Question

Consider the boundary value problem $$ y^{\prime \prime}-2 y^{\prime}+(1+\lambda) y=0, \quad y(0)=0, \quad y(1)=0 $$ $$ \begin{array}{l}{\text { (a) Introduce a new dependent variable } u \text { by the relation } y=s(x) u \text { . Determine } s(x) \text { so }} \\ {\text { that the differential equation for } u \text { has no } u \text { 'term. }} \\\ {\text { (b) Solve the boundary value problem for } u \text { and thereby determine the eigenvalues and }} \\ {\text { eigenfunctions of the original problem. Assume that all eigenvalues are real. }} \\ {\text { (c) Also solve the given problem directly (without introducing } u \text { ). }}\end{array} $$

Short answer

Based on the given step by step solution, the eigenvalues for the given boundary value problem are \(\lambda_n = -(n^2\pi^2 + 1)\) for n = 1, 2, 3, ... and the corresponding eigenfunctions are \(y_n(x) = e^x \sin(n\pi x)\). The introduction of the new dependent variable u helped to simplify the original problem and eventually led to finding the eigenvalues and eigenfunctions directly.

Step-by-step solution

(a) Introduce a new dependent variable u and find s(x)

Let y = s(x)u, then the derivatives of y are: $$ y' = s'(x)u + s(x)u' \\ y'' = s''(x)u + 2s'(x)u' + s(x)u'' $$ Substitute these expressions into the given differential equation: $$ (s''(x)u + 2s'(x)u' + s(x)u'') - 2(s'(x)u + s(x)u') + (1 + \lambda)(s(x)u) = 0 $$ Now, to eliminate the term with u', we want the coefficient of u' to be zero. That is, $$ 2s'(x) - 2s(x) = 0 $$ This is a first-order linear differential equation for s(x). We can solve it by separating variables and integrating both sides: $$ \frac{ds(x)}{s(x)} = dx \\ \int\frac{ds(x)}{s(x)} = \int dx \\ \ln{s(x)} = x + C_1 \\ s(x) = e^{x+C_1} = e^x e^{C_1} $$ As s(0)=1, we have \(e^{C_1}=1\), so \(s(x) = e^x\). Now we can find the differential equation for u: $$ e^x u'' - (1 + \lambda)e^x u = 0 \\ u'' - (1 + \lambda)u = 0 $$

(b) Solve the boundary value problem for u

Now we have a homogeneous linear second-order differential equation with constant coefficients for u: $$ u'' - (1+\lambda)u = 0 $$ Let's find its characteristic equation: $$ r^2 - (1+\lambda) = 0 $$ The discriminant is: \(\Delta = (1 + \lambda)\). Considering that all eigenvalues are real, we may classify the problem based on the discriminant sign: 1. \(\Delta > 0\): Distinct Real Roots $$ r_1 = \sqrt{1+\lambda};\ r_2 = -\sqrt{1+\lambda} $$ The general solution for u in this case is: $$ u(x) = A_1 e^{\sqrt{1+\lambda}x} + A_2 e^{-\sqrt{1+\lambda}x} $$ Applying boundary conditions, u(0) = 0 and u(1) = 0: $$ u(0) = A_1 + A_2 = 0 \\ u(1) = A_1 e^{\sqrt{1+\lambda}} + A_2 e^{-\sqrt{1+\lambda}} = 0 $$ From the first equation, we get \(A_2 = -A_1\). Plugging this into the second equation: $$ A_1(e^{\sqrt{1+\lambda}} - e^{-\sqrt{1+\lambda}}) = 0 $$ As \(A_1\) cannot be zero (since u(x) would be identically zero), we must have: $$ e^{\sqrt{1+\lambda}} - e^{-\sqrt{1+\lambda}} = 0 $$ This equation has no real solution for \(\lambda\), so this case does not give any eigenvalues or eigenfunctions. 2. \(\Delta < 0\): Complex Conjugate Roots $$ r_1 = i\sqrt{-(1+\lambda)},\ r_2 = -i\sqrt{-(1+\lambda)} $$ The general solution for u in this case is: $$ u(x) = A_1 e^{i\sqrt{-(1+\lambda)}x} + A_2 e^{-i\sqrt{-(1+\lambda)}x} $$ Using Euler's formula, we can rewrite this as: $$ u(x) = B_1 \cos(\sqrt{-(1+\lambda)}x) + B_2 \sin(\sqrt{-(1+\lambda)}x) $$ Applying boundary conditions, u(0) = 0 and u(1) = 0: $$ u(0) = B_1 = 0 \\ u(1) = B_2 \sin(\sqrt{-(1+\lambda)}) = 0 $$ From the second equation, we find that for nontrivial solutions for u(x) we must have: $$ \sin(\sqrt{-(1+\lambda)}) = 0 \\ \sqrt{-(1+\lambda)} = n\pi \\ -(1+\lambda) = n^2\pi^2 \\ \lambda = -(n^2\pi^2 + 1) $$ So, the eigenvalues are \(\lambda_n = -(n^2\pi^2 + 1)\) for n = 1, 2, 3, ... and the corresponding eigenfunctions are u_n(x) = \(\sin(n\pi x)\). Now, we recover the eigenfunctions for the original problem y(x): $$ y_n(x) = e^x u_n(x) = e^x \sin(n\pi x) $$

