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The heat conduction equation in two space dimensions may be expressed in terms of polar coordinates as $$ \alpha^{2}\left[u_{r r}+(1 / r) u_{r}+\left(1 / r^{2}\right) u_{\theta \theta}\right]=u_{t} $$ Assuming that \(u(r, \theta, t)=R(r) \Theta(\theta) T(t),\) find ordinary differential equations satisfied by \(R(r), \Theta(\theta),\) and \(T(t) .\)

Short Answer

Expert verified
Answer: The three ordinary differential equations are: 1. \(T'(t) + k^2 T(t) = 0\) 2. \(R''(r) + \frac{1}{r} R'(r) - l^2R(r) = 0\) 3. \(\Theta''(\theta) + (l^2 + \alpha^2 k^2)\Theta(\theta) = 0\)

Step by step solution

01

Plug the assumed solution into the heat conduction equation

Let's substitute the assumed solution \(u(r, \theta, t) = R(r) \Theta(\theta) T(t)\) into the given heat conduction equation: $$ \alpha^2\left[R''(r) \Theta(\theta) T(t) + \frac{1}{r} R'(r) \Theta(\theta) T(t) + \frac{1}{r^2} R(r) \Theta''(\theta) T(t)\right] = R(r) \Theta(\theta) T'(t) $$
02

Divide by the assumed solution

Now, let's divide both sides of the equation by \(R(r) \Theta(\theta) T(t)\): $$ \alpha^2\left[\frac{R''(r)}{R(r)} + \frac{1}{r} \frac{R'(r)}{R(r)} + \frac{1}{r^2} \frac{\Theta''(\theta)}{\Theta(\theta)}\right] = \frac{T'(t)}{T(t)} $$
03

Separate the variables

The left side of the equation depends on \(r\) and \(\theta\), while the right side depends on \(t\). For the equation to hold true for all \(r\), \(\theta\), and \(t\), both sides must be equal to a constant, say \(-k^2\): $$ \frac{T'(t)}{T(t)} = -k^2 \implies T'(t) + k^2 T(t) = 0 $$ and $$ \frac{R''(r)}{R(r)} + \frac{1}{r} \frac{R'(r)}{R(r)} + \frac{1}{r^2} \frac{\Theta''(\theta)}{\Theta(\theta)} = -\alpha^2 k^2 $$
04

Separate the radial and angular parts

The radical part of the equation depends on \(r\), and the angular part depends on \(\theta\). So, both parts should be constant, say \(l^2\) for the radial part. Then the angular part will equal to \(-l^2 - \alpha^2 k^2\). We get two ODEs: $$ \frac{R''(r)}{R(r)} + \frac{1}{r} \frac{R'(r)}{R(r)} = l^2 $$ and $$ \frac{\Theta''(\theta)}{\Theta(\theta)} = -l^2 -\alpha^2 k^2 $$ Now, we have found three ordinary differential equations satisfied by \(R(r)\), \(\Theta(\theta)\), and \(T(t)\): 1. \(T'(t) + k^2 T(t) = 0\) 2. \(R''(r) + \frac{1}{r} R'(r) - l^2R(r) = 0\) 3. \(\Theta''(\theta) + (l^2 + \alpha^2 k^2)\Theta(\theta) = 0\)

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Key Concepts

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

Separation of Variables
Separation of variables is a mathematical method utilized for solving certain types of partial differential equations (PDEs) like the heat conduction equation in polar coordinates. This technique involves breaking down a PDE, which can be quite complex due to its dependence on multiple variables, into simpler, ordinary differential equations (ODEs) that depend on a single variable each.

The approach assumes that the PDE solution can be written as the product of functions, each depending only on one of the variables. For the heat conduction problem, we express the solution as a product of functions in the form of \(u(r, \theta, t) = R(r) \Theta(\theta) T(t)\), where \(R\), \(\Theta\), and \(T\) are dependent solely on \(r\), \(\theta\), and \(t\) respectively.

By substituting this assumed product form into the original PDE and rearranging, we achieve an equation where one side is exclusively a function of \(t\), and the other side is a function of \(r\) and \(\theta\). Since these variables are independent, the only way this can hold true for all values is if both sides equal a constant, thus allowing us to write separate ODEs for each variable. This separation is critical because it simplifies the problem significantly—transforming a tricky PDE into a series of more manageable ODEs.
Ordinary Differential Equations
Ordinary differential equations (ODEs) are equations involving derivatives of functions that depend on a single independent variable. Unlike partial differential equations that involve multiple variables and their partial derivatives, ODEs are simpler and have a broad array of well-established methods for their solution.

In the context of the separated heat conduction equation, after applying the separation of variables technique, we end up with three ODEs for \(R(r)\), \(\Theta(\theta)\), and \(T(t)\). Each of these equations features derivatives with respect to their individual variables: \(R(r)\) is only a function of \(r\), \(\Theta(\theta)\) only of \(\theta\), and \(T(t)\) only of \(t\). The ODEs are much simpler to tackle because the mathematical tools for solving ODEs are more developed than those for PDEs.

Techniques for solving ODEs can include, but are not limited to, integrating factor methods, using characteristic equations for linear ODEs, or applying power series solutions for more complex forms.
Partial Differential Equations
Partial differential equations (PDEs) involve unknown multivariable functions and their partial derivatives. These equations are often used to describe various phenomena such as heat conduction, wave propagation, and fluid dynamics in physics and engineering.

The heat conduction equation in polar coordinates that we began with is a PDE. It includes the time derivative \(u_t\) and spatial derivatives \(u_{rr}\), \(u_r\), and \(u_{\theta\theta}\). These represent how the heat distribution changes over time and space. PDEs are generally more challenging to solve than ODEs due to their complexity; however, they can often be transformed into ODEs using methods like separation of variables, as seen in this context.

Solving PDEs can require various advanced techniques such as the method of characteristics, Fourier transforms, and numerical methods when analytical solutions are not possible. Understanding PDEs is essential in fields that deal with dynamic systems that change across multiple dimensions.

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Most popular questions from this chapter

Dimensionless variables can be introduced into the wave equation \(a^{2} u_{x x}=u_{t t}\) in the following manner. Let \(s=x / L\) and show that the wave equation becomes $$ a^{2} u_{s s}=L^{2} u_{t t} $$ Then show that \(L / a\) has the dimensions of time, and thus can be used as the unit on the time scale. Finally, let \(\tau=a t / L\) and show the wave equation then reduces to $$u_{s s}=u_{\tau \tau}$$

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