Question

By substituting \(x=\exp t\) find the normalized eigenfunctions \(y_{n}(x)\) and the eigenvalues \(\lambda_{n}\) of the operator \(\mathcal{L}\) defined by $$ \mathcal{L} y=x^{2} y^{\prime \prime}+2 x y^{\prime}+\frac{1}{4} y, \quad 1 \leq x \leq e $$ with \(y(1)=y(e)=0 .\) Find, as a series \(\sum a_{n} y_{n}(x)\), the solution of \(\mathcal{L} y=x^{-1 / 2}\).

Short answer

The normalized eigenfunctions and eigenvalues are zero. Series solution is invalid due to trivial basis.

Step-by-step solution

- Substitute the Variable

Substitute the given transformation for the variable: let \(x = \exp(t)\), then \( dx = \exp(t) dt = x dt\). Transform the operator \( \mathcal{L} \) to the new variable \(t\).

- Transform the Differential Operator

Given: \[\mathcal{L} y = x^{2} y^{\prime \prime} + 2 x y^{\prime} + \frac{1}{4} y.\] Using \(x = e^t\), we have:\[ \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{1}{x} \frac{dy}{dt}\] and\[ \frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{1}{x} \frac{dy}{dt} \right) = \frac{d}{dt} \left( \frac{1}{x} \frac{dy}{dt} \right) \cdot \frac{dt}{dx} = \frac{1}{x^2} \left( \frac{d^2 y}{dt^2} - \frac{dy}{dt} \right).\]

- Substitute into the Operator

Substitute the expressions for \( y^{\prime} \) and \( y^{\prime \prime} \) from step 2 back into the operator \( \mathcal{L} y \): \[\mathcal{L}y = x^2 \left( \frac{1}{x^2} \left( \frac{d^2 y}{dt^2} - \frac{dy}{dt} \right) \right) + 2x \left( \frac{1}{x} \frac{dy}{dt} \right) + \frac{1}{4} y = \frac{d^2 y}{dt^2} - \frac{dy}{dt} + 2 \frac{dy}{dt} + \frac{1}{4} y = \frac{d^2 y}{dt^2} + \frac{dy}{dt} + \frac{1}{4} y.\]

- Solve the Transformed Equation

Solve the differential equation \( \frac{d^2 y}{dt^2} + \frac{dy}{dt} + \frac{1}{4} y = 0 \). Use the characteristic equation: \[ r^2 + r + \frac{1}{4} = 0.\]

- Find Characteristic Roots

Solve the characteristic equation for \( r \):\[ r = \frac{-1 \pm \sqrt{1 - 1}}{2} = \frac{-1 \pm 0}{2} = -\frac{1}{2}.\] Thus, the general solution is\[ y(t) = (c_1 + c_2 t) e^{-t/2}.\]

- Convert Back to Original Variable

Return to the original variable using \( x = e^t \):\[ y(x) = (c_1 + c_2 \ln(x)) x^{-1/2}.\]

- Apply Boundary Conditions

Apply the boundary conditions \( y(1) = 0 \) and \( y(e) = 0 \):\[ y(1) = c_1 + c_2 \ln(1) = c_1 = 0 \Rightarrow c_1 = 0.\] For the condition at \( x = e \): \[ y(e) = c_2 \ln(e) e^{-1/2} = c_2 e^{-1/2} = 0 \Rightarrow c_2 = 0.\]

- Verify Result

Since both constants are zero, we verify normalized eigenfunctions and eigenvalues: \( y_n(x) \) are eigenfunctions and \( \lambda_n \) are eigenvalues, but since both are zero, our functions and values are trivial.

- Find Series Solution

To find the series solution of \( \mathcal{L} y = x^{-1/2} \), use the eigenfunctions found above. Since the eigenfunctions are trivial, the series solution is invalid.

Key concepts

Differential Operator

A differential operator is a tool used in mathematics, especially in differential equations, to describe a function and its derivatives. In this exercise, the differential operator is given as: \[ \mathcal{L} y = x^{2} y^{\prime \prime} + 2 x y^{\prime} + \frac{1}{4} y \qquad \text{for} \; 1 \leq x \leq e. \] This operator showcases the derivatives of a function \( y \) multiplied by powers of the variable \(x\). Differential operators are vital in finding solutions to differential equations, specifically when dealing with eigenvalues and eigenfunctions. They help transform the equation into a more manageable form. Understanding how to work with differential operators is crucial in solving complex differential equations.

Boundary Conditions

Boundary conditions specify the values of a function at certain points, which are essential for finding unique solutions to differential equations. In the given exercise, the boundary conditions are: \[ y(1) = 0 \] and \[ y(e) = 0. \] These specify that the function \( y(x) \) must be zero at \( x = 1 \) and \( x = e \). When solving differential equations, boundary conditions help in determining the constants within the general solution, ensuring that the solution is specific to the problem. Without boundary conditions, we would have an infinite number of solutions.

Characteristic Equation

The characteristic equation is derived from the differential equation to find the roots that help shape the solution. For the equation at hand, the characteristic equation comes from the transformed differential equation: \[ \frac{d^2 y}{dt^2} + \frac{dy}{dt} + \frac{1}{4} y = 0. \] By assuming a solution of the form \( y = e^{rt} \), we obtain the characteristic equation: \[ r^2 + r + \frac{1}{4} = 0. \] Solving this quadratic equation gives us the roots: \[ r = \frac{-1 \pm \sqrt{1 - 1}}{2} = \frac{-1 \pm 0}{2} = -\frac{1}{2}. \] These roots are crucial because they define the exponential functions that form the solution to the differential equation. Understanding the characteristic equation and how to solve it is essential for tackling linear differential equations.

Series Solution

In some cases, the solution to a differential equation can be represented as a series. This exercise asks for the series solution of the differential operator applied to a function: \[ \mathcal{L} y = x^{-1/2}. \] A series solution involves expressing the function \( y(x) \) as a sum of eigenfunctions \( \sum a_{n} y_{n} (x) \). However, because the eigenfunctions in this exercise turned out to be trivial (i.e., zero), finding a meaningful series solution isn't possible. Proper series solutions rely on non-trivial eigenfunctions, which accurately represent the behavior of \( y(x) \) over the given range. Exploring series solutions is invaluable in mathematical analysis and physics as they can approximate functions that are otherwise difficult to solve.