Problem 8 - Advanced Engineering Mathematics

Question

(a) The roots of the auxiliary equation \(m^{2}+m+\lambda=0\) are \(\frac{1}{2}(-1 \pm \sqrt{1-4 \lambda}) .\) When \(\lambda=0\) the general solution of the differential equation is \(c_{1}+c_{2} e^{-x} .\) The boundary conditions imply \(c_{1}+c_{2}=0\) and \(c_{1}+c_{2} c^{-2}=0 .\) since the determinant of the coefficients is not \(0,\) the only solution of this homogeneous system is \(c_{1}=c_{2}=0\) in which case \(y=0 .\) When \(\lambda=\frac{1}{4},\) the general solution of the differential equation is \(c_{1} e^{-x / 2}+c_{2} x e^{-x / 2}\) The boundary conditions imply \(c_{1}=0\) and \(c_{1}+2 c_{2}=0,\) so \(c_{1}=c_{2}=0\) and \(y=0 .\) Similarly, if \(0 < \lambda < \frac{1}{4},\) the general solution is $$y=c_{1} e^{\frac{1}{2}(-1+\sqrt{1-4 \lambda}) x}+c_{2} c^{\frac{1}{2}(-1-\sqrt{1-4 \lambda}) x}.$$ In this case the boundary conditions again imply \(c_{1}=c_{2}=0,\) and so \(y=0 .\) Now, for \(\lambda>\frac{1}{4},\) the general solution of the differential equation is $$y=c_{1} e^{-x / 2} \cos \sqrt{4 \lambda-1} x+c_{2} e^{-x / 2} \sin \sqrt{4 \lambda-1} x.$$ The condition \(y(0)=0\) implies \(c_{1}=0\) so \(y=c_{2} e^{-x / 2} \sin \sqrt{4 \lambda-1} x .\) From $$y(2)=c_{2} e^{-1} \sin 2 \sqrt{4 \lambda-1}=0$$ we see that the eigenvalues are determined by \(2 \sqrt{4 \lambda-1}=n \pi\) for \(n=1,2,3, \ldots .\) Thus, the eigenvalues are \(n^{2} \pi^{2} / 4^{2}+1 / 4\) for \(n=1,2,3, \ldots,\) with corresponding eigenfunctions \(e^{-x / 2} \sin (n \pi x / 2)\). (b) The self-adjoint form is $$\frac{d}{d x}\left[e^{x} y^{\prime}\right]+\lambda e^{x} y=0.$$ (c) An orthogonality relation is $$\int_{0}^{2} e^{x}\left(e^{-x / 2} \sin \frac{m \pi}{2} x\right)\left(e^{-x / 2} \cos \frac{n \pi}{2} x\right) d x=\int_{0}^{2} \sin \frac{m \pi}{2} x \cos \frac{n \pi}{2} x d x=0.$$

Step-by-Step Solution

Answer

The eigenvalues are \( \lambda = \frac{n^2\pi^2}{16} + \frac{1}{4} \) for \( n=1,2,3,\ldots \), and corresponding eigenfunctions are orthogonal.

Step 1: Identify and Solve the Auxiliary Equation

The auxiliary equation given is \( m^2 + m + \lambda = 0 \). To solve for \( m \), use the quadratic formula: \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1, b = 1, c = \lambda \). Hence, \( m = \frac{-1 \pm \sqrt{1 - 4\lambda}}{2} \).

Step 2: Analyze the Behavior for Different \( \lambda \) Values

- **\( \lambda = 0 \)**: The roots are real and distinct; the general solution is \( y = c_1 + c_2 e^{-x} \). Applying boundary conditions, \( c_1 + c_2 = c_1 + c_2 \cdot c^{-2} = 0 \) leads to \( c_1 = c_2 = 0 \), thus \( y = 0 \).- **\( \lambda = \frac{1}{4} \)**: The roots become \( m = -\frac{1}{2} \) (repeated); the general solution is \( y = c_1 e^{-x/2} + c_2 x e^{-x/2} \). Applying boundary conditions, \( c_1 = 0, c_1 + 2c_2 = 0 \) leads to \( c_1 = c_2 = 0 \), hence \( y = 0 \).- **\( 0 < \lambda < \frac{1}{4} \)**: Roots are complex; solution is \( y = c_1 e^{\frac{1}{2}(-1 + \sqrt{1 - 4\lambda})x} + c_2 e^{\frac{1}{2}(-1 - \sqrt{1 - 4\lambda})x} \). Boundary conditions again imply \( c_1 = c_2 = 0 \), so \( y = 0 \).- **\( \lambda > \frac{1}{4} \)**: Roots are complex with imaginary part; general solution is \( y = c_1 e^{-x/2} \cos(\sqrt{4\lambda - 1} x) + c_2 e^{-x/2} \sin(\sqrt{4\lambda - 1} x) \). The condition \( y(0) = 0 \) implies \( c_1 = 0 \), so \( y = c_2 e^{-x/2} \sin(\sqrt{4\lambda-1} x) \). Given \( y(2) = 0 \), eigenvalues are determined by \( 2\sqrt{4\lambda - 1} = n\pi \), leading to eigenvalues \( \lambda = \frac{n^2\pi^2}{16} + \frac{1}{4} \).

