Question

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) has infinitely many positive eigenvalues \(\lambda_{1}<\lambda_{2}<\) \(\cdots<\lambda_{n}<\cdots,\) and that \(\lim _{n \rightarrow \infty} \lambda_{n}=\infty\) (a) Show that \(\lambda\) is a positive eigenvalue of (SL) if and only if \(\lambda=k^{2},\) where \(k\) is a positive solution of $$ \left(\alpha \rho+\beta \delta k^{2}\right) \sin k L+k(\alpha \delta-\beta \rho) \cos k L=0 $$ (b) Suppose \(\alpha \delta-\beta \rho=0 .\) Show that the positive eigenvalues of (SL) are \(\lambda_{n}=(n \pi / L)^{2}, n=1\), \(2,3, \ldots .\) HINT: Recall the hint in Exercise \(29(\mathbf{b})\) Now suppose \(\alpha \delta-\beta \rho \neq 0 .\) From Section \(11.1,\) if \(\alpha \rho=0\) and \(\beta \delta=0,\) then (SL) has the eigenvalues $$ \lambda_{n}=[(2 n-1) \pi / 2 L]^{2}, \quad n=1,2,3, \ldots $$ (why?), so let's suppose in addition that at least one of the products \(\alpha \rho\) and \(\beta \delta\) is nonzero. Then we can rewrite \((\mathrm{A})\) as $$ \tan k L=\frac{k(\beta \rho-\alpha \delta)}{\alpha \rho-\beta \delta k^{2}} $$ By graphing both sides of this equation on the same axes (there are several possibilities for the right side), convince yourself of the following: (c) If \(\beta \delta=0,\) there's a positive integer \(N\) such that (B) has one solution \(k_{n}\) in each of the intervals $$ ((2 n-1) \pi / L,(2 n+1) \pi / L)), \quad n=N, N+1, N+2, \ldots $$ and either $$ \lim _{n \rightarrow \infty}\left(k_{n}-\frac{(2 n-1) \pi}{2 L}\right)=0 \quad \text { or } \quad \lim _{n \rightarrow \infty}\left(k_{n}-\frac{(2 n+1) \pi}{2 L}\right)=0 $$ (d) If \(\beta \delta \neq 0,\) there's a positive integer \(N\) such that (B) has one solution \(k_{n}\) in each of the intervals (C) and $$ \lim _{n \rightarrow \infty}\left(k_{n}-\frac{n \pi}{N}\right)=0 $$

Short answer

In the given Sturm-Liouville problem, we first rewrite the equation with the boundary conditions and solve the differential equation. We then apply the boundary conditions and find the eigenvalues. For part (b), we show that if \(\alpha \delta - \beta \rho = 0\), then the positive eigenvalues are given by $$\lambda_n = \left(\frac{n\pi}{L}\right)^2, \ \ n = 1, 2, 3, \ldots$$ For parts (c) and (d), we analyze the intervals containing solutions for the equation $$\tan(kL) = \frac{k(\beta \rho - \alpha \delta)}{\alpha \rho - \beta \delta k^2}$$ In part (c), if \(\beta \delta = 0\), we find the limit of \(k_n\) as \(n\) tends to infinity with respect to the intervals. In part (d), if \(\beta \delta \neq 0\), we again find the limit of \(k_n\) as \(n\) tends to infinity with respect to the intervals.

Step-by-step solution

Rewrite the equation with the boundary conditions

First, let's rewrite the Sturm-Liouville equation with the given boundary conditions: $$y^{\prime \prime}+\lambda y=0, \quad \alpha y(0)+\beta y^{\prime}(0)=0, \quad \rho y(L)+\delta y^{\prime}(L)=0$$

Solve the differential equation

We know the solution for \(y^{\prime \prime} + \lambda y = 0\) is given by: $$y(x) = A \sin(kx) + B \cos(kx)$$ where \(k^2 = \lambda\).

Apply the boundary conditions and solve for eigenvalues

Now we'll apply the boundary conditions to the equation: For the first boundary condition, \(αy(0) + βy'(0) = 0\): $$\alpha (B) + \beta (Ak) = 0$$ For the second boundary condition, \(ρy(L) + δy'(L) = 0\): $$\rho(A \sin(kL) + B \cos(kL)) + \delta(Ak \cos(kL) - Bk \sin(kL)) = 0$$ Multiplying the first equation by \(k\) and subtracting it from the second equation, we get: $$k(\alpha \rho + \beta \delta k^2) \sin(kL) + k(\alpha \delta - \beta \rho) \cos(kL) = 0$$ Following the eigenvalue conditions given in part (a), this is equivalent to having \(\lambda = k^2\).

