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

Short Answer

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Question: Determine the normalized eigenfunctions associated with the eigenvalue problem given by the following second-order homogeneous ordinary differential equation with boundary conditions: $$ y^{\prime \prime}+\lambda y=0,\quad y^{\prime}(0)=0,\quad y(1)=0 $$

Step by step solution

01

Find the general solution of the ODE

First, let's find the general solution of the given second-order homogeneous ODE without considering the boundary conditions: $$ y^{\prime \prime}+\lambda y=0 $$ Assume that the solution is of the form \(y(x) = c \cdot e^{r x}\), where \(c\) and \(r\) are constants. Substituting \(y(x)\) into the ODE gives the characteristic equation: $$ r^2 + \lambda = 0 $$ Solving for \(r\), we get two distinct solutions: $$ r_{1,2} = \pm \sqrt{-\lambda} $$ Thus, the general solution of the ODE is given by: $$ y(x) = c_1 e^{\sqrt{-\lambda}x} + c_2 e^{-\sqrt{-\lambda}x} $$ with \(c_1\) and \(c_2\) being constants.
02

Apply the boundary conditions

Now let's apply the given boundary conditions to the general solution in order to find the eigenvalues \(\lambda\) and the associated eigenfunctions \(y(x)\): 1. \(y^{\prime}(0)=0\) 2. \(y(1)=0\) Differentiating the general solution with respect to \(x\), we get: $$ y^\prime(x) = c_1 \sqrt{-\lambda} e^{\sqrt{-\lambda} x} - c_2 \sqrt{-\lambda} e^{-\sqrt{-\lambda} x} $$ Applying the first boundary condition: $$ 0 = c_1 \sqrt{-\lambda} e^{\sqrt{-\lambda}\cdot 0} - c_2 \sqrt{-\lambda} e^{-\sqrt{-\lambda}\cdot 0} \implies c_1=c_2 $$ Substitute \(c_1=c_2\) into the general solution, we have: $$ y(x) = c_1 (e^{\sqrt{-\lambda}x} + e^{-\sqrt{-\lambda}x}) $$ Now apply the second boundary condition: $$ 0 = c_1 (e^{\sqrt{-\lambda}} + e^{-\sqrt{-\lambda}}) \implies e^{\sqrt{-\lambda}} + e^{-\sqrt{-\lambda}}=0 \implies e^{2\sqrt{-\lambda}}=-1 $$ Since \(e^{2\sqrt{-\lambda}}\,-1=0\), the eigenvalues \(\lambda\) are given by: $$ \lambda_n = -\frac{1}{4} (2n \pi)^2 ,\quad n=\pm 1, \pm 2, \pm 3, \dots $$ Thus, the eigenfunctions for the given problem are: $$ y_n(x) = c_n (e^{i n \pi x} + e^{-i n \pi x}),\quad n = 1, 2, 3, \dots $$ with \(c_n\) being a constant.
03

Normalize the eigenfunctions

Now normalize the eigenfunctions by ensuring the integral of their squared magnitudes is equal to 1: $$ 1 = \int_0^1 |y_n(x)|^2 dx = \int_0^1 |c_n (e^{i n \pi x} + e^{-i n \pi x})|^2 dx $$ Solving for \(c_n\): $$ c_n = \frac{1}{\sqrt{\int_0^1 |e^{i n \pi x} + e^{-i n \pi x}|^2 dx}} $$ Computing the integral and taking the reciprocal square root of the result, we get: $$ c_n = \frac{1}{\sqrt{ \frac{1}{2} }} $$ Hence, the normalized eigenfunctions are: $$ y_n(x) = \frac{1}{\sqrt{ \frac{1}{2} }} (e^{i n \pi x} + e^{-i n \pi x}),\quad n = 1, 2, 3, \dots $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ordinary Differential Equations (ODEs)
Ordinary Differential Equations, or ODEs, are equations involving derivatives of a function with respect to one independent variable. They are fundamental in expressing dynamics and changes in various scientific and engineering problems. An ODE like \( y'' + \lambda y = 0 \) is a linear homogenous second-order differential equation, where \( y'' \) denotes the second derivative of \( y \) with respect to an independent variable, usually time or space, and \( \lambda \) is a parameter.

When solving such an ODE, the first step is to find the general solution, which involves finding a family of solutions that includes every particular solution satisfying the equation. This could be achieved, for example, by assuming a trial solution and then determining constants through the characteristic equation. As noted in the step by step solution, this method yields a general solution with constants \( c_1 \) and \( c_2 \) dependent on boundary conditions.
Boundary Value Problems
Boundary Value Problems (BVPs) are situations where we are interested in finding a solution to an ordinary differential equation subject to certain fixed conditions, known as 'boundary conditions,' at the extremes or boundaries of the independent variable's domain. These kinds of problems are common in physics and engineering, for example, in determining the deflection of a beam fixed at both ends or the temperature distribution along a rod.

In the given exercise, the boundary conditions are \( y'(0) = 0 \) and \( y(1) = 0 \), expressing the behavior of the solution at the endpoints \( x=0 \) and \( x=1 \). This results in a system that only accepts specific values of \( \lambda \) (eigenvalues) and corresponding functions (eigenfunctions), which meet the imposed restrictions. Upon applying these conditions to the general solution of the ODE, one narrows down the possible solutions to a discrete set that corresponds to the natural frequencies or modes of the system being examined.
Normalization of Eigenfunctions
Normalization is a process to adjust the magnitude of eigenfunctions so that they satisfy a given condition—commonly, that the integral of their square over a certain range is 1. This condition arises from the probabilistic interpretation in quantum mechanics and other fields where the total probability must equal 1. However, normalization also plays a role in simplifying problems in mathematics and engineering by imposing a standard form.

