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### Convolutions

#### Discussion

Since the Laplace transform is linear, the Laplace transform of a sum is the sum of the Laplace transforms. This is*not*true for products. To study products of Laplace transforms, we introduce a generalized product called the convolution. The convolution of two functions $f$ and $g$ is denoted $f\star g$ and is defined to be $$f\star g(t)=\int_0^t f(t-T)g(T)\,dT$$ For our purposes, the most important fact about convolutions is

*Theorem:*
${\mathcal L}\{f\star g\} = {\mathcal L}\{f\}{\mathcal L}\{g\}$.

*Proof:*
$$ \begin{align}
{\mathcal L}\{f\star g\}&=\int_0^{\infty}f\star g(t)e^{-st}\,dt \\
&=\int_0^{\infty}\biggl(\int_0^t f(t-T)g(T)\,dT\biggl)e^{-st}\,dt \\
&=\int_0^{\infty}\biggl(\int_T^{\infty}f(t-T)e^{-st}\,dt\biggl)g(T)\,dT \\
&=\int_0^{\infty}\biggl(\int_0^{\infty}f(u)e^{-s(u+T)}\,du\biggl)g(T)\,dT \\
&=\int_0^{\infty}{\mathcal L}\{f\}(s)e^{-sT}g(T)\,dT \\
&={\mathcal L}\{f\}(s)\int_0^{\infty}e^{-sT}g(T)\,dT \\
&={\mathcal L}\{f\}{\mathcal L}\{g\}
\end{align} $$
This will be very useful in
finding integral
representations for solutions of differential equations where the forcing
function is either unknown or something unusual that isn't in our tables. It
is possible to obtain integral representations of the solution in these cases
by using variation of parameters, but Laplace transforms are easier to handle
and usually lead to simpler integrals.

#### Paradigm

$$ \begin{align} x'' + x &= \sqrt t \\ x(0)&=0 \\ x'(0)&=0 \end{align} $$ We again use exactly the same techniques as in the two previous sections.
*Step 1:* Take the Laplace transform of both sides.
$$ \eqalign {
{\mathcal L}\{x'' + x\} &= s^2 {\mathcal L}\{x\} - sx(0) - x'(0)
+ {\mathcal L}\{x\} \cr
&= (s^2 + 1){\mathcal L}\{x\} \cr
{\mathcal L}\{\sqrt{t}\} &= \Gamma(3/2)/s^{3/2} \cr
} $$
So we have
$$(s^2 + 1){\mathcal L}\{x\} = \Gamma(3/2)/s^{3/2} $$
*Step 2:* Solve for ${\mathcal L}\{x\}$.
$${\mathcal L}\{x\} = [\Gamma(3/2)/s^{3/2}]\times[1/(s^2 + 1)]$$
*Step 3:* Take the inverse Laplace transform to find the solution.

At first glance there is nothing like this in the table. So we find the inverse Laplace transform in the form of a convolution. $$\begin{align} {\mathcal L}^{-1} \{\Gamma(3/2)/s^{3/2}\}&=\sqrt{t} \\ {\mathcal L}^{-1} \{1/(s^2 +1)\} &= \sin(t) \end{align} $$ So $$ x(t) = \int_0^t\sqrt{t-T}\sin(T)\, dT $$ Unfortunately, I can't evaluate this integral. But then from the very beginning we have run into problems where the best we could do was find an integral version for the solution.

EXAMPLE: $$ \begin{align} x'' + 3x' + 2x &= f(t) \\ x(0)=0\quad&\quad x'(0)=0 \end{align} $$ Step 1: $$ (s^2 + 3s + 2){\mathcal L}\{x\} = {\mathcal L}\{f\} $$ STEP 2: $$ {\mathcal L}\{x\} = {\mathcal L}\{f\}[1/(s^2 + 3s + 2)] $$ STEP 3: $1/(s^2 + 3s + 2) = 1/(s+1) - 1/(s+2)$ so $$ {\mathcal L}^{-1} \{1/(s^2 + 3s + 2)\} = {\mathcal L}^{-1}\{1/(s+1)\}- {\mathcal L}^{-1}\{1/(s+2)\}= e^{-t} - e^{-2t} $$ and $$ x(t) = f(t)\star [e^{-t} - e^{-2t} ] $$

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©2010, 2014 Andrew G. Bennett