Convolution of $f(x)={1 \over 2} \chi_{[-1,1]}*\chi_{[-5,5]}$

Question

Convolution of $f(x)={1 \over 2} \chi_{[-1,1]}*\chi_{[-5,5]}$. This is my first exercise, it's basically an interpolation between the two functions. So here my results:

Calling $u={1 \over 2} \chi_{[-1,1]}$ and $v=\chi_{[-5,5]}$ I can calculate the different values:

$u*v(0)={ 3 \over 2}$

$u*v({1 \over 2})={ 3 \over 2}$

$u*v(-{1 \over 2})={ 3 \over 2}$

So for the borders, we have:

• $0$ for $x<-5,x>5$

• $1$ for $x \geq -5,x \leq -{ 3\over 4}$

• $-x+{ 3 \over 4}$ for $-{ 3\over 4} \leq x \leq -{ 1 \over 4}$

• ${ 3 \over 2}$ for $-{ 1 \over 4} \leq x \leq { 1 \over 4}$

• $x-{ 3 \over 4}$ for ${ 1 \over 4} \leq x \leq { 3 \over 4}$

• $1$ for $x \geq { 3 \over 4},x\leq5$

I hope it's right (I can't draw the graphic here); if there is some error or not rigorous passage, please let me know I want to be capable to solve this at the best of possibilities

Answer 1

Is known that

$$\mathcal L(h(x-b)) = \dfrac1s e^{-bs},$$ $$\mathcal L((x-b)h(x-b)) = \dfrac1{s^2} e^{-bs},$$ $$\mathcal L(u(x)*v(x)) = \mathcal L(u(x)) \mathcal L(v(x)),$$

where $\mathcal L(f(x))$ is Laplace transform, $$h(t)= \begin{cases} 0,\quad\text{if}\quad t\in(-\infty,0]\\ 1,\quad\text{otherwize} \end{cases}$$ is the Heaviside function.

So $$u(x)=\dfrac12(h(x-1)-h(x+1)),\quad v(x)=h(x-5)-h(x+5),$$ $$\mathcal L(u(x)*v(x)) = \dfrac1{2s^2}(e^{-5s}-e^{5s})(e^{-s}-e^s) = \dfrac1{2s^2}(e^{-6s}-e^{-4s}-e^{4s}+e^{6s}),$$ $$u(x)*v(x) = \dfrac12\left((t+6)h(t+6)- (t+4)h(t+4)- (t-4)h(t-4)+ (t-6)h(t-6)\right),$$ $$u(x)*v(x)=\begin{cases} 0,\quad\text{if}\quad x \in(-\infty, -6]\\ \dfrac t2+3,\quad\text{if}\quad x \in(-6, -4]\\ 1,\quad\text{if}\quad x \in(-4, 4]\\ 3- \dfrac t2,\quad\text{if}\quad x \in(4, 6]\\ 0,\quad\text{if}\quad x \in(6,\infty). \end{cases}$$