Question

Consider the problem $$ y^{\prime \prime}+\lambda y=0, \quad y(0)=0, \quad y^{\prime}(L)=0 $$ $$ \begin{array}{l}{\text { Show that if } \phi_{\infty} \text { and } \phi_{n} \text { are eigenfunctions, corresponding to the eigenvalues } \lambda_{m} \text { and } \lambda_{n},} \\ {\text { respectively, with } \lambda_{m} \neq \lambda_{n} \text { , then }}\end{array} $$ $$ \int_{0}^{L} \phi_{m}(x) \phi_{n}(x) d x=0 $$ $$ \text { Hint. Note that } $$ $$ \phi_{m}^{\prime \prime}+\lambda_{m} \phi_{m}=0, \quad \phi_{n}^{\prime \prime}+\lambda_{n} \phi_{n}=0 $$ $$ \begin{array}{l}{\text { Multiply the first of these equations by } \phi_{n}, \text { the second by } \phi_{m}, \text { and integrate from } 0 \text { to } L,} \\ {\text { using integration by parts. Finally, subtract one equation from the other. }}\end{array} $$

Short answer

Question: Prove that two eigenfunctions, corresponding to distinct eigenvalues, in a Sturm-Liouville problem are orthogonal. Answer: Two eigenfunctions, corresponding to distinct eigenvalues, in a Sturm-Liouville problem are orthogonal if the integral of the product of the two eigenfunctions is equal to zero. Following a series of steps involving multiplying the given differential equations by their respective eigenfunctions, integrating, and using integration by parts, we showed that the integral of the product of two eigenfunctions corresponding to distinct eigenvalues is indeed zero, proving that they are orthogonal.

Step-by-step solution

Multiply the equations and integrate

Multiply the first equation by \(\phi_n\), the second equation by \(\phi_m\), and integrate from 0 to L: $$ \int_{0}^{L}\phi_{m}^{\prime \prime}(x)\phi_{n}(x)dx + \lambda_m \int_{0}^{L}\phi_{m}(x)\phi_{n}(x)dx = 0 \\ \int_{0}^{L}\phi_{n}^{\prime \prime}(x)\phi_{m}(x)dx + \lambda_n \int_{0}^{L}\phi_{n}(x)\phi_{m}(x)dx = 0 $$

Integration by parts

Apply integration by parts to the integrals with second derivatives. Let \(u = \phi_m(x)\) and \(v' = \phi_n'(x)\) for the first integral of the first equation, and \(u = \phi_n(x)\) and \(v' = \phi_m'(x)\) for the first integral of the second equation. Integration by parts tells us that \(\int u \cdot v' = uv - \int u' \cdot v\). Thus, we have: $$\left.\phi_{m}(x) \phi_{n}^{\prime}(x)\right|_{0}^{L} - \int_{0}^{L}\phi_{m}^{\prime}(x)\phi_{n}^{\prime}(x)dx + \lambda_m \int_{0}^{L}\phi_{m}(x)\phi_{n}(x)dx = 0 \\ \left.\phi_{n}(x) \phi_{m}^{\prime}(x)\right|_{0}^{L} - \int_{0}^{L}\phi_{n}^{\prime}(x)\phi_{m}^{\prime}(x)dx + \lambda_n \int_{0}^{L}\phi_{n}(x)\phi_{m}(x)dx = 0$$ Note that \(u'= \phi_m'(x)\) and \(v=\phi_n(x)\) for the first equation and \(u'=\phi_n'(x)\) and \(v=\phi_m(x)\) for the second equation.

Subtract equations

Subtract the second equation from the first equation: $$ \left(\left.\phi_{m}(x) \phi_{n}^{\prime}(x)\right|_{0}^{L} - \left.\phi_{n}(x) \phi_{m}^{\prime}(x)\right|_{0}^{L}\right) + (\lambda_m - \lambda_n) \int_{0}^{L}\phi_{m}(x)\phi_{n}(x)dx = 0 $$

