Problem 13 - Applied Partial Differential Equations

Question

Consider a function \(f(x)\) that is even around \(x=L / 2\). Show that the even coefficients ( \(n\) even) of the Fourier sine series of \(f(x)\) on \(0 \leqslant x \leqslant L\) are zero.

Step-by-Step Solution

Answer

The even coefficients of the Fourier sine series are zero. This can be shown by noting that for even numbered coefficients, the integral in the coefficient formula simplifies to zero due to the property of even functions, hence making the coefficients themselves equal to zero.

Step 1: Definition of General Coefficients

Remember that the \(n\)th coefficient \(b_{n}\) for the Fourier sine series can be calculated using the equation \[b_{n} = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n \pi x}{L}\right) \, dx\]. Here \(f(x)\) is the given function, \(\sin\left(\frac{n \pi x}{L}\right)\) is the basis function (Fourier basis), \(n\) is the order of the coefficient, and \(L\) is the length of the period.

Step 2: Apply Properties of Even Functions

Since \(f(x)\) is even around \(x=L/2\), this means that \(f(L-x) = f(x)\) for all \(x\) in the interval [0, L]. Therefore, the integral in the coefficient formula can be rewritten by substitution \(u=L-x\). The integral then becomes \[\int_{0}^{L} f(x) \sin\left(\frac{n \pi x}{L}\right) \, dx = \int_{L}^{0} f(L-u) \sin\left(\frac{n \pi (L-u)}{L}\right) \, du\] and flipping the limits we get \[-\int_{0}^{L} f(L-u) \sin\left(\frac{n \pi (L-u)}{L}\right) \, du\]. Combining the two integrals, we get \[\int_{0}^{L} f(x) \sin\left(\frac{n \pi x}{L}\right) \, dx - \int_{0}^{L} f(L-x) \sin\left(\frac{n \pi x}{L}\right) \, dx = 0\].

Step 3: Apply Property of Even Function to Integral

Replacing \(f(L-x)\) by \(f(x)\) because of the even function property, this equation simplifies to \[\int_{0}^{L} f(x) \left(\sin\left(\frac{n \pi x}{L}\right) - \sin\left(\frac{n \pi x}{L}\right) \right) \, dx = 0\]. This simplifies to an integral of zero, which is zero.

Step 4: Coefficient Calculation

Finally, substituting this result back in the coefficient formula, we find the coefficient \(b_{n}\) (for \(n\) even) will be zero, because it is multiplied by zero (the result of our integral). Thus, all even-numbered coefficients of the Fourier sine series for \(f(x)\) are zero. This completes the proof.

Key Concepts

Even Function

An even function is symmetrical around a central point. Specifically, for a function to be even, it must satisfy the condition where the function value at negative x is the same as that at positive x: given as \(f(-x) = f(x)\) for all x in the domain. In our exercise, the function is even around \(x = L/2\), adjusting this symmetry point.
- Even functions mirror themselves around this midpoint.- This property greatly simplifies calculations in Fourier analysis, where symmetry can lead to certain coefficients being zero.
When we encounter an even function, we immediately think about how this symmetry impacts integral calculations and Fourier series, often leading to simplifications that highlight the zero coefficients in sine series.

Fourier Coefficients

Fourier coefficients are constants calculated to approximate a function using trigonometric functions, specifically sine and cosine. In a Fourier sine series, we focus on the sine part:
- The coefficient \(b_n\) is given by the integral \( b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n \pi x}{L}\right) \, dx \).- Here, \(b_n\) reflects the magnitude and phase of a sine wave at the harmonic frequency \( \frac{n \pi}{L}\).
Each coefficient captures a component of the original function, allowing it to be reconstructed by summing all components. With even functions, certain coefficients, particularly the ones aligned with odd harmonics when translated, might simplify or cancel, as observed in this exercise.

Integral Calculation

The calculation of integrals is crucial in determining Fourier coefficients. Let’s delve into the integration step involved in our task:
- We utilize symmetry properties by substituting \(u = L-x\) which transforms the limits and variable of the integral: \(\int_{0}^{L} f(x) \sin\left(\frac{n \pi x}{L}\right) \, dx\) becomes a new integral setup.- Changing variables highlights the symmetry, modifying the integral essentially into its negative form when re-evaluated over the same limits.
The symmetry, as seen with our even function about \(x = L/2\), allows us to simplify the integrals involved. Two identical terms of opposite signs result in a combined integral of zero. This ingenuity shows where calculus and symmetrical properties of functions serve as powerful tools for simplification.

Properties of Even Functions

Properties of even functions play a significant role in simplifying our exercise solution. These properties tell us that:
- Even functions \(f\) satisfy \(f(x) = f(-x)\).- For functions even about different axes or points, this property can adjust to match the symmetry around that particular axis or point.
Such properties mean that certain integrals involve terms canceling each other out, resulting naturally in zeroes. These computations directly impact Fourier series, especially when deriving Fourier coefficients for sine terms, such as the zeroing of coefficients for even \(n\). This integral simplification due to symmetry is a cornerstone of using even functions, making complex calculations feasible with algebraic ease.