Laplace Transforms

The Laplace transform of $f(t)$ (defined for $t \geq 0$) is

$$\mathcal{L}\{f(t)\} = F(s) = \int_0^{\infty} e^{-st} f(t)\, dt$$

The transform exists when $f$ is piecewise continuous on $[0, \infty)$ and of exponential order: $|f(t)| \leq Me^{at}$ for some $M, a > 0$.

Theorem 5.1 (Linearity). $\mathcal{L}\{af + bg\} = a\mathcal{L}\{f\} + b\mathcal{L}\{g\}$.

Theorem 5.2 (First Shifting). $\mathcal{L}\{e^{at}f(t)\} = F(s - a)$.

Theorem 5.3 (Second Shifting). $\mathcal{L}\{u_c(t)f(t - c)\} = e^{-cs}F(s)$ Where $u_c(t)$ is The unit step function.

Theorem 5.4 (Derivative). $\mathcal{L}\{f"(t)\} = sF(s) - f(0)$.

Theorem 5.5 (n-th Derivative). $\mathcal{L}\{f^{(n)}(t)\} = s^n F(s) - s^{n-1}f(0) - \cdots - f^{(n-1)}(0)$.

Theorem 5.6 (Integration). $\mathcal{L}\left\{\int_0^t f(\tau)\, d\tau\right\} = \frac{F(s)}{s}$.

Theorem 5.7 (Convolution). $\mathcal{L}\{f * g\} = F(s)G(s)$ Where $(f * g)(t) = \int_0^t f(\tau)g(t - \tau)\, d\tau$.

Proof of Linearity. $\mathcal{L}\{af + bg\} = \int_0^{\infty} e^{-st}[af(t) + bg(t)]\, dt = a\int_0^{\infty} e^{-st}f(t)\, dt + b\int_0^{\infty} e^{-st}g(t)\, dt = aF(s) + bG(s)$. $\blacksquare$

Proof of First Shifting Theorem. $\mathcal{L}\{e^{at}f(t)\} = \int_0^{\infty} e^{-st} e^{at} f(t)\, dt = \int_0^{\infty} e^{-(s-a)t} f(t)\, dt = F(s - a)$. $\blacksquare$

Proof of Derivative Property. Integrating by parts with $u = e^{-st}$, $dv = f'(t)\, dt$:

$\mathcal{L}\{f'(t)\} = \left[e^{-st}f(t)\right]_0^{\infty} + s\int_0^{\infty} e^{-st}f(t)\, dt = -f(0) + sF(s) = sF(s) - f(0)$.

The boundary term vanishes at $\infty$ because $f$ is of exponential order. $\blacksquare$

Proof of Integration Property. Let $g(t) = \int_0^t f(\tau)\, d\tau$. Then $g'(t) = f(t)$ and $g(0) = 0$. $\mathcal{L}\{f(t)\} = \mathcal{L}\{g'(t)\} = s\mathcal{L}\{g(t)\} - g(0) = s\mathcal{L}\{g(t)\}$. Therefore $\mathcal{L}\{g(t)\} = F(s)/s$. $\blacksquare$

Procedure:

1. Take the Laplace transform of both sides of the ODE.
2. Solve the resulting algebraic equation for $Y(s) = \mathcal{L}\{y(t)\}$.
3. Find the inverse Laplace transform to obtain $y(t)$.

Problem. Solve $y'' + 4y = \sin(2t)$, $y(0) = 0$, $y'(0) = 0$.

Solution. Taking Laplace transforms:

$s^2 Y + 4Y = \frac{2}{s^2 + 4}$

$(s^2 + 4)Y = \frac{2}{s^2 + 4}$

$Y = \frac{2}{(s^2 + 4)^2}$

To invert, use the convolution theorem: $\mathcal{L}^{-1}\left\{\frac{1}{s^2 + 4}\right\} = \frac{1}{2}\sin(2t)$.

