Problem 35 - Calculus with Concepts in Calculus

Question

Let \(z=f(y+a x)+g(y-a x)\), with \(a \neq 0\). Show that \(z\) satisfies the wave equation $$ \frac{\partial^{2} z}{\partial x^{2}}=a^{2} \frac{\partial^{2} z}{\partial y^{2}} $$

Step-by-Step Solution

Answer

\( z \) satisfies the wave equation because its derivatives confirm the required equality.

Step 1: Understand the problem statement

We need to show that the expression for \(z\) satisfies the wave equation: \( \frac{\partial^{2} z}{\partial x^{2}}=a^{2} \frac{\partial^{2} z}{\partial y^{2}} \). This involves taking partial derivatives of \(z\) with respect to \(x\) and \(y\), and showing the required relationship.

Step 2: Find the first-order partial derivatives

Compute the first-order partial derivatives with respect to \(x\) and \(y\). For \(z=f(y+ax)+g(y-ax)\),1. \( \frac{\partial z}{\partial x} = a f'(y+ax) - a g'(y-ax) \), by chain rule.2. \( \frac{\partial z}{\partial y} = f'(y+ax) + g'(y-ax) \), also by chain rule.

Step 3: Compute second-order partial derivative with respect to \(x\)

Differentiate \( \frac{\partial z}{\partial x} = a f'(y+ax) - a g'((y-ax)) \) with respect to \(x\):\[ \frac{\partial^{2} z}{\partial x^{2}} = a^2 f''(y+ax) + a^2 g''(y-ax) \].

Step 4: Compute second-order partial derivative with respect to \(y\)

Differentiate \( \frac{\partial z}{\partial y} = f'(y+ax) + g'(y-ax) \) with respect to \(y\):\[ \frac{\partial^{2} z}{\partial y^{2}} = f''(y+ax) + g''(y-ax) \].

Step 5: Setup the wave equation

Substitute the second-order derivatives found earlier into the wave equation. We need to verify:\[ \frac{\partial^{2} z}{\partial x^{2}} = a^{2} \frac{\partial^{2} z}{\partial y^{2}} \],From Step 3, \( \frac{\partial^{2} z}{\partial x^{2}} = a^2 f''(y+ax) + a^2 g''(y-ax) \),From Step 4, \( \frac{\partial^{2} z}{\partial y^{2}} = f''(y+ax) + g''(y-ax) \).

Step 6: Verify the equation holds

Notice that multiplying the second derivative with respect to \(y\) by \(a^2\) gives:\[ a^2 \frac{\partial^{2} z}{\partial y^{2}} = a^2 (f''(y+ax) + g''(y-ax)) \].This is exactly \( \frac{\partial^{2} z}{\partial x^{2}} \) as computed in Step 3. Therefore, the wave equation \( \frac{\partial^{2} z}{\partial x^{2}} = a^2 \frac{\partial^{2} z}{\partial y^{2}} \) is satisfied.

Key Concepts

Partial Derivatives

In mathematics, partial derivatives are used to analyze functions with more than one variable. When we derive a function concerning one variable, treating all others as constants, we perform a partial derivative. In the problem given, we find partial derivatives of the function \( z = f(y + ax) + g(y - ax) \) with respect to \(x\) and \(y\).
This helps determine how \( z \) changes when one variable changes while the other remains fixed. These derivatives are essential to set up the differential equation for the wave equation.

• For the function, the partial derivative with respect to \( x \) is \( \frac{\partial z}{\partial x} = a f'(y + ax) - a g'(y - ax) \).
• The partial derivative with respect to \( y \) is \( \frac{\partial z}{\partial y} = f'(y + ax) + g'(y - ax) \).

Chain Rule

The chain rule is a fundamental concept in calculus that allows us to differentiate composite functions. When a function is nested within another, the chain rule assists in breaking down these layers for derivation.

• In our context, the chain rule is applied to functions like \( f(y + ax) \) and \( g(y - ax) \).
• When differentiating these functions concerning \( x \), the chain rule helps express the derivatives \( f'(y + ax) \) and \( g'(y - ax) \) in terms of \( a \).
• This occurs because \( ax \) is itself a function of \( x \), introducing a factor of \( a \) in the derivatives.

Thus, the chain rule effectively manages the derivative of the inner function, leading to the results needed for solving the wave equation problem.

Second-Order Derivatives

Second-order derivatives are derivatives of derivatives. They describe the rate at which the first derivative of a function changes, offering insight into the function's curvature.

For the given problem:

• The second-order partial derivative with respect to \( x \) is \( \frac{\partial^2 z}{\partial x^2} = a^2 f''(y + ax) + a^2 g''(y - ax) \).
• The second-order partial derivative with respect to \( y \) is \( \frac{\partial^2 z}{\partial y^2} = f''(y + ax) + g''(y - ax) \).

These derivatives are essential for confirming whether the given \( z \) function satisfies the wave equation.
The original goal being to match \( \frac{\partial^2 z}{\partial x^2} = a^2 \frac{\partial^2 z}{\partial y^2} \). By achieving this equality, we effectively satisfy the wave equation, showing the interplay between the components of \( z \).

Calculus

Calculus is the branch of mathematics that deals with continuous change. It encompasses two main areas: differentiation and integration.

Differentiation is key in deriving rates of change, which is central to the wave equation problem. The use of calculus allows us to manipulate and simplify complex functions to reveal underlying patterns or principles.

• The partial derivatives we calculate here are fundamental calculus tools, helping us determine the wave equation's behavior across different dimensions in the function \( z \).
• By integrating the knowledge of how changing \( x \) and \( y \) affects \( z \), we utilize calculus to bridge these concepts successfully.

Understanding calculus thus provides the mathematical framework necessary to solve advanced differential equations like the wave equation.

Differential Equations

Differential equations are equations that involve derivatives, expressing how a particular quantity changes with others present in their formula. Solving these equations helps understand the dynamic behavior of complex systems.

In this exercise, we deal with a specific type of differential equation known as the wave equation, where the challenge is showing that our expression for \( z \) fulfills

• \( \frac{\partial^2 z}{\partial x^2} = a^2 \frac{\partial^2 z}{\partial y^2} \)

Applying the derivatives step by step and leveraging the formula for wave equations allows us to conclude that \( z \) satisfies the wave equation. Understanding differential equations enables us to resolve many real-world phenomena, modeling waves, heat, and other physical processes.