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A function \(f\) of \(x\) and \(y\) is harmonic if $$ \frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}=0 $$ throughout the domain of \(f\). Prove that the given function is harmonic. $$ f(x, y)=e^{-x} \cos y+e^{-y} \cos x $$

Short Answer

Expert verified
The function is harmonic because the sum of second partial derivatives is zero.

Step by step solution

01

Find the First Partial Derivative with Respect to x

Calculate \(\frac{\partial f}{\partial x}\) for \(f(x, y)=e^{-x} \cos y+e^{-y} \cos x\). \begin{align*} \frac{\partial f}{\partial x} & = \frac{\partial}{\partial x}(e^{-x} \cos y) + \frac{\partial}{\partial x}(e^{-y} \cos x)\ & = -e^{-x} \cos y - e^{-y} \sin x. \end{align*}
02

Find the Second Partial Derivative with Respect to x

Differentiate \(\frac{\partial f}{\partial x}\) with respect to \(x\) again. \begin{align*} \frac{\partial^2 f}{\partial x^2} & = \frac{\partial}{\partial x}(-e^{-x} \cos y - e^{-y} \sin x)\ & = e^{-x} \cos y - e^{-y} \cos x. \end{align*}
03

Find the First Partial Derivative with Respect to y

Calculate \(\frac{\partial f}{\partial y}\) for \(f(x, y)=e^{-x} \cos y+e^{-y} \cos x\). \begin{align*} \frac{\partial f}{\partial y} & = \frac{\partial}{\partial y}(e^{-x} \cos y) + \frac{\partial}{\partial y}(e^{-y} \cos x) \ & = -e^{-x} \sin y - e^{-y} \sin x. \end{align*}
04

Find the Second Partial Derivative with Respect to y

Differentiate \(\frac{\partial f}{\partial y}\) with respect to \(y\) again. \begin{align*} \frac{\partial^2 f}{\partial y^2} & = \frac{\partial}{\partial y}(-e^{-x} \sin y - e^{-y} \sin x)\ & = -e^{-x} \cos y + e^{-y} \cos x. \end{align*}
05

Check if Function is Harmonic

Sum the second partial derivatives to check the harmonic condition. \[ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = (e^{-x} \cos y - e^{-y} \cos x) + (-e^{-x} \cos y + e^{-y} \cos x) = 0. \] The sum is zero, thus \(f(x, y)\) is a harmonic function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
When dealing with functions of several variables like \( f(x, y) = e^{-x} \cos y + e^{-y} \cos x \), it's critical to understand the concept of partial derivatives. Partial derivatives measure how the function changes as one of the variables changes, while keeping the other variable constant.

Think of it like this: if you have a surface represented by the function, partial derivatives describe the slope of the surface in a specific direction (either along the x-axis or the y-axis).

To find a partial derivative with respect to a variable, you differentiate the function as if the other variable is a constant. For example, when computing \( \frac{\partial f}{\partial x} \), you treat \( y \) as a constant and differentiate with respect to \( x \).

  • First Partial Derivative with respect to \( x \): \( \frac{\partial f}{\partial x} = -e^{-x} \cos y - e^{-y} \sin x \)
  • Second Partial Derivative with respect to \( x \): \( \frac{\partial^2 f}{\partial x^2} = e^{-x} \cos y - e^{-y} \cos x \)
These derivatives help in analyzing and solving functions that involve multiple variables.
Functions of Several Variables
Functions like \( f(x, y) = e^{-x} \cos y + e^{-y} \cos x \) are called functions of several variables because they depend on more than one variable, in this case, both \( x \) and \( y \). Unlike single-variable functions, these offer a richer framework for modeling and problem-solving in various fields such as physics, engineering, and economics.

Understanding these functions involves delving into the interactions between the various variables. They are typically visualized as surfaces or planes in a 3D space, which offers a graphical perspective to comprehend the rates of change along each direction.

For instance, consider how the function changes as both \( x \) and \( y \) vary simultaneously:
  • The function might increase with \( x \) while decreasing with \( y \).
  • Grasping these simultaneous changes is key to solving complex problems where multiple factors interact.
This makes understanding their derivatives crucial, both for understanding the function itself and for applications such as optimizing processes or determining critical points.
Laplace's Equation
Laplace's equation is a significant concept in the study of harmonic functions. A function satisfies this equation if \[\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0 \]

Laplace's equation is fundamental in various areas, including physics and engineering, where it is used to model steady-state heat distribution, electrostatic potential, and fluid flow.

In the given problem, the function \( f(x, y) = e^{-x} \cos y + e^{-y} \cos x \) was tested against this equation.

  • By calculating the second partial derivatives:\( \frac{\partial^2 f}{\partial x^2} = e^{-x} \cos y - e^{-y} \cos x \)
    and \( \frac{\partial^2 f}{\partial y^2} = -e^{-x} \cos y + e^{-y} \cos x \),we see how they interact.
  • When summed, they satisfy the equation: \( \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0 \).
This confirms that the function \( f(x, y) \) is indeed harmonic, showing the practical application of Laplace's equation in verifying harmonic status.

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Most popular questions from this chapter

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