Problem 53 - Advanced Engineering Mathematics

Question

If \(u=f(x, y)\) and \(x=r \cos \theta, y=r \sin \theta\), show that Laplace's equation \(\partial^{2} u / \partial x^{2}+\partial^{2} u / \partial y^{2}=0\) becomes $$ \frac{\partial^{2} u}{\partial r^{2}}+\frac{1}{r} \frac{\partial u}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} u}{\partial \theta^{2}}=0 $$

Step-by-Step Solution

Answer

Laplace’s equation in polar coordinates is \( \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0 \).

Step 1: Identify the Chain Rule for Partial Derivatives

To transform the Laplace equation, we need to express derivatives with respect to \(x\) and \(y\) in terms of \(r\) and \(\theta\). We use the chain rule: \( \frac{\partial}{\partial x} = \frac{\partial r}{\partial x}\frac{\partial}{\partial r} + \frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta} \) and similarly for \(\frac{\partial}{\partial y}\). Determine \( \frac{\partial r}{\partial x}, \frac{\partial \theta}{\partial x}, \frac{\partial r}{\partial y}, \, \text{and} \, \frac{\partial \theta}{\partial y} \).

Step 2: Calculate Partial Derivatives

Using the relations \(x = r \cos \theta\) and \(y = r \sin \theta\), find: \( \frac{\partial r}{\partial x} = \cos \theta \), \( \frac{\partial \theta}{\partial x} = -\frac{\sin \theta}{r} \), \( \frac{\partial r}{\partial y} = \sin \theta \), and \( \frac{\partial \theta}{\partial y} = \frac{\cos \theta}{r} \).

Step 3: Apply Chain Rule to First Order Derivatives

Express \(\frac{\partial u}{\partial x}\) and \(\frac{\partial u}{\partial y}\) using the chain rule: \( \frac{\partial u}{\partial x} = \cos \theta \frac{\partial u}{\partial r} - \frac{\sin \theta}{r} \frac{\partial u}{\partial \theta} \) and \( \frac{\partial u}{\partial y} = \sin \theta \frac{\partial u}{\partial r} + \frac{\cos \theta}{r} \frac{\partial u}{\partial \theta} \).

Step 4: Second Order Partial Derivatives

Calculate the second order partial derivatives:\(\frac{\partial^{2} u}{\partial x^{2}}\) and \(\frac{\partial^{2} u}{\partial y^{2}}\). Use product and chain rules to expand second derivatives.

Step 5: Expand Second Order Derivatives

Expand the second derivatives: \( \frac{\partial^{2} u}{\partial x^{2}} = \cos^2 \theta \frac{\partial^2 u}{\partial r^2} - \frac{2 \sin \theta \cos \theta}{r} \frac{\partial^2 u}{\partial r \partial \theta} + \frac{\sin^2 \theta}{r^2} \frac{\partial^2 u}{\partial \theta^2} - \frac{\cos \theta}{r} \frac{\partial u}{\partial r} \) and \( \frac{\partial^{2} u}{\partial y^{2}} = \sin^2 \theta \frac{\partial^2 u}{\partial r^2} + \frac{2 \sin \theta \cos \theta}{r} \frac{\partial^2 u}{\partial r \partial \theta} + \frac{\cos^2 \theta}{r^2} \frac{\partial^2 u}{\partial \theta^2} - \frac{\sin \theta}{r} \frac{\partial u}{\partial r} \).

Step 6: Combine Derivatives and Simplify

Add \(\frac{\partial^2 u}{\partial x^2}\) and \(\frac{\partial^2 u}{\partial y^2}\). The terms involving \(\frac{\partial^2 u}{\partial r \partial \theta}\) and \(\frac{\partial^2 u}{\partial r^2}\) simplify, and the other terms cancel. Simplify the resulting expression to \( \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0 \).

Step 7: Conclude the Transformation

Conclude by stating that the Laplace equation in Cartesian coordinates has transformed into its polar form. Show that: \( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0 \).

Key Concepts

Polar Coordinates

When we talk about polar coordinates, we're diving into a different world compared to Cartesian coordinates. Instead of the standard \(x\) and \(y\) axes used in geometry, polar coordinates use a radius \(r\) and an angle \(\theta\). This is useful when dealing with circular or rotational symmetries. In polar coordinates, every point is described by how far away it is from the origin \((r)\) and the angle \((\theta)\) it makes with the positive x-axis.

This system is particularly handy in cases like Laplace's Equation where circular symmetry can simplify the mathematics involved. In our problem, the transformation from \(x\) and \(y\) to \(r\) and \(\theta\) allows us to express functions more intuitively and solve equations like Laplace’s with less effort. Polar coordinates often turn complex differential equations into simpler forms, which are easier to work with.

Partial Derivatives

Partial derivatives involve taking the derivative of a function with respect to one variable while treating other variables as constants. This is super useful when the function depends on multiple variables.

For example, if we have \(u = f(x, y)\), then the partial derivative of \(u\) with respect to \(x\) is noted as \(\frac{\partial u}{\partial x}\). This measures how \(u\) changes as \(x\) changes, while keeping \(y\) constant. Similarly, \(\frac{\partial u}{\partial y}\) measures the change in \(u\) as \(y\) varies with \(x\) held fixed.

In the context of Laplace’s Equation and transformation to polar coordinates, partial derivatives help us express the change in \(u\) in terms of \(r\) and \(\theta\) instead of \(x\) and \(y\). These derivatives are key to rewriting the equation in a form that is solvable when symmetry is present.

Chain Rule

The chain rule is a fundamental concept in calculus for handling derivatives of composite functions. It allows us to differentiate a function based on the derivatives of its components.

In the context of transforming Laplace's equation, we use the chain rule to connect derivatives in Cartesian coordinates to those in polar coordinates. This involves expressing derivatives like \(\frac{\partial}{\partial x}\) and \(\frac{\partial}{\partial y}\) in terms of \(\frac{\partial}{\partial r}\) and \(\frac{\partial}{\partial \theta}\).

More specifically, the chain rule tells us that to find \(\frac{\partial u}{\partial x}\), we need the directions for \( \frac{\partial r}{\partial x} \) and \( \frac{\partial \theta}{\partial x} \). By determining these, we can rewrite the equation accordingly. This application simplifies the process and is essential to converting the Laplace equation from one coordinate system to another.

Transformation of Equations

Transforming equations is all about changing the way an equation is represented. In the case of Laplace's Equation, we're moving from Cartesian to polar coordinates. This process involves expressing the differentials in terms of the new system variables.

Once you start applying transformations, especially with partial differential equations, this step often involves using the chain rule to express changes from one variable set to another. For Laplace’s equation, this transformation not only simplifies the math but often reveals new insights into the problems, such as symmetries.

By converting the equation into polar form, like in our exercise, what was once a complex system in \(x\) and \(y\) becomes much more manageable and reveals the inherent symmetries involved. This transformation leads to simpler solutions and is often crucial for solving many real-world physics and engineering problems.