Question

Consider the Sturm-Liouville problem $$ y^{\prime \prime}+\lambda y=0, \quad y(0)+\alpha y^{\prime}(0)=0, \quad y(1)+(\alpha-1) y^{\prime}(1)=0 $$ where \(0<\alpha<1\) (a) Show that \(\lambda=0\) is an eigenvalue of (A), and find an associated eigenfunction. (b) Show that (A) has a negative eigenvalue, and find the form of an associated eigenfunction. (c) Give a graphical argument to show that (A) has infinitely many positive eigenvalues \(\lambda_{1}<\lambda_{2}<\) \(\cdots<\lambda_{n}<\cdots,\) and state the form of the associated eigenfunctions.

Short answer

Now, let's summarize our findings: 1. We've shown that \(\lambda = 0\) is an eigenvalue, with the associated eigenfunction being \(y = (1-\alpha)x-\alpha(1-\alpha)\). 2. We've demonstrated that there must be at least one negative eigenvalue, and its associated eigenfunction has the form \(y(x) = C_1\cosh(\mu x) + C_2\sinh(\mu x)\). 3. We've used graphical arguments to show that there are infinitely many positive eigenvalues, with the associated eigenfunctions having the form \(y(x) = C_1\cos(\sqrt{\lambda}x) + C_2\sin(\sqrt{\lambda}x)\). These results address all three parts of the given Sturm-Liouville problem and provide a comprehensive understanding of the eigenvalues and eigenfunctions for the given boundary conditions.

Step-by-step solution

Part (a) Solution for \(\lambda = 0\)

Let's consider the case \(\lambda = 0\). The given equation becomes: $$ y^{\prime\prime} = 0 $$ To find the eigenfunction, we integrate twice: $$ \int y^{\prime\prime} \, dx = \int 0 \, dx \Rightarrow y^{\prime} = C_1 \Rightarrow y = C_1x + C_2 $$ Now we apply the boundary conditions: 1. \(y(0)+\alpha y^{\prime}(0)=0\) $$ C_2 + \alpha C_1 = 0 \Rightarrow C_2 = -\alpha C_1 $$ 2. \(y(1)+(\alpha-1)y^{\prime}(1) = 0\) $$ (-\alpha +1)C_1 + (-\alpha +1)=0 \Rightarrow C_1 = 1-\alpha $$ Now we have $$ y=C_1x + C_2 = (1-\alpha)x-\alpha(1-\alpha) $$ Hence, \(\lambda=0\) is an eigenvalue with the associated eigenfunction being \(y = (1-\alpha)x-\alpha(1-\alpha)\).

Part (b) Solution for a negative eigenvalue

Now we will consider the case when \(\lambda < 0\), and let \(\lambda = -\mu^2\) where \(\mu > 0\). The equation becomes: $$ y^{\prime\prime}-\mu^2y = 0 $$ The general solution is: $$ y(x) = C_1\cosh(\mu x) + C_2\sinh(\mu x) $$ We apply the boundary conditions: 1. \(y(0)+\alpha y^{\prime}(0)=0\) $$ C_1 + \alpha\mu C_2 = 0 \Rightarrow C_1 = -\alpha\mu C_2 $$ 2. \(y(1)+(\alpha-1)y^{\prime}(1) = 0\) $$ -\alpha\mu C_2\cosh(\mu) + C_2\sinh(\mu) + (\alpha-1)(-\alpha\mu C_2\sinh(\mu)+C_2\cosh(\mu)) = 0 $$ We can simplify the expression above and solve for \(\mu\): $$ \mu\cosh(\mu)\sinh(\mu)+(\alpha-1)\sinh^2(\mu)=0 $$ Since \(\sinh(\mu)\neq 0\), we can divide by it: $$ \mu\cosh(\mu)+(\alpha-1)\sinh(\mu)=0 \Rightarrow \mu \left( \cosh(\mu) + \frac{\alpha-1}{\mu} \sinh(\mu) \right)=0 $$ Since \(\mu > 0\), there must exist at least one root of $$ \cosh(\mu) + \frac{\alpha-1}{\mu} \sinh(\mu) = 0 $$ Hence, there is at least one negative eigenvalue, and the associated eigenfunction has the form: $$ y(x) = C_1\cosh(\mu x) + C_2\sinh(\mu x) $$

