Math 555: Differential Equations

6.6 The Convolution Integral

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6.6.1 Definition of The Convolution Integral

Let us solve the initial value problem

$$y'' + 4y = g(t);\quad y(0)=3,\ y'(0)=-1.$$
Since we do not know either the solution of the differential equation or the input function (forcing function), computing the Laplace transform of both sides results in

$$\begin{align*} s^2Y(s) - s(3) - (-1) + 4Y(s) &= G(s) \\\ \\ \left(s^2 + 4\right)Y(s) &= 3s - 1 + G(s) \\ \\ Y(s) &= \dfrac{3s-1}{s^2 + 4} + \dfrac{G(s)}{s^2 + 4} \\ \\ &= 3\dfrac{s}{s^2 + 4} - \dfrac{1}{2}\dfrac{2}{s^2 + 4} + \dfrac{1}{2}\dfrac{2}{s^2 + 4}G(s) \end{align*}$$
Using our Laplace Transform Table to find the inverse Laplace transform results in the solution,

$$y(t) = 3\cos(2t) - \dfrac{1}{2}\sin(2t) + \mathcal{L}^{-1}\left\{\dfrac{1}{2}\dfrac{2}{s^2 + 4}\cdot G(s)\right\}.$$

How do we compute the inverse Laplace transform of $\,\dfrac{1}{2}\dfrac{2}{s^2 + 4}\cdot G(s)$?

The lazy answer is that we use the convolution of the inverse Laplace transforms of $F(s) = \dfrac{1}{2}\dfrac{2}{s^2 + 4}$ and $G(s)$.

What does it mean to convolve two functions?

The convolution of two integrable functions $f$ and $g$ on the interval $[0, \infty)$ is a new operation usually denoted with the splat or asterisk * that results in a new function $f*g$ defined by

$$(f*g)(t) = \displaystyle\int_0^t f(\sigma)g(t-\sigma)\,d\sigma.$$
Notice that the integral in the definition has a variable upper limit and the variable upper limit is the independent variable of result. Notice also that the independent variable of our new function also appears in the integrand. Let us look closely at the factor $g(t-\sigma)$.

For function $g(\sigma)$ defined on the interval $[0,\infty)$, $g(-\sigma)$ reflects the function with respect to $0$, or reflects the graph of $g$ with respect to the vertical axis. Moreover $g(t-\sigma) = g\left(-(\sigma - t)\right)\,$ shifts $\,g(-\sigma)$ to the right $t$. Thus for a fixed input $t$, our convolution function $f*g$ has the value

$$(f*g)(t) = \displaystyle\int_0^t f(\sigma)g(t-\sigma)\,d\sigma.$$
There are several animations of convolution that you can find on the internet using your favorite search engine. For our mathematics, the most important property of our new operation is the following identity.

If we compute the Laplace transform of the convolution of two functions, we obtain

$$\begin{align*} \mathcal{L}\left[(f*g)\right] &= \displaystyle\int_0^{\infty} (f*g)(t)e^{-st}\,dt \\ \\ &= \displaystyle\int_0^{\infty} \left[\displaystyle\int_0^t f(\sigma)g(t-\sigma)\,d\sigma\right]\,e^{-st}\,dt \\ \\ &= \displaystyle\int_0^{\infty}\displaystyle\int_0^t f(\sigma)g(t-\sigma)e^{-st}\,d\sigma dt. \\ \end{align*}$$
The last step was due to Fubini's Theorem , which states that the order of integration may be switched under appropriate conditions. We are going to use Fubini's Theorem often in this derivation. The next step is to reverse the order of the integration so that instead of integrating first with respect to $\sigma$ and then $t$, we integrate first with respect to $t$ and then $\sigma$. This requires more Calculus III and an understanding of the domain of integration.