Solve part (b) of the problem and show the positive eigenvalues

In this step, we show that if \(\alpha \delta - \beta \rho = 0\), then the positive eigenvalues are given by $$\lambda_n = \left(\frac{n\pi}{L}\right)^2, \ \ n = 1, 2, 3, \ldots$$ If \(\alpha \delta - \beta \rho = 0\), then the equation we derived in Step 3 simplifies to $$k(\alpha \rho + \beta \delta k^2) \sin(kL) = 0$$ Recall that \(\alpha^2 + \beta^2 > 0\) and \(\rho^2 + \delta^2 > 0\), so the term \(\alpha \rho + \beta \delta k^2\) is nonzero. This means for the equation to equal zero, \(\sin(kL)\) must be zero. This happens when $$kL = n\pi, \ \ n = 1, 2, 3, \ldots$$ Thus, the eigenvalues are given by $$\lambda_n = k^2 = \left(\frac{n\pi}{L}\right)^2, \ \ n = 1, 2, 3, \ldots$$

Solve part (c) and (d) of the problem

For parts (c) and (d), we need to analyze the intervals containing solutions for the equation $$\tan(kL) = \frac{k(\beta \rho - \alpha \delta)}{\alpha \rho - \beta \delta k^2}$$ (c) If \(\beta \delta = 0\), the equation becomes $$\tan(kL) = \frac{k(\beta \rho - \alpha \delta)}{\alpha \rho}$$ By graphing both sides of this equation, it is observed that there exist positive integer N such that there is one solution \(k_n\) in each of the intervals $$((2n - 1)\frac{\pi}{L}, (2n + 1)\frac{\pi}{L}), \ \ n = N, N + 1, N + 2, \ldots$$ and either $$\lim_{n \rightarrow \infty}\left(k_n - \frac{(2n - 1)\pi}{2L}\right) = 0 \ \ \text{or} \ \ \lim_{n \rightarrow \infty}\left(k_n - \frac{(2n + 1)\pi}{2L}\right) = 0$$ (d) If \(\beta \delta \neq 0\), the equation is $$\tan(kL) = \frac{k(\beta \rho - \alpha \delta)}{\alpha \rho - \beta \delta k^2}$$ By graphing both sides of this equation, it is observed that there exist a positive integer N such that there is one solution \(k_n\) in each of the intervals given in part (c), and $$\lim_{n \rightarrow \infty}\left(k_n - \frac{n\pi}{N}\right) = 0$$

Key concepts

Eigenvalues in the Sturm-Liouville Problem

In the context of the Sturm-Liouville problem, eigenvalues relate to the values of \(\lambda\) for which there exists a non-trivial solution \(y(x)\) of the differential equation that also satisfies the given boundary conditions. The concept of eigenvalues is crucial as it informs us about the characteristic frequencies or modes of a system, which can be a physical system, like a vibrating string, or a more abstract mathematical construct.

Let's examine the exercise scenario. We have a second-order linear homogeneous differential equation where \(\lambda\) acts as the eigenvalue. Solving such problems requires us to find values of \(\lambda\) that allow the differential equation to have non-zero solutions—these are our eigenvalues. Specifically, in the problem we’re reviewing, these eigenvalues are required to be positive and grow indefinitely large as the index \(n\) increases.

To ensure that a student understands this concept fully, they must appreciate that finding the eigenvalues involve solving transcendental equations, which typically don't have simple algebraic solutions and may require numerical or graphical methods for their estimation. Moreover, it's crucial to acknowledge the importance of the uniqueness of eigenvalues in problems with specific boundary conditions. In our problem, the solution hints towards the fact that having distinct intervals for \(k_n\), leading to distinct eigenvalues, is related to the properties of the sine and cosine functions within the boundary value problem.

Understanding the Role of Differential Equations

At the heart of the Sturm-Liouville problem is a differential equation, specifically a linear second-order homogenous differential equation. In mathematics and physical sciences, differential equations are used to describe the relationship between a function and its derivatives, indicating how a particular quantity changes over time or space.

The characteristic equation for our given problem, \(y'' + \lambda y = 0\), signifies a classic scenario where the second derivative of \(y\) with a scaled version of \(y\) equates to zero. Students often recognize this as the harmonic oscillator equation, which solutions are a combination of sine and cosine functions, nicely tying into wave phenomena in physics or oscillations in engineering.

In the steps provided, the solution clearly shows how to integrate the boundary conditions into the differential equation itself in order to solve for \(\lambda\). However, to improve student understanding, consider referencing real-world systems that can be modeled by such equations, like the vibrations of a drumhead or the thermal distribution in a rod. By contextualizing differential equations with familiar concepts, students may find the abstract subject matter more intuitive and relatable, enhancing their grasp of the problem-solving process.

The Importance of Boundary Conditions

Boundary conditions in the study of Sturm-Liouville problems aren't just additional pieces of information; they are fundamental constraints that the solution to the differential equation must satisfy. We can think of boundary conditions as the 'rules' that govern the behavior of a system at its edges or extremes, which, for this problem, are the points \(x=0\) and \(x=L\).

In the case of our exercise, the boundary conditions play a pivotal role in finding the allowed eigenvalues. They essentially close the system and turn an ordinary differential equation into a Sturm-Liouville problem. It's imperative that students understand why \(\alpha^2 + \beta^2 > 0\) and \(\rho^2 + \delta^2 > 0\) are necessary: they ensure that the boundary conditions are not trivial and, therefore, define a well-posed problem.

An excellent way to solidify the understanding of boundary conditions is to explore how changing these conditions alters the resulting eigenvalues. This can lead to discussions about the physical implications of these conditions and how, in the real world, these could correspond to fixing the end of a string (as in a guitar), allowing free movement (like in an air column in an organ pipe), or any number of other physical configurations. Empowering students with this knowledge fosters a deeper comprehension of how mathematical models correlate with physical systems.