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: $$ \begin{align} x'' + 2x' + 5x &= f(t) \\ x(0)=0\quad&\quad x'(0)=0 \end{align} $$ Step 1: $$ (s^2 + 2s + 5){\mathcal L}\{x\} = {\mathcal L}\{f\} $$ STEP 2: $$ {\mathcal L}\{x\} = {\mathcal L}\{f\}[1/(s^2 + 3s + 2)] $$ STEP 3: $1/(s^2 + 2s + 5) = 1/((s+1)^2 + 2^2)$ and $$ {\mathcal L}^{-1} \{1/(s^2 + 2s + 5)\} = (1/2)e^{-t}\sin(2t) $$ and $$ x(t) = f(t)\star (1/2)e^{-t}\sin(2t) $$ If you compare this to the paradigm in the section on Impulse Functions, you'll see that we have the same equation here, except that we have $f(t)$ on the right-hand side instead of $\delta(t)$. And our solution is $f(t)$ convolved with the solution we got when we used $\delta(t)$. This is another reason why we might want to solve a differential equation with a Dirac delta function on the right-hand side. The solution when you have a $\delta(t)$ is called the "fundamental solution" (though this terminology is more common when working with partial differential equations). Once you have the fundamental solution to a linear ordinary differential equation, you can find the solution for any right-hand side by convolution.If you have any problems with this page, please contact bennett@ksu.edu.
©1994-2025 Andrew G. Bennett