Convolution Integrals — Question 6

Question 6 – Solution

Strategy. Equal exponential factors leave polynomial integrals inside the convolution. Unit mass is preserved while the response is spread over time.

Step 1: Integrate the cascade. Directly, $$(k_a*k_a)(t)=a^2e^{-at}\int_0^t1\,du=a^2t e^{-at}, \qquad r_a(t)=a^3e^{-at}\int_0^t u\,du =\boxed{\tfrac 12a^3t^2e^{-at}}.$$

Step 2: Check transforms and grouping. For real $s>-a$, $R_a(s)=[a/(s+a)]^3$. Both groupings integrate the same product $k_a(t-u-v)k_a(u)k_a(v)$ over $u,v\ge 0$, $u+v\le t$. The continuous integrand is integrable on this bounded triangle, so reversing the integration order proves associativity directly.

Step 3: Compare the peak and onset. For $t>0$, $r_a'=\tfrac 12a^3e^{-at}t(2-at)$. Thus the unique maximum is at $t=2/a$, with height $2ae^{-2}$. Also $r_a(0)=r_a'(0)=0$. A single stage has maximum $a$ at zero and right derivative $-a^2$ there. The cascade begins more smoothly, reaches its peak later and has a lower peak.

Step 4: Compute mass and mean time. Integration by parts yields $\int_0^\infty t^n e^{-at}\,dt=n!/a^{n+1}$ for nonnegative integers $n$. Hence $$\int k_a=1,\quad\int r_a=1,\qquad \int t k_a(t)\,dt=\frac 1a,\quad \int t r_a(t)\,dt=\frac 3a.$$ Since both masses are one, these first moments are their mean times. The total area is preserved; the mean time triples. The cascade remains positive at every $t>0$, so its later peak is not a literal waiting interval with zero output. The figure uses $a=1$ and the same vertical scale for both.