Table Of Laplace Transforms — Question 9

Question 9 – Solution

Strategy. The assumed modes turn derivative data into a finite linear system. A negative coefficient need not make their sum negative.

Step 1: Solve and prove uniqueness. The data give $$c_1+c_2+c_3=0,\quad -c_1-2c_2-3c_3=0,\quad c_1+4c_2+9c_3=2.$$ Adding the first two equations gives $c_2=-2c_3$, and the first then gives $c_1=c_3$. Substitution into the third gives $2c_3=2$. Every coefficient is forced, proving uniqueness: $\boxed{(c_1,c_2,c_3)=(1,-2,1)}$.

Step 2: Assemble the rational transform. The table yields $$F=\frac 1{s+1}-\frac 2{s+2}+\frac 1{s+3} =\boxed{\frac 2{(s+1)(s+2)(s+3)}}.$$ Near zero, $f=t^2+O(t^3)$, so its leading transform is $2/s^3$, matching the quotient. The $1/s$ and $1/s^2$ terms cancel exactly.

Step 3: Establish positivity and the peak. Factor the function as $f=e^{-t}(1-e^{-t})^2$. It is positive for every $t>0$ and zero at $0$. Put $x=e^{-t}\in(0,1]$. The function $x(1-x)^2$ has derivative $(1-x)(1-3x)$ with respect to $x$; along increasing $t$ its unique interior maximum is at $x=1/3$. Therefore $$\boxed{t_{\max}=\ln 3,\qquad f(t_{\max})=4/27.}$$

Step 4: Check area, domain and identifiability. The area is $1-2/2+1/3=\boxed{1/3}$. The nonzero $e^{-t}$ tail gives exact real domain $s>-1$. Without the mode restriction, adding any multiple of $t^3e^{-t}$ would preserve all three measured derivatives but change the response. Thus the identification is unique within the stated model, not among all smooth functions.