Laplace Transforms: Question 5-11.8 & Solution

Question 8

For $\varepsilon\ge 0$ , let $x'=-\varepsilon x-y+1,\qquad y'=x-\varepsilon y, \qquad x(0)=y(0)=0,\quad t\ge 0.$ A student evaluates $\lim_{s\to 0^+}s\mathcal L\{x\}$ and $\lim_{s\to 0^+}s\mathcal L\{y\}$ and declares these the limiting state for every $\varepsilon$ . For rational transforms, the final-value theorem requires all poles of the simplified $s\mathcal L\{f\}$ to have negative real part.

Tasks

Find both transformed components for general $\varepsilon$ and calculate the student’s two proposed limits.
Solve the case $\varepsilon=0$ . Check the theorem’s hypothesis and determine whether either component has a limit as $t\to\infty$ .
For $\varepsilon>0$ , invert the transforms and establish the true limiting state. Explain why the theorem now applies.
Show that $y_\varepsilon\to y_0$ at each fixed time as $\varepsilon\to 0^+$ but not uniformly on $[0,\infty)$ . Use times $2\pi n$ to obtain a lower bound on the uniform error. Sketch $y_0$ and $y_{1/2}$ .

Original worksheet page 1: question and worked solution for 5-11-008

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Question 8 – Solution

Strategy. Check pole locations before applying the final-value theorem, and distinguish fixed-time convergence from an infinite-time assertion.

Step 1: Solve the transformed system. With $D=(s+\varepsilon)^2+1$ , the equations give $F=\frac{s+\varepsilon}{sD},\qquad G=\frac 1{sD}.$ A sufficient convergence half-plane is $\operatorname{Re}s>0$ . The proposed final values are $\varepsilon/(1+\varepsilon^2)$ and $1/(1+\varepsilon^2)$ .

Step 2: Examine the undamped case. At $\varepsilon=0$ , inversion yields $\boxed{x=\sin t,\ y=1-\cos t}$ . Neither has a limit: the values along successive maxima and minima differ. Both $sF$ and $sG$ have poles at $\pm i$ , so the stated theorem’s hypothesis fails. The finite algebraic values $0$ and $1$ do not prove convergence in time.

Step 3: Resolve the damped response. For $\varepsilon>0$ , decompose, for example, $G=\frac 1{1+\varepsilon^2}\left(\frac 1s-\frac{s+\varepsilon}{D}-\frac{\varepsilon}{D}\right).$ Inverting this and the analogous decomposition of $F$ gives $x_\varepsilon=\frac{\varepsilon-e^{-\varepsilon t}(\varepsilon\cos t-\sin t)}{1+\varepsilon^2}, \qquad y_\varepsilon=\frac{1-e^{-\varepsilon t}(\cos t+\varepsilon\sin t)}{1+\varepsilon^2}.$ Both start at zero. The exponentially decaying terms leave the limit $\boxed{(\varepsilon,1)^T/(1+\varepsilon^2)}$ . Now the poles of $sF$ and $sG$ are $-\varepsilon\pm i$ , strictly in the left half-plane, as required. The limit also solves the equilibrium equations.

Step 4: Test uniformity on the whole half-line. At fixed $t$ , the formula gives $y_\varepsilon(t)\to 1-\cos t=y_0(t)$ . But $y_0(2\pi n)=0$ , whereas $y_\varepsilon(2\pi n)=(1-e^{-2\pi n\varepsilon})/(1+\varepsilon^2)$ . Consequently $\boxed{\sup_{t\ge 0}|y_\varepsilon(t)-y_0(t)|\ge\frac 1{1+\varepsilon^2}.}$ This lower bound tends to $1$ , not $0$ . Fixed-time convergence therefore cannot be used to pass a limiting-state conclusion to the undamped system. The dotted guide is the damped limit $4/5$ for $\varepsilon=1/2$ .

See the diagram in the original worksheet below.

Original worksheet page 2: question and worked solution for 5-11-008

Laplace Transforms — Question 8

Question 8

Question 8 – Solution