Question

Consider the problem $$ y^{\prime \prime}+\lambda y=0, \quad \alpha y(0)+y^{\prime}(0)=0, \quad y(1)=0 $$ $$ \begin{array}{l}{\text { where } \alpha \text { is a given constant. }} \\\ {\text { (a) Show that for all values of } \alpha \text { there is an infinite sequence of positive eigenvalues. }} \\ {\text { (b) If } \alpha<1, \text { show that all (real) eigenvalues are positive. Show the smallest eigenvalue }} \\\ {\text { approaches zero as } \alpha \text { approaches } 1 \text { from below. }} \\ {\text { (c) Show that } \lambda=0 \text { is an eigenvalue only if } \alpha=1} \\ {\text { (d) If } \alpha>1 \text { , show that there is exactly one negative eigenvalue and that this eigenvalue }} \\ {\text { decreases as } \alpha \text { increases. }}\end{array} $$

Short answer

Question: Determine the properties of the eigenvalues λ of the given boundary-value problem for each value of α. Based on the solution we derived: $$ \cos\sqrt{\lambda} - \frac{\alpha}{\sqrt{\lambda}} \sin\sqrt{\lambda} = 0 $$ (a) α = 1 For α = 1, the condition becomes: $$ \cos\sqrt{\lambda} - \frac{1}{\sqrt{\lambda}} \sin\sqrt{\lambda} = 0 $$ In this case, the eigenvalues λ are a set of positive real numbers, as the equation can be non-trivially satisfied for different values of λ. (b) α = 0 For α = 0, the condition becomes: $$ \cos\sqrt{\lambda} = 0 $$ In this case, the eigenvalues λ are such that the cosine term equals zero, which occurs when λ = (2n + 1)^2π^2/4 for integer values of n. The eigenvalues are discrete and positive. (c) α = -1 For α = -1, the condition becomes: $$ \cos\sqrt{\lambda} + \frac{1}{\sqrt{\lambda}} \sin\sqrt{\lambda} = 0 $$ In this case, the eigenvalues λ will again be a set of positive real numbers, as the equation can be non-trivially satisfied for different values of λ. (d) α is an arbitrary constant For an arbitrary α, the equation is: $$ \cos\sqrt{\lambda} - \frac{\alpha}{\sqrt{\lambda}} \sin\sqrt{\lambda} = 0 $$ In this general case, the properties of the eigenvalues λ will depend on the specific value of α. For certain values of α, the eigenvalues may be positive real numbers or discrete and positive, similar to cases (a), (b), and (c).

Step-by-step solution

I. Solve the given ODE

The given ODE is $$ y^{\prime \prime}+\lambda y=0 $$ This is the simple form of a second-order homogenous ODE. The general solution for this ODE can be written as follows: $$ y(x) = A \cos\sqrt{\lambda} x + B \sin\sqrt{\lambda} x $$ Now let's apply the boundary conditions.

II. Apply the first boundary condition

The first boundary condition is $$ \alpha y(0)+y^{\prime}(0)=0 $$ Let's compute y(0) and y'(0) and substitute the values into the boundary condition: $$ y(0) = A \cos(0) + B \sin(0) = A $$ $$ y^{\prime}(x) = -A\sqrt{\lambda} \sin\sqrt{\lambda} x + B\sqrt{\lambda} \cos\sqrt{\lambda} x $$ $$ y^{\prime}(0) = -A\sqrt{\lambda} \sin(0) + B\sqrt{\lambda} \cos(0) = B\sqrt{\lambda} $$ Substituting these values gives: $$ \alpha A + B\sqrt{\lambda} = 0 $$ Now let's apply the second boundary condition.

