Problem 16 - 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 \(\quad \frac{\partial^{2} u}{\partial t^{2}}=\sum_{n=1}^{\infty} u_{n}^{\prime \prime}(t) \sin n x\). Then\\[u_{t t}-u_{x x}=\sum_{n=1}^{\infty}\left[u_{n}^{\prime \prime}(t)+n^{2} u_{n}(t)\right] \sin n x=\cos t \sin x\\]. Equating coefficients, we obtain \(u_{1}^{\prime \prime}(t)+u_{1}(t) \cos t\) and \(u_{n}^{\prime \prime}(t)+n^{2} u_{n}(t)=0\) for \(n=2,3,4, \ldots .\) Solving the first differential equation we obtain \(u_{1}(t)=A_{1} \cos t+B_{1} \sin t+\frac{1}{2} t \sin t .\) From the second differential equation we obtain \(u_{n}(t)=A_{n} \cos n t+B_{n} \sin n t\) for \(n=2,3,4, \ldots .\) Thus \(u(x, t)=\left(A_{1} \cos t+B_{1} \sin t+\frac{1}{2} t \sin t\right) \sin x+\sum_{n=2}^{\infty}\left(A_{n} \cos n t+B_{n} \sin n t\right) \sin n x\). From \(u(x, 0)=A_{1} \sin x+\sum_{n=2}^{\infty} A_{n} \sin n x=0\) we see that \(A_{n}=0\) for \(n=1,2,3, \ldots .\) Thus \(u(x, t)=\left(B_{1} \sin t+\frac{1}{2} t \sin t\right) \sin x+\sum_{n=2}^{\infty} B_{n} \sin n t \sin n x\) and \(\frac{\partial u}{\partial t}=\left(B_{1} \cos t+\frac{1}{2} t \cos t+\frac{1}{2} \sin t\right) \sin x+\sum_{n=2}^{\infty} n B_{n} \cos n t \sin n x\), so \(\left.\frac{\partial u}{\partial t}\right|_{t=0}=B_{1} \sin x+\sum_{n=2}^{\infty} n B_{n} \sin n x=0\). We see that \(B_{n}=0\) for all \(n\) so \(u(x, t)=\frac{1}{2} t \sin t \sin x\).

Step-by-Step Solution

Answer

The solution is \( u(x,t) = \frac{1}{2} t \sin(t) \sin(x) \).

Step 1: Identify the Problem

The problem involves solving a second-order partial differential equation given by \(X'' + \lambda X = 0\) with boundary conditions \(X(0) = 0\) and \(X(\pi) = 0\). Here, \(k=1\) and \(L=\pi\), which implies the eigenfunctions are \(\sin(nx)\) for \(n = 1, 2, 3, \ldots\).

Step 2: Setup of u(x,t)

We assume a solution in the form \(u(x, t) = \sum_{n=1}^{\infty} u_{n}(t) \sin(nx)\). We'll compute the partial derivatives \(\frac{\partial^{2} u}{\partial x^{2}}\) and \(\frac{\partial^{2} u}{\partial t^{2}}\).

Step 3: Calculate Partial Derivatives

By taking the partial derivatives, we get \(\frac{\partial^{2} u}{\partial x^{2}} = \sum_{n=1}^{\infty} u_{n}(t)(-n^{2}) \sin(nx)\) and \(\frac{\partial^{2} u}{\partial t^{2}} = \sum_{n=1}^{\infty} u_{n}''(t) \sin(nx)\).

Step 4: Formulate the Differential Equation

Substitute the partial derivatives into the equation \(u_{tt} - u_{xx} = \cos(t) \sin(x)\), which results in \(\sum_{n=1}^{\infty}[u_{n}''(t) + n^{2} u_{n}(t)] \sin(nx) = \cos(t) \sin(x)\).

Step 5: Equate Coefficients

Since the equation must hold for all \(x\), equate coefficients for each \(\sin(nx)\). We obtain two equations: \(u_{1}''(t) + u_{1}(t) = \cos(t)\) for \(n=1\) and \(u_{n}''(t) + n^{2} u_{n}(t) = 0\) for \(n=2, 3, 4, \ldots\).

