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Magazine Chapter 9: Problem 7
Solve the boundary value problem.
$$
\begin{aligned}
&3 u_{t}=u_{x x}, 0
Short Answer
Expert verified
The solution is \( u(x,t) = \frac{1}{2} + \frac{1}{4} e^{-\frac{16\pi^2}{3}t} \cos(4 \pi x) \).
Step by step solution
01
Understand the Problem Type
This is a boundary value problem involving a partial differential equation (PDE) for the heat equation. The PDE given is \( 3u_{t} = u_{xx} \), which indicates a diffusion process with specific boundary conditions \( u_{x}(0, t) = u_{x}(2, t) = 0 \). The initial condition is \( u(x, 0) = \cos^{2}(2\pi x) \).
02
Separation of Variables
We use separation of variables assuming a solution of the form \( u(x, t) = X(x)T(t) \). Substituting this into the PDE gives \( 3XT' = X''T \). Dividing by \( XT \) and rearranging, we get \( 3 \frac{T'}{T} = \frac{X''}{X} = -\lambda \), where \( \lambda \) is a separation constant.
03
Solve the Temporal Part
From \( 3 \frac{T'}{T} = -\lambda \), solve the ordinary differential equation: \( T'(t) + \frac{\lambda}{3}T(t) = 0 \). The solution is \( T(t) = Ae^{-\frac{\lambda}{3}t} \), where \( A \) is a constant.
04
Solve the Spatial Part
Solve \( X'' + \lambda X = 0 \) with boundary conditions \( X'(0) = X'(2) = 0 \). The general solution is \( X(x) = C_1 \cos(\sqrt{\lambda} x) + C_2 \sin(\sqrt{\lambda} x) \). Differentiating and applying boundary conditions gives \( C_2 = 0 \), and \( \sqrt{\lambda} \) must satisfy \( \sin(2\sqrt{\lambda}) = 0 \), implying \( \sqrt{\lambda} = n\pi \), \( n \) even.
05
Establish Eigenvalues and Eigenfunctions
Substitute back to have eigenvalues \( \lambda = (n\pi)^2 \), so eigenfunctions are \( X_n(x) = C_n \cos(n\pi x) \) for even \( n \). Hence, \( u_n(x, t) = C_n \cos(n\pi x) e^{-\frac{(n\pi)^2}{3}t} \).
06
Apply Initial Condition
Express the initial condition \( u(x,0) = \cos^2(2\pi x) = \frac{1 + \cos(4\pi x)}{2} \) in terms of eigenfunctions. Write as a series: \( u(x,0) = \sum_{n=0}^{\infty} C_n \cos(n\pi x) \). Coefficients are determined using Fourier series: \( C_0 = \frac{1}{2} \) and \( C_4 = \frac{1}{4} \), with all other \( C_n = 0 \) since expansion matches exactly.
07
Form the Solution
Combine expansion coefficients to establish: \( u(x,t) = \frac{1}{2} \cos(0 \cdot \pi x) e^{-\frac{(0\cdot \pi)^2}{3}t} + \frac{1}{4} \cos(4 \pi x) e^{-\frac{16\pi^2}{3}t} \). Simplify to \( u(x,t) = \frac{1}{2} + \frac{1}{4} \cos(4 \pi x) e^{-\frac{16\pi^2}{3}t} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Differential Equations
Partial differential equations (PDEs) are a fundamental concept in mathematical physics that involves equations with multiple independent variables and partial derivatives. They are essential for modeling various physical phenomena. In our exercise, we're dealing with a PDE in the form of the heat equation:
\( 3 u_{t} = u_{xx} \). This equation describes the change over time of a function that depends on both time \( t \) and space \( x \).
\( 3 u_{t} = u_{xx} \). This equation describes the change over time of a function that depends on both time \( t \) and space \( x \).
- Independent Variables: In this context, the variables are time \( t \) and space \( x \).
- Partial Derivatives: These are rates of change with respect to one variable while keeping others constant, like \( u_t \) for time and \( u_{xx} \) for space.
Heat Equation
A central feature of many physical systems, the heat equation models the distribution of heat (or variation in temperature) in a given region over time. The general form is
\( u_t = u u_{xx} \), where \( u \) is the diffusion coefficient. In our problem,
\( u = \frac{1}{3} \).
\( u_t = u u_{xx} \), where \( u \) is the diffusion coefficient. In our problem,
\( u = \frac{1}{3} \).
- Diffusion: The heat equation represents the process of heat diffusing through a medium.
- Boundary Conditions: These are constraints such as \( u_{x}(0, t)=u_{x}(2, t)=0 \), indicating no heat flow through the boundaries.
- Initial Condition: This gives the temperature distribution at time \( t = 0 \), here \( u(x, 0)=\cos^2 2\pi x \).
Separation of Variables
The method of separation of variables is a powerful technique for solving PDEs. It assumes that the solution can be broken down into functions that depend solely on each independent variable. For example, for our heat equation,
assume \( u(x, t) = X(x)T(t) \).
assume \( u(x, t) = X(x)T(t) \).
- Product Solution: By writing the PDE in terms of \( X(x) \) and \( T(t) \), each part is handled separately.
- Separation Constant: When separating, a constant \( \lambda \) is introduced to balance both sides of the equation.
- Simplification: This simplifies solving into two ordinary differential equations (ODEs), one for \( X(x) \) and one for \( T(t) \).
Fourier Series
The Fourier series is essential for expressing functions as an infinite sum of sine and cosine terms. It aids in solving boundary value problems when functions need to fit boundary conditions over intervals. In our problem:
- Trigonometric Functions: Represent the solution \( u(x, t) \) as a series of terms like \( \cos(n\pi x) \) for various \( n \).
- Eigenfunctions and Eigenvalues: These are derived from the boundary conditions and are used to expand the initial condition into the series.
- Coefficients: Each term's coefficient is determined by integrating the initial condition over the interval, here \( C_0 = \frac{1}{2} \), \( C_4 = \frac{1}{4} \).