III. Apply the second boundary condition

The second boundary condition is $$ y(1)=0 $$ Let's compute y(1) and substitute the value: $$ y(1) = A \cos\sqrt{\lambda} + B \sin\sqrt{\lambda} = 0 $$ Now we have two simultaneous equations for A and B: (1) $$\alpha A + B\sqrt{\lambda} = 0$$ (2) $$A \cos\sqrt{\lambda} + B \sin\sqrt{\lambda} = 0$$

IV. Determine the eigenvalues

To find the eigenvalues λ satisfying the two equations we can combine the equations and consider how they depend on α: From Equation (1): $$ B = -\frac{\alpha A}{\sqrt{\lambda}} $$ Plugging B into Equation (2) gives: $$ A \cos\sqrt{\lambda} - \frac{\alpha A}{\sqrt{\lambda}} \sin\sqrt{\lambda} = 0 $$ Now we can divide both sides by A: $$ \cos\sqrt{\lambda} - \frac{\alpha}{\sqrt{\lambda}} \sin\sqrt{\lambda} = 0 $$ This is the necessary and sufficient condition for eigenvalues λ after applying the boundary conditions for a given α. Finally, we can analyze the properties of the eigenvalues based on the values of α as given in parts (a) to (d) of the problem.

Key concepts

Understanding Boundary Value Problems

Boundary value problems (BVPs) are a class of differential equations that involve finding a solution which satisfies certain fixed conditions, known as boundary conditions, at the endpoints of a given interval.

In the context of the given exercise, we are dealing with a second-order homogeneous ordinary differential equation (ODE) that is defined on the interval from 0 to 1. The solution to the ODE is required to meet two boundary conditions: one relates to the value of the function and its derivative at the left endpoint, and the other specifies the value of the function at the right endpoint.

The primary challenge in solving BVPs lies not only in solving the ODE itself but also in ensuring that the solution satisfies the boundary conditions. This is achieved by finding a set of eigenvalues and corresponding eigenfunctions that fulfill the boundary conditions, which in turn helps us understand the behavior of the system described by the BVP.

Analyzing the behavior of the solutions as the parameters, such as the constant \( \alpha \), change is a key aspect of boundary value problems. This kind of analysis can provide insights into the stability of systems and the spectrum of possible states a system may exhibit.

Solving Second-Order Homogeneous ODEs

A second-order homogeneous ordinary differential equation (ODE) has the general form \( y'' + p(x)y' + q(x)y = 0 \), where \( p(x) \) and \( q(x) \) are functions of \( x \). Homogeneity refers to the absence of a non-zero term independent of \( y \) on the right-hand side of the equation.

The solution method includes finding two linearly independent solutions that form a basis for the solution space, and the general solution is a linear combination of these basis solutions. In the exercise we are considering, the differential equation is \( y'' + \lambda y = 0 \) which is homogeneous and has constant coefficients. The characteristic roots of this equation can be complex, real, or zero, and they determine the form of the solution.

The characteristic equation derived from this ODE is \( r^2 + \lambda = 0 \), whose roots are \( r = \pm\sqrt{-\lambda} \). Depending on the sign of \( \lambda \), the general solution can consist of exponential functions, sine and cosine functions, or polynomial functions.

To meet the boundary conditions, we adjust the constants in the general solution accordingly. The nature of the boundary conditions significantly influences the permissible values of \( \lambda \), leading us to the concept of eigenvalues in the context of differential equations.

The Role of Sturm-Liouville Theory

Sturm-Liouville theory is a powerful framework in the field of differential equations, particularly for solving second-order linear differential equations with boundary conditions. This theory assists in characterizing the eigenvalues and the corresponding eigenfunctions of such problems.

A Sturm-Liouville problem typically involves a differential operator of the form \( L[y] = -(py')' + qy \) and is accompanied by boundary conditions that must be fulfilled by the solutions. When the boundary conditions are applied, only certain discrete values of \( \lambda \) called eigenvalues will allow the corresponding solutions, or eigenfunctions, to exist.

In the problem presented, we deduce the necessary condition for eigenvalues \( \lambda \), which gives rise to a transcendental equation involving trigonometric functions. The transcendental nature of this equation implies the existence of an infinite number of eigenvalues, and these eigenvalues are determined by the parameter \( \alpha \).

Sturm-Liouville theory not only confirms the existence of an infinite sequence of eigenvalues under certain conditions but also helps in ordering them and understanding their behavior as parameters vary. This theory is employed in various areas of physics and engineering, such as in the analysis of vibrations, heat conduction, and quantum mechanics, due to its robustness in handling a wide range of boundary conditions.