Eigenvalues and Eigenfunctions — Question 5

Question 5 – Solution

Strategy. Treat value and derivative matching together as a two-dimensional endpoint map.

Step 1: Handle nonpositive parameters. The boundary term $[yy']_0^{2\pi}$ is zero because both values and slopes match. Thus $\lambda\int y^2=\int(y')^2\geq 0$. At zero, $y=A+Bx$; value matching forces $B=0$ and slope matching is then automatic. Hence $\boxed{E_0=\operatorname{span}\{1\}}$, of dimension one.

Step 2: Find the positive eigenspaces. For $\lambda=k^2>0$, write $y=A\cos(kx)+B\sin(kx)$. With $\theta=2\pi k$, the two conditions are $$\begin{pmatrix}\cos\theta-1&\sin\theta\\-\sin\theta&\cos\theta-1\end{pmatrix} \binom AB=\binom 00.$$ The determinant is $2(1-\cos\theta)$, so nonzero solutions require $k=n$, a positive integer. Then the entire matrix is zero and both constants are free: $$\boxed{\lambda_n=n^2,\quad E_n=\operatorname{span}\{\cos nx,\sin nx\}, \quad n\geq 1.}$$ Each positive eigenspace has dimension two. One equation alone would generally leave a nonzero vector even when the other endpoint equation fails.

Step 3: Apply the initial normalization. The data $y(0)=0,y'(0)=1$ require $A=0,nB=1$, giving $\boxed{y_n=\sin(nx)/n}$. At zero every eigenfunction is constant, so its initial derivative cannot be one. These extra data select a representative from each positive eigenspace, not from the zero eigenspace.

Step 4: Interpret the midpoint measurements. For every positive integer $n$, $y_n(\pi)=\sin(n\pi)/n=0$, so that measurement adds no restriction. But $y_n'(\pi)=\cos(n\pi)=(-1)^n$ equals $y_n'(0)=1$ exactly for even $n$. The second measurement selects eigenvalues $\boxed{4,16,36,\ldots}$ from the original positive spectrum.