Heaviside step function properties $\theta(t-t_1) \theta(t_1-t_2) + \theta(t-t_2) \theta(t_2-t_1) = \theta(t-t_1) \theta(t-t_2)$

Question

I want to prove the following Heaviside theta function property :

\begin{align} \theta(t-t_1) \theta(t_1-t_2) + \theta(t-t_2) \theta(t_2-t_1) = \theta(t-t_1) \theta(t-t_2) \end{align}

Recalling the definitions of theta \begin{align} \theta(x) = \begin{cases} 1 , x\geq 0\\ 0, x<0 \end{cases} \end{align} In some sense, i got the feeling but I want to show this property with explicitly.

Answer 1

Pick $t=t_1=t_2$. Then your relation becomes $2=1$ which is false.

Assume now that $t_1\neq t_2$. More precisely $t_1<t_2$ (the other case is treated the same way). The right hand side is non zero if and only if $t\geq t_1$ and $t\geq t_2$.

Then the left hand side becomes $$\theta(t-t_2)\theta(t_2-t_1)=\theta(t-t_2)$$ which is alzo non-zero if and only if $t \geq t_2>t_1$.