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Chapter 13: Problem 9

A circular plate is a composite of two different materials in the form of concentric circles. See Figure 13.12. The temperature in the plate is determined from the boundary-value problem $$ \begin{array}{ll} \frac{\partial^{2} u}{\partial r^{2}}+\frac{1}{r} \frac{\partial u}{\partial r}=\frac{\partial u}{\partial t}, & 00 \\ u(2, t)=100, & t>0 \\ u(r, 0)= \begin{cases}200, & 0

Short Answer

Expert verified
Solve using \(u(r, t) = v(r, t) + \psi(r)\) to meet both boundary and initial conditions.

Step by step solution

01

Understand the Problem

We are given a partial differential equation (PDE) that models the temperature in a circular plate with concentric circles having different initial conditions. The plate's temperature is described as a function of both radius \(r\) and time \(t\).
02

Decompose the PDE

Let's decompose the solution by introducing a change of variables, \(u(r, t) = v(r, t) + \psi(r)\). Here, \(\psi(r)\) is a steady-state component, and \(v(r, t)\) will adjust to meet initial conditions.
03

Identify Steady-State Solution

The steady-state solution \(\psi(r)\) should satisfy the boundary condition and should be chosen such that \(\frac{\partial \psi}{\partial t} = 0\). It's given that \(u(2, t) = 100\) for all \(t\), so we can set \(\psi(2) = 100\). To further define \(\psi(r)\), assume \(\psi\) is a simple function that adheres to boundary conditions.
04

Set Up the Initial Conditions for v

Using \(u(r, 0) = v(r, 0) + \psi(r)\), we can find \(v(r, 0) = u(r, 0) - \psi(r)\). This leads to \(v(r, 0) = 200 - \psi(r)\) for \(0<r<1\) and \(v(r, 0) = 100 - \psi(r)\) for \(1<r<2\). \(v(r, 0)\) ensures that the transient behavior accounts for the differences in initial conditions.
05

Solve for v(r, t)

The function \(v(r, t)\) should satisfy the PDE with the initial condition derived in the previous step. This step requires solving the resulting PDE for \(v(r, t)\) with appropriate techniques, potentially using a method like separation of variables or another applicable technique based on the boundary conditions.
06

Combine Solutions

Once \(v(r, t)\) and \(\psi(r)\) are determined, the full solution to the original problem is \(u(r, t) = v(r, t) + \psi(r)\). This satisfies both the boundary and initial conditions of the problem.

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Key Concepts

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

Partial Differential Equation
A partial differential equation (PDE) is an equation that involves multiple independent variables, unknown functions, and their partial derivatives. In this problem, the PDE "\( \frac{\partial^{2} u}{\partial r^{2}}+\frac{1}{r} \frac{\partial u}{\partial r}=\frac{\partial u}{\partial t} \)" represents the distribution of temperature in a circular plate composed of two materials. Half of the equation deals with spatial change (changes in radius \(r\)), while the other half deals with changes over time \(t\). Understanding and solving PDEs involves:
  • Identifying variables: Here, \( u \) is a function of \(r\) and \(t\), representing the temperature at those points.
  • Recognizing boundary conditions: These conditions provide values of \( u \) at specific points to help find unique solutions.
  • Applying initial conditions: These conditions give the state of \( u \) at \(t = 0\), supplying necessary information for time-based predictions.
Such equations are crucial in modeling physical processes where changes depend on more than one factor, such as heat distribution.
Steady-State Solution
A steady-state solution refers to a state where the system's behavior becomes constant over time. In the context of this exercise, \( \psi(r) \) is the steady-state component of the solution for temperature \( u(r, t) \). This part of the solution signifies that the change over time is zero, i.e., \( \frac{\partial \psi}{\partial t} = 0 \). It represents:
  • The state in which the system doesn't change with time.
  • Satisfies boundary conditions without time dependence.
In the problem statement, given \( u(2, t) = 100 \) for all \( t \), \( \psi(2) = 100 \). This ensures the temperature at the boundary \( r = 2 \) is stable and consistent, forming the foundation for solving the transient or time-varying part \( v(r, t) \). Steady-state solutions are vital as they simplify the analysis by isolating the time-invariant aspects of a system.
Initial Conditions
In a boundary-value problem, initial conditions are used to define the state of the system at the starting point of observation, usually \( t = 0 \). In this task:
  • We have: \( u(r, 0) = 200 \text{ for } 0
  • And \( u(r, 0) = 100 \text{ for } 1
These initial conditions convey how temperature is distributed across the circular plate initially. They are key for:
  • Setting up conditions from which the transient solution \( v(r, t) \) is derived.
  • Ensuring the model matches the physical situation at the beginning of the observation.
These conditions work together with boundary conditions to provide a detailed picture of the problem environment, ensuring unique and precise predictions for \( u(r, t) \). Understanding these is essential for forming correct solutions and interpretations.
Circular Plate
The shape and structure of the object are critical for formulating the boundary-value problem. A circular plate, in this context, refers to a disk-like surface made of concentric circles of different materials. This configuration impacts the calculation and boundaries:
  • The PDE accounts for cylindrical coordinates due to the circular symmetry, hence the terms involving \( r \) (radius).
  • Different materials imply differing thermal properties, influencing how the temperature equation behaves.
These factors contribute to constructing an accurate model and solutions that appropriately fit the physics of heat distribution within the plate. Understanding the geometrical setup is crucial for properly defining and applying initial and boundary conditions, and ensuring they reflect the true physical constraints and behaviors expected in reality.

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Most popular questions from this chapter

Find the steady-state temperature \(u(r, \theta)\) in a circular plate of radius \(r=1\) if the temperature on the circumference is as given. \(u(1, \theta)=\theta, \quad 0<\theta<2 \pi\)

Find the steady-state temperature \(u(r, \theta)\) in a circular plate of radius \(c\) if the temperature on the circumference is given by $$ u(c, \theta)=\left\\{\begin{aligned} u_{0}, & 0<\theta<\pi \\ -u_{0}, & \pi<\theta<2 \pi . \end{aligned}\right. $$

Find the steady-state temperature \(u(r, \theta)\) in a semicircular plate of radius \(r=1\) if $$ \begin{array}{ll} u(1, \theta)=u_{0}, & 0<\theta<\pi \\ u(r, 0)=0, & u(r, \pi)=u_{0}, \quad 0

Find the steady-state temperature \(u(r, \theta)\) in a semicircular plate of radius \(r=2\) if $$ u(2, \theta)= \begin{cases}u_{0} & 0<\theta<\pi / 2 \\ 0, & \pi / 2<\theta<\pi\end{cases} $$ \(u_{0}\) a constant, and the edges \(\theta=0\) and \(\theta=\pi\) are insulated.

A conducting sphere of radius \(r=c\) is grounded and placed in a uniform electric field that has intensity \(E\) in the \(z\)-direction. The potential \(u(r, \theta)\) outside the sphere is determined from the boundary. value problem $$ \begin{aligned} &\frac{\partial^{2} u}{\partial r^{2}}+\frac{2}{r} \frac{\partial u}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} u}{\partial \theta^{2}}+\frac{\cot \theta}{r^{2}} \frac{\partial u}{\partial \theta}=0, \quad r>c, \quad 0<\theta<\pi \\ &u(c, \theta)=0, \quad 0<\theta<\pi \\ &\lim _{r \rightarrow \infty} u(r, \theta)=-E z=-E r \cos \theta . \end{aligned} $$ Show that \(\quad u(r, \theta)=-E r \cos \theta+E \frac{c^{3}}{r^{2}} \cos \theta\).
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