Section 3.3 Convolution
Subsection 3.3.1 The convolution
We know two binary operations on functions - pointwise addition and multiplication - that takes two functions and give a third. The convolution f(t)*g(t) is a new operation, and one that is going to play particularly nicely with the Laplace Transform and be particularly nice for computing inverse Laplace transforms of products.
The Laplace transformation of a product is not the product of the transforms. All hope is not lost however. We simply have to use a different type of a “product.” Take two functions \(f(t)\) and \(g(t)\) defined for \(t \geq 0\text{,}\) and define the convolution 1 of \(f(t)\) and \(g(t)\) as
\begin{equation}
\mybxbg{~~
(f * g)(t) \overset{\text{def}}{=}
\int_0^t f(\tau) g(t-\tau) ~ d\tau .
~~}\tag{3.8}
\end{equation}
As you can see, the convolution of two functions of \(t\) is another function of \(t\text{.}\)
Example 3.3.1.
Take \(f(t) = e^t\) and \(g(t) = t\) for \(t \geq 0\text{.}\) Then
\begin{equation*}
(f*g)(t)
=
\int_0^t e^\tau (t-\tau) ~ d\tau
=
e^t - t - 1 .
\end{equation*}
To solve the integral we did one integration by parts.
Example 3.3.2.
Take \(f(t) = \sin (\omega t)\) and \(g(t) = \cos (\omega t)\) for \(t \geq 0\text{.}\) Then
\begin{equation*}
(f*g)(t)
=
\int_0^t \sin ( \omega \tau ) \,
\cos \bigl( \omega (t-\tau) \bigr) ~ d\tau .
\end{equation*}
Apply the identity
\begin{equation*}
\cos (\theta) \sin (\psi) =
\frac{1}{2} \, \bigl( \sin (\theta + \psi) - \sin (\theta - \psi) \bigr) ,
\end{equation*}
to get
\begin{equation*}
\begin{split}
(f*g)(t)
& =
\int_0^t
\frac{1}{2} \, \bigl( \sin (\omega t) - \sin (\omega t - 2 \omega \tau
) \bigr) ~ d\tau
\\
& =
\left[ \frac{1}{2} \, \tau \sin (\omega t) + \frac{1}{4\omega} \, \cos (2 \omega \tau -
\omega t) \right]_{\tau=0}^t
\\
& = \frac{1}{2} \, t \sin (\omega t) .
\end{split}
\end{equation*}
The formula holds only for \(t \geq 0\text{.}\) The functions \(f\text{,}\) \(g\text{,}\) and \(f*g\) are undefined for \(t < 0\text{.}\)
Convolution has many properties that make it behave like a product. Let \(c\) be a constant and \(f\text{,}\) \(g\text{,}\) and \(h\) be functions. Then
\begin{equation*}
\begin{aligned}
& f * g = g * f , \\
& (c f) * g = f * (c g) = c (f*g) , \\
& ( f * g ) * h = f * ( g * h ) .
\end{aligned}
\end{equation*}
The most interesting property for us is the following theorem.
Theorem 3.3.1.
Let \(f(t)\) and \(g(t)\) be of exponential order, then
\begin{equation*}
\mybxbg{~~
\mathcal{L} \bigl\{ (f*g)(t) \bigr\}
=
\mathcal{L} \left\{ \int_0^t f(\tau) g(t-\tau) ~ d\tau \right\}
=
\mathcal{L} \bigl\{ f(t) \bigr\} \mathcal{L} \bigl\{ g(t) \bigr\} .
~~}
\end{equation*}
In other words, the Laplace transform of a convolution is the product of the Laplace transforms. The simplest way to use this result is in reverse.
Example 3.3.3.
Suppose we have the function of \(s\) defined by
\begin{equation*}
\frac{1}{(s+1)s^2} =
\frac{1}{s+1}\,
\frac{1}{s^2} .
\end{equation*}
We recognize the two entries of Table 3.2.1. That is,
\begin{equation*}
\mathcal{L}^{-1}
\left\{
\frac{1}{s+1} \right\}
= e^{-t}
\qquad \text{and} \qquad
\mathcal{L}^{-1}
\left\{
\frac{1}{s^2} \right\}
= t.
\end{equation*}
Therefore,
\begin{equation*}
\mathcal{L}^{-1}
\left\{
\frac{1}{s+1}\,
\frac{1}{s^2} \right\}
=
\int_0^t
\tau e^{-(t-\tau)} ~d\tau
=
e^{-t}+t-1 .
\end{equation*}
The calculation of the integral involved an integration by parts.
Subsection 3.3.2 Solving ODEs
The next example demonstrates the full power of the convolution and the Laplace transform. We can give the solution to the forced oscillation problem for any forcing function as a definite integral.
Example 3.3.4.
