Problem 10 - Advanced Engineering Mathematics

Question

Using the Fourier sine transform we obtain $$U(\alpha, t)=c_{1} \cos \alpha a t+c_{2} \sin \alpha a t.$$ Now $$\mathscr{F}_{S}\\{u(x, 0)\\}=\mathscr{F}_{S}\left\\{x e^{-x}\right\\}=\int_{0}^{\infty} x e^{-x} \sin \alpha x d x=\frac{2 \alpha}{\left(1+\alpha^{2}\right)^{2}}=U(\alpha, 0).$$ Also, $$\mathscr{F}_{S}\left\\{u_{t}(x, 0)\right\\}=\left.\frac{d U}{d t}\right|_{t=0}=0.$$ This last condition gives \(c_{2}=0 .\) Then \(U(\alpha, 0)=2 \alpha /\left(1+\alpha^{2}\right)^{2}\) yields \(c_{1}=2 \alpha /\left(1+\alpha^{2}\right)^{2}\). Therefore $$U(\alpha, t)=\frac{2 \alpha}{\left(1+\alpha^{2}\right)^{2}} \cos \alpha a t$$ and $$u(x, t)=\frac{4}{\pi} \int_{0}^{\infty} \frac{\alpha \cos \alpha a t}{\left(1+\alpha^{2}\right)^{2}} \sin \alpha x d \alpha.$$

Step-by-Step Solution

Answer

The solution is derived by finding \( U(\alpha, t) \) and using the inverse Fourier sine transform to find \( u(x, t) \).

Step 1: Understanding the Fourier Sine Transform

The Fourier Sine Transform is used to convert a time-space function into its frequency domain representation using sine functions. In this case, the Fourier sine transform of \( u(x, 0) = x e^{-x} \) is given as \( \frac{2 \alpha}{(1+\alpha^2)^2} \). We are given that this represents \( U(\alpha, 0) \).

Step 2: Determine \( U(\alpha, t) \)

We have \( U(\alpha, t) = c_1 \cos(\alpha a t) + c_2 \sin(\alpha a t) \). Since the condition \( \mathscr{F}_S\{u_t(x,0)\} = 0 \) and \( \left.\frac{dU}{dt}\right|_{t=0} = 0 \), this implies that \( c_2 = 0 \). Thus, \( U(\alpha, t) = c_1 \cos(\alpha a t) \).

Step 3: Find \( c_1 \)

Using the initial condition \( U(\alpha, 0) = \frac{2 \alpha}{(1+\alpha^2)^2} \) and the expression \( U(\alpha, 0) = c_1 \), it follows that \( c_1 = \frac{2 \alpha}{(1+\alpha^2)^2} \). Substituting this back, we have \( U(\alpha, t) = \frac{2 \alpha}{(1+\alpha^2)^2} \cos(\alpha a t) \).

Step 4: Find \( u(x, t) \) via Inverse Transform

The inverse Fourier sine transform is used to convert \( U(\alpha, t) \) back to \( u(x, t) \). The formula to do this involves integrating over \( \alpha \):\[u(x, t) = \frac{4}{\pi} \int_0^\infty \frac{\alpha \cos(\alpha a t)}{(1+\alpha^2)^2} \sin(\alpha x) \, d\alpha\]

Key Concepts

Frequency Domain Representation

The frequency domain representation of a function is a way to express that function in terms of its frequency components rather than its original time or spatial values. In the context of the Fourier sine transform, we shift from observing a function over the spatial dimension to analyzing its behavior across different frequencies. This approach is particularly useful in solving partial differential equations, where understanding how different frequencies contribute to the solution can simplify the problem-solving process.

In this exercise, the Fourier sine transform was applied to the function \( u(x, 0) = xe^{-x} \). The transform gives us \( U(\alpha, 0) = \frac{2 \alpha}{(1+\alpha^2)^2} \), which is the frequency domain representation. This means that every frequency \( \alpha \) contributes to the overall shape of \( u(x, 0) \) with a proportional weight defined by this expression.

Here's what happens in this representation:

• Frequency components \( \alpha \) are highlighted, showing how each impacts the initial wave function.
• Understanding contributions at each frequency helps in reverting back to the time-space function through inverse transforms.

Initial Condition

Initial conditions are essential in solving differential equations as they provide specific information about the system at the starting point. They ensure the uniqueness of the solution by reducing the infinite possibilities to a single one that fits the given conditions. In our task, the initial condition provided was \( U(\alpha, 0) = \frac{2 \alpha}{(1+\alpha^2)^2} \). This means that at time \( t = 0 \), the frequency representation of the function is predetermined, guiding us in constructing the solution over time.

By relating this initial condition to the expression for \( U(\alpha, t) \), we determined the value of the constant \( c_1 \), which plays a critical role in defining how the solution evolves over time. It is the key ingredient that ties together the frequency domain representation with dynamic behavior. Without the initial condition:

• We would lack information on how the system behaves at \( t = 0 \).
• It would be impossible to determine the constants in the representation \( U(\alpha, t) \).

Inverse Fourier Transform

The inverse Fourier transform is the mathematical process that takes us back from the frequency domain to the original time or space domain. It reassembles the function from its frequency components, effectively turning data expressed in terms of frequency back into its original form. This is crucial for interpreting results in terms of the physical systems we study, whether these are mechanical vibrations, sound waves, or other phenomena.

For this problem, we used the inverse Fourier sine transform to find \( u(x, t) \) from \( U(\alpha, t) \). The inverse transform is expressed by the integral\[u(x, t) = \frac{4}{\pi} \int_0^\infty \frac{\alpha \cos(\alpha a t)}{(1+\alpha^2)^2} \sin(\alpha x) \, d\alpha\]

• This process integrates over all frequency components \( \alpha \), weighted and shaped by the function \( \sin(\alpha x) \), to recreate the original spatial domain function.
• The specific integration limits and formulas depend on the type of Fourier transform (sine, cosine, etc.) used initially.

Partial Differential Equations

Partial differential equations (PDEs) are mathematical equations that involve rates of change with respect to continuous variables. They are used to formulate problems involving functions of several variables and are either solved by analytical means (exact solutions) or by numerical methods. PDEs appear in many branches of physics and engineering, describing a wide range of phenomena, such as sound, heat, electrostatics, and fluid dynamics.

In this exercise, the Fourier sine transform is used because it simplifies the problem by converting the PDE into an ordinary differential equation (ODE) in the frequency domain. Solving the resulting ODE is typically more straightforward.

Key points about PDEs relevant here include:

• The form of the PDE can determine the applicability of different transform methods.
• Transform methods like the Fourier sine transform help in handling complex boundary and initial conditions.
• They are crucial for understanding how solutions evolve with time and space, given the specific conditions and constraints of the system.