Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Rewrite the equation in Sturm-Liouville form (with \(\lambda=0) .\) Assume that \(b, c, \alpha,\) and \(\nu\) are contants. $$ x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-\nu^{2}\right) y=0 $$

Short Answer

Expert verified
Question: Rewrite the given differential equation in Sturm-Liouville form with λ = 0: $$ x^2y'' + xy' + (x^2 - \nu^2)y = 0 $$ Answer: The Sturm-Liouville form of the given equation is: $$ \frac{d}{dx}\left[x^2 \frac{dy}{dx}\right] + (x^2 - \nu^2)y = 0 $$ Where p(x) = x², q(x) = x² - ν², and r(x) = 1.

Step by step solution

01

Matching terms with Sturm-Liouville form

Let's notice that which terms in the Sturm-Liouville form match with the given equation: For the first term in the Sturm-Liouville form: $$ \frac{d}{dx}\left[p(x) \frac{dy}{dx}\right] \Rightarrow x^{2} y^{\prime \prime}+x y^{\prime} $$ For the second term: $$ q(x)y \Rightarrow (x^2 - \nu^2)y $$ Now let's find p(x), q(x), and r(x) functions.
02

Identify p(x)

To identify p(x), we will look at the first term of the Sturm-Liouville form, which is: $$ \frac{d}{dx}\left[p(x) \frac{dy}{dx}\right] $$ The equation is already given in the form where we can identify p(x): $$ x^{2} y^{\prime \prime} + x y^{\prime} $$ We can rewrite the above equation as: $$ \frac{d}{dx}\left[x^{2} \frac{dy}{dx}\right] $$ Therefore, p(x) is: $$ p(x) = x^2 $$
03

Identify q(x)

To identify q(x), we will look at the second term of the Sturm-Liouville form, which is: $$ q(x) y \Rightarrow (x^2 - \nu^2) y $$ Comparing this with the given equation term (x² - ν²)y, we can directly identify q(x) as: $$ q(x) = x^2 - \nu^2 $$
04

Identify r(x) and write the final equation

Since λ = 0, r(x) doesn't have an effect on the equation, but for the sake of completeness, let's identify r(x). In the Sturm-Liouville form, r(x)y term should include lambda: $$ \lambda r(x)y $$ Since λ=0, r(x) can be any function, for example: $$ r(x) = 1 $$ Now we can rewrite the original equation in Sturm-Liouville form: $$ \frac{d}{dx}\left[x^2 \frac{dy}{dx}\right] + (x^2 - \nu^2)y = 0 $$ Where p(x) = x², q(x) = x² - ν², and r(x) = 1.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Differential equations
Differential equations are equations that involve a function and its derivatives. They play a crucial role in describing physical phenomena such as motion, heat, light, and more. In general, solving a differential equation means finding the function or set of functions that satisfy the equation.In our particular example, we have a second-order differential equation given by:\[ x^2 y'' + x y' + (x^2 - u^2) y = 0 \]This type of equation is common in physics and engineering, and requires specific methods for solution depending on its form and constraints. Differential equations can be quite complex, so it's important to identify their order (the highest derivative that appears) and type. This helps determine the best approach to find their solutions.
Sturm-Liouville form
The Sturm-Liouville form is a specific way of writing linear second-order differential equations that helps in solving them using standard methods. It is particularly useful in problems involving eigenvalues and eigenfunctions, which are important concepts in many areas of physics and engineering.An equation is in Sturm-Liouville form if it can be expressed as:\[ \frac{d}{dx}\left[ p(x) \frac{dy}{dx} \right] + q(x)y = \lambda r(x)y \]Here, \( p(x) \), \( q(x) \), and \( r(x) \) are known functions, and \( \lambda \) is a parameter often related to eigenvalues. By rewriting our given equation in this form with \( \lambda = 0 \), we simplify the process of finding its solutions while preserving essential properties of the function \( y \).
Eigenvalue problems
Eigenvalue problems in differential equations involve finding characteristic values, called eigenvalues, for which the differential equation has non-trivial solutions, known as eigenfunctions.These problems often arise in the context of the Sturm-Liouville form of differential equations, where \( \lambda \) represents the eigenvalue. Solutions to such problems are significant in various scientific fields as they can represent modes of vibration, frequencies, or energy levels in quantum mechanics.Though the step-by-step solution in this exercise assumes \( \lambda = 0 \), making it a special case, it highlights the general method of how these problems are studied: by first identifying the standard form, and then resolving or simplifying the equation to discover relevant eigenvalues and eigenfunctions.
p(x), q(x), r(x) identification
Identifying the functions \( p(x) \), \( q(x) \), and \( r(x) \) is crucial to reformulating a differential equation into the Sturm-Liouville form. In this exercise, these functions are derived directly from comparing the terms of the given differential equation with the standard Sturm-Liouville form.
  • \( p(x) \): This is the function that precedes the second derivative term. In our example, the expression \( \frac{d}{dx}\left[x^2 \frac{dy}{dx}\right] \) allows us to identify \( p(x) = x^2 \).
  • \( q(x) \): This function is associated with the \( y \) term itself. By identifying \((x^2 - u^2)y\) as the term in question, we set \( q(x) = x^2 - u^2 \).
  • \( r(x) \): This is typically a positive function included with \( \lambda y \). Since \( \lambda = 0 \) in our problem, \( r(x) \) can be any constant, hence it is taken as \( r(x) = 1 \) to keep things simple.
Understanding these identifications helps in re-writing the differential equation in a manner that is simpler to solve and analyze.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Deal with the Sturm-Liouville problem $$ y^{\prime \prime}+\lambda y=0, \quad \alpha y(0)+\beta y^{\prime}(0), \quad \rho y(L)+\delta y^{\prime}(L)=0 $$ where \(\alpha^{2}+\beta^{2}>0\) and \(\rho^{2}+\delta^{2}>0\) The point of this exercise is that (SL) can't have more than two negative eigenvalues. (a) Show that \(\lambda\) is a negative eigenvalue of (SL) if and only if \(\lambda=-k^{2}\), where \(k\) is a positive solution of $$ \left(\alpha \rho-\beta \delta k^{2}\right) \sinh k L+k(\alpha \delta-\beta \rho) \cosh k L $$ (b) Suppose \(\alpha \delta-\beta \rho=0\). Show that (SL) has a negative eigenvalue if and only if \(\alpha \rho\) and \(\beta \delta\) are both nonzero. Find the negative eigenvalue and an associated eigenfunction. HINT: Show that in this case \(\rho=p \alpha\) and \(s=q \beta,\) where \(q \neq 0\) (c) Suppose \(\beta \rho-\alpha \delta \neq 0\). We know from Section 11.1 that (SL) has no negative eigenvalues if \(\alpha \rho=0\) and \(\beta \delta=0\). Assume that either \(\alpha \rho \neq 0\) or \(\beta \delta \neq 0\). Then we can rewrite (A) as $$ \tanh k L=\frac{k(\beta \rho-\alpha \delta)}{\alpha \rho-\beta \delta k^{2}} $$ By graphing both sides of this equation on the same axes (there are several possibilities for the right side), show that it has at most two positive solutions, so (SL) has at most two negative eigenvalues.

