Chapter 12: Problem 21
Represent \(f(x, y)\) \((0
Short Answer
Expert verified
The function is represented by a double Fourier series with coefficients found through integration.
Step by step solution
01
Set Up Double Fourier Series
We need to express the function \( f(x, y) = x y (a^2 - x^2) (b^2 - y^2) \) as a double Fourier series. The general form of a double Fourier series is:\[f(x, y) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} C_{mn} \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right)\]Our task is to find the coefficients \( C_{mn} \).
02
Determine Coefficients Formula
The coefficients \( C_{mn} \) are given by:\[C_{mn} = \frac{4}{ab} \int_0^a \int_0^b f(x, y) \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) \, dy \, dx\]This formula comes from projecting the function onto the basis of sines in the appropriate region.
03
Evaluate Inner Integral Over y
Substitute \( f(x, y) \) into the coefficient formula and perform the integration over \( y \):\[\int_0^b y (b^2 - y^2) \sin\left(\frac{n\pi y}{b}\right) \, dy\]This integral can be solved using standard techniques for integration, often involving parts and trigonometric identities.
04
Evaluate Outer Integral Over x
After solving the integral over \( y \), substitute back and evaluate the next integral over \( x \):\[\int_0^a x (a^2 - x^2) \text{Result from Step 3} \sin\left(\frac{m\pi x}{a}\right) \, dx\]Again, this integral typically requires methods such as integration by parts to simplify.
05
Express Full Fourier Series
After determining the coefficients \( C_{mn} \) by solving both integrals, insert them back into the double Fourier series expression:\[f(x, y) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} C_{mn} \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right)\]This completes the representation of the function as a double Fourier series.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Double Fourier Series
A double Fourier series is a way to represent a function of two variables as an infinite sum of sine functions. This concept extends the idea of a single-variable Fourier series to functions depending on two parameters, often denoted as \( x \) and \( y \). In this context, each term in the series takes the form of a product of sine functions: \( \sin \left( \frac{m\pi x}{a} \right) \sin \left( \frac{n\pi y}{b} \right) \). Here, \( m \) and \( n \) are positive integers representing the harmonic frequencies along the \( x \) and \( y \) directions respectively. The main idea is to express the target function \( f(x, y) \) as a superposition of these basis functions. The double Fourier series for \( f(x, y) \) is given by the expression:
- \( \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} C_{mn} \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) \)
Integration Techniques in Double Fourier Series
The calculation of coefficients \( C_{mn} \) in a double Fourier series relies heavily on integration techniques. For each coefficient, the process involves evaluating a double integral. The given formula:
First, we tackle the integral over \( y \):
- \( C_{mn} = \frac{4}{ab} \int_0^a \int_0^b f(x, y) \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) \, dy \, dx \)
First, we tackle the integral over \( y \):
- \( \int_0^b y(b^2-y^2) \sin\left(\frac{n\pi y}{b}\right) \, dy \)
- \( \int_0^a x(a^2-x^2) \cdot \text{Result from Step 3} \sin\left(\frac{m\pi x}{a}\right) \, dx \)
Coefficients Calculation for Double Fourier Series
The coefficients \( C_{mn} \) play a critical role in constructing the double Fourier series. These coefficients are calculated by integrating the product of the function \( f(x, y) \) with the sine basis functions. Each \( C_{mn} \) is found using:
- \( C_{mn} = \frac{4}{ab} \int_0^a \int_0^b f(x, y) \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) \, dy \, dx \)
- Substituting \( f(x, y) \) into the integral.
- Performing the integration over \( y \) first to simplify the expression.
- Using the result to evaluate the integral over \( x \).