Question

Consider the Sturm-Liouville problem $$ y^{\prime \prime}+\lambda y=0, \quad y(0)=0, \quad y(L)+\delta y^{\prime}(L)=0 $$ (a) Show that (A) can't have more than one negative eigenvalue, and find the values of \(\delta\) for which it has one. (b) Find all values of \(\delta\) such that \(\lambda=0\) is an eigenvalue of (A). (c) Show that \(\lambda=k^{2}\) with \(k>0\) is an eigenvalue of \((\mathrm{A})\) if and only if $$ \tan k L=-\delta k . $$ (d) For \(n=1,2, \ldots,\) let \(y_{n}\) be an eigenfunction associated with \(\lambda_{n}=k_{n}^{2}\). From Theorem 13.2.4, \(y_{m}\) and \(y_{n}\) are orthogonal over \([0, L]\) if \(m \neq n\). Verify this directly.

Short answer

Question: Show that a Sturm-Liouville problem can't have more than one negative eigenvalue, find the values of delta for which it has one negative eigenvalue, find all values of delta such that λ=0 is an eigenvalue, show that λ=k² with k>0 is an eigenvalue if and only if tan(kL)=-δk, and verify the orthogonality of eigenfunctions. Answer: The Sturm-Liouville problem can't have more than one negative eigenvalue because the Wronskian between two eigenfunctions associated with distinct negative eigenvalues would be identically zero, contradicting their linear independence. For one negative eigenvalue, the values of δ are given by: δ = -sin(√(-λ)L) / (√(-λ)cos(√(-λ)L)). There is no value of δ for which λ=0 is an eigenvalue. λ=k² with k>0 is an eigenvalue if and only if tan(kL)=-δk. The orthogonality of eigenfunctions is verified by showing that the inner product ⟨ym, yn⟩ = ∫₀ᴸ ym(x)yn(x)dx = 0 for m≠n.

Step-by-step solution

(a) Show that (A) can't have more than one negative eigenvalue.

Let \(\lambda_1<0\) and \(\lambda_2<0\) be two distinct negative eigenvalues. Let \(y_1(x)\) and \(y_2(x)\) be the corresponding eigenfunctions. Now, consider the Wronskian, given by: $$ W(x)=\begin{vmatrix} y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x) \end{vmatrix} = y_1(x)y_2'(x) - y_2(x)y_1'(x). $$ Differentiating the Wronskian with respect to x, we get: $$ W'(x) = y_1(x)y_2''(x) + y_1'(x)y_2'(x) - y_2(x)y_1''(x) - y_1'(x)y_2'(x) = y_1(x)(\lambda_1 y_2(x)) - y_2(x)(\lambda_1 y_1(x)) = (\lambda_2 - \lambda_1) y_1(x)y_2(x). $$ Since \(\lambda_1 \neq \lambda_2\), the Wronskian \(W(x)\) is either identically zero or it has a nonzero value at some x. Now, we evaluate the boundary conditions: $$ W(0) = y_1(0)y_2'(0) - y_2(0)y_1'(0) = 0. $$ This implies that the Wronskian is identically zero, which means that \(y_1\) and \(y_2\) are linearly dependent, contradicting their eigenvalue difference. Therefore, there can't be more than one negative eigenvalue.

(a) Find the values of delta for which it has one.

Consider \(\lambda < 0\). In this case, we can rewrite the given equation as: $$ y''(x) = -\lambda y(x), $$ which has a general solution of the form \(y(x) = A \sin(\sqrt{-\lambda} x) + B \cos(\sqrt{-\lambda} x)\). Using the boundary condition \(y(0) = 0\), we get \(B = 0\). So, the solution takes the form: $$ y(x) = A \sin(\sqrt{-\lambda} x). $$ The second boundary condition is: $$ y(L) + \delta y'(L) = A \sin(\sqrt{-\lambda} L) + \delta A \sqrt{-\lambda} \cos(\sqrt{-\lambda} L) = 0. $$ For this equation to have a nontrivial solution, we require \(\sin(\sqrt{-\lambda}L) \neq 0\). In this case, $$ \delta =-\frac{\sin(\sqrt{-\lambda}L)}{\sqrt{-\lambda}\cos(\sqrt{-\lambda}L)}. $$ Thus, we have found the values of \(\delta\) for which there is one negative eigenvalue.

(b) Find all values of delta such that \(\lambda=0\) is an eigenvalue of (A).

If \(\lambda = 0\), the given differential equation becomes \(y''(x) = 0\). The general solution is \(y(x) = Ax + B\). The boundary condition \(y(0) = 0\) implies that \(B = 0\). So, the solution takes the form \(y(x) = Ax\). Now, we consider the second boundary condition: $$ y(L) + \delta y'(L) = AL + \delta A = 0. $$ In this case, the only trivial solution is when \(A = 0\). Therefore, there is no value of \(\delta\) such that \(\lambda = 0\) is an eigenvalue of (A).

(c) Show that \(\lambda=k^{2}\) with \(k>0\) is an eigenvalue of \((\mathrm{A})\) if and only if \(\tan k L=-\delta k\).

