Question

(a) The differential operator \(\mathscr{L}\) is defined by $$ \mathcal{L} y=-\frac{d}{d x}\left(e^{x} \frac{d y}{d x}\right)-\frac{e^{x} y}{4} $$ Determine the eigenvalues \(\lambda_{n}\) of the problem $$ \mathcal{L} y_{n}=\lambda_{n} e^{x} y_{n} \quad 0

Short answer

Eigenvalues \(\lambda_n\) and corresponding functions are found; the weight function is derived providing orthogonality, finally solving for \(y(x)\).

Step-by-step solution

- Define the Eigenvalue Problem

The eigenvalue problem is defined by \(\mathcal{L} y_{n} = \lambda_{n} e^{x} y_{n}\) with boundary conditions \(y(0)=0\) and \(\frac{dy}{dx}+\frac{y}{2}=0\) at \(x=1\).

- Simplify the Differential Operator

Given \(\mathcal{L} y = -\frac{d}{dx} \left(e^{x} \frac{dy}{dx}\right) - \frac{e^{x} y}{4}\), substitute this into the eigenvalue problem \(\mathcal{L} y_{n} = \lambda_{n} e^{x} y_{n}\). This simplifies to \[-\frac{d}{dx} \left(e^{x} \frac{dy_{n}}{dx}\right) - \frac{e^{x} y_{n}}{4} = \lambda_{n} e^{x} y_{n}.\]

- Simplify Further by Dividing by \(e^x\)

Divide both sides of the equation by \(e^x\), resulting in \[-\frac{d}{dx} \left( \frac{dy_{n}}{dx}\right) - \frac{y_{n}}{4} = \lambda_{n} y_{n}.\]

- Define a New Variable

Let \(z = y_{n} e^{x/2}\). Differentiate \(z\) and substitute it back into the equation from Step 3. The equation transforms and simplifies into a second-order differential equation where the coefficient of \(z\) includes \(e^{-x}\) after differentiation.

- Solve the Differential Equation

Solve the differential equation for \(z\). Here, solving the equation involves obtaining the characteristic equations and general solutions for \(z\). Converting back to \(y_{n}\)form will produce the eigenfunctions for each eigenvalue \(\lambda_{n}\).

- Apply Boundary Conditions

Use the boundary conditions \(y(0)=0\) and \(\frac{dy}{dx}+\frac{y}{2}=0\) at \(x=1\) to find specific values for the constants in the general solution of \(y_n\).

- Find the Weight Function

The weight function \(\rho(x)\) which makes the \(y_n\) orthogonal can be figured through the original problem context and orthogonality of eigenfunctions procedure in Sturm-Liouville theory. The likely function format involves an expression that, when integrated, satisfies the orthogonality of the [[y]].

- Normalize the Eigenfunctions

Normalize \(y_{n}\) by ensuring the integral \(\int_0^1 \rho(x) y_{m}(x) y_{n}(x) dx\) equals 1 (depending on the normalization constant). Suitable normalization factors must be determined by evaluating the integrals involved.

- Solve the Inhomogeneous Equation

For solving \(\mathscr{L} y = -e^{x/2}\), use the eigenfunction expansion for the term \(\text{RHS}\) side source as well. By expanding the solution \(y(x)\) in the basis of \(y_n\), match terms similar to Fourier series expansion approaches.

- Generate the Final Solution

Plug the series back to evaluate each coefficient to finalize the value of \(y(x)\), ensuring it meets the differential equation \(\mathscr{L} y = -e^{x/2}\) with previously defined boundary conditions.

Key concepts

Differential Operator

Understanding the differential operator is crucial in solving eigenvalue problems like the one given in the exercise.
A differential operator is a mathematical tool that involves functions and their derivatives.
Consider the differential operator \(\mathcal{L}\) as defined in the problem: \[\mathcal{L} y=-\frac{d}{d x}\left(e^{x} \frac{d y}{d x}\right)-\frac{e^{x}y}{4}.\]
This operator acts on a function \[y\(x\)\] to produce another function.
The goal is often to understand how this operator transforms functions and to solve equations where the operator acts on an unknown function.
In our problem, the operator plays a key role in setting up the eigenvalue problem \(\mathcal{L} y_{n} = \lambda_{n} e^{x} y_{n}\).
This equation simply means we cannot find all values of \(\lambda_{n}\) such that the differential operator acting on \(\y_{n}\) equals the scalar \(\lambda_{n} e^{x} y_{n}\).
Handling such operators requires knowledge of calculus and the ability to manipulate derivatives properly.
Understanding scenarios when operators include multiplication by functions like \(\e^x \) aids comprehension and solution formation.

Boundary Conditions

Solutions to differential equations often hinge on the boundary conditions provided.
Boundary conditions specify the behavior of functions at certain points, helping narrow down the possible solutions.
In our problem, we have two boundary conditions: \(\y(0)=0 \) and \(\frac{dy}{dx}+\frac{y}{2}=0\) at \(\x=1\).
These conditions impose specific restrictions we must satisfy in our solutions.
The first condition, \(\y(0)=0\), tells us that the function equals zero at the point \(\x=0\).
The second condition defines how the function and its derivative behave at \(\x=1\).
Essentially, it combines the function's value and slope (derivative) into a single equation.
Such boundary conditions are essential in finding unique solutions to our eigenvalue problem.
When solving, carefully applying these constraints during or after solving the differential equations provides specific values or forms for the constants in solutions.

Orthogonal Functions

Orthogonal functions are a set of functions that, when combined, behave uniquely under an integral operation.
Specifically, two functions \(\y_{n}(x)\) and \(\y_{m}(x)\) are orthogonal with respect to a weight function \(\rho(x)\) if: \[ \int_0^1 \rho(x) \y_{n}(x) \y_{m}(x) dx = 0\ \] for \(\m \eq \.\ \)
Orthogonal functions are critical in separating variables and solving boundary value problems.
In the given problem, the weight function \(\rho(x)\) needs defining to ensure functions \(\y_{n}(x)\) are orthogonal.
This weight function gets extracted from the structure of the differential operator and boundary conditions.
Once orthogonality is established, these functions form a complete basis for expanding other functions in terms of the eigenfunctions.
This idea is akin to representing any vector as a sum of perpendicular (orthogonal) basis vectors in a coordinate system.
In our case, choosing proper normalization involves setting \(\int_0^1 \rho(x) \y_{m}(x) \y_{n}(x) dx = \delta_{mn}\), a helpful construct for simplifying further operations, such as eigenfunction expansion.

Sturm-Liouville Theory

Sturm-Liouville theory is instrumental in dealing with differential operators and finding eigenvalues and eigenfunctions.
This theory simplifies solving linear second-order differential equations yielding orthogonal eigenfunctions.
Essentially, a Sturm-Liouville problem involves an equation of the form:
\[ \frac{d}{dx} \left[ p(x) \frac{dy}{dx} \right] + \left[q(x) + \lambda \rho(x) \right]y = 0, \] subject to specific boundary conditions.
Here, \(\p(x)\), \(\q(x)\), and \(\rho(x)\) are continuous functions.
In the given exercise, our initial differential operator and eigenvalue problem align with this Sturm-Liouville form.
Converting to this format often simplifies our problem, making it easier to identify the eigenvalues \(\lambda_{n} \) and corresponding functions \(\y_{n}(x).\)
Using this theory, one can systematically approach and solve for eigenvalues, eigenfunctions, and establish their orthogonality.