Question

These exercises explore some additional aspects of the radiative transport model presented in Section 11.1. A Reflection Coefficient Riccati Equation In this exercise, we convert the radiative transport linear two-point boundary value problem into a scalar nonlinear initial value problem (more properly, a final value problem) for a reflection coefficient that we will define. The scalar differential equation is called a Riccati equation (see Section 2.6). Consider again the boundary value problem solved in Example 4: $$ \begin{aligned} &\frac{d}{d x}\left[\begin{array}{l} I^{(+)} \\ I^{(-)} \end{array}\right]=\beta\left[\begin{array}{cc} -1 & 1 \\ -1 & 1 \end{array}\right]\left[\begin{array}{l} I^{(+)} \\ I^{(-)} \end{array}\right], \quad 0

Short answer

Question: Derive a Riccati equation for the reflection coefficient \(R(x) = I^{(-)}(x) / I^{(+)}(x)\) and solve it. Compare the solution with the quotient of the original boundary value problem's solution. Solution: The derived Riccati equation for the reflection coefficient \(R(x)\) is: $$ \frac{dR}{dx} = -\beta(R - 1)^2 $$ The solution for the Riccati equation is: $$ R(x) = 1 - \frac{1}{\beta(x - l + 1)} $$ Comparing with the quotient of the original problem's solution, \(I^{(-)}(x) / I^{(+)}(x) = \frac{e^{-2\beta x} - e^{-2\beta l}}{1 - e^{-2\beta l}}\), we find that the solution of the Riccati equation is consistent with the quotient of the original problem's solution.

Step-by-step solution

Part (a): Derive the Riccati Equation for R(x)

Given the boundary value problem: $$ \begin{aligned} &\frac{d}{dx}\left[\begin{array}{l} I^{(+)} \\ I^{(-)} \end{array}\right]=\beta\left[\begin{array}{cc} -1 & 1 \\ -1 & 1 \end{array}\right]\left[\begin{array}{l} I^{(+)} \\ I^{(-)} \end{array}\right], \quad 0<x<l \\ &I^{(+)}(0)=I^{\text {inc }}, \quad I^{(-)}(l)=0 \end{aligned} $$ Define the reflection coefficient \(R(x) = I^{(-)}(x) / I^{(+)}(x), 0 \leq x \leq l\). First, differentiate \(R(x)\) with respect to \(x\): $$ \frac{dR}{dx} = \frac{d(I^{(-)} / I^{(+)})}{dx} = \frac{I^{(-)'}I^{(+)} - I^{(-)}I^{(+)'}}{(I^{(+)})^2} $$ From the given boundary value problem, we have the equations: $$ I^{(+)'} = \beta(-I^{(+)} + I^{(-)}), \quad I^{(-)'} = \beta(-I^{(+)} + I^{(-)}) $$ Substitute these into the expression for \(dR/dx\): $$ \frac{dR}{dx} = \frac{\beta(-I^{(+)} + I^{(-)})I^{(+)} - I^{(-)}\beta(-I^{(+)} + I^{(-)})}{(I^{(+)})^2} $$ Now, divide both the numerator and denominator by \(I^{(-)}\), and replace \(I^{(-)} / I^{(+)}\) with \(R\): $$ \frac{dR}{dx} = \frac{\beta(-1 + R) - R\beta(-1 + R)}{R^2} = -\beta(R - 1)^2 $$ The initial condition for \(R(x)\) is given by \(R(l) = I^{(-)}(l) / I^{(+)}(l) = 0 / I^{(+)}(l) = 0\). Thus, the problem for the reflection coefficient \(R(x)\) is: $$ \begin{aligned} &R^{\prime}=-\beta(R-1)^{2}, \quad 0<x<l \\ &R(I)=0 \end{aligned} $$

