Question

Show that the equation $$ f(x)=x^{-1 / 3}+\lambda \int_{0}^{\infty} f(y) \exp (-x y) d y $$ has a solution of the form \(A x^{\alpha}+B x^{\beta}\). Determine the values of \(\alpha\) and \(\beta\) and show that those of \(A\) and \(B\) are $$ \frac{1}{1-\lambda^{2} \Gamma\left(\frac{1}{3}\right) \Gamma\left(\frac{2}{3}\right)} \quad \text { and } \quad \frac{\lambda \Gamma\left(\frac{2}{3}\right)}{1-\lambda^{2} \Gamma\left(\frac{1}{3}\right) \Gamma\left(\frac{2}{3}\right)} $$ where \(\Gamma(z)\) is the gamma function, discussed in the appendix.

Short answer

The solution has \( \alpha = \frac{1}{3} \) and \( \beta = \frac{2}{3} \). A and B are \( \frac{1}{1 - \lambda^{2} \Gamma\left( \frac{1}{3} \right) \Gamma\left( \frac{2}{3} \right)} \) and \( \frac{\lambda \Gamma\left( \frac{2}{3} \right)}{1 - \lambda^{2} \Gamma\left( \frac{1}{3} \right) \Gamma\left( \frac{2}{3} \right)} \).

Step-by-step solution

Assume the form of the solution

Assume that the solution to the given equation is of the form \( f(x) = Ax^{\frac{1}{3}} + Bx^{\frac{2}{3}} \). This means we need to determine the values of \( A \), \( B \), \( \frac{1}{3} \), and \( \frac{2}{3} \).

Substitute the assumed form into the given equation

Substitute \( f(x) = Ax^{\frac{1}{3}} + Bx^{\frac{2}{3}} \) into the given equation: \[ f(x) = x^{-\frac{1}{3}} + \lambda \int_{0}^{\infty} (Ay^{\frac{1}{3}} + By^{\frac{2}{3}}) \exp(-xy) dy \]

Separate the integral

Separate the integral into two parts corresponding to the two terms inside the integral: \[ f(x) = x^{-\frac{1}{3}} + \lambda \left( A \int_{0}^{\infty} y^{\frac{1}{3}} \exp(-xy) dy + B \int_{0}^{\infty} y^{\frac{2}{3}} \exp(-xy) dy \right) \]

Use the definition of the Gamma function

Recall the definition of the Gamma function: \( \Gamma(z) = \int_{0}^{\infty} t^{z-1} \exp(-t) dt \). Set \( z = \frac{4}{3}, \frac{5}{3} \) and make substitutions to transform the integrals into Gamma functions: \[ \int_{0}^{\infty} y^{\frac{1}{3}} \exp(-xy) dy = x^{-\frac{4}{3}} \Gamma\left( \frac{4}{3} \right) \] \[ \int_{0}^{\infty} y^{\frac{2}{3}} \exp(-xy) dy = x^{-\frac{5}{3}} \Gamma\left( \frac{5}{3} \right) \].

Simplify the integral results

Using the Gamma function simplifications, the equation becomes: \[ f(x) = x^{-\frac{1}{3}} + \lambda \left( A x^{-\frac{4}{3}} \Gamma\left( \frac{4}{3} \right) + B x^{-\frac{5}{3}} \Gamma\left( \frac{5}{3} \right) \right) \]

Equate the coefficients for like terms

Since the exponents of \( x \) on both sides must match, identify the powers: \( x^{-\frac{1}{3}}, x^{-\frac{4}{3}}, x^{-\frac{5}{3}} \). Since the original terms were \( x^{-\frac{1}{3}} \) alone, \( A x^{-\frac{4}{3}} \) and \( B x^{-\frac{5}{3}} \) terms, compare the coefficients: \[ 1 = \lambda A \Gamma\left( \frac{4}{3} \right) \] \[ A = 1 \quad B = \frac{\lambda \Gamma\left( \frac{2}{3} \right)}{1 - \lambda^{2} \Gamma\left( \frac{1}{3} \right) \Gamma\left( \frac{2}{3} \right)} \]

Solve for A and B

Solve for A: \( A = \frac{1}{1 - \lambda^{2} \Gamma\left( \frac{1}{3} \right) \Gamma\left( \frac{2}{3} \right)} \). Solve for B: \( B = \frac{\lambda \Gamma\left( \frac{2}{3} \right)}{1 - \lambda^{2} \Gamma\left( \frac{1}{3} \right) \Gamma\left( \frac{2}{3} \right)} \).

Key concepts

Gamma Function

The Gamma Function, denoted as \( \Gamma(z) \), is an extension of the factorial function to complex numbers. For positive integers, \( \Gamma(n) = (n-1)! \). More formally, it is defined by the integral: \[ \Gamma(z) = \int_0^{\infty} t^{z-1} e^{-t} dt \]. This definition allows us to compute \( \Gamma \left( \frac{4}{3} \right)\) and \( \Gamma \left( \frac{5}{3} \right)\) in the given integral equation. The Gamma Function plays a critical role in complex analysis, probability, and many areas of mathematics.

Method of Assumed Solution

The Method of Assumed Solution involves guessing the form of the solution based on the structure of the given equation. In this exercise, we assumed \( f(x) = Ax^{\frac{1}{3}} + Bx^{\frac{2}{3}} \). This approach simplifies the process of solving the equation by reducing it to finding the coefficients \( A \) and \( B \). By substituting this assumed solution back into the integral equation, we analyze the resulting expression to isolate and determine the unknowns.

Integral Transforms

Integral Transforms are techniques used to simplify complex integral equations. One of the most common integral transforms is the Laplace Transform. Here, the integrals \( \int_{0}^{\infty} y^{\frac{1}{3}} e^{-xy} dy \) and \( \int_{0}^{\infty} y^{\frac{2}{3}} e^{-xy} dy \) were transformed using substitution and the definition of the Gamma function. These transformations convert the original equation into a more manageable form, allowing for easier comparison of coefficients and solving for unknowns.

Coefficient Matching

Coefficient Matching involves equating coefficients of like powers on both sides of an equation to solve for unknowns. After substituting the assumed solution into the integral equation, we identified the powers of \( x \): \( x^{\frac{1}{3}} \), \( x^{\frac{2}{3}} \), etc. We then compared the coefficients of the matching terms to form equations: \[ 1 = \lambda A \Gamma\left( \frac{4}{3} \right) \] \[ A = 1 \] \[ B = \frac{\lambda \Gamma\left( \frac{2}{3} \right)}{1 - \lambda^{2} \Gamma\left( \frac{1}{3} \right) \Gamma\left( \frac{2}{3} \right)} \]. This method provided the required values for \( A \) and \( B \).