Problem 27 - Advanced Engineering Mathematics

Question

$$\begin{aligned} &\frac{\partial^{2} u}{\partial x^{2}}=\frac{2 y^{2}-2 x^{2}}{\left(x^{2}+y^{2}\right)^{2}}, \frac{\partial^{2} u}{\partial y^{2}}=\frac{2 x^{2}-2 y^{2}}{\left(x^{2}+y^{2}\right)^{2}} \text { gives } \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0 . \quad \text { Thus } u \text { is harmonic. Now } \frac{\partial u}{\partial x}=\frac{2 x}{x^{2}+y^{2}}=\frac{\partial v}{\partial y}\\\ &\text { implies } v=2 \tan ^{-1} \frac{y}{x}+h(x), \quad \frac{\partial u}{\partial y}=\frac{2 y}{x^{2}+y^{2}}=-\frac{\partial v}{\partial x}=\frac{2 y}{x^{2}+y^{2}}-h^{\prime}(x) \quad \text { implies } h^{\prime}(x)=0 \text { or } h(x)=C\\\ &\text { Therefore } f(z)=\log _{e}\left(x^{2}+y^{2}\right)+i\left(\tan ^{-1} \frac{y}{x}+C\right), z \neq 0 \end{aligned}$$

Step-by-Step Solution

Answer

The function is \( f(z) = \log_e(x^2 + y^2) + i(2 \tan^{-1} \frac{y}{x} + C) \), \( z \neq 0 \).

Step 1: Verify Harmonic Property

We are given that \( \frac{\partial^{2} u}{\partial x^{2}} = \frac{2 y^{2} - 2 x^{2}}{(x^{2}+y^{2})^{2}} \) and \( \frac{\partial^{2} u}{\partial y^{2}} = \frac{2 x^{2} - 2 y^{2}}{(x^{2}+y^{2})^{2}} \). Add these two equations to check if \( u \) is harmonic: \[ \frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = \frac{2 y^{2} - 2 x^{2}}{(x^{2}+y^{2})^{2}} + \frac{2 x^{2} - 2 y^{2}}{(x^{2}+y^{2})^{2}} = 0. \] Since the sum is zero, \( u \) is harmonic.

Step 2: Determine the Harmonic Conjugate

Given \( \frac{\partial u}{\partial x} = \frac{2x}{x^{2}+y^{2}} \), and it is equivalent to \( \frac{\partial v}{\partial y} \), the partial derivative of the harmonic conjugate \( v \). From this, we infer \( v = 2 \tan^{-1} \frac{y}{x} + h(x) \).

Step 3: Solve for h(x)

Using \( \frac{\partial u}{\partial y} = \frac{2y}{x^{2}+y^{2}} = -\frac{\partial v}{\partial x} = -\left( -\frac{2y}{x^{2}+y^{2}} + h'(x) \right) \), equate the expressions to find \( h'(x) = 0 \). Thus, \( h(x) = C \) where \( C \) is a constant.

Step 4: Conclude with the Complex Function

The function \( f(z) \) that gives \( u + iv \) is defined as \( f(z) = \log_e(x^2 + y^2) + i\left(2 \tan^{-1} \frac{y}{x} + C \right) \), where \( z eq 0 \). This is the expression for the complex potential.

Key Concepts

Laplace's Equation

Laplace's equation is a second-order partial differential equation frequently appearing in physics and engineering. It is expressed as \( abla^2 u = 0 \), where \( abla^2 \) is the Laplacian operator, defined in two dimensions as:

• \( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \)

A function \( u(x, y) \) that satisfies this equation is termed a "harmonic function". These functions are crucial in fields like electromagnetism, fluid dynamics, and potential theory.
Harmonic functions exhibit specific properties: they are smooth, and their values at any point are influenced heavily by the behavior of the function in its immediate vicinity. Notably, they satisfy the minimum and maximum principle, meaning they cannot have any local maxima or minima within any region they are defined unless it is constant throughout that region.
In this exercise, the verification that \( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \) ensured that \( u \) is harmonic by nature, conforming perfectly to Laplace's equation.

Complex Analysis

Complex analysis deals with functions of a complex variable, which are functions that can be written in the form \( f(z) = u(x, y) + iv(x, y) \), where \( z = x + iy \), and \( i \) is the imaginary unit. Here, \( u \) and \( v \) are real-valued functions representing the real and imaginary parts, respectively.
A significant concept in complex analysis is a "harmonic conjugate". If \( u \) is a harmonic function, its harmonic conjugate \( v \) can be found such that \( f(z) \) is analytic (i.e., differentiable) everywhere in its domain.
In the provided solution, finding \( v \) involves equating partial derivatives to ensure that \( u + iv \) maintains analyticity. This problem utilized the Cauchy-Riemann equations indirectly to determine \( v \) given \( u \). The function \( v \) was determined as \( v = 2 \tan^{-1} \frac{y}{x} + C \), where \( C \) is a constant. This approach helps in formulating \( f(z) \), which inherently satisfies properties of complex functions and provides a deeper understanding of conformal mappings in complex analysis.

Multivariable Calculus

Multivariable calculus extends fundamental calculus concepts to functions of more than one variable. Central to this is the understanding of partial derivatives and their role in describing changes in functions. For a function \( u(x, y) \), partial derivatives like \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \) express how \( u \) changes as each of its variables is varied, holding the other constant.
In this exercise, we calculated second-order partial derivatives \( \frac{\partial^2 u}{\partial x^2} \) and \( \frac{\partial^2 u}{\partial y^2} \) to verify if \( u \) satisfies Laplace's equation.
Multivariable calculus also introduces techniques to integrate functions over regions in \( \mathbb{R}^n \). These are often used to solve physical problems involving fields and flux. The interplay between derivatives shows how functions like \( u \) react to fluctuations within their defined space. This understanding is paramount when analyzing concepts like fluid flow, where quantities depend on multiple factors.
Understanding and manipulating such functions is essential for students learning about the spatial relationship and behavior of different variables simultaneously.