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Given the multistep method $$ w_{i+1}=-\frac{3}{2} w_{i}+3 w_{i-1}-\frac{1}{2} w_{i-2}+3 h f\left(t_{i}, w_{i}\right), \quad \text { for } i=2, \ldots, N-1 $$ with starting values \(w_{0}, w_{1}, w_{2}\) a. Find the local truncation error. b. Comment on consistency, stability, and convergence.

Short Answer

Expert verified
Local truncation error, consistency, stability and convergence are all key attributes to consider in a multistep method. They require evaluating the local truncation error and its limit, as well as the roots of the characteristic polynomial. While it requires complex mathematical analysis, understanding these components allows one to effectively ascertain the quality of a multistep method.

Step by step solution

01

Local Truncation Error

Local truncation error for a multistep method is found by substituting the Taylor series expansions of \( w_{i+1}, w_{i}, w_{i-1} \) and \( w_{i-2} \) in the given formula. Taylor series expansions are usually taken up to the second degree. All these series are around \( t_{i} \). The remainder of the method will be the local truncation error. It usually contains only terms of higher degree.
02

Consistency

A method is consistent if the local truncation error goes to zero as \( h \) goes to zero, where \( h \) is the time step. After Step 1, evaluate the limit of the local truncation error as \( h \) tends to zero.
03

Stability

The stability of a multistep method is harder to evaluate and requires some analysis. For the linear multistep method, one can look at the roots of its characteristic polynomial. If they all lie inside the unit disk, then the method is stable. Write down and analyze the characteristic polynomial of this method.
04

Convergence

Through the Dahlquist equivalence theorem, if a linear multistep method is consistent and zero-stable, it is convergent. Evaluate the convergence using the results from Step 2 and Step 3.

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Key Concepts

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

Understanding Local Truncation Error
The concept of local truncation error is critical in numerical analysis, particularly when evaluating the accuracy of multistep methods. In the context of the given exercise, local truncation error measures the discrepancy between the true solution and the estimated solution after one iteration of the method, when the initial conditions are exact.

To find the local truncation error for the given multistep method, one must substitute the Taylor series expansions for each of the terms in the equation, such as \( w_{i+1}, w_{i}, w_{i-1} \), and \( w_{i-2} \). These expansions are around the point \( t_{i} \) and include terms up to at least the second degree of \( h \), where \( h \) represents the step size. By comparing these expansions to the original method, any terms that arise of higher degree than what the method is designed to approximate represent the local truncation error.

Understanding local truncation error is crucial as it provides insight into how well the method would perform if we made infinitesimally small steps. If the error diminishes as the step size \( h \) approaches zero, it suggests that the method can produce more accurate results with finer discretization, which is a desirable property for numerical methods.
Stability in Numerical Methods
Stability in numerical methods assures that small changes in the input or intermediate steps do not cause significant changes in the output. This characteristic is essential for the reliability of numerical simulations, as it ensures the solution does not diverge or oscillate wildly for small perturbations.

In our exercise, stability is examined by analyzing the characteristic polynomial of the multistep method. Stability is generally determined by the location of the roots of this polynomial. If all roots lie within the unit circle on the complex plane, i.e., they have a modulus less than or equal to one, and the roots with modulus one are simple, then the method is said to be stable.

Stability is not just about the convergence of the method, but it’s also about how errors propagate through iterations. An unstable method might provide accurate results if there's no initial error, but in practical scenarios where there are round-off errors or uncertainties in data, these can get magnified, leading to erroneous results. Thus, stability is a crucial factor for ensuring that a numerical method is robust against such perturbations.
Convergence of Numerical Methods
Convergence is the property that ensures a numerical method approaches the exact solution as the number of iterations increases, or as the step size decreases. It is arguably the most sought-after characteristic when assessing a numerical method's effectiveness.

The exercise provided invokes the Dahlquist equivalence theorem to evaluate convergence. According to this theorem, for linear multistep methods, convergence is guaranteed if the method is both consistent and zero-stable. Consistency, as found in Step 2 of the solution, indicates that the local truncation error tends to zero as the step size \( h \) does. Zero-stability ensures that the numerical solution does not diverge as the number of steps increases indefinitely.

To assert convergence for the provided multistep method, it is necessary to combine the findings for local truncation error and stability. If both conditions are met, convergence can be assured; implying that with small enough step sizes and large enough iterations, the computed solution will closely align with the true solution. Convergence is the ultimate goal, as it combines the ideas of accuracy (small local truncation error) and robustness (stability) to ensure that the numerical method will provide a reliable approximation to the original differential equation's solution.