Step 6: Solve First Differential Equation

Solve \(u_{1}''(t) + u_{1}(t) = \cos(t)\), obtaining \(u_{1}(t) = A_{1} \cos(t) + B_{1} \sin(t) + \frac{1}{2} t \sin(t)\).

Step 7: Solve Remaining Equations

Solve \(u_{n}''(t) + n^{2} u_{n}(t) = 0\) for \(n=2, 3, 4, \ldots\), giving \(u_{n}(t) = A_{n} \cos(nt) + B_{n} \sin(nt)\).

Step 8: Apply Initial Conditions

Given \(u(x, 0) = 0\), it follows \(A_{n} = 0\) for all \(n\), and from \(\frac{\partial u}{\partial t}|_{t=0} = 0\), find \(B_{n} = 0\) for all \(n\).

Step 9: Substitute and Simplify

Substitute the constants into the expression for \(u(x,t)\), yielding the simplified solution \(u(x,t) = \frac{1}{2} t \sin(t) \sin(x)\).

Key Concepts

Eigenfunctions

In solving partial differential equations (PDEs), identifying eigenfunctions is a crucial step. They are fundamental solutions used to construct a more complex solution tailored to the problem's unique conditions. An eigenfunction is a function that satisfies a differential equation, usually along with some boundary conditions. For the equation given, * the differential equation is \(X^{\prime \prime} + \lambda X = 0\), * with boundary conditions \(X(0) = 0\) and \(X(\pi) = 0\).From the exercise, it is determined that the eigenfunctions for \(k = 1\) and \(L = \pi\) are \(\sin(nx)\). The functions \(\sin(nx)\) satisfy both the differential equation and the given boundary conditions for \(n = 1, 2, 3, \ldots\). This means that these eigenfunctions will be the building blocks for the overall solution of the original PDE.

Boundary Conditions

Boundary conditions define the behavior of a solution at specific points along the boundary of the domain. They ensure that the solution is not just any arbitrary function, but one that matches prescribed conditions at these boundaries. For the problem at hand: * The equation \(X(0) = 0\) specifies that the function must be zero at \(x = 0\).* Similarly, \(X(\pi) = 0\) imposes that the solution is also zero at \(x = \pi\).These conditions allow us to reject any solutions that do not meet them, narrowing down the possible solutions to those that satisfy these requirements. In mechanistic terms, the boundary conditions ensure that the solution behaves properly at the borders of the system, fitting the physical interpretation one might expect from such a problem.

Series Solutions

Series solutions are a powerful method to solve differential equations, especially when we seek to express a function as an infinite sum of simpler trigonometric functions. In the problem, the solution takes the form: \[ u(x, t) = \sum_{n=1}^{\infty} u_{n}(t) \sin(nx) \]This approach uses a series of sine functions, each multiplied by a time-dependent coefficient \(u_n(t)\), representing the amplitude corresponding to each mode \(\sin(nx)\). By assuming a series solution, we can handle oscillations and wave-like phenomena dynamically, making it suitable for solving the partial differential equations like wave or heat equations. Each term in the series can independently satisfy the boundary conditions, and the infinite sum allows tailoring the solution to fit the specific initial conditions given.

Initial Conditions

Initial conditions define how the solution starts, often necessary to determine unique solutions amongst many possible ones. For this exercise:* \( u(x, 0) = 0 \) ensures the solution begins in a specific state at \(t = 0\).* This condition leads to the conclusion that \( A_n = 0 \) for all \( n\), as each term of the series solution must be zero at this initial time.Likewise, when we apply \( \frac{\partial u}{\partial t}|_{t=0} = 0 \), we restrict the rate of change at the beginning, which further refines the solution by finding that \( B_n = 0 \) too, ensuring the movement starts from rest. These conditions help lock down the constants in the series solution, ensuring the behavior of \(u(x, t)\) aligns with the scenario's physical constraints right from the outset.