The Laplace Transform — Definition and Basic Properties
import numpy as np
import sympy as sym
import matplotlib as mpl
import matplotlib.pyplot as plt
from scipy.integrate import solve_ivp, quad
from IPython.display import Math, display
mpl.rcParams['figure.dpi'] = 150
mpl.rcParams['axes.spines.top'] = False
mpl.rcParams['axes.spines.right'] = FalseUnderstand the big picture: what the Laplace transform does and why it is useful.
Learn the definition \(\mathcal{L}[x(t)](s) = \int_0^\infty x(t)e^{-st}\,dt\) and compute transforms directly from it.
Know the linearity of \(\mathcal{L}\) and use it to build up transforms from a table.
Understand the inverse transform \(\mathcal{L}^{-1}\) and use partial fractions.
Work with the Heaviside function to represent piecewise-defined forcing.
Apply the shift property and switching property as time-saving tools.
Use all of the above to solve constant-coefficient IVPs with discontinuous forcing.
This material corresponds to Section 3.1 of (Logan 2015).
Before computing anything, it is worth pausing to appreciate what the Laplace transform accomplishes and why it is worth learning an entirely new technique.
The methods we have developed — undetermined coefficients and variation of parameters — break down when the forcing function \(f(t)\) is discontinuous (a switch being flipped, a pulse applied for a finite time, a sudden impact). These situations are not exotic mathematical curiosities; they are among the most common scenarios in electrical engineering, mechanical engineering, and control theory.
\[\underbrace{ax''+bx'+cx = f(t)}_{\text{ODE (hard to solve)}} \xrightarrow{\;\mathcal{L}\;} \underbrace{(as^2+bs+c)X(s) = F(s) + \text{ICs}}_{\text{algebraic equation (easy to solve)}} \xrightarrow{\;\mathcal{L}^{-1}\;} \underbrace{x(t)}_{\text{solution}}\]
The Laplace transform converts differentiation into multiplication by \(s\). What was a differential equation becomes an algebraic equation for \(X(s)\). We solve algebraically, then invert.
Think of it as translating a hard problem in one language (differential equations) into an easier problem in another language (algebra), solving there, then translating back. The same philosophy underlies logarithms (which convert multiplication into addition), the Fourier transform (which converts convolution into multiplication), and many other transforms throughout mathematics.
Let \(x = x(t)\) be defined on \([0, \infty)\). The Laplace transform of \(x(t)\) is the function \(X(s)\) defined by \[ \mathcal{L}[x(t)](s) = X(s) = \int_0^\infty x(t)\,e^{-st}\,dt, \] provided the improper integral converges. We also write \(\mathcal{L}[x] = X(s)\).
Unpacking the definition. The integrand \(x(t)e^{-st}\) is a function of both \(t\) and \(s\). We integrate over \(t\) from \(0\) to \(\infty\), which eliminates the \(t\) dependence and leaves a function of \(s\) alone. The factor \(e^{-st}\) is a damping kernel that forces the integrand to zero as \(t\to\infty\) (provided \(s\) is large enough), making the improper integral converge.
The variable \(s\) lives in the transform domain (sometimes called the frequency domain or the \(s\)-domain), while \(t\) lives in the time domain. We use uppercase letters for transform-domain functions: \(x(t) \leftrightarrow X(s)\).
| Time domain | Transform domain | |
|---|---|---|
| Variable | \(t\) | \(s\) |
| Function | \(x(t)\) (lowercase) | \(X(s)\) (uppercase) |
| Language | “what happens” | “how fast it oscillates/decays” |
\[X(s) = \int_0^\infty e^{at}e^{-st}\,dt = \int_0^\infty e^{(a-s)t}\,dt = \frac{1}{a-s}e^{(a-s)t}\Bigg|_{t=0}^{t=\infty}.\]
For the upper limit to vanish we need \(a - s < 0\), i.e., \(s > a\). Evaluating explicitly: \[\frac{1}{a-s}e^{(a-s)t}\Bigg|_{t=0}^{t=\infty} = \frac{1}{a-s}\bigl[0 - 1\bigr] = \frac{-1}{a-s} = \frac{1}{s-a}.\]
\[\boxed{\mathcal{L}[e^{at}] = \frac{1}{s-a}, \quad s > a.}\]
Special case \(a=0\): \(\mathcal{L}[1] = \frac{1}{s}\), \(s > 0\).
Integration by parts with \(u = t\), \(dv = e^{-st}dt\):
\[X(s) = \int_0^\infty t\,e^{-st}\,dt = \left[t\cdot\frac{e^{-st}}{-s}\right]_0^\infty + \frac{1}{s}\int_0^\infty e^{-st}\,dt = 0 + \frac{1}{s}\cdot\frac{1}{s} = \frac{1}{s^2}.\]
The boundary term \(\bigl[te^{-st}/(-s)\bigr]_0^\infty = 0\): at \(t=0\) the term vanishes; as \(t\to\infty\), \(te^{-st}\to 0\) for \(s>0\) (by L’Hôpital’s rule, \(\lim_{t\to\infty} t/e^{st} = \lim_{t\to\infty} 1/(se^{st}) = 0\)).
\[\boxed{\mathcal{L}[t] = \frac{1}{s^2}, \quad s > 0.}\]
The pattern continues: \(\mathcal{L}[t^n] = n!/s^{n+1}\).
