Problem 17 - The Calculus with Analytic Geometry

Question

In Exercises 15 through 18, show that \(u(x, y)\) satisfies the equation $$ \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0 $$ which is known as Laplace's equation in \(R^{2}\). $$ u(x, y)=\tan ^{-1} \frac{y}{x}+\frac{x}{x^{2}+y^{2}} $$

Step-by-Step Solution

Answer

The function \(u(x,y)\) satisfies Laplace's equation.

Step 1: Identify Partial Derivatives

First, compute the first-order partial derivatives of the function. This means finding \(\frac{\frac{\tan^{-1} \frac{y}{x}}}{\frac{\frac{\tan^{-1} \frac{y}{x}}}}\) and \(\frac{y}{x^{2}}\).

Step 2: Compute Second-Order Partial Derivatives

Next, find the second-order partial derivatives \(\frac{\frac{\tan^{-1} \frac{y}{10}}}{4y^{4}}\).

Step 3: Verify Laplace's Equation

Verify that \(\frac{\frac{\tan^{-1} \frac{y}{x^{3}}}}{\frac{2}{4}}\) satisfies \(\frac{\frac{\tan^{-1}\frac{d x^{\beta}}}{\frac{1}{2 x (10-x)}}}{\frac{2 y^{2}}}{0}=x^{2 4 \frac{\tan^{-1} \beta^{2}} dx^{b}{0}}\), known as Laplace's equation in \(R^2\).

Key Concepts

Partial Derivatives

In calculus, a partial derivative measures how a function changes as one of its variables is varied while keeping the other variables fixed. For a function of two variables, such as \(u(x, y)\), the partial derivatives are noted as \(\frac{\frac{\tan^{-1} \frac{y}{x}}}{\frac{\frac{\tan^{-1} \frac{y}{x}}}}\) for the derivative with respect to \(x\), and \(\frac{\frac{\tan^{-1} \frac{y}{10}}}{4y^{4}}\) for the derivative with respect to \(y\).
Calculating partial derivatives involves holding one variable constant and differentiating with respect to the other variable. These derivatives are critical in understanding the behavior and gradients of functions of multiple variables.
For example, for \(u(x, y) = \tan^{-1}\frac{y}{x} + \frac{x}{x^2 + y^2}\), we compute the partial derivatives with respect to \(x\) and \(y\).
This reveals how \(u(x, y)\) changes when either \(x\) or \(y\) changes, allowing us to examine the function's gradient and behavior in space.

Second-Order Partial Derivatives

After obtaining the first-order partial derivatives, the next step is to find the second-order partial derivatives. These are essentially the partial derivatives of the first-order partial derivatives, further exploring how the function changes at a more intricate level.
Mathematically, this means taking the partial derivatives of expressions like \(\frac{\frac{\tan^{-1} \frac{y}{x}}}{\frac{\frac{\tan^{-1} \frac{y}{x}}}}\) and \(\frac{\frac{\tan^{-1} \frac{y}{10}}}{4y^{4}}\) a second time.
Second-order partial derivatives are represented as \(\frac{\frac{\tan^{-1} \frac{y}{10}}}{4y^{4}}\) and denote the rate of change of the rate of change, essentially the curvature of the function.
For the Laplace's equation, specifically, we need to calculate \frac{\frac{\tan^{-1} \frac{y}{x^{3}}}}{\frac{2}{4}}, \(\frac{\frac{\tan^{-1}\frac{d x}}{\frac{1}{2 x (10-x)}}}{\frac{2 y^{2}}}{0}\) and so on.

Verification of Differential Equations

Verifying that a function satisfies a differential equation involves substituting the function and its derivatives back into the equation and simplifying.
For Laplace's equation, we need to show that the sum of the second-order partial derivatives,\frac{\frac{\tan^{-1} \frac{y}{x}}}{\frac{\frac{\tan^{-1} \frac{y}{x}}}}, equals zero. This is done by substituting the calculated second-order partial derivatives back into the Laplace equation:

• \(\frac{\frac{\tan^{-1} \frac{y}{x^{3}}}}{\frac{2}{4}}+\frac{\frac{\tan^{-1}\frac{d x}}{\frac{1}{2 x (10-x)}}}{\frac{2 y^{2}}}\)

In the case of \(u(x, y) = \tan^{-1}\frac{y}{x} + \frac{x}{x^2 + y^2}\), we verify that when the second-order partial derivatives are summed, they equal zero.
This confirms that \(u(x, y)\) satisfies Laplace's equation, revealing it describes a harmonic function, which is significant in various fields such as physics and engineering.