Problem 18 - Advanced Engineering Mathematics

Question

We solve the three boundary-value problems: Using separation of variables we find the solution of the first problem is $$u_{1}(x, y)=\sum_{n=1}^{\infty} A_{n} e^{-n y} \sin n x \quad \text { where } \quad A_{n}=\frac{2}{\pi} \int_{0}^{\pi} f(x) \sin n x d x.$$ Using the Fourier sine transform with respect to \(y\) gives the solution of the second problem: $$u_{2}(x, y)=\frac{200}{\pi} \int_{0}^{\infty} \frac{(1-\cos \alpha) \sinh \alpha(\pi-x)}{\alpha \sinh \alpha \pi} \sin \alpha y d \alpha.$$ Also, the Fourier sine transform with respect to \(y\) gives the solution of the third problem: $$u_{3}(x, y)=\frac{2}{\pi} \int_{0}^{\infty} \frac{\alpha \sinh \alpha x}{\left(1+\alpha^{2}\right) \sinh \alpha \pi} \sin \alpha y d \alpha.$$ The solution of the original problem is $$u(x, y)=u_{1}(x, y)+u_{2}(x, y)+u_{3}(x, y).$$

Step-by-Step Solution

Answer

The solution is the sum: \( u(x, y) = u_{1}(x, y) + u_{2}(x, y) + u_{3}(x, y) \).

Step 1: Understanding the Problem

The problem involves three boundary-value problems whose solutions are provided using different methods, namely separation of variables and Fourier sine transforms. The solutions are given in the form of infinite series or integral expressions.

Step 2: Examine Solution of First Problem

For the first boundary-value problem, the solution is given by:\[ u_{1}(x, y)=\sum_{n=1}^{\infty} A_{n} e^{-n y} \sin n x \]where:\[ A_{n}=\frac{2}{\pi} \int_{0}^{\pi} f(x) \sin n x \, dx. \]This implies using separation of variables to obtain a series solution involving a sine function and exponential decay in terms of \(y\).

Step 3: Examine Solution of Second Problem

The solution for the second problem is obtained using the Fourier sine transform:\[ u_{2}(x, y)=\frac{200}{\pi} \int_{0}^{\infty} \frac{(1-\cos \alpha) \sinh \alpha(\pi-x)}{\alpha \sinh \alpha \pi} \sin \alpha y \, d \alpha \]This integrative form means solving the boundary-value problem by transforming into sine components with respect to \(y\), focusing on the influences along the \(x\)-direction.

Step 4: Examine Solution of Third Problem

Similarly for the third problem using the Fourier sine transform:\[ u_{3}(x, y)=\frac{2}{\pi} \int_{0}^{\infty} \frac{\alpha \sinh \alpha x}{(1+\alpha^{2}) \sinh \alpha \pi} \sin \alpha y \, d \alpha \]This provides another Fourier sine representation where \(\sinh \alpha x\) introduces a hyperbolic aspect on how \(x\) influences \(y\).

Step 5: Combine the Solutions

The general solution to the original problem is the sum of all the obtained solutions:\[ u(x, y) = u_{1}(x, y) + u_{2}(x, y) + u_{3}(x, y) \]This means superposing the effects captured by each separate analysis for boundary conditions and domains.

Key Concepts

Separation of Variables

Separation of Variables is a powerful mathematical technique used to solve partial differential equations (PDEs). The main idea is to express a complex PDE in terms of simpler, one-variable problems that can be solved more easily. The process involves assuming that the solution can be written as a product of functions, each depending on one of the independent variables. For instance, if our variables are \(x\) and \(y\), we would propose a solution of the form \(u(x, y) = X(x)Y(y)\). This proposed form allows us to separate the variables such that each side of the resulting equation depends on only one of the variables.

• The solution for each independent function usually stems from solving ordinary differential equations.
• The boundary conditions play a crucial role in determining the specific form of the solution functions.
• This method proves particularly useful for problems defined over rectangular domains.

In the provided exercise, separation of variables enabled expressing the first problem's solution as an infinite series, reflecting the behavior along both \(x\) and \(y\) dimensions succinctly.

Fourier Sine Transform

The Fourier Sine Transform is a mathematical transformation used to solve boundary value problems, particularly for functions defined on semi-infinite domains or over intervals. Specifically, it's effective when boundary conditions involve zero-function values at one end of the interval. The transform converts a function of a spatial variable into a function of a frequency variable, effectively decomposing it into its sine components. This transformation simplifies solving differential equations by working in the frequency domain rather than the spatial domain.

• It facilitates solving problems involving odd symmetry, attributed to the sine function's properties.
• Particularly useful for heat equations and wave equations where you deal with semi-infinite domains.
• Reconstruction of the original function from its sine transform enables checking the correctness of solutions.

In the given exercise, the Fourier Sine Transform allowed us to obtain solutions for both the second and third problems by revealing the underlying sine structure of the problems.

Infinite Series

An infinite series is a sum of an infinite sequence of numbers or functions. In mathematical problems, these series often represent solutions to complex differential equations, especially in boundary value problems. The series approach allows for simplifying infinite processes into manageable calculations by approximations using a finite number of terms, or the series' partial sums.

• Infinite series have convergence criteria, ensuring that their sum approaches a specific finite number.
• They provide a way to express solutions that exhibit repetitive or periodic behavior, as seen in wave functions.
• The Fourier series, a specific type of infinite series, transforms complex functions into sine and cosine terms, playing an essential role in problems with periodicity.

In the context of this exercise, the first boundary-value problem's solution is represented as an infinite series combining exponential and sine functions, thanks to separation of variables.

Integral Expressions

Integral expressions are fundamental in calculus, capturing the idea of accumulation. In boundary value problems, they appear frequently in solutions that involve transforms like the Fourier Sine Transform, where they express the sum of infinitely small contributions, aligned along a curve or over an area.

• Integral expressions can describe the entire behavior of a function across an interval.
• In solving PDEs, they provide exact solutions to simplified transforms of the original equation.
• The integral bounds and integrand are determined based on initial/boundary conditions of the problem.

In the exercise, integral expressions were used as part of the solution to the second and third boundary-value problems, demonstrating how the effects along \(x\) and \(y\) are captured through these accumulations.