Question

Consider the initial-value problem $$ \begin{aligned} y^{\prime} &=1+x+2 y, \\ y(0) &=1 \end{aligned} $$ with \(h=0.1\). Using the Runge-Kutta method for starting values and predictor- corrector formulae, find the solution by the Milne method at \(x=0.4\) and \(x=0.5 .\)

Short answer

Question: Solve the given initial value problem using the Runge-Kutta method and the Milne method to determine the approximate values of the function y(x) at x = 0.4 and x = 0.5. Given: $$ \begin{aligned} y^{\prime} &= 1 + x + 2y, \\ y(0) &= 1. \end{aligned} $$ Solution: Using the Runge-Kutta method to obtain the first three steps and the Milne method with Predictor-Corrector formulae, we find the following approximate values: $$ y(0.4) \approx 1.5593062 \\ y(0.5) \approx 1.7045664 $$

Step-by-step solution

Given initial-value problem functions

We are given the initial value problem with the following functions: $$ \begin{aligned} y^{\prime} &= 1 + x + 2y, \\ y(0) &= 1. \end{aligned} $$

Apply Runge-Kutta method to find starting values

We will use the Runge-Kutta method of order 4 with step size \(h = 0.1\) to obtain the first three steps in the solution: 1. Calculate \(k_1, k_2, k_3, k_4\) for each step, based on the following formulas: $$ \begin{aligned} k_1 &= h\cdot f(x_n, y_n) \\ k_2 &= h\cdot f(x_n + \frac{h}{2}, y_n + \frac{k_1}{2}) \\ k_3 &= h\cdot f(x_n + \frac{h}{2}, y_n + \frac{k_2}{2}) \\ k_4 &= h\cdot f(x_n + h, y_n + k_3) \\ \end{aligned} $$ 2. Update the solution with: $$ y_{n+1} = y_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) $$ First three steps with Runge-Kutta method: - \(y_1\): \(k_1=0.1(1+0+2\cdot1)=0.3\), \(k_2=0.1(1+0.05+2\cdot1.15)=0.335\), \(k_3=0.1(1+0.05+2\cdot1.1675)=0.3365\), \(k_4=0.1(1+0.1+2\cdot1.16917)=0.33883\). Then, \(y_1 = 1+\frac{1}{6}(0.3+2\cdot0.335+2\cdot0.3365+0.33883)=1.1169267\). - \(y_2\): \(k_1=0.1(1+0.1+2\cdot1.1169267)=0.3438534\), \(k_2=0.1(1+0.15+2\cdot1.2397585)=0.3822652\), \(k_3=0.1(1+0.15+2\cdot1.2417246)=0.3829656\), \(k_4=0.1(1+0.2+2\cdot1.2436772)=0.3840685\). Then, \(y_2 = 1.1169267+\frac{1}{6}(0.3438534+2\cdot0.3822652+2\cdot0.3829656+0.3840685)=1.2428455\). - \(y_3\): \(k_1=0.1(1+0.2+2\cdot1.2428455)=0.386691\), \(k_2=0.1(1+0.25+2\cdot1.397689)=0.4296778\), \(k_3=0.1(1+0.25+2\cdot1.3995839)=0.4300419\), \(k_4=0.1(1+0.3+2\cdot1.401506)=0.4301811\). Then, \(y_3 = 1.2428455+\frac{1}{6}(0.3866910+2\cdot0.4296778+2\cdot0.4300419+0.4301811)=1.3945637\).

