Question

determine whether the given boundary value problem is self-adjoint. $$ y^{\prime \prime}+\lambda y=0, \quad y(0)=0, \quad y(\pi)+y^{\prime}(\pi)=0 $$

Short answer

In summary, the given boundary value problem is self-adjoint since its ODE is already in self-adjoint form and its boundary conditions are also in self-adjoint form.

Step-by-step solution

Transform the ODE to the self-adjoint form if needed.

The given ODE is already in the self-adjoint form: $$ \frac{d}{dx} \left[ 1 \cdot \frac{dy}{dx} \right] + 0 \cdot \frac{dy}{dx} + \lambda y = 0 \Rightarrow y'' + \lambda y = 0. $$

Check if the given boundary conditions are self-adjoint.

The given boundary conditions for the problem are: $$ y(0) = 0 \quad \text{and} \quad y(\pi) + y'(\pi) = 0. $$ These boundary conditions can be written as $$ 1 \cdot y(0) + 0 \cdot y'(0) = 0 \quad \text{and} \quad 1 \cdot y(\pi) + 1 \cdot y'(\pi) = 0. $$ They are both in the self-adjoint form: $$ \alpha_1 y(a) + \alpha_2 y'(a) = 0 \quad \text{and} \quad \beta_1 y(b) + \beta_2 y'(b) = 0. $$ Since the given ODE is already in self-adjoint form and the boundary conditions are also self-adjoint, the given boundary value problem is self-adjoint.