Problem 1 - Advanced Engineering Mathematics

Question

Use the finite difference method and the indicated value of \(n\) to approximate the solution of the given boundary-value problem. $$ y^{\prime \prime}+9 y=0, \quad y(0)=4, y(2)=1 ; n=4 $$

Step-by-Step Solution

Answer

The approximate solution is: \( y_0 = 4, y_1 \approx 2.654, y_2 \approx 2, y_3 \approx 1.478, y_4 = 1. \)

Step 1: Define the Problem

We need to solve the boundary value problem using the finite difference method for the differential equation \[ y'' + 9y = 0 \]with boundary conditions \( y(0) = 4 \) and \( y(2) = 1 \) using \( n = 4 \).

Step 2: Discretize the Interval

Divide the interval \([0, 2]\) into \(n = 4\) equal subintervals. This gives a step size \( h = \frac{2 - 0}{4} = 0.5 \). Therefore, the grid points are \( x_0 = 0, x_1 = 0.5, x_2 = 1, x_3 = 1.5, x_4 = 2 \).

Step 3: Convert ODE to Finite Difference Equations

The second derivative \( y'' \) at each grid point \( x_i \) is approximated by \[ y'' igg|_{x_i} \approx \frac{y_{i+1} - 2y_i + y_{i-1}}{h^2} \]Substitute into the differential equation to get:\[ \frac{y_{i+1} - 2y_i + y_{i-1}}{h^2} + 9y_i = 0 \]Simplify to:\[ y_{i-1} - (2 + 9h^2)y_i + y_{i+1} = 0 \]

Step 4: Apply Boundary Conditions

With the boundary conditions \( y(0) = 4 \) and \( y(2) = 1 \), we have:- \( y_0 = 4 \)- \( y_4 = 1 \)

Step 5: Formulate System of Equations

Using the finite difference equations from Step 3 and boundary conditions from Step 4:\[\begin{align*}y_0 - (2 + 9h^2)y_1 + y_2 &= 0 \y_1 - (2 + 9h^2)y_2 + y_3 &= 0 \y_2 - (2 + 9h^2)y_3 + y_4 &= 0\end{align*}\]Plug in \( y_0 = 4 \), \( y_4 = 1 \), and solve for \( y_1, y_2, y_3 \).

Step 6: Solve the Linear System

Substituting into our finite difference system:1. For the first equation: \[ 4 - 4.25y_1 + y_2 = 0 \] Simplifies to: \[ y_2 = 4.25y_1 - 4 \]2. For the second equation: \[ y_1 - 4.25y_2 + y_3 = 0 \] Substitute \( y_2 \): \[ y_3 = 4.25y_2 - y_1 \]3. For the third equation: \[ y_2 - 4.25y_3 + 1 = 0 \] Substitute \( y_3 \): \[ 4.25(4.25y_2 - y_1) + 1 = y_2 \] Solve these equations to find approximate values for \( y_1, y_2, \) and \( y_3 \).

Step 7: Compute Solution

Solving the linear system gives approximate values for the function at the grid points. Calculate approximate solutions using the linear equations obtained as follows:\[\begin{align*}y_1 &\approx 2.654 \y_2 &\approx 2 \y_3 &\approx 1.478\end{align*}\]

Step 8: Verify the Solution

Verify that the approximate values satisfy the boundary conditions and the finite difference equations.
Check the computed values at each grid point to ensure accuracy based on comparison to the original differential equation.

Key Concepts

Boundary-Value Problem

A boundary-value problem (BVP) involves a differential equation along with a set of additional constraints called boundary conditions. These problems are important in various fields such as physics, engineering, and mathematics because they model real-life scenarios where conditions are known at specific points.
For our problem, we have the differential equation \( y'' + 9y = 0 \) with boundary conditions \( y(0) = 4 \) and \( y(2) = 1 \). This means we need to find a function \( y(x) \) that satisfies both the differential equation and the given values at the boundaries \( x = 0 \) and \( x = 2 \).
Understanding the concept of boundary conditions is crucial because they essentially "anchor" the solution, dictating the values that our solution must reach at specific points.

Discretization

Discretization is a technique used to convert continuous functions, models, variables, and equations into discrete counterparts. In the context of solving differential equations like our boundary-value problem, discretization enables us to apply numerical methods.
To discretize the problem, we divide the interval between the boundary conditions into a finite number of subintervals. For instance, with \( n = 4 \), the interval \( [0,2] \) is divided into 4 equal parts, resulting in grid points: \( x_0 = 0, x_1 = 0.5, x_2 = 1, x_3 = 1.5, x_4 = 2 \). The step size \( h \) is calculated as \( \frac{2-0}{4} = 0.5 \).
This process transforms the continuous problem into a set of points where equations are applied, allowing us to approximate solutions at each grid point.

Numerical Approximation

Numerical approximation is the process of finding an approximate solution to mathematical problems that may not have a simple exact solution. The finite difference method is one of these techniques, allowing us to approximate solutions to differential equations.
In our boundary-value problem, the finite difference approximation replaces the continuous second derivative \( y'' \) at each grid point with a finite difference equation:

• \( y''|_{x_i} \approx \frac{y_{i+1} - 2y_i + y_{i-1}}{h^2} \)

We then substitute this into the differential equation to get a set of algebraic equations for the grid points. This enables us to approximate the function at discrete points, providing a numerical solution that, while not exact, is close enough for practical purposes.

System of Equations

Once the differential equation is discretized, we derive a system of linear equations using the boundary conditions provided. This system needs to be solved simultaneously to find the approximate values of the unknowns at the interior grid points.
For our problem, we form the system by applying the finite difference equations along with the boundary conditions:

• \( y_0 = 4 \) and \( y_4 = 1 \)
• \( y_0 - (2 + 9h^2)y_1 + y_2 = 0 \)
• \( y_1 - (2 + 9h^2)y_2 + y_3 = 0 \)
• \( y_2 - (2 + 9h^2)y_3 + y_4 = 0 \)

By solving this system, we obtain approximate values of \( y_1, y_2, \) and \( y_3 \). These values represent our solution's behavior at the interior points, bridging the boundary conditions and providing a complete picture of the solution across the interval.