(a)
Let \(f(t),g(t)\) be functions. The convolution of \(f\) and \(g\) is the function
\[\displaystyle (f\ast g)(t)=\int _{0}^{t}f(x)g(t-x)\,\mathrm {d}x.\]
(i)
Compute the convolution of \(f(t)=t^2\) and \(g(t)=e^{-t}\) directly from the definition above. Put the answer in the box provided.

(ii)
Show that the convolution of \(f(t)\) and \(g(t)\) does not depend on order, that is, \(f\ast g=g\ast f\).

Hint: use the \(u\)-substitution \(u=t-x\).

Recall: the Laplace transform of a function \(f(t)\) is \(\displaystyle \mathcal {L}[f]=\int _{0}^{\infty }f(t)e^{-st}\,\mathrm {d}t\).

(b)
Let \(f(t)\) and \(g(t)\) be continuous functions for \(t\geq 0\). Fill-in the missing details below, after each of the ellipses, which will demonstrate that
\(\mathcal {L}[f\ast g]=\mathcal {L}[f]\mathcal {L}[g]\):

First note that \(\displaystyle \mathcal {L}[f]\mathcal {L}[g]=\int _0^{\infty }f(t)e^{-st}\,\mathrm {d}t\int _0^{\infty }g(t)e^{-st}\,\mathrm {d}t\). By replacing all instances of \(``\)t"with \(``\)x"in the first integral and \(``\)y"in the second integral, we get that

\(\displaystyle \mathcal {L}[f]\mathcal {L}[g]=\) ... .

Bringing the second integral into the integrand of the first integral,

\(\displaystyle \mathcal {L}[f]\mathcal {L}[g]=\) ... .

Then by bringing the integrand of the first integral into the second integral, we get the iterated double integral

\(\displaystyle \mathcal {L}[f]\mathcal {L}[g]=\int _0^{\infty }\int _0^{\infty }\) ... \(\mathrm {d}y\,\mathrm {d}x\).

Using the substitution \(t=x+y\) in the \(``\)inner"integral, (\(x\) can be regarded as a constant).

Then \(\mathrm {d}t=\mathrm {d}y\), and \(t=\) ... when \(y=0\), and \(t\rightarrow ...\) when \(y\rightarrow \infty \),

so we get \(\displaystyle \mathcal {L}[f]\mathcal {L}[g]=\int _0^{\infty }\int _x^{\infty }\) ... \(\mathrm {d}t\,\mathrm {d}x\).

The region of the \(xt\)-plane for this improper iterated double integral is ...
(Shade the region!)

The identity function through the first quadrant of a plane. [Picture]
Figure 1: A line in a a plane

Let’s reverse the order of the integration, from \(\mathrm {d}t\,\mathrm {d}x\) to \(\mathrm {d}x\,\mathrm {d}t\). The region shaded can be described as

\(\{(t,x)|0\leq t\leq \infty ,\) ... \(\leq x\leq \) ... \(\}\) so we get that

\(\displaystyle \mathcal {L}[f]\mathcal {L}[g]=\) ...

and the inner integral is not improper! Pulling \(e^{-st}\) out of the inner integral and using \(\displaystyle (f\ast g)(t)=\int _{0}^{t}f(x)g(t-x)\,\mathrm {d}x\) we get

\(\displaystyle \mathcal {L}[f]\mathcal {L}[g]=\) ... \(=\mathcal {L}[...\)      \(]\).

(c)
Now compute the convolution of \(f(t)=t^2\) and \(g(t)=e^{-t}\) as follows:
  • Determine the product \(\mathcal {L}(f)\mathcal {L}(g)\).
  • Perform a partial fraction decomposition on the above function.
  • Take the inverse Laplace transform of the result.
(d)
Solve the “integral equation"\(y(t)=2+\displaystyle \int _0^t (t-x)y(x)\,\mathrm {d}x\). Put the answer in the box provided.

Hint: the integral is the convolution of two functions.