Problem 15 - Advanced Engineering Mathematics

Question

We use the Fourier sine transform with respect to \(x\) to obtain $$U(\alpha, y)=c_{1} \cosh \alpha y+c_{2} \sinh \alpha y.$$ The transforms of \(u(x, 0)=f(x)\) and \(u(x, 2)=0\) give, in turn, \(U(\alpha, 0)=F(\alpha)\) and \(U(\alpha, 2)=0 .\) The first condition gives \(c_{1}=F(\alpha)\) and the second condition then yields $$c_{2}=-\frac{F(\alpha) \cosh 2 \alpha}{\sinh 2 \alpha}.$$ Hence $$\begin{aligned} U(\alpha, y) &=F(\alpha) \cosh \alpha y-\frac{F(\alpha) \cosh 2 \alpha \sinh \alpha y}{\sinh 2 \alpha} \\ &=F(\alpha) \frac{\sinh 2 \alpha \cosh \alpha y-\cosh 2 \alpha \sinh \alpha y}{\sinh 2 \alpha} \\ &=F(\alpha) \frac{\sinh \alpha(2-y)}{\sinh 2 \alpha} \end{aligned}$$ and $$u(x, y)=\frac{2}{\pi} \int_{0}^{\infty} F(\alpha) \frac{\sinh \alpha(2-y)}{\sinh 2 \alpha} \sin \alpha x d \alpha.$$

Step-by-Step Solution

Answer

The solution is derived by applying boundary conditions to find constants, simplifying with hyperbolic identities, and using inverse transform to express \(u(x, y)\).

Step 1: Understanding the Problem

We need to apply the Fourier sine transform to solve for the solution to a partial differential equation involving boundary conditions. The given conditions are transformed for the functions at specific values of \(y\). Our task is to derive the equation for \(U(\alpha, y)\) and subsequently find \(u(x, y)\).

Step 2: Analyze Boundary Conditions

The boundary conditions given are \(U(\alpha, 0)=F(\alpha)\) and \(U(\alpha, 2)=0\). These translate to the following: 1. \(c_1 = F(\alpha)\) 2. \(c_1 \cosh 2 \alpha + c_2 \sinh 2 \alpha = 0\).From these, we can directly obtain the values for \(c_1\) and solve for \(c_2\).

Step 3: Solve for Constants

From \(U(\alpha, 0)=F(\alpha)\), it follows that \(c_1 = F(\alpha)\). Using \(U(\alpha, 2)=0\), substitute \(c_1\) to get the equation for \(c_2\): \(0 = F(\alpha) \cosh 2\alpha + c_2 \sinh 2\alpha\). Solving for \(c_2\), we find \(c_2 = -\frac{F(\alpha) \cosh 2\alpha}{\sinh 2\alpha}\).

Step 4: Derive the Expression for \(U(\alpha, y)\)

Substitute \(c_1\) and \(c_2\) into the general solution for \(U(\alpha, y)\). This gives:\(U(\alpha, y) = F(\alpha) \cosh \alpha y - \frac{F(\alpha) \cosh 2 \alpha \cdot \sinh \alpha y}{\sinh 2 \alpha}\).Factor out \(F(\alpha)\) and simplify using the hyperbolic identity: \(U(\alpha, y) = F(\alpha) \cdot \frac{\sinh \alpha (2-y)}{\sinh 2\alpha}\).

Step 5: Use Inverse Fourier Transform

Finally, to determine \(u(x, y)\), inverse the Fourier sine transform. This provides:\(u(x, y) = \frac{2}{\pi} \int_{0}^{\infty} F(\alpha) \frac{\sinh \alpha (2 - y)}{\sinh 2 \alpha} \sin \alpha x \, d\alpha\). This equation gives \(u(x, y)\) as a function of \(x\) and \(y\).

Key Concepts

Partial Differential Equation

Partial differential equations (PDEs) are fundamental in mathematics and physics because they describe systems which involve rates of change in multiple variables. In this context, PDEs are used to model phenomena such as heat conduction, wave propagation, and fluid dynamics. A classic example of a PDE is the heat equation, which describes how temperature evolves over time:\[ \frac{\partial u}{\partial t} = \alpha abla^2 u \]where \(u\) is the temperature and \(\alpha\) is the thermal diffusivity.
This class of equations typically involves partial derivatives, meaning derivatives with respect to more than one variable, making their solutions more complex. Dealing with PDEs often requires numerical methods or advanced analytical techniques, like Fourier transforms, to find solutions.
Understanding PDEs requires knowledge of calculus and the ability to manipulate derivatives.

• They describe how a quantity, such as temperature or pressure, changes with respect to time and space.
• Solutions to PDEs can provide valuable information about the behavior of physical systems.

Boundary Conditions

Boundary conditions are essential when solving partial differential equations because they specify the constraints or values that the solution must satisfy on the boundaries of the domain. These conditions ensure the solution is unique and matches the physical scenario being modeled.
In the exercise provided, the boundary conditions are given as \(U(\alpha, 0) = F(\alpha)\) and \(U(\alpha, 2) = 0\). This means the function \(u(x, y)\) takes specific values at \(y=0\) and \(y=2\). Such conditions help determine the form of the solution and the constants within it.
There are common types of boundary conditions:

• Dirichlet boundary conditions: Specify values of the solution on the boundary, like \(u(x,0) = f(x)\).
• Neumann boundary conditions: Specify values of the derivative of the solution on the boundary.
• Mixed boundary conditions: Involve a combination of both values of the solution and its derivative.

These boundary conditions impact how the solution behaves and are key to transitioning from a mathematical model to a real-world application.

Hyperbolic Functions

Hyperbolic functions are similar to trigonometric functions but are based on hyperbolas rather than circles. They arise frequently in the study of PDEs, particularly in hyperbolic PDEs which model wave-like phenomena.
The key hyperbolic functions are the hyperbolic sine and cosine, \( \sinh x \) and \( \cosh x \), defined as:\[ \sinh x = \frac{e^x - e^{-x}}{2} \]\[ \cosh x = \frac{e^x + e^{-x}}{2} \]These functions have properties analogous to the sine and cosine functions, like addition formulas, and can be used to transform and solve equations.
In this problem, we use them as part of the solution to the PDE:

• \( \cosh x \) and \( \sinh x \) describe the hyperbolic shape of solutions.
• They are vital in representing solutions working under particular forms of boundary conditions.

They simplify complex expressions and enable solutions to be expressed in terms of exponential functions.

Inverse Fourier Transform

The inverse Fourier transform is a mathematical process used to retrieve a function from its frequency domain representation back to its original spatial or time domain. This is crucial in the field of signal processing and the solution of PDEs.
In our context, the inverse Fourier sine transform is used to obtain \(u(x, y)\) from \(U(\alpha, y)\), providing a complete solution across space and verifying boundary conditions were appropriately employed.
The process involves integrating over all possible frequencies, weighted by a specific kernel associated with the type of transform:\[ u(x, y) = \frac{2}{\pi} \int_{0}^{\infty} F(\alpha) \frac{\sinh \alpha (2 - y)}{\sinh 2 \alpha} \sin \alpha x \, d\alpha \]This solution gives us the desired physical interpretation and spatial behavior. It's a powerful technique because it converts problems from a differential form into an algebraic form, making them easier to handle.

• The inverse Fourier transform converts frequency domain solutions back to real space.
• It requires a good understanding of both integration techniques and transform pairs.

The inverse transform is often the final step in validating whether our approach matches reality.