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

Short answer

Answer: Yes, the function u(x, y) is harmonic because it satisfies Laplace's equation, i.e., the sum of its second partial derivatives with respect to x and y equals zero.

Step-by-step solution

Finding the first partial derivatives of u(x, y)

First, we need to find the first partial derivatives of u(x, y) with respect to x and y: $$u(x, y) = \tan^{-1}\left(\frac{y}{x-1}\right) - \tan^{-1}\left(\frac{y}{x+1}\right)$$ Using the chain rule, we have: $$\frac{\partial u}{\partial x} = \frac{- y}{(x-1)^2 + y^2} - \frac{y}{(x+1)^2 + y^2}$$ $$\frac{\partial u}{\partial y} = \frac{x-1}{(x-1)^2 + y^2} - \frac{x+1}{(x+1)^2 + y^2}$$

Finding the second partial derivatives of u(x, y)

Next, we need to find the second partial derivatives of u(x, y) with respect to x and y: $$\frac{\partial^2 u}{\partial x^2} = \frac{2y^2}{[(x-1)^2 + y^2]^2} - \frac{2y^2}{[(x+1)^2 + y^2]^2}$$ $$\frac{\partial^2 u}{\partial y^2} = \frac{2(x-1)^2}{[(x-1)^2 + y^2]^2} - \frac{2(x+1)^2}{[(x+1)^2 + y^2]^2}$$

Plugging the second partial derivatives into Laplace's equation

Now, we plug the second partial derivatives into the Laplace's equation and check if it equals zero: $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \left(\frac{2y^2}{[(x-1)^2 + y^2]^2} - \frac{2y^2}{[(x+1)^2 + y^2]^2}\right) + \left(\frac{2(x-1)^2}{[(x-1)^2 + y^2]^2} - \frac{2(x+1)^2}{[(x+1)^2 + y^2]^2}\right)$$

Simplify the expression

Let's simplify the expression by combining the terms with the same denominator: $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{2y^2[(x+1)^2+y^2] + 2(x-1)^2[(x+1)^2+y^2]}{[(x-1)^2+y^2]^2[(x+1)^2+y^2]^2} - \frac{2y^2[(x-1)^2+y^2] + 2(x+1)^2[(x-1)^2+y^2]}{[(x-1)^2+y^2]^2[(x+1)^2+y^2]^2}$$ We can see that the expressions in the numerator are the same, so the result is: $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$$ Since the Laplace's equation is satisfied, the given function u(x, y) is harmonic.

Key concepts

Harmonic Function

A harmonic function is a twice continuously differentiable function that satisfies Laplace's equation. In two dimensions, this equation can be written as \[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\]
Harmonic functions have several properties that make them fundamental in various fields such as physics and engineering. They represent potential fields in electrostatics, gravitational fields, and steady-state heat distribution. In the given exercise, the function \[u(x, y)=\tan ^{-1}\left(\frac{y}{x-1}\right)-\tan^{-1}\left(\frac{y}{x+1}\right)\]is proven to be harmonic because it satisfies Laplace's equation after going through the steps of finding the first and second order partial derivatives, ultimately leading to a sum of zero when these derivatives are used in the Laplace's equation. This outcome confirms the harmonic nature of the function, highlighting the connection between abstract mathematical concepts and their physical interpretations.

Partial Derivatives

In multivariate calculus, partial derivatives play a crucial role in analyzing the behavior of functions of several variables. When we compute a partial derivative of a function, we are finding the rate at which the function changes with respect to one of its variables while holding the others constant.
For example, in our exercise with function \[u(x, y)\], the partial derivatives \[\frac{\partial u}{\partial x}\] and \[\frac{\partial u}{\partial y}\]
are calculated to determine how \(u\) changes in the \(x\)-direction and \(y\)-direction respectively. These calculations are crucial in solving Laplace's equation because they provide the second-order derivatives required for the equation. The process of finding these involves applying the chain rule and careful algebraic manipulation, as demonstrated in the step-by-step solution. Understanding partial derivatives is fundamental not just for solving this problem but also for diving into deeper areas of mathematical analysis and applied disciplines like physics and engineering.

Chain Rule

The chain rule is a vital tool in differential calculus used when dealing with composite functions — functions composed of other functions. It provides a method to calculate the derivative of a composite function based on the derivatives of its constituent functions.
In practice, for a composite function \[f(g(x))\], the chain rule states that the derivative of \(f\) with respect to \(x\) is the derivative of \(f\) with respect to \(g\), multiplied by the derivative of \(g\) with respect to \(x\), or expressed formulaically as \[\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\].
This becomes invaluable when finding the first and, as we've seen in the exercise, the second partial derivatives of more complex functions. When \(u(x, y)\) involved taking the arctangent of a fraction, the chain rule helped us find the necessary derivatives efficiently and accurately. Mastering the chain rule facilitates solving not only basic calculus problems but also more complex equations involving partial derivatives, such as Laplace's equation.