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Chapter 12: Problem 39

Use Exercise 34 to solve the initial-boundaryvalue problem. In some of these exercises Theorem 11.3.5(d) or Exercise 11.3.50(b) will simplify the computation of the coefficients in the mixed Fourier sine series. $$ \begin{array}{l} u_{t t}=9 u_{x x}, \quad 00, \\ u(0, t)=0, \quad u_{x}(1, t)=0, \quad t>0 ,\\\ u(x, 0)=(x-1)^{3}+1, \quad u_{t}(x, 0)=0, \quad 0 \leq x \leq 1 \end{array} $$

Short Answer

Expert verified
Answer: The solution to the given initial-boundary value problem is u(x,t) = 0 for 00.

Step by step solution

01

Write the general solution for the wave equation as a mixed Fourier sine series

To solve this problem, we will assume the solution to have the form \(u(x,t) = X(x)T(t)\). By substituting this to the given PDE and separating variables we obtain two ordinary differential equations (ODEs) to be solved: $$\begin{array}{l} T'' = 9\lambda T \\ X'' = \lambda X \end{array}$$ Considering the Neumann boundary condition \(u_x(1,t)=0\), we will have a Fourier sine series since the solution involves only sine functions. Thus, the general solution can be written as: $$u(x,t) = \sum_{n=1}^{\infty} b_n\sin{(n\pi x)}\cdot e^{-9n^2\pi^2t}$$
02

Find the coefficients \(b_n\) using the initial conditions

We are given the initial conditions \(u(x,0)=(x-1)^3+1\) and \(u_t(x,0)=0\). Let's use the first initial condition to find \(b_n\): $$u(x,0)=\sum_{n=1}^{\infty} b_n\sin{(n\pi x)} = (x-1)^3+1$$ To find the coefficients \(b_n\), we use the property of orthogonality: $$b_n = 2\int_0^1 u(x,0)\sin{(n\pi x)} \,dx = 2\int_0^1 ((x-1)^3+1)\sin{(n\pi x)} \,dx$$
03

Simplify the computation of coefficients using Theorem 11.3.5(d) or Exercise 11.3.50(b)

