Question

Solve the initial-boundaryvalue problem. In some of these exercises, Theorem 11.3.5(b) or Exercise 11.3.35 will simplify the computation of the coefficients in the Fourier sine series. $$ \begin{array}{l} u_{t t}=5 u_{x x}, \quad 00, \\ u(0, t)=0, \quad u(\pi, t)=0, \quad t>0, \\ u(x, 0)=x \sin x, \quad u_{t}(x, 0)=0, \quad 0 \leq x \leq \pi \end{array} $$

Short answer

Question: Determine the solution u(x,t) of the given initial-boundary value problem: Partial Differential Equation (PDE): $$u_{tt} = 5u_{xx}, \quad 0 \leq x \leq \pi, \quad t > 0$$ Boundary conditions: $$u(0, t) = 0, \quad u(\pi, t) = 0, \quad t > 0$$ Initial conditions: $$u(x, 0) = x \sin x, \quad u_t(x, 0) = 0, \quad 0 \leq x \leq \pi$$ Using the Fourier series method, we obtain: Answer: The solution of the initial-boundary value problem is $$u(x,t) = \frac{2}{\pi} \sin{x} e^{\frac{-5 \pi^2 t}{4}}.$$

Step-by-step solution

Identify the given equations

We are given the following partial differential equation, boundary conditions, and initial conditions: Partial Differential Equation (PDE): $$u_{tt} = 5u_{xx}$$ Boundary conditions: $$u(0, t) = 0, \quad u(\pi, t) = 0, \quad t > 0$$ Initial conditions: $$u(x, 0) = x \sin x, \quad u_t(x, 0) = 0, \quad 0 \leq x \leq \pi$$

Calculate the 2L value

Since the problem is given on the interval 0 < x < π, the value of 2L is simply 2π.

Find the Fourier sine series coefficients (B_n)

(Using Theorem 11.3.5(b) or Exercise 11.3.35 will simplify the computation) To find the coefficients \(B_n\), we must solve for them using the initial condition \(u(x, 0) = x \sin x\). Since \(u_t(x, 0) = 0\), the coefficients for the cosine series (\(A_n\)) are all zero. Calculate \(B_n\) using the formula for Fourier sine coefficient: $$B_n = \frac{2}{\pi} \int_{0}^{\pi} x \sin{x} \sin{nx} dx$$ Using the technique from Exercise 11.3.35, for n=1 and for odd n > 1, we find: $$B_1 = \frac{2}{\pi} \quad and \quad B_n = 0 \text{ for } n>1 \text{ odd}$$ Since the given problem has homogeneous boundary conditions, all even-indexed coefficients \(B_n\) are zero.

Write the solution (u(x,t)) as a Fourier series using the coefficients

The general solution of a homogeneous wave equation with non-zero temperature is given by: $$u(x, t) = \sum_{n=1}^{\infty} b_n \sin{\frac{n \pi x}{2L}} e^{\frac{-5 n^2 \pi^2 t}{4 L^2}}$$ Recall that 2L = 2π and that B_n is given for n = 1 and n > 1 with odd indices. Plug in those values and simplify the solution: $$u(x,t) = \frac{2}{\pi} \sin{x} e^{\frac{-5 \pi^2 t}{4}}$$

Key concepts

Fourier Sine Series

The Fourier sine series is a way to represent functions using sine functions. It works well for problems defined on finite intervals with specific boundary conditions. For example, suppose we have a function \( f(x) \) defined on the interval \( 0 < x < \pi \). The Fourier sine series can be expressed as:
\[ f(x) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n \pi x}{L}\right) \]
where \( B_n \) are the coefficients determined by:
\[ B_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n \pi x}{L}\right) dx \]
In our problem, we calculated \( B_n \) using the initial condition \( u(x, 0) = x \sin x \) and found that:

• \( B_1 = \frac{2}{\pi} \)
• \( B_n = 0 \) for \( n > 1 \) (odd \( n \))

The sine series is particularly useful when the boundary conditions are homogeneous, such as \( u(0, t) = 0 \) and \( u(\pi, t) = 0 \). These conditions lead to only sine terms being present in the series, as cosine terms are zero due to symmetry.

Initial-Boundary Value Problems

Initial-boundary value problems involve finding a function that satisfies a differential equation with specific initial and boundary conditions. These problems are common in fields like physics and engineering, where systems evolve over time within fixed spatial constraints.

In our exercise, we worked with a second-order wave equation, which describes how a wave propagates through a medium. The given initial conditions were:

• \( u(x, 0) = x \sin x \)
• \( u_t(x, 0) = 0 \)

The boundary conditions were:

• \( u(0, t) = 0 \)
• \( u(\pi, t) = 0 \)

These conditions specify the state of the system at the beginning and at the edges of the interval. Solving such problems requires techniques that include separations of variables, Fourier series, or numerical methods. Each method utilizes the uniqueness and structure provided by the conditions at hand.

Wave Equation

The wave equation is a fundamental partial differential equation that describes the behavior of waves, such as sound waves, light waves, or waves in a string. It is expressed as:
\[ u_{tt} = c^2 u_{xx} \]
where \( u(x, t) \) is the wave function, and \( c \) is the wave speed. In our case, the wave equation is \( u_{tt} = 5 u_{xx} \) with a specific speed factor of \( c^2 = 5 \).

To solve the wave equation, one generally uses the method of separation of variables, which separates the function into spatial and temporal parts. After separation, the spatial part often leads to a Fourier series expansion:
\[ u(x, t) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n \pi x}{L}\right) e^{-\frac{5 n^2 \pi^2 t}{4 L^2}} \]
In the context of our problem, the solution is simplified using the known values of \( B_n \). Given that many \( B_n \) are zero, the resulting expression reduces significantly, showing how Fourier series can turn complex differential equations manageable and insightful.