Section 6.3 Convolution
Subsection 6.3.1 The convolution
Video 6.3.1. Convolutions.
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
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
Apply the identity
to get
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
The most interesting property for us 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 6.2.1. 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. We get
or in other words
We know
Therefore,
or if we reverse the order
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 6.1.5 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.
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 6.3.4.
Find the solution to
for an arbitrary function \(f(t)\text{,}\) where \(m > 0\text{,}\) \(c > 0\text{,}\) \(k > 0\text{,}\) and \(c^2 - 4km > 0\) (the system is overdamped). 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 \(c^2 - 4km < 0\) (the system is underdamped). Write the solution as a definite integral.
Exercise 6.3.6.
Find the solution to
for an arbitrary function \(f(t)\text{,}\) where \(m > 0\text{,}\) \(c > 0\text{,}\) \(k > 0\text{,}\) and \(c^2 = 4km\) (the system is critically damped). Write the solution as a definite integral.
Exercise 6.3.7.
Solve \(x''+x = \sin t\text{,}\) \(x(0) = 0\text{,}\) \(x'(0)=0\) using convolution.
\(\frac{1}{2}(\sin t - t \cos t)\)
Exercise 6.3.8.
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.
\(\int_0^t f(\tau) \bigl( 1 - \cos (t-\tau)\bigr)~ d\tau\)
Exercise 6.3.9.
Find the Laplace transform of: \(t\int_0^t \cos(t-\tau) sin(\tau) ~ d\tau\)
Using the property of derivatives of the transform and the Convolution property, we write \[\mathcal{L}\left[t\int_0^t \cos(t-\tau) sin(\tau) d\tau\right]=-\frac{d}{ds}\mathcal{L}\left[\int_0^t \cos(t-\tau) sin(\tau) d\tau\right]=-\frac{d}{ds}\left[\mathcal{L}(\cos t)\cdot\mathcal{L}(\sin t)\right]\] \[\hspace*{5in}=-\frac{d}{ds}\left[\frac{s}{(s^{2}+1)}\right]=\frac{3 s^2-1}{\left(s^2+1\right)^3}\]
Exercise 6.3.10.
Find the Laplace transform of: \(\int_0^t \cos(t-\tau) sin(\tau) ~ d\tau\)
\(\frac{s}{(s^{2}+1)^{2}}\)
Exercise 6.3.11.
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 \(e^{-t^2}\text{.}\)
Exercise 6.3.12.
Solve
Exercise 6.3.13.
Solve
Exercise 6.3.14.
Compute \({\mathcal{L}}^{-1} \left\{ \frac{s}{{(s^2+4)}^2} \right\}\) using convolution.
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 6.3.15.
Compute \({\mathcal{L}}^{-1} \left\{\frac{2}{s^{3}(s-4)}\right\}\) using convolution.
\(\frac{1}{32}\left(e^{4t}-1-4t-8t^{2}\right)\)
Exercise 6.3.16.
Compute \({\mathcal{L}}^{-1} \left\{ \frac{5}{s^4+s^2} \right\}\) using convolution.
\(5t-5\sin t\)