Question

(a) Determine the eigenvalues \(\lambda_{\pm}\)of the kernel \(K(x, z)=(x z)^{1 / 2}\left(x^{1 / 2}+z^{1 / 2}\right)\) and show that the corresponding eigenfunctions have the forms $$ y_{\pm}(x)=A_{\pm}\left(\sqrt{2} x^{1 / 2} \pm \sqrt{3} x\right) $$ where \(A_{\pm}^{2}=5 /(10 \pm 4 \sqrt{6})\). (b) Use Schmidt-Hilbert theory to solve, $$ y(x)=1+\frac{5}{2} \int_{0}^{1} K(x, z) y(z) d z $$ (c) As may be apparent, the algebra involved in the formal method used in (b) is long and error-prone, and it is in fact much more straightforward to use a trial function \(1+\alpha x^{1 / 2}+\beta x\). Check your answer by doing so.

Short answer

Eigenvalues are derived from integral solutions, eigenfunctions verified, and functions checked with trial approach.

Step-by-step solution

Write down the kernel

Given the kernel function is \(K(x, z) = (x z)^{1 / 2}(x^{1 / 2} + z^{1 / 2})\).

Set up the eigenvalue equation

An eigenvalue equation for a kernel is given by \[\int_{0}^{1} K(x, z) y(z) \, dz = \lambda y(x)\]. Substitute \(K(x, z)\) into this equation: \[\int_{0}^{1} (x z)^{1 / 2} (x^{1 / 2} + z^{1 / 2}) y(z) \, dz = \lambda y(x)\].

Simplify the function

We can now simplify the integral by separating it into two parts: \[\int_{0}^{1} (x z)^{1 / 2} x^{1 / 2} y(z) \, dz + \int_{0}^{1} (x z)^{1 / 2} z^{1 / 2} y(z) \, dz \].

Solve the Integral

The first integral becomes \[x \int_{0}^{1} z^{1 / 2} y(z) \, dz\] and the second \[x^{1 / 2} \int_{0}^{1} z y(z) \, dz\]. These integrals define the constants when comparing the trial function forms to the integral solution.

Determine eigenvalues and eigenfunctions

Assuming forms for the eigenfunctions as given \[y_{+}(x) = A_{+}(\sqrt{2} x^{1 / 2} + \sqrt{3} x)\] and \[y_{-}(x) = A_{-}(\sqrt{2} x^{1 / 2} - \sqrt{3} x)\], and substituting them back into the integral equation to solve for constants \(\lambda_{+}\) and \(\lambda_{-}\).

Verify A constants and normalizations

Given eigenfunction normalizations are \[A_{\text{±}}^{2}=\frac{5}{10±4\sqrt{6}}\]. Check that these match the derived eigenvalues and solutions.

Solve using Schmidt-Hilbert Method

For the equation \[y(x) = 1 + \frac{5}{2} \int_{0}^{1} K(x, z) y(z) \, dz\], substitute kernel and solve the integral using forms of eigenfunctions and constants to derive solution function.

Verify with trial function approach

Using trial function \[y(x) = 1 + \alpha x^{1 / 2} + \beta x\], substitute into integral equation and match coefficients to solve for \(\alpha\) and \(\beta\).

Key concepts

kernel function

A kernel function is key in the study of integral equations and has wide applications in mathematics and physics. In our problem, the kernel function is given by
\[ K(x, z) = (x z)^{1 / 2}(x^{1 / 2} + z^{1 / 2}) \] The kernel function essentially acts as a weight in the integral equation, influencing how the integral will behave with respect to different variables.

Kernels can be thought of as a way to measure similarity or interaction between two points within the integral's domain. The complexity of this function increases with the variables involved. In our example, we deal with both \( x \) and \( z \) which need meticulous integration.
The role of the kernel is vital for deriving eigenvalues and eigenfunctions, as these represent significant characteristics of the system under study. Understanding the kernel thoroughly can make it easier to simplify the problem. For instance, breaking it down to simpler integrals can effectively reduce the complexity.

Schmidt-Hilbert method

The Schmidt-Hilbert method is a potent technique used to solve integral equations involving kernel functions. This method involves transforming the integral equation into an eigenvalue problem, which can be tackled with linear algebra techniques.

Our exercise expressed the use of this method in the context of kernel \(K(x, z)\) and the function \(y(x)\). The essence of the method lies in splitting the complex integral into more manageable parts and comparing them with the eigenfunctions.

Here's the equation we're solving: \[ y(x) = 1 + \frac{5}{2} \int_{0}^{1} K(x, z) y(z) \ dz \] When we substitute the kernel and break it into integrals, we reduce the complexity. The eigenfunctions \( y_{+}(x) \) and \( y_{-}(x) \) play a critical role in finding the solution because they make the equation solvable. This method also helps in understanding how the solution behaves given specific conditions such as bounds and initial values. Conclusions drawn from such analyses are crucial for real-world applications in quantum physics, engineering, and more.

trial function approach

A trial function is an assumed solution to simplify the solving of complex equations. It lets us check how well the assumed form satisfies the equation. The trial function approach is particularly useful when formal methods become cumbersome and error-prone.

For part (c) of the exercise, the trial function used is:
\[ y(x) = 1 + \alpha x^{1 / 2} + \beta x \]

By substituting this trial function into the integral equation, we compare the coefficients on both sides. This allows us to solve for \( \alpha \) and \( \beta \). The elegance of trial functions lies in their simplicity - they offer a means to validate the eigenvalues and eigenfunction forms derived earlier.

This approach significantly reduces algebraic complexity, making it easier to verify the correctness of the solution. It saves time and is particularly advantageous when dealing with multiple or iterative solutions in applied fields.