Solving IVP’s with Laplace Transforms — Question 6

Question 6 – Solution

Strategy. Solve once with symbolic parameters, then evaluate the resulting basis functions at the measurement times.

Step 1: Transform with unknown initial velocity. The transformed equation is $$[(s+1)^2+4]Y=v+\frac A{s+1}.$$ Writing $p=s+1$ and using $1/[p(p^2+4)]=(1/4)[1/p-p/(p^2+4)]$ gives $$\boxed{y(t)=e^{-t}[\tfrac v2\sin 2t+\tfrac A4(1-\cos 2t)].}$$ Transforming the sine and constant-minus-cosine terms verifies the algebraic equation for sufficiently large $s$.

Step 2: Recover the parameters. At $t=\pi/2$, the local sine is zero and the cosine is $-1$, so $A/2=2$, giving $A=4$. At $t=\pi/4$, the sine is one and the cosine is zero, so $v/2+A/4=2$, giving $v=2$. Thus $$\boxed{A=4,\quad v=2,\qquad y=e^{-t}(1-\cos 2t+\sin 2t).}$$ The first measurement fixes $A$ independently, then the second fixes $v$ with a nonzero coefficient. This proves uniqueness for the specified two-parameter model.

Step 3: Verify calibration and the IVP. Set $w=1-\cos 2t+\sin 2t$, so $y=e^{-t}w$. The product rule gives $$y''+2y'+5y=e^{-t}(w''+4w)=4e^{-t}.$$ Also $w(0)=0$ and $w'(0)=2$, so $y(0)=0$ and $y'(0)=2$. The exact transform domain is $s>-1$. At its boundary the weighted signal is the nonzero periodic function $w$, with nonzero mean; below it fixed-sign lobe intervals fail the Cauchy criterion. Above it the exponential envelope gives absolute convergence.

Step 4: Identify blind sampling times. At every $t=n\pi$, both $\sin 2t=0$ and $1-\cos 2t=0$. Therefore $$\boxed{y(n\pi)=0\quad\text{for every }A,v.}$$ Even infinitely many such samples carry no information about either parameter. The issue is where the response basis vanishes, not merely how many measurements are collected. The two given times avoid this degeneracy.