Question

These exercises explore some additional aspects of the radiative transport model presented in Section 11.1. Consider the two-point boundary value problem $$ \begin{aligned} &\frac{d}{d x}\left[\begin{array}{l} I^{(+)} \\ I^{(-)} \end{array}\right]=\beta\left[\begin{array}{ll} -1 & 1 \\ -1 & 1 \end{array}\right]\left[\begin{array}{l} I^{(+)} \\ I^{(-)} \end{array}\right], \quad 0

Short answer

Question: Show that the difference between the radiative intensities \(I^{(+)}(x)\) and \(I^{(-)}(x)\) is constant and transform the boundary value problem with a variable scattering coefficient \(\beta(x)\) in terms of \(\xi\). Answer: The difference between the radiative intensities \(I^{(+)}(x)\) and \(I^{(-)}(x)\) is constant, as the derivative of their difference is equal to zero. The boundary value problem can be transformed by changing the independent variable \(x\) to \(\xi\) through the expressions \(\frac{dI^{(\pm)}(\xi)}{d\xi} = \frac{1}{\beta(x)} \frac{dI^{(\pm)}(x)}{dx}\) and the boundary conditions in terms of \(\xi\). The resulting boundary value problem depends only on \(\xi\) and \(\xi_{l}\).

Step-by-step solution

Part (a) : Show that the difference between \(I^{(+)}(x)\) and \(I^{(-)}(x)\) is constant

Firstly, we must observe the given system of differential equations: $$ \frac{d}{d x}\left[\begin{array}{l} I^{(+)} \\\ I^{(-)} \end{array}\right]=\beta\left[\begin{array}{ll} -1 & 1 \\\ -1 & 1 \end{array}\right]\left[\begin{array}{l} I^{(+)} \\\ I^{(-)} \end{array}\right], \quad 0<x<l . $$ To find out whether the difference \(I^{(+)}(x) - I^{(-)}(x)\) is constant, we have to differentiate it with respect to \(x\): $$ \frac{d}{dx}(I^{(+)}(x) - I^{(-)}(x)). $$ Using chain rule, we have: $$ \frac{dI^{(+)}(x)}{dx} - \frac{dI^{(-)}(x)}{dx}. $$ Now, let's recall the original system of differential equations and substitute their expressions: $$ \beta(-I^{(+)} + I^{(-)}) - \beta(-I^{(+)} + I^{(-)}) = 0. $$ It is clear that the difference \(I^{(+)}(x) - I^{(-)}(x)\) is constant as its derivative is equal to zero.

Part (b) : Transform the boundary value problem with variable scattering coefficient

Firstly, let's define \(\xi\): $$ \xi=\int_{0}^{x} \beta(\lambda) d \lambda, $$and let \(\xi_{l}=\int_{0}^{l} \beta(\lambda) d \lambda .\) To transform the boundary value problem, we need to change the independent variable \(x\) to \(\xi\). For this, we'll find the derivative of \(I^{(\pm)}(\xi)\) with respect to \(\xi\): $$ \frac{dI^{(\pm)}(\xi)}{d\xi} = \frac{dI^{(\pm)}(x)}{dx} \frac{dx}{d\xi}. $$ We need to find the expression for \(\frac{dx}{d\xi}\). Since we have \(\xi = \int_{0}^{x} \beta(\lambda) d \lambda\), we can differentiate both sides with respect to \(x\) to get: $$ \frac{d\xi}{dx} = \beta(x). $$ Thus, we have: $$ \frac{dx}{d\xi} = \frac{1}{\beta(x)}. $$ Now, we can substitute this into the expression for \(\frac{dI^{(\pm)}(\xi)}{d\xi}\): $$ \frac{dI^{(\pm)}(\xi)}{d\xi} = \frac{1}{\beta(x)} \frac{dI^{(\pm)}(x)}{dx}. $$ Recall the differential equation system for \(I^{(\pm)}(x)\): $$ \frac{d}{d x}\left[\begin{array}{l} I^{(+)} \\\ I^{(-)} \end{array}\right]=\beta\left[\begin{array}{ll} -1 & 1 \\\ -1 & 1 \end{array}\right]\left[\begin{array}{l} I^{(+)} \\\ I^{(-)} \end{array}\right], \quad 0<x<l . $$ With our transformations, we get: $$ \frac{d}{d \xi}\left[\begin{array}{l} I^{(+)} \\\ I^{(-)} \end{array}\right]=\left[\begin{array}{ll} -1 & 1 \\\ -1 & 1 \end{array}\right]\left[\begin{array}{l} I^{(+)} \\\ I^{(-)} \end{array}\right], \quad 0<\xi<\xi_{l}. $$ The boundary conditions are transformed as well: $$ \left.I^{(+)}\right|_{\xi=0}=I^{\text {inc }},\left.\quad I^{(-)}\right|_{\xi-\xi_{l}}=0 . $$ The resulting boundary value problem is shown to depend only on \(\xi\) and \(\xi_{l}\). In the case where \(\beta\) is constant, \(\xi=\beta x\) and \(\xi_{l}=\beta l\).

