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The Ordinary Differential Equations Project

Thomas W. Judson

Section 6.4 Convolution

When solving an initial value problem using Laplace transforms, we employed the strategy of converting the differential equation to an algebraic equation. Once the the algebraic equation is solved, we can recover the solution to the initial value problem using the inverse Laplace transform. While our strategy is straight forward, finding an inverse Laplace tranform can be a hindrance. Often finding the inverse transform involves decomposing a complicated product into the sum for partial fractions. Fortunately, there is a product rule for inverse Laplace transforms. This product rule will allow us to quickly compute solutions of a harmonic oscillator with different forcing functions.

Subsection 6.4.1 Convolution

If \(f\) and \(g\) are two piecewise continuous exponentially bounded functions, then we define the convolution product of \(f\) and \(g\) to be
\begin{equation*} (f*g)(t) = \int_0^t f(t - \tau) g(\tau) \, d\tau = \int_0^t f(\tau) g(t - \tau) \, d\tau. \end{equation*}
The convolution product has many properties similar to those of ordinary multiplication.
  • Commutivity: \(f*g = g*f\text{.}\)
  • Distribution: \(f*(g + h) = f*g + f*h\text{.}\)
  • Associativity: \(f*(g*h) = (f*g)*h\text{.}\)
  • \(0 *f = f*0 = 0\text{.}\)
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
\begin{equation*} (f*1)(t) = \int_0^t \cos(t - \tau) \cdot 1 \, d\tau = \sin t. \end{equation*}
Also, it may not be the case that \(f*f\) is a nonnegative function.
A key property of the Laplace transform is how convolution products behave.

Proof.

If
\begin{align*} F(s) & = \int_0^\infty e^{-s\xi} f(\xi) \, d\xi\\ G(s) & = \int_0^\infty e^{-s\tau} g(\tau) \, d\tau, \end{align*}
then
\begin{align*} F(s) G(s) & = \int_0^\infty e^{-s\xi} f(\xi) \, d\xi \int_0^\infty e^{-s\tau} g(\tau) \, d\tau\\ & = \int_0^\infty g(\tau) \left( \int_\tau^\infty e^{-st} f(t - \tau) \, dt \right) \, d\tau, \end{align*}
where \(\xi = t - \tau\) is the change of variable. Reversing the order of integration, we have
\begin{equation*} \int_0^\infty g(\tau) \left( \int_\tau^\infty e^{-st} f(t - \tau) \, dt \right) \, d\tau = \int_0^\infty e^{-st} \left( \int_0^t f(t - \tau) g(\tau) \, d\tau \right) \, dt. \end{equation*}
However, this last expression is just the Laplace transform of \(f*g\text{.}\)

Subsection 6.4.2 Applying the Convolution Theorem

The Convolution Theorem (Theorem 6.4.1) can be very useful for finding the inverse transforms of products, since
\begin{equation*} F(s) G(s) = {\mathcal L}(f*g)(s). \end{equation*}
For example, suppose that we wish to find the inverse transform of
\begin{equation*} \frac{1}{s^2(s + 1)} = \frac{1}{s^2} \cdot \frac{1}{s + 1}. \end{equation*}
Since \({\mathcal L}[t] = 1/s^2\) and \({\mathcal L}[e^{-t}] = 1/(s + 1)\text{,}\) we know that
\begin{align*} {\mathcal L}^{-1} \left[ \frac{1}{s^2(s + 1)}\right] & = {\mathcal L}^{-1} \left[ \frac{1}{s^2} \right] * {\mathcal L}^{-1} \left[ \frac{1}{s + 1} \right]\\ & = \int_0^t (t - u) e^{-u}\, du\\ & = t + e^{-t} - 1. \end{align*}
Of course, we could have used partial fractions to decompose \(1/(s^2 (s + 1))\text{.}\)

Example 6.4.2.