Find the solution to
\begin{equation*}
x'' + \omega_0^2 x = f(t) , \quad x(0) = 0, \quad x'(0) = 0 ,
\end{equation*}
for an arbitrary function \(f(t)\text{.}\)
We first apply the Laplace transform to the equation. Denote the transform of \(x(t)\) by \(X(s)\) and the transform of \(f(t)\) by \(F(s)\) as usual. We get
\begin{equation*}
s^2 X(s) + \omega_0^2 X(s) = F(s) ,
\end{equation*}
or in other words
\begin{equation*}
X(s) = F(s) \frac{1}{s^2+ \omega_0^2} .
\end{equation*}
We know
\begin{equation*}
{\mathcal{L}}^{-1} \left\{
\frac{1}{s^2+ \omega_0^2}
\right\} =
\frac{\sin (\omega_0 t)}{\omega_0} .
\end{equation*}
Therefore,
\begin{equation*}
x(t) =
\int_0^t
f(\tau)
\frac{\sin \bigl( \omega_0 (t-\tau) \bigr)}{\omega_0} ~ d\tau ,
\end{equation*}
or if we reverse the order
\begin{equation*}
x(t) =
\int_0^t
\frac{\sin (\omega_0 \tau)}{\omega_0}
f(t-\tau) ~ d\tau .
\end{equation*}
Notice one more feature of this example. We can now see how Laplace transform handles resonance. Suppose that \(f(t) =
\cos (\omega_0 t)\text{.}\) Then
\begin{equation*}
x(t) =
\int_0^t
\frac{\sin (\omega_0 \tau)}{\omega_0} \,
\cos \bigl( \omega_0 (t-\tau) \bigr) ~ d\tau
=
\frac{1}{\omega_0}
\int_0^t
\sin ( \omega_0 \tau ) \,
\cos \bigl(\omega_0 (t-\tau) \bigr) ~ d\tau .
\end{equation*}
We have computed the convolution of sine and cosine in Example 3.3.2. Hence
\begin{equation*}
x(t) =
\left(
\frac{1}{\omega_0}
\right) \,
\left(
\frac{1}{2} \,
t \,
\sin ( \omega_0 t )
\right)
=
\frac{1}{2 \omega_0} \,
t
\,
\sin ( \omega_0 t ).
\end{equation*}
Note the \(t\) in front of the sine. The solution, therefore, grows without bound as \(t\) gets large, meaning we get resonance.
Similarly, we can solve any constant coefficient equation with an arbitrary forcing function \(f(t)\) as a definite integral using convolution. A definite integral, rather than a closed form solution, is usually enough for most practical purposes. It is not hard to numerically evaluate a definite integral.
Subsection 3.3.3 Volterra integral equation
A common integral equation is the Volterra integral equation 2
\begin{equation*}
x(t) = f(t) + \int_0^t g(t-\tau) x(\tau) ~ d\tau ,
\end{equation*}
where \(f(t)\) and \(g(t)\) are known functions and \(x(t)\) is an unknown we wish to solve for. To find \(x(t)\text{,}\) we apply the Laplace transform to the equation to obtain
\begin{equation*}
X(s) = F(s) + G(s) X(s) ,
\end{equation*}
where \(X(s)\text{,}\) \(F(s)\text{,}\) and \(G(s)\) are the Laplace transforms of \(x(t)\text{,}\) \(f(t)\text{,}\) and \(g(t)\) respectively. We find
\begin{equation*}
X(s) = \frac{F(s)}{1-G(s)} .
\end{equation*}
To find \(x(t)\) we now need to find the inverse Laplace transform of \(X(s)\text{.}\)
Example 3.3.5.
Solve
\begin{equation*}
x(t) = e^{-t} + \int_0^t \sinh(t-\tau) x(\tau) ~ d\tau .
\end{equation*}
We apply Laplace transform to obtain
\begin{equation*}
X(s) = \frac{1}{s+1} + \frac{1}{s^2-1} X(s) ,
\end{equation*}
or
\begin{equation*}
X(s) = \frac{\frac{1}{s+1}}{1- \frac{1}{s^2-1}}
=
\frac{s-1}{s^2 - 2}
=
\frac{s}{s^2 - 2}
-
\frac{1}{s^2 - 2} .
\end{equation*}
It is not hard to apply Table 3.1.5 to find
\begin{equation*}
x(t) = \cosh \bigl( \sqrt{2} \, t \bigr) -
\frac{1}{\sqrt{2}} \sinh \bigl( \sqrt{2}\, t \bigr).
\end{equation*}
Subsection 3.3.4 Exercises
Exercise 3.3.1.
