Laplace Transforms — Question 7

Question 7 – Solution

Strategy. Positive inverse kernels give all-time bounds; elimination then turns the same system into an exact input-recovery formula.

Step 1: Invert the transfer kernels. If $U=\mathcal L\{u\}$, the zero initial values give $F=U/(s+1)$ and $G=U/[(s+1)(s+3)]$. Partial fractions yield $$x(t)=\int_0^t e^{-(t-r)}u(r)\,dr,\qquad \boxed{y(t)=\tfrac 12\int_0^t\bigl(e^{-(t-r)}-e^{-3(t-r)}\bigr)u(r)\,dr.}$$ These identities hold on any common right half-plane of convergence and, after inversion, for all $t\ge 0$.

Step 2: Use positivity to obtain a sharp bound. The kernel $k(v)=(e^{-v}-e^{-3v})/2$ is nonnegative for $v\ge 0$. For $0\le u\le 1$, $$0\le y(t)\le\int_0^t k(v)\,dv =\tfrac 13-\tfrac 12e^{-t}+\tfrac 16e^{-3t}<\tfrac 13\quad(t<\infty).$$ The strict upper inequality follows from the positive remaining tail of $k$; at $t=0$ it also holds. With $u\equiv 1$, $y(t)\to 1/3$. Thus $1/3$ is the least uniform constant upper bound, although it is not attained at finite time from zero.

Step 3: Recover the input and its compatibility conditions. The second equation forces $x=y'+3y$. Substitution into the first gives $$\boxed{u=y''+4y'+3y,\qquad y(0)=0,\quad y'(0)=0.}$$ The conditions are necessary for both initial states to vanish. Conversely, under the stated regularity, these formulas satisfy both equations and the zero initial values whenever the conditions hold. They prove uniqueness: any input producing that $y$ must give exactly this $x$ and $u$.

Step 4: Test the proposed target. For $y=t^2e^{-t}$, the compatibility conditions hold, and direct calculation gives $\boxed{x=(2t+2t^2)e^{-t},\ u=(2+4t)e^{-t}}$. The unique input is nonnegative but already has $u(0)=2>1$; its maximum is $4e^{-1/2}>1$ at $t=1/2$. Independently, the target peaks at $t=2$ with value $4e^{-2}>1/3$, contradicting the kernel bound. It is therefore achievable by a continuous input, but not by an input confined to $[0,1]$.