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.

Convolution 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

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

Counterexample 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.

Convolution theorem 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^{2 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^{2 t}# and #g(t) = t^2# using the Laplace transform.

#(f\ast g) (t)=# #{{\euler^{2\cdot t}}\over{4}}-{{t^2}\over{2}}-{{t}\over{2}}-{{1}\over{4}}#

For starters, we determine #\laplace{(f)}# and #\laplace{(g)}# using the special cases of the Calculation rules for Laplace transforms:
\[\begin{array}{rclcl} \laplace{(f)}(s) &=& \laplace{(\e^{2 t})}(s) &=& \displaystyle {{1}\over{s-2}} \\
\laplace{(g)}(s) &=& \laplace{(t^2)}(s) &=& \displaystyle {{2}\over{s^3}} \\
\end{array}\]
We now complete the calculation with the aid of the Convolution theorem and the Inverse Laplace transforms of rational functions: \[\begin{array}{rclcl}(f\ast g) (t) &=&\displaystyle \laplace^{-1}{\left(\laplace{f}\cdot \laplace{g}\right)}(t) \\
&&\phantom{xx}\color{blue}{\text{Convolution theorem}}\\
&=&\displaystyle \laplace^{-1}{\left({{1}\over{s-2}}\cdot {{2}\over{s^3}}\right)}(t) \\
&&\phantom{xx}\color{blue}{\text{calculated above}}\\
&=&\displaystyle \laplace^{-1}{\left(-{{1}\over{4\cdot s}}-{{1}\over{2\cdot s^2}}-{{1}\over{s^3}}+{{1}\over{4\cdot \left(s-2\right)}}\right)}(t) \\
&&\phantom{xx}\color{blue}{\text{partial fraction decomposition}}\\
&=& \mathcal{L}^{-1}\left( -{{1}\over{4\cdot s}} \right)(t)+\mathcal{L}^{-1}\left( -{{1}\over{2\cdot s^2}} \right)(t)+\mathcal{L}^{-1}\left( -{{1}\over{s^3}} \right)(t)+\mathcal{L}^{-1}\left( {{1}\over{4\cdot \left(s-2\right)}} \right) (t) \\
&&\phantom{xx}\color{blue}{\text{linearity of }\mathcal{L}^{-1}}\\
&=& \displaystyle {{\euler^{2\cdot t}}\over{4}}-{{t^2}\over{2}}-{{t}\over{2}}-{{1}\over{4}} \\
&&\phantom{xx}\color{blue}{\text{known inverse Laplace transforms}}\\
\end{array}\]

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