Evaluate boundary terms

Evaluate the boundary terms resulting from the integration by parts at x=0 and x=L: $$ (\phi_m(L)\phi_n'(L) - \phi_n(L)\phi_m'(L)) - (\phi_m(0)\phi_n'(0) - \phi_n(0)\phi_m'(0)) + (\lambda_m - \lambda_n) \int_{0}^{L}\phi_{m}(x)\phi_{n}(x)dx = 0 $$ We are given that \(y(0)=0\), and \(\phi_m(0)=\phi_n(0)=0\). Additionally, \(y'(L)=0\) and \(\phi_m'(L)=\phi_n'(L)=0\). Using these boundary conditions, we simplify the above equation to: $$ 0 + (\lambda_m - \lambda_n) \int_{0}^{L}\phi_{m}(x)\phi_{n}(x)dx = 0 $$

Orthogonality

Since \(\lambda_m \neq \lambda_n\), divide both sides of the equation by \(\lambda_m - \lambda_n\): $$ \int_{0}^{L}\phi_{m}(x)\phi_{n}(x)dx = 0 $$ We have shown that the integral of the product of two eigenfunctions corresponding to distinct eigenvalues is zero, proving that they are orthogonal.

Key concepts

Eigenfunctions

In the context of Sturm-Liouville theory, eigenfunctions are special solutions to differential equations that, when multiplied by a value called an eigenvalue, provide the same shape of the function after the differential equation is applied. In the given problem, \( \phi_m(x) \) and \( \phi_n(x) \) represent eigenfunctions. These functions are crucial because they form a basis over which any reasonable function (satisfying certain conditions) can be expanded within the domain of the Sturm-Liouville problem, similar to how you might use sine and cosine functions in a Fourier series.

It's important to note that eigenfunctions have unique properties related to the specific boundary conditions of the problem. Here, the boundary conditions are \( y(0) = 0 \) and \( y'(L) = 0 \). These conditions play a role in determining the exact form of the eigenfunctions.

Eigenvalues

The eigenvalues \( \lambda_m \) and \( \lambda_n \) in the exercise are the special constants associated with the eigenfunctions \( \phi_m(x) \) and \( \phi_n(x) \) respectively. These eigenvalues are determined as part of solving the boundary value problem and are essential for constructing the equations that eigenfunctions satisfy. In the context of this problem, they say that each function resonates at a specific frequency set by its eigenvalue in the differential equation \( y'' + \lambda y = 0 \).

The significance of these values is that they are required for the function to satisfy both the differential equation and the boundary conditions simultaneously. Different eigenvalues will lead to different eigenfunctions, which can be thought of as different vibration modes in a physical system, like harmonics on a string.

Orthogonality

Orthogonality is a property that indicates two functions are perpendicular in a certain sense when integrated against each other over a given domain. What does this mean in practice? Well, it ensures that when you multiply two orthogonal functions and integrate over the domain, the result is zero. This property is fundamental in solving and simplifying a wide variety of problems in mathematics.

Within the exercise, orthogonality is shown between two eigenfunctions \( \phi_m(x) \) and \( \phi_n(x) \) by proving that their inner product (the integral of their product over the domain from 0 to L) is zero, which only happens when \( \lambda_m eq \lambda_n \). This property allows us to decompose complex functions into a sum of eigenfunctions, which is extremely useful, for example, in solving partial differential equations using separation of variables.

Boundary Value Problems

Boundary value problems are differential equations that, alongside a set of boundary conditions, dictate the possible solutions for functions defined on certain domains. These problems are common in physical situations where conditions at the borders, like temperature or pressure, are known.

In the exercise, we're asked to consider a second-order differential equation with specific boundary conditions, and it's these conditions at \( x = 0 \) and \( x = L \) that lead to the eigenvalues and eigenfunctions. Successfully solving such boundary value problems is crucial in many branches of engineering and physics, such as determining the modes of vibration of a drumhead or the temperature distribution in a metal bar.

Differential Equations

A differential equation is an equation involving derivatives which describes a relationship between a function and its derivatives. In physics and engineering, differential equations are the bread and butter for modeling the way systems evolve over time or space.

The equation presented in the exercise, \( y'' + \lambda y = 0 \), is an example of a linear homogeneous second-order differential equation. This particular form is known as the classical simple harmonic oscillator equation, which appears frequently in mechanics and electrical engineering. The general solutions to such equations involve trigonometric functions and exponential functions. The uniqueness of solutions to differential equations heavily depends on the accompanying boundary or initial conditions, which often guide the problem-solving approach in practical applications.