Sturm-Liouville Problems Questions and Answers

Answer: A

The weight function is \(e^{-3x}\).

Bring the equation to Sturm-Liouville form (p y')' + ... = \(\lambda w y\). Multiply by an integrating factor \(\mu(x)\) chosen so that \(\mu(y'' - 3y') = (\mu y')'\), which needs \(\mu' = -3\mu\), so \(\mu = e^{-3x}\):

\[ \begin{aligned} &\frac{d}{dx}\left(e^{-3x}\frac{dy}{dx}\right) \\ &= \lambda e^{-3x} y \end{aligned} \]

With p = \(e^{-3x}\) and the boundary conditions y(0) = y(\(\pi\)) = 0, eigenfunctions for distinct eigenvalues are orthogonal with weight \(w(x) = e^{-3x}\).

\(e^{3x}\) is the reciprocal, from taking the factor with the wrong sign, and \(e^{\pm 2x}\) does not turn \(y'' - 3y'\) into an exact derivative.

Answer: A

The eigenvalues are \(\lambda_{n} = \left[\dfrac{(2n - 1)\pi}{2}\right]^{2}\).

For \(\lambda \le 0\) the boundary conditions force \(X = 0\). For \(\lambda = k^{2} \gt 0\), \(X = A\cos kt + B\sin kt\). \(X(0) = 0\) gives \(A = 0\), and \(X'(1) = Bk\cos k = 0\) gives \(\cos k = 0\), so \(k = (2n - 1)\pi/2\), \(n = 1, 2, \ldots\), and

\[ \lambda_{n} = \left(\frac{(2n - 1)\pi}{2}\right)^{2} \]

\((n\pi/2)^{2}\) would include the even multiples, where \(\cos k = \pm 1\), not 0. \((2n\pi)^{2}\) belongs to periodic conditions, and \((n + 1)\pi\) is not even a square.