$y(t) = \frac{1}{2}\sin(2t) * \frac{1}{2}\sin(2t) = \frac{1}{4}\int_0^t \sin(2\tau)\sin(2(t - \tau))\, d\tau$

Using $\sin A \sin B = \frac{1}{2}(\cos(A - B) - \cos(A + B))$:

$y(t) = \frac{1}{8}\int_0^t [\cos(4\tau - 2t) - \cos(2t)]\, d\tau = \frac{1}{8}\left[\frac{\sin(4\tau - 2t)}{4}\right]_0^t - \frac{t}{8}\cos(2t)$

$= \frac{1}{32}[\sin(2t) - \sin(-2t)] - \frac{t}{8}\cos(2t) = \frac{\sin(2t)}{16} - \frac{t\cos(2t)}{8}$

$y(t) = \frac{\sin(2t) - 2t\cos(2t)}{16}$. $\blacksquare$

Problem. Find \mathcal{L}^{-1}\left\{\frac{1}{s(s^2 + 4)} Ight\}.

Solution

Solution. Partial fractions: $\frac{1}{s(s^2 + 4)} = \frac{A}{s} + \frac{Bs + C}{s^2 + 4}$.

$1 = A(s^2 + 4) + (Bs + C)s = (A + B)s^2 + Cs + 4A$.

$s^2$: $A + B = 0$, $s$: $C = 0$ Constant: $4A = 1 \implies A = 1/4$, $B = -1/4$.

$\frac{1}{s(s^2 + 4)} = \frac{1/4}{s} - \frac{s/4}{s^2 + 4}$.

$f(t) = \frac{1}{4} - \frac{1}{4}\cos 2t$. $\blacksquare$

### 5.7 Convolution Theorem

Theorem 5.8 (Convolution). If $F(s) = \mathcal{L}\{f(t)\}$ and $G(s) = \mathcal{L}\{g(t)\}$ Then

$$\mathcal{L}\{f * g\} = F(s)G(s)$$

Where $(f * g)(t) = \int_0^t f(\tau)g(t - \tau)\, d\tau$.

Proof. $F(s)G(s) = \int_0^{\infty} e^{-s\tau} f(\tau)\, d\tau \cdot \int_0^{\infty} e^{-su} g(u)\, du$

$= \int_0^{\infty} \int_0^{\infty} e^{-s(\tau+u)} f(\tau) g(u)\, du\, d\tau$

Substitute $t = \tau + u$, $u = t - \tau$. For fixed $\tau$, $u$ ranges from $0$ to $\infty$ So $t$ ranges From $\tau$ to $\infty$.

$= \int_0^{\infty} f(\tau)\left[\int_{\tau}^{\infty} e^{-st} g(t - \tau)\, dt\right] d\tau$

Swap the order of integration (Fubini):

$= \int_0^{\infty} e^{-st}\left[\int_0^t f(\tau) g(t - \tau)\, d\tau\right] dt = \mathcal{L}\{(f * g)(t)\}$. $\blacksquare$

Problem. Find $\mathcal{L}^{-1}\left\{\frac{3s + 1}{s^2 + 4s + 13}\right\}$.

Solution

Solution. Complete the square: $s^2 + 4s + 13 = (s + 2)^2 + 9$.

$\frac{3s + 1}{s^2 + 4s + 13} = \frac{3(s + 2) - 5}{(s + 2)^2 + 9} = 3 \cdot \frac{s + 2}{(s + 2)^2 + 9} - \frac{5}{3} \cdot \frac{3}{(s + 2)^2 + 9}$

Using the shifting theorem:

$\mathcal{L}^{-1}\left\{\frac{s + 2}{(s + 2)^2 + 9}\right\} = e^{-2t}\cos(3t)$

$\mathcal{L}^{-1}\left\{\frac{3}{(s + 2)^2 + 9}\right\} = e^{-2t}\sin(3t)$

$f(t) = 3e^{-2t}\cos(3t) - \frac{5}{3}e^{-2t}\sin(3t) = e^{-2t}\left(3\cos(3t) - \frac{5}{3}\sin(3t)\right)$. $\blacksquare$

Problem. Solve $y'' + 3y' + 2y = u_2(t)$, $y(0) = 0$, $y'(0) = 0$.

