Question

Consider the initial value problem \(x^{\prime}=f(t, x, y)\) and \(y^{\prime}=g(t, x, y)\) with \(x\left(t_{0}\right)=x_{0}\) and \(y\left(t_{0}\right)=y_{0} .\) The generalization of the Adams-Moulton predictor-corrector method of Section 8.4 is $$ \begin{array}{l}{x_{n+1}=x_{n}+\frac{1}{24} h\left(55 f_{n}-59 f_{n-1}+37 f_{n-2}-9 f_{n-3}\right)} \\ {y_{n+1}=y_{n}+\frac{1}{24} h\left(55 g_{n}-59 g_{n-1}+37 g_{n-2}-9 g_{n-1}\right)}\end{array} $$ and $$ \begin{array}{l}{x_{n+1}=x_{n}+\frac{1}{24} h\left(9 f_{n+1}+19 f_{n}-5 f_{n-1}+f_{n-2}\right)} \\ {y_{n+1}=y_{n}+\frac{1}{24} h\left(9 g_{n+1}+19 g_{n}-5 g_{n-1}+g_{n-2}\right)}\end{array} $$ Determine an approximate value of the solution at \(t=0.4\) for the example initial value problem \(x^{\prime}=x-4 y, y^{\prime}=-x+y\) with \(x(0)=1, y(0)=0\). Take \(h=0.1 .\) Correct the predicted value once. For the values of \(x_{1}, \ldots, y_{3}\) use the values of the exact solution rounded to six digits: \(x_{1}=1.12883, x_{2}=1.32042, x_{3}=1.60021, y_{1}=-0.110527, y_{2}=\) \(-0.250847,\) and \(y_{3}=-0.429696\)

Short answer

The approximate solution at t=0.4 for the given initial value problem is x ≈ 1.827808 and y ≈ -0.532750.

Step-by-step solution

Define the given functions and initial conditions

To begin, let's set up the differential equations and the initial values: - \(f(t, x, y) = x-4y\) - \(g(t, x, y) = -x+y\) - \(x(0) = x_0 = 1\), \(y(0) = y_0 = 0\)

Calculate \(f_n\) and \(g_n\) values

To compute the solution at t=0.4 (which corresponds to \(n = 4\)), we need to calculate the values \(f_4\) and \(g_4\). To do so, we will use the given exact solution values for \(x_1, x_2, x_3, y_1, y_2\) and \(y_3\). Since \(f(t, x, y) = x - 4y\) and \(g(t, x, y) = -x + y\), we have: - \(f_1 = x_1 - 4y_1 = 1.12883 - 4(-0.110527) = 1.12883 + 0.442108 = 1.570938\) - \(f_2 = x_2 - 4y_2 = 1.32042 - 4(-0.250847) = 1.32042 + 1.003388 = 2.323808\) - \(f_3 = x_3 - 4y_3 = 1.60021 - 4(-0.429696) = 1.60021 + 1.718784 = 3.318994\) - \(g_1 = -x_1 + y_1 = -1.12883 - 0.110527 = -1.239357\) - \(g_2 = -x_2 + y_2 = -1.32042 - 0.250847 = -1.571267\) - \(g_3 = -x_3 + y_3 = -1.60021 - 0.429696 = -2.029906\)

Compute the predicted values \(x_4\) and \(y_4\)

According to the Adams-Moulton predictor method, the predicted values for \(x_4\) and \(y_4\) can be calculated with the following equations: \(x_4 = x_3 + \frac{1}{24}h(55f_3 - 59f_2 + 37f_1 - 9f_0)\) \(y_4 = y_3 + \frac{1}{24}h(55g_3 - 59g_2 + 37g_1 - 9g_0)\) We already have the values of \(f_1, f_2, f_3, g_1, g_2, g_3\). For \(f_0\) and \(g_0\), we can use the initial conditions: - \(f_0 = x_0 - 4y_0 = 1 - 4(0) = 1\) - \(g_0 = -x_0 + y_0 = -1 + 0 = -1\) Now, we can calculate the predicted values of \(x_4\) and \(y_4\): \(x_4 = 1.60021 + \frac{1}{24}(0.1)(55(3.318994) - 59(2.323808) + 37(1.570938) - 9(1)) \approx 1.60021 + 0.232862 = 1.833072\) \(y_4 = -0.429696 + \frac{1}{24}(0.1)(55(-2.029906) - 59(-1.571267) + 37(-1.239357) - 9(-1)) \approx -0.429696 - 0.100930 = -0.530626\)

