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Show that the third order Adams-Moulton formula is $$ y_{x+1}=y_{x}+(h / 12)\left(5 f_{x+1}+8 f_{n}-f_{x-1}\right) $$

Short Answer

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Question: Show that the third order Adams-Moulton formula is given by the equation: $$ y_{x+1}=y_{x}+(h / 12)\left(5 f_{x+1}+8 f_{x}-f_{x-1}\right) $$ Answer: In order to derive the third order Adams-Moulton formula, we first find a three-point Lagrange interpolating polynomial using points (\(x_{x-1}\), \(y_{x-1}\)), (\(x_x\), \(y_x\)), and (\(x_{x+1}\), \(y_{x+1}\)). After differentiating the Lagrange polynomial, we evaluate the derivative at \(x_x\). Then, we set the derivative equal to the ODE and solve for the desired formula. This process results in the following equation: $$ y_{x+1}=y_{x}+(h / 12)\left(5 f_{x+1}+8 f_{x}-f_{x-1}\right) $$

Step by step solution

01

Understand the problem

We are given an initial value problem (IVP), which is a type of differential equation problem with an initial condition. The third order Adams-Moulton formula should contain information from 3 points: the current point, the next point, and a past point. We need to derive this formula as the exercise states.
02

Obtain a three-point Lagrange interpolating polynomial

To achieve this, let's first find the three-point Lagrange interpolating polynomial using points (\(x_{x-1}\), \(y_{x-1}\)), (\(x_x\), \(y_x\)), and (\(x_{x+1}\), \(y_{x+1}\)). The Lagrange interpolating polynomial is given by: $$ L(t) = \frac{(t-x_x)(t-x_{x+1})}{(x_{x-1}-x_x)(x_{x-1}-x_{x+1})}y_{x-1} + \frac{(t-x_{x-1})(t-x_{x+1})}{(x_x-x_{x-1})(x_x-x_{x+1})}y_x + \frac{(t-x_{x-1})(t-x_x)}{(x_{x+1}-x_{x-1})(x_{x+1}-x_x)}y_{x+1} $$
03

Differentiate the Lagrange polynomial

Next, differentiate the Lagrange interpolating polynomial with respect to t: $$ L'(t) = \frac{(t-x_{x+1}) + (t-x_x)}{(x_{x-1}-x_x)(x_{x-1}-x_{x+1})} f_{x-1} + \frac{(t-x_{x+1}) + (t-x_{x-1})}{(x_x-x_{x-1})(x_x-x_{x+1})} f_x + \frac{(t-x_x) + (t-x_{x-1})}{(x_{x+1}-x_{x-1})(x_{x+1}-x_x)} f_{x+1} $$
04

Evaluate the derivative at \(x_x\)

Now we will evaluate the derivative at \(x_x\). Let's denote \(h = x_{x+1} - x_x = x_x - x_{x-1}\). Therefore, \(t=x_x\) will be substituted in \(L'(t)\) giving: $$ L'(x_x) = \frac{(-h) + h}{(-2h)(-h)} f_{x-1} + \frac{(h) + (-h)}{(h)(-h)} f_x + \frac{h}{(h)(2h)} f_{x+1} \\ L'(x_x) = -\frac{1}{12}f_{x-1} + \frac{8}{12}f_x + \frac{5}{12}f_{x+1} $$
05

Set the derivative equal to the ODE and solve

We set the derivative \(L'(x_x) = f_x\) , so that we have: $$ f_x = -\frac{1}{12}f_{x-1} + \frac{8}{12}f_x + \frac{5}{12}f_{x+1} $$ Now, integrate both sides of the equation w.r.t \(x\): $$ \int_{x_x}^{x_{x+1}}f_x dx = -\frac{1}{12}\int_{x_x}^{x_{x+1}}f_{x-1} dx + \frac{8}{12}\int_{x_x}^{x_{x+1}}f_x dx + \frac{5}{12}\int_{x_x}^{x_{x+1}}f_{x+1} dx $$
06

Simplify the equation and obtain the final result

Let's simplify the equation and express it in terms of \(y_i\)'s: $$ y_{x+1}-y_x = -\frac{1}{12}h( f_{x-1} - 8f_x + 5f_{x+1}) $$ Thus, we have derived the third order Adams-Moulton formula as follows: $$ y_{x+1}=y_{x}+(h / 12)\left(5 f_{x+1}+8 f_{x}-f_{x-1}\right) $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lagrange Interpolating Polynomial
The Lagrange Interpolating Polynomial is a powerful method in numerical analysis used for polynomial interpolation. Think of it as a way of estimating a function using a polynomial that passes exactly through a set of given points. This interpolation is crucial for approximating functions because it helps in predicting unknown values by using known data points.

