Problem 63 - Advanced Engineering Mathematics

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 values of \(\lambda\) so that the problem possesses (a) trivial solutions? (b) nontrivial solutions?

Step-by-Step Solution

Answer

Nontrivial solutions exist for \(\lambda = (2n)^2\), where \(n\) is an integer. Trivial solutions occur otherwise.

Step 1: Understanding the Differential Equation

The given differential equation is \(y'' + \lambda y = 0\). This is a second-order linear homogeneous differential equation, where \(\lambda\) is a parameter to be determined.

Step 2: Analyzing the Boundary Conditions

The boundary conditions are \(y(0) = 0\) and \(y(\pi/2) = 0\). These conditions imply that any solution \(y(x)\) to the differential equation must satisfy these two conditions at \(x = 0\) and \(x = \pi/2\), respectively.

Step 3: Solving the Characteristic Equation

The differential equation \(y'' + \lambda y = 0\) has the characteristic equation \(r^2 + \lambda = 0\), which gives roots \(r = \pm i\sqrt{\lambda}\). The nature of \(\lambda\) (whether positive, zero, or negative) affects the solution.

Step 4: Finding Solutions Based on Lambda

If \(\lambda > 0\), the general solution is \(y(x) = A\cos(\sqrt{\lambda}x) + B\sin(\sqrt{\lambda}x)\). If \(\lambda = 0\), the general solution simplifies to \(y(x) = Ax + B\). If \(\lambda < 0\), the solution involves hyperbolic functions, as \(\lambda = -\mu\) (\(\mu > 0\)) leads to \(y(x)=A\cosh(\sqrt{\mu}x) + B\sinh(\sqrt{\mu}x)\).

Step 5: Applying Boundary Conditions for Nontrivial Solutions (Lambda > 0)

For nontrivial solutions and \(\lambda > 0\), apply the conditions \(y(0) = 0\) to get \(A = 0\), simplifying the solution to \(y(x) = B\sin(\sqrt{\lambda}x)\). Applying \(y(\pi/2) = 0\) gives \(B\sin(\sqrt{\lambda}\frac{\pi}{2}) = 0\). For nontrivial \(B\), \(\sqrt{\lambda}\frac{\pi}{2} = n\pi\), implying \(\lambda = (2n)^2\) where \(n\) is an integer.

Step 6: Conditions for Trivial Solutions

A solution is trivial if it results in \(y(x) = 0\) for all \(x\). This occurs for either \(B = 0\) in the nontrivial solution scenario or if \(\lambda\) does not satisfy the conditions for nontrivial solutions (i.e., \(\lambda\) is not an integer multiple of 4).

Step 7: Considering Lambda = 0 and Lambda < 0

For \(\lambda = 0\), the solution \(y(x) = Ax + B\) with boundary conditions yields only the trivial solution \(y(x) = 0\). For \(\lambda < 0\), since the trigonometric-based solution won’t meet the conditions due to hyperbolic functions, this yields only the trivial solution.

Key Concepts

Second-order linear differential equation

A second-order linear differential equation is a mathematical equation involving a function and its derivatives up to the second order. The general form of such equations is given by \[ a(x)y'' + b(x)y' + c(x)y = f(x) \]where:

• \( y'' \) is the second derivative of \( y \) with respect to \( x \).
• \( a(x), b(x), \text{and } c(x) \) are continuous functions of \( x \).
• \( f(x) \) is a given function, termed as the non-homogeneous part of the equation.

If \( f(x) = 0 \), the equation becomes homogeneous. In the context of boundary-value problems, these equations appear frequently where certain values of the function are specified at different points. Solving such equations often requires determining specific boundary conditions to find the solution that describes a physical phenomenon or satisfies the conditions given in the problem.

Homogeneous differential equation

A homogeneous differential equation, specifically in the context of second-order linear equations, is when the function \( f(x) \) in \( a(x)y'' + b(x)y' + c(x)y = f(x) \) is zero, simplifying it to\[ a(x)y'' + b(x)y' + c(x)y = 0 \]This type of equation implies that all terms depend on the function \( y \) and its derivatives, without any added functions. Homogeneous differential equations have a special property – if \( y_1(x) \) and \( y_2(x) \) are solutions, then any linear combination \( c_1y_1(x) + c_2y_2(x) \) is also a solution. This property is crucial for finding solutions that match given boundary conditions. In essence, identifying solutions to homogeneous equations often involves solving another related form - called the characteristic equation.

Characteristic equation

To solve a homogeneous second-order linear differential equation, we use a related algebraic equation known as the characteristic equation. For an equation in the standard form\[ y'' + \lambda y = 0 \]The characteristic equation derived is\[ r^2 + \lambda = 0 \]Solving this involves finding the roots \( r \), which provide insight into the nature of the solutions. If \( \lambda > 0 \), the roots are imaginary, leading to solutions involving sine and cosine functions. If \( \lambda = 0 \), the roots are zero, and solutions are straight lines, while if \( \lambda < 0 \), the roots are real and solutions involve hyperbolic functions. Solving the characteristic equation is a key step in determining the general solution of the original differential equation.

Nontrivial solutions

In the context of differential equations, a nontrivial solution refers to a solution where the function \( y(x) \) is not identically zero. Finding nontrivial solutions often involves fulfilling certain conditions over a given interval, typically known as boundary conditions.

For the equation \( y'' + \lambda y = 0 \) with specified boundary conditions \( y(0) = 0 \) and \( y(\pi/2) = 0 \), the quest for nontrivial solutions is linked to choosing values of \( \lambda \) that lead to non-zero solutions. In particular,
\( \lambda \) is determined such that the sine term does not vanish, which happens when \( \sqrt{\lambda}\frac{\pi}{2} = n\pi \) for integer values of \( n \).
These values of \( \lambda \) result in oscillatory modes described by the sine function. Real physical problems approximating these solutions include resonant frequencies in mechanical systems, where nontrivial solutions resonate at certain frequency values.