Convolution of a Convolution

Question

I am confusing myself more and more with this question and require some assistance. The question reads:

Let: $$g(x) = \begin{cases}1 & \text{ if } |x|<\frac{1}{2}\\ 0 & \text{ otherwise }\end{cases}$$ Compute $(g*g)(x)$ and $(g*g*g)$ explicitly.

Now I have found that: $$(g*g)(x) = \begin{cases} 1-x & 0\leq x\leq 1\\ x+1 & -1\leq x\leq 0\\ 0 & \text{ otherwise}\end{cases}$$

But now I am trying to compute $g*g*g$. I have started: $$g*(g*g) = \int_{-\infty}^{\infty}g(x-y)\cdot (g*g)(y)dy$$ Thus, we require $-1\leq y\leq 1$ and $x-\frac{1}{2}\leq y\leq x+\frac{1}{2}$ for the integrand to not vanish. This says that $-\frac{3}{2}\leq x\leq \frac{3}{2}$. But now I'm confusing the hell out of myself. How do I divide this part into cases like I did with $(g*g)$? Do I separate it into two intervals $[-\frac{3}{2},0]$ and $[0,\frac{3}{2}]$? And treat each case separately?

Answer 1

At first we calculate $g\star g$ and then we use this result and proceed similarly to calculate $g\star g \star g$.

We obtain \begin{align*} \color{blue}{(g\star g)(x)}&=\int_{-\infty}^\infty g(y)g(x-y)\,dy\tag{1}\\ &=\int_{-\frac{1}{2}}^{\frac{1}{2}} g(x-y)\,dy\tag{2}\\ &=\int_{x-\frac{1}{2}}^{x+\frac{1}{2}} g(u)\,du\tag{3}\\ &=\begin{cases} \int_{x-\frac{1}{2}}^{x+\frac{1}{2}}1\,du &\qquad-\frac{1}{2}\leq u\leq \frac{1}{2}\\ 0&\qquad\text{else} \end{cases}\tag{4}\\ &=\begin{cases} \int_{-\frac{1}{2}}^{x+\frac{1}{2}}1\,du &\qquad-\frac{1}{2}\leq x+\frac{1}{2}\leq \frac{1}{2}\\ \int_{x-\frac{1}{2}}^{\frac{1}{2}}1\,du &\qquad-\frac{1}{2}\leq x-\frac{1}{2}\leq \frac{1}{2}\\ 0&\qquad\text{else} \end{cases}\tag{5}\\ &=\begin{cases} u\big|_{-\frac{1}{2}}^{x+\frac{1}{2}} &\qquad\qquad-1\leq x\leq 0\\ u\big|_{x-\frac{1}{2}}^{\frac{1}{2}} &\qquad\qquad0\leq x \leq 1\\ 0&\qquad\qquad\text{else} \end{cases}\\ &\color{blue}{=\begin{cases} x+1 &\qquad\qquad-1\leq x\leq 0\\ 1-x &\qquad\qquad0\leq x \leq 1\\ 0&\qquad\qquad\text{else} \end{cases}}\tag{6}\\ \end{align*}

Comment:

• In (1) we use the definition of the convolution $\star$.

• In (2) we use that $g(x)=1$ if $-\frac{1}{2}<x<\frac{1}{2}$ and $0$ else.

• In (3) we apply the substitution: $u=x-y, du=-dy$.

• In (4) we split the integral according to the support of $g(u)$.

• In (5) we note the support of $g$ is the interval $\left[-\frac{1}{2},\frac{1}{2}\right]$ with length $1$. The integral has limits of length $1$ and is non-zero in the interval $\left[x-\frac{1}{2},x+\frac{1}{2}\right]$. We consider this interval sliding over $[-\frac{1}{2},\frac{1}{2}]$. When the upper limit $x+\frac{1}{2}$ of this interval is part of the support of $g$ we have to consider $-\frac{1}{2}\leq x+\frac{1}{2}\leq \frac{1}{2}$ and when the lower limit $x-\frac{1}{2}$ of this interval is part of the support of $g$ we have to consider $-\frac{1}{2}\leq x-\frac{1}{2}\leq \frac{1}{2}$. Everything else has no contribution.

• In (6) we do the integration and write the ranges of validity in simplified form.

