Sturm-Liouville Problems Questions and Answers

Q: If eigenfunctions corresponding to distinct eigenvalues \(\lambda\) of the Sturm-Liouville problem\[ \begin{aligned} &\frac{d^{2}y}{dx^{2}} - 3\frac{dy}{dx} = \lambda y, \\ &0 \lt x \lt \pi, \end{aligned} \]\[ y(0) = y(\pi) = 0 \]are orthogonal with respect to the weight function w(x), then w(x) is

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.

Q: Eigenvalues corresponding to the following Sturm-Liouville problem :\[ \frac{d^{2}X}{dt^{2}} + \lambda X = 0,\ X(0) = 0,\ X'(1) = 0 \]are \(\lambda_{n}\), \(n \in \mathbf{N}\), where \(\lambda_{n} =\)

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.

Q: Let \(y_{1}(x)\) and \(y_{2}(x)\) be two solutions of :\[ (r(x)y')' + p(x)y = 0 \]such that \(y_{1}(x)\) and \(y_{2}(x)\) have a common zero in \([a, b]\). Then :

The Wronskian vanishes: \(W(y_{1}, y_{2}) = 0\).For the self-adjoint equation \((r y')' + p y = 0\), Abel's formula gives \(r(x)\,W(y_{1}, y_{2})(x) = \text{constant}\). So, with \(r\) nonzero on [a, b], the Wronskian is either identically zero or never zero.At the common zero \(x_{0}\), \(W = y_{1}(x_{0}) y_{2}'(x_{0}) - y_{1}'(x_{0}) y_{2}(x_{0}) = 0 - 0 = 0\). Hence \(W \equiv 0\) on [a, b].That makes \(W \neq 0\) false, and it also means \(y_{1}\) and \(y_{2}\) are linearly dependent, so 'linearly independent' is false. (Two independent solutions of such an equation can never share a zero; this is the starting point of the Sturm separation theorem.)

If eigenfunctions corresponding to distinct eigenvalues \(\lambda\) of the Sturm-Liouville problem

\[ \begin{aligned} &\frac{d^{2}y}{dx^{2}} - 3\frac{dy}{dx} = \lambda y, \\ &0 \lt x \lt \pi, \end{aligned} \]

\[ y(0) = y(\pi) = 0 \]

are orthogonal with respect to the weight function w(x), then w(x) is

A.\(e^{-3x}\)
B.\(e^{-2x}\)
C.\(e^{2x}\)
D.\(e^{3x}\)

Exam Year(s)

GATE Mathematics - 2022,

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.

Discuss in Forum

Eigenvalues corresponding to the following Sturm-Liouville problem :

\[ \frac{d^{2}X}{dt^{2}} + \lambda X = 0,\ X(0) = 0,\ X'(1) = 0 \]

are \(\lambda_{n}\), \(n \in \mathbf{N}\), where \(\lambda_{n} =\)

A.\(\left[\dfrac{(2n - 1)\pi}{2}\right]^{2}\)
B.\((2n\pi)^{2}\)
C.\(\left(\dfrac{n\pi}{2}\right)^{2}\)
D.\((n + 1)\pi\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

Discuss in Forum

Let \(y_{1}(x)\) and \(y_{2}(x)\) be two solutions of :

\[ (r(x)y')' + p(x)y = 0 \]

such that \(y_{1}(x)\) and \(y_{2}(x)\) have a common zero in \([a, b]\). Then :

A.\(y_{1}, y_{2}\) are linearly independent on \([a, b]\)
B.\(W(y_{1}, y_{2}) = 0\)
C.\(W(y_{1}, y_{2}) \neq 0\)
D.None of the above

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The Wronskian vanishes: \(W(y_{1}, y_{2}) = 0\).

For the self-adjoint equation \((r y')' + p y = 0\), Abel's formula gives \(r(x)\,W(y_{1}, y_{2})(x) = \text{constant}\). So, with \(r\) nonzero on [a, b], the Wronskian is either identically zero or never zero.

At the common zero \(x_{0}\), \(W = y_{1}(x_{0}) y_{2}'(x_{0}) - y_{1}'(x_{0}) y_{2}(x_{0}) = 0 - 0 = 0\). Hence \(W \equiv 0\) on [a, b].

That makes \(W \neq 0\) false, and it also means \(y_{1}\) and \(y_{2}\) are linearly dependent, so 'linearly independent' is false. (Two independent solutions of such an equation can never share a zero; this is the starting point of the Sturm separation theorem.)

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