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Show that the fourth-order Runge-Kutta method, $$ \begin{aligned} k_{1} &=h f\left(t_{i}, w_{i}\right) \\ k_{2} &=h f\left(t_{i}+h / 2, w_{i}+k_{1} / 2\right) \\ k_{3} &=h f\left(t_{i}+h / 2, w_{i}+k_{2} / 2\right) \\ k_{4} &=h f\left(t_{i}+h, w_{i}+k_{3}\right) \\ w_{i+1} &=w_{i}+\frac{1}{6}\left(k_{1}+2 k_{2}+2 k_{3}+k_{4}\right) \end{aligned} $$ when applied to the differential equation \(y^{\prime}=\lambda y\), can be written in the form $$ w_{i+1}=\left(1+h \lambda+\frac{1}{2}(h \lambda)^{2}+\frac{1}{6}(h \lambda)^{3}+\frac{1}{24}(h \lambda)^{4}\right) w_{i}. $$

Short Answer

Expert verified
The process involves computing the values of \( k_1, k_2, k_3\), and \( k_4 \) as per the Runge-Kutta method's system and substituting them into the overall formula for the solution \( w_{i+1} \). After expanding and simplifying the result, it leads to the desired formula.

Step by step solution

01

Apply the initial formula for \(k_1, k_2, k_3, k_4\)

Apply the given Runge-Kutta method formulas:\n\(k_{1} = h f(t_{i}, w_{i}) = h \lambda w_{i} \)\n\(k_{2} = h f(t_{i} + h/2, w_{i}+k_{1} / 2) = h \lambda (w_i + \frac{h \lambda w_i}{2}) = h \lambda w_i ( 1 + \frac{h \lambda }{2})\)\n\(k_{3} = h f(t_{i} + h/2, w_{i} + k_{2} / 2) = h \lambda (w_i + \frac{h \lambda w_i ( 1 + \frac{h \lambda }{2})}{2}) = h \lambda w_i(1 + \frac{h \lambda}{2} + \frac{(h \lambda)^2}{4})\)\n\(k_{4} = h f(t_{i}+h, w_{i} + k_{3} = h \lambda ( w_i + h \lambda w_i(1 + \frac{h \lambda}{2} + \frac{(h \lambda)^2}{4})) = h \lambda w_i(1 + h\lambda + \frac{(h\lambda)^2}{2} + \frac{(h\lambda)^3}{6})\)
02

Calculate \(w_{i+1}\)

Apply: \(w_{i+1} = w_{i} + \frac{1}{6}(k_{1}+2 k_{2}+2 k_{3}+k_{4})\) which simplifies to:\n \( w_{i+1} = w_{i} + \frac{1}{6}\left[ h \lambda w_i+2h \lambda w_i(1 + \frac{h \lambda}{2})+2h \lambda w_i(1 + \frac{h\lambda}{2} + \frac{(h\lambda)^2}{4}) + h \lambda w_i(1 + h \lambda + \frac{(h\lambda)^2}{2} + \frac{(h\lambda)^3}{6}) \right] \)
03

Simplify the result

Simplify the resulting expression for \(w_{i+1}\) to obtain: \(w_{i+1} = (1+ h \lambda + \frac{1}{2}(h \lambda)^2 + \frac{1}{6}(h \lambda)^3 + \frac{1}{24}(h \lambda)^4) w_{i}\) which is the desired result.

