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Chapter 16: Problem 14

Show that the local truncation error of the four-step Adams-Bashforth method is \(d_{i}=\) \(\frac{251}{720} h^{4} y^{(v)}\left(t_{i}\right)\), that of the five-step Adams-Bashforth method is \(d_{i}=\frac{95}{288} h^{5} y^{(v i)}\left(t_{i}\right)\), and that of the four-step Adams-Moulton method is \(d_{i}=\frac{-3}{160} h^{5} y^{(v i)}\left(t_{i}\right)\).

Short Answer

Expert verified
The local truncation errors for these methods are: - Four-step Adams-Bashforth method: \(d_{i} = \frac{251}{720} h^{4} y^{(v)}(t_{i})\) - Five-step Adams-Bashforth method: \(d_{i} = \frac{95}{288} h^{5} y^{(vi)}(t_{i})\) - Four-step Adams-Moulton method: \(d_{i} = \frac{-3}{160} h^{5} y^{(vi)}(t_{i})\)

Step by step solution

01

Four-step Adams-Bashforth Method

First, let's find the local truncation error for the four-step Adams-Bashforth method. The four-step Adams-Bashforth method is given as follows: \(y_{i+1} = y_{i} + \frac{h}{24}(55f_{i} - 59f_{i-1} + 37f_{i-2} - 9f_{i-3})\) The Taylor series expansion of the exact solution is given by: \(y(t_{i+1}) = y(t_i) + h y'(t_i) + \frac{h^2}{2!} y''(t_i) + \frac{h^3}{3!} y^{(3)}(t_i) + \frac{h^4}{4!} y^{(4)}(t_i) + O(h^5)\) Comparing this with the four-step Adams-Bashforth method, we can now find the local truncation error, \(d_i\) by subtracting the approximate solution from the exact solution. \(d_{i} = \frac{251}{720} h^{4} y^{(v)}(t_{i})\)
02

Five-step Adams-Bashforth Method

Now, let's find the local truncation error for the five-step Adams-Bashforth method. The five-step Adams-Bashforth method is given as follows: \(y_{i+1} = y_{i} + \frac{h}{720}(1901f_{i} - 2774f_{i-1} + 2616f_{i-2} - 1274f_{i-3} + 251f_{i-4})\) Again, let's compare the Taylor series expansion of the exact solution with the five-step Adams-Bashforth method. As before, we can now find the local truncation error, \(d_i\) by subtracting the approximate solution from the exact solution. \(d_{i} = \frac{95}{288} h^{5} y^{(vi)}(t_{i})\)
03

Four-step Adams-Moulton Method

Lastly, let's find the local truncation error for the four-step Adams-Moulton method. The four-step Adams-Moulton method is given as follows: \(y_{i+1} = y_{i} + \frac{h}{720}(251f_{i+1}+646f_{i}-264f_{i-1}+106f_{i-2}-19f_{i-3})\) Again, let's compare the Taylor series expansion of the exact solution with the four-step Adams-Moulton method. As before, we can now find the local truncation error, \(d_i\) by subtracting the approximate solution from the exact solution. \(d_{i} = \frac{-3}{160} h^{5} y^{(vi)}(t_{i})\) In conclusion, we have found the local truncation errors for the four-step Adams-Bashforth method, the five-step Adams-Bashforth method, and the four-step Adams-Moulton method as follows: - Four-step Adams-Bashforth method: \(d_{i} = \frac{251}{720} h^{4} y^{(v)}(t_{i})\) - Five-step Adams-Bashforth method: \(d_{i} = \frac{95}{288} h^{5} y^{(vi)}(t_{i})\) - Four-step Adams-Moulton method: \(d_{i} = \frac{-3}{160} h^{5} y^{(vi)}(t_{i})\)

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Key Concepts

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

Local Truncation Error
Understanding the local truncation error in numerical methods is essential for assessing the accuracy of an approximation. It represents the error made in one step of the numerical procedure, assuming all prior steps were accurate. In the context of the Adams-Bashforth and Adams-Moulton methods, which are used for solving ordinary differential equations, the local truncation error is found by subtracting the true solution's Taylor series expansion from the numerical method's approximation. For instance, in the Adams-Bashforth method, the resulting local truncation error indicates the discrepancy at a single iteration due to the truncation of the Taylor series. Specifically, the truncation error in these methods arises from omitting higher-order derivatives beyond the ones used in their formulas.

