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 \([0, L]\). $$ f(x)=x^{2}(L-x) $$

Short answer

Question: Write the Fourier sine series representation of the given function $$f(x) = x^2(L - x)$$ on the interval $$[0, L]$$. Answer: The Fourier sine series representation of the function $$f(x) = x^2(L - x)$$ on the interval $$[0, L]$$ is given by: $$ f(x) = \sum_{n=1}^{\infty} b_n \sin \frac{n \pi x}{L} = \sum_{n=1}^{\infty} \left[- \frac{4}{n^3 \pi^3} \int_0^L (L-2x) \sin \frac{n \pi x}{L} dx\right] \sin \frac{n \pi x}{L} $$

Step-by-step solution

Find the sine series coefficients (b_n)

To find the coefficients \(b_n\) of the Fourier sine series, we'll use Theorem 11.3.5(b) which states that $$b_n = \frac{2}{L} \int_0^L f(x) \cdot \sin \frac{n \pi x}{L} dx$$. Given the function \(f(x) = x^2(L - x)\), we can write: $$b_n = \frac{2}{L} \int_0^L x^2(L - x) \sin \frac{n \pi x}{L} dx$$We'll use integration by parts twice to solve this integral.

Integration by Parts (1st Time)

Let's use the integration by parts formula: $$\int udv = uv - \int vdu$$ First, let's choose our u and dv for first integration: $$u = x^2(L-x), dv = \sin \frac{n \pi x}{L} dx$$ Now find du and v: $$du = (2x(L - x) - x^2) dx, v = -\frac{L}{n\pi} \cos \frac{n \pi x}{L}$$ Now using the formula: $$\int u dv = uv - \int v du = -\frac{L}{n\pi} x^2(L-x) \cos \frac{n \pi x}{L} + \frac{L}{n\pi} \int (2x(L - x) - x^2) \cos \frac{n \pi x}{L} dx$$

Integration by Parts (2nd Time)

Second integration by parts, choose u and dv: $$u = (2x(L - x) - x^2), dv = -\frac{L}{n\pi} \cos \frac{n \pi x}{L} dx$$ Now find du and v: $$du = (2(L - 2x)) dx, v = -\frac{L^2}{n^2 \pi^2} \sin \frac{n \pi x}{L}$$ Now using the formula again: $$\int u dv = uv - \int v du = -\frac{L^2}{n^2 \pi^2} (2x(L - x) - x^2) \sin \frac{n \pi x}{L} - \frac{2L^3}{n^3 \pi^3} \int (L - 2x) \sin \frac{n \pi x}{L} dx$$ Substitute this back into our integral in step 2: $$b_n = \frac{2}{L} \left[-\frac{L}{n\pi} x^2(L-x) \cos \frac{n \pi x}{L} - \frac{L^2}{n^2 \pi^2} (2x(L - x) - x^2) \sin \frac{n \pi x}{L} \right]_{x = 0}^{x = L}$$ $$- \frac{4}{n^3 \pi^3} \int_0^L (L-2x) \sin \frac{n \pi x}{L} dx$$

Final Solution (Fourier sine series of f)

Now we have found all sine series coefficients. They can be expressed as $$b_n = - \frac{4}{n^3 \pi^3} \int_0^L (L-2x) \sin \frac{n \pi x}{L} dx$$ (for n > 0). Hence, the Fourier sine series representation of the function $$f(x) = x^2(L - x)$$ on the interval \([0, L]\) is given by: $$ f(x) = \sum_{n=1}^{\infty} b_n \sin \frac{n \pi x}{L} = \sum_{n=1}^{\infty} \left[- \frac{4}{n^3 \pi^3} \int_0^L (L-2x) \sin \frac{n \pi x}{L} dx\right] \sin \frac{n \pi x}{L} $$

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In context to the Fourier sine series problem, differential equations help us understand how a particular function, here described as f(x) = x^2(L - x), behaves across a certain domain. Specifically, Fourier methods, including the sine series, are utilized to solve a wide range of differential equations, especially those with periodic or boundary conditions.

Understanding the nature of differential equations is crucial because they are central to modeling the behavior of physical systems. For example, heat distribution over a rod, vibrations of a string, and many other phenomena can be described using differential equations. The Fourier sine series aids in expressing such phenomena in terms of sinusoidal functions, which are easier to analyze and solve.

Integration by Parts

Why Use Integration by Parts?

Integration by parts is a technique derived from the product rule for differentiation and is used for integrating products of functions. It is particularly handy when we are dealing with products of polynomial and trigonometric functions, as seen in the Fourier sine series exercise. The repeated use of integration by parts simplifies the problem into a form where the integral can be evaluated more easily.

When finding the coefficients b_n, the technique requires us to identify parts of our function to set as u and dv. This strategy, although sometimes involving complex algebra, leads to a situation where the integrals are solvable, allowing us to progress towards finding the coefficients for the series solution.

Boundary Value Problems

Boundary value problems (BVPs) arise in the context of differential equations where the solution is to be determined in a domain subject to certain conditions at the boundaries. This Fourier sine series exercise reflects a BVP since the function f(x) is defined on the closed interval [0, L], providing the boundaries for the problem.

Importance of Boundaries in BVPs

The roles of these boundaries are pivotal as they provide the necessary constraints that allow the unique solution of the problem. The Fourier series takes into account these boundary conditions to develop a series representation of the solution that is valid within these limits. Specifically, a sine series is employed when the function satisfies certain symmetry and boundary conditions, leading to a representation of the original function in terms of its sine components.