Currently the order of integration uses the vertical magenta bar as an example of our differential element. We want to change the order of integration and that will use the blue bar as an example of our differential element. This gives us

$$\begin{align*} \mathcal{L}\left[(f*g)\right] &= \displaystyle\int_0^{\infty}\displaystyle\int_t^{\infty} f(\sigma)g(t-\sigma)e^{-st}\,dt d\sigma \\ \\ &= \displaystyle\int_0^{\infty}\left[\displaystyle\int_t^{\infty} f(\sigma)g(t-\sigma)e^{-st}\,dt\right] d\sigma \\ \\ &= \displaystyle\int_0^{\infty}f(\sigma)\left[\displaystyle\int_t^{\infty} g(t-\sigma)e^{-st}\,dt\right] d\sigma. \\ \end{align*}$$
Substituting

$$\begin{matrix} \tau = t - \sigma & \qquad & \left.\tau\right|_{t=\sigma} = \sigma - \sigma = 0 \\ \ & & \left.\tau\right|_{t=\infty} = \infty - \sigma = \infty \\ d\tau = dt & & \color{#ec008c}{t = \tau + \sigma} \end{matrix}$$
results in the double integral

$$\begin{align*} \mathcal{L}\left\{(f*g)\right\} &= \displaystyle\int_0^{\infty}f(\sigma)\left[\displaystyle\int_0^{\infty} g(\tau)e^{-s(\tau + \sigma)}\,d\tau\right] d\sigma. \\ \\ &= \displaystyle\int_0^{\infty}f(\sigma)e^{-s\sigma}\left[\displaystyle\int_0^{\infty} g(\tau)e^{-s\tau}\,d\tau\right] d\sigma. \\ \\ &= \displaystyle\int_0^{\infty}f(\sigma)e^{-s\sigma}\,d\sigma\,\cdot\,\displaystyle\int_0^{\infty} g(\tau)e^{-s\tau}\,d\tau. \\ \\ &= F(s)\,\cdot\,G(s). \\ \end{align*}$$
We use this identity to find the inverse Laplace transform of a function in the $s$-domain.

$$\mathcal{L}^{-1}\left\{F(s)\cdot G(s)\right\} = f(t)\,*\,g(t) = (f*g)(t) = \displaystyle\int_0^t f(\sigma)g(t-\sigma)\,d\sigma.$$
This means that we can now determine a solution to our original initial value problem,

$$\begin{align*} y(t) &= 3\cos(2t) - \dfrac{1}{2}\sin(2t) + \mathcal{L}^{-1}\left\{\dfrac{1}{2}\dfrac{2}{s^2 + 4}\cdot G(s)\right\} \\ \\ &= 3\cos(2t) - \dfrac{1}{2}\sin(2t) + \dfrac{1}{2}\left(\mathcal{L}^{-1}\left\{\dfrac{2}{s^2 + 4}\right\}\,*\, \mathcal{L}^{-1}\left\{G(s)\right\} \right) \\ \\ &= 3\cos(2t) - \dfrac{1}{2}\sin(2t) + \dfrac{1}{2}\left( \sin(2t)\,*\,g(t) \right) \\ \\ &= 3\cos(2t) - \dfrac{1}{2}\sin(2t) + \dfrac{1}{2}\displaystyle\int_0^t \sin(2\sigma)g(t-\sigma)\,d\sigma. \end{align*}$$

6.6.2 Properties of the Convolution Integral

The convolution integral has several nice algebraic properties.

Properties of the Convolution Integral

1. Commutative

$$ f\,*\,g = g\,*\,f $$
2. Associative

$$ \left(f\,*\,g\right)\,*\,h = f\,*\,\left(g\,*\,h\right) $$
3. Distributive Over Addition

$$ f\,*\,\left(g + h\right) = f\,*\,g + f\,*\,h $$
4. Convolution with Zero is Zero

$$ f\,*\,0 = 0\,*\,f = 0 $$

Let us prove the first identity; all are exercises in the homework.

$$\left(f\,*\,g\right)(t) = \int_0^t f(\sigma)g(t-\sigma)\,d\sigma $$
To compute the integral, we integrate by substitution

$$ \int_0^t f(\sigma)g(t-\sigma)\,d\sigma = \left.\color{#307fe2}{\begin{vmatrix} u = t-\sigma & \quad & \left.u\right|_{\sigma=0} = t-0 = t \\ \ & \quad & \left.u\right|_{\sigma=t} = t-t = 0 \\ du = -d\sigma & \quad & \color{#ec008c}{\sigma = t-u} \end{vmatrix}}\right._{\Large .} $$
Under the substitution, our integral changes to

$$\begin{align*} \left(f\,*\,g\right)(t) &= \int_t^0 f(t-u)g(u)\,\left(-du\right) \\ \\ &= \displaystyle\int_0^t g(u)f(t-u)\,du \\ \\ &= \left(g\,*\,f\right)(t). \end{align*}$$

6.6.3 Using the Convolution Integral

Try the following exercises on your own before looking at the solution.

