Problem 13 - Advanced Engineering Mathematics

Question

Identifying \(k=1\) and \(L=\pi\) we see that the eigenfunctions of \(X^{\prime \prime}+\lambda X=0, X(0)=0, X(\pi)=0\) are \(\sin n x\), \(n=1,2,3, \ldots .\) Assuming that \(u(x, t)=\sum_{n=1}^{\infty} u_{n}(t) \sin n x,\) the formal partial derivatives of \(u\) are \(\frac{\partial^{2} u}{\partial x^{2}}=\sum_{n=1}^{\infty} u_{n}(t)\left(-n^{2}\right) \sin n x \quad\) and \(\quad \frac{\partial u}{\partial t}=\sum_{n=1}^{\infty} u_{n}^{\prime}(t) \sin n x\). Assuming that \(x e^{-3 t}=\sum_{n=1}^{\infty} F_{n}(t) \sin n x\) we have \(F_{n}(t)=\frac{2}{\pi} \int_{0}^{\pi} x e^{-3 t} \sin n x d x=\frac{2 e^{-3 t}}{\pi} \int_{0}^{\pi} x \sin n x d x=\frac{2 e^{-3 t}(-1)^{n+1}}{n}\). Then \(x e^{-3 t}=\sum_{n=1}^{\infty} \frac{2 e^{-3 t}(-1)^{n+1}}{n} \sin n x\) and \(u_{t}-u_{x x}=\sum_{n=1}^{\infty}\left[u_{n}^{\prime}(t)+n^{2} u_{n}(t)\right] \sin n x=x e^{-3 t}=\sum_{n=1}^{\infty} \frac{2 e^{-3 t}(-1)^{n+1}}{n} \sin n x\). Equating coefficients we obtain \(u_{n}^{\prime}(t)+n^{2} u_{n}(t)=\frac{2 e^{-3 t}(-1)^{n+1}}{n}\). This is a linear first-order differential equation whose solution is \(u_{n}(t)=\frac{2(-1)^{n+1}}{n\left(n^{2}-3\right)} e^{-3 t}+C_{n} e^{-n^{2} t}\). Thus \(u(x, t)=\sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n\left(n^{2}-3\right)} e^{-3 t} \sin n x+\sum_{n=1}^{\infty} C_{n} e^{-n^{2} t} \sin n x\) and \(u(x, 0)=0\) implies \(\sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n\left(n^{2}-3\right)} \sin n x+\sum_{n=1}^{\infty} C_{n} \sin n x=0\) so that \(C_{n}=2(-1)^{n} / n\left(n^{2}-3\right) .\) Therefore \(u(x, t)=2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n\left(n^{2}-3\right)} e^{-3 t} \sin n x+2 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n\left(n^{2}-3\right)} e^{-n^{2} t} \sin n x\).

Step-by-Step Solution

Answer

The solution is \(u(x, t)=2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n^{2}-3)} e^{-3 t} \sin n x + 2 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n(n^{2}-3)} e^{-n^{2} t} \sin n x\).

Step 1: Understanding the Problem

We have a differential equation \(X'' + \lambda X = 0\) with boundary conditions \(X(0) = 0\) and \(X(\pi) = 0\). We identify eigenfunctions \(\sin nx\) where \(n = 1, 2, 3, \ldots\). The function \(u(x,t)\) is represented as a series solution using these eigenfunctions.

Step 2: Expressing Partial Derivatives

We express \(u(x,t)\) using the series \(u(x, t)=\sum_{n=1}^{\infty} u_{n}(t) \sin n x\). The partial derivatives are \(\frac{\partial^{2} u}{\partial x^{2}}=\sum_{n=1}^{\infty} u_{n}(t)(-n^{2}) \sin n x\) and \(\frac{\partial u}{\partial t}=\sum_{n=1}^{\infty} u_{n}^{\prime}(t) \sin n x\).

Step 3: Identifying Coefficients of Series

Given \(x e^{-3 t}=\sum_{n=1}^{\infty} F_{n}(t) \sin n x\), the coefficients \(F_n(t)\) are found as \(\frac{2 e^{-3 t}(-1)^{n+1}}{n}\). This gives the series \(x e^{-3 t}=\sum_{n=1}^{\infty} \frac{2 e^{-3 t}(-1)^{n+1}}{n} \sin n x\).

