Differential Equations - Laplace Transforms: Convolution Integrals

On occasion we will run across transforms of the form,

$$H\left(s\right)=F\left(s\right)G\left(s\right)$$

that can’t be dealt with easily using partial fractions. We would like a way to take the inverse transform of such a transform. We can use a convolution integral to do this.

Convolution Integral

If $f\left(t\right)$ and $g\left(t\right)$ are piecewise continuous function on$\left[0,\infty \right)$ then the convolution integral of $f\left(t\right)$ and $g\left(t\right)$ is,
$$\left(f\ast g\right)\left(t\right)={\int }_0^{t}f\left(t-\tau \right)g\left(\tau \right)d\tau$$

A nice property of convolution integrals is.

$$\left(f\ast g\right)\left(t\right)=\left(g\ast f\right)\left(t\right)$$

Or,

$${\int }_0^{t}f\left(t-\tau \right)g\left(\tau \right)d\tau ={\int }_0^{t}f\left(\tau \right)g\left(t-\tau \right)d\tau$$

The following fact will allow us to take the inverse transforms of a product of transforms.

Fact

$$L\left\{f\ast g\right\}=F\left(s\right)G\left(s\right)L^{-1}\left\{F\left(s\right)G\left(s\right)\right\}=\left(f\ast g\right)\left(t\right)$$

Let’s work a quick example to see how this can be used.

Example 1 Use a convolution integral to find the inverse transform of the following transform.
$$H\left(s\right)=\frac{1}{{\left(s^2+a^2\right)}^2}$$

First note that we could use #11 from out table to do this one so that will be a nice check against our work here. Also note that using a convolution integral here is one way to derive that formula from our table.
Now, since we are going to use a convolution integral here we will need to write it as a product whose terms are easy to find the inverse transforms of. This is easy to do in this case.
$$H\left(s\right)=\left(\frac{1}{s^2+a^2}\right)\left(\frac{1}{s^2+a^2}\right)$$So, in this case we have,
$$F\left(s\right)=G\left(s\right)=\frac{1}{s^2+a^2}\Rightarrow f\left(t\right)=g\left(t\right)=\frac{1}{a}\sin \left(at\right)$$Using a convolution integral, along with a massive use of trig formulas, $h\left(t\right)$ is,
$$\begin{array}{cc}h\left(t\right)& =\left(f\ast g\right)\left(t\right)\\ & =\frac{1}{a^2}{\int }_0^t\sin \left(at-a\tau \right)\sin \left(a\tau \right)d\tau \\ & =\frac{1}{a^2}{\int }_0^t\left[\sin \left(at\right)\cos \left(a\tau \right)-\cos \left(at\right)\sin \left(a\tau \right)\right]\sin \left(a\tau \right)d\tau \\ & =\frac{1}{a^2}{\int }_0^t\sin \left(at\right)\cos \left(a\tau \right)\sin \left(a\tau \right)-\cos \left(at\right){\sin }^2\left(a\tau \right)d\tau \\ & =\frac{1}{2a^2}\sin \left(at\right){\int }_0^t\sin \left(2a\tau \right)d\tau -\frac{1}{2a^2}\cos \left(at\right){\int }_0^{t}1-\cos \left(2a\tau \right)d\tau \\ & =\frac{1}{2a^2}\sin \left(at\right){\left(-\frac{1}{2a}\cos \left(2a\tau \right)\right)|}_0^t-\frac{1}{2a^2}\cos \left(at\right){\left(\tau -\frac{1}{2a}\sin \left(2a\tau \right)\right)|}_0^t\\ & =\frac{1}{4a^3}\left(\sin \left(at\right)-\sin \left(at\right)\cos \left(2at\right)\right)-\frac{1}{2a^2}\cos \left(at\right)\left(t-\frac{1}{2a}\sin \left(2at\right)\right)\\ & =\frac{1}{4a^3}\sin \left(at\right)-\frac{1}{2a^2}t\cos \left(at\right)-\frac{1}{4a^3}\sin \left(at\right)\cos \left(2at\right)+\frac{1}{4a^3}\cos \left(at\right)\sin \left(2at\right)\\ & =\frac{1}{4a^3}\sin \left(at\right)-\frac{1}{2a^2}t\cos \left(at\right)-\frac{1}{8a^3}\left[\sin \left(3at\right)+\sin \left(-at\right)\right]+\frac{1}{8a^3}\left[\sin \left(3at\right)-\sin \left(-at\right)\right]\\ & =\frac{1}{4a^3}\sin \left(at\right)-\frac{1}{2a^2}t\cos \left(at\right)+\frac{1}{8a^3}\sin \left(at\right)+\frac{1}{8a^3}\sin \left(at\right)\\ & =\frac{1}{2a^3}\sin \left(at\right)-\frac{1}{2a^2}t\cos \left(at\right)\\ & =\frac{1}{2a^3}\left(\sin \left(at\right)-at\cos \left(at\right)\right)\end{array}$$This is exactly what we would have gotten by using #11 from the table.
Note however, that this did require a massive use of trig formulas that many do not readily recall. Also, while technically the integral was “simple”, in reality it was a very long and messy integral and illustrates why convolution integrals are not always done even when they technically can be.
One final note about the integral just to make a point clear. In the fourth step we factored the $\sin \left(at\right)$ and $\cos \left(at\right)$ out of the integrals. We could do that, in this case, because the integrals are with respect to $\tau$ and so, as for as the integrals were concerned, any function of $t$ is a constant. We can’t, of course, generally factor variables out of integrals. We can only do that when the variables do not, in any way, depend on the variable of integration.

