Question

Use Theorem \(11.3 .5(\mathbf{b})\) or where applicable, Exercise \(11.1 .35(\mathbf{b}),\) to find the Fourier sine series of \(f\) on \([0, L]\). $$ f(x)=x\left(x^{3}-2 L x^{2}+L^{3}\right) $$

Short answer

The Fourier sine series representation is given by: \(S_f(x) = \sum_{n=1}^{\infty} \frac{2(-1)^n}{n^2\pi^2}L^3 \sin\left(\frac{n\pi x}{L}\right)\).

Step-by-step solution

Determine the Fourier sine series representation

First, we need to determine the Fourier sine series representation. A Fourier sine series of a function \(f(x)\) is given by $$ S_f(x) = \sum_{n=1}^{\infty} b_n \sin \left(\frac{n\pi x}{L}\right), $$ where $$ b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin \left(\frac{n\pi x}{L}\right) dx. $$ Our goal is to find the coefficients \(b_n\).

Calculate the coefficients \(b_n\)

To find the coefficients, we will integrate the product of the function \(f(x)\) and the sine term from \(0\) to \(L\) and multiply by \(\frac{2}{L}\): $$ b_n = \frac{2}{L} \int_{0}^{L} x\left(x^3 - 2Lx^2 + L^3\right) \sin \left(\frac{n\pi x}{L}\right) dx. $$

Evaluate the integral

To simplify, we use integration by parts, focusing on the term inside the integral: $$ u = x^4 - 2Lx^3 + L^3x, \quad dv = \sin \left(\frac{n\pi x}{L}\right) dx, $$ hence $$ du = (4x^3 - 6Lx^2 + L^3) dx, \quad v = -\frac{L}{n\pi} \cos \left(\frac{n\pi x}{L}\right). $$ Now, we apply integration by parts formula: $$ \int udv = uv - \int vdu, $$ which gives \begin{align*} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx &= \left[ -\frac{Lx}{n\pi} (x^4 - 2Lx^3 + L^3x)\cos \left(\frac{n\pi x}{L}\right) \right]_{0}^{L} \\ &\quad + \frac{L}{n\pi} \int_{0}^{L} (4x^3 - 6Lx^2 + L^3) \cos\left(\frac{n\pi x}{L}\right) dx. \end{align*}

Evaluate the definite integral

Notice that at \(x = 0\) and \(x = L\), the cosine term is always non-zero, while the polynomial term becomes zero. Hence, the first term becomes zero. Let's now focus on the second term and compute the integral: $$ \int_{0}^{L} (4x^3 - 6Lx^2 + L^3) \cos\left(\frac{n\pi x}{L}\right) dx. $$ We apply integration by parts again, and use \(u = 4x^3 - 6Lx^2 + L^3\), \(dv = \cos\left(\frac{n\pi x}{L}\right)dx\). We find $$ du = (12x^2 - 12Lx) dx \quad \text{and} \quad v = \frac{L}{n\pi} \sin\left(\frac{n\pi x}{L}\right). $$ Applying the integration by parts formula gives \begin{align*} &\int_{0}^{L} (4x^3 - 6Lx^2 + L^3) \cos\left(\frac{n\pi x}{L}\right) dx \\ &= \left[ \frac{L}{n\pi}(4x^3 - 6Lx^2 + L^3) \sin\left(\frac{n\pi x}{L}\right) \right]_{0}^{L}\\ &\quad - \frac{L}{n\pi} \int_{0}^{L} (12x^2 - 12Lx) \sin\left(\frac{n\pi x}{L}\right) dx. \end{align*} Once again, at \(x = 0\) and \(x = L\), the sine term is always zero, hence the first term becomes zero. We are now left with evaluating the integral: $$ \int_{0}^{L} (12x^2 - 12Lx) \sin\left(\frac{n\pi x}{L}\right) dx. $$ We break this integral into two parts and solve them separately: $$ \frac{-L^2}{n^2\pi^2} \int_{0}^{L} (12x^2 - 12Lx) \frac{d}{dx}\cos\left(\frac{n\pi x}{L}\right) dx = \frac{-L^2}{n^2\pi^2}\left [ (12L^2 -12L^2) \cos\left(\frac{n\pi}{1}\right) - (0) \right]. $$ Since cos(nπ) = (-1)^n, we have: $$ b_n = \frac{2}{L} \times \frac{-L^2}{n^2\pi^2}(0 - (-1)^nL^2) = \frac{2(-1)^n}{n^2\pi^2}L^3. $$

Write the Fourier sine series of f(x)

Now we can write the Fourier sine series of \(f(x)\) using the coefficients \(b_n\): $$ S_f(x) = \sum_{n=1}^{\infty} \frac{2(-1)^n}{n^2\pi^2}L^3 \sin\left(\frac{n\pi x}{L}\right). $$ This is the Fourier sine series of the given function \(f(x) = x(x^3 - 2Lx^2 + L^3)\) on the interval \([0, L]\).

Key concepts

Fourier Coefficients

Fourier coefficients are the building blocks that allow us to express a function using a sine or cosine series. To find these coefficients, particularly in a Fourier sine series, involves an integration process. This is essential for decomposing a function into sine component parts that can approximate the function across a certain interval.

In the context of the Fourier sine series, the coefficients are calculated using the integral:

• \( b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin \left(\frac{n\pi x}{L}\right) dx \)

Here, \( b_n \) are the Fourier sine coefficients for the function \( f(x) \). This integration gives the "weight" of each sine function within the series. Each coefficient captures how much the corresponding sine function contributes to the overall shape of \( f(x) \) in the interval \([0, L]\).

The goal when calculating Fourier coefficients is to accurately represent periodic functions using trigonometric series, which leads to numerous applications in engineering and physics, including signal processing and vibration analysis.

Integration by Parts

Integration by parts is a powerful technique used to simplify the integration process, especially when dealing with products of functions. Here, it plays a crucial role in solving the integral needed to find the Fourier coefficients.

The integration by parts formula is given by:

• \( \int u \, dv = uv - \int v \, du \)

Choosing which function to differentiate (\( u \)) and which to integrate (\( dv \)) is critical for simplifying the integral. In the context of the solution, we selected functions such as \( u = x^4 - 2Lx^3 + L^3x \) and \( dv = \sin \left(\frac{n\pi x}{L}\right) dx \), then calculated \( du \) and \( v \).

This technique was iteratively applied to break down the complex polynomial multiplied by a sine function into more manageable integrals. By systematically reducing the order of polynomial integration using this method, you track each step's contribution more clearly to the resulting series expansion. Integration by parts is especially useful in scenarios where integrals are not straightforward, which is typically the case in mathematical and engineering problems.

Boundary Value Problems

Boundary value problems are vital in understanding how complex systems behave under specified conditions at the boundaries of the domain. In the context of Fourier series, they help define how functions should behave when evaluated at specific points, such as the boundaries of a given interval.

When constructing a Fourier sine series, such as in our exercise, the function is extended such that its values at the boundaries are zero, i.e., it satisfies Dirichlet boundary conditions.

Boundary value conditions are essential in physics and engineering applications. By ensuring the conditions hold at the interval's edges, we can model phenomena like heat distribution, wave propagation, or even quantum states, where constraints at boundaries dictate possible solutions.

In essence, handling boundary value problems correctly ensures that the Fourier series used accurately models the real-world situation being studied. It emphasizes the relevance of mathematical abstraction in solving practical engineering and scientific problems.