Inverse Laplace Transforms — Question 7

Question 7 – Solution

Strategy. Residues, including their possible zeros, decide which modes are actually present. An unreduced denominator can mislead.

Step 1: Recover both mode coefficients. Write $F_\lambda=A/(s-2)+B/(s+1)$. Clearing denominators gives $A+B=1$ and $A-2B=-2+\lambda$, so $$\boxed{f_\lambda(t)=\frac\lambda 3e^{2t} +(1-\tfrac\lambda 3)e^{-t}.}$$ Forward transformation recombines to $(A+B)s+(A-2B)=s-2+\lambda$, as required, for sufficiently large $s$.

Step 2: Classify boundedness and domains. If $\lambda=0$, the function is $e^{-t}$, bounded and decaying, with exact transform domain $s>-1$. If $\lambda\ne 0$, then $e^{-2t}f_\lambda(t)\to\lambda/3\ne 0$. Thus it is unbounded and its weighted tail at $s\le 2$ has an eventual fixed sign and nonintegrable magnitude. For $s>2$, all remaining terms converge absolutely. Consequently $$\boxed{\text{bounded exactly when }\lambda=0;}\qquad \boxed{\begin{array}{ll}s>-1,&\lambda=0,\\s>2,&\lambda\ne 0.\end{array}}$$ This includes $\lambda=3$, when the decaying term disappears.

Step 3: Interpret the canceled point. At $\lambda=0$, cancellation gives $F_0=1/(s+1)$. The apparent pole at two was never a pole of the reduced transform. The actual integral at $s=2$ is $$\boxed{\int_0^\infty e^{-2t}e^{-t}\,dt=1/3.}$$ The continuous algebraic extension agrees with it. The inverse is determined by the whole expression, not by a list of uncanceled denominator factors.

Step 4: Measure the delayed dominance. For $\lambda=.03$, the inverse is $.01e^{2t}+.99e^{-t}$. Equality of the contributions requires $e^{3t}=99$, so $$\boxed{t_c=\tfrac 13\ln 99\approx 1.5317.}$$ The ratio of growing to decaying contributions is $e^{3t}/99$, strictly increasing through one. The initially small growing coefficient eventually dominates. At fixed finite time the inverse approaches $e^{-t}$ as $\lambda\to 0$, but for any nonzero $\lambda$ its eventual boundedness is different. That distinction does not justify discarding a small nonzero residue.