Theorem6.4.1
Let \(f\) and \(g\) be two piecewise continuous exponentially bounded functions, and suppose that \({\mathcal L}(f)(s) = F(s)\) and \({\mathcal L}(g)(s) = G(s)\) for \(s \geq a \gt 0\text{.}\) Then
for \(s \gt a\text{.}\)
If \(f\) and \(g\) are two piecewise continuous exponentially bounded functions, then we define the convolution product of \(f\) and \(g\) to be
The convolution product has many properties similar to those of ordinary multiplication.
All of these properties can be proven using the definition of convolution and the properties of integration.
There is, however, no multiplicative identity. In other words, \(f*1 \neq f\text{.}\) For example, suppose that \(f(t) = \cos t\text{.}\) Then
Also, it may not be the case that \(f*f\) is a nonnegative function.
One important property of the Laplace transform is how convolution products behave.
Let \(f\) and \(g\) be two piecewise continuous exponentially bounded functions, and suppose that \({\mathcal L}(f)(s) = F(s)\) and \({\mathcal L}(g)(s) = G(s)\) for \(s \geq a \gt 0\text{.}\) Then
for \(s \gt a\text{.}\)
If
then
where \(\xi = t - \tau\) is the change of variable. Reversing the order of integration, we have
However, this last expression is just the Laplace transform of \(f*g\text{.}\)
Let us use the Convolution Theorem (Theorem 6.4.1) to calculate the inverse Laplace transform of
instead of using partial fractions. The inverse Laplace transform of \(1/s^2\) is \(t\text{,}\) and the inverse Laplace transform of \(a/(s^2 + a^2)\text{.}\) is \(\sin at\text{.}\) Using the Convolution Theorem, the inverse Laplace transform of \(H(s)\) is
We can also use the Convolution Theorem to solve initial value problems. Consider the IVP
Taking the Laplace transform of both sides of the differential equation and applying the initial conditions, we obtain
where \(G(s)\) is the Laplace transform of \(g(t)\text{.}\) Solving for \(Y(s)\text{,}\) we have
The last term corresponds to the forcing term of our differential equation. Taking the inverse Laplace transform of both sides and applying the Convolution Theorem, we get
It is possible to write a solution for the initial value problem
using the Convolution Theorem. Taking the Laplace transform of both sides of the differential equation and using the initial conditions, we have
or
where
Therefore,
where \(\phi(t) = {\mathcal L}^{-1}( \Phi(s))\) and \(\psi(t) = {\mathcal L}^{-1}( \Psi(s))\text{.}\) Observe that \(\phi(t)\) is the solution to the initial value problem
Once we have values for \(a\text{,}\) \(b\text{,}\) and \(c\text{,}\) the function \(\phi(t)\) is easy to find. To find \(\psi(t)\text{,}\) we first write
where \(H(s) = 1/(as^2 + bs + c)\text{.}\) If \(h(t)\) is the inverse Laplace transform of \(H(s)\text{,}\) then
Let us consider the case where \(G(s) = 1\text{.}\) Consequently, \(g(t) = \delta(t)\) and \(\psi(s) = H(s)\text{.}\) This means that \(h(t)\) is a solution to the initial value problem
For this reason, \(h(t)\) is sometimes called the impulse response of the system.
The convolution product has many properties similar to those of ordinary multiplication.
There is, however, no multiplicative identity. In other words, \(f*1 \neq f\text{.}\) Also, it may not be the case that \(f*f\) is a nonnegative function.
Using convolution of functions, find the solution to the initial value problem
and \(g(t)\) is a piecewise continuous function.