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 \(11.3 .50(\mathbf{b})\). $$ \begin{array}{ll} u_{t}=16 u_{x x}, & 00 \\ u_{x}(0, t)=0, & u(2 \pi, t)=0, \quad t>0 \\ u(x, 0)=4, & 0 \leq x \leq 2 \pi \end{array} $$

Short answer

After evaluating the integral, we obtain: $$ A_n = \begin{cases} 0, & \text{if} \; n \geq 2 \text{ and even} \\ \frac{8}{n\pi}, & \text{if} \; n \text{ is odd} \\ 4, & \text{if} \; n = 0 \end{cases} $$ And since the initial condition is an even function, \(B_n = 0\) for all \(n\). #tag_title#Step 5: Combine the Results to Obtain the Solution#tag_content# Now we have the coefficients \(A_n\) and \(B_n\), so we can write the combined solution for \(u(x,t)\) as follows: $$ u(x, t) = X(x)T(t) = \sum_{n=1}^{\infty} A_n \cos(nx) e^{-16n^2 t} $$ Replacing \(A_n\) with the values obtained in Step 4: $$ u(x, t) = \frac{8}{\pi} \sum_{n\; \text{odd}}^{\infty} \frac{\cos(nx)}{n} e^{-16n^2 t} $$ This is the solution to the given initial-boundary value problem for the heat equation.

Step-by-step solution

Apply Separation of Variables

To use separation of variables, we will assume that the solution is of the form \(u(x,t)=X(x)T(t)\). Now substitute this expression into the given PDE: $$ (X(x)T'(t)) = 16(X''(x)T(t)) $$ Next, divide by \(X(x)T(t)\) to separate the variables: $$ \frac{T'(t)}{16 T(t)} = \frac{X''(x)}{X(x)} $$ Since the left side of the equation only depends on \(t\) and the right side only depends on \(x\), both sides must be equal to a constant, say \(\lambda\).

Solve the ODEs for X(x) and T(t)

Now we have two ODEs to solve: 1. \(T'(t)= 16 \lambda T(t)\) 2. \(X''(x) = \lambda X(x)\) First, we'll solve the ODE for \(T(t)\): \(T'(t)-16\lambda T(t) = 0\). This is a first-order linear ODE and the general solution is given by: $$ T(t) = C_1 e^{16\lambda t} $$ where \(C_1\) is a constant. Now, we'll solve the ODE for \(X(x)\): \(X''(x) - \lambda X(x) = 0\). This is a second-order linear ODE, and depending on the value of \(\lambda\), we have three cases: 1. If \(\lambda > 0\), the ODE has the general solution \(X(x) = C_2 e^{\sqrt{\lambda}x} + C_3 e^{-\sqrt{\lambda}x}\) 2. If \(\lambda = 0\), the ODE has the general solution \(X(x) = C_4 + C_5 x\) 3. If \(\lambda < 0\), the ODE has the general solution \(X(x) = C_6 \cos(\sqrt{-\lambda}x) + C_7 \sin(\sqrt{-\lambda}x)\)

Apply the Boundary Conditions to Determine λ and X(x)

Now let's apply the boundary conditions \(u_x(0,t)=0\) and \(u(2\pi,t)=0\). We get: 1. \(X'(0)T(t)=0 \Rightarrow X'(0)=0\) 2. \(X(2\pi)T(t)=0 \Rightarrow X(2\pi)=0\) From the first boundary condition \(X'(0)=0\), we can already see that case 1 and case 2 are not suitable for our problem because we cannot guarantee \(X'(0)=0\). So we are left with case 3: $$ X(x) = C_6 \cos(\sqrt{-\lambda}x) + C_7 \sin(\sqrt{-\lambda}x) $$ Now let's apply the second boundary condition \(X(2\pi)=0\) to determine \(\lambda\) and \(C_7\) (after substituting \(x=2\pi\) into \(X(x)\): $$ C_6 \cos(2\pi \sqrt{-\lambda}) = 0 $$ This equation has non-trivial solutions if \(\sqrt{-\lambda} = n\) for some integer \(n \geq 1\). Therefore, \(\lambda = -n^2\), and we can write \(X_n(x) = C_{6n} \cos(nx) + C_{7n} \sin(nx)\).

