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 \(11.3 .50(\mathbf{b})\). $$ \begin{array}{l} u_{t}=u_{x x}, \quad 00 \\ u_{x}(0, t)=0, \quad u_{x}(\pi, t)=0, \quad t>0 \\ u(x, 0)=x^{2}(x-\pi)^{2}, \quad 0 \leq x \leq \pi \end{array} $$

Short answer

Question: Solve the initial-boundary value problem for the heat equation with Neumann boundary conditions and given initial condition. Heat equation: \(u_t = u_{xx}\), on the domain \((0, \pi)\) and time \(t>0\) Neumann boundary conditions: \(u_x(0,t) = 0\) and \(u_x(\pi, t)=0\), for \(t>0\) Initial condition: \(u(x, 0)=x^2(x-\pi)^2\) for \(0 \leq x \leq \pi\) Answer: The solution of the heat equation is given by: $$ u(x,t) = x^2(x-\pi)^2 -\sum_{n=1}^{\infty} B_n\cosh(nx)e^{-n^2t} $$ where the coefficients \(B_n\) are given by: $$ B_n = \frac{2}{\pi}\int_0^\pi x^2(x-\pi)^2\cosh(nx)dx $$

Step-by-step solution

Check if \(u(x,0)\) satisfies the boundary conditions

Calculate the derivative of the initial condition, \(u_x(x,0)\): $$ u_x(x,0) = \frac{d}{dx}(x^2(x-\pi)^2) $$ Now, we check whether these values satisfy the Neumann boundary conditions: $$ u_x(0,0)=\frac{d}{dx}(0^2(0-\pi)^2)=0 $$ $$ u_x(\pi,0)=\frac{d}{dx}(\pi^2(\pi-\pi)^2)=0 $$ Both boundary conditions are satisfied by the initial condition.

Apply separation of variables

Assume that the solution has the form \(u(x,t)=X(x)T(t)\). We will now substitute this into the heat equation and separate the variables: $$ X(x)T'(t)=X''(x)T(t) $$ Now divide both sides by \(XT\), we get: $$ \frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)} = -\lambda $$ It follows two ODEs: $$ X''(x) + \lambda X(x) = 0 $$ $$ T'(t) + \lambda T(t) = 0 $$

Solve the spatial ODE

Given that \(X''(x) + \lambda X(x) = 0\), assume three cases for \(\lambda\): Case 1: \(\lambda < 0\), let \(\lambda=-\mu^2\). The solution becomes: $$ X(x) = C_1\cos(\mu x) + C_2\sin(\mu x) $$ Apply the Neumann boundary conditions: $$ X'(0) = 0 \Longrightarrow -C_1\mu\sin(0) + C_2\mu\cos(0) = 0 \Longrightarrow C_2 = 0 $$ $$ X'(\pi) = 0 \Longrightarrow -C_1\mu\sin(\mu\pi) = 0 $$ Thus, no non-trivial solution exists in this case. Case 2: \(\lambda = 0\). The solution becomes: $$ X(x) = C_1 + C_2x $$ Apply the Neumann boundary conditions: $$ X'(0) = 0 \Longrightarrow C_2 = 0 $$ $$ X'(\pi) = 0 \Longrightarrow 0 = 0 $$ This case leads to a trivial solution. Case 3: \(\lambda > 0\), let \(\lambda=\mu^2\). The solution becomes: $$ X(x) = C_1\cosh(\mu x) + C_2\sinh(\mu x) $$ Apply the Neumann boundary conditions: $$ X'(0) = 0 \Longrightarrow C_2\mu\cosh(0) = 0 \Longrightarrow C_2 = 0 $$ $$ X'(\pi) = 0 \Longrightarrow C_1\mu\sinh(\mu\pi) = 0 \Longrightarrow \mu\pi = n\pi, \quad n \in \mathbb{Z} $$ So, the solution becomes: $$ X_n(x) = C_n\cosh(n x) $$

Solve the temporal ODE

The temporal ODE is given by \(T'(t)+\lambda T(t)=0\). For \(\lambda = n^2\) the solution is: $$ T_n(t) = B_ne^{-n^2t} $$

Compute the coefficients

Now, we write the general solution as: $$ u_n(x,t) = X_n(x)T_n(t) = B_n\cosh(nx)e^{-n^2t} $$ To satisfy the initial condition, we need to find \(B_n\) using Fourier series expansion: $$ x^2(x-\pi)^2 = \sum_{n=1}^{\infty} B_n\cosh(nx) $$ Compute the Fourier coefficients in the usual way: $$ B_n = \frac{2}{\pi}\int_0^\pi x^2(x-\pi)^2\cosh(nx)dx $$ Solve the integral for \(B_n\) to find the coefficients.

Write the final solution

After finding the coefficients \(B_n\), the final solution of the heat equation is given by: $$ u(x,t) = x^2(x-\pi)^2 -\sum_{n=1}^{\infty} B_n\cosh(nx)e^{-n^2t} $$

Key concepts

Separation of Variables

When confronted with partial differential equations (PDEs), such as in the heat equation, the technique of separation of variables is a powerful method to find solutions. Simplifying the problem, this method assumes that the solution can be represented as a product of functions, each depending on a single variable. For the heat equation, we express the temperature u(x, t) as a product X(x)T(t), where X is a spatial function and T is a temporal function.

Upon substitution into the heat equation, the complexity reduces as each variable can be isolated to either side of the equation, leading to ordinary differential equations (ODEs) with respect to time and space separately. Since the spatial and temporal variables are independent, their ratio must equal a constant, which is denoted by -λ. This results in two separate ODEs: one for X(x) and one for T(t). Finding solutions to these ODEs is the next critical step and brings us towards forming a complete solution to the original PDE.

Neumann Boundary Conditions

Neumann boundary conditions specify the derivative of the function rather than its value at the boundary of the domain. In the context of our problem, they are used to describe the rate of change of temperature at the edges of a rod, which is particularly relevant for heat transfer problems.

For the given heat equation, the Neumann boundary conditions state that the spatial derivative of temperature (u_x) must be zero at both ends of the rod (x=0 and x=π). Physically, this implies there is no heat flux passing through the ends of the rod. Mathematically, it means that for any acceptable solution X(x), its first derivative must be zero at these boundaries. These conditions shape the form of potential solutions and play a crucial role in determining the constants when solving the spatial ODE using separation of variables.

Fourier Series Expansion

After separating the initial-boundary value problem into spatial and temporal parts and applying the Neumann boundary conditions, we are left needing to satisfy the initial heat distribution, u(x, 0). One way to express this initial distribution is by using a Fourier series expansion, which allows us to represent a periodic function as an infinite sum of sines and cosines. In cases where the function is not inherently periodic, as with the initial temperature distribution in our rod, this series may consist of hyperbolic sine or cosine terms.

A Fourier series is particularly valuable because it can approximate complex functions with a series of simpler ones, and it converges to the actual function under certain conditions. In our problem, we find Fourier coefficients that provide the weights for the series expansion of the initial temperature distribution. By carefully calculating these coefficients, we can describe the initial heat distribution as a sum of the eigenfunctions obtained through the separation of variables process. This final step reconciles the spatially and temporally separated solutions with the original conditions of the problem.