Problem 64 - A First Course in Differential Equations with Modeling Applications

Question

Consider the boundary-value problem \(y^{\prime \prime}+\lambda y=0, y(0)=0\) \(y(\pi / 2)=0 .\) Discuss: Is it possible to determine real values of \(\lambda\) so that the problem possesses (a) trivial solutions? (b) nontrivial solutions? the general solution, simplify the output and, if necessary, write the solution in terms of real functions.

Step-by-Step Solution

Answer

Trivial solutions occur for any real \(\lambda\); nontrivial solutions occur for \(\lambda = \frac{4n^2}{\pi^2}\), \(n\in\mathbb{Z}\).

Step 1: Understand the Boundary-Value Problem

We are given the differential equation \( y'' + \lambda y = 0 \) with boundary conditions \( y(0) = 0 \) and \( y(\pi/2) = 0 \). We need to discuss solutions for different values of \( \lambda \).

Step 2: General Solution of the Differential Equation

The characteristic equation for \( y'' + \lambda y = 0 \) is \( r^2 + \lambda = 0 \), giving roots \( r = \pm i \sqrt{\lambda} \). Thus, the general solution is \( y(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x) \).

Step 3: Apply Boundary Condition at \( x = 0 \)

Using \( y(0) = 0 \), we find \( A \cos(0) + B \sin(0) = A = 0 \). So the solution is \( y(x) = B \sin(\sqrt{\lambda} x) \).

Step 4: Apply Boundary Condition at \( x = \frac{\pi}{2} \)

Applying \( y(\frac{\pi}{2}) = 0 \) gives \( B \sin(\sqrt{\lambda} \frac{\pi}{2}) = 0 \). This implies either \( B = 0 \) (trivial solution) or \( \sin(\sqrt{\lambda} \frac{\pi}{2}) = 0 \) (for nontrivial solutions).

Step 5: Trivial Solutions

If \( B = 0 \), the solution is \( y(x) = 0 \) for any \( \lambda \), which is a trivial solution.

Step 6: Nontrivial Solutions Condition

For nontrivial solutions, we need \( \sin(\sqrt{\lambda} \frac{\pi}{2}) = 0 \). This holds when \( \sqrt{\lambda} \frac{\pi}{2} = n\pi \) for some integer \( n \). Solving for \( \lambda \), we get \( \lambda = \frac{4n^2}{\pi^2} \).

Step 7: Conclusions on \( \lambda \) Values

(a) For trivial solutions, any real \( \lambda \) is acceptable. (b) For nontrivial solutions, \( \lambda \) must be \( \frac{4n^2}{\pi^2} \) where \( n \) is an integer.

Key Concepts

Understanding Differential Equations

Differential equations are equations that involve unknown functions and their derivatives. They are used to describe various phenomena such as physical systems, economics, and biological processes. In our given problem, we have a second-order linear differential equation:

• This takes the form \( y'' + \lambda y = 0 \).
• The term \( y'' \) represents the second derivative of the function \( y \), indicating the acceleration if interpreted in a physical context.
• The parameter \( \lambda \) is a real number that acts as a scaling factor or coefficient in the equation.

The solution to a differential equation is a function or a set of functions that satisfy the equation. The process involves finding the general solution, which contains constants or parameters, and then applying initial or boundary conditions to find particular solutions.

Identifying Nontrivial Solutions

Nontrivial solutions to a differential equation are solutions other than the zero solution. In our boundary-value problem, this means finding solutions for \( y(x) \) that aren't simply zero for all \( x \). For the problem to possess nontrivial solutions, specific conditions must be met:

• The boundary condition \( y(0) = 0 \) simplifies the general solution to \( y(x) = B \sin(\sqrt{\lambda} x) \).
• Applying \( y(\pi/2) = 0 \) leads to \( B \sin(\sqrt{\lambda} \pi/2) = 0 \).
• For \( B eq 0 \) (thus, nontrivial), it requires that \( \sin(\sqrt{\lambda} \pi/2) = 0 \), meaning \( \sqrt{\lambda} \pi/2 \) must be an integer multiple of \( \pi \), or \( \lambda = \frac{4n^2}{\pi^2} \).

Thus, \( \lambda \) must take specific discrete values, allowing the sine function to complete whole cycles, hence preventing the solution from triviality.

The Characteristic Equation

The characteristic equation helps us to find the general solution of a linear differential equation system. It is derived from the differential equation by assuming a solution of a specific form, typically exponential. For our problem:

• Starting with \( y'' + \lambda y = 0 \), assume a solution of the form \( y = e^{rx} \).
• Substituting gives the characteristic equation: \( r^2 + \lambda = 0 \), leading to roots \( r = \pm i \sqrt{\lambda} \).
• The roots suggest solutions in terms of trigonometric functions, expressed as \( y(x) = A \cos(\sqrt{\lambda} x) + B \sin(\sqrt{\lambda} x) \).

The characteristic equation is crucial for determining the nature of the solutions, such as oscillatory behavior from the sine and cosine functions, dependent on the nature of \( \lambda \) being positive.

Applying Boundary Conditions

Boundary conditions are essential elements in solving differential equations as they provide specific values the solution must satisfy. These conditions help in determining the constants present in the general solution, tailoring it to fit the particular problem.

• For our problem, boundary conditions are given as \( y(0) = 0 \) and \( y(\pi/2) = 0 \).
• The first condition \( y(0) = 0 \) eliminated the cosine term, as \( A = 0 \).
• The second condition \( y(\pi/2) = 0 \) further refines the solution to conclude on possible values of \( \lambda \), especially to discern trivial from nontrivial solutions.

The boundary conditions are a practical approach to specify the function at the boundaries, ensuring the general solution conforms to the problem's requirements. Without these conditions, the solution might remain ambiguous.