Problem 150 - Calculus 3

Question

Show that \(f(x, y)=\ln \left(x^{2}+y^{2}\right)\) solves Laplace's equation \(\frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}=0\)

Step-by-Step Solution

Answer

Yes, \(f(x, y)=\ln(x^2+y^2)\) satisfies Laplace's equation.

Step 1: Identify the function

We are given the function \( f(x, y) = \ln(x^2 + y^2) \). We need to check if this function satisfies Laplace's equation, \(\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 0\).

Step 2: Compute the first partial derivatives

Find the first derivatives of \( f(x, y) \) with respect to \(x\) and \(y\):- \( \frac{\partial f}{\partial x} = \frac{2x}{x^2 + y^2} \)- \( \frac{\partial f}{\partial y} = \frac{2y}{x^2 + y^2} \)

Step 3: Compute the second partial derivatives

Find the second derivatives by differentiating the first derivatives:- \( \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{2x}{x^2 + y^2}\right) \) Compute: \[ \frac{\partial^2 f}{\partial x^2} = \frac{2(y^2 - x^2)}{(x^2 + y^2)^2} \]- \( \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}\left(\frac{2y}{x^2 + y^2}\right) \) Compute: \[ \frac{\partial^2 f}{\partial y^2} = \frac{2(x^2 - y^2)}{(x^2 + y^2)^2} \]

Step 4: Check Laplace's equation

Add the second partial derivatives:\[\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{2(y^2 - x^2)}{(x^2 + y^2)^2} + \frac{2(x^2 - y^2)}{(x^2 + y^2)^2} = 0\]Since the sum is zero, the function \(f(x, y) = \ln(x^2 + y^2)\) satisfies Laplace's equation.

Key Concepts

Partial Derivatives

Partial derivatives are an extension of the concept of a derivative to functions of multiple variables. In simpler terms, a partial derivative measures how a function changes as only one of the variables changes, keeping the others constant.
For example, if you have a function of two variables, such as \( f(x, y) \), and you want to find out how this function changes as you increase \( x \) while keeping \( y \) constant, you would compute the partial derivative of \( f \) with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \).
This process is similar when you take the derivative with respect to \( y \), resulting in \( \frac{\partial f}{\partial y} \). By using partial derivatives, we can analyze multivariable functions more precisely by focusing on how each input affects the output independently. This is crucial when solving equations like Laplace's equation.

Function of Two Variables

A function of two variables, such as \( f(x, y) = \ln(x^2 + y^2) \), assigns a real number value based on two input variables, \( x \) and \( y \). This type of function can be thought of as a surface in three-dimensional space.
To visualize this, consider the entire coordinate plane where each point \((x, y)\) represents an input to the function, and the output \( f(x, y) \) determines the 'height' of the surface at that point.

• Functions of two variables often deal with real-world phenomena, such as temperature over a geographical area or elevation over a landscape.
• These functions help us understand complex relationships by breaking them down into simple, dependent variables.

Analyzing functions of multiple variables often requires calculus tools like partial derivatives to uncover the hidden intricacies of these relationships.

Logarithmic Functions

Logarithmic functions are inverse operations of exponentiation and are critical in many areas of mathematics and science. The natural logarithm denoted as \( \ln(x) \) specifically deals with the base of mathematical constant \( e \), approximately equal to 2.718.
These functions help us describe processes that grow rapidly yet slow down over time, like population growth or radioactive decay.

• Logarithmic functions also play a key role in simplifying complex multiplicative relationships into additive forms.
• For example, the function \( \ln(x^2 + y^2) \) can transform geometric growth into linear growth, which can be easier to manage.

In the context of solving Laplace's equation, a logarithmic transformation can reveal symmetries and properties of the function that are not immediately obvious, aiding in the simplification and solution of complex problems.