Consider the following ODE, which is solved using a numerical method.

dy

dx
=y4+3x2+log(y)

Let

Eeuler be the global truncation error of the Forward Euler method when solving the equation above
Emidpoint be the global truncation error of the midpoint method when solving the equation above
EHeun be the global truncation error of the Heun method when solving the equation above
ERK4 be the global truncation error of the RK4 method when solving the equation above

Which of the following inequalities is FALSE?

Question

Consider the following ODE, which is solved using a numerical method.

dy

dx
=y4+3x2+log(y)

Let

Eeuler be the global truncation error of the Forward Euler method when solving the equation above
Emidpoint be the global truncation error of the midpoint method when solving the equation above
EHeun be the global truncation error of the Heun method when solving the equation above
ERK4 be the global truncation error of the RK4 method when solving the equation above

Which of the following inequalities is FALSE?

BlackTom AI · Accepted Answer

To tackle the question, first recall the general behavior of global truncation error for classical time-stepping methods when solving an ODE numerically. The Forward Euler method is first-order accurate, so its global error Eeuler scales like O(h). The Midpoint method (a second-order Runge–Kutta method) has a global error Emidpoint that scales like O(h^2). The Heun method (also a second-order RK method) similarly yields EHeun that scales like O(h^2). The RK4 method (classical fourth-order Runge–Kutta) has a global error ERK4 that scales like O(h^4). Given a fixed step size h, higher-order methods produce smaller errors in general, so their errors satisfy ERK4 < EHeun and ERK4 < Emidpoint, and all of these are less than Eeuler (ERK4 < Emidpoint < Eeuler, for sufficiently small h). Option analysis: - Option: ERK4 > EHeun This statement claims that the RK4 method has a larger global error than the Heun method. That contradicts the standard order arguments, since RK4 is a higher-order method with ERK4 ~ C4 h^4 while EHeun ~ C2 h^2 for small h, so ERK4 should be smaller, not larger. Therefore this option is false. In clearer terms, the expected inequality sequence for small step size h is ERK4 < EHeun, and similarly ERK4 < Emidpoint and ERK4 < Eeuler. Saying ERK4 > EHeun would imply RK4 is less accurate than RK2, which contradicts their respective truncation error orders.

类似问题

When employing a computational method to solve an Ordinary Differential Equation (ODE), the choice of a smaller time step will result in

When employing a computational method to solve an Ordinary Differential Equation (ODE), the choice of a smaller time step will result in

A project has a required return of 12.6 percent, an initial cash outflow of $42,100, and cash inflows of $16,500 in Year 1, $11,700 in Year 2, and $10,400 in Year 4. What is the net present value?

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单