Instead of using partial fractions, let us use the Convolution Theorem (Theorem 6.4.1) to calculate the inverse Laplace transform of
\begin{equation*} H(s)= \frac{a}{s^2(s^2 + a^2)} = \frac{1}{s^2} \cdot \frac{a}{s^2 + a^2}. \end{equation*}
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
\begin{equation*} h(t) = \int_0^t (t - \tau) \sin a \tau \, d \tau = \frac{at - \sin at}{a^2}. \end{equation*}
We can also use the Convolution Theorem to solve initial value problems.

Example 6.4.3.

Consider the initial value problem
\begin{align*} y'' + 4y & = g(t)\\ y(0) & = 3\\ y'(0) & = -1. \end{align*}
Taking the Laplace transform of both sides of the differential equation and applying the initial conditions, we obtain
\begin{equation*} s^2 Y(s) - 3s + 1 + 4Y(s) = G(s), \end{equation*}
where \(G(s)\) is the Laplace transform of \(g(t)\text{.}\) Solving for \(Y(s)\text{,}\) we have
\begin{equation*} Y(s) = \frac{3s-1}{s^2 + 4} + \frac{G(s)}{s^2 + 4} = 3\frac{s}{s^2 + 4} - \frac{1}{2} \frac{2}{s^2 + 4} + \frac{1}{2} \frac{2}{s^2 + 4} G(s). \end{equation*}
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
\begin{equation*} y = 3 \cos 2t - \frac{2}{2} \sin 2t + \frac{1}{2} \int_0^t \sin 2(t - \tau) g(\tau) \, d \tau. \end{equation*}
Notice that we did not specify a particular forcing function in Example 6.4.3. In fact, it is possible to write a solution for the initial value problem
\begin{align*} ay'' + by' + cy & = g(t)\\ y(0) & = y_0\\ y'(0) & = y_1 \end{align*}
using the Convolution Theorem without knowing what the actual forcing function \(g(t)\text{.}\) Taking the Laplace transform of both sides of the differential equation and using the initial conditions, we have
\begin{equation*} (as^2 + bs + c)Y(s) - (as + b)y_0 - ay_1 = G(s) \end{equation*}
or
\begin{equation*} Y(s) = \Phi(s) + \Psi(s), \end{equation*}
where
\begin{align*} \Phi(s) & = \frac{(as + b)y_0 + ay_1 }{as^2 + bs + c}\\ \Psi(s) & = \frac{G(s)}{as^2 + bs + c}. \end{align*}
Therefore,
\begin{equation*} y = \phi(t) + \psi(t), \end{equation*}
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
\begin{align*} ay'' + by' + cy & = 0\\ y(0) & = y_0\\ y'(0) & = y_1. \end{align*}
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
\begin{equation*} \Psi(s) = \frac{1}{as^2 + bs + c} G(s) = H(s) G(s), \end{equation*}
where \(H(s) = 1/(as^2 + bs + c)\text{.}\) If \(h(t)\) is the inverse Laplace transform of \(H(s)\text{,}\) then
\begin{equation*} \psi(t) = {\mathcal L}^{-1}(H(s)G(s)) = \int_0^t h(t - \tau) g(\tau) \, d\tau. \end{equation*}
If we consider the case where \(G(s) = 1\text{,}\) then \(g(t) = \delta(t)\) and \(\psi(s) = H(s)\text{.}\) This means that \(h(t)\) is a solution to the initial value problem
\begin{align*} ay'' + by' + cy & = \delta(t)\\ y(0) & = y_0\\ y'(0) & = y_1. \end{align*}
For this reason, \(h(t)\) is sometimes called the impulse response of the system.

