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Show that the first order Adams-Bashforth method is the Euler method, and that the first order Adams-Moulton method is the backward Euler method.

Short Answer

Expert verified
Question: Prove that the first order Adams-Bashforth method is the Euler method and that the first order Adams-Moulton method is the backward Euler method. Answer: We have shown that the first order Adams-Bashforth method yields the same equation as the Euler method, and the first order Adams-Moulton method yields the same equation as the backward Euler method. Therefore, we can conclude that the first order Adams-Bashforth method is indeed the Euler method, and the first order Adams-Moulton method is the backward Euler method.

Step by step solution

01

Define the Euler and Backward Euler methods

First, let's recall the definition of the Euler method and the backward Euler method for solving an ordinary differential equation (ODE) of the form: dy/dt = f(t, y(t)). Euler method is given by: y_{n+1} = y_n + h*f(t_n, y_n), where y_n is the approximate solution at time t_n, and h is the time step. Backward Euler method is given by: y_{n+1} = y_n + h*f(t_{n+1}, y_{n+1}). Now let's define the Adams-Bashforth and Adams-Moulton methods.
02

Define the Adams-Bashforth and Adams-Moulton methods

The k-th order Adams-Bashforth method can be expressed as: y_{n+1} = y_n + h * \sum_{i=0}^{k-1} a_i * f(t_{n-i}, y_{n-i}), where the coefficients a_i depend on the order k. The k-th order Adams-Moulton method can be expressed as: y_{n+1} = y_n + h * \sum_{i=0}^{k} b_i * f(t_{n+1-i}, y_{n+1-i}), where the coefficients b_i also depend on the order k.
03

Show that first order Adams-Bashforth method is Euler method

For the first order Adams-Bashforth method, we have k = 1, so we get: y_{n+1} = y_n + h * a_0 * f(t_n, y_n), Since there is only one coefficient a_0, we have a_0 = 1, so the equation becomes: y_{n+1} = y_n + h * f(t_n, y_n). This is identical to the Euler method, so we have proved that the first order Adams-Bashforth method is the Euler method.
04

Show that first order Adams-Moulton method is backward Euler method

For the first order Adams-Moulton method, we have k = 1, so we obtain: y_{n+1} = y_n + h * (b_0 * f(t_{n+1}, y_{n+1}) + b_1 * f(t_n, y_n)). Since there are two coefficients, we have b_0 + b_1 = 1. In the first-order Adams-Moulton method, b_0 = 1 and b_1 = 0, so the equation becomes: y_{n+1} = y_n + h * f(t_{n+1}, y_{n+1}). This is identical to the backward Euler method, so we have proved that the first order Adams-Moulton method is the backward Euler method.

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Key Concepts

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

Euler Method
The Euler method is a straightforward numerical approach for solving ordinary differential equations (ODEs), specifically initial value problems. It was named after the prolific Swiss mathematician Leonhard Euler. Consider an ODE of the form \( \frac{{dy}}{{dt}} = f(t, y(t)) \), where \( f(t, y) \) describes the rate of change of \( y \) with respect to \( t \). The aim is to find the function \( y(t) \) that satisfies this equation given an initial condition \( y(t_0) = y_0 \).

The Euler method begins by dividing the time domain into small steps of size \( h \). For each step, it assumes that the rate of change \( f(t, y) \) is constant over that step, and therefore the solution can be approximated linearly. The formula for this is \( y_{n+1} = y_n + h \cdot f(t_n, y_n) \), where \( y_{n+1} \) is the estimated value of \( y \) at time \( t_{n+1} \) and \( y_n \) is the estimated value at time \( t_{n} \).

This method is simple and often easy to implement, which makes it a popular choice for introductory classes on numerical analysis. However, its accuracy is dependent on the size of the step \( h \), with smaller steps generally leading to more accurate results.
Backward Euler Method
The Backward Euler method, also known as implicit Euler method, is another technique used to numerically solve ODEs. This method is particularly useful when dealing with stiff equations, where more explicit methods like the standard Euler method can become unstable unless the step size \( h \) is taken to be very small.

While the standard Euler method is explicit and calculates the solution at the next step based only on information from the current step, the Backward Euler method determines \(y_{n+1}\) by solving the equation \(y_{n+1} = y_{n} + h \cdot f(t_{n+1}, y_{n+1})\). Here, \(f(t_{n+1}, y_{n+1})\) is the rate of change at the next step, making this method implicit as it requires solving an equation involving the unknown \(y_{n+1}\) itself.