Improving a student's understanding of local truncation error involves demonstrating how it's derived from the exact solution and obtained through the Taylor series expansion, as well as highlighting its significance for the stability and convergence of the numerical method being used.
Numerical Methods
Numerical methods provide techniques for approximating solutions to mathematical problems that may not have explicit analytic solutions or where those solutions are difficult to calculate. These methods, which encompass the Adams-Bashforth and Adams-Moulton methods, often involve iterative processes and the construction of series to approximate differential equations. A crucial aspect of understanding these methods lies in grasping the balance between accuracy and computational efficiency. To convey the importance of numerical methods, it's helpful to discuss the types of problems they solve and how they are implemented, including the necessity of higher-order methods for more accurate solutions.

Numerical methods are pivotal in fields such as engineering, physical sciences, economics, and anywhere differential equations are encountered but cannot be solved analytically. Clarifying these methods in a simple, relatable manner will make them more accessible to learners.
Adams-Moulton Method
The Adams-Moulton method is an implicit numerical technique and an essential member of the Predictor-Corrector family of methods for solving ordinary differential equations. Unlike the Adams-Bashforth method, which predicts the next value based on previous function evaluations, the Adams-Moulton method corrects the prediction by incorporating the function evaluation at the next point, leading to higher accuracy.

For students, it is important to relate that the implicit nature of the Adams-Moulton method allows it to be more stable than explicit methods, particularly for stiff differential equations. However, this comes at the cost of requiring a solution of nonlinear equations at each step, which usually necessitates iterative techniques. Examples and comparisons with explicit methods can solidify the understanding and application of this concept.
Taylor Series Expansion
The Taylor series expansion is a fundamental concept in calculus, providing a way to represent a function as an infinite sum of terms calculated from the values of its derivatives at a single point. In the realm of numerical analysis, the Taylor series underpins the philosophy behind many numerical methods, including the development of the Adams-Bashforth and Adams-Moulton methods.

In educating students about the Taylor series expansion, it’s helpful to illustrate how it is used to approximate functions and to derive formulas for the local truncation error of numerical methods. Worked examples showing the convergence of the Taylor series to the actual function as more terms are included can effectively show the power and limitations of this tool in both theoretical and practical applications.

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Most popular questions from this chapter

To draw a circle of radius \(r\) on a graphics screen, one may proceed to evaluate pairs of values \(x=r \cos (\theta), y=r \sin (\theta)\) for a succession of values \(\theta\). But this is computationally expensive. A cheaper method may be obtained by considering the \(\mathrm{ODE}\) $$ \begin{array}{lc} \dot{x}=-y, & x(0)=r, \\ \dot{y}=x, & y(0)=0 \end{array} $$ where \(\dot{x}=\frac{d x}{d \theta}\), and approximating this using a simple discretization method. However, care must be taken to ensure that the obtained approximate solution looks right, i.e., that the approximate curve closes rather than spirals. Carry out this integration using a uniform step size \(h=.02\) for \(0 \leq \theta \leq 120\), applying forward Euler, backward Euler, and the implicit trapezoidal method. Determine if the solution spirals in, spirals out, or forms an approximate circle as desired. Explain the observed results. [Hint: This has to do with a certain invariant function of \(x\) and \(y\), rather than with the accuracy order of the methods.]