Convolution integrals are very useful in the following kinds of problems.

Example 2 Solve the following IVP$$4y^{\prime \prime }+y=g\left(t\right),y\left(0\right)=3y^{\prime }\left(0\right)=-7$$

First, notice that the forcing function in this case has not been specified. Prior to this section we would not have been able to get a solution to this IVP. With convolution integrals we will be able to get a solution to this kind of IVP. The solution will be in terms of $g\left(t\right)$ but it will be a solution.
Take the Laplace transform of all the terms and plug in the initial conditions.
$$\begin{array}{cc}4\left(s^{2}Y\left(s\right)-sy\left(0\right)-y^{\prime }\left(0\right)\right)+Y\left(s\right)& =G\left(s\right)\\ \left(4s^2+1\right)Y\left(s\right)-12s+28& =G\left(s\right)\end{array}$$Notice here that all we could do for the forcing function was to write down $G\left(s\right)$ for its transform. Now, solve for $Y\left(s\right)$.
$$\begin{array}{cc}\left(4s^2+1\right)Y\left(s\right)& =G\left(s\right)+12s-28\\ Y\left(s\right)& =\frac{12s-28}{4\left(s^2+\frac{1}{4}\right)}+\frac{G\left(s\right)}{4\left(s^2+\frac{1}{4}\right)}\end{array}$$We factored out a 4 from the denominator in preparation for the inverse transform process. To take inverse transforms we’ll need to split up the first term and we’ll also rewrite the second term a little.
$$\begin{array}{cc}Y\left(s\right)& =\frac{12s-28}{4\left(s^2+\frac{1}{4}\right)}+\frac{G\left(s\right)}{4\left(s^2+\frac{1}{4}\right)}\\ & =\frac{3s}{s^2+\frac{1}{4}}-\frac{7\frac{2}{2}}{s^2+\frac{1}{4}}+\frac{1}{4}G\left(s\right)\frac{\frac{2}{2}}{s^2+\frac{1}{4}}\end{array}$$Now, the first two terms are easy to inverse transform. We’ll need to use a convolution integral on the last term. The two functions that we will be using are,
$$g\left(t\right)f\left(t\right)=2\sin \left(\frac{t}{2}\right)$$We can shift either of the two functions in the convolution integral. We’ll shift $g\left(t\right)$ in our solution. Taking the inverse transform gives us,
$$y\left(t\right)=3\cos \left(\frac{t}{2}\right)-14\sin \left(\frac{t}{2}\right)+\frac{1}{2}{\int }_0^t\sin \left(\frac{\tau }{2}\right)g\left(t-\tau \right)d\tau$$So, once we decide on a $g\left(t\right)$ all we need to do is to an integral and we’ll have the solution.

As this last example has shown, using convolution integrals will allow us to solve IVP’s with general forcing functions. This could be very convenient in cases where we have a variety of possible forcing functions and don’t know which one we’re going to use. With a convolution integral all that we need to do in these cases is solve the IVP once then go back and evaluate an integral for each possible $g\left(t\right)$. This will save us the work of having to solve the IVP for each and every $g\left(t\right)$.