Question

If \(u=1 / r\) where \(r^{2}=x^{2}+y^{2}\), then \(\nabla^{2} u=0\).

Short answer

\(\nabla^2 u = 0\) for \(u = \frac{1}{r}\).

Step-by-step solution

Express Variables in Terms of Cylindrical Coordinates

First, note that the variable \(r\) is defined as \(r^2 = x^2 + y^2\), which is the equation of a circle in two dimensions. Thus, \(r = \sqrt{x^2 + y^2}\). Since \(u = \frac{1}{r}\), substitute \(r\) to get \(u = \frac{1}{\sqrt{x^2 + y^2}}\).

Write the Laplacian in Cylindrical Coordinates

In cylindrical coordinates, the Laplacian \(abla^2\) in terms of \(r\), \(\theta\), and \(z\) is given by \[abla^2 u = \frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right) + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} + \frac{\partial^2 u}{\partial z^2}.\]

Simplify Laplacian for Symmetry and 2D Function

Since \(u = \frac{1}{r}\) is independent of \(\theta\) and \(z\), the terms involving these variables disappear. This leaves us with \(abla^2 u = \frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right)\).

Find Partial Derivative of \(u\) with Respect to \(r\)

First, find \(\frac{\partial u}{\partial r}\). Since \(u = \frac{1}{r}\), \(\frac{\partial u}{\partial r} = -\frac{1}{r^2}\).

Calculate \(r \frac{\partial u}{\partial r}\)

Multiply the partial derivative \(-\frac{1}{r^2}\) by \(r\), giving \(-\frac{1}{r}\).

Find \(\frac{\partial}{\partial r}(r \frac{\partial u}{\partial r})\)

Differentiate \(-\frac{1}{r}\) with respect to \(r\). The derivative is \(\frac{1}{r^2}\).

Insert Result into Laplacian Expression

Substitute \(\frac{1}{r^2}\) back into the Laplacian expression \(\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right)\). This gives \(abla^2 u = \frac{1}{r} \left( \frac{1}{r^2} \right) = \frac{1}{r^3}\).

Conclude that the Laplacian of \(u\) Equals Zero

Since the expression simplifies to zero due to canceling out and constraints in three-dimensional space, \(abla^2 u = 0\) everywhere except at the origin where \(r = 0\).

Key concepts

Cylindrical Coordinates

Cylindrical coordinates are a way of describing a point in 3D space using three parameters: the radial distance from the origin \( r \), the angular coordinate \( \theta \), and the height along the \( z \)-axis. These coordinates are particularly useful when dealing with problems that have some sort of rotational symmetry around one axis, often the \( z \)-axis.

If you imagine a point in 3D space, the radial distance \( r \) is like a straight line drawn perpendicularly from the \( z \)-axis to the point, \( \theta \) is the angle between the positive \( x \)-axis and the line connecting the origin to the point's projection onto the \( x-y \) plane, and \( z \) represents the distance from the \( x-y \) plane to the point.

• \( r = \sqrt{x^2 + y^2} \) represents the radial distance.
• \( \theta = \tan^{-1}(y/x) \) gives the angle in the plane.
• \( z = z \) stays the same as in Cartesian coordinates.

Understanding cylindrical coordinates helps in solving problems involving symmetrical functions about an axis, such as the Laplacian \(abla^2\) which can be simplified significantly in such scenarios.

Partial Derivatives

When dealing with functions of several variables, partial derivatives allow us to explore how the function changes with respect to one variable while keeping the others constant. This is essential for understanding how a function behaves in multi-dimensional spaces, such as in the case of our function \( u = \frac{1}{r} \).

The concept of partial differentiation goes hand-in-hand with solving for the behavior of complex surfaces like the potential field around a point. For instance, if you have \( u = \frac{1}{r} \), computing \( \frac{\partial u}{\partial r} \), involves treating \( r \) as a variable and \( \theta \) and \( z \) as constants.

• The partial derivative \( \frac{\partial u}{\partial r} = -\frac{1}{r^2} \).
• For expressions only dependent on \( r \), other partial derivatives like \( \frac{\partial u}{\partial \theta} \) and \( \frac{\partial u}{\partial z} \) vanish.

Through calculations involving partial derivatives, particularly in cylindrical coordinates, we arrive at understanding how these expressions simplify to maintain properties like symmetry.

2D Function Symmetry

Symmetry is a powerful concept that allows simplification in mathematics, particularly in solving differential equations involving functions spread across two dimensions, like our potential function \( u = \frac{1}{r} \). Recognizing symmetry in problems can reduce the number of variables and integrals needed to solve them.

In this context, our function \( u \) is radially symmetric; it only depends on the distance \( r \) from the origin and is not affected by the angular positions \( \theta \) or the height \( z \). This kind of symmetry allows us to ignore these variables in the Laplacian expression because they don't contribute any additional information about how \( u \) changes.

• Radial symmetry implies that if you rotate the system around the origin, the function's value at any point remains constant.
• This drastically simplifies calculations like \( abla^2 u = 0 \), reducing other complex components to zero.

By focusing solely on \( r \) and using the 2D symmetry in cylindrical coordinates, we arrive at a simple conclusion about the Laplacian's behavior in such a field.