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Chapter 11: Problem 18

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

Short Answer

Expert verified
In summary, the given boundary value problem is self-adjoint since its ODE is already in self-adjoint form and its boundary conditions are also in self-adjoint form.

Step by step solution

01

Transform the ODE to the self-adjoint form if needed.

The given ODE is already in the self-adjoint form: $$ \frac{d}{dx} \left[ 1 \cdot \frac{dy}{dx} \right] + 0 \cdot \frac{dy}{dx} + \lambda y = 0 \Rightarrow y'' + \lambda y = 0. $$
02

Check if the given boundary conditions are self-adjoint.

The given boundary conditions for the problem are: $$ y(0) = 0 \quad \text{and} \quad y(\pi) + y'(\pi) = 0. $$ These boundary conditions can be written as $$ 1 \cdot y(0) + 0 \cdot y'(0) = 0 \quad \text{and} \quad 1 \cdot y(\pi) + 1 \cdot y'(\pi) = 0. $$ They are both in the self-adjoint form: $$ \alpha_1 y(a) + \alpha_2 y'(a) = 0 \quad \text{and} \quad \beta_1 y(b) + \beta_2 y'(b) = 0. $$ Since the given ODE is already in self-adjoint form and the boundary conditions are also self-adjoint, the given boundary value problem is self-adjoint.

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Key Concepts

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

Ordinary Differential Equations
Ordinary Differential Equations (ODEs) are equations involving functions and their derivatives. Unlike partial differential equations, which involve multiple independent variables, ODEs typically involve just one. In the simplest form, ODEs like the one in our exercise represent mathematical relationships between varying quantities that can be expressed using derivatives.

ODEs are crucial in understanding natural phenomena and appear in fields such as physics, biology, and finance. They allow us to describe the behavior of dynamic systems. For example:
  • The growth rate of a population in biology.
  • The motion of objects in physics.
  • The rate of change in financial markets.
In our exercise, we work with the equation \(y'' + \lambda y = 0\). Here, \(y''\) is the second derivative of \(y\) with respect to \(x\), indicating how the rate of change itself is changing. Solving such ODEs helps us predict and understand complex behaviors.
Boundary Conditions
Boundary conditions are additional constraints provided in differential equations to determine specific solutions from a general solution. These conditions are essential because many differential equations have infinitely many solutions, and boundary conditions help select the one that fits a particular physical scenario.

In the given problem, the boundary conditions are \(y(0) = 0\) and \(y(\pi) + y'(\pi) = 0\). These conditions specify the behavior of functions at the boundaries—\(x = 0\) and \(x = \pi\) in this case.
  • First condition: \(y(0) = 0\) signifies that at \(x = 0\), the function \(y\) must be zero.
  • Second condition: \(y(\pi) + y'(\pi) = 0\) imposes that at \(x = \pi\), the sum of the function and its derivative is zero.
Boundary conditions dictate the specific shape and nature of the solution curve across the interval, ensuring it adheres to these prescribed conditions at the endpoints.
Eigenvalue Problems
Eigenvalue problems in differential equations involve finding the characteristic values (eigenvalues) for which there are non-trivial solutions (eigenfunctions) to the equations. These problems are essential in mathematical physics and engineering because they often describe natural frequencies of systems, resonances, or stability contours.

In the equation \(y'' + \lambda y = 0\), \(\lambda\) plays the role of an eigenvalue. Solving such problems leads to a deeper understanding of systems such as vibrating strings or heat transfer domains.
  • Eigenvalues, \(\lambda\), represent special parameter values where the system exhibits particular resonant behaviors.
  • Finding the correct \(\lambda\) means determining which values lead to valid, permissible solutions under the given boundary conditions.
Eigenvalue problems often lead to quantization, where only specific values are possible, fundamentally linking the abstract mathematics to real-world phenomena. In our boundary value problem, determining whether it is self-adjoint is a step toward confirming that our approach will lead to legitimate eigenvalue solutions.

