Question

determine the normalized eigenfunctions of the given problem. $$ y^{\prime \prime}+\lambda y=0, \quad y^{\prime}(0)=0, \quad y^{\prime}(1)=0 $$

Short answer

Answer: The normalized eigenfunctions are \(\sqrt{2} \cos(n\pi x) \), where \(n \in \mathbb{Z}\).

Step-by-step solution

Solve the differential equation for each case of λ

There are three cases for the value of λ: λ > 0, λ = 0, and λ < 0. We need to solve the given differential equation for each case and see which case satisfies the given boundary conditions. Case 1: λ > 0 Let \(\lambda = k^2\) for some \(k> 0\), then the given differential equation becomes $$y^{\prime\prime} + k^2 y = 0$$ The general solution for this equation is $$y(x) = A\cos(kx) + B\sin(kx)$$ Case 2: λ = 0 For this case, the given differential equation becomes $$y^{\prime\prime} = 0$$ The general solution for this equation is $$y(x) = Ax + B$$ Case 3: λ < 0 Let \(\lambda = -k^2\) for some \(k> 0\), then the given differential equation becomes $$y^{\prime\prime} - k^2 y = 0$$ The general solution for this equation is $$y(x) = A\cosh(kx) + B\sinh(kx)$$

Apply the boundary conditions and find the eigenvalues

Now we need to apply the boundary conditions \(y'(0) = 0\) and \(y'(1) = 0\) to the general solutions found in Step 1 and determine the eigenvalues. Case 1: λ > 0 $$y'(x) = -Ak\sin(kx) + Bk\cos(kx)$$ Applying boundary conditions: $$ -Ak\sin(0) + Bk\cos(0) = 0 \Rightarrow Bk = 0$$ $$ -Ak\sin(k) + Bk\cos(k) = 0 $$ Since k > 0, \(B = 0\). Therefore, $$y(x) = A\cos(kx)$$ Plugging \(B = 0\) into the second boundary condition, $$ -A\sin(k) = 0 $$ For nontrivial solutions, \(A \neq 0\), so we have $$\sin(k) = 0 \Rightarrow k = n\pi, n \in \mathbb{Z}$$ Case 2: λ = 0 Since both boundary conditions are derivatives, they won't change the result in this case. Therefore, this case doesn't contain any eigenvalues. Case 3: λ < 0 The boundary conditions won't affect the result in this case since it requires positive eigenvalues solution. Therefore, this case doesn't contain any eigenvalues.

Find the unnormalized eigenfunctions

We found the eigenvalues as \(k = n\pi\), \(n \in \mathbb{Z}\), which correspond to \(\lambda = n^2 \pi^2\). Now we need to find the eigenfunctions. Since \(B = 0\) for the case \(\lambda > 0\), $$y(x) = A\cos(n\pi x)$$ These are the unnormalized eigenfunctions for the given problem.

Normalize the eigenfunctions

To normalize the eigenfunctions, we need to determine the constant A so that the norm of the eigenfunction equals one: $$\int_{0}^{1} (A\cos(n\pi x))^2 dx = 1$$ First, calculate the integral: $$\int_{0}^{1} A^2\cos^2(n\pi x) dx = A^2\int_{0}^{1}\frac{1 + \cos(2n\pi x)}{2} dx$$ $$=A^2\left[\frac{x}{2}+\frac{\sin(2n\pi x)}{4n\pi}\right]_{0}^{1}=\frac{A^2}{2}$$ Now, let's find the constant \(A\) such that the integral result equals 1: $$\frac{A^2}{2} = 1 \Rightarrow A = \sqrt{2}$$

Write down the normalized eigenfunctions

Now that we have A, the normalized eigenfunctions are expressed as: $$y(x) = \sqrt{2} \cos(n\pi x), \quad n \in \mathbb{Z}$$ These are the normalized eigenfunctions of the given problem.

Key concepts

Boundary Value Problem

A boundary value problem (BVP) consists of a differential equation coupled with a set of additional restraints called boundary conditions. In the context of physical phenomena, BVPs are crucial since they allow one to calculate the behavior of a system based on its conditions at the boundaries. For instance, in the given exercise, we are dealing with the second-order linear differential equation \(y'' + \lambda y = 0\) where \(y''\) is the second derivative of the function \(y\) with respect to the variable, usually representing space or time. The boundary conditions provided are \(y'(0) = 0\) and \(y'(1) = 0\), which specify the derivative of the function at both ends of the interval [0, 1]. These are known as Neumann boundary conditions because they specify values of the derivative rather than the function itself.
Solving a BVP involves finding a function \(y(x)\) which satisfies both the differential equation and the boundary conditions. It is important to note that while some boundary value problems have a unique solution, others have either no solution or infinitely many solutions.

Sturm-Liouville Problem

The Sturm-Liouville problem is a particular type of boundary value problem that arises frequently in physics and engineering, especially in the fields of vibration analysis, heat transfer, and quantum mechanics. It's a cornerstone of understanding how physical systems behave under various conditions. Such problems can be represented as \( -\frac{d}{dx}(p(x)\frac{dy}{dx}) + q(x)y = \lambda r(x)y \), where \(p(x)\), \(q(x)\), and \(r(x)\) are given functions, and \(\lambda\) is the eigenvalue to be determined.In this type of problem, we are not only looking for the solutions of a differential equation but also the special values of \(\lambda\), these are called eigenvalues. The corresponding solutions for these values of \(\lambda\) are known as eigenfunctions. The process of solving involves finding pairs \((\lambda, y(x))\) that satisfy the differential equation along with the boundary conditions. The eigenfunctions are orthogonal with respect to the weight function \(r(x)\) over the interval of the problem, which is crucial for their normalization and subsequent physical interpretation.

Normalization of Eigenfunctions

Normalization is a process to adjust the scale of an eigenfunction so that it has a unit norm. This is a critical step for functions that represent physical states, such as in quantum mechanics where the square of the absolute value of an eigenfunction represents a probability density.To normalize an eigenfunction, we adjust the amplitude of the function so that its inner product with itself over the domain of interest equals one. This is equivalent to ensuring the area under the square of the absolute value of the function (over the domain) equals one. In mathematical terms for the function \(y(x)\), normalization requires that \(\int (y(x))^2 \,dx = 1\).For the exercise given, we first found the eigenfunctions in the form of \(y(x) = A\cos(n\pi x)\) which are not yet normalized. Applying the normalization condition results in calculating the integral over the domain, which leads us to find the constant \(A\). Once we calculated the integral and set it equal to one, the appropriate scale factor \(A = \sqrt{2}\) ensures that the eigenfunction \(y(x)\) is normalized. Consequently, the normalized eigenfunctions of our problem are \(y(x) = \sqrt{2} \cos(n\pi x)\), which are now appropriately scaled for any physical interpretations that might follow.