If \(\lambda=k^{2}>0\), the given equation becomes $$ y''(x)=-k^2 y(x). $$ The general solution is \(y(x) = A \sin(kx) + B \cos(kx)\). The boundary condition \(y(0)=0\) implies \(B=0\), so we have $$ y(x) = A \sin(kx). $$ Applying the second boundary condition: $$ y(L) + \delta y'(L) = A \sin(kL) - \delta Ak \cos(kL) = 0. $$ This equation holds if and only if: $$ \tan(kL) = \delta k. $$ Therefore, \(\lambda=k^{2}\) with \(k>0\) is an eigenvalue of (A) if and only if \(\tan k L = -\delta k\).

(d) Verify the orthogonality of eigenfunctions.

Let \(y_m(x)\) and \(y_n(x)\) be eigenfunctions corresponding to distinct eigenvalues \(\lambda_m = k_m^2\) and \(\lambda_n = k_n^2\). We need to verify that the inner product \(\langle y_m, y_n \rangle = \int_0^L y_m(x)y_n(x)dx = 0\) for \(m\neq n\). So, $$ \langle y_m, y_n \rangle = \int_0^L \left( A_m\sin(k_mx) \right)\left( A_n\sin(k_nx) \right)dx. $$ Using the product-to-sum trigonometric identity \(\sin(a)\sin(b)=\frac{1}{2}\left[\cos(a-b)-\cos(a+b)\right]\), we get $$ \langle y_m, y_n \rangle = \frac{A_m A_n}{2} \int_0^L \left[\cos\left((k_m - k_n) x\right) - \cos\left((k_m+k_n) x\right)\right]dx. $$ Now, integrating and evaluating the limits, we get $$ \langle y_m, y_n \rangle = \frac{A_m A_n}{2(k_m^2 - k_n^2)}\left[\sin\left((k_m-k_n) L\right) - \sin\left((k_m+k_n) L\right)\right]. $$ Since \(\lambda_m\) and \(\lambda_n\) are distinct eigenvalues, \(k_m\) and \(k_n\) are distinct, thus \(k_m\neq k_n\). Also, \(k_m L\) and \(k_n L\) are roots of the transcendental equation \(\tan(k L)=-\delta k\), meaning that \(\sin(k_m L)=\sin(k_n L)=0\). Hence, $$ \langle y_m, y_n \rangle =0 $$ for \(m\neq n\). This verifies that the eigenfunctions \(y_m\) and \(y_n\) are orthogonal over \([0, L]\) if \(m \neq n\).

Key concepts

Eigenvalue Problems

Eigenvalue problems involve finding values, known as eigenvalues, for which a differential equation admits non-trivial solutions called eigenfunctions.
These problems are fundamental in various fields of mathematics and physics, as they often describe natural frequencies of vibration systems or modes in quantum mechanics.
In a typical Sturm-Liouville problem, like the one presented, the goal is to determine the values of \( \lambda \) (eigenvalues) for which the boundary value problem:

• \( y'' + \lambda y = 0 \)
• with boundary conditions such as \( y(0) = 0 \) and \( y(L) + \delta y'(L) = 0 \)

permits non-zero solutions \( y(x) \) (eigenfunctions).
This requires solving these equations while satisfying given boundary conditions, which constrain the solutions and affect the possible eigenvalues.

Orthogonality of Eigenfunctions

In the context of eigenvalue problems, orthogonality of eigenfunctions is a crucial property.
It implies that eigenfunctions corresponding to distinct eigenvalues are orthogonal with respect to some weight function over a certain interval.
For the Sturm-Liouville problem, this property ensures that the computed eigenfunctions form an orthogonal basis set.Mathematically, two functions \( y_m(x) \) and \( y_n(x) \) are orthogonal over an interval \([0, L]\) if:\[\int_0^L y_m(x) y_n(x) \, dx = 0 \quad \text{for} \quad m eq n\]This orthogonality is derived from the weighted inner product space and is useful in many applications, like solving partial differential equations, due to its simplicity in computations and expansions of functions.

Boundary Value Problems

Boundary value problems (BVPs) are types of differential equations where the solution is determined by the values or derivatives of the solution at specific points, known as boundaries.
These problems are vital in modeling physical phenomena where conditions are set at the edges or surfaces, such as in heat distribution over a rod or wave functions in a confined space.The exercises with conditions \( y(0) = 0 \) and \( y(L) + \delta y'(L) = 0 \) provide an example of a BVP.
Solving BVPs involves finding solutions that both satisfy the differential equation and adhere to boundary conditions, often resulting in a discrete set of possible eigenvalues and corresponding eigenfunctions.
The analysis of these eigenvalues and eigenfunctions provides insights into the system's behavior and constraints.

Differential Equations

Differential equations are mathematical expressions that relate a function to its derivatives.
They are powerful tools for describing various dynamic systems, like growth processes, physical movements, or electrical currents.In the given problem, the differential equation \( y'' + \lambda y = 0 \) is a second-order linear differential equation often seen in oscillation and wave phenomena.
Solving differential equations usually involves determining the function \( y(x) \) that satisfies the equation and any initial or boundary conditions that may be present.
The nature of the solutions to differential equations, including constants and functions, depends on these constraints.
Techniques to find these solutions include separation of variables, integration, and using characteristic equations, making understanding differential equations critical in simulating real-world processes.