Convolution

Differential equations and Laplace transforms: Laplace-transformations

Convolution

The inverse Laplace transform of the product of the Laplace transforms of two functions $f$ and $g$ is a convolution of the functions $f$ and $g$ . Before dealing with this result, we define the convolution of two functions.

Let $f$ and $g$ be two functions defined on $\ivco{0}{\infty}$ . The convolution product or just convolution, of $f$ and $g$ , indicated by $f\ast g$ , is the function on $\ivco{0}{\infty}$ given by the rule $( f\ast g) (t) = \int_0^t f(u)\cdot g(t-u)\,\dd u$

The convolution products $f\ast 1$ and $1\ast f$ are antiderivatives of $f$ . After all, the first is equal to $\int_0^t f(t)\,\dd t$ and the second to $\int_0^t f(t-u)\,\dd t = -\int_t^0f(v)\,\dd v$ (thanks to the substitution $t-u =v$ ).

The convolution product has the following properties in common with the ordinary product:

Calculation rules for convolutions

$\begin{array}{lcrcl}\color{blue}{\text{commutativity:}}&\phantom{x}& f\ast g &=& g\ast f\\\color{blue}{\text{associativity:}}&\phantom{x}&(f\ast g)\ast h&=& f\ast (g\ast h)\\\color{blue}{\text{distributivity:}}&\phantom{x}& f\ast (g+ h)&=&f\ast g+f \ast h\end{array}$

The convolution does not have all the properties of the regular product. For instance, $1\ast 1 \ne 1$ because

$\left(1\ast1\right)(t) = \int_0^t 1\cdot 1\,\dd u = t \ne 1$

Commutativity:

$\begin{array}{rcl}(f\ast g)(t) &=& \int_0^t f(u)\cdot g(t-u)\dd u\\ &=&-\int_t^0 f(t-v)\cdot g(v)\dd v\\ &&\phantom{xx}\color{blue}{\text{substitution } u=t-v}\\ &=&\int_0^t g(v)\cdot f(t-v) \dd v\\ &=& (g\ast f)(t) \end{array}$

Associativity:

$\begin{array}{rcl}((f\ast g)\ast h)(t) &=& \int_0^t (f\ast g)(u)\cdot h(t-u)\dd u\\ &=&\int_0^t\int_0^u f(v)\cdot g(u-v) h(t-u)\dd v\,\dd u\\ &=&\int_0^t\int_v^t f(v)\cdot g(u-v) h(t-u)\dd u\,\dd v\\ &&\phantom{xx}\color{blue}{\text{integrals interchanged}}\\ &=&\int_0^t\int_0^{t-v} f(v)\cdot g(w) h(t-w-v)\dd w\,\dd v\\ &&\phantom{xx}\color{blue}{\text{substitution } w=u-v}\\ &=&\int_0^t f(v)\cdot (g\ast h)(t-v)\dd v\\ &=& (f\ast(g\ast h))(t) \end{array}$

Distributivity:

$\begin{array}{rcl}(f\ast (g+ h))(t) &=& \int_0^t f(u)\cdot (g+h)(t-u)\dd u\\ &=&\int_0^t f(u)\cdot g(t-u)\dd u+\int_0^t f(u)\cdot h(t-u)\dd u\\ &&\phantom{xx}\color{blue}{\text{linearity of the definite integral }}\\ &=&(f\ast g)(t)+(f \ast h)(t)\\ &=& (f\ast g+f \ast h)(t) \end{array}$

The following theorem says that the Laplace transform of a convolution in the $t$ -domain carries over to an ordinary product into $s$ -domain.

The Laplace transformation $\mathcal{L}$ satisfies the following equality for all piece-wise continuous functions $f$ and $g$ on $\ivco{0}{\infty}$ whose Laplace transforms are defined on $\ivco{a}{\infty}$ . $\mathcal{L}(f\ast g) ={\mathcal L}(f) \cdot {\mathcal L}(g)\phantom{xx} \text{ on }\phantom{xx}\ivco{a}{\infty}$

