Problem 17 - Advanced Engineering Mathematics

Question

In Problems, find the eigenvalues and eigenfunctions for the given boundary- value problem. $$ y^{\prime \prime}+2 y^{\prime}+(\lambda+1) y=0, y(0)=0, y(5)=0 $$

Step-by-Step Solution

Answer

Eigenvalues: \( \lambda_n = \frac{n^2\pi^2}{25} \), Eigenfunctions: \( y_n(x) = B e^{-x} \sin\left(\frac{n\pi}{5} x\right) \).

Step 1: Characteristic Equation

To solve the differential equation, we start by finding the characteristic equation associated with the differential operator. The form of the proposed solution is \( y = e^{rx} \). Substituting into the differential equation: \( r^2 e^{rx} + 2r e^{rx} + (\lambda + 1) e^{rx} = 0 \). Factoring out \( e^{rx} \), we get the characteristic equation: \( r^2 + 2r + (\lambda + 1) = 0 \).

Step 2: Solve the Characteristic Equation

The characteristic equation \( r^2 + 2r + (\lambda + 1) = 0 \) is solved using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 2 \), and \( c = \lambda+1 \). Substituting these values, \( r = \frac{-2 \pm \sqrt{4 - 4(\lambda+1)}}{2} \) simplifying to \( r = \frac{-2 \pm \sqrt{-4\lambda}}{2} \). Therefore, the roots are \( r = -1 \pm i\sqrt{\lambda} \).

Step 3: General Solution Formation

The roots are complex, indicating an oscillatory solution. The general solution of the differential equation is given by \( y(x) = e^{-x}(A \cos(\sqrt{\lambda}x) + B \sin(\sqrt{\lambda}x)) \), where \( A \) and \( B \) are constants determined by boundary conditions.

Step 4: Apply Boundary Conditions

Apply the first boundary condition \( y(0) = 0 \): \( e^{0}(A \cos(0) + B \sin(0)) = 0 \Rightarrow A = 0 \). The solution simplifies to \( y(x) = B e^{-x} \sin(\sqrt{\lambda} x) \). Apply the second boundary condition \( y(5) = 0 \): \( B e^{-5} \sin(5\sqrt{\lambda}) = 0 \). Since \( B eq 0 \), it must be that \( \sin(5\sqrt{\lambda}) = 0 \).

Step 5: Eigenvalue Calculation

The condition \( \sin(5\sqrt{\lambda}) = 0 \) implies that \( 5\sqrt{\lambda} = n\pi \) for \( n = 0, 1, 2, \ldots \). Solving for \( \lambda \), we have \( \sqrt{\lambda} = \frac{n\pi}{5} \) leading to \( \lambda = \frac{n^2\pi^2}{25} \). These are the eigenvalues for the problem.

Step 6: Determine Eigenfunctions

Substitute the expression for \( \lambda \) back into the form of \( y \): \( y(x) = B e^{-x} \sin\left(\frac{n\pi}{5} x\right) \). The eigenfunctions corresponding to the eigenvalues \( \lambda_n = \frac{n^2\pi^2}{25} \) are \( y_n(x) = B e^{-x} \sin\left(\frac{n\pi}{5} x\right) \). The constant \( B \) is typically determined through normalization or given initial conditions.

Key Concepts

Boundary Value Problems

Boundary value problems (BVPs) are a type of differential equation problem where the solution is required to satisfy conditions at different values of the independent variable. These conditions are called boundary conditions. For instance, in the given problem, we want to solve a differential equation under the requirements that the solution equals zero at two distinct points, specifically at 0 and 5.

BVPs are common in physics and engineering because they often describe systems where conditions are known at different locations or times, like the temperature at both ends of a rod. Solving these types of problems typically involves finding a solution to the differential equation that simultaneously satisfies multiple boundary conditions.

• Initial conditions specify the state of a system at a single point.
• Boundary conditions specify the state of a system at different points.

The solutions to boundary value problems are deeply connected to the eigenvalues and eigenfunctions of the system, as these often represent stable states or modes of a system.

Differential Equations

Differential equations are mathematical equations that involve derivatives of a function or functions. They are used to model various phenomena in engineering, physics, and other sciences. Essentially, a differential equation relates a function to its derivatives, describing how one quantity changes with respect to another.

In the exercise, we deal with a second-order linear differential equation of the form: \[y'' + 2y' + (lambda + 1)y = 0\]This equation involves the second derivative \( y'' \) and first derivative \( y' \) of the function \( y \). Solving such an equation typically involves finding a function that satisfies the equation and any given boundary conditions.

• Ordinary differential equations (ODEs) involve functions of a single variable and their derivatives.
• Partial differential equations (PDEs) involve multiple variables and partial derivatives.

Differential equations are fundamental in predicting the future behavior of dynamic systems from classical mechanics to population dynamics.

Characteristic Equation

The characteristic equation is a key step in solving linear ordinary differential equations, especially those with constant coefficients. It derives from assuming a solution in the exponential form and substituting it into the differential equation.
For the given problem, substitute \( y = e^{rx} \) into the equation to get:\[r^2 + 2r + (lambda + 1) = 0\]This is the characteristic equation, a quadratic that can be solved to find the roots \( r \). Solving the characteristic equation provides insight into the behavior of the solutions.

• If the roots are real and distinct, solutions are exponential functions.
• If the roots are real and repeated, solutions include a polynomial function multiplied by an exponential function.
• If the roots are complex, solutions involve oscillatory functions like sine and cosine.

Understanding the roots helps in forming the general solution of the differential equation.

Complex Roots

When the solution of a characteristic equation yields complex roots, it indicates that the solution to the differential equation involves oscillatory components. In mathematical terms, complex roots occur when the discriminant \( b^2 - 4ac \) of the characteristic equation is negative, resulting in roots of the form \( \alpha \pm i\beta \).
For the characteristic equation:\[r^2 + 2r + (lambda + 1) = 0 \]the roots are complex: \( r = -1 \pm i \sqrt{\lambda} \). This means that the differential equation's solution will involve imaginary and thus oscillatory functions.
The general solution is expressed as:\[y(x) = e^{-x}(A \cos(\sqrt{\lambda}x) + B \sin(\sqrt{\lambda}x))\]

• \( e^{-x} \) represents the exponential decay.
• \( \cos(\sqrt{\lambda}x) \) and \( \sin(\sqrt{\lambda}x) \) describe the oscillatory behavior.

Oscillatory solutions are prevalent in systems that exhibit periodic motion, such as vibrating strings or electrical circuits.