# Problem of the Week #17 - July 23rd, 2012

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#### Chris L T521

##### Well-known member
Staff member
Thanks to those who participated in last week's POTW!! Here's this week's problem.

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Problem: Recall that the convolution of $f$ and $g$ is defined by the integral

$(f\ast g)(t) = \int_0^{t}f(t-\tau)g(\tau)\,d\tau.$

Establish the commutative, distributive, and associative properties of convolution, i.e.

(1) $f\ast g = g\ast f$
(2) $f\ast (g_1 + g_2) = f\ast g_1 + f\ast g_2$
(3) $f\ast(g\ast h) = (f\ast g)\ast h$.

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Last edited:

#### Chris L T521

##### Well-known member
Staff member
This week's question was correctly answered by Sudharaka and Badhi. You can find their solutions below.

Sudharaka's solution:

$(f*g)(t)=\int_{0}^{t}f(t-\tau)g(\tau)\,d\tau$

Substituting $$u=t-\tau$$ we get,

\begin{eqnarray}

(f*g)(t)&=&-\int_{t}^{0}f(u)g(t-u)\,du\\

&=&\int_{0}^{t}f(u)g(t-u)\,du\\

&=&(g*f)(t)

\end{eqnarray}

$\therefore f*g=g*f~~~~~~~~~~~~~(1)$

\begin{eqnarray}

[f*(g_{1}+g_{2})](t)&=&\int_{0}^{t}f(t-\tau)(g_{1}+g_{2})(\tau)\,d\tau\\

&=&\int_{0}^{t}f(t-\tau)[g_{1}(\tau)+g_{2}(\tau)]\,d\tau\\

&=&\int_{0}^{t}f(t-\tau)g_{1}(\tau)\,d\tau+\int_{0}^{t}f(t-\tau)g_{2}(\tau)\,d\tau\\

&=&(f*g_{1})(t)+(f*g_{2})(t)

\end{eqnarray}

$\therefore f*(g_{1}+g_{2})=(f*g_{1})+(f*g_{2})~~~~~~~~~~~~~(2)$

\begin{eqnarray}

[(f*g)*h](t)&=&\int_{0}^{t}(f*g)(t-\tau)h(\tau)\,d\tau\\

&=&\int_{0}^{t}\left[\int_{0}^{t-\tau}f(t-\tau-u)g(u)\,du\right]h(\tau)\,d\tau\\

\end{eqnarray}

Substituting $$w=u+\tau$$ we get,

$[(f*g)*h](t)=\int_{0}^{t}\left[\int_{\tau}^{t}f(t-w)g(w-\tau)\,dw\right]h(\tau)\,d\tau$

Changing the order of integration,

\begin{eqnarray}

[(f*g)*h](t)&=&\int_{0}^{t}\left[\int_{0}^{w}f(t-w)g(w-\tau)h(\tau)\,d\tau\right]\,dw\\

&=&\int_{0}^{t}f(t-w)\left[\int_{0}^{w}g(w-\tau)h(\tau)\,d\tau\right]\,dw\\

&=&\int_{0}^{t}f(t-w)(g*h)(w)\,dw\\

&=&[f*(g*h)](t)

\end{eqnarray}

$\therefore (f*g)*h=f*(g*h)~~~~~~~~~~~~~(3)$

from the definition,

$$(f\ast g)(t)=\int_0^tf(t-\tau)g(\tau)\,d \tau$$

(1) Proving that $f\ast g=g\ast f$

$f\ast g =\int_0^tf(t-\tau )g(\tau )\,d \tau$

with the substitution,
\begin{align*}t-\tau=v\\ \,dv=-\,d\tau\\ \end{align*}

then,
$$\tau=0 \implies v=t, \qquad \tau=t \implies v=0$$
\begin{align*} f\ast g &=\int_0^tf(t-\tau )g(\tau )\,d \tau\\ &=-\int_t^0f(v)g(t-v)\,dv\\ \end{align*}

Since the definite integral is independent from the variable, we can substitute $v$ with $\tau$

then the equation becomes,

\begin{align*}f\ast g &=-\int_t^0f(\tau )g(t-\tau )\, d\tau\\ &=\int_0^tf(\tau )g(t-\tau )\, d\tau\\ &=\underbrace{\int_0^tg(t-\tau )f(\tau )\, d\tau }_{g\ast f}\\ \end{align*}

thus proves the commutative property,
$$f\ast g=g\ast f$$

(2) proving, $f\ast (g_1+g_2)=f\ast g_1 +f\ast g_2$

\begin{align*} f\ast(g_1+g_2)&=\int_0^tf(t-\tau )\left[g_1(\tau )+g_2(\tau )\right]\, d\tau \\ &=\int_0^t f(t-\tau )g_1(\tau ) + f(t-\tau)g_2(\tau )\, d\tau \\ &=\underbrace{\int_0^t f(t-\tau )g_1(\tau )\, d\tau }_{f\ast g_1} + \underbrace{\int_0^t f(t-\tau )g_2(\tau )\, d\tau }_{f\ast g_2}\\ \end{align*}

thus proves the distributive property,
$$f\ast (g_1+g_2)=f\ast g_1 +f\ast g_2$$

(3) proving that, $f\ast (g\ast h)=(f\ast g)\ast h$

\begin{align*} f\ast (g\ast h) &= (g\ast h)\ast f \qquad \qquad \left(from \;proof: (1)\right)\\ &=\int_0^t\left[ (g\ast h)(t-\tau )\right] f(\tau ) \,d\tau \\ &=\int_0^tf(\tau )\left[ \int_0^{t-\tau} g(t- \tau -u)h(u) \,du\right]\,d\tau\\ &=\int_0^t\int_0^{t-\tau}f(\tau )g(t-\tau -u)h(u)\,du\,d\tau \\ \end{align*}

by changing the order of the integral, the change of boundaries of $\tau$ and $u$ are,

$0<\tau<t,\: 0<u<(t-\tau) \longrightarrow 0<u<t, \: 0<\tau<(t-u)$

then,

\begin{align*} f\ast (g\ast h) &=\int_0^t\int_0^{t-u}f(\tau )g(t-\tau -u)h(u)\,d\tau\,du\\ \end{align*}

since $h(u)$ is independent from $\tau$,

\begin{align*} f\ast (g\ast h) &=\int_0^th(u)\underbrace{\left[ \int_0^{t-u}f(\tau )g(t-u-\tau )\,d\tau\right]}_{(g\ast f)(t-u)} \,du\\ &=\int_0^th(u)[(g\ast f)(t-u)]\,du\\ &=(g\ast f)\ast h \end{align*}

from $proof : (1)$,

$f\ast (g\ast h)=(f\ast g)\ast h$

Thus proves the associative property,

$$f\ast (g\ast h)=(f\ast g)\ast h$$

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