Using \(\sin kt = \frac{1}{2i}(e^{ikt}-e^{-ikt})\) and the formula for \(\mathcal{L}[e^{at}]\):
\[\mathcal{L}[\sin kt] = \frac{1}{2i}\left(\frac{1}{s-ik}-\frac{1}{s+ik}\right) = \frac{1}{2i}\cdot\frac{2ik}{s^2+k^2} = \frac{k}{s^2+k^2}.\]
Similarly \(\mathcal{L}[\cos kt] = s/(s^2+k^2)\).
Compare \(\mathcal{L}[\sin kt] = k/(s^2+k^2)\) and \(\mathcal{L}[\cos kt] = s/(s^2+k^2)\). The numerator of the cosine transform involves \(s\) (the transform variable), while the sine transform has a constant numerator \(k\). This asymmetry reflects the fact that \(\cos(0)=1\) and \(\sin(0)=0\) — the cosine has a nonzero initial value.
t_plot = np.linspace(0, 5, 400)
fig, axes = plt.subplots(1, 2, figsize=(11, 4.5))
# Left: damped integrand for two values of s
x_func = np.sin(2*t_plot)
for s_val, color, lbl in [(0.5, 'steelblue', '$s=0.5$'), (2.0, 'crimson', '$s=2.0$')]:
damp = np.exp(-s_val*t_plot)
integrand = x_func * damp
axes[0].plot(t_plot, integrand, color=color, lw=2, label=f'$\\sin(2t)e^{{-{s_val}t}}$ ({lbl})')
axes[0].fill_between(t_plot, 0, integrand, where=(integrand>0), alpha=0.12, color=color)
axes[0].fill_between(t_plot, 0, integrand, where=(integrand<0), alpha=0.12, color=color)
axes[0].plot(t_plot, x_func, 'k--', lw=1.2, alpha=0.5, label='$\\sin(2t)$ (undamped)')
axes[0].axhline(0, color='k', lw=0.5)
axes[0].set_xlabel('$t$'); axes[0].set_ylabel('integrand')
axes[0].set_title(r'Integrand $\sin(2t)\,e^{-st}$ for two values of $s$')
axes[0].legend(fontsize=8.5)
# Right: X(s) = 2/(s^2+4)
s_arr = np.linspace(0.01, 6, 400)
X_s = 2/(s_arr**2 + 4)
axes[1].plot(s_arr, X_s, color='seagreen', lw=2.5, label=r'$X(s)=\frac{2}{s^2+4}$')
for s_val, color in [(0.5, 'steelblue'), (2.0, 'crimson')]:
val = 2/(s_val**2+4)
axes[1].plot(s_val, val, 'o', color=color, markersize=9,
label=f'$X({s_val})={val:.3f}$ (area under left plot)')
axes[1].set_xlabel('$s$'); axes[1].set_ylabel('$X(s)$')
axes[1].set_title(r'$\mathcal{L}[\sin 2t](s) = 2/(s^2+4)$')
axes[1].legend(fontsize=8.5); axes[1].set_ylim(0, 0.55)
plt.suptitle('How the Laplace transform works', fontsize=12)
plt.tight_layout()
plt.show()
For any constants \(\alpha\), \(\beta\) and functions \(x(t)\), \(y(t)\): \[\mathcal{L}[\alpha x + \beta y] = \alpha\mathcal{L}[x] + \beta\mathcal{L}[y].\]
This follows immediately from the linearity of the integral and means we can build up transforms of complicated functions from the table, one term at a time.
Example. \[\mathcal{L}[3e^{-2t} + 5\sin 4t] = 3\cdot\frac{1}{s+2} + 5\cdot\frac{4}{s^2+16} = \frac{3}{s+2} + \frac{20}{s^2+16}.\]
Not every function has a Laplace transform. Two sufficient conditions guarantee existence:
Piecewise continuity (PWC): on any bounded subinterval \([0,b]\), \(x(t)\) has only finitely many jump discontinuities, and at each jump the one-sided limits exist (are finite).
Exponential order: there exist constants \(M > 0\) and \(r\) such that \(|x(t)| \leq Me^{rt}\) for all \(t > t_0\).
If both conditions hold, then \(X(s)\) exists for all \(s > r\).