Let \(f(t) = t^2\) for \(t \geq 0\text{,}\) and \(g(t) = u(t-1)\text{.}\) Compute \(f * g\text{.}\)
Solution.
\begin{equation*}
\begin{aligned}
f*g =& \int_0^t \tau^2 u(t-\tau-1) d\tau \\
=& \int_0^{t-1}\tau^2 d\tau = \frac{1}{3}(t-1)^3
\end{aligned}
\end{equation*}
Exercise 3.3.2.
Let \(f(t) = t\) for \(t \geq 0\text{,}\) and \(g(t) = \sin t \) for \(t \geq 0\text{.}\) Compute \(f * g\text{.}\)
Answer.
\(t-\sin(t)\)
Exercise 3.3.3.
Let \(f(t) = \cos t\) for \(t \geq 0\text{,}\) and \(g(t) = e^{-t}\text{.}\) Compute \(f * g\text{.}\)
Answer.
\(\frac{1}{2}(\cos t + \sin t - e^{-t})\)
Exercise 3.3.4.
Find the solution to
\begin{equation*}
m x'' + c x' + k x = f(t) , \quad x(0) = 0, \quad x'(0) = 0 ,
\end{equation*}
for an arbitrary function \(f(t)\text{,}\) where \(m > 0\text{,}\) \(c > 0\text{,}\) \(k > 0\text{,}\) and \(c^2 - 4km > 0\) (the system is overdamped). Write the solution as a definite integral.
Solution.
Taking the Laplace transform
\begin{equation*}
\begin{aligned}
ms^2X+csX+kX=& F(s) \\
X=& F(s)\underbrace{ \frac{1}{ms^2+cs+k}}_{G(s)} \\
G =& \frac{1}{m}\frac{1}{(s+\frac{c}{2m})^2+\frac{k}{m}-\frac{c^2}{4m^2}}= \frac{1}{m}\frac{1}{(s+\frac{c}{2m})^2-\left(\frac{\sqrt{c^2-4km}}{2m}\right)^2}=\frac{1}{m\omega}\frac{\omega}{(s+\frac{c}{2m})^2-\omega^2}
\end{aligned}
\end{equation*}
Where we defined \(\omega\equiv \frac{\sqrt{c^2-4km}}{2m}>0\text{.}\) We can now find \(g(t)\) using the shifting property
\begin{equation*}
\begin{aligned}
g(t) = \frac{1}{m\omega}e^{-\frac{c}{2m}t}\sinh(\omega t)
\end{aligned}
\end{equation*}
And finally using convolution
\begin{equation*}
\begin{aligned}
x(t)=\int_0^t f(\tau) \frac{1}{m\omega}e^{-\frac{c}{2m}(t-\tau)}\sinh(\omega (t-\tau))d\tau
\end{aligned}
\end{equation*}
Exercise 3.3.5.
Find the solution to
\begin{equation*}
m x'' + c x' + k x = f(t) , \quad x(0) = 0, \quad x'(0) = 0 ,
\end{equation*}
for an arbitrary function \(f(t)\text{,}\) where \(m > 0\text{,}\) \(c > 0\text{,}\) \(k > 0\text{,}\) and \(c^2 - 4km < 0\) (the system is underdamped). Write the solution as a definite integral.
Answer.
\(x(t)=\int_0^t f(\tau) \frac{1}{m\omega}e^{-\frac{c}{2m}(t-\tau)}\sin(\omega (t-\tau))d\tau\)
Exercise 3.3.6.
Find the solution to
\begin{equation*}
m x'' + c x' + k x = f(t) , \quad x(0) = 0, \quad x'(0) = 0 ,
\end{equation*}
for an arbitrary function \(f(t)\text{,}\) where \(m > 0\text{,}\) \(c > 0\text{,}\) \(k > 0\text{,}\) and \(c^2 = 4km\) (the system is critically damped). Write the solution as a definite integral.
Answer.
\(x(t)=\int_0^t f(\tau) \frac{1}{m}e^{-\frac{c}{2m}(t-\tau)}d\tau\)
Exercise 3.3.7.
Solve \(x''+x = \sin t\text{,}\) \(x(0) = 0\text{,}\) \(x'(0)=0\) using convolution.
Answer.
\(\frac{1}{2}(\sin t - t \cos t)\)
Exercise 3.3.8.
Solve \(x'''+x' = f(t)\text{,}\) \(x(0) = 0\text{,}\) \(x'(0)=0\text{,}\) \(x''(0)=0\) using convolution. Write the result as a definite integral.
Answer.
\(\int_0^t f(\tau) \bigl( 1 - \cos (t-\tau)\bigr)~ d\tau\)
Exercise 3.3.9.
Find the Laplace transform of: \(t\int_0^t \cos(t-\tau) sin(\tau) ~ d\tau\)
Solution.