Exercise 6.6.1

Use the convolution integral to find the inverse Laplace transform of

$$F(s) = \dfrac{1}{s^4\left(s^2 + 1\right)}.$$

View Solution

$$\begin{align*} F(s) &= \dfrac{1}{s^4\left(s^2 + 1\right)} \\ \\ &= \dfrac{1}{s^4}\cdot\dfrac{1}{s^1+1} \\ \\ &= G(s)\cdot H(s) \end{align*}$$
We have expressed the function to invert as a product of two functions whose inverse Laplace transforms we know.

$$\begin{align*} G(s) &= \dfrac{1}{s^4} & H(s) &= \dfrac{1}{s^1+1} \\ \\ g(t) &= \dfrac{1}{6}t^3 & h(t) &= \sin(t) \end{align*}$$
This means that we can express the inverse transform of the product as a convolution.

$$\begin{align*} f(t) &= g(t)\,*\,h(t) \\ \\ &= \displaystyle\int_0^t \dfrac{1}{6}\sigma^3\sin(t-\sigma)\,d\sigma \end{align*}$$
Integration by parts several times gives us an expression for $f(t)$ which simplifies greatly after applying the limits of the integration. Doing so gives us our final result.

$$\begin{align*} f(t) &= \left[ \dfrac{\sigma^3}{6}\cos(t-\sigma) + \dfrac{\sigma^2}{2}\sin(t-\sigma) - \sigma\cos(t-\sigma) - \sin(t-\sigma) \right]_0^t \\ \\ &= \dfrac{t^3}{6} + 0 - t - 0 - \left( 0 + 0 - 0 - \sin(t) \right) \\ \\ &= \dfrac{t^3}{6} - t - \sin(t). \end{align*}$$

Exercise 6.6.2

Use the convolution integral to find the solution to the initial value problem

$$y'' + 3y' + 2y = \cos(\alpha t);\quad y(0)=1,\ y'(0)=0$$

View Solution

We begin by noting that $\alpha$ is a arbitrary constant, so it will carry through all of our calculations. Computing the Laplace transform of our expression, we have

$$\begin{align*} s^2Y(s) - s(1) - 0 + 3\left(sY(s) - 1\right) + 2Y(s) &= \dfrac{s}{s^2 + \alpha^2} \\ \\ \left(s^2 + 3s + 2\right)Y(s) &= s + 3 + \dfrac{s}{s^2 + \alpha^2}. \end{align*}$$
Next we seek to isolate $Y(s)$ to find an expression to invert.

$$\begin{align*} Y(s) &= \dfrac{s + 3}{s^2 + 3s + 2} + \dfrac{s}{\left(s^2 + 3s + 2\right)\left(s^2 + \alpha^2\right)} \\ \\ &= \dfrac{s + 3}{(s + 1)(s + 1)} + \dfrac{1}{s + 3}\cdot\dfrac{1}{s + 1}\cdot\dfrac{s}{s^2 + \alpha^2} \\ \\ &= \frac{-1}{s+2} + \dfrac{2}{s+1} + \left(\dfrac{-1}{s+2} + \dfrac{1}{s+1}\right)\left(\dfrac{s}{s^2 + \alpha^2}\right) \end{align*} $$
The first two terms of this expression may be inverted using the Laplace transform table. The product of the two factors at the end will need to be written as a convolution.

$$\begin{align*} y(t) &= -e^{-2t} + 2e^{-t} + f(t)\,*\,g(t) \\ \\ f(t) &= -e^{-2t} + e^{-t} \qquad\qquad\qquad g(t) = \cos(\alpha t) \end{align*}$$
The final expression takes for form

$$ y(t) = -e^{-2t} + 2e^{-t} + \displaystyle\int_0^t \left(e^{-\sigma} - e^{-2\sigma}\right)\cos\left(\alpha(t-\sigma)\right)\,d\sigma. $$