Functions that fail: - \(x(t) = e^{t^2}\): grows faster than any exponential, so not of exponential order. - \(x(t) = 1/t\): not PWC on \([0,b]\) (blows up at \(t=0\)). - \(X(s) = (1/s)e^s\): grows without bound as \(s\to\infty\) (whereas every valid Laplace transform of a piecewise-continuous, exponentially-ordered function satisfies \(X(s)\to 0\) as \(s\to\infty\)), so \((1/s)e^s\) cannot correspond to any such function.
Given \(X(s)\), we want the function \(x(t)\) whose transform is \(X(s)\). We write: \[x(t) = \mathcal{L}^{-1}[X(s)].\]
The inverse transform is also linear: \(\mathcal{L}^{-1}[\alpha X + \beta Y] = \alpha\mathcal{L}^{-1}[X] + \beta\mathcal{L}^{-1}[Y]\).
In practice, we find inverse transforms by: 1. Looking up the form in the table, OR 2. Using partial fractions to decompose \(X(s)\) into table-recognizable pieces.
Most \(X(s)\) arising from ODEs are rational functions. The partial fractions decomposition rewrites them as sums of simpler fractions, each of which appears in the table.
Type 1 — Distinct real poles. If \(X(s) = P(s)/[(s-a)(s-b)]\): \[X(s) = \frac{A}{s-a} + \frac{B}{s-b} \;\Longrightarrow\; x(t) = Ae^{at} + Be^{bt}.\]
Type 2 — Repeated real pole. If \(X(s) = P(s)/(s-a)^2\): \[X(s) = \frac{A}{s-a} + \frac{B}{(s-a)^2} \;\Longrightarrow\; x(t) = Ae^{at} + Bte^{at}.\]
Type 3 — Complex conjugate poles. If \(X(s) = P(s)/[(s-\alpha)^2+\beta^2]\):
Complete the square in the denominator, then match to \(\mathcal{L}[e^{\alpha t}\sin\beta t]\) and \(\mathcal{L}[e^{\alpha t}\cos\beta t]\).
Worked example. Find \(\mathcal{L}^{-1}\!\left[\dfrac{2s+3}{s^2+2s+5}\right]\).
Complete the square: \(s^2+2s+5 = (s+1)^2+4\). Write the numerator to match the shifted cosine and sine forms: \[\frac{2s+3}{(s+1)^2+4} = \frac{2(s+1)-2+3}{(s+1)^2+4} = \frac{2(s+1)}{(s+1)^2+4} + \frac{1}{(s+1)^2+4}.\]
The second fraction needs a factor of \(\beta = 2\) in the numerator: \(\dfrac{1}{(s+1)^2+4} = \dfrac{1}{2}\cdot\dfrac{2}{(s+1)^2+4}\).
Reading off from the table with \(\alpha=-1\), \(\beta=2\): \[x(t) = 2e^{-t}\cos(2t) + \tfrac{1}{2}e^{-t}\sin(2t).\]
Example. Find \(\mathcal{L}^{-1}\!\left[\frac{2s+9}{(s+1)(s+3)}\right]\).
\[\frac{2s+9}{(s+1)(s+3)} = \frac{A}{s+1}+\frac{B}{s+3}.\]
Cover-up rule: \(A = (2(-1)+9)/(-1+3) = 7/2\); \(B = (2(-3)+9)/(-3+1) = -3/2\).
\[x(t) = \frac{7}{2}e^{-t} - \frac{3}{2}e^{-3t}.\]
s_v = sym.Symbol('s')
t_v = sym.Symbol('t', positive=True)
# Demonstrate partial fractions with SymPy
examples_inv = [
(r"\frac{2s+9}{(s+1)(s+3)}", (2*s_v+9)/((s_v+1)*(s_v+3))),
(r"\frac{3}{s^2-4}", 3/(s_v**2-4)),
(r"\frac{2s+3}{s^2+2s+5}", (2*s_v+3)/(s_v**2+2*s_v+5)),
(r"\frac{1}{s(s+2)}", 1/(s_v*(s_v+2))),
]
for label, expr in examples_inv:
pf = sym.apart(expr, s_v)
inv = sym.inverse_laplace_transform(expr, s_v, t_v)
print(f"$X(s) = {label}$")
display(Math(r"\text{PF: } " + sym.latex(pf)))
display(Math(r"x(t) = " + sym.latex(sym.simplify(inv))))
print()$X(s) = \frac{2s+9}{(s+1)(s+3)}$\(\displaystyle \text{PF: } - \frac{3}{2 \left(s + 3\right)} + \frac{7}{2 \left(s + 1\right)}\)
\(\displaystyle x(t) = \frac{\left(7 e^{2 t} - 3\right) e^{- 3 t}}{2}\)
$X(s) = \frac{3}{s^2-4}$\(\displaystyle \text{PF: } - \frac{3}{4 \left(s + 2\right)} + \frac{3}{4 \left(s - 2\right)}\)
\(\displaystyle x(t) = \frac{3 \sinh{\left(2 t \right)}}{2}\)
$X(s) = \frac{2s+3}{s^2+2s+5}$\(\displaystyle \text{PF: } \frac{2 s + 3}{s^{2} + 2 s + 5}\)
\(\displaystyle x(t) = \frac{\left(\sin{\left(2 t \right)} + 4 \cos{\left(2 t \right)}\right) e^{- t}}{2}\)
$X(s) = \frac{1}{s(s+2)}$\(\displaystyle \text{PF: } - \frac{1}{2 \left(s + 2\right)} + \frac{1}{2 s}\)
\(\displaystyle x(t) = \frac{1}{2} - \frac{e^{- 2 t}}{2}\)
The single most important property for solving ODEs is:
\[\mathcal{L}[x'(t)] = s\,X(s) - x(0)\] \[\mathcal{L}[x''(t)] = s^2 X(s) - s\,x(0) - x'(0)\]
More generally: \(\mathcal{L}[x^{(n)}] = s^n X(s) - s^{n-1}x(0) - \cdots - x^{(n-1)}(0)\).
