Problem 4 - Applied Partial Differential Equations

Question

Suppose that \(\cosh x \sim \sum_{n=1}^{\infty} b_{n} \sin n \pi x / L\). (a) Determine \(b_{n}\) by correctly differentiating this series twice. (b) Determine \(b\), by integrating this series twice.

Step-by-Step Solution

Answer

After differentiating the series twice in part (a), the coefficients \(b_n\) are found to be \(- (cosh(L) -1) / (pi^2n^2)\). In part (b) by integrating the series twice, the coefficients \(b_n\) are \(- (L^2 (1-cosh(L))/(n^2 \pi^2))\).

Step 1: Differentiating the series twice for part (a)

The series can be differentiated term by term, so \((\sin(n\pi x/L))'' = - (n\pi /L)^2 \sin(n\pi x/L)\). Then the entire series becomes \( -\sum (n\pi /L)^2 b_n \sin(n\pi x/L)\) after differentiating twice. But \(\cosh(x)'' = \cosh(x)\), thus equating the two gives \(b_n = - (cosh(L) -1) / (pi^2n^2)\).

Step 2: Integrating the series twice for part (b)

According to the rules of integration, the indefinite integral of \(\sin(n\pi x/L)\) with respect to \(x\) is \(-(L/n\pi) \cos(n\pi x/L)\), and the second integral gives \(-(L^2/n^2\pi^2) \sin(n\pi x/L)\). Thus, \(b\) can be evaluated by comparing this result to the double integral of \(\cosh(x)\), which is \(\sinh(x) + C\), where \(C\) is the constant of integration. Solving \(b_{n}=- (L^2 (1-cosh(L))/(n^2 \pi^2))\), where \(C\) has been set to zero (which is valid because it represents an arbitrary constant and can be absorbed into the \(\sinh(x)\) term).

Step 3: Summary

To summarise, we first find the derivative(s) of our given function, then compare it to the derivative(s) of our arbitrary function. Then, by integrating it we find the second set of coefficients. This process makes use of fundamental calculus concepts such as differentiation, integration, and series representations of functions.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It involves finding the derivative of a function, which measures how the function value changes as its input changes. In this exercise, we applied differentiation to a series representing a function. Differentiating the series twice involves taking the derivative of each term twice.

When you see a term like \( \sin(n\pi x/L) \), its second derivative is \( -(n\pi /L)^2 \sin(n\pi x/L) \). This is because the derivative of \( \sin \) is \( \cos \), and differentiating \( \cos \) returns us to \( \sin \), introducing a factor of \(-1\) each time. Term by term differentiation of the series reveals the changes imposed by these factors.

• First, differentiate each term to get \( (n\pi /L) \cos(n\pi x/L) \).
• Second, differentiate again to introduce \(-(n\pi /L)^2 \sin(n\pi x/L) \).

Through this process, we equate it with the double derivative of \( \cosh(x) \), since \( \cosh(x)'' = \cosh(x) \), allowing us to solve for the coefficients \( b_n \). This involves balancing each differentiated term to match the target function's properties.

Integration

Integration is the inverse process of differentiation. It's used to find areas under curves or to recover a function from its derivative. In this exercise, we integrate a trigonometric series representation of a function twice. Integrating \( \sin(n\pi x/L) \) gives a term involving \( \cos \) functions.

Indefinite integration of \( \sin(n\pi x/L) \) first gives \(-(L/n\pi) \cos(n\pi x/L) \). When integrated a second time, this results in \(-(L^2/n^2\pi^2) \sin(n\pi x/L) \).

• Each integration reduces the power of \( n \) in the coefficients because of the division by \( n \pi \).
• The integration constants can be ignored or set to zero in specific contexts, like matching function forms.

This process shows how integrating affects the series components, and allows the reconstruction of the coefficients by comparing with \( \sinh(x) + C \), the double integral of \( \cosh(x) \). By comparing the series' form after integration, we derive expressions for the coefficients related to \( b \).

Trigonometric Functions

Trigonometric functions like \( \sin \) and \( \cos \) are fundamental in both calculus and the Fourier series. They describe periodic phenomena and allow complex waveforms to be expressed as sums of simple sine and cosine waves. In this exercise, they form the building blocks of our series representation.

Each coefficient in a Fourier series multiplies a sine or cosine term. These functions have specific differentiation and integration properties used to manipulate and transform the series. For example:

• \( \sin(x) \) becomes \( \cos(x) \) when differentiated.
• Integrating \( \sin(x) \) results in \(-\cos(x)\) up to a constant.

By carefully applying these rules, you can transform a series for \( \cosh(x) \) into a sum of sine terms. Understanding these transformations is key to generating the expressions needed for Fourier coefficients, translating smooth function behavior into periodic representations. Trigonometric identities and properties play a critical role in modeling and problem-solving within this framework.