Complex Roots — Question 8

Question 8 – Solution

Strategy. An exponential prefactor never vanishes, so solvability and sensitivity are governed by the sine factor in the endpoint data map.

Step 1: Reduce to one coefficient. The roots are $-1\pm 2i$. The first endpoint fixes the cosine coefficient, giving $y=e^{-t}(A\cos 2t+C\sin 2t)$. The second requires $$C\sin 2L=Be^L-A\cos 2L.$$ If $\sin 2L\ne 0$, there is exactly one $C$, so unique solvability for arbitrary data holds exactly when $$\boxed{L\ne n\pi/2\quad(n=1,2,3,\ldots)}.$$

Step 2: Classify the exceptional lengths. At $L=n\pi/2$, the condition becomes $0=Be^L-(-1)^n A$. Thus $$\boxed{B=(-1)^ne^{-L}A}$$ is necessary and sufficient for existence. If it holds, every real $C$ works, giving infinitely many solutions. If it fails, there is no solution. The initial-value uniqueness theorem does not promise endpoint uniqueness at these lengths.

Step 3: Apply the two specified lengths. For $A=0$, $B=1$, $L=\pi/4$, we have $C=e^{\pi/4}$ and $$\boxed{y=e^{\pi/4-t}\sin 2t}.$$ At $L=\pi/2$, compatibility would require $B=0$, contradicting $B=1$, so no solution exists.

Step 4: Quantify sensitivity near an exceptional length. For $L=\pi/2+\varepsilon$, $\sin 2L=-\sin 2\varepsilon\ne 0$. Hence $$y(t)=-\frac{e^{L-t}\sin 2t}{\sin 2\varepsilon},\qquad \boxed{y(\pi/4)=-\frac{e^{\pi/4+\varepsilon}}{\sin 2\varepsilon}}.$$ Using $\sin 2\varepsilon\sim 2\varepsilon$ and $e^\varepsilon\to 1$ gives $$\boxed{y(\pi/4)\sim-\frac{e^{\pi/4}}{2\varepsilon}\longrightarrow-\infty}.$$ For each positive $\varepsilon$ there is one finite coefficient, but the denominator approaches zero as the length approaches the incompatible endpoint problem. Uniqueness at each parameter value is not a uniform bound on the dependence on that parameter.