Step 3: Find Self-Adjoint Form

To express the differential equation in self-adjoint form, consider: \( \frac{d}{dx}\left[e^x y'\right] + \lambda e^x y = 0 \). This changes the original equation to emphasize self-adjoint properties by including \( e^x \) as a weighting function.

Step 4: Verify Orthogonality Relation

Check that the given orthogonality relation holds: \( \int_0^2 e^x \left( e^{-x/2} \sin \frac{m\pi}{2} x \right) \left( e^{-x/2} \cos \frac{n\pi}{2} x \right) dx = \int_0^2 \sin \frac{m\pi}{2} x \cos \frac{n\pi}{2} x dx = 0 \). This ensures orthogonality of the eigenfunctions, a crucial property in eigenfunction expansions.

Key Concepts

Differential Equations

Differential equations are equations that involve a function and its derivatives. They are essential in modeling various physical phenomena, such as sound, heat, electricity, and more. The goal is to find a function that satisfies the given relation between its derivatives.

In our exercise, the differential equation is linked with the concept of eigenvalues and eigenfunctions. It involves solving an auxiliary equation of the form:

• \( m^2 + m + \lambda = 0 \)

Understanding how to solve this auxiliary equation using the quadratic formula is critical for finding potential solutions. The solutions to the differential equations involve cases based on the parameter \(\lambda\):

• \(\lambda = 0\): Real distinct roots.
• \(\lambda = \frac{1}{4}\): Repeated roots.
• \(0 < \lambda < \frac{1}{4}\): Complex roots.
• \(\lambda > \frac{1}{4}\): Complex roots with imaginary part.

Each case leads to a different form of the general solution, making the study of boundary conditions pivotal.

Boundary Conditions

Boundary conditions are the additional constraints needed to find specific solutions to differential equations for physical problems. These conditions specify the values of the function or its derivatives at specific points, determining a unique solution.

In the exercise, boundary conditions like \(y(0) = 0\) and \(y(2) = 0\) are crucial. They help determine constants in the general solution, effectively reducing the degrees of freedom of the solution. For instance:

• With \(\lambda = 0\), boundary conditions led to \(c_1 = c_2 = 0\), resulting in \(y=0\).
• For \(\lambda > \frac{1}{4}\), they helped find eigenvalues based on the periodicity constraint of the trigonometric functions involved.

These boundaries ensure that the solutions are aligned with the physical interpretation or system constraints pursued.

Orthogonality Relation

Orthogonality in the context of eigenfunctions indicates that two functions are perpendicular to each other in terms of an inner product. This property is pivotal when expanding functions into series, providing a set of unique basis functions.

In this exercise, the orthogonality relation is:

• \[ \int_{0}^{2} e^{x}(e^{-x / 2} \sin \frac{m \pi}{2} x)(e^{-x / 2} \cos \frac{n \pi}{2} x) dx = 0 \]

This implies that the sine and cosine functions (with specified frequencies) are orthogonal over the interval from 0 to 2. In practical terms, it means that these functions do not "interfere" with each other. Orthogonal functions simplify solving systems and are foundational in Fourier series and other methods for solving differential equations.

Self-Adjoint Form

A differential equation is said to be in self-adjoint form if it can be expressed in such a way that encompasses symmetry properties. This form makes various mathematical analyses and operations easier and more intuitive, especially when dealing with eigenvalue problems.

The self-adjoint form for the differential equation in the exercise is:

• \[ \frac{d}{d x}\left[e^{x} y^{\prime}\right]+\lambda e^{x} y=0 \]

The inclusion of \(e^x\) as a weighting function gives it this symmetry. This adjustment to the equation structure highlights how physical properties like conservation and stability might manifest in mathematical form. It also provides a foundation for proving theorems related to eigenvalues and eigenfunctions, ensuring that the solutions respect inherent symmetry.