Problem 150 - Calculus Volume 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, \( f(x, y) = \ln(x^2 + y^2) \) solves Laplace's equation.

Step 1: Identify the Problem Structure

We are given a function \( f(x, y) = \ln(x^2 + y^2) \) and need to show that it satisfies Laplace's equation: \( \frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 0 \). We must compute the second partial derivatives with respect to \(x\) and \(y\), then sum them to evaluate the equation.

Step 2: First Order Partial Derivatives

Calculate the first order partial derivatives of \( f(x, y) \). For \( \frac{\partial f}{\partial x} \), we get:\[ \frac{\partial}{\partial x}(\ln(x^2 + y^2)) = \frac{2x}{x^2 + y^2} \]Similarly, for \( \frac{\partial f}{\partial y} \), the result is:\[ \frac{\partial}{\partial y}(\ln(x^2 + y^2)) = \frac{2y}{x^2 + y^2} \]

Step 3: Second Order Partial Derivatives

Next, calculate the second partial derivatives. Starting with \( \frac{\partial^2 f}{\partial x^2} \):Differentiate \( \frac{2x}{x^2 + y^2} \) with respect to \(x\), using the quotient rule:\[ \frac{\partial^2 f}{\partial x^2} = \frac{2(x^2 + y^2) - 4x^2}{(x^2 + y^2)^2} = \frac{2y^2 - 2x^2}{(x^2 + y^2)^2} \]For \( \frac{\partial^2 f}{\partial y^2} \), differentiate \( \frac{2y}{x^2 + y^2} \) with respect to \(y\):\[ \frac{\partial^2 f}{\partial y^2} = \frac{2(x^2 + y^2) - 4y^2}{(x^2 + y^2)^2} = \frac{2x^2 - 2y^2}{(x^2 + y^2)^2} \]

Step 4: Sum of Second Order Derivatives

Now, sum the second order partial derivatives to verify Laplace's equation:\[ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{2y^2 - 2x^2}{(x^2 + y^2)^2} + \frac{2x^2 - 2y^2}{(x^2 + y^2)^2} \]This simplifies to:\[ \frac{0}{(x^2 + y^2)^2} = 0 \]

Step 5: Conclusion

Since the sum of the second partial derivatives is 0, \( f(x, y) = \ln(x^2 + y^2) \) satisfies Laplace's equation. This confirms \( f(x, y) \) is a solution.

Key Concepts

Partial Derivatives

Partial derivatives extend the concept of a derivative to functions of multiple variables. For a function of two variables, say \( f(x, y) \), the partial derivative with respect to \( x \) measures how \( f \) changes as \( x \) changes while keeping \( y \) constant. Similarly, the partial derivative with respect to \( y \) measures how \( f \) changes with \( y \) constant.

To compute a first-order partial derivative, we differentiate the function as if the variable with respect to which we're differentiating is the only variable. For example:

• First partial derivative of \( f(x, y) = \ln(x^2 + y^2) \) with respect to \( x \) is \( \frac{2x}{x^2 + y^2} \).
• With respect to \( y \), it is \( \frac{2y}{x^2 + y^2} \).

Second-order partial derivatives involve differentiating a first-order partial derivative again. Each variable gets its own second-order partial derivative, such as \( \frac{\partial^2 f}{\partial x^2} \) and \( \frac{\partial^2 f}{\partial y^2} \). These help us analyze the function's curvature or concavity and are crucial in solving Laplace's equation.

Multivariable Calculus

Multivariable calculus extends calculus to functions with more than one variable. While single-variable calculus focuses on curves, multivariable calculus deals with surfaces and the rates of change along them.

In problems involving functions such as \( f(x, y) \), we consider how changes in both \( x \) and \( y \) affect the output of \( f \). Concepts like gradients, directional derivatives, and double integrals become pivotal. Here:

• Gradients point in the direction of the greatest rate of increase of a function.
• Directional derivatives measure how a function changes in any given direction.
• Double integrals extend the idea of integration to two dimensions, allowing us to compute areas and volumes under surfaces.

Laplace's equation belongs to this field because it involves partial derivatives of multivariable functions, like in this exercise with the function \( f(x, y) = \ln(x^2 + y^2) \). Many concepts combine in multivariable calculus to solve such differential equations.

Differential Equations

Differential equations involve unknown functions and their derivatives. They can describe real-world phenomena like heat, sound, and electricity.

Laplace's equation, \( \frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 0 \), is a type of partial differential equation (PDE). Here, the unknown function \( z \) depends on two variables, \( x \) and \( y \), and the equation specifies a condition the second derivatives must satisfy.

This equation often arises in physics and engineering, representing steady-state heat distribution or potential fields in electrostatics. To find a function that satisfies it:

• Calculate the necessary second partial derivatives.
• Sum them up to see if they equal zero.

Our task in this exercise was to verify that for \( f(x, y) = \ln(x^2 + y^2) \), this condition holds true. Thus, showing that it is indeed a solution to Laplace's equation.