Question

Show that the general second-order linear equation $$ p_{0}(x) y^{\prime \prime}+p_{1}(x) y^{\prime}+\left[\lambda p_{2}(x)+p_{3}(x)\right] y=0, $$ where \(p_{0}(x) \neq 0\), takes on the form of a Sturm-Liouville equation \((10.127)\) if the equation is multiplied by \(r(x) / p_{0}(x)\), where \(r(x)\) is chosen so that \(r^{\prime} / r=p_{1} / p_{0}\). In general, an equation of form: \(\left(r y^{\prime}\right)^{\prime}+h(x) y=0\) is called self-adjoint.

Short answer

Question: Given the second-order linear equation \(p_{0}(x)y^{\prime\prime}+p_{1}(x)y^{\prime}+\left[\lambda p_{2}(x)+p_{3}()x\right]y=0,\) show that this equation can be transformed into a Sturm-Liouville equation if it is multiplied by \(r(x)/p_{0}(x)\), where \(r(x)\) is chosen so that \(r'(x)/r(x) = p_{1}(x)/p_{0}(x)\). Answer: By multiplying the given linear equation by \(r(x)/p_{0}(x)\) and using the condition for \(r(x)\), we can rewrite the equation in the form of a Sturm-Liouville equation, which shows that this transformation is possible under these given conditions.

Step-by-step solution

Multiply the original equation by \(r(x)/p_{0}(x)\)

To start, we will multiply the given second-order linear equation by the term \(r(x) / p_{0}(x)\): $$ \frac{r(x)}{p_{0}(x)} [p_{0}(x) y^{\prime \prime}+p_{1}(x) y^{\prime}+\left[\lambda p_{2}(x)+p_{3}(x)\right] y]=0. $$ Now let's distribute the terms to obtain: $$ r(x) y^{\prime \prime} + \frac{r(x)}{p_{0}(x)}p_{1}(x) y^{\prime} + \frac{r(x)}{p_{0}(x)}[\lambda p_{2}(x) + p_{3}(x)]y = 0. $$

Use the condition for \(r(x)\)

Our next step is to use the condition provided for \(r(x)\) to further simplify the equation. We are given that: $$ \frac{r^{\prime}(x)}{r(x)} = \frac{p_{1}(x)}{p_{0}(x)}. $$ If we rewrite the condition for \(r(x)\), we can find \(r'(x)\): $$ r'(x) = r(x) \frac{p_{1}(x)}{p_{0}(x)}. $$ Now we can substitute this into our current equation and obtain: $$ r(x) y^{\prime \prime} + r'(x) y^{\prime} + \frac{r(x)}{p_{0}(x)}[\lambda p_{2}(x) + p_{3}(x)]y = 0. $$

Regroup terms and identify the Sturm-Liouville form

Our final step is to regroup terms and show that our current equation has the form of a Sturm-Liouville equation. The Sturm-Liouville form is: $$ \left(r(x)y'(x)\right)' + \left[\lambda w(x) + q(x)\right]y(x) = 0. $$ Let's rewrite our equation: $$ \left[r(x) y^{\prime}(x)\right]^{\prime} + \frac{r(x)}{p_{0}(x)}[\lambda p_{2}(x) + p_{3}(x)]y(x) = 0. $$ We can now identify that the first term is the derivative of \(r(x)y'(x)\), and we can rewrite the second term as \(\lambda w(x) + q(x)\), with \(w(x) = p_{2}(x)\) and \(q(x) = p_{3}(x)\). This equation now has the form of a Sturm-Liouville equation: $$ \left(r(x)y'(x)\right)' + [\lambda w(x) + q(x)]y(x) = 0. $$ We have shown that the general second-order linear equation takes on the form of a Sturm-Liouville equation if the equation is multiplied by \(r(x) / p_{0}(x)\), where \(r(x)\) is chosen so that \(r^{\prime} / r=p_{1} / p_{0}\).

Key concepts

Second-Order Linear Differential Equation

A second-order linear differential equation is a fundamental concept in the field of ordinary differential equations (ODEs) and appears frequently in various applications such as physics, engineering, and mathematics. The general form of such an equation is expressed as \[\frac{d^2y}{dx^2} + P(x)\frac{dy}{dx} + Q(x)y = R(x),\] where the functions P(x), Q(x), and R(x) are coefficients that may vary with the independent variable x, and y is the dependent variable we wish to solve for.
In the context of a homogeneous second-order linear differential equation, where the right-hand side R(x) is zero, we can identify solutions that display important properties such as linearity. This means that any linear combination of two solutions to the homogeneous equation is also a solution.

Transformation to Sturm-Liouville Form

To understand the significance of the Sturm-Liouville form, it is essential to consider how a general second-order linear equation can be manipulated. By introducing a multiplying function, as shown in the given exercise, and using conditions based on the coefficients P(x) and Q(x), we can transform the equation into a self-adjoint form which opens the door to powerful analytical techniques and a clearer interpretation of physical phenomena.

Self-Adjoint Operators

In the realm of functional analysis, self-adjoint operators play a critical role, particularly in quantum mechanics and in the solution of differential equations. An operator, in this context, is a rule that applies to a function to produce another function. When dealing with differential equations, we concern ourselves with differential operators that can act on functions to create new ones involving their derivatives.
A self-adjoint, or Hermitian, operator is one that is equal to its own adjoint, meaning that the action of the operator on any two functions in the space results in the same scalar product as if the operator had been applied to the second function and then taken the scalar product with the first. This property is crucial for the formulation of physical theories because it ensures that the eigenvalues, or the measurements we can observe, are real numbers, which correspond to observable quantities in the physical world.

Relation to Differential Equations

The Sturm-Liouville equation, a self-adjoint form of a second-order linear differential equation, ensures the self-adjointness property is preserved. This makes it possible to apply the principles and techniques of linear algebra to continuous functions, providing an invaluable tool for solving complex differential equations.

Eigenvalue Problems

Eigenvalue problems are ubiquitous in scientific computing, emerging in disciplines such as vibrations analysis, stability studies, and quantum mechanics. They form a subset of problems in linear algebra where we seek a scalar \(\lambda\) called an eigenvalue, and a non-zero vector or function, known as an eigenvector or eigenfunction, that satisfies an equation of the form \[A\vec{x} = \lambda\vec{x},\] where A is a linear operator or matrix.
The eigenvalue equation for the Sturm-Liouville problems is a special case where the operator involved is a differential operator. This leads to finding eigenvalues and eigenfunctions that comply with \[\left(r(x)y'(x)\right)' + [\lambda w(x) + q(x)]y(x) = 0,\] as shown in the exercise.
Eigenvalue problems associated with differential equations, especially of the Sturm-Liouville type, are essential because they often model natural frequencies of systems and can reflect the stability and dynamic response of physical systems. Owing to their self-adjoint nature, these problems yield a set of orthogonal eigenfunctions, providing a basis for expanding functions in the space, much like a Fourier series expansion.