Question

Solve the initial-boundaryvalue problem. Where indicated by \([\mathrm{C}]\), perform numerical experiments. To simplify the computation of coefficients in some of these problems, check first to see if \(u(x, 0)\) is a polynomial that satisfies the boundary conditions. If it does, apply Theorem 11.3.5; also, see Exercises \(11.3 .35(\mathbf{b}), 11.3 .42(\mathbf{b}),\) and \(11.3 .50(\mathbf{b})\). $$ \begin{array}{l} u_{t}=4 u_{x x}, \quad 00 \\ u(0, t)=0, \quad u(2, t)=0, \quad t>0 \\ u(x, 0)=\left\\{\begin{array}{cl} x, & 0 \leq x \leq 1 \\ 2-x, & 1 \leq x \leq 2 \end{array}\right. \end{array} $$

Short answer

Question: Determine the function u(x, t) that solves the initial-boundary value problem given by the heat equation \(u_t = 4u_{xx}\) with boundary conditions u(0, t) = 0, u(2, t) = 0, and initial condition u(x, 0) = x for 0≤x≤1, and u(x, 0) = 2-x for 1≤x≤2. Answer: The function u(x, t) that solves the given initial-boundary value problem is: $$ u(x, t) = \sum_{n=1}^{\infty} \frac{5(-1)^n - 4(-1)^{n+1} - 5}{n^2\pi^2} e^{-\frac{n^2\pi^2}{4}t} \sin\left(\frac{n\pi x}{2}\right) $$

Step-by-step solution

Compute the Fourier series of the initial condition

Given the initial condition: $$ u(x, 0) = \left\\{ \begin{array}{cl} x, & 0 \le x \le 1\\ 2-x, & 1 \le x \le 2 \end{array} \right. $$ The function is odd about \(x=\pi\), and the domain is \((0, 2)\), so we can compute the Fourier sine series: $$ u(x, 0) \approx \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{2}\right) $$ with $$ B_n = \frac{2}{2}\left[\int_0^1 x \sin\left(\frac{n\pi x}{2}\right) dx + \int_1^2 (2-x) \sin\left(\frac{n\pi x}{2}\right) dx\right] $$

Calculate the coefficients \(B_n\)

Using integration by parts, we find: $$ \int_0^1 x \sin\left(\frac{n\pi x}{2}\right) dx = \frac{4}{n^2\pi^2}((-1)^n - 1) $$ $$ \int_1^2 (2-x) \sin\left(\frac{n\pi x}{2}\right) dx = \frac{4}{n^2\pi^2}((-1)^n - 4) $$ Therefore, the Fourier sine coefficients are given by: $$ B_n = \frac{(-1)^{n+1} - 1}{n^2\pi^2} + \frac{(-1)^n - 4}{n^2\pi^2} = \frac{5(-1)^n - 4(-1)^{n+1} - 5}{n^2\pi^2} $$

Apply Theorem 11.3.5

By Theorem 11.3.5, we know that the solution to the heat equation with given Fourier coefficients is: $$ u(x, t) = \sum_{n=1}^{\infty} B_n e^{-\lambda_n t} \sin\left(\frac{n\pi x}{2}\right) $$ with \(\lambda_n = \frac{n^2\pi^2}{4}\), so the solution is given by: $$ u(x, t) = \sum_{n=1}^{\infty} \frac{5(-1)^n - 4(-1)^{n+1} - 5}{n^2\pi^2} e^{-\frac{n^2\pi^2}{4}t} \sin\left(\frac{n\pi x}{2}\right) $$ Now we have found the solution \(u(x, t)\) of the initial-boundary value problem that satisfies all given conditions.

Key concepts

Fourier Series

The Fourier series is a powerful mathematical tool that expands a periodic function in terms of sines and cosines. This series allows us to represent complex waveforms as the sum of simple oscillatory functions, making it particularly useful in solving partial differential equations such as the heat equation.

For a given function defined on an interval, the Fourier series is written as:\[\begin{equation}\begin{split}& a_0 + \sum_{n=1}^{\infty}(a_n \cos(nx) + b_n \sin(nx)), \quad \text{where} \end{split}\end{equation}\] and the coefficients are calculated using specific integrals of the function times sines and cosines.

In the context of the initial-boundary value problem, we use a Fourier sine series when the function has odd symmetry and is defined on a finite interval. This was indeed the case for the function in our problem, given by the initial condition. The Fourier sine series expansion then allows us to address the odd extension of the function and solve the heat equation on the defined spatial domain.

Heat Equation

The heat equation, typically denoted as \[\begin{equation}\begin{split}u_{t} = k u_{xx} \end{split}\end{equation}\], models the distribution of heat (or variation in temperature) in a given region over time. In one dimension, it is a second-order partial differential equation (PDE) where \[\begin{equation}u = u(x, t) \end{equation}\] is the temperature distribution function, \[\begin{equation}k \end{equation}\] is a constant proportional to the thermal diffusivity of the material, \[\begin{equation}x \end{equation}\] is the spatial dimension, and \[\begin{equation}t \end{equation}\] is time.

For a rod with ends held at a fixed temperature, as in our exercise, the boundary conditions are that \[\begin{equation}u(0, t) = u(L, t) = 0 \end{equation}\], where \[\begin{equation}L \end{equation}\] is the length of the rod. The initial condition, \[\begin{equation}u(x, 0) \end{equation}\], represents the temperature distribution at time \[\begin{equation}t=0 \end{equation}\]. The goal is to find the function \[\begin{equation}u(x, t) \end{equation}\] that describes how the initial temperature distribution evolves over time, satisfying both the boundary and initial conditions.

Integration by Parts

Integration by parts is a technique derived from the product rule for differentiation and is used to integrate the product of two functions. The formula, \[\begin{equation}\begin{split}\int u dv = uv - \int v du \end{split}\end{equation}\], where \[\begin{equation}u \end{equation}\] and \[\begin{equation}dv \end{equation}\] are differentiable functions of \[\begin{equation}x \end{equation}\], is frequently used in finding Fourier coefficients. This technique simplifies the integration process by reducing complex integrals into simpler ones.

In the step-by-step solution for our initial-boundary value problem, we used integration by parts to calculate the Fourier coefficients \[\begin{equation}B_n \end{equation}\]. By cleverly choosing \[\begin{equation}u \end{equation}\] and \[\begin{equation}dv \end{equation}\], we integrated the product of \[\begin{equation}x \end{equation}\](or \[\begin{equation}2 - x \end{equation}\]) and a sine function, yielding a more manageable expression for the coefficients.

Theorem 11.3.5

In the context of solving PDEs like the heat equation, certain theorems provide a simplified framework for finding solutions. Theorem 11.3.5 particularly concerns the solution of the heat equation with specified boundary conditions and an initial condition given by a Fourier series.

The theorem states that if the initial temperature distribution \[\begin{equation}u(x, 0) \end{equation}\] can be represented by a Fourier series, then the solution \[\begin{equation}u(x, t) \end{equation}\] at any later time \[\begin{equation}t>0 \end{equation}\] can be found by multiplying each term in the Fourier series by a factor that includes \[\begin{equation}e^{-\lambda_n t} \end{equation}\]. Here, \[\begin{equation}\lambda_n \end{equation}\] is a constant that is related to the spatial frequencies of the sine functions used in the Fourier series and the properties of the material through which heat is being conducted.

In our example, we used Theorem 11.3.5 to express the solution to the heat equation. The exponential term represented the cooling effect over time, and each term of the series corresponded to a mode of the temperature distribution within the rod. Thus, we ended up with an infinite series that approximates the evolution of the temperature over time for the entire rod.