(c) Solve the given problem directly

The given problem is a second-order linear homogeneous differential equation with variable coefficients: $$ y'' - 2y' + (1 + \lambda)y = 0 $$ Assume a solution of the form: $$ y(x) = e^{rx} $$ Plug this into the differential equation to get: $$ r^2 e^{rx} - 2re^{rx} + (1+\lambda)e^{rx} = 0 \\ (r^2 - 2r + 1 + \lambda)e^{rx} = 0 $$ Since \(e^{rx}\) is never zero, we must have: $$ r^2 - 2r + 1 + \lambda = 0 $$ This is a quadratic equation for r. Its discriminant is: $$ \Delta = 4 - 4(1 + \lambda) = -4\lambda $$ Again, the eigenvalues are assumed to be real, so we only consider the case where the discriminant is negative. Let \(r = \alpha + i\beta\), so we have $$ y(x) = e^{(\alpha + i\beta)x} = e^{\alpha x}(A \cos(\beta x) + B \sin(\beta x)) $$ Comparing this with y(x) = e^x u(x) = e^x(u_n(x)): 1. \(e^{\alpha x} = e^x\), i.e., \(\alpha = 1\). 2. \(\beta^2 = -\lambda\), so the eigenvalues are given by \(\lambda = -\beta^2\). Since \(\beta = n\pi\), we have \(\lambda_n = -(n^2\pi^2)\) for n = 1, 2, 3, ... Therefore, the eigenfunctions for the original problem y(x) directly obtained are: $$ y_n(x) = e^x \sin(n\pi x), $$ which is the same as the ones obtained through the introduction of the dependent variable u.

Key concepts

Eigenvalues

Eigenvalues are special numbers related to linear transformations and matrices. In the context of differential equations, eigenvalues help identify the characteristics and behavior of solutions. When solving boundary value problems, finding eigenvalues allows us to understand the particular modes at which a system can oscillate or stabilize. Here, we determined the eigenvalues by addressing transformations in the differential equation that resulted in a characteristic equation. The eigenvalues were found to be in the form \(\lambda_n = -(n^2\pi^2 + 1)\). This arises due to the complex roots in the differential equation, dictating the frequency and behavior of solutions. Detecting these values is crucial for establishing the stability and dynamic characteristics of the solutions.

Eigenfunctions

Eigenfunctions are functions associated with eigenvalues that provide solutions to differential equations satisfying specific boundary conditions. When dealing with differential equations, eigenfunctions demonstrate consistent patterns or modes of oscillation. They essentially describe the spatial part of the solution that works with any given eigenvalue. In this exercise, the eigenfunctions were determined as \(u_n(x) = \sin(n\pi x)\). These functions arise naturally when solving the boundary conditions \(u(0)=0\) and \(u(1)=0\). The solutions in this form ensure that each eigenfunction matches its corresponding eigenvalue, providing a complete description of potential solutions. This pairing of eigenvalues and eigenfunctions is vital to solve and interpret boundary value problems.

Differential Equations

Differential equations involve functions and their derivatives and are fundamental in expressing physical phenomena such as motion, heat, and waves. These equations define how a quantity changes, serving as a mathematical model for many real-world systems. In this problem, we dealt with a second-order linear homogeneous differential equation: \[ y'' - 2y' + (1 + \lambda)y = 0 \]This equation describes a system influenced by a parameter \(\lambda\), which affects its behavior significantly. Solving such equations often require determining characteristic equations, which provide insight into possible solution behaviors by identifying eigenvalues and eigenfunctions. Differential equations are solved using initial or boundary conditions to ensure the solution fits the system's physical constraints.

Variable Substitution

Variable substitution is a method used to simplify differential equations by introducing new variables or functions. This technique aims to transform the equation into an easier form to solve. In this exercise, the dependent variable \(y\) was expressed as a combination \(y = s(x)u\). Consequently, the choice of \(s(x) = e^x\) eliminated terms involving \(u'\) in the equation, essentially simplifying the problem to focus solely on terms with \(u\) and its second derivative. Variable substitution is strategic; it streamlines complex differential equations, making analysis and solution derivation more approachable. This simplification uncovers deeper insights into the nature and solutions of the equation.