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Chapter 6: Problem 26

Show that any differentiable function \(f\left(t^{\prime}\right)\) of \(t^{\prime}=\) \(t-(z / v)\) satisfies the classical wave equation, i.e., show $$ \frac{\partial^{2} f\left(t^{\prime}\right)}{\partial t^{2}}=v^{2} \frac{\partial^{2} f\left(t^{\prime}\right)}{\partial z^{2}} $$ Show also that any differentiable function \(g\left(t^{\prime}\right)\) of \(t^{\prime \prime}=t+(z / v)\) satisfies the classical wave equation. Make up an example of a function \(f\left(t^{\prime}\right)\) and show explicitly that it satisfies the classical wave equation.

Short Answer

Expert verified
Both functions \(f(t')\) and \(g(t'')\) satisfy the wave equation.

Step by step solution

01

Define Variables

Identify the transformed variable, in this case, \( t' = t - \frac{z}{v} \). This simplifies the function to \( f(t') \) where it depends only on \( t' \). Similarly, define \( t'' = t + \frac{z}{v} \) for the function \( g(t'') \).
02

Recognize Required Derivatives

Notice that we need to compute the second partial derivatives \( \frac{\partial^2 f}{\partial t^2} \) and \( \frac{\partial^2 f}{\partial z^2} \). These derivatives will relate through the classical wave equation.
03

Calculate First Partial Derivatives

Calculate the first partial derivatives for both \( t \) and \( z \). When taking \( \frac{\partial f}{\partial t} \), note that \( \frac{\partial t'}{\partial t} = 1 \). Similarly, for \( \frac{\partial f}{\partial z} \), we have \( \frac{\partial t'}{\partial z} = -\frac{1}{v} \). This yields: \( \frac{\partial f}{\partial t} = f'(t') \) and \( \frac{\partial f}{\partial z} = -\frac{1}{v}f'(t') \).
04

Calculate Second Partial Derivatives

Differentiate the first derivatives again. For \( \frac{\partial^2 f}{\partial t^2} \), we get \( \frac{\partial^2 f}{\partial t^2} = f''(t') \). For \( \frac{\partial^2 f}{\partial z^2} \), differentiate the first derivative with respect to \( z \), giving \( \frac{\partial^2 f}{\partial z^2} = \frac{1}{v^2}f''(t') \).
05

Verify the Wave Equation for \(f(t')\)

Substitute the second derivatives back into the wave equation: \( f''(t') = v^2 \times \frac{1}{v^2}f''(t') \). This simplifies to the identity \( f''(t') = f''(t') \), confirming that our function satisfies the wave equation.
06

Verify the Wave Equation for \(g(t'')\)

Repeat the same process as for \( f(t') \) but with \( g(t'') \), resulting in similar calculations. The two derivatives will satisfy \( g''(t'') = v^2 \times \frac{1}{v^2}g''(t'') \), confirming \( g(t'') \) also satisfies the wave equation.
07

Create an Example Function \( f(t') = \sin(t') \)

Choose the function \( f(t') = \sin(t') \). We know the derivatives: \( f'(t') = \cos(t') \) and \( f''(t') = -\sin(t') \).
08

Verify Example with Wave Equation

Compute \( \frac{\partial^2 \, \sin(t')}{\partial t^2} = -\sin(t') \) and \( \frac{\partial^2 \, \sin(t')}{\partial z^2} = \frac{1}{v^2}(-\sin(t')) \). Substituting in, we find that \( -\sin(t') = v^2 \times \frac{-1}{v^2}\sin(t') \), confirming equality.

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Key Concepts

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

Differentiable Functions
Differentiable functions play a crucial role in calculus and are central to understanding the wave equation. A function is said to be differentiable if it has a derivative at every point in its domain. This means the function is smooth and its rate of change is well defined at each point.
In the context of the wave equation, we have two differentiable functions, \( f(t') \) and \( g(t'') \), where the transformed variables are \( t' = t - \frac{z}{v} \) and \( t'' = t + \frac{z}{v} \) respectively. These transformations help in simplifying the analysis of wave propagation.
To show that these functions satisfy the wave equation, we must demonstrate that the second partial derivatives exist and are related according to the equation. Differentiability ensures the existence of these derivatives, allowing us to perform the necessary calculus operations.
Partial Derivatives
Partial derivatives extend the concept of differentiation to functions of several variables. They show how a function changes as one particular variable changes, keeping others constant. This is pivotal in solving and understanding the wave equation.

For a function \( f(t') \), taking partial derivatives with respect to which variable makes sense given its dependency. For \( f(t') \), we focus on \( t \) and \( z \) with \( t' = t - \frac{z}{v} \).
  • First, compute \( \frac{\partial f}{\partial t} \), yielding \( f'(t') \) as the transformation gives \( \frac{\partial t'}{\partial t} = 1 \).
  • Then compute \( \frac{\partial f}{\partial z} \), resulting in \( -\frac{1}{v}f'(t') \) because \( \frac{\partial t'}{\partial z} = -\frac{1}{v} \).
Further, taking the second partial derivatives will verify if they meet the criteria set by the wave equation. The symmetry in these derivatives reflects the wave nature of the solutions.
Transformations
Transformations in mathematics are operations that convert a function into another form, easing computation or comprehension. In our wave equation example, the transformations \( t' = t - \frac{z}{v} \) and \( t'' = t + \frac{z}{v} \) simplify the differential analysis.
These transformations suggest a way to understand wave travel. The subtraction operation in \( t' \) represents a shift backward over time, akin to a wave moving toward a source. Conversely, \( t'' \) depicts forward motion, a wave moving away from a source.
Applying transformations facilitates easier differentiation and integration, a crucial step in verifying whether \( f(t') \) and \( g(t'') \) satisfy the wave equation. They allow computation by reducing the complexity of multi-variable calculus down to single-variable equations.
Sine Function
The sine function is a fundamental trigonometric function, characterized by its periodic nature鈥攊deal for modeling waves. Selecting \( f(t') = \sin(t') \) for our example illustrates how a simple trigonometric function can satisfy the wave equation.
Compute the necessary derivatives for \( \sin(t') \):
  • The first derivative is \( \cos(t') \).
  • The second derivative, which we use to verify the wave equation, is \( -\sin(t') \).
Substituting into the wave equation, compute both \( \frac{\partial^2 \sin(t')}{\partial t^2} = -\sin(t') \) and \( \frac{\partial^2 \sin(t')}{\partial z^2} = \frac{1}{v^2}(-\sin(t')). \)
Consistently, the equation balances, demonstrating that \( \sin(t') \) naturally fulfills the criteria of the classical wave equation. This proves the utility and simplicity of the sine function in wave dynamics.

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