Subsection 6.4.3 Important Lessons

  • If \(f\) and \(g\) are two piecewise continuous exponentially bounded functions, then we define the convolution product of \(f\) and \(g\) to be
    \begin{equation*} (f*g)(t) = \int_0^t f(t - \tau) g(\tau) \, d\tau = \int_0^t f(\tau) g(t - \tau) \, d\tau. \end{equation*}
  • The convolution product has many properties similar to those of ordinary multiplication.
    • Commutivity: \(f*g = g*f\)
    • Distribution: \(f*(g + h) = f*g + f*h\)
    • Associativity: \(f*(g*h) = (f*g)*h\)
    • \(\displaystyle 0 *f = f*0 = 0\)
    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.
  • 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
    \begin{equation*} F(s) G(s) = {\mathcal L}(f*g)(s) \end{equation*}
    for \(s \gt a\text{.}\)
  • It is possible to write a solution for the initial value problem
    \begin{align*} ay'' + by' + cy & = g(t)\\ y(0) & = y_0\\ y'(0) & = y_1. \end{align*}
    using the Convolution Theorem.

Reading Questions 6.4.4 Reading Questions

1.

Explain how the convolution product of two functions is useful when solving an initial value problem.

Exercises 6.4.5 Exercises

Convolution of Two Functions.

Calculate the convolution product of \(f(t)\) and \(g(t)\) in Exercise Group 6.4.5.1–6 using the definition of the convolution of two functions.
1.
\(f(t) = 2\) and \(g(t) = t\)
2.
\(f(t) = t\) and \(g(t) = t\)
3.
\(f(t) = t\) and \(g(t) = e^{3t}\)
4.
\(f(t) = t\) and \(g(t) = \sin t\)
5.
\(f(t) = e^{2t}\) and \(g(t) = e^{3t}\)
6.
\(f(t) = e^{3t}\) and \(g(t) = e^{2t}\)

Convolution and the Inverse Transform.

Calculate the inverse Laplace transform of \(F(s)\) in Exercise Group 6.4.5.7–12 using the convolution product.
7.
\(F(s) = \dfrac{1}{(s - 1)(s - 3)}\)
8.
\(F(s) = \dfrac{3}{s(s - 1)}\)
9.
\(F(s) = \dfrac{2}{s^2(s^2 + 2)}\)
10.
\(F(s) = \dfrac{1}{s^2 + 4s - 12}\)
11.
\(F(s) = \dfrac{1}{(s - 1)(s - 3)}\)
12.
\(F(s) = \dfrac{1}{(s - 1)(s - 3)}\)

Solving Initial Value Problems.

Solve the initial problems in Exercise Group 6.4.5.13–18 using the method described in Subsection 6.4.2.
13.
\(2 y'' + y' + 2y = g(t)\text{,}\) \(y(0) = 0\text{,}\) \(y'(0) = 0\)
14.
\(y'' - y' - 2y = g(t)\text{,}\) \(y(0) = 0\text{,}\) \(y'(0) = 0\)
15.
\(\dfrac{d^2x}{dx^2} - 6 \dfrac{dx}{dt} + 25 x = g(t)\text{,}\) \(x(0) = 1\text{,}\) \(x'(0) = -2\)
16.
\(y'' + 16y = g(t)\text{,}\) \(y(0) = 1\text{,}\) \(y'(0) = 0\)
17.
\(y'' + 16y = g(t)\text{,}\) \(y(0) = 1\text{,}\) \(y'(0) = 0\)
18.
\(y'' + 2y' + y = g(t)\text{,}\) \(y(0) = -1\text{,}\) \(y'(0) = 3\)

19.

Prove that the distributive property holds for the convolution product. That is, show that \(f*g = g*f\text{.}\)

20.

Prove that the convolution product of two functions \(f\) and \(g\) distributes multiplication over addition. That is, show that \(f*(g + h) = f*g + f*h\) for functions \(f\text{,}\) \(g\text{,}\) and \(h\text{.}\)

21.

Prove that the convolution product is associative. In other words, show that \(f*(g*h) = (f*g)*h\) for functions \(f\text{,}\) \(g\text{,}\) and \(h\text{.}\)

22.

Prove that \(0 *f = f*0 = 0\text{.}\)
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