As demonstrated by the solution, normalizing the eigenfunctions of the BVP involves integrating the squared magnitude of these functions over the relevant domain, and adjusting the constants \( c_n \) so the integral evaluates to 1. Through normalization, the eigenfunctions become orthonormal, meaning they are not only normalized, but also orthogonal—a property which states that the integral of the product of any two distinct eigenfunctions over the domain is zero. This property of orthonormality is quite useful in expanding functions in terms of eigenfunctions, as it simplifies the computation of coefficients.

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Most popular questions from this chapter

Indicate a way of solving nonhomogeneous boundary value problems that is analogous to using the inverse matrix for a system of linear algebraic equations. The Green's function plays a part similar to the inverse of the matrix of coefficients. This method leads to solutions expressed as definite integrals rather than as infinite series. Except in Problem 35 we will assume that \(\mu=0\) for simplicity. (a) Show by the method of variation of parameters that the general solution of the differential equation $$ -y^{\prime \prime}=f(x) $$ can be written in the form $$ y=\phi(x)=c_{1}+c_{2} x-\int_{0}^{x}(x-s) f(s) d s $$ where \(c_{1}\) and \(c_{2}\) are arbitrary constants. (b) Let \(y=\phi(x)\) also be required to satisfy the boundary conditions \(y(0)=0, y(1)=0\) Show that in this case $$ c_{1}=0, \quad c_{2}=\int_{0}^{1}(1-s) f(s) d s $$ (c) Show that, under the conditions of parts (a) and (b), \(\phi(x)\) can be written in the form $$ \phi(x)=\int_{0}^{x} s(1-x) f(s) d s+\int_{x}^{1} x(1-s) f(s) d s $$ (d) Defining $$ G(x, s)=\left\\{\begin{array}{ll}{s(1-x),} & {0 \leq s \leq x} \\ {x(1-s),} & {x \leq s \leq 1}\end{array}\right. $$ show that the solution takes the form $$ \phi(x)=\int_{0}^{1} G(x, s) f(s) d s $$ The function \(G(x, s)\) appearing under the integral sign is a Green's function. The usefulness of a Green's function solution rests on the fact that the Green's function is independent of the nonhomogencous term in the differential equation. Thus, once the Green's function is determined, the solution of the boundary value problem for any nonhomogeneous term \(f(x)\) is obtained by a single integration. Note further that no determination of arbitrary constants is required, since \(\phi(x)\) as given by the Green's function integral formula automatically satisfies the boundary conditions.

determine whether the given boundary value problem is self-adjoint. $$ y^{\prime \prime}+\lambda y=0, \quad y(0)=0, \quad y(\pi)+y^{\prime}(\pi)=0 $$

In the spherical coordinates \(\rho, \theta, \phi(\rho>0,0 \leq \theta<2 \pi, 0 \leq \phi \leq \pi)\) defined by the equations $$ x=\rho \cos \theta \sin \phi, \quad y=\rho \sin \theta \sin \phi, \quad z=\rho \cos \phi $$ Laplace's equation is $$ \rho^{2} u_{\rho \rho}+2 \rho u_{\rho}+\left(\csc ^{2} \phi\right) u_{\theta \theta}+u_{\phi \phi}+(\cot \phi) u_{\phi}=0 $$ (a) Show that if \(u(\rho, \theta, \phi)=\mathrm{P}(\rho) \Theta(\theta) \Phi(\phi),\) then \(\mathrm{P}, \Theta,\) and \(\Phi\) satisfy ordinary differential equations of the form $$ \begin{aligned} \rho^{2} \mathrm{P}^{\prime \prime}+2 \rho \mathrm{P}^{\prime}-\mu^{2} \mathrm{P} &=0 \\ \Theta^{\prime \prime}+\lambda^{2} \Theta &=0 \\\\\left(\sin ^{2} \phi\right) \Phi^{\prime \prime}+(\sin \phi \cos \phi) \Phi^{\prime}+\left(\mu^{2} \sin ^{2} \phi-\lambda^{2}\right) \Phi &=0 \end{aligned} $$ The first of these equations is of the Euler type, while the third is related to Legendre's equation. (b) Show that if \(u(\rho, \theta, \phi)\) is independent of \(\theta,\) then the first equation in part (a) is unchanged, the second is omitted, and the third becomes $$ \left(\sin ^{2} \phi\right) \Phi^{\prime \prime}+(\sin \phi \cos \phi) \Phi^{\prime}+\left(\mu^{2} \sin ^{2} \phi\right) \Phi=0 $$ (c) Show that if a new independent variable is defined by \(s=\cos \phi\), then the equation for \(\Phi\) in part (b) becomes $$ \left(1-s^{2}\right) \frac{d^{2} \Phi}{d s^{2}}-2 s \frac{d \Phi}{d s}+\mu^{2} \Phi=0, \quad-1 \leq s \leq 1 $$ Note that this is Legendre's equation.

Find the steady-state temperature \(u(\rho, \phi)\) in a sphere of unit radius if the temperature is independent of \(\theta\) and satisfies the boundary condition $$ u(1, \phi)=f(\phi), \quad 0 \leq \phi \leq \pi $$ Hint: Refer to Problem 9 and to Problems 22 through 29 of Section \(5.3 .\) Use the fact that the only solutions of Legendre's equation that are finite at both \(\pm 1\) are the Legendre polynomials.

In each of Problems I through 6 state whether the given boundary value problem is homogeneous or non homogeneous. $$ y^{\prime \prime}+4 y=0, \quad y(-1)=0, \quad y(1)=0 $$

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