Question

Estimate the lowest eigenvalue \(\lambda_{0}\) of the equation $$ \frac{d^{2} y}{d x^{2}}-x^{2} y+\lambda y=0, \quad y(-1)=y(1)=0 $$ using a quadratic trial function.

Short answer

The lowest eigenvalue is \( \lambda_{0} = 2 \).

Step-by-step solution

- Choose a Trial Function

Select a quadratic trial function that satisfies the boundary conditions. A suitable choice is: \[ y(x) = a(1 - x^2) \] Here, the constant factor 'a' will be determined later.

- Compute the First and Second Derivatives

Compute the first derivative: \[ \frac{dy}{dx} = -2ax \] and the second derivative: \[ \frac{d^2y}{dx^2} = -2a \]

- Substitute into the Differential Equation

Substitute \( y(x) \), \( \frac{dy}{dx} \), and \( \frac{d^2y}{dx^2} \) back into the differential equation: \[ -2a - x^2a(1 - x^2) + \lambda a(1 - x^2) = 0 \] Simplify to: \[ -2a - ax^2 + ax^4 + \lambda a - \lambda ax^2 = 0 \]

- Simplify the Equation

Collect terms to obtain: \[ a(-2 - x^2 + x^4 + \lambda - \lambda x^2) = 0 \] For the equation to hold for all x, the coefficient of \( a \) in each term must be zero. Focus first on terms without \( x \): \[ -2 + \lambda = 0 \]

- Solve for the Eigenvalue

From the above equation, solve for the eigenvalue \( \lambda \): \[ \lambda = 2 \]

Key concepts

Differential Equations

Differential equations involve derivatives of a function and connect these derivatives to the function itself.
In our exercise, the differential equation is: \[ \frac{d^{2} y}{d x^{2}}-x^{2} y+\beta y=0 \], where
\( \beta \) is the eigenvalue we need to estimate. This is a second-order differential equation.
Second-order because the highest derivative is a second derivative. We use these equations to describe various physical phenomena, such as wave patterns or heat distribution.
In solving differential equations, we often seek solutions that satisfy given boundary conditions.

Trial Functions

A trial function is a guessed solution that we use to start solving a differential equation.
It should be a simple, known function that satisfies the boundary conditions of the problem.
For our problem, we chose \[ y(x) = a(1 - x^2) \], a quadratic function. This means it has terms up to \( x^2 \).
Our choice is guided by the equation’s boundary conditions. By substituting this back into the original equation, we can test and refine our guessed solution.
This iterative approach helps us approximate complex solutions.

Boundary Conditions

Boundary conditions are constraints on the values that the solution to a differential equation can take at certain boundaries.
In this exercise, the boundary conditions are: \( y(-1)=0 \) and \( y(1)=0 \).
These conditions tell us the value of function \( y \) at the points \( x = -1 \) and \( x = 1 \).
When we select our trial function, it must satisfy these boundary conditions. For our quadratic trial function \[ y(x) = a(1 - x^2) \], we see that \( y(-1) = a(1 - (-1)^2) = a(1 - 1) = 0 \), which meets the boundary condition. Similarly, \( y(1) = a(1 - 1) = 0 \), which also fits the requirement.
Ensuring that our trial function meets these conditions helps us find a valid solution.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are crucial in understanding systems of differential equations.
For our problem, eigenvalues \( \beta \) give us essential characteristics of the differential equation itself.
To find them, we applied our chosen trial function and boundary conditions. In the simplified equation \[ a(-2 + \beta) = 0 \],
we found that \( \beta = 2 \). This tells us that the lowest eigenvalue is 2.
Eigenvectors, although not discussed in detail in this problem, would be the respective solutions of the differential equation.