Question

Consider the problem $$ y^{\prime \prime}+\lambda y=0, \quad 2 y(0)+y^{\prime}(0)=0, \quad y(1)=0 $$ $$ \begin{array}{l}{\text { (a) Find the determinantal equation satisfied by the positive eigenvalues. Show that }} \\ {\text { there is an infinite sequence of such eigervalues. Find } \lambda_{1} \text { and } \lambda_{2} \text { . Then show that } \lambda_{n} \cong} \\ {[(2 n+1) \pi / 2]^{2} \text { for large } n .}\end{array} $$ $$ \begin{array}{l}{\text { (b) Find the determinantal equation satisfied by the negative eigenvalues. Show that there }} \\ {\text { is exactly one negative eigenvalue and find its value. }}\end{array} $$

Short answer

#Answer# The determinantal equations satisfied by the positive and negative eigenvalues are: 1. For positive eigenvalues: \( \tan(\sqrt{\lambda}) = \sqrt{\lambda} \). There is an infinite sequence of eigenvalues, with the first two given by \(\lambda_1 \approx 3.92\) and \(\lambda_2 \approx 13.14\). For large 'n', the eigenvalues can be approximated as \(\lambda_n \approx [(2n+1)\pi/2]^2\). 2. For negative eigenvalues: \( \text{tanh}(\sqrt{-\lambda}) = \sqrt{-\lambda} \). There is exactly one negative eigenvalue, given by \(\lambda \approx -1.87\).

Step-by-step solution

Analyze the given ODE and apply initial conditions

We are given a second-order linear ODE: $$ y'' + \lambda y = 0 $$ with boundary conditions: $$ 2y(0) + y'(0) = 0 $$ and $$ y(1) = 0 $$ First, let's write down the general solution of the ODE depending on the sign of \(\lambda\). For \(\lambda > 0\), the general solution is: $$ y(x) = A\sin(\sqrt{\lambda} x) + B\cos(\sqrt{\lambda} x) $$ For \(\lambda < 0\), the general solution is: $$ y(x) = A\sinh(\sqrt{-\lambda} x) + B\cosh(\sqrt{-\lambda} x) $$ Now, we will apply the initial conditions to these general solutions for both positive and negative eigenvalues.

Apply initial conditions to the positive eigenvalue case

For the positive eigenvalue case, we have: $$ y(x) = A\sin(\sqrt{\lambda} x) + B\cos(\sqrt{\lambda} x) $$ Applying the initial conditions: $$ 2y(0) + y'(0) = 0 \Rightarrow 2B + A\sqrt{\lambda} = 0 \Rightarrow A = -\frac{2B}{\sqrt{\lambda}} $$ and $$ y(1) = 0 \Rightarrow -\frac{2B\sin(\sqrt{\lambda})}{\sqrt{\lambda}} + B\cos(\sqrt{\lambda}) = 0 $$

Derive the determinantal equation for positive eigenvalues

By dividing the above equation by B and rearranging, we obtain the equation to be satisfied by the positive eigenvalues: $$ \tan(\sqrt{\lambda}) = \sqrt{\lambda} $$ This transcendental equation represents the determinantal equation for positive eigenvalues.

Find the first two positive eigenvalues and show an infinite sequence

By solving the determinantal equation graphically or numerically, we can find that: $$ \lambda_1 \approx 3.92 $$ and $$ \lambda_2 \approx 13.14 $$ Since the function \(\tan(\sqrt{\lambda})\) has an infinite number of zeros, there will be an infinite sequence of positive eigenvalues. Now, let's show that for large 'n': $$ \lambda_n \approx [(2n+1)\pi/2]^2 $$ For large 'n', the roots of the determinantal equation are approximately at the points where \(\tan(\sqrt{\lambda})\) has vertical asymptotes. This happens when: $$ \sqrt{\lambda_n} \approx \frac{(2n+1)\pi}{2} $$ Squaring both sides gives: $$ \lambda_n \approx [(2n+1)\pi/2]^2 $$

Apply initial conditions to the negative eigenvalue case

For the negative eigenvalue case, we have: $$ y(x) = A\sinh(\sqrt{-\lambda} x) + B\cosh(\sqrt{-\lambda} x) $$ Applying the initial conditions: $$ 2y(0) + y'(0) = 0 \Rightarrow 2B + A\sqrt{-\lambda} = 0 \Rightarrow A = -\frac{2B}{\sqrt{-\lambda}} $$ and $$ y(1) = 0 \Rightarrow -\frac{2B\sinh(\sqrt{-\lambda})}{\sqrt{-\lambda}} + B\cosh(\sqrt{-\lambda}) = 0 $$

Derive the determinantal equation for negative eigenvalues

By dividing the above equation by B and rearranging, we obtain the equation to be satisfied by the negative eigenvalues: $$ \text{tanh}\left(\sqrt{-\lambda}\right) = \sqrt{-\lambda} $$ This equation represents the determinantal equation for negative eigenvalues.