Correct the predicted values once

In the final step, we can correct the predicted values once using the Adams-Moulton corrector method: \(x_4 = x_3 + \frac{1}{24}h(9f_4 + 19f_3 - 5f_2 + f_1)\) \(y_4 = y_3 + \frac{1}{24}h(9g_4 + 19g_3 - 5g_2 + g_1)\) First, we need to compute \(f_4\) and \(g_4\). As we have the predicted values for \(x_4\) and \(y_4\), we can use them to find \(f_4\) and \(g_4\): \(f_4 = x_4 - 4y_4 = 1.833072 - 4(-0.530626) \approx 4.052924\) \(g_4 = -x_4 + y_4 = -1.833072 - 0.530626 \approx -2.363698\) Now, we correct the predicted values: \(x_4 = 1.60021 + \frac{1}{24}(0.1)(9(4.052924) + 19(3.318994) - 5(2.323808) + 1.570938) \approx 1.60021 + 0.227598 = 1.827808\) \(y_4 = -0.429696 + \frac{1}{24}(0.1)(9(-2.363698) + 19(-2.029906) - 5(-1.571267) - 1.239357) \approx -0.429696 - 0.103054 = -0.532750\) The approximate solution of the given initial value problem at \(t = 0.4\) is \(x \approx 1.827808\) and \(y \approx -0.532750\).

Key concepts

Initial Value Problem

In mathematics, an initial value problem is a key concept when dealing with differential equations. It involves finding a function that satisfies a differential equation and an initial condition. Let's break it down: - A **differential equation** is an equation that involves a function and its derivatives. - The **initial condition** provides a specific value of the function at a certain point. For example, in our exercise, it is given that at time \( t_0 = 0 \), \( x(0) = 1 \) and \( y(0) = 0 \). This information helps establish the starting point of the solution.The initial value problem essentialy sets the stage for solving differential equations numerically. It allows us to proceed with methods like the Adams-Moulton method to approximate the solution, especially when finding an exact solution is difficult or impossible.In practical terms, initial value problems are crucial in modeling real-world scenarios where conditions at a single starting time dictate the behavior of a system over time.

Predictor-Corrector Method

The predictor-corrector method is a powerful technique used for solving initial value problems for ordinary differential equations. It consists of two main phases:- **Predictor Phase:** This phase first predicts the value of the solution at a new point. In our problem, we use the Adams-Moulton predictor formula to estimate these initial predictions. For example, with values like \( x_4 \) and \( y_4 \), this phase gives a rough estimate of the solution.- **Corrector Phase:** Here, the initial prediction is refined to improve accuracy. The Adams-Moulton corrector formula then adjusts the predicted values, often leading to a much more accurate solution.
This two-step process serves a crucial purpose: providing a balance between computational efficiency and accuracy. By predicting first and then correcting, the method effectively handles errors that may occur due to approximations. Additionally, the use of known and computed past values further strengthens the reliability of the results. This way, predictor-corrector methods are essential in providing approximate solutions where exact ones are hard to compute.

Numerical Solution of Differential Equations

When it comes to solving differential equations, especially complex or non-linear ones, purely analytical methods may fall short. This is where numerical methods come into play. These methods use discrete approximations to find solutions to differential equations. Numerical solutions are:

• **Approximate:** Unlike exact solutions, they provide an estimate of the solution.
• **Iterative:** They typically involve iterative processes that successively refine the solution until a satisfactory approximation is reached.
• **Computationally Efficient:** They leverage computational power to handle complex computations that would be impractical to solve by hand.

The numerical approach is indispensable in the real world. Many physical, chemical, and biological processes are modeled using differential equations which cannot be solved exactly. Methods like the Adams-Moulton procedure enable us to tackle these by providing powerful tools that convert complex equations into manageable computations. This ultimately enables practical decision-making based on mathematical models.