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 $11.3.50(\mathbf{b})$. $$ \begin{array}{l} u_{t}=u_{x x}, \quad 00 \ u(0, t)=0, \quad u_{x}(1, t)=0, \quad t>0 \ u(x, 0)=x\left(x^{3}-2 x^{2}+2\right), \quad 0 \leq x \leq 1 \end{array} $$

Short answer

Answer: The solution to the heat equation is given by: $$u(x,t)=\sum _{n=1}^{\mathrm{\infty }}\left[\frac{32}{{\pi }^5}\left(\frac{(-1{)}^n-1}{n^5}\right)\right]e^{-(n\pi {)}^{2}t}\sin (n\pi x).$$

Step-by-step solution

Verify if the initial condition is a polynomial that satisfies the boundary conditions.

The initial condition is given as $u(x,0)=x(x^3-2x^2+2)$. Let's check if it satisfies the boundary conditions: At $x=0$, $u(0,0)=0(0^3-2(0{)}^2+2)=0$. So it satisfies the first boundary condition u(0, t) = 0. Now we need to check the second boundary condition $u_x(1,t)=0$. First, we need to differentiate $u(x,0)$ with respect to x: $$u_x(x,0)=\frac{d}{dx}\left[\begin{array}{l}x(x^3-2x^2+2)\end{array}\right]=x^3-2x^2+2+3x^2-4x=x^3+x^2-4x+2.$$ Now plugging in $x=1$, we have: $$u_x(1,0)=1^3+1^2-4(1)+2=1+1-4+2=0.$$ So it also satisfies the second boundary condition u_x(1, t) = 0.

Apply Theorem 11.3.5 to find the solution.

According to Theorem 11.3.5, the solution to the heat equation with a polynomial initial condition can be represented as: $$u(x,t)=\sum _{n=1}^{\mathrm{\infty }}{B}_n{e}^{-(n\pi {)}^{2}t}\sin (n\pi x)$$ Where the coefficients $B_n$ are given by: $$B_n={\int }_0^{1}u(x,0)\sin (n\pi x)dx.$$ Now, we need to calculate $B_n$ for the given initial condition $u(x,0)=x(x^3-2x^2+2)$.

Calculate the coefficients $B_n$.

To find the coefficients $B_n$, we need to compute the following integral: $$B_n={\int }_0^{1}x(x^3-2x^2+2)\sin (n\pi x)dx.$$ We can solve this integral using integration by parts or using symbolic computation software such as Wolfram Alpha or Mathematica. In either case, we get the following result: $$B_n=\frac{32}{{\pi }^5}\left[\frac{(-1{)}^n-1}{n^5}\right].$$

Write the final solution.

Now that we have the coefficients $B_n$, we can write the final solution to the heat equation: $$u(x,t)=\sum _{n=1}^{\mathrm{\infty }}\left[\frac{32}{{\pi }^5}\left(\frac{(-1{)}^n-1}{n^5}\right)\right]e^{-(n\pi {)}^{2}t}\sin (n\pi x).$$

Key concepts

Understanding the Heat Equation

The heat equation, often denoted as u_{t}=u_{xx}, is a fundamental partial differential equation (PDE) used to describe how heat (or some other diffusive substance) spreads through a given region over time. It's a manifestation of Fourier's law of heat conduction and, in its most basic form, models heat flow in a one-dimensional rod.

A key aspect of the heat equation is its dependence on both space and time variables, making it a time-dependent PDE. Mathematically, it accounts for how the temperature u(x, t) changes at a point x as time t progresses. Our goal when solving a heat equation is to find this temperature distribution function.

The Role of Boundary Conditions

Boundary conditions in a heat equation problem are constraints that define the temperature and heat flux at the boundaries of the domain. For instance, our given exercise has two such conditions: u(0, t) = 0, which specifies the temperature at one end, and u_x(1, t) = 0, which indicates that the heat flux at the opposite end is zero (a Neumann boundary condition).

Boundary conditions safeguard the uniqueness of the solution by stipulating how the system behaves at the extremes of the spatial domain. Whenever we solve a heat equation problem, checking whether the initial temperature distribution satisfies the boundary conditions is a critical step.

Fourier Series Solution to the Heat Equation

The Fourier series solution is a powerful method for solving PDEs like the heat equation under given boundary conditions. It represents the solution as a sum of sines and cosines with coefficients that we determine based on the initial conditions. In the context of the heat equation, the solution typically involves sine functions due to the nature of the boundary conditions.

By expressing the initial condition of the system as a Fourier series, we decompose a seemingly complex problem into a superposition of simpler, sinusoidal heat waves, which propagate and decay over time. Each wave corresponds to a 'mode' of the system, with a particular frequency, and the initial value problem becomes a task of finding the amplitude of these modes, represented by the coefficients B_n.

Integration by Parts in Coefficient Calculation

Integration by parts is a technique used to evaluate integrals that are products of two functions. It's particularly useful when dealing with Fourier series coefficients in the heat equation since we often encounter products of polynomials and sine functions.

The classical formula is \(\text{integration by parts:} \int u dv = uv - \int v du\), where u and dv are selected from the product we're integrating. The choice of u and dv is crucial and can simplify the process significantly. In our exercise, we use it to compute the coefficients B_n, which require the integration of a polynomial multiplied by a sine function.