Introduction
The Laplace transform of the solution of a differential equation is sometimes recognizable as the product of the transforms of two known functions.
Let \( F(s) = \mathcal{L}\{f(t)\} \) and \( G(s) = \mathcal{L}\{g(t)\} \). Suppose \( H(s) = F(s) \cdot G(s) \).
Question: Is \( h(t) = f(t) \cdot g(t) \)?
Answer: No!
Example
Let \( F(s) = \frac{1}{s^2} \) and \( G(s) = \frac{1}{s} \).
- \( f(t) = t \)
- \( g(t) = 1 \)
- \( h(t) = f(t) \cdot g(t) = t \cdot 1 = t \)
However, \( H(s) = F(s)G(s) = \frac{1}{s^3} \), and the inverse transform of \( H(s) \) is \( \frac{t^2}{2} \), not \( t \).
(I) Product of Transforms and Convolution
Definition: The Convolution of Two Functions
The convolution of two piecewise continuous functions \( f \) and \( g \) is defined for \( t \ge 0 \) as follows:
\[ (f * g)(t) = \int_{0}^{t} f(\tau) g(t - \tau) \, d\tau \]
Convolution is commutative: \( f * g = g * f \).
Theorem 1: The Convolution Property
Suppose that \( f \) and \( g \) are piecewise continuous for \( t \ge 0 \) and that \( |f(t)| \) and \( |g(t)| \) are bounded by \( Me^{ct} \) as \( t \to \infty \). Then \( \mathcal{L}\{f(t) * g(t)\} \) exists for \( s > c \), and:
\[ \mathcal{L}\{f(t) * g(t)\} = F(s)G(s) \]
Or, equivalently:
\[ \mathcal{L}^{-1}\{F(s)G(s)\} = f(t) * g(t) \]
Example 1: Using Convolution
Find \( \mathcal{L}^{-1}\left\{ \frac{s}{(s^2+1)^2} \right\} \).
Solution: We can split the transform into a product of two known transforms:
\(
\begin{aligned}
\mathcal{L}^{-1}\left\{ \frac{s}{(s^2+1)^2} \right\} &= \mathcal{L}^{-1}\left\{ \frac{s}{s^2+1} \cdot \frac{1}{s^2+1} \right\} \\
&= \mathcal{L}^{-1}\left\{ \frac{s}{s^2+1} \right\} * \mathcal{L}^{-1}\left\{ \frac{1}{s^2+1} \right\} \\
&= \cos(t) * \sin(t)
\end{aligned}
\)
Now, apply the convolution definition:
\(
\begin{aligned}
\cos(t) * \sin(t) &= \int_{0}^{t} \cos(\tau) \sin(t-\tau) \, d\tau \\
&= \int_{0}^{t} \frac{1}{2} (\sin(t) + \sin(t - 2\tau)) \, d\tau \\
&= \frac{1}{2} \left[ \tau\sin(t) + \frac{1}{2}\cos(t - 2\tau) \right]_{0}^{t}
\end{aligned}
\)
\[ \mathcal{L}^{-1}\left\{ \frac{s}{(s^2+1)^2} \right\} = \frac{1}{2}t\sin(t) \]
Example 2: Initial Value Problem
Solve the Initial Value Problem: \( x'' + 4x' + 13x = f(t), \quad x(0) = x'(0) = 0 \).
Show that \( x(t) = \frac{1}{3} \int_{0}^{t} f(t-\tau)e^{-2\tau}\sin(3\tau) \, d\tau \).
Solution: Take the Laplace transform:
\(
\begin{aligned}
(s^2 + 4s + 13)X(s) &= F(s) \\
X(s) &= \frac{1}{s^2 + 4s + 13} F(s) \\
X(s) &= \frac{1}{(s+2)^2 + 3^2} F(s)
\end{aligned}
\)
Take the inverse transform using the Convolution Theorem:
\(
\begin{aligned}
x(t) &= \mathcal{L}^{-1}\left\{ \frac{1}{(s+2)^2 + 3^2} \right\} * \mathcal{L}^{-1}\{F(s)\} \\
&= \left( \frac{1}{3}e^{-2t}\sin(3t) \right) * f(t)
\end{aligned}
\)
\[ x(t) = \frac{1}{3} \int_{0}^{t} e^{-2\tau}\sin(3\tau)f(t-\tau) \, d\tau \]
(II) Differentiation / Integration of Transforms
From Section 7.2, we know that \( \mathcal{L}\{f'(t)\} = sF(s) \) if \( f(0) = 0 \).
- Differentiation of \( f(t) \) corresponds to multiplication of its transform by \( s \).
- Integration of \( f(t) \) corresponds to division of its transform by \( s \).
Now, what about differentiation and integration of the transform \( F(s) \)?
- Differentiation of \( F(s) \) corresponds to multiplication of \( f(t) \) by \( -t \).
- Integration of \( F(s) \) corresponds to division of \( f(t) \) by \( t \).
Theorem 2: Differentiation of Transforms
\[ \mathcal{L}\{-t f(t)\} = F'(s) \quad \text{or} \quad \mathcal{L}\{t^n f(t)\} = (-1)^n F^{(n)}(s) \]
Equivalently, for inverse transforms:
\[ f(t) = \mathcal{L}^{-1}\{F(s)\} = -\frac{1}{t}\mathcal{L}^{-1}\{F'(s)\} \]
This is helpful in finding an inverse transform when the derivative of the transform is easier to work with than the transform itself.