$\begin{array}{rcl} {\mathcal L}(f\ast g) (s) &=&\displaystyle \int_0^{\infty} (f\ast g)(t) \ee^{-s\cdot t}\dd t \\ &=&\displaystyle \int_0^{\infty} \int_0^t f(u)\cdot g(t-u)\,\dd u \,\ee^{-s\cdot t}\dd t\\ &=&\displaystyle \int_0^{\infty} \int_0^t f(u)\cdot g(t-u)\,\,\ee^{-s\cdot t}\dd u\dd t\\ &=&\displaystyle \int_0^{\infty} \int_u^{\infty} f(u)\cdot g(t-u)\,\ee^{-s\cdot t} \dd t \,\dd u\\ &&\phantom{xx}\color{blue}{\text{order of integration changed}}\\ &=&\displaystyle \int_0^{\infty} \int_u^{\infty} f(u)\cdot\ee^{-su} g(t-u)\,\ee^{-s\cdot( t-u)} \dd t \,\dd u\\ &&\phantom{xx}\color{blue}{\ee^{-su}\cdot \ee^{-s(t-u)} = \ee^{-st}}\\ &=&\displaystyle \int_0^{\infty} \int_0^{\infty} f(u)\cdot\ee^{-su} g(v)\,\ee^{-s\cdot v} \dd v \,\dd u\\ &&\phantom{xx}\color{blue}{\text{substitution }t-u = v}\\ &=&\displaystyle \int_0^{\infty} f (u)\cdot \ee^{-su}\,\dd u \cdot \int_0^{\infty} g (v)\cdot \ee^{-sv}\,\dd v \\ &&\phantom{xx}\color{blue}{\text{second term is constant in }u}\\&=& \displaystyle {\mathcal L}(f) (s)\cdot {\mathcal L}(g) (s)\\&=& \displaystyle \left({\mathcal L}(f) \cdot {\mathcal L}(g) \right)(s)\end{array}$

Application of $\mathcal{L}^{-1}$ to this formula shows that, for two Laplace-transforms $y(s)$ and $z(s)$ ,

$\mathcal{L}^{-1}(y)\ast \mathcal{L}^{-1}(z)={\mathcal L}^{-1}(y\cdot z)$

With $f= \mathcal{L}^{-1}(y)$ and $g = \mathcal{L}^{-1}(z)$ , this can be rewritten as ${\mathcal L}^{-1}\left({\mathcal L}(f)\cdot{\mathcal L}(g) \right)=f\ast g$ , which is the version stated at the beginning of this page.

The convolution product of $f(t) = \e^{5 t}$ and $g(t) = t^2$ can be determined directly by determining an antiderivative of the function $f(\tau)\cdot g(t-\tau)$ . This can be found by means of partial integration.

Determine the convolution product of $f(t) = \e^{5 t}$ and $g(t) = t^2$ using the Laplace transform.

$(f\ast g) (t)=$ ${{2\cdot \euler^{5\cdot t}}\over{125}}-{{t^2}\over{5}}-{{2\cdot t}\over{25}}-{{2}\over{125}}$

For starters, we determine $\laplace{(f)}$ and $\laplace{(g)}$ using the special cases of the Calculation rules for Laplace transforms:
$\begin{array}{rclcl} \laplace{(f)}(s) &=& \laplace{(\e^{5 t})}(s) &=& \displaystyle {{1}\over{s-5}} \\ \laplace{(g)}(s) &=& \laplace{(t^2)}(s) &=& \displaystyle {{2}\over{s^3}} \\ \end{array}$
We now complete the calculation with the aid of the Convolution theorem and the Inverse Laplace transforms of rational functions: $\begin{array}{rclcl}(f\ast g) (t) &=&\displaystyle \laplace^{-1}{\left(\laplace{f}\cdot \laplace{g}\right)}(t) \\ &&\phantom{xx}\color{blue}{\text{Convolution theorem}}\\ &=&\displaystyle \laplace^{-1}{\left({{1}\over{s-5}}\cdot {{2}\over{s^3}}\right)}(t) \\ &&\phantom{xx}\color{blue}{\text{calculated above}}\\ &=&\displaystyle \laplace^{-1}{\left(-{{2}\over{125\cdot s}}-{{2}\over{25\cdot s^2}}-{{2}\over{5\cdot s^3}}+{{2}\over{125\cdot \left(s-5\right)}}\right)}(t) \\ &&\phantom{xx}\color{blue}{\text{partial fraction decomposition}}\\ &=& \mathcal{L}^{-1}\left( -{{2}\over{125\cdot s}} \right)(t)+\mathcal{L}^{-1}\left( -{{2}\over{25\cdot s^2}} \right)(t)+\mathcal{L}^{-1}\left( -{{2}\over{5\cdot s^3}} \right)(t)+\mathcal{L}^{-1}\left( {{2}\over{125\cdot \left(s-5\right)}} \right) (t) \\ &&\phantom{xx}\color{blue}{\text{linearity of }\mathcal{L}^{-1}}\\ &=& \displaystyle {{2\cdot \euler^{5\cdot t}}\over{125}}-{{t^2}\over{5}}-{{2\cdot t}\over{25}}-{{2}\over{125}} \\ &&\phantom{xx}\color{blue}{\text{known inverse Laplace transforms}}\\ \end{array}$

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