24.1 Overview of the Procedure

The Laplace transform pairs a function of a real coordinate, call it $t$, with 0 $\raisebox{.3pt}{$<$}$ $t$ $\raisebox{.3pt}{$<$}$ $\infty$, with a different function of a complex coordinate $s$:

$$u(t,\cdot) \begin{array}{c} {\cal L} \\ \Longrighta... ...leftarrow \\ {\cal L}^{-1} \end{array} \hat u(s,\cdot)$$

The pairing is designed to get rid of derivatives with respect to $t$ in equations for the function $u$. This works as long as the coefficients do not depend on $t$ (or at the very most are low degree powers of $t$.) The transformation is convenient since pairings can be looked up in tables.

24.1.1 Typical procedure

Use tables to find the equations satisfied by $\hat u$ from these satisfied by $u$. Solve for $\hat u$ and look up the corresponding $u$ in the tables.

Table 24.1 lists important properties of the Laplace transform and table 24.2 gives example Laplace transform pairs. In the tables, $k$ $\raisebox{.3pt}{$>$}$ 0, $a$, $b$, and $c$ are constants, normally positive, $n$ is a natural number, and

$$\hbox{erfc}(x) \equiv \frac2{\sqrt{\pi}}\int_x^\infty e^{-\xi^2} { \rm d}\xi$$

Table 24.1 assumes that $a$ and $b$ are positive.

Table 24.1: Properties of the Laplace transform.

$$u(t)_{\strut}^{\strut}$$	$$\hat u(s)$$
0. $$\frac{1}{2 \pi{\rm i}} \int_{{\strut}c-{\rm i}\infty}^{{\strut}c+{\rm i}\infty} e^{-st} \hat u(s){ \rm d}s$$	$$\int_0^\infty u(t)e^{-st}{ \rm d}t$$
1. $$C_1 u_1(t) + C_2 u_2(t)$$	$$C_1 \hat u_1(s) + C_2 \hat u_2(s)$$
2. $$u(at)$$	$$a^{-1} \hat u(s/a)$$
3. $$\frac{\partial^{{\strut}n} u}{\partial_{\strut} t^n}(t)$$	$$s^n \hat u(s) - s^{n-1} u(0) - \ldots - \frac{\partial^{n-1} u}{\partial t^{n-1}}(0)$$
4. $$t^n u(t)$$	$$(-1)^n \frac{\partial^{{\strut}n} \hat u}{\partial_{\strut} s^n}$$
5. $$e^{ct} u(t)$$	$$\hat u(s-c)$$
6. $$\bar u(t-b) \equiv H(t-b) u(t-b)$$ $$= \bigg\{_{\strut}^{\strut} \begin{array}{lr} u(t-b) & (t-b>0) 0 & (t-b<0) \end{array}$$	$$e^{-bs} \hat u(s)$$
7. $$\int_{{\strut}0}^{{\strut}t} f(t-\tau)g(\tau){ \rm d}\tau$$	$$\hat f(s)\hat g(s)$$

Table 24.2: Selected Laplace transform pairs.

$$u(t)$$	$$\hat u(s)$$
1. $$1$$	$$\frac{1}{s}_{\strut}$$
2. $$t^n$$	$$\frac{n!}{s^{n+1}}_{\strut}$$
3. $$e^{bt}$$	$$\frac1{s-b}_{\strut}$$
4. $$\sin(at)$$	$$\frac{a}{s^2+a^2}_{\strut}$$
5. $$\cos(at)$$	$$\frac{s}{s^2+a^2}_{\strut}$$
6. $$\frac{1}{\sqrt{\pi t}}_{\strut}$$	$$\frac{1}{\sqrt{s}}$$
7. $$\frac{1}{\sqrt{\pi t}}_{\strut} e^{-k^2/(4t)}$$	$$\frac{1}{\sqrt{s}} e^{-k\sqrt{s}}$$
8. $$\frac{k}{\sqrt{4\pi t^3}}_{\strut} e^{-k^2/(4t)}$$	$$e^{-k\sqrt{s}}$$
9. $$\hbox{erfc}\left(\frac{k}{2\sqrt{t}}\right)_{\strut}$$	$$\frac1s e^{-k\sqrt{s}}$$

24.1.2 About the coordinate to be transformed

In many cases, $t$ is physically time, since time is most likely to satisfy the constraints 0 $\raisebox{.3pt}{$<$}$ $t$ $\raisebox{.3pt}{$<$}$ $\infty$ and coefficients independent of $t$. Also, the Laplace transform likes initial conditions at $t$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, not boundary conditions at both $t$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and $t$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\infty$.