Apply the Milne method with Predictor-Corrector formulae

Having obtained the starting values \(y_1, y_2, y_3\) using the Runge-Kutta method, we will apply the Milne method to find the solution at \(x=0.4\) and \(x=0.5\) using the Predictor-Corrector formulae: 1. Predictor formula: $$ P_{n+3} = y_{n-1} + \frac{4h}{3}(2f(x_n, y_n)-f(x_{n-1}, y_{n-1})+2f(x_{n-2}, y_{n-2})) $$ 2. Corrector formula: $$ C_{n+3} = y_{n-1} + \frac{h}{3}(f(x_{n}, y_{n})+4f(x_{n-1}, y_{n-1})+f(x_{n-2}, y_{n-2})) $$ - \(x=0.4\): Using Predictor formula, \(P_4 = 1.1169267+\frac{4(0.1)}{3}(2(1+0.3+2\cdot1.2428455)-(1+0.2+2\cdot1.1169267)+2(1+0.1+2\cdot1))=1.5666995\). Then, using Corrector formula, \(C_4 = 1.1169267+\frac{1(0.1)}{3}(1+0.5+2\cdot1.5666995+4(1+0.3+2\cdot1.2428455)+(1+0.2+2\cdot1.1169267))=1.5593062\). - \(x=0.5\): Using Predictor formula, \(P_5 = 1.2428455+\frac{4(0.1)}{3}(2(1+0.4+2\cdot1.3945637)-(1+0.3+2\cdot1.2428455)+2(1+0.2+2\cdot1.1169267))=1.7117522\). Then, using Corrector formula, \(C_5 = 1.2428455+\frac{1(0.1)}{3}(1+0.6+2\cdot1.7117522+4(1+0.4+2\cdot1.3945637)+(1+0.3+2\cdot1.2428455))=1.7045664\). So, the solution of the initial-value problem at \(x=0.4\) is \(y(0.4) \approx 1.5593062\) and at \(x=0.5\) is \(y(0.5) \approx 1.7045664\).

Key concepts

Understanding the Runge-Kutta Method

The Runge-Kutta method is a powerful tool when it comes to solving ordinary differential equations (ODEs), specifically initial-value problems. This numerical method enables us to estimate solutions to ODEs without producing a general formula, which can be particularly useful when such formulas are difficult to obtain.

Think of it as a 'stepping' method, where we calculate the solution at successive small intervals, going from a known initial state to the desired point. At each step, the method employs a series of calculations (using the given function and its derivatives) to predict the value of the variable of interest at the next point, which gradually builds up a solution, step by step.

As described in the exercise, the Runge-Kutta method of fourth order was used to establish starting points for the Milne method. The fourth order Runge-Kutta method calculates four slopes (k-terms) that take into account the value of the derivative (rate of change) at different points within the current interval. These are combined in a weighted average that gives a prediction for the solution at the next point. The weights are chosen to provide high accuracy for each step, while keeping the computational load reasonable.

Solving Initial-Value Problems

An initial-value problem in mathematics is a type of problem that includes both a differential equation and a specific starting condition, called the 'initial value'. To solve such a problem, one seeks a function that satisfies the differential equation and, at the same time, the initial condition.

The initial condition, such as the one given in our exercise where the function y(0) is known, provides a starting point for solving the differential equation. Since many physical phenomena can be described by differential equations, initial-value problems are fundamental in fields like engineering, physics, and finance.

When you use numerical methods, like the Runge-Kutta or Milne method, to solve these problems, the main idea is to start from the initial condition and iteratively predict what happens next, at small increments, relying on the structure provided by the differential equation. Thus, 'initial value' isn't just a condition, it's the foundation upon which the entire subsequent solution is built.

Employing Predictor-Corrector Formulae

Predictor-corrector formulae consist of a pair of algorithms: one predicts the solution, the other corrects it. When tackling complex differential equations, these formulae minimize errors that accumulate over multiple steps in numerical methods.

As the terms suggest, the predictor formula is used first to make an initial guess at the solution (predicting where the solution curve will go next). This is then followed by the corrector formula, which improves upon the prediction by adjusting the solution, using more information from the differential equation.

In the context of the Milne method, these formulae work in tandem. The predictor provides an estimate based on previous steps, while the corrector refines this estimate using both the original differential equation and the predicted values. This combined approach allows the Milne method to achieve a more accurate solution than either formula could alone.