Example 3: Verifying the Rule
Check the rule using the transform for sine:
\(
\begin{aligned}
\mathcal{L}\{t\sin(t)\} &= -\frac{d}{ds}\mathcal{L}\{\sin(t)\} \\
&= -\frac{d}{ds}\left( \frac{1}{s^2+1} \right) \\
&= \frac{2s}{(s^2+1)^2}
\end{aligned}
\)
This matches our result from Example 1.
Example 4: Higher Order Derivatives
Find \( \mathcal{L}\{t^2\cos(2t)\} \).
Solution: Let \( f(t) = \cos(2t) \), so \( F(s) = \frac{s}{s^2+4} \). We need to find \( F''(s) \).
\(
\begin{aligned}
F'(s) &= \frac{(s^2+4)(1) - s(2s)}{(s^2+4)^2} = \frac{4-s^2}{(s^2+4)^2} \\
F''(s) &= \frac{(s^2+4)^2(-2s) - (4-s^2) \cdot 2(s^2+4) \cdot 2s}{(s^2+4)^4} = \frac{2s^3 - 24s}{(s^2+4)^3}
\end{aligned}
\)
\[ \mathcal{L}\{t^2\cos(2t)\} = \frac{2s^3 - 24s}{(s^2+4)^3} \]
Example 5: Inverse Transform via Differentiation
Find \( \mathcal{L}^{-1}\left\{ \tan^{-1}\left(\frac{1}{s}\right) \right\} \).
Solution: Let \( F(s) = \tan^{-1}(1/s) \). The derivative \( F'(s) \) is easier to invert:
\(
\begin{aligned}
f(t) &= -\frac{1}{t} \mathcal{L}^{-1}\{F'(s)\} \\
&= -\frac{1}{t} \mathcal{L}^{-1}\left\{ \frac{-1}{1+s^2} \right\} \\
&= -\frac{1}{t} (-\sin(t))
\end{aligned}
\)
\[f(t) = \frac{\sin(t)}{t} \]
Theorem 3: Integration of Transforms
Suppose that \( f(t) \) is piecewise continuous for \( t \ge 0 \), that \( \lim_{t \to 0^+} \frac{f(t)}{t} \) exists and is finite, and that \( |f(t)| \le Me^{ct} \) as \( t \to \infty \). Then:
\[ \mathcal{L}\left\{ \frac{f(t)}{t} \right\} = \int_{s}^{\infty} F(\sigma) \, d\sigma \quad \text{for } s > c \]
Or, equivalently for inverse transforms:
\[ f(t) = \mathcal{L}^{-1}\{F(s)\} = t\mathcal{L}^{-1}\left\{ \int_{s}^{\infty} F(\sigma) \, d\sigma \right\} \]
This is helpful in finding an inverse transform when the indefinite integral of the transform is easier to handle than the transform itself.
Example 1 (Revisited - Method 2)
Find \( \mathcal{L}^{-1}\left\{ \frac{s}{(s^2+1)^2} \right\} \) using Theorem 3.
Solution: We recognize the expression as a derivative, making it easy to integrate.
\(
\begin{aligned}
\mathcal{L}^{-1}\left\{ \frac{s}{(s^2+1)^2} \right\} &= t\mathcal{L}^{-1}\left\{ \int_{s}^{\infty} \frac{\sigma}{(\sigma^2+1)^2} \, d\sigma \right\} \\
&= t\mathcal{L}^{-1}\left\{ \left[ -\frac{1}{2} \frac{1}{\sigma^2+1} \right]_{s}^{\infty} \right\} \\
&= t\mathcal{L}^{-1}\left\{ \frac{1}{2}\frac{1}{s^2+1} \right\}
\end{aligned}
\)
\[ \mathcal{L}^{-1}\left\{ \frac{s}{(s^2+1)^2} \right\} = \frac{1}{2}t\sin(t) \]
Example 6: Evaluating an Improper Integral
Find \( \mathcal{L}\left\{ \frac{e^t - e^{-t}}{t} \right\} \).
Solution: First, check the limit condition: \( \lim_{t \to 0} \frac{e^t - e^{-t}}{t} = 2 \) (exists and is finite). Apply Theorem 3:
\(
\begin{aligned}
\mathcal{L}\left\{ \frac{e^t - e^{-t}}{t} \right\} &= \int_{s}^{\infty} \mathcal{L}\{e^t - e^{-t}\} \, d\sigma \\
&= \int_{s}^{\infty} \left( \frac{1}{\sigma-1} - \frac{1}{\sigma+1} \right) \, d\sigma \\
&= \left[ \ln\left(\frac{\sigma-1}{\sigma+1}\right) \right]_{s}^{\infty} \\
&= \lim_{\sigma \to \infty} \ln\left(\frac{\sigma-1}{\sigma+1}\right) - \ln\left(\frac{s-1}{s+1}\right) \\
&= 0 - \ln\left(\frac{s-1}{s+1}\right)
\end{aligned}
\)
\[\mathcal{L}\left\{ \frac{e^t - e^{-t}}{t} \right\} = \ln\left(\frac{s+1}{s-1}\right) \]