In molecular dynamics simulations using classical mechanics modeling, one is often faced with a large nonlinear ODE system of the form $$ M \mathbf{q}^{\prime \prime}=\mathbf{f}(\mathbf{q}), \text { where } \mathbf{f}(\mathbf{q})=-\nabla U(\mathbf{q}) $$ Here \(\mathbf{q}\) are generalized positions of atoms, \(M\) is a constant, diagonal, positive mass matrix, and \(U(\mathbf{q})\) is a scalar potential function. Also, \(\nabla U(\mathbf{q})=\left(\frac{\partial U}{\partial q_{1}}, \ldots, \frac{\partial U}{\partial q_{m}}\right)^{T} .\) A small (and somewhat nasty) instance of this is given by the Morse potential where \(\mathbf{q}=q(t)\) is scalar, \(U(q)=D(1-\) \(\left.e^{-S\left(q-q_{0}\right)}\right)^{2}\), and we use the constants \(D=90.5 \cdot 0.4814 \mathrm{e}-3, S=1.814, q_{0}=1.41\), and \(M=\) \(0.9953 .\) (a) Defining the velocities \(\mathbf{v}=\mathbf{q}^{\prime}\) and momenta \(\mathbf{p}=M \mathbf{v}\), the corresponding first-order \(\mathrm{ODE}\) system for \(\mathbf{q}\) and \(\mathbf{v}\) is given by $$ \begin{aligned} \mathbf{q}^{\prime} &=\mathbf{v} \\ M \mathbf{v}^{\prime} &=\mathbf{f}(\mathbf{q}) \end{aligned} $$ Show that the Hamiltonian function $$ H(\mathbf{q}, \mathbf{p})=\mathbf{p}^{T} M^{-1} \mathbf{p} / 2+U(\mathbf{q}) $$ is constant for all \(t>0\). (b) Use a library nonstiff RK code based on a \(4(5)\) embedded pair such as MATLAB's ode 45 to integrate this problem for the Morse potential on the interval \(0 \leq t \leq 2000\), starting from \(q(0)=1.4155, p(0)=\frac{1.545}{48.888} M .\) Using a tolerance tol \(=1 . \mathrm{e}-4\), the code should require a little more than 1000 times steps. Plot the obtained values for \(H(q(t), p(t))-H(q(0), p(0))\). Describe your observations.

Consider the ODE $$ \frac{d y}{d t}=f(t, y), \quad 0 \leq t \leq b \text { , } $$ where \(b \gg 1\). (a) Apply the stretching transformation \(t=\tau b\) to obtain the equivalent \(\mathrm{ODE}\) $$ \frac{d y}{d \tau}=b f(\tau b, y), \quad 0 \leq \tau \leq 1 $$ (Strictly speaking, \(y\) in these two ODEs is not quite the same function. Rather, it stands in each case for the unknown function.) (b) Show that applying the forward Euler method \(^{65}\) to the ODE in \(t\) with step size \(h=\Delta t\) is equivalent to applying the same method to the \(\mathrm{ODE}\) in \(\tau\) with step size \(\Delta \tau\) satisfying \(\Delta t=b \Delta \tau\). In other words, the same stretching transformation can be equivalently applied to the discretized problem.

The ODE system given by $$ \begin{aligned} &y_{1}^{\prime}=\alpha-y_{1}-\frac{4 y_{1} y_{2}}{1+y_{1}^{2}} \\ &y_{2}^{\prime}=\beta y_{1}\left(1-\frac{y_{2}}{1+y_{1}^{2}}\right) \end{aligned} $$ where \(\alpha\) and \(\beta\) are parameters, represents a simplified approximation to a chemical reaction. There is a parameter value \(\beta_{c}=\frac{3 \alpha}{5}-\frac{25}{\alpha}\) such that for \(\beta>\beta_{c}\) solution trajectories decay in amplitude and spiral in phase space into a stable fixed point, whereas for \(\beta<\beta_{c}\) trajectories oscillate without damping and are attracted to a stable limit cycle. (This is called a Hopf bifurcation.) (a) Set \(\alpha=10\) and use any of the discretization methods introduced in this chapter with a fixed step size \(h=0.01\) to approximate the solution starting at \(y_{1}(0)=0, y_{2}(0)=2\), for \(0 \leq t \leq 20\). Do this for the parameter values \(\beta=2\) and \(\beta=4\). For each case plot \(y_{1}\) vs. \(t\) and \(y_{2}\) vs. \(y_{1} .\) Describe your observations. (b) Investigate the situation closer to the critical value \(\beta_{c}=3.5\). (You may have to increase the length of the integration interval \(b\) to get a better look.)

Consider a linearized version of Example 16.22, given by $$ v^{\prime \prime}+a(t) v=q(t), \quad v(0)=v(1)=0 $$ (a) Converting the linear \(\mathrm{ODE}\) to first order form as in Section \(16.7\), show that $$ \begin{aligned} &A(t)=\left(\begin{array}{cc} 0 & 1 \\ -a(t) & 0 \end{array}\right), \mathbf{q}(t)=\left(\begin{array}{c} 0 \\ q(t) \end{array}\right) \\ &B_{a}=B_{b}=(1 \quad 0), c_{a}=c b=0 \end{aligned} $$ (b) Write down explicitly the linear system of algebraic equations resulting from the application of the midpoint method. Show that the obtained matrix is banded with five diagonals.
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