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Note: 1 or 1.5 lectures, §7.2 in [EP], §6.6 in [BD]

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 and deﬁned for , and deﬁne the convolution^{3} of and as

(6.2) |

As you can see, the convolution of two functions of is another function of .

Example 6.3.2: Take and for . Then

We apply the identity

Hence,

The formula holds only for . We assumed that and are zero (or simply not deﬁned) for negative .

The convolution has many properties that make it behave like a product. Let be a constant and , , and be functions then

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

We recognize the two entries of Table 6.2. 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.

Example 6.3.4: Find the solution to

for an arbitrary function .

We ﬁrst apply the Laplace transform to the equation. Denote the transform of by and the transform of by 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 . Then

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

Note the in front of the sine. The solution, therefore, grows without bound as gets large, meaning we get resonance.

Similarly, we can solve any constant coeﬃcient equation with an arbitrary forcing function 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^{4}

where and are known functions and is an unknown we wish to solve for. To ﬁnd , we apply the Laplace transform to the equation to obtain

where , , and are the Laplace transforms of , , and respectively. We ﬁnd

To ﬁnd we now need to ﬁnd the inverse Laplace transform of .

Exercise 6.3.3: Find the solution to

for an arbitrary function , where , , , and (system is overdamped). Write the solution as a deﬁnite integral.

Exercise 6.3.4: Find the solution to

for an arbitrary function , where , , , and (system is underdamped). Write the solution as a deﬁnite integral.

Exercise 6.3.5: Find the solution to

for an arbitrary function , where , , , and (system is critically damped). Write the solution as a deﬁnite integral.

Exercise 6.3.9: Write down the solution to , , as a deﬁnite integral. Hint: Do not try to compute the Laplace transform of .

^{3}For those that have seen convolution deﬁned before, you may have seen it deﬁned as . This deﬁnition agrees with (6.2) if you deﬁne and to be zero for . 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.

^{4}Named for the Italian mathematician Vito Volterra (1860–1940).