Laplace Transforms — Question 2

Question 2 – Solution

Strategy. Invert the triangular matrix before substituting data; its powers of $(s+2)^{-1}$ display how a chain transmits the initial state.

Step 1: Compute the resolvent. Put $q=s+2$. For $q\ne 0$, $$\boxed{(sI-A)^{-1}= \begin{pmatrix}q^{-1}&q^{-2}&q^{-3}\\0&q^{-1}&q^{-2}\\0&0&q^{-1}\end{pmatrix}.}$$ Multiplication by the upper triangular matrix with diagonal $q$ and superdiagonal $-1$ gives diagonal entries $1$; each off-diagonal entry cancels. The inverse exists exactly for $s\ne-2$. This algebraic domain is larger than the convergence half-plane $\operatorname{Re}s>-2$ for the nonzero responses.

Step 2: Multiply by the initial vector. The transforms are $a/q+b/q^2+c/q^3$, $b/q+c/q^2$, and $c/q$. Since $\mathcal L\{t^ne^{-2t}\}=n!/(s+2)^{n+1}$, $$\boxed{x=e^{-2t}(a+bt+ct^2/2),\quad y=e^{-2t}(b+ct),\quad z=ce^{-2t}.}$$ The factorial is essential: the inverse of $q^{-3}$ is $t^2e^{-2t}/2$.

Step 3: Check transmission along the chain. For the specified initial state, $$Z=(t^2e^{-2t}/2,te^{-2t},e^{-2t})^T.$$ Directly, $z'=-2z$, $y'=-2y+z$ and $x'=-2x+y$, with the required initial state. After multiplication by $e^{2t}$, the equations become $(e^{2t}z)'=0$, $(e^{2t}y)'=e^{2t}z$ and $(e^{2t}x)'=e^{2t}y$. Two successive integrations produce the quadratic factor.

Step 4: Classify actual pole orders. The numerator of the first transform over $q^3$ is $aq^2+bq+c$. Its order at $q=0$ gives the exhaustive classification: $$\begin{array}{c|c} \text{initial-data condition}&\text{pole order of }\mathcal L\{x\}\\ \hline c\ne 0&3\\ c=0,\ b\ne 0&2\\ c=b=0,\ a\ne 0&1\\ a=b=c=0&\text{no pole} \end{array}$$ Thus the characteristic multiplicity supplies a possible order, not an order every response must attain. Even for $c\ne 0$, the third component has only a simple pole, whereas the first has a triple pole.