Because the Backward Euler method involves an implicit equation, it can be more challenging to implement compared to the standard Euler method. It usually requires solving a nonlinear equation at each step, often with iterative methods like Newton's method. Nevertheless, its advantageous properties for stiff problems make it an essential tool in the numerical analysis toolbox.
Ordinary Differential Equations
Ordinary differential equations (ODEs) are equations involving a function and its derivatives. These equations play a critical role in modeling various natural phenomena such as motion, growth, decay, and many others across science and engineering disciplines.

An ODE is typically expressed in the form \( \frac{{dy}}{{dt}} = f(t, y) \), where \(y = y(t)\) is the unknown function of the independent variable \(t\), and \(f(t, y)\) is a given function that relates the derivative of \(y\) to \(t\) and \(y\) itself. The solution of an ODE is the function \(y(t)\) that satisfies this relationship for given initial conditions, often expressed as \(y(t_0) = y_0\).

Solving an ODE analytically involves finding an exact expression for \(y(t)\), but many ODEs do not have simple closed-form solutions. This is where numerical methods, like the Euler method and Backward Euler method, come into play, allowing for an approximate solution by discretizing the problem.
Numerical Methods for ODEs
Numerical methods for ordinary differential equations (ODEs) are algorithms used to find approximate solutions to ODEs when analytical methods are impractical or impossible. These methods convert continuous problems into discrete problems that can be solved using computational algorithms.

Popular numerical methods for solving ODEs include:
  • The Euler method, which takes small steps forward using the slope calculated from the current point.
  • The Backward Euler method, which, unlike the standard Euler method, takes into account the slope at the next point, thus requiring the solution of an implicit equation.
  • The Runge-Kutta methods, which are a more advanced set of methods (including the classic fourth-order Runge-Kutta method) that offer increased accuracy by evaluating the slope at several points within each step.
  • The Adams-Bashforth and Adams-Moulton methods, which form a family of predictor-corrector methods that use values from previous steps to predict (Adams-Bashforth) and correct (Adams-Moulton) the solution at the current step.

The choice of method depends on factors like the stiffness of the equation, desired accuracy, and computational resources. For stiff problems where explicit methods may require overly small time steps to maintain stability, implicit methods like the Backward Euler method can be more appropriate despite their computational complexity.

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Most popular questions from this chapter

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

Using a step size \(h=0.05\) and the Euler method, but retaining only three digits throughout the computations, determine approximate values of the solution at \(t=0.1,0.2,0.3,\) and 0.4 for each of the following initial value problems. $$ \begin{array}{ll}{\text { (a) } y^{\prime}=1-t+4 y,} & {y(0)=1} \\ {\text { (b) } y^{\prime}=3+t-y,} & {y(0)=1} \\ {\text { (c) } y^{\prime}=2 y-3 t,} & {y(0)=1}\end{array} $$ Compare the results with those obtained in Example 1 and in Problems 1 and \(3 .\) The small differences between some of those results rounded to three digits and the present results are due to round-off error. The round-off error would become important if the computation required many steps.

Consider the initial value problem $$ y^{\prime}=t^{2}+y^{2}, \quad y(0)=1 $$ (a) Draw a direction field for this equation. (b) Use the Runge-Kutta or other methods to find approximate values of the solution at \(t=0.8,0.9,\) and 0.95 . Choose a small enough step size so that you believe your results are accurate to at least four digits. (c) Try to extend the calculations in part (b) to obtain an accurate approximation to the solution at \(t=1\) If you encounter difficulties in doing this, explain why you think this happens. The direction field in part (a) may be helpful.

Consider the example problem \(x^{\prime}=x-4 y, y^{\prime}=-x+y\) with the initial conditions \(x(0)=1\) and \(y(0)=0\). Use the Runge-Kutta method to solve this problem on the interval \(0 \leq t \leq 1\). Start with \(h=0.2\) and then repeat the calculation with step sizes \(h=0.1,0.05, \ldots\), each half as long as in the preceding case. Continue the process until the first five digit of the solution at \(t=1\) are unchanged for successive step sizes Determine whether these digits are accurate by comparing them with the exact solution given in Eqs. ( 10 ) in the text.

In each of Problems 1 through 6 find approximate values of the solution of the given initial value problem at \(t=0.1,0.2,0.3,\) and \(0.4 .\) Compare the results with those obtained by using other methods and with the exact solution (if available). (a) Use the Runge-Kutta method with \(h=0.1\) (b) Use the Runge-Kutta method with \(h=0.05\) $$ y^{\prime}=\left(t^{2}-y^{2}\right) \sin y, \quad y(0)=-1 $$

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