Chapter 10: Problem 18
Consider the wave equation
$$
a^{2} u_{x x}=u_{t t}
$$
in an infinite one-dimensional medium subject to the initial conditions
$$
u(x, 0)=0, \quad u_{t}(x, 0)=g(x), \quad-\infty
Short Answer
Expert verified
Based on the given wave equation with initial conditions, we derived the final expression for u(x, t) as:
$$
u(x, t) = \frac{1}{2a}\int_{x-at}^{x+at} g(\xi) d\xi
$$
Step by step solution
01
(a) Find the expressions for \(\phi(x)\) and \(\psi(x)\)
To start this part, recall the solution's form from Problem 13, which is:
$$
u(x, t) = \phi(x+at) + \psi(x-at)
$$
First, analyze the initial conditions. We know that \(u(x, 0) = 0\), so let's substitute t = 0 into the solution equation:
$$
u(x, 0) = \phi(x) + \psi(x) = 0
$$
This gives us the equation for the sum of \(\phi\) and \(\psi\):
$$
\phi(x) + \psi(x) = 0
$$
Next, let's consider the second initial condition, \(u_t(x, 0) = g(x)\). Firstly, compute \(u_t(x,t)\), the partial derivative of u with respect to time:
$$
u_t(x, t) = a\phi'(x+at) - a\psi'(x-at)
$$
Now, substitute t = 0 to obtain:
$$
u_t(x, 0) = a\phi'(x) - a\psi'(x) = g(x)
$$
This gives us the second equation:
$$
a\phi'(x) - a\psi'(x) = g(x)
$$
So, we have the following two equations:
$$
\begin{array}{c}{\phi(x)+\psi(x)=0} \\\ {a \phi^{\prime}(x)-a\psi^{\prime}(x)=g(x)}\end{array}
$$
02
(b) Derive expression for \(\phi(x)\) and \(\psi(x)\) from the equations obtained in part (a)
First, let's use the first equation of part (a) to show that \(\psi^{\prime}(x)=-\phi^{\prime}(x)\). We have:
$$
\phi(x) + \psi(x) = 0
$$
Take derivative with respect to x,
$$
\phi'(x) + \psi'(x) = 0
$$
This gives us:
$$
\psi'(x) = -\phi'(x)
$$
Now, use the second equation from part (a) to show that \(-2 a \phi^{\prime}(x)=g(x)\):
$$
a \phi^{\prime}(x)-a\psi^{\prime}(x)=g(x)
$$
Substitute \(\psi^{\prime}(x)=-\phi^{\prime}(x)\):
$$
a \phi^{\prime}(x)+ a\phi^{\prime}(x)=g(x)
$$
$$
-2 a \phi^{\prime}(x) = g(x)
$$
This implies:
$$
\phi'(x) = -\frac{1}{2a}g(x)
$$
Integrate by x from \(x_0\) to \(x\):
$$
\phi(x) = -\frac{1}{2a}\int_{x_0}^x g(\xi)d\xi+\phi(x_0)
$$
Now, we can determine \(\psi(x)\) using the relation we have between \(\phi(x)\) and \(\psi(x)\). We know that \(\psi(x) = - \phi(x)\), so:
$$
\psi(x) = \frac{1}{2a}\int_{x_0}^x g(\xi)d\xi - \phi(x_0)
$$
03
(c) Derive expression for u(x, t) using the expressions for \(\phi(x)\) and \(\psi(x)\) found in part (b)
We're now ready to determine \(u(x, t)\). Recall the solution's form from Problem 13:
$$
u(x, t) = \phi(x+at) + \psi(x-at)
$$
Using our expressions for \(\phi(x)\) and \(\psi(x)\), we have:
$$
u(x, t) = -\frac{1}{2a}\int_{x_0}^{x+at} g(\xi)d\xi + \phi(x_0) + \frac{1}{2a}\int_{x_0}^{x-at} g(\xi)d\xi - \phi(x_0)
$$
Combine the integrals and simplify:
$$
u(x, t) = \frac{1}{2a}\int_{x-at}^{x+at} g(\xi) d\xi
$$
This is the final expression for \(u(x, t)\) based on the given initial conditions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Conditions
The term *initial conditions* refers to specific values or states established at the beginning of a problem involving differential equations. For the wave equation, these conditions determine how the wave behaves at the initial time, t = 0.
These initial conditions allow us to solve the wave equation and predict the future behavior of the solution:
These initial conditions allow us to solve the wave equation and predict the future behavior of the solution:
- The first initial condition, \( u(x, 0) = 0 \), tells us the wave's displacement is zero at t = 0, i.e., the wave starts from a state of rest.
- The second initial condition, \( u_t(x, 0) = g(x) \), provides the initial speed or velocity of the wave.
Partial Differential Equations
Partial differential equations (PDEs) involve functions and their partial derivatives. They describe various physical phenomena, such as sound, heat, and waves in the engineering and physics fields.
The wave equation given revolves around PDEs:
The wave equation given revolves around PDEs:
- We see \(a^2 u_{xx} = u_{tt}\), representing the second derivatives in space and time.
- The PDE states that the acceleration of the wave (described by the second time derivative, \(u_{tt}\)) is proportionate to how it curves in space (given by the second spatial derivative, \(u_{xx}\)).
Method of Characteristics
The method of characteristics is a technique used to solve partial differential equations. It simplifies PDEs by reducing them to ordinary differential equations (ODEs) along certain paths called characteristics.
For the wave equation, this method helps solve by turning the problem into something more manageable:
For the wave equation, this method helps solve by turning the problem into something more manageable:
- Characteristics are paths along which the solution is constant or evolves simply. For a wave equation, they correspond to paths traveled by wave fronts.
- By introducing functions like \( \phi(x+at)\) and \( \psi(x-at)\), this technique allows separating the original problem into parts that can be handled individually.
Integration by Parts
Integration by parts is a mathematical technique used to transform the integral of a product of functions into other (possibly more manageable) integrals. This is particularly useful when dealing with functions in partial differential equations.
In this exercise, once relationships between derivatives like \( \phi'(x) = -\frac{1}{2a}g(x) \) are found, we use integration techniques:
In this exercise, once relationships between derivatives like \( \phi'(x) = -\frac{1}{2a}g(x) \) are found, we use integration techniques:
- The process includes integrating these derivative forms from an initial point \(x_0\) to the desired point \(x\), allowing calculation of \( \phi(x) \)
- The formula used: \( \phi(x) = -\frac{1}{2a} \int_{x_0}^x g(\xi) d\xi + \phi(x_0) \) demonstrates how integrating a known function gives the solution for \( \phi \).