Problem 25 - Advanced Engineering Mathematics

Question

Find the steady-state temperatures \(u(r, z)\) in the semi-infinite cylinder in Problem 24 if the base of the cylinder is insulated and $$ u(1, z)=\left\\{\begin{array}{ll} u_{0}, & 01 \end{array}\right. $$

Step-by-Step Solution

Answer

The steady-state temperature is found by solving Bessel's and trigonometric equations with piecewise boundary conditions.

Step 1: Understanding the Problem

We need to find the steady-state temperature distribution \(u(r, z)\) in a semi-infinite cylinder with \(r = 1\) as the boundary where the temperature is specified. The base is insulated, which means there is no heat flux through \(z = 0\). We have piecewise boundary conditions on the surface \(r = 1\) given by \(u_0\) for \(0 < z < 1\) and 0 for \(z > 1\).

Step 2: Setting Up the Boundary Conditions

The boundary conditions are given by:1. \(\frac{\partial u}{\partial z}(r,z) \bigg|_{z=0} = 0\) due to insulation at the base.2. \(u(1, z)=\begin{cases} u_0, & 01 \end{cases}\) on the surface of the cylinder.

Step 3: Formulate the Problem in Cylindrical Coordinates

Since the problem deals with a symmetry in \(r\), \(z\), and we are not given any \(\theta\) dependence, we consider the heat equation in radial symmetry:\[ \frac{1}{r}\frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right) + \frac{\partial^2 u}{\partial z^2} = 0 \] which simplifies using separation of variables under given conditions.

Step 4: Solve Using Separation of Variables

Assume \( u(r, z) = R(r)Z(z) \). Substituting into the heat equation, we get \(\frac{r}{R} \frac{d}{dr}\left(r \frac{dR}{dr}\right) = -\frac{1}{Z} \frac{d^2Z}{dz^2} = \lambda \). This splits into two equations:\[\frac{1}{r}\frac{d}{dr}\left(r \frac{dR}{dr}\right) - \lambda R = 0 \ \frac{d^2Z}{dz^2} + \lambda Z = 0\]

Step 5: Solve Radial Part of the Differential Equation

The radial equation is \( r^2 \frac{d^2R}{dr^2} + r \frac{dR}{dr} - \lambda r^2 R = 0 \), which is a Bessel's differential equation. For \(r=1\), it needs \(J_0(\sqrt{\lambda}r)\) forms as solution that fits boundary conditions.

Step 6: Solve Axial Part of the Differential Equation

The axial equation is \( \frac{d^2Z}{dz^2} + \lambda Z = 0 \) which gives solutions:\\( Z(z) = A \cos(\sqrt{\lambda} z) + B \sin(\sqrt{\lambda} z) \). Use boundary conditions to find coefficients.

Step 7: Satisfy Boundary Conditions

The boundary conditions are then applied to specify any coefficients in our solutions. Particularly, match piecewise conditions at \( z=0 \) with \( \partial Z/\partial z(0)=0 \) and continuity along \(z=1\) and decay at infinite \(z\). Adjust eigenvalues and coefficients \(A\) and \(B\).

Step 8: Combine Solutions and Normalize

Combine radial and axial solutions into a general form solution,\[ u(r,z) = \sum_{n} C_n J_0(\sqrt{\lambda_n} r) \left( A_n \cos(\sqrt{\lambda_n} z) + B_n \sin(\sqrt{\lambda_n} z) \right) \] and determine \( C_n \) using orthogonality with respect to modified conditions.

Key Concepts

cylindrical coordinates

In mathematics and physics, cylindrical coordinates are a coordinate system that extends the two-dimensional polar coordinates to three dimensions by adding an additional height (usually denoted as \( z \)). Cylindrical coordinates are particularly useful for problems that have rotational symmetry about an axis, often simplifying the mathematics involved compared to Cartesian coordinates.

• The coordinate \( (r, \theta, z) \) represents a point in space, where:
• \( r \) is the radial distance from the z-axis.
• \( \theta \) is the angular coordinate (not needed in this problem as we assume symmetry).
• \( z \) is the height along the axis.

In the context of this exercise, the coordinate \( \theta \) is ignored due to the symmetry of the problem around the z-axis, leaving us with coordinates \( (r, z) \) to describe the steady-state temperature distribution. The use of cylindrical coordinates allows us to specifically address the radial nature of the temperature changes along a semi-infinite cylinder.

Bessel's differential equation

Bessel's differential equation is a second-order ordinary differential equation that appears in various scientific fields when dealing with radial symmetry, especially in cylindrical and spherical problems.

• The general form is: \[ x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + (x^2 - n^2)y = 0 \] where \( n \) is a constant called the order of the equation.
• In our cylindrical problem, it simplifies to: \[ r^2 \frac{d^2R}{dr^2} + r \frac{dR}{dr} - \lambda r^2 R = 0 \] where solutions involve the Bessel functions \( J_n(x) \).

Particularly useful are the Bessel functions of the first kind, such as \( J_0 \), which are used for solutions that are finite at the origin, as is often required in physical contexts.In this exercise, we use \( J_0(\sqrt{\lambda} r) \) because we are dealing with a zero-order Bessel function which corresponds to the radial part of the problem. This choice fits seamlessly within the boundary conditions at \( r = 1 \), ensuring the solution behaves well in the cylindrical geometry.

boundary conditions

Boundary conditions are crucial to solving differential equations as they define the settings that any solution must satisfy. They can be values, slopes, or combinations at the boundaries of the domain where the solution is defined.

• In our cylindrical problem, the boundary conditions are defined on the surface and base of the cylinder.
• The surface condition at \( r = 1 \) is piecewise: \( u(1, z) = \begin{cases} u_0, & 01 \end{cases} \).
• The condition at the base \( z = 0 \) is insulated, meaning no heat flux across: \( \frac{\partial u}{\partial z}(r,z) \bigg|_{z=0} = 0 \).

These conditions allow us to determine unknown constants and ensure that the mathematical solution corresponds to the physical scenario. For the insulated base, the derivative condition ensures continuity and energy conservation.

separation of variables

Separation of variables is a technique used to solve partial differential equations (PDEs). By assuming that each part of the variable product can be described independently, it's easier to tackle complex equations.The process involves:

• Assuming a solution can be written as the product of functions, each in a separate coordinate, such as \( u(r, z) = R(r)Z(z) \).
• Substituting this form into the PDE to separate the variables based on which they depend (usually results in two or more ordinary differential equations).

For our problem:

• We separate the radial and axial components in the heat equation, leading to: \[ \frac{r}{R} \frac{d}{dr}\left(r \frac{dR}{dr}\right) = -\frac{1}{Z} \frac{d^2Z}{dz^2} = \lambda \].

This technique not only simplifies the complex PDE into manageable ordinary differential equations but also allows the exploitation of the symmetry and boundary conditions in the dimensionality of the problem. Through it, we transform the physical problem into a mathematically solvable form, using known solutions like Bessel functions.