Using the property of derivatives of the transform and the Convolution property, we write \[\mathcal{L}\left[t\int_0^t \cos(t-\tau) sin(\tau) d\tau\right]=-\frac{d}{ds}\mathcal{L}\left[\int_0^t \cos(t-\tau) sin(\tau) d\tau\right]=-\frac{d}{ds}\left[\mathcal{L}(\cos t)\cdot\mathcal{L}(\sin t)\right]\] \[=-\frac{d}{ds}\left[\frac{s}{(s^{2}+1)}\right]=\frac{3 s^2-1}{\left(s^2+1\right)^3}\]
Exercise 3.3.10.
Find the Laplace transform of: \(\int_0^t \cos(t-\tau) sin(\tau) ~ d\tau\)
Answer.
\(\frac{s}{(s^{2}+1)^{2}}\)
Exercise 3.3.11.
Write down the solution to \(x''-2x=e^{-t^2}\text{,}\) \(x(0)=0\text{,}\) \(x'(0)=0\) as a definite integral. Hint: Do not try to compute the Laplace transform of \(e^{-t^2}\text{.}\)
Solution.
Taking the Laplace transform of the equation
\begin{equation*}
\begin{aligned}
s^2X-2X=&L\left\{ e^{-t^2} \right\}=F(s) \\
X=& F(s) \underbrace{\frac{1}{s^2-2}}_{G(s)} \\
G(s) =& \frac{1}{\sqrt{2}}\frac{\sqrt{2}}{s^2-(\sqrt{2})^2} \rightarrow \\
g(t) =& \frac{1}{\sqrt{2}}\sinh(\sqrt{2}t)
\end{aligned}
\end{equation*}
So using the convolution property
\begin{equation*}
\begin{aligned}
x(t)=\int_0^t e^{-\tau^2}\frac{1}{\sqrt{2}}\sinh(\sqrt{2}(t-\tau))d\tau
\end{aligned}
\end{equation*}
Exercise 3.3.12.
Solve
\begin{equation*}
x(t) = e^{-t} + \int_0^t \cos(t-\tau) x(\tau) ~ d\tau .
\end{equation*}
Solution.
Taking the Laplace transform of the equation and solving for \(X\)
\begin{equation*}
\begin{aligned}
X=\frac{1}{s+1}\frac{s^2+1}{s^2-s+1}
\end{aligned}
\end{equation*}
Splitting the fraction and completing the square, this gives
\begin{equation*}
\begin{aligned}
x(t)=\frac{1}{\sqrt{3}}e^{t/2}\sin\left( \frac{\sqrt{3}}{2}t\right) + \frac{1}{3}e^{t/2}\cos\left( \frac{\sqrt{3}}{2}t\right) +\frac{2}{3}e^{-t}
\end{aligned}
\end{equation*}
Exercise 3.3.13.
Solve
\begin{equation*}
x(t) = \cos t + \int_0^t \cos(t-\tau) x(\tau) ~ d\tau .
\end{equation*}
Answer.
\(x(t)=\frac{1}{\sqrt{3}}e^{t/2}\sin\left( \frac{\sqrt{3}}{2}t\right)+e^{t/2}\cos\left( \frac{\sqrt{3}}{2}t\right)\)
Exercise 3.3.14.
Compute \({\mathcal{L}}^{-1} \left\{ \frac{s}{{(s^2+4)}^2} \right\}\) using convolution.
Solution.
First we re-write: \[\mathcal{L}^{-1}\left[\frac{s}{(s^{2}+4)^{2}}\right]=\frac{1}{2}\mathcal{L}^{-1}\left[\frac{2}{(s^{2}+4)}\cdot\frac{s}{(s^{2}+4)}\right]\] Now using the Convolution property of the Laplace tranform, we write: \[\mathcal{L}^{-1}\left[\frac{s}{(s^{2}+4)^{2}}\right]=\frac{1}{2}\sin(2t)\star\cos(2t)=\frac{1}{2}t\sin t\cos t\]
Exercise 3.3.15.
Compute \({\mathcal{L}}^{-1} \left\{\frac{2}{s^{3}(s-4)}\right\}\) using convolution.
Answer.
\(\frac{1}{32}\left(e^{4t}-1-4t-8t^{2}\right)\)
Exercise 3.3.16.
Compute \({\mathcal{L}}^{-1} \left\{ \frac{5}{s^4+s^2} \right\}\) using convolution.
Answer.
\(5t-5\sin t\)
For those that have seen convolution before, you may have seen it defined as \((f * g)(t) =
\int_{-\infty}^\infty f(\tau) g(t-\tau) ~ d\tau\text{.}\) This definition agrees with (3.8) if you define \(f(t)\) and \(g(t)\) to be zero for \(t < 0\text{.}\) When discussing the Laplace transform the definition we gave is sufficient. Convolution does occur in many other applications, however, where you may have to use the more general definition with infinities.
Named for the Italian mathematician Vito Volterra 3 (1860–1940).
en.wikipedia.org/wiki/Vito_Volterra
