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Chapter 22: Problem 20

Estimate the lowest eigenvalue, \(\lambda_{0}\), of the equation $$ \frac{d^{2} y}{d x^{2}}-x^{2} y+\lambda y=0, \quad y(-1)=y(1)=0, $$ using a quadratic trial function.

Short Answer

Expert verified
The lowest eigenvalue, \( \lambda_0 \), is approximately 2.5.

Step by step solution

01

Choose a quadratic trial function

Select a trial function that satisfies the boundary conditions. A suitable quadratic function is \[ y(x) = a(1 - x^2) \] where the coefficient 'a' ensures the boundary conditions are met.
02

Substitute the trial function into the equation

Substitute \[ y(x) = a(1 - x^2) \] into the differential equation \[ \frac{d^2 y}{d x^2} - x^2 y + \lambda y = 0 \].
03

Compute the second derivative of the trial function

Calculate \( \frac{d^2 y}{dx^2} \):\[ \frac{d y}{d x} = -2ax \]\[ \frac{d^2 y}{d x^2} = -2a \]
04

Form the integral for Rayleigh quotient

Form the Rayleigh quotient \[ \lambda = \frac{ \int_{-1}^{1} \left( \left( \frac{d^2 y}{dx^2} - x^2 y \right) y \right) dx }{ \int_{-1}^{1} y^2 dx }. \]
05

Evaluate the numerator of the Rayleigh quotient

Evaluate \[ \int_{-1}^{1} \left( -2a - x^2(a - ax^2) + \lambda a (1 - x^2) \right) a(1 - x^2) dx \].
06

Evaluate the denominator of the Rayleigh quotient

Evaluate \[ \int_{-1}^{1} a^2(1 - x^2)^2 dx \].
07

Simplify and solve for \( \lambda_0 \)

Simplify the integral expressions to find \[ \lambda_0 = 2.5 \].

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Key Concepts

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

differential equations
A differential equation relates a function to its derivatives. In our case, we're dealing with a second-order differential equation: \[ \frac{d^2 y}{d x^2} - x^2 y + \frac{\text{Tyhlambda}}{{rho}_y-intersection} = 0 \]
This equation involves the function \(y\) as well as its second derivative \( \frac{d^2 y}{dx^2} \).
The goal is to find values of \( \frac{interrog-y} \) that satisfy the equation for non-trivial \(y\).
Differential equations are foundational in many fields, explaining everything from natural phenomena to engineering systems.
We solve these equations to understand how functions change and interact based on their rates of change.
trial _unction
A trial function is a guessed solution used to approximate the true solution to a differential equation.
In this problem, we choose a quadratic trial function to simplify calculations while meeting the boundary conditions:
\[y(x) = a(1 - x^2) \]
  • The function must be simple enough to handle mathematically but versatile enough to approximate the true solution.
    The coefficient \(a\) ensures the function fits within specified conditions, making it workable for further computations.

The trial function helps us simplify the complex original differential equation.
Rayleigh quotient
The Rayleigh quotient is used to estimate eigenvalues of differential operators.
The general form is:
\[ \frac{RDDydynamic-go-upu-intergration}{{Upper-integral }} \]
  • Here, we substitute our trial function into this quotient to convert a differential equation into an integral equation.
  • The numerator is an integral involving the differential equation and the trial function.
  • The denominator is an integral of the trial function squared.

By solving these integrals, we can estimate the eigenvalue, providing a route from function approximation to specific numerical solutions.
boundary conditions
Boundary conditions specify the values that a solution to a differential equation must satisfy at certain points.
For our problem, these conditions are:
\[ y(-1) = 1 = y(0) \]
By integrating the trial function with boundary conditions, we determine coefficients and further simplify the problem.
The boundary conditions ensure our chosen trial function suitably models the original problem's constraints.
This leads to more accurate solutions and validates our trial function choice.

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Most popular questions from this chapter

The Hamiltonian \(H\) for the hydrogen atom is $$ -\frac{\hbar^{2}}{2 m} \nabla^{2}-\frac{q^{2}}{4 \pi \epsilon_{0} r} $$ For a spherically symmetric state, as may be assumed for the ground state, the only relevant part of \(\nabla^{2}\) is that involving differentiation with respect to \(r\). (a) Define the integrals \(J_{n}\) by $$ J_{n}=\int_{0}^{\infty} r^{n} e^{-2 \beta r} d r $$ and show that, for a trial wavefunction of the form \(\exp (-\beta r)\) with \(\beta>0\), \(\int \psi^{*} H \psi d v\) and \(\int \psi^{*} \psi d v\) (see exercise \(\left.22.24(\mathrm{a})\right)\) can be expressed as \(a J_{1}-b J_{2}\) and \(c J_{2}\) respectively, where \(a, b\) and \(c\) are factors which you should determine. (b) Show that the estimate of \(E\) is minimised when \(\beta=m q^{2} /\left(4 \pi \epsilon_{0} \hbar^{2}\right)\). (c) Hence find an upper limit for the ground-state energy of the hydrogen atom. In fact, \(\exp (-\beta r)\) is the correct form for the wavefunction and the limit gives the actual value.

You are provided with a line of length \(\pi a / 2\) and negligible mass and some lead shot of total mass \(M\). Use a variational method to determine how the lead shot must be distributed along the line if the loaded line is to hang in a circular arc of radius \(a\) when its ends are attached to two points at the same height. Measure the distance \(s\) along the line from its centre.

The upper and lower surfaces of a film of liquid, which has surface energy per unit area (surface tension) \(\gamma\) and density \(\rho\), have equations \(z=p(x)\) and \(z=q(x)\), respectively. The film has a given volume \(V\) (per unit depth in the \(y\)-direction) and lies in the region \(-L

The refractive index \(n\) of a medium is a function only of the distance \(r\) from a fixed point \(O\). Prove that the equation of a light ray, assumed to lie in a plane through \(O\), travelling in the medium satisfies (in plane polar coordinates) $$ \frac{1}{r^{2}}\left(\frac{d r}{d \phi}\right)^{2}-\frac{r^{2}}{a^{2}} \frac{n^{2}(r)}{n^{2}(a)}-1. $$ where \(a\) is the distance of the ray from \(O\) at the point at which \(d r / d \phi=0\) If \(n=\left[1+\left(\alpha^{2} / r^{2}\right)\right]^{1 / 2}\) and the ray starts and ends far from \(O\), find its deviation (the angle through which the ray is turned), if its minimum distance from \(O\) is \(a\).

A dam of capacity \(V\) (less than \(\left.\pi b^{2} h / 2\right)\) is to be constructed on level ground next to a long straight wall which runs from \((-b, 0)\) to \((b, 0)\). This is to be achieved by joining the ends of a new wall, of height \(h\), to those of the existing wall. Show that, in order to minimise the length \(L\) of new wall to be built, it should form part of a circle, and that \(L\) is then given by $$ \int_{-b}^{b} \frac{d x}{\left(1-\lambda^{2} x^{2}\right)^{1 / 2}} $$ where \(\lambda\) is found from $$ \frac{V}{h b^{2}}=\frac{\sin ^{-1} \mu}{\mu^{2}}-\frac{\left(1-\mu^{2}\right)^{1 / 2}}{\mu} $$ and \(\mu=\lambda \underline{\underline{b}}\).
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