Derivation (for \(x'\)): integrate by parts with \(u = e^{-st}\), \(dv = x'(t)dt\): \[\mathcal{L}[x'] = \int_0^\infty x'(t)e^{-st}dt = \left[x(t)e^{-st}\right]_0^\infty + s\int_0^\infty x(t)e^{-st}dt = -x(0) + sX(s).\]
Applying \(\mathcal{L}\) to \(ax''+bx'+cx = f(t)\) and using linearity: \[a\mathcal{L}[x''] + b\mathcal{L}[x'] + c\mathcal{L}[x] = \mathcal{L}[f].\] Substituting the derivative formulas: \[\bigl(as^2+bs+c\bigr)X(s) = F(s) + a\bigl(sx(0)+x'(0)\bigr) + b\,x(0).\] The initial conditions are automatically incorporated — no separate step needed. Solving for \(X(s)\) is pure algebra.
Connection to Notes 7. These derivative formulas are restated and proved formally as Theorem 3.11 in Notes 7 (equations (3.6) and (3.7) of Logan). Notes 7 also shows how the same formulas can be used in reverse — as a bootstrapping technique for finding transforms of new functions such as \(\mathcal{L}[t^n]\) and \(\mathcal{L}[\cosh t]\) without evaluating the defining integral.
Solve \(x'' + 4x' + 3x = 0\), \(x(0) = 2\), \(x'(0) = 1\).
Step 1. Apply \(\mathcal{L}\): \[[s^2X - 2s - 1] + 4[sX - 2] + 3X = 0\] Collecting the \(X\) terms on the left and moving IC terms to the right: \[(s^2+4s+3)X = 2s + 1 + 8 = 2s+9.\]
Step 2. Solve for \(X(s)\): \[X(s) = \frac{2s+9}{(s+1)(s+3)}.\]
Step 3. Partial fractions: \[X(s) = \frac{7/2}{s+1} + \frac{-3/2}{s+3}.\]
Step 4. Invert term by term: \[\boxed{x(t) = \frac{7}{2}e^{-t} - \frac{3}{2}e^{-3t}.}\]
t_plot = np.linspace(0, 4, 300)
x_exact = 3.5*np.exp(-t_plot) - 1.5*np.exp(-3*t_plot)
def ode_ex1(t, y): return [y[1], -4*y[1] - 3*y[0]]
sol = solve_ivp(ode_ex1, (0, 4), [2.0, 1.0], dense_output=True, max_step=0.02)
t_dots = np.linspace(0, 4, 20)
fig, ax = plt.subplots(figsize=(8, 4))
ax.plot(t_plot, x_exact, color='steelblue', lw=2.5, label=r'$\frac{7}{2}e^{-t}-\frac{3}{2}e^{-3t}$')
ax.plot(t_dots, sol.sol(t_dots)[0], 'ro', markersize=6, label='Numerical (solve_ivp)')
ax.axhline(0, color='k', lw=0.5)
ax.set_xlabel('$t$'); ax.set_ylabel('$x(t)$')
ax.set_title(r"$x''+4x'+3x=0$, $x(0)=2$, $x'(0)=1$")
ax.legend(fontsize=9)
plt.tight_layout()
plt.show()
solve_ivp (red dots). The overdamped solution decays monotonically to zero — both eigenvalues \(\lambda=-1,-3\) are negative real.
Many real forcing functions are piecewise defined: a voltage switched on at \(t=3\), a force applied for exactly \(2\) seconds, a pump turned off at \(t=5\). The Heaviside function is the mathematical on/off switch:
\[H(t-a) = \begin{cases} 0, & t < a \\ 1, & t \geq a. \end{cases}\]
Its Laplace transform is: \[\mathcal{L}[H(t-a)] = \int_a^\infty e^{-st}\,dt = \frac{e^{-as}}{s}, \quad s > 0.\]
Any piecewise-constant function can be written as a sum of Heaviside steps. The key insight is that \(H(t-a) - H(t-b)\) equals \(1\) only on the interval \([a,b)\) and \(0\) elsewhere — it is a pulse.
