Question

The smallest positive real number \( \lambda \), for which the following problem \[ y''(x) + \lambda y(x) = 0, y'(0) = 0, y(1) = 0 \] has a non-zero solution, is

• A. \( \pi^2 \)
• B. \( \frac{\pi^2}{2} \)
• C. \( \frac{\pi^2}{4} \)
• D. \( \frac{\pi^2}{8} \)

Hint:

For Sturm-Liouville problems, the eigenvalues correspond to the values of \( \lambda \) that satisfy the boundary conditions. In this case, the smallest eigenvalue corresponds to the smallest value of \( \sqrt{\lambda} \) for which \( \cos(\sqrt{\lambda}) = 0 \).

Answer 1

The Correct Option is C

We are given the differential equation \( y''(x) + \lambda y(x) = 0 \), along with the boundary conditions \( y'(0) = 0 \) and \( y(1) = 0 \). This is a standard Sturm-Liouville problem, which is common in various applications, including vibration analysis and quantum mechanics. The goal is to find the smallest positive value of \( \lambda \) such that the given boundary value problem has a non-trivial (non-zero) solution.

Step 1: Solve the homogeneous equation.
The given differential equation is: \[ y''(x) + \lambda y(x) = 0. \] This is a second-order linear homogeneous differential equation with constant coefficients. The general solution to this equation depends on the value of \( \lambda \). If \( \lambda > 0 \), the solution is: \[ y(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x), \] where \( A \) and \( B \) are constants determined by the boundary conditions.

Step 2: Apply the first boundary condition.
We are given that \( y'(0) = 0 \). Taking the derivative of the general solution: \[ y'(x) = -A \sqrt{\lambda} \sin(\sqrt{\lambda} x) + B \sqrt{\lambda} \cos(\sqrt{\lambda} x). \] Substituting \( x = 0 \) into this equation gives: \[ y'(0) = B \sqrt{\lambda} = 0. \] For a non-zero solution, we must have \( B = 0 \). Therefore, the solution reduces to: \[ y(x) = A \cos(\sqrt{\lambda} x). \]

Step 3: Apply the second boundary condition.
The second boundary condition is \( y(1) = 0 \). Substituting into the solution: \[ y(1) = A \cos(\sqrt{\lambda}) = 0. \] For a non-zero solution, we require \( \cos(\sqrt{\lambda}) = 0 \). This occurs when: \[ \sqrt{\lambda} = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots. \] Thus, the smallest positive value of \( \sqrt{\lambda} \) is \( \frac{\pi}{2} \), which gives: \[ \lambda = \left( \frac{\pi}{2} \right)^2 = \frac{\pi^2}{4}. \]

Final Answer: \[ \boxed{\frac{\pi^2}{4}}. \]