Exercise 6.6.3

Use the convolution integral to solve the initial value problem

$$4y'' + 4y' + 17y = g(t),\qquad y(0)=0,\ y'(0)=0.$$

View Solution

$$\begin{align*} 4\left(s^2Y(s) - 0 - 0\right) + 4\left(sY(s) - 0\right) + 17Y(s) &= G(s) \\ \\ \left(4s^2 + 4s + 17\right)Y(s) &= G(s) \\ \end{align*}$$
$$\begin{align*} Y(s) &= \dfrac{1}{4s^2 + 4s + 17}\cdot G(s) \\ \\ &= \dfrac{1}{4}\dfrac{1}{(s+1)^2 + 4}\cdot G(s) \\ \\ &= \dfrac{1}{8}\dfrac{2}{(s+1)^2 + 4}\cdot G(s) \\ \\ y(t) &= \dfrac{1}{8}\,\left(e^{-t/2}\sin(2t)\right)\,*\,g(t) \\ \\ &= \dfrac{1}{8}\displaystyle\int_0^t e^{-\sigma/2}\sin(2\sigma)g(t-\sigma)\,d\sigma \end{align*}$$

6.6.4 More Examples with Convolutions

Using the convolution may seem a bit convoluted at first, but its value comes in being able to write expressions for unknown functions. This is particularly useful for scientific computing, as computers are exceptional at numerical integration. Having an explicit expression for an integrand means that we can write down a formula for our solution even if the functions involved do not have easily found antiderivatives.

Example 6.6.1

Use the convolution integral to compute the Laplace transform of both sides of the equation

$$\phi(t) + \displaystyle\int_0^t k(t-\xi)\phi(\xi)\,d\xi = f(t).$$

Solution

Computing the Laplace transform results in

$$\begin{align*} \Phi(s) + \mathcal{L}\left\{ k(t)\,*\,\phi(t)\right\} &= F(s) \\ \\ \Phi(s) + K(s)\cdot\Phi(s) &= F(s). \end{align*}$$
The expression we just found the Laplace transfor of is an example of an integral equation . If we were to differentiate both sides of the original expression using the fundamental theorem of calclus, we would recover the differential equation

$$ \phi'(t) + k(t)\phi(t) = g(t) $$
(at least approximately, we would need to carefully apply the limits of integration).

Seeing that Laplace transforms are effective on integral equations as well as differential equations is quite exciting, because it allows us to start thinking about the situation where two ideas are mixed with one another.

Example 6.6.2

Use the convolution integral to solve the integro-differential initial value problem

$$\phi'(t) - \dfrac{1}{2}\displaystyle\int_0^t(t-\xi)^2\phi(\xi)\,d\xi = -t,\qquad\phi(0)=1.$$

Solution

Computing the Laplace transform of both sides of the equation yields

$$\begin{align*} s\Phi(s) - 1 - \dfrac{1}{2}\left( \mathcal{L}\left[t^2\right]\,\cdot\,\mathcal{L}\left[\phi(t)\right] \right) &= -\dfrac{1}{s^2} \\ \\ s\Phi(s) - \dfrac{1}{2}\left[\ \dfrac{2}{s^3}\cdot\Phi(s)\ \right] &= 1 - \dfrac{1}{s^2} \\ \\ \left(s - \dfrac{1}{s^3}\right)\Phi(s) &= \dfrac{s^2-1}{s^2} \end{align*} $$
We may now isolate our function $\Phi(s)$ and simplify it algebraicly to match a form from the Laplace transform table


$$\begin{align*} \Phi(s) &= \dfrac{s^2-1}{s^2}\cdot\dfrac{s^3}{s^4-1} \\ \\ &= \dfrac{s^2-1}{1}\cdot\dfrac{s}{\left(s^2-1\right)\left(s^2+1\right)} \\ \\ &= \dfrac{s}{s^2+1}. \end{align*} $$
The expression simplified tremendously. Inverting the transform gives us that

$$ \phi(t) = \cos(t). $$

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