Step 4: Setting Up the Differential Equation

By equating coefficients from \(u_t - u_{xx} = x e^{-3t}\), we have \(u_{n}^{\prime}(t)+n^{2} u_{n}(t)=\frac{2 e^{-3 t}(-1)^{n+1}}{n}\). Each \(u_n\) satisfies a first-order linear differential equation.

Step 5: Solving the Differential Equation

We solve the equation \(u_{n}^{\prime}(t)+n^{2} u_{n}(t)=\frac{2 e^{-3 t}(-1)^{n+1}}{n}\) with solution \(u_{n}(t)=\frac{2(-1)^{n+1}}{n(n^{2}-3)} e^{-3 t}+C_{n} e^{-n^{2} t}\). This integrates the homogeneous and particular solution components.

Step 6: Applying Initial Conditions

With \(u(x,0) = 0\), we force initial conditions on the series giving \(\sum_{n=1}^{\infty} \frac{2(-1)^{n+1}}{n\left(n^{2}-3\right)} \sin n x + \sum_{n=1}^{\infty} C_{n} \sin n x = 0\), leading to \(C_{n}=2(-1)^n / n(n^{2}-3)\).

Step 7: Constructing Final Solution

Substitute \(C_n\) into the expression for \(u_n(t)\), resulting in \(u(x, t)=2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n^{2}-3)} e^{-3 t} \sin n x+2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n(n^{2}-3)} e^{-n^{2} t} \sin n x\), combining both the terms from the particular and homogeneous solutions.

Key Concepts

Eigenfunctions

In mathematics, eigenfunctions are a set of functions associated with a differential equation that satisfies specific conditions. When you have a boundary value problem, like the one given in the exercise, eigenfunctions help you solve the equation by forming an orthogonal basis for the function space involved.
For our problem, we see that the differential equation is given as \(X'' + \lambda X = 0\), with boundary conditions \(X(0) = 0\) and \(X(\pi) = 0\). These boundary conditions are crucial in defining the eigenfunctions. Here, the solutions are of the form \(\sin nx\), where \(n = 1, 2, 3, \ldots\).
These solutions are periodic and satisfy both the differential equation and the boundary conditions. When we plug in these solutions, they help us break down complex problems into simpler ones, making them easier to solve.

Series Solution

A series solution refers to expressing a function as the sum of an infinite series of functions. In this exercise, we expressed the function \(u(x, t)\) as an infinite sum of eigenfunctions multiplied by time-dependent coefficients, \(u_n(t)\):

• The series representation is \(u(x, t) = \sum_{n=1}^{\infty} u_{n}(t) \sin nx\).
• Breaking down into series simplifies calculation and analysis by transforming a complex problem into manageable parts.

Such representation is powerful because it allows us to efficiently handle each term separately. Each term in this series is influenced by the eigenfunctions \(\sin nx\), which neatly fit our boundary conditions. By solving for the series coefficients \(u_n(t)\), the entire function \(u\) can be determined, effectively providing a comprehensive solution to the dynamic problem at hand.

Boundary Conditions

Boundary conditions are constraints necessary to solve differential equations that describe physical phenomena. They define how solutions behave at the boundaries, or edges, of the domain we are working in.
In our exercise, we have \(X(0) = 0\) and \(X(\pi) = 0\). These constraints limit the permissible form of solutions, dictating that they must cancel out to zero at these particular points.

• They ensure the solution's uniqueness and stability.
• They make the mathematical model correspond to the physical setting or phenomena being described.

The use of boundary conditions is what guides us to use eigenfunctions like \(\sin nx\) because they naturally satisfy these stated conditions, providing a practical and meaningful solution to the problem.

First-Order Linear Differential Equation

A first-order linear differential equation is a type of equation that involves derivatives and linear terms. These equations are significant as they are relatively easier to handle and solve than higher-order ones.
In our example, the derived equation \(u_n'(t) + n^2u_n(t) = \frac{2e^{-3t}(-1)^{n+1}}{n}\) is a classic example of such an equation. It can be solved using well-known methods designed for first-order linear differential equations.

• The general solution involves finding a particular solution and a homogeneous solution.
• These make it possible to describe how the solution behaves under given initial or boundary conditions.
• Particular solutions account for non-homogeneous parts (forcing terms) while homogeneous solutions reflect the natural response of the system.

By applying the appropriate techniques, like integration factors or substitution, one can solve these equations smoothly and tailor the solution to specific initial conditions.