Question

Let \(\phi\) and \(\psi\) be two linearly independent solutions of the ordinary differential equation \[ y'' + (2 - \cos x) y = 0, \quad x \in \mathbb{R}. \] Let \(\alpha, \beta \in \mathbb{R}\) be such that \(\alpha<\beta\), \(\phi(\alpha) = \phi(\beta) = 0\) and \(\phi(x) \neq 0\) for all \(x \in (\alpha, \beta)\).
Consider the following statements:
P: \(\phi'(\alpha)\phi'(\beta)>0\).
Q: \(\psi(x) \neq 0\) for all \(x \in (\alpha, \beta)\).
Then

• A. P is TRUE and Q is FALSE
• B. P is FALSE and Q is TRUE
• C. both P and Q are FALSE
• D. both P and Q are TRUE

Hint:

For any equation \(y''+q(x)y=0\), remember two key results from Sturm-Liouville theory: 1. The slopes at consecutive zeros have opposite signs. 2. Sturm Separation Theorem: Zeros of linearly independent solutions interlace. These two principles answer many qualitative questions about solutions to second-order ODEs.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question deals with Sturm's separation and comparison theorems for second-order linear homogeneous ODEs. The equation is of the form \(y'' + q(x)y = 0\). Key properties relate the zeros of linearly independent solutions and the behavior of solutions at their zeros.
Step 3: Detailed Explanation:
Analysis of Statement P:
We are given that \(\phi(\alpha) = \phi(\beta) = 0\) and \(\phi(x) \neq 0\) for \(x \in (\alpha, \beta)\). This means \(\alpha\) and \(\beta\) are consecutive zeros of the solution \(\phi(x)\).
Since \(\phi(x)\) is non-zero between \(\alpha\) and \(\beta\), it must be either strictly positive or strictly negative on \((\alpha, \beta)\).
Case 1: \(\phi(x)>0\) for \(x \in (\alpha, \beta)\).
Since \(\phi(x)\) increases from 0 at \(x=\alpha\), its derivative at \(\alpha\) must be positive, so \(\phi'(\alpha)>0\).
Since \(\phi(x)\) decreases to 0 at \(x=\beta\), its derivative at \(\beta\) must be negative, so \(\phi'(\beta)<0\).
In this case, the product \(\phi'(\alpha)\phi'(\beta)<0\).
Case 2: \(\phi(x)<0\) for \(x \in (\alpha, \beta)\).
Since \(\phi(x)\) decreases from 0 at \(x=\alpha\), its derivative at \(\alpha\) must be negative, so \(\phi'(\alpha)<0\).
Since \(\phi(x)\) increases to 0 at \(x=\beta\), its derivative at \(\beta\) must be positive, so \(\phi'(\beta)>0\).
In this case, the product \(\phi'(\alpha)\phi'(\beta)<0\).
In both cases, the product of the derivatives at consecutive zeros is negative. Thus, the statement P: \(\phi'(\alpha)\phi'(\beta)>0\) is FALSE.
Analysis of Statement Q:
This statement relates to the Sturm Separation Theorem.
Sturm Separation Theorem: Let \(y_1(x)\) and \(y_2(x)\) be two linearly independent solutions of the ODE \(y'' + q(x)y = 0\), where \(q(x)\) is continuous. Then between any two consecutive zeros of \(y_1(x)\), there is exactly one zero of \(y_2(x)\).
Here, \(\phi\) and \(\psi\) are linearly independent solutions. \(\alpha\) and \(\beta\) are consecutive zeros of \(\phi\). According to the Sturm Separation Theorem, there must be exactly one zero of \(\psi\) in the open interval \((\alpha, \beta)\). The statement Q says \(\psi(x) \neq 0\) for all \(x \in (\alpha, \beta)\), which means \(\psi\) has no zeros in the interval. This directly contradicts the theorem.
Therefore, statement Q is FALSE.
Step 4: Final Answer:
Both P and Q are FALSE.
Step 5: Why This is Correct:
Statement P is false due to the behavior of a continuous, differentiable function between two of its roots; the slope must have opposite signs at these roots. Statement Q is false as it violates the Sturm Separation Theorem, which guarantees that the zeros of two linearly independent solutions must interlace.