Fundamental Sets of Solutions — Question 5

Question 5 – Solution

Strategy. Distinguish a fundamental set of functions from the invertibility of a particular measurement map.

Step 1: Solve at ordinary lengths. Every solution is $A\cos t+B\sin t$. The position measurements give $$A=a,\qquad a\cos L+B\sin L=b.$$ If $\sin L\ne 0$, the unique solution is $$\boxed{y=a\cos t+\frac{b-a\cos L}{\sin L}\sin t.}$$ Thus uniqueness holds exactly at lengths that are not positive integer multiples of $\pi$.

Step 2: Classify exceptional data. If $L=n\pi$, $n=1,2,\ldots$, the second equation reduces to $b=(-1)^n a$. If this fails, there is no solution. If it holds, $B$ is arbitrary and all solutions are $a\cos t+B\sin t$; there are infinitely many.

Step 3: Apply the classification. At $L=\pi$, the data $(1,-1)$ give $$\boxed{y=\cos t+B\sin t,\qquad B\in\mathbb R.}$$ The data $(1,1)$ fail the necessary condition $b=-a$ and give no solution. The figure shows three members of the compatible family.

Step 4: Explain what failed. The data vectors of $\cos t,\sin t$ using value and slope at zero are $(1,0)$ and $(0,1)$, so the pair remains fundamental. At $L=n\pi$, both endpoint positions are insensitive to the coefficient $B$. Measuring $y'(0)=c$ fixes $B=c$ uniquely; the endpoint data must still be compatible with that solution.