## Section 6.3 Convolution

¶### Subsection 6.3.1 The convolution

We said that 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

As you can see, the convolution of two functions of \(t\) is another function of \(t\text{.}\)

###### Example 6.3.1.

Take \(f(t) = e^t\) and \(g(t) = t\) for \(t \geq 0\text{.}\) Then

To solve the integral we did one integration by parts.

###### Example 6.3.2.

Take \(f(t) = \sin (\omega t)\) and \(g(t) = \cos (\omega t)\) for \(t \geq 0\text{.}\) Then

We apply the identity

Hence,

The formula holds only for \(t \geq 0\text{.}\) We assumed that \(f\) and \(g\) are zero (or simply not defined) for negative \(t\text{.}\)

The 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

The most interesting property for us, and the main result of this section is the following theorem.

###### Theorem 6.3.1.

Let \(f(t)\) and \(g(t)\) be of exponential order, then

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 6.3.3.

Suppose we have the function of \(s\) defined by

We recognize the two entries of Table 7.2.1. That is

Therefore,

The calculation of the integral involved an integration by parts.

### Subsection 6.3.2 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 6.3.4.

Find the solution to

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.

or in other words

We know

Therefore,

or if we reverse the order

Let us 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

We have computed the convolution of sine and cosine in Example 6.3.2. Hence

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 6.3.3 Volterra integral equation

A common integral equation is the *Volterra integral equation*^{ 2 }

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

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

To find \(x(t)\) we now need to find the inverse Laplace transform of \(X(s)\text{.}\)

###### Example 6.3.5.

Solve

We apply Laplace transform to obtain

or

It is not hard to apply Table 7.1.5 to find

### Subsection 6.3.4 Exercises

###### Exercise 6.3.1.

Let \(f(t) = t^2\) for \(t \geq 0\text{,}\) and \(g(t) = u(t-1)\text{.}\) Compute \(f * g\text{.}\)

###### Exercise 6.3.2.

Let \(f(t) = t\) for \(t \geq 0\text{,}\) and \(g(t) = \sin t \) for \(t \geq 0\text{.}\) Compute \(f * g\text{.}\)

###### Exercise 6.3.3.

Find the solution to

for an arbitrary function \(f(t)\text{,}\) where \(m > 0\text{,}\) \(c > 0\text{,}\) \(k > 0\text{,}\) and \(c^2 - 4km > 0\) (system is overdamped). Write the solution as a definite integral.

###### Exercise 6.3.4.

Find the solution to

for an arbitrary function \(f(t)\text{,}\) where \(m > 0\text{,}\) \(c > 0\text{,}\) \(k > 0\text{,}\) and \(c^2 - 4km < 0\) (system is underdamped). Write the solution as a definite integral.

###### Exercise 6.3.5.

Find the solution to

for an arbitrary function \(f(t)\text{,}\) where \(m > 0\text{,}\) \(c > 0\text{,}\) \(k > 0\text{,}\) and \(c^2 = 4km\) (system is critically damped). Write the solution as a definite integral.

###### Exercise 6.3.6.

Solve

###### Exercise 6.3.7.

Solve

###### Exercise 6.3.8.

Compute \({\mathcal{L}}^{-1} \left\{ \frac{s}{{(s^2+4)}^2} \right\}\) using convolution.

###### Exercise 6.3.9.

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 \(e^{-t^2}\text{.}\)

###### Exercise 6.3.101.

Let \(f(t) = \cos t\) for \(t \geq 0\text{,}\) and \(g(t) = e^{-t}\text{.}\) Compute \(f * g\text{.}\)

\(\frac{1}{2}(\cos t + \sin t - e^{-t})\)

###### Exercise 6.3.102.

Compute \({\mathcal{L}}^{-1} \left\{ \frac{5}{s^4+s^2} \right\}\) using convolution.

\(5t-5\sin t\)

###### Exercise 6.3.103.

Solve \(x''+x = \sin t\text{,}\) \(x(0) = 0\text{,}\) \(x'(0)=0\) using convolution.

\(\frac{1}{2}(\sin t - t \cos t)\)

###### Exercise 6.3.104.

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.

\(\int_0^t f(\tau) \bigl( 1 - \cos (t-\tau)\bigr)~ d\tau\)