We now use $g\star g$ in order to calculate $g\star g \star g$ and proceed similarly. We obtain \begin{align*} \color{blue}{(g \star (g\star g))}&=\int_{-\infty}^\infty g(y)(g\star g)(x-y)\,dy\\ &=\int_{-\frac{1}{2}}^{\frac{1}{2}}(g\star g)(x-y)\,dy\\ &=\int_{x-\frac{1}{2}}^{x+\frac{1}{2}}(g\star g)(u)\,du\\ &=\begin{cases} \int_{x-\frac{1}{2}}^{x+\frac{1}{2}}(u+1)\,du&\qquad\qquad\qquad\qquad\qquad-1\leq u\leq 0\\ \int_{x-\frac{1}{2}}^{x+\frac{1}{2}}(1-u)\,du&\qquad\qquad\qquad\qquad\qquad0\leq u\leq 1\\ 0&\qquad\qquad\qquad\qquad\qquad\text{else} \end{cases}\tag{7}\\ &=\begin{cases} \int_{-1}^{x+\frac{1}{2}}(u+1)\,du&\qquad-1\leq x+\frac{1}{2}\leq 0\\ \int_{x-\frac{1}{2}}^0(u+1)\,du+\int_{0}^{x+\frac{1}{2}}(1-u)\,du&\qquad-1\leq x-\frac{1}{2}\leq 0,\\ &\qquad 0\leq x+\frac{1}{2}\leq 1\\ \int_{x-\frac{1}{2}}^1(1-u)\,du&\qquad0\leq x-\frac{1}{2}\leq 1\\ 0&\qquad\text{else} \end{cases}\tag{8}\\ &=\begin{cases} \left.\left(\frac{1}{2}u^2+u\right)\right|_{-1}^{x+\frac{1}{2}}&\qquad-\frac{3}{2}\leq x\leq -\frac{1}{2}\\ \left.\left(\frac{1}{2}u^2+u\right)\right|_{x-\frac{1}{2}}^{0}+\left.\left(u-\frac{1}{2}u^2\right)\right|_0^{x+\frac{1}{2}} &\qquad-\frac{1}{2}\leq x\leq \frac{1}{2}\\ \left.\left(u-\frac{1}{2}u^2\right)\right|_{x-\frac{1}{2}}^1&\qquad\frac{1}{2}\leq x\leq \frac{3}{2}\\ 0&\qquad\text{else} \end{cases}\\ &\color{blue}{=\begin{cases} \frac{1}{2}x^2+\frac{3}{2}x+\frac{9}{8}&\qquad\qquad\qquad\qquad\qquad-\frac{3}{2}\leq x\leq -\frac{1}{2}\\ -x^2+\frac{3}{4}&\qquad\qquad\qquad\qquad\qquad-\frac{1}{2}\leq x\leq \frac{1}{2}\\ \frac{1}{2}x^2-\frac{3}{2}x+\frac{9}{8}&\qquad\qquad\qquad\qquad\qquad\frac{1}{2}\leq x\leq \frac{3}{2} \end{cases}}\tag{9}\\ \end{align*}

Comment:

• In (7) we split the integral into three parts according to the representation of $(g\star g)(x)$ in (5).

• In (8) we observe that both integrals in (7) are similar to that in (4) and we treat each of them correspondingly. Note that both integrals have a contribution in the interval $-\frac{1}{2}\leq x\leq \frac{1}{2}$ and the ranges of validity coincide in that case.

• In (9) we do the integration and write the ranges of validity in simplified form.

Answer 2

Using Heaviside step function $g(x)={\bf H}(x+\dfrac12)-{\bf H}(x-\dfrac12)$ then with Laplace transform $${\mathcal L}(g)=\dfrac1s(e^{-\frac12s}-e^{\frac12s})=G(s)$$ \begin{align} g*g &= {\mathcal L}^{-1}(G^2(s)) \\ &= {\mathcal L}^{-1}\left(\dfrac1s(e^{-\frac12s}-e^{\frac12s})\right)^2 \\ &= {\mathcal L}^{-1}\left(\dfrac{1}{s^2}(e^{-s}+e^{s}-e^{0s})\right) \\ &= (x+1){\bf H}(x+1)+(x-1){\bf H}(x-1)-2x{\bf H}(x) \end{align} \begin{align} g*g*g &= {\mathcal L}^{-1}(G^3(s)) \\ &= {\mathcal L}^{-1}\left(\dfrac1s(e^{-\frac12s}-e^{\frac12s})\right)^3 \\ &= {\mathcal L}^{-1}\left(\dfrac{1}{s^3}(e^{-\frac32s}+3e^{-\frac12s}+3e^{\frac12s}+e^{\frac32s})\right) \\ &= (x+\dfrac32)^2{\bf H}(x+\dfrac32)+3(x+\dfrac12){\bf H}(x+\dfrac12)+3(x-\dfrac12)^2{\bf H}(x-\dfrac12)+(x-\dfrac32)^2{\bf H}(x-\dfrac32) \end{align}

Answer 3

Hint:

You want to do the convolution the hard way (using the first principles i.e., the integral form), so here it is made simpler with crude images:

$g$ looks like

$g*g$ looks like

Due to symmetry for $g*g*g$ it suffices to look only in the domain of $[-\frac{3}{2},0]$

$g*g*g$ in the domain $-3/2$ to $-1/2$ increases exponentially (in power of $2$ to be precise) and is depicted in the where the area of the shaded region is what you are after.

Similarly, $g*g*g$ in the domain $-1/2$ to $0$ increases in quadratic sense and is depicted in the where the area of the shaded region is what you are after.

Answer 4

First, notice that convolving two even functions results in an even function. Your $g(x)$ is an even function and hence all of its convolution powers are even functions. This saves us half of the work.

Second, notice that convolving piecewise polynomial functions results in a piecewise polynomial function whose degree is one more than the sum of the degrees of the polynomials we are convolving. Your $g(x)$ is a piecewise degree $0$ polynomial, $g\star g$ is a piecewise degree $1$ polynomial, and $g^{\star 3}:=g\star g\star g$ is a piecewise degree $2$ polynomial.

Third, notice that convolving two functions with finite support results in a function with finite support. Your $g(x)$ has support on the interval $(-1/2,1/2)$, $g\star g$ has support on the interval $(-1,1)$, and $g^{\star 3}$ has support on the interval $(-3/2,3/2)$.

Putting these three observations together we conclude that $g^{\star 3}$ is an even function which is a quadratic polynomial on $(-1/2,1/2)$ while it is a different quadratic polynomial on both $(1/2,3/2)$ and $(-3/2,-1/2)$. Notice that a quadratic polynomial is determined by its values at three different points so our work is greatly eased by this fact.