) \rvert \le Me^{\gamma t}$ for $t \ge T_0$, and let $y'$ be piecewise continuous on $[0, \infty)$. Then $\mathcal{L}\{y'\}(s)$ exists for $\operatorname{Re} s > \gamma$ and

If also $y'$ is continuous and of exponential order $\gamma$, and $y''$ is piecewise continuous, then for $\operatorname{Re} s > \gamma$

The first rule is the integration by parts above as $T \to \infty$, because $\lvert e^{-sT}y(T) \rvert \le Me^{-(\operatorname{Re} s - \gamma)T}$ for $T \ge T_0$, and this bound tends to $0$. The second applies the first rule to $y'$. In the same way, if $y, y', \ldots, y^{(n-1)}$ are continuous and of exponential order $\gamma$ and $y^{(n)}$ is piecewise continuous, then $\mathcal{L}\{y^{(n)}\} = s^nY - s^{n-1}y(0) - \cdots - y^{(n-1)}(0)$ for $\operatorname{Re} s > \gamma$.

Definition The initial-value terms are the known quantities built from $y(0), y'(0), \ldots$ that the derivative rules attach to $Y$. For the expression $y'' + py' + qy$ with constants $p$ and $q$, and $y$ meeting the hypotheses of the second-derivative rule,

The initial data enter the calculation only through these terms. Dropping them gives the Laplace transform of the solution with $y(0) = 0$ and $y'(0) = 0$, which is in general a different function.

Definition Consider $y'' + py' + qy = g(t)$ with constants $p$ and $q$, initial values $y(0) = y_0$ and $y'(0) = y_1$, and a piecewise continuous forcing $g$ of exponential order with Laplace transform $G$. Applying $\mathcal{L}$ to both sides gives, on a half-plane where all the Laplace transforms exist,

where

The polynomial $P$ is the characteristic polynomial of the equation, written in the variable $s$, and $I$ collects the initial-value terms. For $\operatorname{Re} s$ large enough $P(s) \ne 0$, and

The first part is the Laplace transform of the response that starts from rest, and the second is that of the free motion from the initial state with no forcing.

Definition Consider $y'' + py' + qy = g(t)$ for $t \ge 0$, with constants $p$ and $q$, data $y(0) = y_0$ and $y'(0) = y_1$, and $g$ piecewise continuous and of exponential order. Then the solution $y$ and its derivative $y'$ are continuous and of exponential order, so the derivative rules apply. First take the Laplace transform of every term and keep the initial-value terms. Then solve the algebraic equation for $Y$, split $Y$ by partial fractions and invert each piece. Finally check the result in the differential equation and in both initial conditions. The answer is the solution for $t \ge 0$, and when $g$ has jumps, $y''$ jumps there too and the equation holds except at those points. Data given at a time $t_0 \ne 0$ need the shift $\tau = t - t_0$ first. The method relies on constant coefficients and linearity. A coefficient that depends on $t$ brings in derivatives with respect to $s$, as in $\mathcal{L}\{ty'\} = -Y - s\,dY/ds$, and a nonlinear term has no rule of this kind, since $\mathcal{L}\{y^2\}$ is not $Y^2$ in general. For $y = 1$, for example, $\mathcal{L}\{y^2\} = 1/s$ while $Y^2 = 1/s^2$.