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Key Concepts

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

Numerical Analysis
Numerical analysis is a branch of mathematics that develops and analyzes methods to solve mathematical problems approximately. It's often coupled with computer algorithms to estimate the solutions of complex mathematical problems that may not be solvable analytically. One of its primary goals is to achieve accurate results through approximations when exact solutions are not feasible. This field is essential in various scientific and engineering applications, where precise mathematical modeling is necessary.
Natalie uses numerical analysis techniques like the Runge-Kutta methods to approximate solutions to differential equations. These methods are crucial when dealing with real-world problems in sectors like physics or finance where equations don't have simple closed-form solutions. Among the different Runge-Kutta methods, the fourth-order method is particularly renowned for its balance between computational simplicity and accuracy.
Differential Equations
Differential equations are mathematical equations that involve derivatives of a function. They describe how a particular quantity changes over time or space, and they are crucial in modeling various natural phenomena such as physics, engineering, and biology. The simple differential equation given in the exercise, \(y' = \lambda y\), is a linear first-order differential equation that represents exponential growth or decay depending on the sign of \(\lambda\).
Solving these using analytical methods might be straightforward for simple forms but can get complicated quickly. The Runge-Kutta methods provide robust numerical techniques to approximate solutions for these equations, especially useful when dealing with more complicated forms not easily solved by traditional means.
Stability Analysis
Stability analysis refers to determining whether an algorithm or a numerical solution will remain consistent over numerous iterations. It's a critical aspect when solving differential equations numerically because it ensures that small changes in initial values or computation errors don't lead to significant errors in the solution.
With the fourth-order Runge-Kutta method, stability is an essential concern because it dictates how the error propagates with each step. For the given differential equation, \(y' = \lambda y\), the method shows stability when the step size, \(h\), and the parameter \(\lambda\) are chosen appropriately. This is highlighted in the resulting expression which outlines how the weights of different \(h \lambda\) powers contribute to the solution stability. Proper stability analysis ensures your numerical solution doesn't suddenly diverge but instead maintains accuracy over time.
Error Analysis
Error analysis in numerical methods involves studying the accuracy of an approximate solution by understanding the nature and source of errors. The goal is to minimize these errors to achieve a solution that closely matches the exact answer.
In applying the fourth-order Runge-Kutta method, error analysis is vital to ensure that the approximation is close to the true solution as much as possible. The explicit step in the algorithm attempts to mitigate the local truncation error, which arises from integrating over a finite time step, \(h\). The sophistication of the Runge-Kutta method lies in its ability to achieve high accuracy by systematically balancing the weights in the evaluation of \(k_1, k_2, k_3,\) and \(k_4\). Understanding how these errors affect your solution is crucial, particularly in long-term computations, so that you can choose the appropriate method, time step, and parameters for your numerical solver.

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Most popular questions from this chapter

Consider the differential equation $$ y^{\prime}=f(t, y), \quad a \leq t \leq b, \quad y(a)=\alpha $$ a. Show that $$ y^{\prime}\left(t_{i}\right)=\frac{-3 y\left(t_{i}\right)+4 y\left(t_{i+1}\right)-y\left(t_{i+2}\right)}{2 h}+\frac{h^{2}}{3} y^{\prime \prime \prime}\left(\xi_{1}\right) $$ for some \(\xi\), where \(t_{i}<\xi_{i}

In a book entitled Looking at History Through Mathematics, Rashevsky [Ra], pp. 103-110, considers a model for a problem involving the production of nonconformists in society. Suppose that a society has a population of \(x(t)\) individuals at time \(t\), in years, and that all nonconformists who mate with other nonconformists have offspring who are also nonconformists, while a fixed proportion \(r\) of all other offspring are also nonconformist. If the birth and death rates for all individuals are assumed to be the constants \(b\) and \(d\), respectively, and if conformists and nonconformists mate at random, the problem can be expressed by the differential equations $$ \frac{d x(t)}{d t}=(b-d) x(t) \quad \text { and } \quad \frac{d x_{n}(t)}{d t}=(b-d) x_{n}(t)+r b\left(x(t)-x_{n}(t)\right) $$ where \(x_{n}(t)\) denotes the number of nonconformists in the population at time \(t\). a. Suppose the variable \(p(t)=x_{n}(t) / x(t)\) is introduced to represent the proportion of nonconformists in the society at time \(t\). Show that these equations can be combined and simplified to the single differential equation $$ \frac{d p(t)}{d t}=r b(1-p(t)) $$ b. Assuming that \(p(0)=0.01, b=0.02, d=0.015\), and \(r=0.1\), approximate the solution \(p(t)\) from \(t=0\) to \(t=50\) when the step size is \(h=1\) year. c. Solve the differential equation for \(p(t)\) exactly, and compare your result in part (b) when \(t=50\) with the exact value at that time.

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 \(1 \mathrm{~min}\).

Solve the following stiff initial-value problems using Euler's method, and compare the results with the actual solution. a. \(\quad y^{\prime}=-5 y+6 e^{t}, \quad 0 \leq t \leq 1, \quad y(0)=2\), with \(h=0.1\); actual solution \(y(t)=e^{-5 t}+e^{t}\). b. \(\quad y^{\prime}=-10 y+10 t+1, \quad 0 \leq t \leq 1, \quad y(0)=e\), with \(h=0.1\); actual solution \(y(t)=e^{-10 t+1}+t\). c. \(y^{\prime}=-15\left(y-t^{-3}\right)-3 / t^{4}, \quad 1 \leq t \leq 3, \quad y(1)=0\), with \(h=0.25 ;\) actual solution \(y(t)=-e^{-15 t}+t^{-3}\) d. \(y^{\prime}=-20 y+20 \cos t-\sin t, \quad 0 \leq t \leq 2, \quad y(0)=0\), with \(h=0.25 ;\) actual solution \(y(t)=-e^{-20 t}+\cos t\).

Derive the Adams-Moulton Two-Step method and its local truncation error by using an appropriate form of an interpolating polynomial.

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