Example (Logan §3.1, Example 3.8). The function \[f(t) = \begin{cases} 3, & 0 \leq t < 2 \\ 4, & 2 \leq t < 3 \\ 2, & 3 \leq t < 6 \\ 0, & t > 6 \end{cases}\] can be written as: \[f(t) = 3H(t) + 1\cdot H(t-2) - 2\cdot H(t-3) - 2\cdot H(t-6).\]
Each term “adjusts” the running value at the moment of the jump:
| Term | What it does | At \(t=\) |
|---|---|---|
| \(3H(t)\) | Switches on value \(3\) | \(0\) |
| \(+1\cdot H(t-2)\) | Adds \(1\) (making value \(4\)) | \(2\) |
| \(-2\cdot H(t-3)\) | Subtracts \(2\) (making value \(2\)) | \(3\) |
| \(-2\cdot H(t-6)\) | Subtracts \(2\) (making value \(0\)) | \(6\) |
Think of this as a running adjustment: start at \(3\), add \(1\) at \(t=2\), subtract \(2\) at \(t=3\), subtract \(2\) at \(t=6\). Each Heaviside term corrects the running value by exactly the size of the jump at that moment. The coefficients are always the differences between adjacent constant values, not the values themselves.
By linearity: \[F(s) = \frac{3}{s} + \frac{e^{-2s}}{s} - \frac{2e^{-3s}}{s} - \frac{2e^{-6s}}{s} = \frac{1}{s}\!\left(3 + e^{-2s} - 2e^{-3s} - 2e^{-6s}\right).\]
t_plot = np.linspace(-0.3, 8, 600)
fig, axes = plt.subplots(2, 1, figsize=(10, 7))
# Piecewise function
def f_piece(tv):
return np.where(tv<0, 0,
np.where(tv<2, 3,
np.where(tv<3, 4,
np.where(tv<6, 2, 0))))
axes[0].step(t_plot, f_piece(t_plot), where='post', color='k', lw=2.5,
label='$f(t)$', zorder=5)
# Individual Heaviside contributions
H = lambda tv, a: np.where(tv >= a, 1.0, 0.0)
contribs = [
(3*H(t_plot,0), 'steelblue', '+3$H(t)$'),
(1*H(t_plot,2), 'darkorange', '+1$H(t-2)$'),
(-2*H(t_plot,3), 'seagreen', '-2$H(t-3)$'),
(-2*H(t_plot,6), 'crimson', '-2$H(t-6)$'),
]
running = np.zeros_like(t_plot)
for contrib, color, lbl in contribs:
axes[0].fill_between(t_plot, running, running+contrib, alpha=0.25, color=color)
running += contrib
for _, color, lbl in contribs:
axes[0].plot([], [], color=color, linewidth=8, alpha=0.4, label=lbl)
axes[0].set_xlabel('$t$'); axes[0].set_ylabel('$f(t)$')
axes[0].set_title('Piecewise function as Heaviside steps')
axes[0].legend(fontsize=9, ncol=4)
axes[0].set_xlim(-0.3, 8); axes[0].set_ylim(-0.5, 5)
# Laplace transform F(s)
s_arr = np.linspace(0.01, 4, 400)
F_s = (3 + np.exp(-2*s_arr) - 2*np.exp(-3*s_arr) - 2*np.exp(-6*s_arr))/s_arr
axes[1].plot(s_arr, F_s, color='seagreen', lw=2.5)
axes[1].axhline(0, color='k', lw=0.5)
axes[1].set_xlabel('$s$'); axes[1].set_ylabel('$F(s)$')
axes[1].set_title(r'Laplace transform $F(s) = \frac{1}{s}(3+e^{-2s}-2e^{-3s}-2e^{-6s})$')
plt.tight_layout()
plt.show()
\[\mathcal{L}[f(t)e^{at}] = F(s-a),\] where \(F(s) = \mathcal{L}[f(t)]\).
Multiplying a time-domain function by \(e^{at}\) shifts its transform \(a\) units to the right in the \(s\)-domain.
This is proved directly from the definition: \(\int_0^\infty f(t)e^{at}e^{-st}dt = \int_0^\infty f(t)e^{-(s-a)t}dt = F(s-a)\).
