Boundary Value Problem Analysis

We are given a two-point boundary value problem:

\[ \left\{ \begin{aligned} y^{\prime \prime}(x)+\lambda y(x) &=0 & \text { for } x \in(0, \pi) \\ y(0) &=0 \\ y(\pi) &=0 \end{aligned} \right. \]

We need to find the values of the parameter \(\lambda\) for which this problem has a non-trivial solution (a solution other than \(y(x) = 0\) for all \(x\)). This type of problem is an eigenvalue problem, where the values of \(\lambda\) for which non-trivial solutions exist are called eigenvalues.

The differential equation is \(y^{\prime \prime}(x)+\lambda y(x) = 0\). The characteristic equation is \(r^2 + \lambda = 0\), which gives \(r^2 = -\lambda\).

Lambda Cases for Non-Trivial Solutions

We consider three cases based on the value of \(\lambda\):

Case 1: \(\lambda < 0\)

Let \(\lambda = -\mu^2\) where \(\mu > 0\). The characteristic equation becomes \(r^2 - \mu^2 = 0\), so \(r = \pm \mu\). The general solution is \(y(x) = c_1 e^{\mu x} + c_2 e^{-\mu x}\).

Applying the boundary conditions:

• \(y(0) = 0 \Rightarrow c_1 e^0 + c_2 e^0 = 0 \Rightarrow c_1 + c_2 = 0 \Rightarrow c_2 = -c_1\).
• So, \(y(x) = c_1 e^{\mu x} - c_1 e^{-\mu x} = c_1 (e^{\mu x} - e^{-\mu x}) = 2 c_1 \sinh(\mu x)\).
• \(y(\pi) = 0 \Rightarrow 2 c_1 \sinh(\mu \pi) = 0\). Since \(\mu > 0\) and \(\pi > 0\), \(\mu \pi > 0\), and \(\sinh(z) = 0\) only when \(z = 0\). Thus, \(\sinh(\mu \pi) \neq 0\).
• Therefore, we must have \(c_1 = 0\). Since \(c_2 = -c_1\), \(c_2 = 0\) as well.

This gives only the trivial solution \(y(x) = 0\). So, there are no non-trivial solutions when \(\lambda < 0\).

Case 2: \(\lambda = 0\)

The differential equation becomes \(y^{\prime \prime}(x) = 0\). The general solution is \(y(x) = Ax + B\).

Applying the boundary conditions:

• \(y(0) = 0 \Rightarrow A(0) + B = 0 \Rightarrow B = 0\).
• So, \(y(x) = Ax\).
• \(y(\pi) = 0 \Rightarrow A\pi = 0 \Rightarrow A = 0\).

This gives only the trivial solution \(y(x) = 0\). So, there are no non-trivial solutions when \(\lambda = 0\).

Case 3: \(\lambda > 0\)

Let \(\lambda = \omega^2\) where \(\omega > 0\). The characteristic equation becomes \(r^2 + \omega^2 = 0\), so \(r = \pm i\omega\). The general solution is \(y(x) = c_1 \cos(\omega x) + c_2 \sin(\omega x)\).

Applying the boundary conditions:

• \(y(0) = 0 \Rightarrow c_1 \cos(0) + c_2 \sin(0) = 0 \Rightarrow c_1(1) + c_2(0) = 0 \Rightarrow c_1 = 0\).
• So, \(y(x) = c_2 \sin(\omega x)\).
• \(y(\pi) = 0 \Rightarrow c_2 \sin(\omega \pi) = 0\).

For a non-trivial solution, we require \(y(x)\) not to be identically zero, which means \(c_2 \neq 0\). If \(c_2 \neq 0\), then we must have \(\sin(\omega \pi) = 0\).

The sine function is zero at integer multiples of \(\pi\). So, \(\omega \pi = n \pi\) for some integer \(n\). Since \(\omega > 0\), we require \(n\) to be a positive integer, i.e., \(n = 1, 2, 3, \ldots\). Thus, \(\omega = n\) for \(n \in \mathbb{N}\).

Substituting back \(\lambda = \omega^2\), we find that non-trivial solutions exist for \(\lambda = n^2\) where \(n \in \mathbb{N}\) (positive integers). The corresponding non-trivial solutions are \(y_n(x) = c_n \sin(nx)\) for some non-zero constant \(c_n\).

Eigenvalues and Options

The values of \(\lambda\) for which non-trivial solutions exist are the eigenvalues, which are \(\lambda = 1^2, 2^2, 3^2, \ldots\), or \(\lambda = 1, 4, 9, \ldots\). These are \(\lambda = n^2\) for \(n \in \mathbb{N}\).

Let's look at the given options:

• Option 1: no values of \(\lambda \in \mathbb{R}\). This is incorrect, as we found non-trivial solutions for \(\lambda = n^2\) with \(n \in \mathbb{N}\).
• Option 2: \(\lambda = 1\). This corresponds to the eigenvalue when \(n=1\). A non-trivial solution exists for \(\lambda = 1\). This option is correct.
• Option 3: \(\lambda = n\)² for all \(n \in \mathbb{N}\) , \(n > 1\). This corresponds to the eigenvalues when \(n=2, 3, 4, \ldots\). Non-trivial solutions exist for these values of \(\lambda\). This option is correct.
• Option 4: \(\lambda \le 0\). As shown in Case 1 and Case 2, for \(\lambda \le 0\), only the trivial solution exists. This option is incorrect.

The boundary value problem has a non-trivial solution for \(\lambda = n^2\), where \(n\) is a positive integer (\(n \in \mathbb{N}\)). This includes \(\lambda = 1\) (when \(n=1\)) and \(\lambda = n^2\) for \(n > 1\).