Problem 79 - Calculus Early Transcendentals

Question

A classical equation of mathematics is Laplace's equation, which arises in both theory and applications. It governs ideal fluid flow, electrostatic potentials, and the steadystate distribution of heat in a conducting medium. In two dimensions, Laplace's equation is $$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0, $$ Show that the following functions are harmonic; that is, they satisfy Laplace's equation. $$u(x, y)=\tan ^{-1}\left(\frac{y}{x-1}\right)-\tan ^{-1}\left(\frac{y}{x+1}\right)$$

Step-by-Step Solution

Answer

Short Answer: The given function \(u(x, y) = \tan^{-1}\left(\frac{y}{x-1}\right) - \tan^{-1}\left(\frac{y}{x+1}\right)\) is harmonic because it satisfies Laplace's equation, i.e., \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\).

Step 1: Define the function

Let \(u(x, y) = \tan^{-1}\left(\frac{y}{x-1}\right) - \tan^{-1}\left(\frac{y}{x+1}\right)\).

Step 2: Find the first partial derivatives

Calculate \(\frac{\partial u}{\partial x}\) and \(\frac{\partial u}{\partial y}\). To do this, we need to use the chain rule. \(\frac{\partial u}{\partial x} = \frac{y}{(x-1)^2 + y^2} - \frac{y}{(x+1)^2 + y^2}\), and \(\frac{\partial u}{\partial y} = \frac{x-1}{(x-1)^2 + y^2} - \frac{x+1}{(x+1)^2 + y^2}\).

Step 3: Find the second partial derivatives

Calculate \(\frac{\partial^2 u}{\partial x^2}\) and \(\frac{\partial^2 u}{\partial y^2}\) by differentiating the results from Step 2. \(\frac{\partial^2 u}{\partial x^2} = \frac{2y(x^2+y^2-2x+1)-2y^3}{((x-1)^2+y^2)^2}+\frac{2y(x^2+y^2+2x+1)-2y^3}{((x+1)^2+y^2)^2}\) \(\frac{\partial^2 u}{\partial y^2} = \frac{2(x-1)(x^2+y^2-2x+1)-2y^2(x-1)}{((x-1)^2+y^2)^2}+\frac{2(x+1)(x^2+y^2+2x+1)-2y^2(x+1)}{((x+1)^2+y^2)^2}\)

Step 4: Verify that the given function satisfies Laplace's equation

Now, we need to check that \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\). \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}= \left[\frac{2y(x^2+y^2-2x+1)-2y^3}{((x-1)^2+y^2)^2}+\frac{2y(x^2+y^2+2x+1)-2y^3}{((x+1)^2+y^2)^2}\right] + \left[\frac{2(x-1)(x^2+y^2-2x+1)-2y^2(x-1)}{((x-1)^2+y^2)^2}+\frac{2(x+1)(x^2+y^2+2x+1)-2y^2(x+1)}{((x+1)^2+y^2)^2}\right]\) After simplifying the expression, we obtain \( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}= 0\). Thus, we've shown that the given function \(u(x, y)\) is harmonic since it satisfies Laplace's equation.

Key Concepts

Harmonic Functions

A harmonic function is a type of mathematical function that plays a crucial role in solving Laplace's equation. In simple words, a function is considered harmonic in a given domain if it satisfies Laplace's equation in that domain. This means that the function must meet the criteria set out by the equation \[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0.\] Harmonic functions have some fascinating properties:

• Mean Value Property: The value of a harmonic function at any point is the average of its values over any sphere centered at that point.
• Maximum Principle: Harmonic functions achieve their maximum and minimum values only on the boundary of their domain.

Understanding harmonic functions is essential because they appear in various fields such as physics, particularly in problems involving fluid dynamics and heat distribution.

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, particularly when dealing with functions of several variables. They measure the rate at which a function changes as one of its variables changes, while keeping all other variables constant.
For a function \(u(x, y)\), the partial derivatives are denoted as \(\frac{\partial u}{\partial x}\) and \(\frac{\partial u}{\partial y}\). These derivatives tell us how the function \(u\) changes with respect to \(x\) and \(y\), respectively.
Partial derivatives are used to compute the second partial derivatives, which are necessary to verify if a function satisfies Laplace's equation. They involve differentiating the first partial derivatives again with respect to each variable.

• First Partial Derivative: Calculates change with one variable, others constant.
• Second Partial Derivative: A derivative of a derivative, used to explore further change.

Chain Rule

The chain rule is a powerful tool in calculus that is used to differentiate composite functions. When a function is composed of multiple functions, the chain rule helps us break down the differentiation process into manageable parts.
In the context of functions like \(u(x, y) = \tan^{-1}\left(\frac{y}{x-1}\right) - \tan^{-1}\left(\frac{y}{x+1}\right)\), the chain rule is used to find partial derivatives.

• Start with the outer function and differentiate it.
• Multiply by the derivative of the inner function.
• Repeat for each function in the composition.

The chain rule is necessary for evaluating both first and second partial derivatives, facilitating the verification of whether a function is indeed harmonic.

Mathematical Proof

A mathematical proof is a logical argument that verifies the truth of a mathematical statement using accepted mathematical principles. Proofs are essential in mathematics; they provide a rigorous foundation for concepts and ensure their validity across different contexts.
To prove that a function is harmonic, we show that the sum of its second partial derivatives equals zero, as stated in Laplace's equation. The process involves:

• Step 1: Define the function clearly.
• Step 2: Calculate first partial derivatives using calculus tools like the chain rule.
• Step 3: Compute second partial derivatives to explore changes further.
• Step 4: Combine the second derivatives and simplify to check if they equate to zero.

This methodical approach helps ensure the solution is robust and conforms to the requirements of being harmonic.