Inverse Laplace Transforms — Question 10

Question 10 – Solution

Strategy. Initial data determine large-parameter coefficients. The fixed denominator converts those coefficients into a finite algebraic reconstruction.

Step 1: Recover the missing numerator. Three integrations by parts, with vanishing upper boundary terms for sufficiently large $s$, give $$F(s)=\frac{f(0)}s+\frac{f'(0)}{s^2}+\frac{f''(0)}{s^3} +\frac 1{s^3}\mathcal L\{f^{(3)}\}(s).$$ The last transform is $O(1/s)$ by an exponential bound. Since the denominator is $s^3+6s^2+11s+6$, expansion of the rational expression gives $$F(s)=\frac A{s}+\frac{B-6A}{s^2}+\frac{C-6B+25A}{s^3}+O(s^{-4}).$$ The prescribed values imply $\boxed{A=0,\ B=0,\ C=6}$.

Step 2: Resolve the three poles and check. Evaluation of the cleared identity at $s=-1,-2,-3$ gives coefficients $3,-6,3$. Hence $$F=\frac 3{s+1}-\frac 6{s+2}+\frac 3{s+3},\qquad \boxed{f(t)=3e^{-t}-6e^{-2t}+3e^{-3t}.}$$ Their forward numerator is six. The initial derivative sums are $3-6+3=0$, $-3+12-9=0$, and $3-24+27=6$, verifying all data.

Step 3: Factor to determine the sign and tail. The useful form is $\boxed{f(t)=3e^{-t}(1-e^{-t})^2}$. It is zero at $t=0$ and strictly positive for $t>0$. Its tail is asymptotic to $3e^{-t}$, so the exact real convergence interval is $\boxed{s>-1}$: above it all terms converge; at or below it the eventually positive weighted tail has infinite integral.

Step 4: Find the sharp peak and scope of reconstruction. Put $u=e^{-t}$, so $0<u\le 1$ and $f=3u(1-u)^2$. Its derivative with respect to $u$ is $3(1-u)(1-3u)$, so the unique interior maximum occurs at $u=1/3$: $$\boxed{t_* =\ln 3,\qquad \max_{t\ge 0}f(t)=4/9.}$$ The endpoint values are zero (at $t=0$ and as $t\to\infty$). The fixed three-parameter numerator made the reconstruction unique. Adding $t^3e^{-t}$ to this inverse preserves all three initial data but changes its transform and leaves the specified rational family. Thus three initial quantities alone do not determine an arbitrary function.