Problem 24 - Advanced Engineering Mathematics

Question

Verify that the given function satisfies Laplace's equation: $$\frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}=0$$ $$ z=e^{x^{2}-y^{2}} \cos 2 x y $$

Step-by-Step Solution

Answer

The function satisfies Laplace's equation.

Step 1: Compute the First Partial Derivatives

To verify Laplace's equation, we first need to compute the first partial derivatives of \( z = e^{x^2 - y^2} \cos(2xy) \). Start by computing \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).- Let \( u = x^2 - y^2 \) and \( v = 2xy \) so that \( z = e^u \cos v \).- The first derivative with respect to \( x \) is \( \frac{\partial z}{\partial x} = \frac{\partial}{\partial x} (e^u \cos v) = e^u \cdot (-\sin v \cdot \frac{\partial v}{\partial x}) + \cos v \cdot \frac{\partial}{\partial x}(e^u) \) where \( \frac{\partial v}{\partial x} = 2y \) and \( \frac{\partial u}{\partial x} = 2x \).- The first derivative with respect to \( y \) is \( \frac{\partial z}{\partial y} = \frac{\partial}{\partial y} (e^u \cos v) = e^u \cdot (-\sin v \cdot \frac{\partial v}{\partial y}) + \cos v \cdot \frac{\partial}{\partial y}(e^u) \) where \( \frac{\partial v}{\partial y} = 2x \) and \( \frac{\partial u}{\partial y} = -2y \).

Step 2: Compute the Second Partial Derivatives

Now compute the second partial derivatives: \( \frac{\partial^2 z}{\partial x^2} \) and \( \frac{\partial^2 z}{\partial y^2} \) using the results from Step 1.- Differentiate \( \frac{\partial z}{\partial x} \) with respect to \( x \) to get \( \frac{\partial^2 z}{\partial x^2} \).- Differentiate \( \frac{\partial z}{\partial y} \) with respect to \( y \) to get \( \frac{\partial^2 z}{\partial y^2} \).

Step 3: Substitute and Verify Laplace's Equation

Substitute the second partial derivatives back into Laplace's equation:\[ \frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 0 \].- If the sum of these derivatives equals zero, then the function satisfies Laplace's equation.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, especially when dealing with functions of multiple variables. When you have a function like \( z = e^{x^2 - y^2} \cos(2xy) \), you need to understand how it changes as each variable changes independently. This is where partial derivatives come in.
First, identify each variable you want to differentiate with respect to. Here, we are dealing with the function in terms of \( x \) and \( y \). The first partial derivative with respect to \( x \), denoted as \( \frac{\partial z}{\partial x} \), shows how \( z \) changes as \( x \) changes, keeping \( y \) constant. Similarly, \( \frac{\partial z}{\partial y} \) is computed by differentiating \( z \) with respect to \( y \,\), while \( x \) is held constant.

• Set \( u = x^2 - y^2 \) and \( v = 2xy \). Express \( z \) in terms of \( u \) and \( v \) to simplify differentiation.
• Use the product rule and chain rule for differentiation to find the first partial derivatives.
• Differentiating further, the second partial derivatives \( \frac{\partial^2 z}{\partial x^2} \) and \( \frac{\partial^2 z}{\partial y^2} \) are found by differentiating the first derivatives again with respect to the same variable.

These calculations help verify whether the function satisfies certain partial differential equations, like Laplace's equation.

Advanced Engineering Mathematics

Advanced engineering mathematics involves complex mathematical concepts and methods that are essential in various engineering fields. Laplace’s equation is one example of these concepts.
This equation appears in many physical problems, such as heat conduction, fluid flow, and electric fields, describing equilibrium states. In practical terms, advanced engineering mathematics helps in:

• Modeling real-world systems with differential equations to predict how they behave under different conditions.
• Utilizing techniques such as Laplace transforms, Fourier analysis, and partial differential equations to solve problems efficiently.
• Understanding fundamental principles, which can then be applied to design and optimize engineering solutions. For example, verifying that a potential function satisfies Laplace’s equation can confirm that it’s valid in describing physical phenomena.

An engineering student must be adept at using mathematical tools to not only solve specific textbook problems but also to tackle challenges in real-world applications.

Laplace's differential operator

Laplace's differential operator, often denoted as \( \Delta \), plays a crucial role in mathematical physics and engineering. This operator is used to find regions where a function can be continuous and smooth, which implies equilibrium in physical processes.
When applied to a function \( f(x, y) \), Laplace’s operator is expressed as:
\[ \Delta f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}. \]
Laplace's equation, \( \Delta z = 0 \), suggests that the function \( z \) exhibits balance across all points. This condition is vital for describing phenomena where a steady state exists, such as:

• Heat distribution in a stationary metal plate.
• Potential fields in electrostatics, where charges are in equilibrium.
• The pressure field in a fluid at rest.

For mathematicians and engineers, the ability to verify that a function satisfies this equation by calculating partial derivatives and ensuring their sum is zero is a critical skill for analyzing systems where perfect balance or equilibrium is essential.