Question

Calculate the Laplacian \(\nabla^{2}\) of each of the following scalar fields.\(\ln \left(x^{2}+y^{2}\right)\)

Short answer

The Laplacian \( abla^{2} \ln (x^{2} + y^{2}) \) is 0

Step-by-step solution

- Define the Scalar Field

The given scalar field is \( f(x,y) = \ln (x^{2} + y^{2}) \)

- Compute the First Derivatives

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

- Compute the Second Derivatives

Calculate the second partial derivatives: \( \frac{\partial^{2} f}{\partial x^{2}} = \frac{2(y^{2} - x^{2})}{(x^{2} + y^{2})^{2}} \) and \( \frac{\partial^{2} f}{\partial y^{2}} = \frac{2(x^{2} - y^{2})}{(x^{2} + y^{2})^{2}} \)

- Sum the Second Derivatives

Apply the definition of the Laplacian: \( abla^{2} f = \frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} \). Summing the second partial derivatives, we get: \( abla^{2} f = \frac{2(y^{2} - x^{2})}{(x^{2} + y^{2})^{2}} + \frac{2(x^{2} - y^{2})}{(x^{2} + y^{2})^{2}} = 0 \)

Key concepts

Scalar field

A scalar field represents a mathematical function that assigns a scalar value—just a single number—to every point in space. Imagine a temperature map; each point on the map has a specific temperature (the scalar value). In our exercise, the function we start with is the scalar field \( f(x,y) = \ln (x^2 + y^2) \). This particular scalar field assigns the natural logarithm of the sum of the squares of the coordinates to each point \( (x, y) \) in the plane.

Partial derivatives

To find how our scalar field changes in specific directions, we use partial derivatives. Imagine you want to understand how the temperature changes if you move just horizontally or just vertically on your map. Here, in our exercise, we start by taking the partial derivatives of the scalar field \( f(x,y) = \ln (x^2 + y^2) \).

• For the direction along the x-axis: \( \frac{\partial f}{\partial x} = \frac{2x}{x^2 + y^2} \)
• For the direction along the y-axis: \( \frac{\partial f}{\partial y} = \frac{2y}{x^2 + y^2} \)

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Second derivatives

The second derivative measures how the first derivative changes as you move in space. Think of it as checking the rate of change in the rate of change. In our scalar field \( f(x,y) \), we calculate the second partial derivatives to delve deeper into the function's behavior over space. For our given function, we have:

• For the second derivative along the x-axis: \( \frac{\partial^2 f}{\partial x^2} = \frac{2(y^2 - x^2)}{(x^2 + y^2)^2} \)
• For the second derivative along the y-axis: \( \frac{\partial^2 f}{\partial y^2} = \frac{2(x^2 - y^2)}{(x^2 + y^2)^2} \)

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Mathematical methods

Mathematical methods often bring everything together, helping us solve complex problems. Here, the Laplacian is a mathematical method that makes use of second partial derivatives to measure the rate at which a quantity changes in all directions around a point.To find the Laplacian \( abla^2 f \), we sum the second partial derivatives:\( abla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} \). For our function, we get:\( \frac{2(y^2 - x^2)}{(x^2 + y^2)^2} + \frac{2(x^2 - y^2)}{(x^2 + y^2)^2} = 0 \) This means that the Laplacian at any point for our function is zero. In physical terms, this could denote a state of equilibrium at that point.