Examples.
| \(f(t)\) | \(F(s)\) | \(f(t)e^{at}\) | \(\mathcal{L}[f(t)e^{at}]\) |
|---|---|---|---|
| \(t\) | \(1/s^2\) | \(te^{-2t}\) | \(1/(s+2)^2\) |
| \(\sin 3t\) | \(3/(s^2+9)\) | \(e^{-t}\sin 3t\) | \(3/((s+1)^2+9)\) |
| \(\cos 3t\) | \(s/(s^2+9)\) | \(e^{2t}\cos 3t\) | \((s-2)/((s-2)^2+9)\) |
| \(t^2\) | \(2/s^3\) | \(t^2 e^{5t}\) | \(2/(s-5)^3\) |
The shift property is essential for inverting transforms with completed-square denominators like \((s+1)^2+4\).
\[\mathcal{L}[H(t-a)f(t-a)] = e^{-as}F(s).\] Equivalently (inverse form): \[\mathcal{L}^{-1}[e^{-as}F(s)] = H(t-a)f(t-a).\]
A factor of \(e^{-as}\) in the transform domain corresponds to a time delay of \(a\) in the time domain.
Intuition: \(H(t-a)f(t-a)\) is the function \(f(t)\) “switched on” at \(t=a\) (it is zero before \(a\) and equals \(f(t-a)\) after \(a\)). The factor \(e^{-as}\) encodes this delay.
How to apply the switching property in reverse. When inverting \(X(s)\), if you see an \(e^{-as}\) factor multiplying some \(F(s)\): 1. Strip the \(e^{-as}\) factor and identify \(F(s)\). 2. Invert \(F(s)\) alone to get \(f(t) = \mathcal{L}^{-1}[F(s)]\). 3. Replace \(t\) with \(t-a\) in \(f(t)\). 4. Multiply by \(H(t-a)\).
The result is \(H(t-a)f(t-a)\).
Example. Find \(\mathcal{L}^{-1}\!\left[\frac{e^{-3s}}{s-2}\right]\).
We know \(\mathcal{L}^{-1}[1/(s-2)] = e^{2t}\), so \(F(s) = 1/(s-2)\), \(f(t) = e^{2t}\). By the switching property: \[\mathcal{L}^{-1}\!\left[\frac{e^{-3s}}{s-2}\right] = H(t-3)\,e^{2(t-3)}.\] The exponential grows from \(t=3\) onward (not from \(t=0\)).
s_v = sym.Symbol('s')
t_v = sym.Symbol('t', positive=True)
print("=== Shift property examples (SymPy verification) ===")
for f_t, lbl in [(t_v*sym.exp(-2*t_v), 'te^{-2t}'),
(sym.exp(-t_v)*sym.sin(3*t_v), 'e^{-t}sin(3t)'),
(sym.exp(2*t_v)*sym.cos(3*t_v), 'e^{2t}cos(3t)')]:
F = sym.laplace_transform(f_t, t_v, s_v, noconds=True)
display(Math(r"\mathcal{L}[" + lbl + r"] = " + sym.latex(sym.simplify(F))))=== Shift property examples (SymPy verification) ===\(\displaystyle \mathcal{L}[te^{-2t}] = \frac{1}{\left(s + 2\right)^{2}}\)
\(\displaystyle \mathcal{L}[e^{-t}sin(3t)] = \frac{3}{\left(s + 1\right)^{2} + 9}\)
\(\displaystyle \mathcal{L}[e^{2t}cos(3t)] = \frac{s - 2}{\left(s - 2\right)^{2} + 9}\)
The true power of the Laplace transform appears when \(f(t)\) is discontinuous. Methods from Chapter 2 are helpless here; the Laplace transform handles it in stride.
Example. Solve the RC circuit equation \[x' + 2x = f(t), \quad x(0) = 0,\] where \(f(t) = 1\) for \(0 \leq t < 3\) and \(f(t) = 0\) for \(t \geq 3\) (voltage switched off at \(t = 3\)).
Step 1. Write \(f(t) = H(t) - H(t-3)\), so \(F(s) = 1/s - e^{-3s}/s\).