Part (c) Graphical argument for positive eigenvalues

To show that there are infinitely many positive eigenvalues, consider now the case when \(\lambda > 0\). The equation becomes: $$ y^{\prime\prime}+\lambda y = 0 $$ The general solution is: $$ y(x) = C_1\cos(\sqrt{\lambda}x) + C_2\sin(\sqrt{\lambda}x) $$ Applying the boundary conditions, we get: 1. \(y(0)+\alpha y^{\prime}(0)=0\) $$ C_1 + \alpha\sqrt{\lambda}C_2 = 0 \Rightarrow C_1 = -\alpha\sqrt{\lambda}C_2 $$ 2. \(y(1)+(\alpha-1)y^{\prime}(1) = 0\) $$ -\alpha\sqrt{\lambda}C_2\cos(\sqrt{\lambda}) + C_2\sin(\sqrt{\lambda})+(\alpha-1)(-\alpha\sqrt{\lambda}C_2\sin(\sqrt{\lambda})+C_2\cos(\sqrt{\lambda}))=0 $$ Simplifying the expression above, we see that the left-hand side is a continuous and differentiable function of \(\lambda\). Moreover, it oscillates and has an infinite number of zeroes. It follows that there are an infinite number of positive eigenvalues. Therefore, there are infinitely many positive eigenvalues \(\lambda_{1}<\lambda_{2}<\cdots<\lambda_{n}<\cdots,\) and the form of the associated eigenfunctions is: $$ y(x) = C_1\cos(\sqrt{\lambda}x) + C_2\sin(\sqrt{\lambda}x) $$

Key concepts

Eigenvalues and Eigenfunctions

Eigenvalues and eigenfunctions are foundational concepts in the study of differential equations, particularly in the context of the Sturm-Liouville problem. In simple terms, an eigenvalue is a special value in which a differential equation has non-trivial solutions called eigenfunctions. These non-trivial solutions imply that when substituted into the differential equation, the outcome is zero without relying solely on zero functions for solutions.

In our given Sturm-Liouville problem, we identify eigenvalues through conditions that lead non-trivial solutions to the differential equation:

• When \(\lambda = 0\), the differential equation simplifies to \(y''=0\). Solving this results in linear functions as eigenfunctions.
• For negative eigenvalues, represented as \(\lambda = -\mu^2\), the solutions involve hyperbolic functions like \(\cosh\) and \(\sinh\).
• Positive eigenvalues lead to trigonometric solutions, involving \(\cos\) and \(\sin\), that complete infinitely due to their oscillatory nature, implying infinite positive eigenvalues.

This interplay between eigenvalues and eigenfunctions forms the core of much analysis in various fields such as quantum mechanics, where systems' properties are often explored through such problems.

Boundary Value Problems

Boundary value problems (BVPs) are a class of differential equations accompanied by conditions that are specified at the boundaries of the interval. In study contexts like the Sturm-Liouville problem, these boundary conditions are key for finding unique solutions corresponding to specific eigenvalues.

The provided problem gives us boundaries at points \(x = 0\) and \(x = 1\):

• At \(x = 0\), \(y(0) + \alpha y'(0) = 0\) establishes a linear condition mixing function values with derivatives.
• At \(x = 1\), \(y(1) + (\alpha-1) y'(1) = 0\) forms another linear condition.

These conditions constrain the solutions to ensure that they are consistent across the whole interval. Essentially, they "pin" the solution in place, ensuring physically meaningful results. Without these boundary conditions, the differential equation might have many more solutions or even unbounded behavior, which can be non-physical for real-world problems.
BVPs regularly arise in fields like engineering and physics, where they might describe a physical phenomenon in a constrained environment, like heat distribution along a rod with fixed temperatures at its ends.

Differential Equations

Differential equations are equations that involve the derivatives of a function. In our context, the Sturm-Liouville problem begins with a second-order differential equation. This type involves the second derivative of a function, indicating how the function's slope changes over its domain.

The differential equation given in the exercise is \(y'' + \lambda y = 0\). This equation type is powerful for modeling various physical systems. It's categorized based on how the eigenvalue \(\lambda\) affects the behavior of its solutions:

• When \(\lambda=0\), the equation reduces to \(y'' = 0\), emphasizing constant acceleration.
• For negative values \(\lambda = -\mu^2\), the equation inserts exponential decay/growth modes like \(\cosh\) and \(\sinh\).
• With positive \(\lambda\), oscillatory solutions emerge, characterized by sine and cosine functions.

Understanding differential equations and their solutions enables the study and prediction of dynamic systems over time or space. From population dynamics in biology to oscillating circuits in electronics, differential equations form the mathematical backbone of countless scientific inquiries.