Deal with the Sturm-Liouville problem $$ y^{\prime \prime}+\lambda y=0, \quad \alpha y(0)+\beta y^{\prime}(0), \quad \rho y(L)+\delta y^{\prime}(L)=0 $$ where \(\alpha^{2}+\beta^{2}>0\) and \(\rho^{2}+\delta^{2}>0\) Show that \(\lambda=0\) is an eigenvalue of (SL) if and only if $$ \alpha(\rho L+\delta)-\beta \rho=0 $$

If the boundary value problem has a solution for every continuous \(F\), then find the Green's function for the problem and use it to write an explicit formula for the solution. Otherwise, if the boundary value problem does not have a solution for every continuous \(F,\) find a necessary and sufficient condition on \(F\) for the problem to have a solution, and find all solutions. Assume that \(a

Take it as given that \(\left\\{x e^{k x}, x e^{-k x}\right\\}\) and \(\\{x \cos k x, x \sin k x\\}\) are fundamental sets of solutions of $$ x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y-k^{2} x^{2} y=0 $$ and $$ x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y+k^{2} x^{2} y=0 $$ respectively. Find the first five eigenvalues of $$ x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y+\lambda x^{2} y=0, \quad y^{\prime}(1)=0, \quad y(2)=0 $$ with errors no greater than \(5 \times 10^{-8}\). State the form of the associated eienfunctions.

In Example 13.2 .4 we found that the eigenvalue problem $$ x^{2} y^{\prime \prime}+x y^{\prime}+\lambda y=0, \quad y(1)=0, \quad y(2)=0 $$ is equivalent to the Sturm-Liouville problem $$ \left(x y^{\prime}\right)^{\prime}+\frac{\lambda}{x} y=0, \quad y(1)=0, \quad y(2)=0 $$ Multiply the differential equation in (B) by \(y\) and integrate to show that $$ \lambda \int_{1}^{2} \frac{y^{2}(x)}{x} d x=\int_{1}^{2} x\left(y^{\prime}(x)\right)^{2} d x $$ Conclude from this that the eigenvalues of (A) are all positive.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free