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Most popular questions from this chapter

Consider the Sturm-Liouville problem $$ -\left[p(x) y^{\prime}\right]^{\prime}+q(x) y=\lambda r(x) y $$ $$ a_{1} y(0)+a_{2} y^{\prime}(0)=0, \quad b_{1} y(1)+b_{2} y^{\prime}(1)=0 $$ where \(p, q,\) and \(r\) satisfy the conditions stated in the text. (a) Show that if \(\lambda\) is an eigenvalue and \(\phi\) a corresponding eigenfunction, then $$ \lambda \int_{0}^{1} r \phi^{2} d x=\int_{0}^{1}\left(p \phi^{2}+q \phi^{2}\right) d x+\frac{b_{1}}{b_{2}} p(1) \phi^{2}(1)-\frac{a_{1}}{a_{2}} p(0) \phi^{2}(0) $$ provided that \(a_{2} \neq 0\) and \(b_{2} \neq 0 .\) How must this result be modified if \(a_{2}=0\) or \(b_{2}=0\) ? (b) Show that if \(q(x) \geq 0\) and if \(b_{1} / b_{2}\) and \(-a_{1} / a_{2}\) are nonnegative, then the eigenvalue \(\lambda\) is nonnegative. (c) Under the conditions of part (b) show that the eigenvalue \(\lambda\) is strictly positive unless \(q(x)=0\) for each \(x\) in \(0 \leq x \leq 1\) and also \(a_{1}=b_{1}=0\)

The wave equation in polar coordinates is $$ u_{r r}+(1 / r) u_{r}+\left(1 / r^{2}\right) u_{\theta \theta}=a^{-2} u_{t t} $$ Show that if \(u(r, \theta, t)=R(r) \Theta(\theta) T(t),\) then \(R, \Theta,\) and \(T\) satisfy the ordinary differential equations $$ \begin{aligned} r^{2} R^{\prime \prime}+r R^{\prime}+\left(\lambda^{2} r^{2}-n^{2}\right) R &=0 \\ \Theta^{\prime \prime}+n^{2} \Theta &=0 \\\ T^{\prime \prime}+\lambda^{2} a^{2} T &=0 \end{aligned} $$

Determine a formal eigenfunction series expansion for the solution of the given problem. Assume that \(f\) satisfies the conditions of Theorem \(11.3 .1 .\) State the values of \(\mu\) for which the solution exists. $$ y^{\prime \prime}+\mu y=-f(x), \quad y(0)=0, \quad y^{\prime}(1)=0 $$

In each of Problems 12 through 15 use the method of Problem 11 to transform the given equation into the form \(\left[p(x) y^{\prime}\right]'+q(x) y=0\) $$ y^{\prime \prime}-2 x y^{\prime}+\lambda y=0, \quad \text { Hermite equation } $$

In this problem we consider a higher order eigenvalue problem. In the study of transverse vibrations of a uniform elastic bar one is led to the differential equation $$ y^{\mathrm{w}}-\lambda y=0 $$ $$ \begin{array}{l}{\text { where } y \text { is the transverse displacement and } \lambda=m \omega^{2} / E I ; m \text { is the mass per unit length of }} \\\ {\text { the rod, } E \text { is Young's modulus, } I \text { is the moment of inertia of the cross section about an }} \\ {\text { axis through the centroid perpendicular to the plane of vibration, and } \omega \text { is the frequency of }} \\ {\text { vibration. Thus for a bar whose material and geometric properties are given, the eigenvalues }} \\ {\text { determine the natural frequencies of vibration. Boundary conditions at each end are usually }} \\ {\text { one of the following types: }}\end{array} $$ $$ \begin{aligned} y=y^{\prime} &=0, \quad \text { clamped end } \\ y=y^{\prime \prime} &=0, \quad \text { simply supported or hinged end, } \\ y^{\prime \prime}=y^{\prime \prime \prime} &=0, \quad \text { free end } \end{aligned} $$ $$ \begin{array}{l}{\text { For each of the following three cases find the form of the eigenfunctions and the equation }} \\ {\text { satisfied by the eigenvalues of this fourth order boundary value problem. Determine } \lambda_{1} \text { and }} \\ {\lambda_{2}, \text { the two eigenvalues of smallest magnitude. Assume that the eigenvalues are real and }} \\ {\text { positive. }}\end{array} $$ $$ \begin{array}{ll}{\text { (a) } y(0)=y^{\prime \prime}(0)=0,} & {y(L)=y^{\prime \prime}(L)=0} \\ {\text { (b) } y(0)=y^{\prime \prime}(0)=0,} & {y(L)=y^{\prime \prime}(L)=0} \\ {\text { (c) } y(0)=y^{\prime}(0)=0,} & {y^{\prime \prime}(L)=y^{\prime \prime \prime}(L)=0 \quad \text { (cantilevered bar) }}\end{array} $$
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