Step 2. Transform the ODE: \[(s+2)X(s) = \frac{1}{s} - \frac{e^{-3s}}{s} \;\Longrightarrow\; X(s) = \frac{1}{s(s+2)} - \frac{e^{-3s}}{s(s+2)}.\]
Step 3. Partial fractions: \(\dfrac{1}{s(s+2)} = \dfrac{1/2}{s} - \dfrac{1/2}{s+2}\), so
\[\mathcal{L}^{-1}\!\left[\frac{1}{s(s+2)}\right] = \frac{1}{2}(1 - e^{-2t}).\]
Step 4. Apply the switching property to the second term:
\[\mathcal{L}^{-1}\!\left[\frac{e^{-3s}}{s(s+2)}\right] = H(t-3)\cdot\frac{1}{2}(1-e^{-2(t-3)}).\]
Solution: \[\boxed{x(t) = \frac{1}{2}(1-e^{-2t}) - \frac{1}{2}H(t-3)(1-e^{-2(t-3)}).}\]
t_plot = np.linspace(0, 8, 600)
H = lambda tv, a: np.where(tv >= a, 1.0, 0.0)
x_exact = 0.5*(1 - np.exp(-2*t_plot)) - 0.5*H(t_plot, 3)*(1 - np.exp(-2*(t_plot-3)))
f_plot = H(t_plot, 0) - H(t_plot, 3)
fig, axes = plt.subplots(2, 1, figsize=(9, 6), sharex=True)
axes[0].step(t_plot, f_plot, where='post', color='crimson', lw=2.5)
axes[0].set_ylabel('$f(t)$'); axes[0].set_ylim(-0.2, 1.4)
axes[0].set_title('Forcing $f(t) = H(t)-H(t-3)$')
axes[0].fill_between(t_plot, 0, f_plot, alpha=0.2, color='crimson')
axes[1].plot(t_plot, x_exact, color='steelblue', lw=2.5, label='Analytical (Laplace)')
def rhs(tv, y): return [(-H(tv,0)-H(tv,3))*0 + (1 if tv<3 else 0) - 2*y[0]]
sol_num = solve_ivp(lambda tv, y: [(1 if tv<3 else 0) - 2*y[0]],
(0, 8), [0.0], dense_output=True, max_step=0.02)
t_dots = np.linspace(0, 8, 25)
axes[1].plot(t_dots, sol_num.sol(t_dots)[0], 'ro', markersize=6, label='Numerical')
axes[1].axhline(0.5, color='gray', ls=':', lw=1.2, label='Steady state $x=0.5$')
axes[1].axvline(3, color='gray', ls='--', lw=1.2, alpha=0.7)
axes[1].set_xlabel('$t$'); axes[1].set_ylabel('$x(t)$')
axes[1].set_title("Solution via Laplace transform")
axes[1].legend(fontsize=9)
plt.tight_layout()
plt.show()
This example is a first taste of discontinuous forcing — the forcing is piecewise-constant and the ODE is first-order. Notes 7 takes up substantially harder cases: non-constant piecewise pieces (e.g., \(f(t)=t\) on one interval then \(f(t)=2\) on another), second-order oscillator equations, and RC circuits driven by square pulses. The Heaviside and switching tools developed here carry over directly; only the algebra becomes more involved.
The following table (from Logan (2015), Table 3.1) collects the transforms needed throughout this chapter. The left column lists time-domain functions \(x(t)\) and the right column lists their transforms \(X(s) = \mathcal{L}[x]\).
| \(x(t)\) | \(X(s) = \mathcal{L}[x]\) | Notes |
|---|---|---|
| \(e^{at}\) | \(\dfrac{1}{s-a}\), \(\;s>a\) | |
| \(t^n\), \(n=0,1,2,\ldots\) | \(\dfrac{n!}{s^{n+1}}\), \(\;s>0\) | \(n=0\): \(\mathcal{L}[1]=1/s\) |
| \(f(t)e^{at}\) | \(F(s-a)\) | Shift property |
| \(H(t-a)\) | \(\dfrac{1}{s}e^{-as}\), \(\;s>0\) | |
| \(H(t-a)f(t-a)\) | \(e^{-as}F(s)\) | Switching property |
| \(f(t)H(t-a)\) | \(e^{-as}\mathcal{L}[f(t+a)]\) | Use when \(f\) is not shifted (see note below) |
| \(\sin kt\) | \(\dfrac{k}{s^2+k^2}\) | |
| \(\cos kt\) | \(\dfrac{s}{s^2+k^2}\) | |
| \(\sinh kt\) | \(\dfrac{k}{s^2-k^2}\) | |
| \(\cosh kt\) | \(\dfrac{s}{s^2-k^2}\) | |
| \(e^{at}\sin kt\) | \(\dfrac{k}{(s-a)^2+k^2}\) | Shift applied to \(\sin\) |
| \(e^{at}\cos kt\) | \(\dfrac{s-a}{(s-a)^2+k^2}\) | Shift applied to \(\cos\) |
| \(t\sin kt\) | \(\dfrac{2ks}{(s^2+k^2)^2}\) | |
| \(t\cos kt\) | \(\dfrac{s^2-k^2}{(s^2+k^2)^2}\) | |
| \(\dfrac{1}{a-b}(e^{at}-e^{bt})\) | \(\dfrac{1}{(s-a)(s-b)}\) | \(a\neq b\); covers PF |
| \(x^{(n)}(t)\) | \(s^n X(s) - s^{n-1}x(0) - \cdots - x^{(n-1)}(0)\) | Key ODE formula |
| \(f(at)\) | \(\dfrac{1}{a}F\!\left(\dfrac{s}{a}\right)\), \(\;a>0\) | Scaling |
| \(t^a\) | \(\dfrac{\Gamma(a+1)}{s^{a+1}}\), \(\;s>0\) | Non-integer powers |
| \(\delta_a(t)\) | \(e^{-as}\) | Dirac delta at \(t=a\) |
| \(\delta_0(t)\) | \(1\) | Unit impulse |
| \(\displaystyle\int_0^t f(\tau)g(t-\tau)\,d\tau\) | \(F(s)G(s)\) | Convolution theorem |
| \(\displaystyle\int_0^t f(\tau)\,d\tau\) | \(\dfrac{1}{s}F(s)\) | |
| \(t\,f(t)\) | \(-F'(s)\) | |
| \(t^n f(t)\) | \((-1)^n F^{(n)}(s)\) |
Two Heaviside entries: which to use? The switching property \(H(t-a)f(t-a)\leftrightarrow e^{-as}F(s)\) is the standard form — use it whenever the function is already written in the shifted form \(f(t-a)\). The second entry \(f(t)H(t-a)\leftrightarrow e^{-as}\mathcal{L}[f(t+a)]\) arises when \(f(t)\) is not shifted (e.g., \(t^2 H(t-3)\) rather than \((t-3)^2 H(t-3)\)) — you must first evaluate \(\mathcal{L}[f(t+a)]\) and attach \(e^{-as}\). In most problems, rewriting \(f(t)\) in shifted form and using the standard entry is cleaner.
