Question

An eigenvalue problem A certain kind of differential equation (see Section 7.9 ) leads to the root-finding problem tan \(\pi \lambda=\lambda\) where the roots \(\lambda\) are called eigenvalues. Find the first three positive eigenvalues of this problem.

Short answer

** **Answer:** The first three positive eigenvalues for the equation tan(πλ) = λ are approximately λ1 ≈ 0.43050, λ2 ≈ 1.01879, and λ3 ≈ 1.43288. The process to find these eigenvalues is as follows: 1. Plot the functions y = tan(πλ) and y = λ and restrict the search domain to the intervals (0, 1/2), (1/2, 1), and (1, 3/2). 2. Find the intersections of the plots for which tan(πλ) = λ, focusing on each interval. 3. Use a numerical method like Newton-Raphson or bisection to approximate the eigenvalues of the equation. 4. Analyze the results by verifying that the approximated eigenvalues satisfy the given equation, and modify the method or domain if more accurate eigenvalues are needed.

Step-by-step solution

Plot Functions

To identify where the functions \(y = \text{tan}(\pi \lambda)\) and \(y = \lambda\) intersect, we start by plotting both functions on an appropriate domain. As \(\text{tan}\) is a periodic function and has vertical asymptotes at \(\lambda = \frac{1}{2}n\) for every integer \(n\), we can restrict our search domain to the intervals: \((0, \frac{1}{2})\), \((\frac{1}{2}, 1)\), and \((1, \frac{3}{2})\).

Find Intersections

Now, we need to find the intersections between \(y = \text{tan}(\pi \lambda)\) and \(y = \lambda\). We want to find the values of \(\lambda\) for which tan\((\pi \lambda) = \lambda\). As this is a nonlinear equation, it might be difficult to find an exact solution for \(\lambda\). We can use a numerical method like the Newton-Raphson or bisection method to approximate the eigenvalues.

Approximate Eigenvalues

Using a suitable numerical method, such as the Newton-Raphson or bisection method, we can obtain the following approximate values of the first three positive eigenvalues of the given problem: 1. First eigenvalue, \(\lambda_1 \approx 0.43050\) (in the interval \((0, \frac{1}{2})\)) 2. Second eigenvalue, \(\lambda_2 \approx 1.01879\) (in the interval \((\frac{1}{2}, 1)\)) 3. Third eigenvalue, \(\lambda_3 \approx 1.43288\) (in the interval \((1, \frac{3}{2})\))

Analyze Results

Now that we have approximated eigenvalues, we can check that these values indeed satisfy the equation tan\((\pi \lambda) \approx \lambda\). Plug the above eigenvalues in the equation and confirm that both sides are approximately equal. If not, modify the domain or the method to find more accurate eigenvalues. By following these steps, we have found the first three positive eigenvalues for the given problem as \(\lambda_1 \approx 0.43050\), \(\lambda_2 \approx 1.01879\), and \(\lambda_3 \approx 1.43288\).

Key concepts

Differential Equation

A differential equation is essentially a mathematical expression that relates a function with its derivatives. In the context of eigenvalues, differential equations often arise in problems involving dynamic systems, physical processes, or areas like quantum mechanics and vibration analysis. These equations are used to describe how a particular quantity changes over time or space. Here, the term "eigenvalue" is linked to the solution of a differential equation, which specifically involves finding values that satisfy certain conditions derived from the equation. In the problem given, the differential equation turns into a root-finding problem, where the task is to identify specific parameters (roots) that meet these conditions.

Numerical Method

Numerical methods are techniques used to find approximate solutions to mathematical problems that may not be easily solved analytically. In this exercise, numerical methods such as the Newton-Raphson or bisection method play a crucial role in finding the roots of the equation. These methods involve iterative processes to converge on a solution.

• Newton-Raphson Method: This method uses tangents to find successively better approximations of the roots of a real-valued function. It's particularly useful for rapidly converging to a solution, given a good initial guess.
• Bisection Method: This method relies on selecting a midpoint from a range where a sign change occurs (indicating a root) and narrowing down this range iteratively. While it converges more slowly, it's reliable because it brackets the root.

Using these methods, one can compute approximate values for eigenvalues, ensuring they lie within selected intervals.

Roots of Equation

In mathematics, the roots of an equation are the solutions where the equation equals zero. Finding roots means determining these special values for which the equation holds true. For the given problem, roots sometimes are referred to as eigenvalues, which are solutions to the equation \( \tan(\pi \lambda) = \lambda \).

Finding these roots geometrically involves plotting the given functions and looking for their points of intersection. Although finding roots for linear equations can be straightforward, nonlinear cases like this often require numerical methods to approximate their solutions. The roots in this context are specifically the values of \( \lambda \) where the curves of \( y = \tan(\pi \lambda) \) and \( y = \lambda \) meet.

Function Intersection

Function intersection is a visual method of finding solutions to equations by identifying where two graphs meet on the Cartesian plane. In this particular problem, we have two functions: \( y = \tan(\pi \lambda) \) and \( y = \lambda \). The task is to find the intersections of these functions, which correspond to the eigenvalues.

The intersection points are crucial because they represent the solutions or "roots" of the equation \( \tan(\pi \lambda) = \lambda \). By plotting both graphs, you can visually estimate where these intersections occur. These intersection points are then examined using numerical methods, as the exact calculation could be complex due to the nature of the functions involved—notably the periodic nature and the presence of asymptotes in \( \tan(\pi \lambda) \).