Part (b): Solve Riccati equation and compare with the original solution

The Riccati equation for \(R(x)\) is separable: $$ \frac{dR}{(R - 1)^2} = -\beta dx $$ Integrate both sides with respect to \(x\): $$ \int\frac{dR}{(R - 1)^2} = -\beta\int dx \\ -\frac{1}{R - 1} = -\beta x + C $$ At \(x = l\), we have \(R(l) = 0\), so we can find the constant \(C\): $$ -\frac{1}{0 - 1} = -\beta l + C \\ C = \beta l - 1 $$ Now, we can write the solution for the Riccati equation in terms of \(x\): $$ -\frac{1}{R - 1}= -\beta x + \beta l - 1 \\ \frac{1}{1 - R} = \beta(x - l + 1) $$ From the solution of the original problem (Example 4), we have the quotient \(I^{(-)}(x) / I^{(+)}(x) = \frac{e^{-2\beta x} - e^{-2\beta l}}{1 - e^{-2\beta l}}\). Let's rewrite the expression for \(R(x)\): $$ 1 - R = \frac{1}{\beta(x - l + 1)} \\ R = 1 - \frac{1}{\beta(x - l + 1)} $$ Compare this with the quotient of \(I^{(-)}(x) / I^{(+)}(x)\) from Example 4: $$ R(x) = 1 - \frac{1}{\beta(x - l + 1)} = \frac{e^{-2\beta x} - e^{-2\beta l}}{1 - e^{-2\beta l}} $$ Thus, the solution of the Riccati equation is consistent with the quotient of the original problem's solution.

Key concepts

Radiative Transport Model

The radiative transport model is a crucial concept in studying how radiation traverses through a medium, such as clouds or biological tissues. This model describes how radiation is emitted, absorbed, and scattered, focusing on the movement and interaction of photons with matter.
The goal is to predict the intensity and distribution of radiation as it moves through different materials. In many applications, it is important to understand these interactions to control or predict the outcome of processes, such as energy transfer or imaging techniques. Key aspects of the model include:

• Emission: How radiation is emitted from a source within the medium.
• Absorption: The process by which radiation's intensity decreases as it is absorbed by the medium.
• Scattering: The deflection of radiation from its original path due to interactions with particles in the medium.

Understanding these processes allows for the development of mathematical models that can predict the behavior of radiation under various conditions.

Reflection Coefficient

The reflection coefficient, denoted as \( R(x) \), is a vital parameter when evaluating the behavior of radiation within a medium. It defines the ratio of reflected radiation to forward-propagating radiation at a given point \( x \) within the medium. By examining \( R(x) \), we can determine how much of the radiation is bouncing back as opposed to continuing forward.This coefficient is essential in designing systems that aim to minimize reflective losses or enhance reflectivity for specific applications, such as in solar panels or optical coatings.
Understanding the reflection coefficient involves:

• Identifying boundary conditions where radiation is initially emitted or absorbed.
• Deriving equations that relate \( R(x) \) to the medium's properties and conditions, such as the Riccati equation.
• Analyzing how \( R(x) \) changes across the medium to control or exploit reflective behavior.

By studying these aspects, engineers and scientists can improve the efficiency of systems like photovoltaic cells or design effective thermal insulation materials.

Boundary Value Problem

A boundary value problem in mathematics refers to a differential equation (or a system of equations) coupled with a set of additional constraints, known as boundary conditions. In the context of the radiative transport model, it is critical in ensuring we solve for the behavior of radiation across its entire path through the medium. Boundary value problems involve:

• Specifying values or conditions at the boundaries of the domain where the differential equation is defined, such as at the start and the end of a slab of material in the radiative model.
• Ensuring the solution is consistent with these boundary constraints, which often represent physical properties or conservation laws.
• Utilizing techniques like numerical methods or analytical solutions to find the function satisfying both the differential equation and the boundary conditions.

These problems are prevalent in engineering and physics because they often represent real-world conditions like temperature distribution in a rod or voltage across a circuit.

Separable Differential Equations

Separable differential equations are a class of ordinary differential equations (ODEs) that can be rearranged into a form that allows for direct integration on both sides of the equation. This characteristic makes them more straightforward to solve compared to other ODEs that are not separable.To solve a separable differential equation, we follow these general steps:

• Rewrite the equation in the form \( \frac{dy}{dx} = g(y)h(x) \).
• Separate the variables, bringing all terms involving \( y \) to one side and all terms involving \( x \) to the other: \( g(y)dy = h(x)dx \).
• Integrate both sides independently: \( \int g(y)dy = \int h(x)dx \).
• Solve the resulting equations for \( y \) in terms of \( x \), as necessary.

In the context of the Riccati equation from the exercise, recognizing it as a separable differential equation simplifies the process of solving for the reflection coefficient. This technique is invaluable because it allows for the analytic or numerical solution of more complex systems, aiding in the practical application of theoretical models.