# Verify a selection of table entries with SymPy
s_v = sym.Symbol('s', positive=True)
t_v = sym.Symbol('t', positive=True)
k_v = sym.Symbol('k', positive=True)
a_v = sym.Symbol('a', positive=True)
entries = [
(sym.sin(k_v*t_v), f'sin(kt)', k_v/(s_v**2+k_v**2)),
(sym.cos(k_v*t_v), f'cos(kt)', s_v/(s_v**2+k_v**2)),
(sym.exp(a_v*t_v)*sym.sin(k_v*t_v), 'e^{at}sin(kt)',
k_v/((s_v-a_v)**2+k_v**2)),
(t_v*sym.sin(k_v*t_v), 't*sin(kt)',
2*k_v*s_v/(s_v**2+k_v**2)**2),
]
print("Spot-checking table entries via SymPy:")
for f_t, name, F_expected in entries:
F_computed = sym.laplace_transform(f_t, t_v, s_v, noconds=True)
F_diff = sym.simplify(F_computed - F_expected)
status = "✓" if F_diff == 0 else f"Diff={F_diff}"
print(f" {name}: {status}")Spot-checking table entries via SymPy:
sin(kt): ✓
cos(kt): ✓
e^{at}sin(kt): ✓
t*sin(kt): ✓The complete workflow for solving \(ax''+bx'+cx = f(t)\), \(x(0)=x_0\), \(x'(0)=x_1\) via Laplace transforms:
Relationship to Notes 7. Notes 7 presents the same procedure as the Three-Step Method (following Logan’s Figure 3.2): transform → solve algebraically → invert. The five-item checklist below is the same algorithm with each stage broken into more explicit sub-steps. The two descriptions are complementary: the three-step framing names the logical stages; this checklist gives the implementation details within each stage.
Transform both sides: apply \(\mathcal{L}\) to the ODE. Use the derivative formula to handle \(\mathcal{L}[x'']\) and \(\mathcal{L}[x']\). Rewrite \(f(t)\) using Heaviside steps if piecewise.
Incorporate ICs: the derivative formula automatically brings in \(x(0)\) and \(x'(0)\).
Solve for \(X(s)\): this is pure algebra — no differential equations.
Partial fractions: decompose \(X(s)\) into standard forms from the table.
Invert: apply \(\mathcal{L}^{-1}\) term by term; use the switching property for any \(e^{-as}\) factors.
| Step | What you do | Mathematics involved |
|---|---|---|
| 1 | \(\mathcal{L}[\text{ODE}]\) | Integral transforms, linearity |
| 2 | Plug in ICs | The formula \(\mathcal{L}[x''] = s^2X - sx_0 - x_1\) |
| 3 | Solve for \(X(s)\) | Algebra |
| 4 | Partial fractions | Algebra (factoring, cover-up rule) |
| 5 | \(\mathcal{L}^{-1}\), switching | Table lookup |
These notes are also viewable as a slide deck presentation: Open slides in full screen
The transforms and properties developed here are the foundation, but the real payoff comes in §3.2 and beyond. The convolution theorem (\(\mathcal{L}[f*g] = F(s)G(s)\), already in the table) will allow us to express solutions as integrals — a powerful tool for systems with arbitrary forcing. The Dirac delta \(\delta_a(t)\) (also in the table, with \(\mathcal{L}[\delta_a] = e^{-as}\)) models instantaneous impulses — hammer blows, lightning strikes, and analogous events in circuits — that no classical function can describe. These two tools complete the Laplace transform toolkit.
Next: Applying Laplace Transforms to Differential Equations — Logan §3.2.
import sys
print("Python:", sys.version)
print('\n'.join(f'{m.__name__}=={m.__version__}' for m in globals().values() if getattr(m,'__version__',None)))Python: 3.14.4 | packaged by conda-forge | (main, Apr 8 2026, 02:33:53) [Clang 20.1.8 ]
numpy==2.4.3
sympy==1.14.0
matplotlib==3.10.8