Background. (Convolution Theorem) Let denote the Laplace transforms of , respectively. Then the product is the Laplace transform of the convolution of , and is denoted by , and has the integral representation

Load Mathematica's built in "LaplaceTransform" subroutine package.

Computer Lab Work.

Example 1. Use convolution to find the inverse Laplace transform of .
Solution: H(s) is the product of ,
which are the Laplace transoms of f(t) = sin(t) and g(t) = 2 cos(t), respectively.

We can check this with Mathematica's result using the InverseLaplaceTransform package.

Example 2. Use the convolution theorem to solve the integral equation
.
Solution: Take the Laplace transform of each of the terms and get:

Use Mathematica to find the inverse Laplace transform, and plot f[t] over the interval 0 <= t <= 10.

More Background. The Laplace transform of the Dirac delta function is F(s) = 1. This can be illustrated with Mathematica.

Example 3. Solve the initial value problem
with .
Solution: Taking transforms results in the equation:

Use Mathematica to find the inverse Laplace transform, and plot f[t] over the interval 0 <= t <= 3.

Example 4. Use convolution to solve the initial value problem
with .
Plot the solution over the interval 0 <= t <= 1.57

Solution. First solve with .

Second solve for with .

Then the desired solution is y(t) = u(t) + v(t).

Now form y[t] and plot the solution over 0 <= t <= 1.57

If you wish to plot the solution over a larger interval, then y[t] will need to be simplified.

Furthermore, we need to take the logarithm of the absolute value, and this must be done by hand.

(c) John H. Mathews, 1998