18.2. Applications of Laplace Transform#

18.2.1. Ordinary differential equations#

Exploiting the properties of transforming derivation into product by the complex variable \(s\), Laplace transform can be a useful tool in solving Ordinary differential equations.

Linear ODEs are treated in details in the section about Linear Time-Invariant Systems.

18.2.1.1. First-order linear differential equation with constant coefficients#

A Cauchy problem governed by a first-order linear differential equation with constant coefficients reads

\[\begin{split}\begin{cases} \dot{u} + a u = f(t) && \text{for $t > 0$} \\ u(0^-) = u_0 \ . \end{cases}\end{split}\]

Comments. 1) For linear differential equations with constant coefficients, a time shift is always possible to set initial conditions in \(t_0 = 0\); 2) dealing with impulsive forces (more on this in LTI systems: impulsive forces), some attention needs to be paid to the time of initial conditions: if impulsive forces at \(t = 0\) may exist, initial conditions need to be set at \(t=0^-\); the effect of the impulsive force is equivalent to a jump in initial conditions from \(t=0^-\) and \(t=0^+\).

General solution in time domain

Multiplying the ODE by \(e^{a t}\),

\[e^{a t} f(t) = e^{a t} \left( \dot{u} + a u \right) = \dfrac{d}{dt} \left( e^{at} u(t) \right)\]

and integrating from \(0^-\) to a generic time \(t\) — after changing the dummy integration variable from \(\tau\) to \(t\) —

\[e^{at} u(t) - u(0^-) = \int_{\tau = 0^-}^{t} e^{a \tau} f(\tau) d \tau \ ,\]

and thus

\[u(t) = e^{-at} u(0^-) + \int_{\tau = 0^-}^{t} e^{-a(t- \tau)} f(\tau) d \tau \ .\]

The solution of the problem can be also computed using Laplace transform:

  1. transforming the problem from time to Laplace domain: the differential problem with unknown \(u(t)\) in time is transformed into an algebraic problem \(\hat{u}(s)\). Using Laplace transform of the time derivative

\[s \hat{u}(s) + u_0 + a \hat{u}(s) = \hat{f}(s) \ ,\]

  1. solving the algebraic problem in Laplace domain for \(\hat{u}(s)\)

\[\hat{u}(s) = \dfrac{1}{s + a} u_0 + \dfrac{1}{s + a} f(s) \ .\]

  1. transforming back the solution in time domain, \(u(t) = \mathscr{L}^{-1}\{ \hat{u}(s) \}\),

    \[\begin{split}\begin{aligned} u(t) & = \mathscr{L}^{-1}\{ \hat{u}(s) \} (t) = \\ & = \mathscr{L}^{-1} \left\{ \dfrac{u_0}{s + a} \right\} + \mathscr{L}^{-1} \left\{ \dfrac{1}{s + a} \hat{f}(s) \right\} = \\ & = e^{-at} u_0 + \int_{0^-}^{t} e^{-a(t-\tau)} f(\tau) d \tau \ , \end{aligned}\end{split}\]

    having used the inverse transform of the exponential \(\mathscr{L}\{e^{ct}\} = \int_{t=0^-}^{+\infty} e^{ct} e^{-st} \, dt = \frac{1}{s-c}\), and the inverse transform of the convolution

    \[\hat{f}(s) \hat{g}(s) = \mathscr{L}\{ f(t) \ast g(t) \}(s) = \int_{t=0^-}^{+\infty} \left\{ \int_{\tau=0^-}^{t} f(\tau) g(t-\tau) d \tau \right\} e^{-st} dt \ .\]