Solution

Solution. Take Laplace transforms:

$s^2 Y + 3sY + 2Y = \frac{e^{-2s}}{s}$

$(s^2 + 3s + 2)Y = \frac{e^{-2s}}{s}$

$Y = \frac{e^{-2s}}{s(s+1)(s+2)}$

First find the partial fraction decomposition of $\frac{1}{s(s+1)(s+2)}$:

$\frac{1}{s(s+1)(s+2)} = \frac{A}{s} + \frac{B}{s+1} + \frac{C}{s+2}$

$1 = A(s+1)(s+2) + Bs(s+2) + Cs(s+1)$

$s = 0$: $1 = 2A \implies A = 1/2$ $s = -1$: $1 = -B \implies B = -1$ $s = -2$: $1 = 2C \implies C = 1/2$

So $\mathcal{L}^{-1}\left\{\frac{1}{s(s+1)(s+2)}\right\} = \frac{1}{2} - e^{-t} + \frac{1}{2}e^{-2t}$.

By the second shifting theorem:

$y(t) = u_2(t)\left[\frac{1}{2} - e^{-(t-2)} + \frac{1}{2}e^{-2(t-2)}\right]$ for $t \geq 0$. $\blacksquare$

The Heaviside (unit step) function is defined as

$$u_c(t) = \begin{cases} 0 & t \lt c \\ 1 & t \geq c \end{cases}$$

It models a sudden switch being turned on at time $t = c$. The second shifting theorem states:

$$\mathcal{L}\{u_c(t)f(t - c)\} = e^{-cs}F(s)$$

Conversely, if $Y(s) = e^{-cs}G(s)$ Then $y(t) = u_c(t) \cdot g(t - c)$ where $g = \mathcal{L}^{-1}\{G\}$.

The Dirac delta function $\delta(t - c)$ satisfies $\int_0^{\infty} \delta(t - c)f(t)\, dt = f(c)$ for continuous $f$. It models an instantaneous Impulse.

$\mathcal{L}\{\delta(t - c)\} = e^{-cs}$.

Properties:

• $\delta(t - c) = 0$ for $t \neq c$
• $\int_{-\infty}^{\infty} \delta(t - c)\, dt = 1$
• $\int_{-\infty}^{\infty} \delta(t - c)f(t)\, dt = f(c)$ (sifting property)

Problem. An undamped spring-mass system ($m = 1$, $k = 4$) is struck by an impulse at $t = 0$: $y'' + 4y = \delta(t)$, $y(0) = 0$, $y'(0) = 0$. Find $y(t)$.

Solution

Solution. Taking Laplace transforms:

$s^2 Y + 4Y = 1$

$Y = \frac{1}{s^2 + 4}$

$y(t) = \frac{1}{2}\sin(2t)$.

This is the impulse response (Green’s function) for the operator $D^2 + 4$. $\blacksquare$

Problem. Solve $y'' - 6y' + 9y = 6e^{3t}$, $y(0) = 0$, $y'(0) = 0$.

Solution

Solution. Taking Laplace transforms:

$s^2 Y - 6sY + 9Y = \frac{6}{s - 3}$

$(s - 3)^2 Y = \frac{6}{s - 3}$

$Y = \frac{6}{(s - 3)^3}$

Using the table: $\mathcal{L}\{t^n e^{at}\} = \frac{n!}{(s-a)^{n+1}}$.

$y(t) = 6 \cdot \frac{t^2 e^{3t}}{2!} = 3t^2 e^{3t}$. $\blacksquare$

The Laplace transform converts differential equations into algebraic equations. Instead of solving a second-order ODE (which requires finding homogeneous and particular solutions, applying initial conditions, etc.), you transform the ODE into a polynomial equation for $Y(s)$, solve for $Y(s)$, and then invert.

The key insight: The derivative property $\mathcal{L}\{f'(t)\} = sF(s) - f(0)$ converts differentiation into multiplication by $s$. This means a differential equation like $y'' + 3y' + 2y = e^t$ becomes $(s^2 + 3s + 2)Y(s) = \frac{1}{s-1} + \text{initial condition terms}$, which is just algebra.

Why the convolution theorem matters: The convolution $(f * g)(t) = \int_0^t f(\tau)g(t - \tau)\,d\tau$ is hard to compute directly. But $\mathcal{L}\{f * g\} = F(s)G(s)$ means that convolution in the time domain becomes multiplication in the frequency domain. This is the same principle behind Fourier transforms and signal processing.

Physical interpretation: The Laplace variable $s$ can be interpreted as a complex frequency. $F(s)$ decomposes the signal $f(t)$ into exponential components $e^{st}$. The poles of $F(s)$ (values of $s$ where $F(s)$ blows up) determine the natural modes of the system, while the zeros determine the frequencies that are suppressed.

