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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

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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.

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

The equation $$ v_{x x}+v_{y y}+k^{2} v=0 $$ is a generalization of Laplace's equation, and is sometimes called the Helmholtz \((1821-1894)\) equation. (a) In polar coordinates the Helmholtz equation is $$v_{r r}+(1 / r) v_{r}+\left(1 / r^{2}\right) v_{\theta \theta}+k^{2} v=0$$ If \(v(r, \theta)=R(r) \Theta(\theta),\) show that \(R\) and \(\Theta\) satisfy the ordinary differential equations $$ r^{2} R^{\prime \prime}+r R^{\prime}+\left(k^{2} r^{2}-\lambda^{2}\right) R=0, \quad \Theta^{\prime \prime}+\lambda^{2} \Theta=0 $$ (b) Consider the Helmholtz equation in the disk \(r

Use the method of Problem 11 to transform the given equation into the form \(\left[p(x) y^{\prime}\right]'+q(x) y=0\) $$ \left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime}+\alpha^{2} y=0, \quad \text { Chebyshev equation } $$

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.

Differ from those in previous problems in that the parameter \(\lambda\) multiplies the \(y^{\prime}\) term as well as the \(y\) term. In each of these problems determine the real eigenvalues and the corresponding eigenfunctions. $$ \begin{array}{l}{x^{2} y^{\prime \prime}-\lambda\left(x y^{\prime}-y\right)=0} \\\ {y(1)=0, \quad y(2)-y^{\prime}(2)=0}\end{array} $$

State whether the given boundary value problem is homogeneous or non homogeneous. $$ -y^{\prime \prime}+x^{2} y=\lambda y, \quad y^{\prime}(0)-y(0)=0, \quad y^{\prime}(1)+y(1)=0 $$

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