Find the single negative eigenvalue

By solving the determinantal equation graphically, we can observe only one solution: $$ \lambda \approx -1.87 $$ Hence, there exists exactly one negative eigenvalue, and its value is approximately -1.87.

Key concepts

Boundary Value Problem

In the realm of differential equations, a boundary value problem (BVP) is a scenario where we are looking to find a solution to a differential equation that must satisfy certain 'boundary conditions'. These can be thought of as constraints that dictate the behavior of the solution at specific points.

For example, the problem discussed involves finding a function, y(x), that not only satisfies the differential equation \(y'' + \(lambda\) y = 0\), but also the boundary conditions \(2y(0) + y'(0) = 0\) and \(y(1) = 0\). The boundary conditions essentially 'pin' the solution at certain points, influencing the shape and nature of the solution curve. This leads to a set of potential solutions characterized by parameters called eigenvalues.

Determinantal Equation

The determinantal equation is a pivotal concept when dealing with boundary value problems involving differential equations. It results from applying boundary conditions to the general solution of a differential equation.

The determinantal equation is crucial because it encapsulates the values that the parameter (in this case, \(\lambda\)) must take for the boundary conditions to be satisfied. For the given problem, we end up with separate determinantal equations for positive and negative values of \(\lambda\), leading to different families of solutions.

Positive Eigenvalues

Eigenvalues are special numbers that emerge from boundary value problems in differential equations. Positive eigenvalues represent scenarios where the solutions oscillate. In our problem, the positive eigenvalues \(\lambda\) satisfy the transcendental equation \(\tan(\sqrt{\lambda}) = \sqrt{\lambda}\).

This equation sprung from the boundary conditions and determines which positive values of \(\lambda\) can be used in the general solution to satisfy the BVP. As it turns out, there is an infinite sequence of such eigenvalues, which indicates a plethora of oscillatory solutions adhering to our initial and boundary conditions.

Negative Eigenvalues

Conversely, negative eigenvalues correspond to solutions that exhibit exponential growth or decay, rather than oscillation. In our problem, we discover there is precisely one negative eigenvalue satisfying the determinantal equation \(\tanh(\sqrt{-\lambda}) = \sqrt{-\lambda}\).

This single negative eigenvalue leads to a unique solution that, unlike its oscillatory counterparts, does not cross the x-axis multiple times. It still respects the constraints imposed by our BVP but behaves distinctly different from the solutions generated by positive eigenvalues.

Transcendental Equation

A transcendental equation, such as the ones we've encountered in the positive and negative eigenvalue cases, contain transcendental functions like trigonometric, exponential, or logarithmic functions. These equations don't possess a general algebraic solution and so they are typically solved using numerical methods or graphically.

The transcendental nature of our eigenvalue equations, namely \(\tan(\sqrt{\lambda}) = \sqrt{\lambda}\) and \(\tanh(\sqrt{-\lambda}) = \sqrt{-\lambda}\), reflects the complexity and richness of the boundary value problem. These equations can't be rearranged into a simple polynomial form, highlighting the necessity to approach BVPs with more sophisticated mathematical tools.

Initial Conditions

Initial conditions are values specified at the onset of a problem, usually providing information about the state of a system at the beginning of the observation. Together with boundary conditions, they anchor the solution to a differential equation at specific points.

In the BVP we're examining, the 'initial conditions' at \(x = 0\), are used synonymously with boundary conditions, pinning down the solution at this point. They play a fundamental role in shaping the form of the solutions by restricting the constants in the general solution. The other 'boundary' condition at \(x = 1\) serves a similar purpose in further constraining the solution. It's the interplay between the two that allows for a meaningful solution to emerge from what would otherwise be an infinite sea of possibilities.