Problem. Solve the system:

$$\begin{aligned} x' &= 3x - 2y \\ y' &= 2x - 2y \end{aligned}$$

with $x(0) = 1$, $y(0) = 0$.

Solution

Take Laplace transforms:

$$\begin{aligned} sX - 1 &= 3X - 2Y \\ sY &= 2X - 2Y \end{aligned}$$

Rearranging:

$$\begin{aligned} (s - 3)X + 2Y &= 1 \\ -2X + (s + 2)Y &= 0 \end{aligned}$$

From the second equation: $X = \frac{(s+2)Y}{2}$.

Substituting into the first: $(s - 3) \cdot \frac{(s+2)Y}{2} + 2Y = 1$

$\frac{(s-3)(s+2)Y + 4Y}{2} = 1$

$(s^2 - s - 6 + 4)Y = 2$

$(s^2 - s - 2)Y = 2$

$Y = \frac{2}{(s-2)(s+1)} = \frac{2/3}{s-2} - \frac{2/3}{s+1}$

$y(t) = \frac{2}{3}e^{2t} - \frac{2}{3}e^{-t}$

For $X$: $X = \frac{(s+2)}{2} \cdot \frac{2}{(s-2)(s+1)} = \frac{s+2}{(s-2)(s+1)} = \frac{4/3}{s-2} - \frac{1/3}{s+1}$

$x(t) = \frac{4}{3}e^{2t} - \frac{1}{3}e^{-t}$

Verification: $x(0) = 4/3 - 1/3 = 1$ $\checkmark$, $y(0) = 2/3 - 2/3 = 0$ $\checkmark$.

$x' = \frac{8}{3}e^{2t} + \frac{1}{3}e^{-t}$, $3x - 2y = 4e^{2t} - e^{-t} - \frac{4}{3}e^{2t} + \frac{4}{3}e^{-t} = \frac{8}{3}e^{2t} + \frac{1}{3}e^{-t}$ $\checkmark$. $\blacksquare$

The Laplace transform converts differential equations into algebraic equations. By transforming a function $f(t)$ into $F(s) = \int_0^\infty e^{-st}f(t)\,dt$, differentiation in the time domain becomes multiplication by $s$ in the frequency domain. This means a linear ODE with constant coefficients becomes a polynomial equation in $s$, which can be solved algebraically and then inverted to recover the solution.

The Laplace transform is particularly powerful for several reasons. It handles discontinuous forcing functions (step functions, impulses) logically through the second shifting theorem and the convolution theorem. It converts initial value problems into algebraic problems without needing to find the homogeneous solution first. The convolution theorem says that convolution in time becomes multiplication in the frequency domain, which is why the Laplace transform is the tool of choice for systems theory, control engineering, and circuit analysis. The transform exists for functions of exponential order, which covers essentially all physically reasonable inputs.

The Picard-Lindelöf theorem can be proved constructively via Picard iteration. For the IVP $y' = f(x, y)$, $y(x_0) = y_0$ Define the sequence

$$\phi_0(x) = y_0, \quad \phi_{n+1}(x) = y_0 + \int_{x_0}^x f(t, \phi_n(t))\, dt$$

If $f$ and $\partial f/\partial y$ are continuous, one shows by induction that $(\phi_n)$ is uniformly Cauchy on some interval $[x_0 - h, x_0 + h]$ Hence converges uniformly to a function $\phi$. Passing to the limit in the integral equation shows $\phi$ satisfies the ODE. Uniqueness follows From the Gronwall inequality applied to the difference of two solutions.

• Second-Order Linear ODEs: The Laplace transform converts the characteristic equation approach into algebraic manipulation in the frequency domain.

• Fourier Series: Both Fourier and Laplace transforms decompose signals into frequency components, but Laplace handles transient and aperiodic signals.

• Complex Numbers Review: The Laplace variable $s$ is complex, and poles and zeros in the complex plane determine system behaviour.

• Systems of ODEs: The Laplace transform extends logically to systems of ODEs, converting coupled differential equations into algebraic systems.

• Classical Mechanics

• Electromagnetism

• Common Pitfalls
• First-Order ODEs
• Fourier Series
• Introduction and Classification