Chapter 13: Problem 2

Find the steady-state temperature distribution in a solid semi-infinite cylinder if the boundary temperatures are \(u=0\) at \(r=1\) and \(u=y=r \sin \theta\) at \(z=0\). Hins: In (5.10) you want the solution containing \(\sin \theta ;\) therefore you want the functions \(J_{1}\). You will need to integrate \(r^{2} J_{1}\); follow the text method of integrating \(r J_{0}\) just before \((5,15)\).

Short Answer

Expert verified

The steady-state temperature is \[ u(r, \theta, z) = \sum_{n=1}^{\infty} A_n J_{1}(\lambda_n r)\sin \theta e^{-\lambda_n^2 \alpha z} \].

Step by step solution

Understanding the Problem

The problem involves finding the steady-state temperature distribution in a semi-infinite cylinder with given boundary conditions: the temperature \(u = 0\) at radius \(r = 1\) and \(u = y = r \sin\theta\) at \(z = 0\). This involves solving the heat equation with appropriate boundary conditions.

Applying Boundary Conditions

Apply the given boundary conditions. At \(r = 1\), the temperature \(u = 0\), and at \(z = 0\), the temperature \(u = y = r \sin\theta\).

Using Bessel Functions

Since the solution should contain \(\sin \theta\), we use the Bessel function of the first kind \(J_1\) which satisfies the boundary conditions.

General Solution Form

The general solution can be written in the form: \[ u(r, \theta, z) = \sum_{n=1}^{\infty} R_n(r) \Theta_n(\theta) Z_n(z) \].

Solving Radial Part

Solve the radial part using the Bessel function \(R(r) = J_{1}(\lambda r)\). Apply the boundary condition \(u=0\) at \(r=1\), giving \[ J_{1}(\lambda) = 0 \].

Applying Fourier-Bessel Series

Expand the boundary condition at \(z=0\), \(u = y = r \sin \theta\), in a Fourier-Bessel Series. This gives: \[ u(r, \theta, 0) = \sum_{n=1}^{\infty} A_n J_{1}(\lambda_n r)\sin \theta \].

Solving for Coefficients

Solve for the coefficients \(A_n\) by integrating the Radial and Angular parts with respect to the boundary conditions.

Final Steady-State Solution

Combine all parts to form the final steady-state temperature distribution solution: \[ u(r, \theta, z) = \sum_{n=1}^{\infty} A_n J_{1}(\lambda_n r)\sin \theta e^{-\lambda_n^2 \alpha z} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

semi-infinite cylinder

A semi-infinite cylinder is a cylindrical object extending infinitely in one direction. Imagine a cylinder that goes endlessly in the positive or negative z-direction. In mathematical problems, this simplifies the boundary conditions we need to consider. For instance, in this exercise, we only focus on how the temperature changes within the part of the cylinder defined by the boundary conditions: at radius r=1 and at the surface z=0.

boundary conditions

Boundary conditions are crucial for solving differential equations. They define the conditions that our solution must satisfy at the boundaries of the domain we're interested in. Here, we have two boundary conditions: At the surface of the cylinder where \(r=1\), the temperature \(u=0\), and at \(z=0\) on the surface of the cylinder, the temperature \(u=y=r\text{sin}\theta\). These conditions help us find and check our solutions to ensure they are physically viable.

Bessel functions

Bessel functions come from solving certain types of differential equations, especially those with cylindrical symmetry like our semi-infinite cylinder. The Bessel function of the first kind, \(J_1\), is particularly important when our boundary conditions involve sines or cosines. In this exercise, we use \(J_1\) to simplify and solve our radial part of the solution. Bessel functions are essential because they help convert complex equations into forms that we can solve more easily.

Fourier-Bessel series

The Fourier-Bessel series combines Fourier series and Bessel functions. This combination is particularly useful when dealing with cylindrical geometries. Fourier series decompose a function into sines and cosines, while the Bessel functions handle the cylindrical coordinates. By expanding our boundary condition at \(z=0\) into a Fourier-Bessel series, \[/\text u(r, \theta, 0) = \sum_{n=1}^{\infty} A_n J_{1}(\text{\lambda_n r})\text{sin} \theta \]. This represents our temperature distribution in a manageable mathematical form.

heat equation

The heat equation describes how heat spreads through a medium over time. For steady-state problems, it simplifies to Laplace's equation, as we assume the temperature doesn't change with time. The equation in cylindrical coordinates is \[\frac{\text{\partial^2 u}}{\text{\partial r^2}} + \frac{1}{\text{r}} \frac{\text{\partial u}}{\text{\partial r}} + \frac{1}{\text{r^2}} \frac{\text{\partial^2 u}}{\text{\partial \theta^2}} + \frac{\text{\partial^2 u}}{\text{\partial z^2}} = 0 \]. Solving this with our given boundary conditions gives us the temperature distribution within the cylinder.