Example6.3.1
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.
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 }For those that have seen convolution defined 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 (2) 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. of \(f(t)\) and \(g(t)\) as
As you can see, the convolution of two functions of \(t\) is another function of \(t\text{.}\)
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.
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.
Let \(f(t)\) and \(g(t)\) be of exponential type, 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.
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.
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.
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.
A common integral equation is the Volterra integral equation^{ 2 }Named for the Italian mathematician Vito Volterra (1860–1940).
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 \(X(s)\text{.}\)
Solve
We apply Laplace transform to obtain
or
It is not hard to apply Table 7.1.5 to find
Let \(f(t) = t^2\) for \(t \geq 0\text{,}\) and \(g(t) = u(t-1)\text{.}\) Compute \(f * g\text{.}\)
Let \(f(t) = t\) for \(t \geq 0\text{,}\) and \(g(t) = \sin t \) for \(t \geq 0\text{.}\) Compute \(f * g\text{.}\)
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.
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.
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.
Solve
Solve
Compute \({\mathcal{L}}^{-1} \left\{ \frac{s}{{(s^2+4)}^2} \right\}\) using convolution.
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 \(e^{-t^2}\text{.}\)
Let \(f(t) = \cos t\) for \(t \geq 0\text{,}\) and \(g(t) = e^{-t}\text{.}\) Compute \(f * g\text{.}\)
\(\frac{1}{2}(\cos t + \sin t - e^{-t})\)
Compute \({\mathcal{L}}^{-1} \left\{ \frac{5}{s^4+s^2} \right\}\) using convolution.
\(5t-5\sin t\)
Solve \(x''+x = \sin t\text{,}\) \(x(0) = 0\text{,}\) \(x'(0)=0\) using convolution.
\(\frac{1}{2}(\sin t - t \cos t)\)
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.
\(\int_0^t f(\tau) \bigl( 1 - \cos (t-\tau)\bigr)~ d\tau\)