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Chapter 5: Problem 2

Consider the cigenvalue problem $$ \frac{d^{\prime} \phi}{d x^{2}}+\left(\lambda-x^{2}\right) \phi=0 $$ subject to \(\frac{d \phi}{d x}(0)=0\) and \(\frac{d \phi}{d x}(1)=0 .\) Show that \(\lambda>0\) (be sure to show that \(\lambda \neq 0\) ).

Short Answer

Expert verified
Through analysis of the differential equation and application of the boundary conditions, it is shown that the eigenvalue \( \lambda \) must be greater than zero for solutions to exist.

Step by step solution

01

Assume a solution and plug it into the given equation

To solve the given differential equation, assume a solution of the form \( \phi(x) = e^{(-x^{2}/2)}u(x) \). Here \( u(x) \) is a function we will determine. Plugging this into the differential equation results in a simpler differential equation, \( \frac{d^{2}u}{dx^{2}} - \lambda u = 0\).
02

Apply boundary conditions

Next, apply the given boundary conditions \( \frac{d \phi}{dx}(0)=0 \) and \( \frac{d \phi}{dx}(1)=0 \). Substituting our \( \phi(x) \) into these gives us \( u'(0) = 0 \) and \( u'(1) + u(1) = 0 \).
03

Analyze the differential equation

Now look at the simpler differential equation to find the conditions for \( \lambda \). The general solutions to the differential equation \( \frac{d^{2}u}{dx^{2}} - \lambda u = 0\) are trigonometric functions for \( \lambda > 0 \), exponential functions for \( \lambda = 0 \) and hyperbolic functions for \( \lambda < 0 \). We need to find which type of solutions can satisfy our boundary conditions.
04

Dismiss the option for \( \lambda = 0 \) and \( \lambda < 0 \)

For \( \lambda = 0 \), the general solutions are exponential functions. However, these will result in diverging functions at \( x = 1 \) which cannot satisfy \( u'(1) + u(1) = 0 \). For \( \lambda < 0 \) the general solutions are hyperbolic functions which also cannot satisfy both boundary conditions. Thus, \( \lambda \) cannot be zero or negative.
05

Show that \( \lambda > 0 \) can work

For \( \lambda > 0 \), the general solutions are sinusoidal functions which can meet both boundary conditions. Thus, \( \lambda \) must be greater than zero.

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Key Concepts

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

Differential Equation
A differential equation is a mathematical equation that involves unknown functions and their derivatives. In the context of physics and engineering, these equations often describe how physical quantities change over space and time. The eigenvalue problem given in the exercise is an example of a second-order linear homogeneous differential equation. In this case, we deal with the function \( \phi(x) \) and its second derivative \( \phi''(x) \).

To solve it, one typically looks for solutions that can be expressed in terms of familiar functions. In this exercise, the assumption is made that the solution can be expressed as \( \phi(x) = e^{-x^2/2}u(x) \), where \( u(x) \) is another function to be determined. This assumption simplifies the differential equation, transforming it into a format that is more straightforward to handle. By replacing the original function with this new product, we greatly simplify the solving process and align it with the prescribed boundary conditions.
Boundary Conditions
Boundary conditions are constraints necessary for determining a unique solution to a differential equation. They are essential when dealing with eigenvalue problems like the one presented, as they provide necessary information about the behavior of the function at specific points.

For our example, the boundary conditions given are \( \frac{d \phi}{d x}(0)=0 \) and \( \frac{d \phi}{d x}(1)=0 \). These are known as Neumann boundary conditions, which specify the gradient of the function at the boundaries. In the context of this problem, they help us to further constrain the form of the function \( u(x) \) that we introduced earlier. When the boundary conditions are applied to the theorized solution of \( \phi(x) \) in the differential equation, they ultimately allow us to determine that the eigenvalue \( \lambda \) must be positive. This is because only certain types of functions, in this case, trigonometric functions, will satisfy both the differential equation and the boundary conditions.
Trigonometric Functions
Trigonometric functions appear frequently as solutions to differential equations in physics and engineering, especially in problems involving wave-like or periodic phenomena. Sinusoidal wave functions are an example of trigonometric functions, characterized by their sines and cosines.

