Question

Use Theorem \(11.3 .5(\mathbf{d})\) or where applicable, Exercise \(50(\mathbf{b}),\) to find the mixed Fourier sine series of the \(f\) on \([0, L]\). $$ f(x)=x\left(x^{2}-3 L^{2}\right) $$

Short answer

#Problem# Determine the mixed Fourier sine series of the function \(f(x)=x(x^2-3L^2)\) on the interval \([0, L]\). #Answer# The mixed Fourier sine series of the function \(f(x)=x(x^2-3L^2)\) on the interval \([0, L]\) is given by: $$ f(x) = \sum_{n=1}^{\infty} \frac{6L}{\pi^2 n^2}\left[\frac{(-1)^nL^2}{2n^2}-\frac{3L^2}{5}\right] \sin\left(\frac{n \pi x}{L}\right) $$

Step-by-step solution

Write down the formula for Fourier sine series

For a function \(f(x)\) defined on the interval \([0, L]\), its mixed Fourier sine series can be represented as: $$ f(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n \pi x}{L}\right) $$ where the sine coefficients \(b_n\) are given by: $$ b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n \pi x}{L}\right) dx $$

Compute the sine coefficients \(b_n\)

In this case, \(f(x)=x(x^2-3L^2)\). So to find the sine coefficients, we need to compute the integral: $$ b_n = \frac{2}{L} \int_{0}^{L} x(x^2-3L^2) \sin\left(\frac{n \pi x}{L}\right) dx $$ First, substitute \(u=\frac{n\pi x}{L}\), so \(x=\frac{Lu}{n\pi}\) and \(dx=\frac{L}{n\pi}du\): $$ b_n = \frac{2}{L} \int_{0}^{\frac{n\pi L}{L}} \frac{L^2u^2}{n^2\pi^2}\left(\frac{L^2u^2}{n^2\pi^2}-3L^2\right) \sin(u) \frac{L}{n\pi} du $$ Simplify the integrand by dividing out any common terms: $$ b_n = \frac{2}{L} \cdot \frac{L^3}{n^3\pi^3} \int_{0}^{n\pi} u^2\left(u^2-\frac{3n^2\pi^2}{L^2}\right) \sin(u) du $$ We want to use integration by parts to compute the integral. Let \(v=\sin(u)\) and \(dv= \cos(u)du\). Now we need to find an expression for \(u^2\left(u^2-\frac{3n^2\pi^2}{L^2}\right)\) to substitute as \(dw\): $$ u^2\left(u^2-\frac{3n^2\pi^2}{L^2}\right) \rightarrow dw $$ Solve for \(w\): $$ w = \int u^2\left(u^2-\frac{3n^2\pi^2}{L^2}\right) du = \frac{u^5}{5} - \frac{3n^2\pi^2}{L^2} \cdot \frac{u^3}{3} $$ Using integration by parts: $$ b_n = \frac{2}{L} \cdot \frac{L^3}{n^3\pi^3} \cdot \left[\left(\frac{u^5}{5} - \frac{3n^2\pi^2}{L^2} \cdot \frac{u^3}{3}\right)\left(-\cos(u)\right) - \int \left(\frac{u^5}{5} - \frac{3n^2\pi^2}{L^2}\cdot \frac{u^3}{3}\right)(-\sin(u)) du \right]_{0}^{n\pi} $$ Following through with the integration and substitution of the limits, we have: $$ b_n = \frac{6L}{\pi^2 n^2}\left[\frac{(-1)^n}{2n^2}\pi^4-\frac{3\pi^2}{5}\right] $$

Write down the final mixed Fourier sine series

Finally, we can now write down the mixed Fourier sine series of the function \(f(x)=x(x^2-3L^2)\) on the interval \([0, L]\) using the sine coefficients we calculated: $$ f(x) = \sum_{n=1}^{\infty} \frac{6L}{\pi^2 n^2}\left[\frac{(-1)^nL^2}{2n^2}-\frac{3L^2}{5}\right] \sin\left(\frac{n \pi x}{L}\right) $$ This is the mixed Fourier sine series of the given function, as required.

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are a fundamental part of mathematical modeling, which allows us to predict the behavior of physical systems over time. In the context of Fourier sine series, we often use them to express the solution to certain types of partial differential equations, particularly those with periodic boundary conditions.

For instance, the heat equation, which models the distribution of temperature in a given region over time, can often be solved by finding its Fourier series representation. The exercise provided demonstrates this application indirectly by breaking down a given function into an infinite sum of sinusoids, which can then be used to solve boundary value problems involving differential equations.

Boundary Value Problems

Boundary value problems (BVPs) are specific types of differential equations that require the solution to satisfy certain conditions at the boundaries of the domain. In the given exercise, the domain is the interval \[0, L\], and the function is seeking a representation that meets the boundary values. Fourier series play a crucial role in solving BVPs because they can represent periodic functions very effectively.

In the exercise's context, the Fourier sine series is used which is particularly suitable for functions with boundary conditions of being zero at both ends of the interval, known as Dirichlet boundary conditions. This makes the Fourier sine series an invaluable tool for physicists and engineers who are often concerned with phenomena that can be modeled using BVPs.

Integration by Parts

Integration by parts is a technique that stems from the product rule of differentiation and is used to integrate products of functions. It translates into the formula \[\int u dv = uv - \int v du\], where \(u\) and \(dv\) are differentiable functions of \(x\).

In the exercise, integration by parts is employed to calculate the coefficients \(b_n\) for the Fourier sine series, which are integrals that include the product of \(x(x^2-3L^2)\) and \(\sin\left(\frac{n \pi x}{L}\right)\). By strategically choosing \(u\) and \(dv\), the integral is transformed into a simpler form that can be integrated directly. The exercise improvement advice highlighted the importance of this choice, as it can greatly simplify the calculation and illuminate the underlying structure of the series.