Key concepts

Radiative Transport Model

The radiative transport model is a mathematical model used to describe how radiation, such as light or heat, moves through a medium. This model is particularly important in fields like astrophysics, atmospheric science, and optical engineering. It helps in understanding how radiation is absorbed, emitted, and scattered within a medium, which can range from the atmosphere to human tissue.

In the context of our exercise, the radiative transport model explains the movement of light within a slab. Here, the light intensities, denoted as \(I^{(+)}\) and \(I^{(-)}\), indicate how radiation is moving in two opposite directions. The complexity of the model often involves solving differential equations that incorporate physical properties like absorption and scattering. These differential equations can be balanced by boundary conditions that specify how the light enters and exits the medium.

Understanding this model allows us to predict how light behaves under various conditions, which is crucial when designing optical instruments or analyzing atmospheric phenomena.

Boundary Value Problem

A boundary value problem is a mathematical problem where we need to determine a function based on its values at the boundary of its domain. Such problems are fundamental in mathematical physics, engineering, and applied mathematics. They often occur in the context of differential equations, which describe how a function behaves across a specified range.

In our case, the boundary value problem is given by differential equations that govern the light intensities \(I^{(+)}\) and \(I^{(-)}\) inside the slab. The problem is guided by the boundary conditions, \(I^{(+)}(0) = I^{\text{inc}}\) and \(I^{(-)}(l) = 0\), meaning that initially, there is an incoming light intensity, and at the other end of the slab, there is no outgoing intensity.

Solving boundary value problems typically involves finding a solution that satisfies both the differential equation and the boundary conditions. Such solutions are crucial for predicting physical behaviors in systems ranging from thermal conduction to wave propagation.

Scattering Coefficient

The scattering coefficient, denoted as \(\beta\), is a measure of how much light or radiation is scattered by a medium per unit path length. It is a critical parameter in the study of radiative transport and greatly affects the movement of light through a material.

In this exercise, the scattering coefficient \(\beta\) could vary with position, meaning that the property of the medium is not uniform across the slab. This variation changes how the light is scattered as it moves, influencing the solution to our problem.

A variable scattering coefficient leads to more complex mathematical models, as seen in our transformation to the variable \(\xi\). Understanding \(\beta\) allows us to model and predict how light will diffuse through different materials, which is key in applications like designing lenses, lighting, and analyzing atmospheric conditions.

Variable Change in Calculus

Variable change is a technique used in calculus to simplify problems, often involving integrals or differential equations. By redefining variables, complex expressions can be transformed into simpler forms, enabling easier analysis or solution.

In this exercise, we introduce a new variable \(\xi\) to simplify the boundary value problem involving the differential equation. The transformation \(\xi = \int_{0}^{x} \beta(\lambda) \, d\lambda\) allows us to deal with a potentially varying scattering coefficient \(\beta\).

This technique is particularly useful when the original variables create complications, such as non-constant coefficients. With the change to \(\xi\), we convert the problem into a form where the differential equation does not depend on the original variables, making it easier to analyze and solve. The ability to perform variable changes is an essential tool in calculus and often provides pathways to solutions that are not immediately obvious in the problem's original form.