Question

Using the results of Problem 4, determine the general solution of the heat conduction problem: $$ \begin{aligned} &\frac{\partial u}{\partial t}-c^{2} \nabla^{2} u=0, \quad x^{2}+y^{2}<1 \\ &u(x, y, t)=0 \quad \text { for } x^{2}+y^{2}=1 \end{aligned} $$

Short answer

Answer: The solution to the heat conduction problem for the unit disk with the given boundary conditions is u(x, y, t) = 0.

Step-by-step solution

Write down the general solution using separation of variables

Assuming Problem 4 provided a solution using separation of variables, we should have a general solution with the form: $$ u(x, y, t) = \sum_{n=1}^{\infty} A_n \sin(n\pi r) \cos(n\pi\theta) e^{-c^2 (n\pi)^2 t} $$ Here, \((r, \theta)\) are the polar coordinates corresponding to \((x, y)\), and \(A_n\) are the coefficients to be determined.

Apply the boundary conditions

When \(x^2 + y^2 = 1\), we have \(u(1, \theta, t) = 0\). Using the general solution, we get: $$ 0 = \sum_{n=1}^{\infty} A_n \sin(n\pi) \cos(n\pi\theta) e^{-c^2 (n\pi)^2 t} $$ Since this is true for all times \(t\), we must have: $$ 0 = \sum_{n=1}^{\infty} A_n \sin(n\pi) \cos(n\pi\theta) $$

Determine the coefficients using orthogonal functions

In order to find the coefficients \(A_n\), we should use the orthogonality of cosine functions. First, multiply both sides of the boundary condition equation by \(\cos(m\pi\theta)\), where \(m\) is a positive integer, and then integrate over the boundary (\(\theta\) from 0 to \(2\pi\)), giving: $$ \int_0^{2\pi} 0 \cdot \cos(m\pi\theta) d\theta = \sum_{n=1}^{\infty} A_n \sin(n\pi) \int_0^{2\pi} \cos(n\pi\theta)\cos(m\pi\theta) d\theta $$ Using the orthogonality of cosine functions, this simplifies to: $$ 0 = A_m \sin(m\pi) \int_0^{2\pi}\cos^2(m\pi\theta)d\theta $$

Solve for the coefficients

Since the integral \(\int_0^{2\pi}\cos^2(m\pi\theta)d\theta > 0\), we must have: $$ A_m \sin(m\pi) = 0 $$ For all odd values of \(m\), \(\sin(m\pi) = 0\). Hence, all coefficients \(A_n\) must be zero for odd \(n\). For even values of \(n\), we can choose any values for the coefficients \(A_n\); however, we can choose them to be zero in order to keep the solution trivial. Thus, all the coefficients \(A_n\) are zero.

Write down the final solution

Since all the coefficients \(A_n\) are zero, the final solution to the heat conduction problem is simply: $$ u(x,y,t) = 0 $$ This solution satisfies both the boundary conditions and the given PDE.

Key concepts

Partial Differential Equations

Partial Differential Equations (PDEs) are mathematical equations that involve the partial derivatives of a function of multiple variables. These functions often represent physical quantities, and the PDEs are used to describe various phenomena such as heat, sound, electrostatics, fluid flow, elasticity, or quantum mechanics.

In the given exercise, we are dealing with a heat conduction problem that uses a PDE to model the temperature distribution over time within a circular region. The equation used is the heat equation, \(\frac{\partial u}{\partial t}-c^{2} abla^{2} u=0\), where \(u\) represents the temperature, \(t\) is time, and \(c^2\) is a constant related to the material's properties. The PDE is subject to the boundary condition \(u(x, y, t)=0\) for \(x^{2}+y^{2}=1\), which implies that along the boundary of the circle, the temperature is maintained at a constant value (often zero, corresponding to a fixed temperature).

Understanding and solving PDEs requires various mathematical techniques, and they are essential for modeling numerous physical and engineering problems.

Separation of Variables Method

The separation of variables method is a powerful technique used to solve PDEs. This method simplifies a PDE by breaking it down into simpler ordinary differential equations (ODEs) that are easier to solve. The idea behind it is to assume that the solution can be written as a product of functions, each depending only on a single variable.

In the heat conduction problem provided, separation of variables assumes a solution of the form \(u(x, y, t) = \sum_{n=1}^{\infty} A_n \sin(n\pi r) \cos(n\pi\theta) e^{-c^2 (n\pi)^2 t}\), where each term in the sum is a product of three functions, each in ones of the variables \(r\), \( \theta\), and \(t\), respectively. This breakdown allows us to deal with the equation in a more structured way and eventually to apply the initial and boundary conditions to find the solution of the original PDE.

It's worth noting that the success of this method highly relies on the initial conditions and boundary conditions being simple enough to apply the superposition principle and for the PDE to be linear and homogeneous.

Orthogonal Functions

Orthogonal functions play a crucial role in the solution of PDEs, especially when using the separation of variables technique. Orthogonality in this context refers to the property that two functions are orthogonal if their integral over a certain range with respect to a weight function is zero. This property is analogous to orthogonality in geometry, where two vectors are orthogonal if their dot product is zero.

In our heat conduction problem, after applying separation of variables, the coefficients \(A_n\) are found using the orthogonality of the cosine functions. This is done by multiplying the boundary condition by \(\cos(m\pi\theta)\) and integrating over the interval \([0, 2\pi]\). The orthogonality principle allows us to isolate each coefficient independently. Since the cosine functions are orthogonal over the given range, only the integral involving the square of the function itself remains, simplifying the computation of the coefficients.

Understanding the concept of orthogonal functions and their properties is fundamental in simplifying complex integrations in many problems across mathematics and theoretical physics, leading to the determination of the unique coefficients in series solutions to PDEs.