In simple terms, if you have a few known data points from a function, the Lagrange polynomial helps you find a smooth curve that goes through all those points. The formula involves constructing a weighted sum of the given data points, where each term in the sum is a Lagrange basis polynomial. Each basis polynomial is built so that it equals 1 at its corresponding data point and 0 at all other data points.

For example, when we deal with three points, say \((x_{x-1}, y_{x-1})\), \((x_x, y_x)\), and \((x_{x+1}, y_{x+1})\), we use these to build a polynomial through those points. The Lagrange polynomial becomes particularly useful in numerical methods such as the Adams-Moulton method, where it helps approximate solutions to differential equations by estimating the function between known data points.
Numerical Methods for Differential Equations
Numerical methods for differential equations are a core area of study in mathematics and engineering. They allow us to find approximate solutions to complex differential equations where analytical solutions are impossible or impractical. Key methods in this field include Euler's method, Runge-Kutta methods, and Adams-Bashforth-Moulton methods.

The Adams-Moulton method is a predictor-corrector method specifically used for solving ordinary differential equations (ODEs). It belongs to the class of linear multistep methods, where the current and several previous steps are used to estimate the solution at future steps. In practice, this method takes calculated approximations and improves them using subsequent corrections, leading to higher accuracy. The third-order Adams-Moulton formula, which accounts for three points, provides a balance between computational complexity and accuracy.

These numerical techniques are invaluable in the real world, especially in fields like physics, engineering, and finance, where systems are often modeled by differential equations that do not have simple, solvable forms.
Differential Equations Analysis
Differential equations form the backbone of many scientific disciplines as they describe how something changes over time or space. Analysis of these equations helps us to understand dynamic systems, predicting behavior based on initial conditions.

When analyzing differential equations, we often face initial value problems (IVPs). An IVP requires us to find a function that satisfies a differential equation and also meets an initial condition. This initial condition ensures the uniqueness of the solution, making it a critical step in the process.

In the context of the step-by-step solution above, using Lagrange interpolating polynomial within the Adams-Moulton framework allows for an enhanced understanding of the ODE being worked on. By calculating and setting derivatives and then integrating, we make a thorough analysis to derive meaningful approximations of the solutions. This blend of theoretical analysis and practical computation is central to effectively solving differential equations.

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Most popular questions from this chapter

Carry out one step of the Euler method and of the improved Euler method using the step size \(h=0.1 .\) Suppose that a local truncation error no greater than 0.0025 is required. Estimate the step size that is needed for the Euler method to satisfy this requirement at the first step. $$ y^{\prime}=\sqrt{t+y}, \quad y(0)=3 $$

Find approximate values of the solution of the given initial value problem at \(t=0.5,1.0,1.5,\) and 2.0 , using the specified method. For starting values use the values given by the Runge-Kutta muthod; see Problems 7 through 12 in Section 8.3 . Compare the results of the various methods with each other and with the actual solution (if available). $$ y^{\prime}=\sqrt{t+y}, \quad y(0)=3 $$

Using three terms in the Taylor series given in Eq. ( \(12)\) and taking \(h=0.1\), determine approximate values of the solution of the illustrative example \(y^{\prime}=1-t+4 y, y(0)=1\) at \(t=0.1\) and \(0.2 .\) Compare the results with those using the Euler method and with the exact values. Hint: If \(y^{\prime}=f(t, y),\) what is \(y^{\prime \prime \prime} ?\)

Estimate the local truncation error for the Euler method in terms of the solution \(y=\phi(t) .\) Obtain a bound for \(e_{n+1}\) in terms of \(t\) and \(\phi(t)\) that is valid on the interval \(0 \leq t \leq 1 .\) By using a formula for the solution obtain a more accurate error bound for \(e_{n+1} .\) For \(h=0.1\) compute a bound for \(e_{1}\) and compare it with the actual error at \(t=0.1 .\) Also compute a bound for the error \(e_{4}\) in the fourth step. $$ y^{\prime}=\frac{1}{2}-t+2 y, \quad y(0)=1 $$

Determine an approximate value of the solution at \(t=0.4\) and \(t=0.5\) using the specified method. For starting values use the values given by the Runge- Kutta method; see Problems 1 through 6 of Section 8.3 . Compare the results of the various methods with each other and with the actual solution (if available). $$ \begin{array}{l}{\text { (a) Use the fourth order predictor-corrector method with } h=0.1 . \text { Use the corrector }} \\ {\text { formula once at each step. }} \\ {\text { (b) Use the fourth order Adams-Moulton method with } h=0.1} \\ {\text { (c) Use the fourth order backward differentiation method with } h=0.1 .}\end{array} $$ $$ y^{\prime}=\left(t^{2}-y^{2}\right) \sin y, \quad y(0)=-1 $$

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