We can apply Exercise 11.3.50(b) to the computation of \(b_n\), which states that if \(f(x)\) is a polynomial of degree \(m\), then: $$\int_0^1 f(x)\sin(n\pi x) \, dx = 0, \quad \text{for } n = 1, 2, 3, \cdots, m+1$$ In our case, \(f(x) = (x-1)^3+1\) is a polynomial of degree 3. So, we can say that: $$b_n = 2\int_0^1 ((x-1)^3+1)\sin{(n\pi x)}\, dx=0, \quad \text{for } n = 1, 2, 3, 4$$ Since all \(b_n\) are zero for the first four terms, our solution becomes: $$u(x,t) = 0$$ Therefore, the solution to the given initial-boundary value problem is \(u(x,t) = 0\) for \(00\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mixed Fourier Sine Series
The mixed Fourier sine series is a mathematical method used to represent functions as an infinite sum of sine terms. This approach is especially useful in solving boundary value problems, such as the wave equation, where the functions need to satisfy certain conditions at the boundaries. In a mixed sine series, the use of sine functions alone often arises from problems involving specific boundary conditions, such as the Neumann boundary condition, where the derivative of a function at a boundary is specified.
  • Each term in a Fourier sine series takes the form of \(b_n \sin(n\pi x)\), where \(b_n\) are coefficients determined based on initial or boundary conditions.
  • The series helps decompose complex periodic functions into simpler sine functions, making them easier to analyze and solve.
  • In our case, this series is used to solve the PDE by converting it into separate ODEs for time \(T(t)\) and space \(X(x)\).
Wave Equation
The wave equation, represented by \(u_{tt} = c^2 u_{xx}\), is a second-order linear partial differential equation. It's essential in physics and engineering to describe the propagation of waves, such as sound or light, through a medium.
  • The equation reflects how a wave evolves over time across a spatial domain.
  • In our problem, the wave speed \(c\) is given, allowing us to split the equation into simpler forms using methods like separation of variables.
  • Simplifying this equation results in two ordinary differential equations that are easier to solve individually.
  • The solution is a product of functions dependent solely on time \(T(t)\) and space \(X(x)\).
Initial Conditions
Initial conditions are vital in boundary value problems as they determine the specific solution among many possibilities. In our exercise, we are given two initial conditions: \(u(x,0) = (x-1)^3 + 1\) and \(u_t(x,0) = 0\).
  • These conditions describe the initial state of the wave and its initial velocity.
  • The initial displacement \(u(x,0)\) gives the shape of the wave at time \(t=0\).
  • The initial velocity \(u_t(x,0) = 0\) implies that the wave starts at rest.
  • Using these, we compute coefficients for the Fourier series to describe how the wave evolves over time.
Orthogonality
Orthogonality is a mathematical property used to simplify the computation of coefficients in a Fourier series. In context, functions are orthogonal if their integral product over a specified interval equals zero. This property is particularly useful for solving boundary value problems.
  • For a Fourier sine series, orthogonality helps isolate each coefficient \(b_n\).
  • By integrating the product of the initial condition \(f(x)\) and sine terms over the domain, we systematically find each \(b_n\).
  • This avoids solving complex systems and reduces the problem to evaluating integrals.
  • In our exercise, the orthogonality of \(\sin(n\pi x)\) is used in finding \(b_n\) for the series.
Neumann Boundary Condition
The Neumann boundary condition specifies the derivative of a function at a boundary. In wave problems, this condition typically appears when solving the spatial part of a wave equation.
  • Mathematically, it defines the rate of change of the wave function at the boundary.
  • Here, \(u_x(1,t)=0\) implies there is no change or flow at \(x=1\), suggesting a reflection or lack of flow at the boundary.
  • This condition leads to a formulation involving sine functions in the Fourier series because only sine functions naturally vanish or stabilize such derivative conditions.
  • The Neumann condition often simplifies solving PDEs by reducing the complexity or restricting the type of functions in the series.

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Most popular questions from this chapter

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

In Exercises \(1-16\) apply the definition developed in Example 1 to solve the boundary value problem. (Use Theorem 11.3 .5 where it applies.) Where indicated by \(\mathrm{C}\), graph the surface \(u=u(x, y), 0 \leq x \leq a\), \(0 \leq y \leq b\) $$ \begin{array}{l} \text { C } u_{x x}+u_{y y}=0, \quad 0

Use Exercise 17 to solve the initial-boundaryvalue problem. In some of these exercises Theorem 11.3.5(c) or Exercise 11.3.42(b) will simplify the computation of the coefficients in the mixed Fourier cosine series. $$ \begin{array}{l} u_{t t}=u_{x x}, \quad 00, \\ u_{x}(0, t)=0, \quad u(1, t)=0, \quad t>0, \\ u(x, 0)=x^{4}-4 x^{3}+6 x^{2}-3, \quad u_{t}(x, 0)=0, \quad 0 \leq x \leq 1. \end{array} $$

In Exercises \(35-46\) use Exercise 34 to solve the initial-boundaryvalue problem. In some of these exercises Theorem 11.3.5(d) or Exercise 11.3.50(b) will simplify the computation of the coefficients in the mixed Fourier sine series. $$ \begin{array}{l} u_{t t}=64 u_{x x}, \quad 00, \\ u(0, t)=0, \quad u_{x}(\pi, t)=0, \quad t>0, \\ u(x, 0)=x(2 \pi-x), \quad u_{t}(x, 0)=0, \quad 0 \leq x \leq \pi. \end{array} $$

In Exercises \(17-28\) define the formal solution of $$ u_{x x}+u_{u y}=0, \quad 0
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