In the context of the eigenvalue problem we're considering, when \( \lambda > 0 \) the solutions to the simplified differential equation are trigonometric. Only solutions of this type, sinusoidal functions, have the oscillatory properties needed to satisfy the specific boundary conditions of the problem. The nature of these functions allows them to have gradients equal to zero at certain points which align with the Neumann boundary conditions provided. It鈥檚 the unique property of the trigonometric functions to oscillate and repeat their values that makes them fitting solutions for the eigenvalue problem, assuming \( \lambda > 0 \).

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Most popular questions from this chapter

Consider $$ L=\frac{d^{2}}{d x^{2}}+6 \frac{d}{d x}+9 $$ (a) Show that \(L\left(e^{r x}\right)=(r+3)^{2} e^{r \pi}\). (b) Use part (a) to obtain solutions of \(L(y)-0\) (a second order constant coefficient differential equation). (c) If \(z\) depends on \(x\) and a parameter \(r\), show that $$ \frac{\partial}{\partial r} L(z)=L\left(\frac{\partial z}{\partial r}\right) \text {. } $$ (d) Using part (c), evaluate \(L(\partial z / \partial r)\) if \(z-e^{r x}\). (e) Obtain a second solution of \(L(y)=0\), using part (d).

Use the Rayleigh quotient to obtain a (reasonably accurate) upper bound for the lowest eigenvalue of (a) \(\frac{d^{2} \phi}{d x^{2}}+\left(\lambda-x^{2}\right) \phi=0 \quad\) with \(\frac{d \phi}{d x}(0)=0 \quad\) and \(\quad \phi(1)=0\) (b) \(\frac{d^{2} \phi}{d x^{2}}+(\lambda-x) \phi=0\) with \(\frac{d \phi}{d x}(0)=0\) and \(\frac{d \phi}{d x}(1)+2 \phi(1)=0\) *(c) \(\frac{d^{2} \phi}{d x^{2}}+\lambda \phi-0\) with \(\phi(0)-0\) and \(\frac{d \phi}{d x}(1)+\phi(1)-0\) (See Exercise 5.8.10.)

Consider the non-Sturm-Liouville differential equation $$ \frac{d^{2} \phi}{d x^{2}}+\alpha(x) \frac{d \phi}{d x}+[\lambda \beta(x)+\gamma(x)] \phi=0 . $$ Multiply this equation by \(H(x)\). Determine \(H(x)\) such that the equation may be reduced to the standard Sturm-Liouville form: $$ \frac{d}{d x}\left[p(x) \frac{d \phi}{d x}\right]+[\lambda \sigma(x)+q(x)] \phi=0 . $$ Given \(\alpha(x), \beta(x)\), and \(\gamma(x)\), what are \(p(x), \sigma(x)\), and \(q(x)\) ?

Consider the boundary value problem $$ \frac{d^{2} \phi}{d x^{2}}+\lambda \phi=0 \quad \text { with } $$ $$ \phi(1)+\frac{d \phi}{d x}(1)=0 . $$ (a) Using the Rayleigh quotient, show that \(\lambda \approx 0 .\) Why is \(\lambda>0\) ? (b) Prove that eigenfunctions corresponding to different eigenvalues arc (c) Show that $$ \frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}} $$ $$ u(0, t)-\frac{\partial u}{\partial x}(0, t)-0 $$ (b) Prove that eigenfunctions corresponding to different eigenvalues are orthogonal. *(c) Show that Determine the eigenvalues graphically. Estimate the large eigenvalues. (d) Solve with $$ \begin{aligned} u(0, t)-\frac{\partial u}{\partial x}(0, t) &=0 \\ u(1, t)+\frac{\partial u}{\partial x}(1, t) &=0 \\ u(x, 0) &=f(x) \end{aligned} $$ You may call the relevant eigenfunctions \(\phi_{n}(x)\) and assume that they are known.

Prove that if \(x\) is a root of a sixth-order polynomial with real coefficients, then \(x\) is also a root.
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