Apply the Initial Condition and Determine the Coefficients C_{6n} and C_{7n}

Now let's apply the initial condition \(u(x,0)=4\). This gives us the following equation: $$ u(x,0)=X(x)T(0)=4 \Rightarrow X(x)=\sum_{n=1}^{\infty}\left( A_n \cos(nx) + B_n \sin(nx)\right)=4 $$ We need to find the coefficients \(A_n\) and \(B_n\) from the Fourier series representation. Since the initial condition is an even function, we only have cosine terms in the series: $$ A_n =\frac{1}{\pi}\int_{0}^{2\pi} 4\cos(nx)dx $$

Key concepts

Initial Boundary Value Problem

An Initial Boundary Value Problem (IBVP) is a type of problem in differential equations where the goal is to find a solution to a differential equation that meets certain initial conditions at a starting time, as well as specific boundary conditions at the spatial boundaries. In our exercise, this means finding a function \(u(x, t)\) that solves the given partial differential equation (PDE) for all times \(t>0\) and all positions \(0 < x < 2\pi\), while also satisfying:

• Initial condition: \(u(x,0) = 4\),
• Boundary conditions: \(u_x(0,t)=0\) and \(u(2\pi,t)=0\).

These conditions ensure that the solution begins in a particular form and evolves over time while obeying certain rules at the edges of the domain. Understanding these constraints is crucial as they define the problem’s physical and mathematical boundaries.

Separation of Variables

Separation of Variables is a powerful technique used to solve partial differential equations (PDEs), transforming them into simpler ordinary differential equations (ODEs). This approach assumes the solution can be expressed as a product of functions, each depending on a single variable. For example, in this exercise, we assume \(u(x,t) = X(x)T(t)\). By substituting this into the PDE, the equation can be split such that one side depends only on \(x\) and the other only on \(t\). For our problem:\[\frac{T'(t)}{16 T(t)} = \frac{X''(x)}{X(x)}\]Both sides become equal to a constant, often denoted as \(\lambda\), allowing the original PDE to be separated into two ODEs that can be solved independently. This separation makes complex PDEs much more manageable and is especially useful when the PDE follows certain symmetry properties or linearity.

Partial Differential Equations (PDEs)

Partial Differential Equations (PDEs) involve multivariable functions and their partial derivatives. Unlike ordinary differential equations, which deal with functions of a single variable, PDEs address situations where changes occur in multiple independent variables, such as time and space. In our problem, the PDE $u_t = 16u_{xx}$ models a heat equation scenario, describing how temperature, represented by $u(x,t)$, changes over time at every point $x$. This equation specifically shows the diffusion of heat over time in a one-dimensional space. PDEs like these are prevalent in physics and engineering, offering insights into wave mechanics, heat flow, and quantum mechanics. Solving PDEs often involves applying methods like separation of variables, where we break down the PDE into a set of simpler equations.

Boundary Conditions

Boundary conditions define how the solution behaves at the endpoints of the domain. They are crucial in ensuring the solution is physically viable and mathematically sound. In our exercise, we have two boundary conditions:

• The derivative condition at \(x=0\): \(u_x(0,t)=0\) implies a Neumann condition, indicating no flux or gradient at this boundary — the rate of change of \(u\) with respect to \(x\) is zero, suggesting thermal insulation or symmetry.
• The value condition at \(x=2\pi\): \(u(2\pi,t)=0\) is a Dirichlet condition, fixing the value of \(u\) to zero — suggesting, for instance, a boundary held at a constant temperature.

These conditions allow us to choose appropriate solutions for \(X(x)\) that satisfy these constraints, such as using trigonometric functions that inherently meet these boundary requirements. Correct adherence to boundary conditions ensures the model accurately describes the real-world situation it represents.