Laplace Transforms — Question 7

Question 7 – Solution

Strategy. Convert memory into a rational factor, then guard against the extra solutions that differentiation can introduce.

Step 1: Transform and factor. The convolution theorem gives $$(s+1)^3Y=\frac{Y}{s+1}+\frac 1{s+1},\qquad \boxed{Y=\frac 1{(s+1)^4-1}.}$$ The denominator is $s(s+2)((s+1)^2+1)$, with roots $0,-2,-1\pm i$. A sufficient convergence half-plane is $\operatorname{Re}s>0$.

Step 2: Invert real quadratic factors. For $z=s+1$, $$\frac 1{z^4-1}=\frac 12\left(\frac 1{z^2-1}-\frac 1{z^2+1}\right), \qquad \boxed{y(t)=\tfrac 12 e^{-t}(\sinh t-\sin t).}$$ Equivalently, $y=(1-e^{-2t})/4-e^{-t}\sin t/2$.

Step 3: Prove equivalence, including the lost condition. If $m(t)=\int_0^t e^{-(t-v)}y(v)\,dv$, then $(D+1)m=y$ and $m(0)=0$. Applying $D+1$ to the original equation gives $$((D+1)^4-1)y=0,\qquad (y,y',y'',y''')(0)=(0,0,0,1).$$ The fourth value follows from the original equation at zero. The displayed solution has $y=t^3/6+O(t^4)$ and lies in the four characteristic modes, so it satisfies this fourth-order IVP. Conversely its residual $R=(D+1)^3y-m-e^{-t}$ obeys $R'+R=0$ and $R(0)=0$. Hence $R\equiv 0$, proving equivalence rather than just necessity.

Step 4: Interpret the surviving mode. The explicit formula gives $\boxed{\lim_{t\to\infty}y(t)=1/4}$. This also equals $\lim_{s\to 0}sY$; after cancellation, the poles of $sY$ are $-2,-1\pm i$, all strictly stable. Without memory, the transform is $1/(s+1)^4$ and the response is $t^3e^{-t}/6\to 0$. Feedback introduces a zero characteristic root and thus a nonzero constant component, although the external forcing itself decays.