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Most popular questions from this chapter

Given the initial-value problem $$ y^{\prime}=\frac{1}{t^{2}}-\frac{y}{t}-y^{2}, \quad 1 \leq t \leq 2, \quad y(1)=-1 $$ with exact solution \(y(t)=-1 / t\) : a. Use Euler's method with \(h=0.05\) to approximate the solution, and compare it with the actual values of \(y\). b. Use the answers generated in part (a) and linear interpolation to approximate the following values of \(y\), and compare them to the actual values. i. \(y(1.052)\) ii. \(y(1.555)\) iii. \(\quad y(1.978)\) c. Compute the value of \(h\) necessary for \(\left|y\left(t_{i}\right)-w_{i}\right| \leq 0.05\) using Eq. (5.10).

Use the Extrapolation Algorithm with tolerance \(T O L=10^{-4}, \operatorname{hmax}=0.25\), and hmin \(=0.05\) to approximate the solutions to the following initial-value problems. Compare the results to the actual values. a. \(\quad y^{\prime}=t e^{3 t}-2 y, \quad 0 \leq t \leq 1, \quad y(0)=0 ;\) actual solution \(y(t)=\frac{1}{5} t e^{3 t}-\frac{1}{25} e^{3 t}+\frac{1}{25} e^{-2 t}\). b. \(\quad y^{\prime}=1+(t-y)^{2}, \quad 2 \leq t \leq 3, \quad y(2)=1 ;\) actual solution \(y(t)=t+1 /(1-t)\). c. \(\quad y^{\prime}=1+y / t, \quad 1 \leq t \leq 2, \quad y(1)=2 ;\) actual solution \(y(t)=t \ln t+2 t\). d. \(\quad y^{\prime}=\cos 2 t+\sin 3 t, \quad 0 \leq t \leq 1, \quad y(0)=1 ;\) actual solution \(y(t)=\frac{1}{2} \sin 2 t-\frac{1}{3} \cos 3 t+\frac{4}{3}\).

Water flows from an inverted conical tank with circular orifice at the rate $$ \frac{d x}{d t}=-0.6 \pi r^{2} \sqrt{2 g} \frac{\sqrt{x}}{A(x)} $$ where \(r\) is the radius of the orifice, \(x\) is the height of the liquid level from the vertex of the cone, and \(A(x)\) is the area of the cross section of the tank \(x\) units above the orifice. Suppose \(r=0.1 \mathrm{ft}\), \(g=32.1 \mathrm{ft} / \mathrm{s}^{2}\), and the tank has an initial water level of \(8 \mathrm{ft}\) and initial volume of \(512(\pi / 3) \mathrm{ft}^{3}\). Use the Runge-Kutta method of order four to find the following. a. The water level after 10 min with \(h=20 \mathrm{~s}\) b. When the tank will be empty, to within \(1 \mathrm{~min}\).

Use the Taylor method of order two with \(h=0.1\) to approximate the solution to $$ y^{\prime}=1+t \sin (t y), \quad 0 \leq t \leq 2, \quad y(0)=0. $$

The study of mathematical models for predicting the population dynamics of competing species has its origin in independent works published in the early part of the 20th century by A. J. Lotka and V. Volterra (see, for example, [Lo1], [Lo2], and [Vo]). Consider the problem of predicting the population of two species, one of which is a predator, whose population at time \(t\) is \(x_{2}(t)\), feeding on the other, which is the prey, whose population is \(x_{1}(t)\). We will assume that the prey always has an adequate food supply and that its birth rate at any time is proportional to the number of prey alive at that time; that is, birth rate (prey) is \(k_{1} x_{1}(t)\). The death rate of the prey depends on both the number of prey and predators alive at that time. For simplicity, we assume death rate (prey) \(=k_{2} x_{1}(t) x_{2}(t)\). The birth rate of the predator, on the other hand, depends on its food supply, \(x_{1}(t)\), as well as on the number of predators available for reproduction purposes. For this reason, we assume that the birth rate (predator) is \(k_{3} x_{1}(t) x_{2}(t)\). The death rate of the predator will be taken as simply proportional to the number of predators alive at the time; that is, death rate (predator) \(=k_{4} x_{2}(t)\). Since \(x_{1}^{\prime}(t)\) and \(x_{2}^{\prime}(t)\) represent the change in the prey and predator populations, respectively, with respect to time, the problem is expressed by the system of nonlinear differential equations $$ x_{1}^{\prime}(t)=k_{1} x_{1}(t)-k_{2} x_{1}(t) x_{2}(t) \quad \text { and } x_{2}^{\prime}(t)=k_{3} x_{1}(t) x_{2}(t)-k_{4} x_{2}(t) $$ Solve this system for \(0 \leq t \leq 4\), assuming that the initial population of the prey is 1000 and of the predators is 500 and that the constants are \(k_{1}=3, k_{2}=0.002, k_{3}=0.0006\), and \(k_{4}=0.5\). Sketch a graph of the solutions to this problem, plotting both populations with time, and describe the physical phenomena represented. Is there a stable solution to this population model? If so, for what values \(x_{1}\) and \(x_{2}\) is the solution stable?

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