6 Laplace Transforms
The Laplace transform converts a function of \(t\) into a function of \(s\) by an integral, and its value is that it converts calculus into algebra: under the transform, differentiation becomes multiplication by \(s\).
That single fact is the whole method. A differential equation in \(t\), with its initial conditions built in, becomes an algebraic equation in \(s\); the algebraic equation is solved by ordinary manipulation; and the answer is carried back by an inverse transform. The initial conditions enter automatically rather than being fitted at the end, which is the practical advantage over the methods of the previous section.
Most of the section builds the equipment needed to make that route usable: the transforms of the standard functions, the two shifting theorems, the transforms of derivatives and integrals, the convolution theorem for products, and partial fractions for taking the inverse. Each is a tool for recognising a function from its transform, which is where the real work of the method lies.
Definition 6.0.1. Let \(f(t)\) be defined for all \(t\geq 0\). The Laplace transform of \(f\) is the function \[F(s) = \mathcal {L}\{f(t)\} = \int _0^{\infty }e^{-st}f(t)\,dt = \lim _{T\to \infty }\int _0^T e^{-st}f(t)\,dt ,\] defined for those \(s\) for which the improper integral converges. The function \(f\) is the inverse transform of \(F\), written \(f(t) = \mathcal {L}^{-1}\{F(s)\}\).
The transform exists for \(s>\alpha \) whenever \(f\) is piecewise continuous on every finite interval and of exponential order \(\alpha \), meaning \(|f(t)|\leq Me^{\alpha t}\) for some constant \(M\): then \(|e^{-st}f(t)|\leq Me^{-(s-\alpha )t}\), whose integral converges. Two continuous functions with the same transform are equal, so the inverse transform is well defined.
Notation
Original functions are denoted by lower-case letters and their transforms by the same letters in capitals, so that \(F(s)\) denotes the transform of \(f(t)\), \(Y(s)\) denotes the transform of \(y(t)\), and so on.
- 1.
- Let \(f(t) = 1\) for \(t\geq 0\). Find \(F(s)\).
- 2.
- Let \(f(t) = e^{at}\) for \(t\geq 0\), where \(a\) is a constant. Find \(\mathcal {L}\{f(t)\}\).
Solution.
- 1.
- For \(s>0\), \[F(s) = \int ^{\infty }_{0} e^{-st}\,dt = \lim _{T \to \infty }\left [-\frac {e^{-st}}{s} \right ]_0^T = \lim _{T \to \infty } -\frac {1}{s}\left (e^{-sT}-1\right ) = \frac {1}{s} ,\] so \(\mathcal {L}\{1\} = \frac {1}{s}\) for \(s>0\). For \(s\leq 0\) the integral diverges.
- 2.
- For \(s>a\), \[\mathcal {L}\{e^{at}\} = \int ^{\infty }_0 e^{-st}e^{at}\, dt = \int ^{\infty }_0 e^{-(s-a)t}\,dt = \left [-\frac {e^{-(s-a)t}}{s - a}\right ]^{\infty }_0 = \frac {1}{s - a} .\]
Example 6.0.3. Find \(\mathcal {L}\{t\}\) and \(\mathcal {L}\{\sin \omega t\}\) from the definition.
Solution. Integrating by parts, for \(s>0\), \[\mathcal {L}\{t\} = \left [-\frac {te^{-st}}{s}\right ]_0^\infty + \frac 1s\int _0^\infty e^{-st}\,dt = 0 + \frac {1}{s^2} .\] For the sine, it is quickest to use \(e^{i\omega t} = \cos \omega t + i\sin \omega t\) and part 2 of Example 6.0.2 with \(a = i\omega \): \[\mathcal {L}\{e^{i\omega t}\} = \frac {1}{s - i\omega } = \frac {s + i\omega }{s^2 + \omega ^2},\] and taking real and imaginary parts, \(\mathcal {L}\{\cos \omega t\} = \frac {s}{s^2 + \omega ^2}\) and \(\mathcal {L}\{\sin \omega t\} = \frac {\omega }{s^2 + \omega ^2}\), for \(s>0\).
6.2 Laplace Transform of the Derivatives of \(f(t)\)
6.3 Laplace Transform of the Integral of a Function
6.4 First Shifting Theorem: the \(s\)-Shift
6.5 Second Shifting Theorem: the \(t\)-Shift
6.6 Differentiation of Transforms
6.7 Convolution Theorem
6.8 Integral Equations
6.9 Integration of Transforms
6.10 Partial Fractions
6.11 Practice Problems
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.