Question

The exact solution of the initial value problem \(\mathbf{y}^{\prime}=\left[\begin{array}{cc}0.5 & 1 \\ 1 & 0.5\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l}1 \\\ 0\end{array}\right] \quad\) is given by \(\quad \mathbf{y}(t)=\frac{1}{2}\left[\begin{array}{c}e^{-t / 2}+e^{3 t / 2} \\\ -e^{-t / 2}+e^{3 t / 2}\end{array}\right] .\) (a) Write a program that applies the Runge-Kutta method (12) to this problem. (b) Run your program on the interval \(0 \leq t \leq 1\), using step size \(h=0.01\). (c) Run your program on the interval \(0 \leq t \leq 1\), using step size \(h=0.005\). (d) Let \(\mathbf{y}_{100}\) and \(\mathbf{y}_{200}\) denote the numerical approximations to \(\mathbf{y}(1)\) computed in parts (b) and (c), respectively. Compute the error vectors \(\mathbf{y}(1)-\mathbf{y}_{100}\) and \(\mathbf{y}(1)-\overline{\mathbf{y}}_{200}\). By roughly what fractional amount is the error reduced when the step size is halved?

Short answer

Question: Approximate the solution of the given initial value problem using Runge-Kutta method (fourth-order) with step sizes \(h = 0.01\) and \(h = 0.005\), and find the error when the step sizes are reduced. \(\mathbf{y}^{\prime}=\left[\begin{array}{cc}0.5 & 1 \\\ 1 & 0.5\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l}1\\\ 0\end{array}\right]\). The exact solution is given by: \(\mathbf{y}(t)=\frac{1}{2}\left[\begin{array}{c}e^{-t / 2}+e^{3 t / 2} \\\ -e^{-t / 2}+e^{3 t / 2}\end{array}\right] .\)

Step-by-step solution

Represent the system of differential equations and exact solution

Given the initial value problem: \(\mathbf{y}^{\prime}=\left[\begin{array}{cc}0.5 & 1 \\\ 1 & 0.5\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l}1\\\ 0\end{array}\right]\). And the exact solution is: \(\mathbf{y}(t)=\frac{1}{2}\left[\begin{array}{c}e^{-t / 2}+e^{3 t / 2} \\\ -e^{-t / 2}+e^{3 t / 2}\end{array}\right] .\)

Implement Runge-Kutta method for the given system

A generic Runge-Kutta method (fourth-order) for a system like this can be implemented using the following equations: \(k_1 = hf(t_n, \mathbf{y}_n)\) \(k_2 = hf(t_n + \frac{1}{2}h, \mathbf{y}_n + \frac{1}{2}k_1)\) \(k_3 = hf(t_n + \frac{1}{2}h, \mathbf{y}_n + \frac{1}{2}k_2)\) \(k_4 = hf(t_n + h, \mathbf{y}_n + k_3)\) \(\mathbf{y}_{n+1} = \mathbf{y}_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4)\) Here, \(f(t_n, \mathbf{y}_n)\) is given by the matrix multiplication of the matrix \(A = \left[\begin{array}{cc}0.5 & 1 \\\ 1 & 0.5\end{array}\right]\) and the vector \(\mathbf{y}_n\). Use this to apply the Runge-Kutta method for the given system with step sizes \(h = 0.01\) and \(h = 0.005\) on the interval \(0 \leq t \leq 1\).

Calculate the error and compare the fractional reduction

Compute the error vectors for the approximations obtained with the two different step sizes: 1. \(\mathbf{y}(1)-\mathbf{y}_{100}\) using step size \(h = 0.01\) 2. \(\mathbf{y}(1)-\overline{\mathbf{y}}_{200}\) using step size \(h = 0.005\) Compare these errors and calculate the fractional reduction in error when the step size is reduced by half.

Key concepts

Numerical Approximation

Numerical approximation is an essential concept in computational mathematics, particularly when finding an analytical solution is difficult or impossible. For this specific problem, we use the Runge-Kutta method, a popular technique for approximating solutions to ordinary differential equations (ODEs). The goal is to approximate the solution of the system of equations over the interval \( 0 \leq t \leq 1 \) using small, constant step sizes.Understanding the fourth-order Runge-Kutta method is crucial here. It improves accuracy by considering the rate of change of the function at several points within each step. More specifically:

• \( k_1 \) represents the slope at the start of the interval.
• \( k_2 \) and \( k_3 \) are slopes at the midpoint of the step, calculated using \( k_1 \).
• \( k_4 \) gives the slope at the end point of the step.

Combining these weighted slopes provides a better approximation for the interval. This method is iterative and calculates approximate values step by step, producing a numerical solution that approximates the real problem solution.The choice of step size \( h \) significantly impacts the accuracy of the result. A smaller step size generally leads to a more accurate approximation but increases the computational cost. Therefore, choosing an appropriate step size is vital for efficient numerical approximation.

Initial Value Problem

An initial value problem (IVP) involves solving a differential equation subject to specific initial conditions. In this exercise, the system of equations is a first-order Ordinary Differential Equation (ODE) expressed as \( \mathbf{y}' = A \mathbf{y} \), where \( A \) is a 2x2 constant matrix, and the initial condition is \( \mathbf{y}(0) = \left[ \begin{array}{c} 1 \ 0 \end{array} \right] \).The role of the initial condition is crucial as it uniquely determines the solution to the differential equation. These problems model a wide range of real-world systems where the state of the system is known at a starting point in time.To solve the initial value problem using the Runge-Kutta method, we begin the computation from the initial condition and iterate it over the desired time interval. This step-by-step process, using the specified method, provides a sequence of approximate values that define the solution curve starting from the initial point. Understanding the initial condition allows us to apply numerical methods efficiently and appropriately to provide critical insights into system behavior.

Error Analysis

Error analysis in numerical methods helps us understand the accuracy of our computational solutions compared to exact solutions. For this Runge-Kutta method, errors occur due to the discretization of the time variable, truncation, and round-off errors.The Runge-Kutta method used here is known for minimizing truncation errors due to its higher order. In each iterative step, by considering multiple slopes, we gain a more accurate result, thus reducing the approximation error. Nonetheless, smaller step sizes lead to improved accuracy because the local truncation error proportionally decreases.In this exercise:

• \( \mathbf{y}_{100} \) is computed with a step size of \( h = 0.01 \).
• \( \mathbf{y}_{200} \) is computed with a smaller step size of \( h = 0.005 \).

We compare these approximations to the exact solution at \( t = 1 \) by calculating the error vectors:

• \( \mathbf{y}(1) - \mathbf{y}_{100} \)
• \( \mathbf{y}(1) - \mathbf{y}_{200} \)

Typically, halving the step size results in a significant reduction in error. In this particular implementation, we expect the error to decrease roughly by a factor of sixteen (since Runge-Kutta is a fourth-order method). Thus, proper error analysis is vital, as it informs the efficiency and reliability of the numerical approximations.