Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 8: Problem 20

Carry out one step of the Euler method and of the improved Euler method using the step size \(h=0.1 .\) Suppose that a local truncation error no greater than 0.0025 is required. Estimate the step size that is needed for the Euler method to satisfy this requirement at the first step. $$ y^{\prime}=\sqrt{t+y}, \quad y(0)=3 $$

Short Answer

Expert verified
Answer: After one step, the Euler's method gives an approximate value of \(y_1 \approx 3.173\), and the Improved Euler's method gives an approximate value of \(y_1 \approx 3.164\). To achieve a local truncation error no greater than 0.0025 at the first step, the estimated step size required for the Euler method is approximately \(h' \approx 0.0037\).

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards
Start your free trial Skip

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Understand the Euler Method Formula and Steps

To perform one step of the Euler Method, use the following formula: $$y_{n+1} = y_n + h \, f(t_n, y_n)$$ Here, \(f(t_n, y_n) = y'(t_n) = \sqrt{t_n+y_n}\), and we have the given step size \(h=0.1\).
02

Use the Euler Method to Find the First Step Solution

Using the initial condition \(y(0) = 3\), we can calculate \(y_1\) as follows: $$y_1 = y_0 + h \, f(t_0, y_0) = 3 + 0.1 \, \sqrt{0 + 3} = 3 + 0.1 \times \sqrt{3}$$ Calculate the value: $$y_1 = 3 + 0.1 \times \sqrt{3} ≈ 3.173$$
03

Understand the Improved Euler Method Formula and Steps

To perform one step of the Improved Euler Method, use the following formula: $$y_{n+1} = y_n + \frac{h}{2} \left( f(t_n, y_n) + f(t_{n+1}, y_n + h \, f(t_n, y_n)) \right)$$ Here, \(f(t_n, y_n) = y'(t_n)=\sqrt{t_n+y_n}\) and we have the given step size \(h=0.1\).
04

Use the Improved Euler Method to Find the First Step Solution

Using the initial condition \(y(0) = 3\), we can calculate \(y_1\) as follows: $$y_1 = y_0 + \frac{h}{2} \left( f(t_0, y_0) + f(t_1, y_0 + h \, f(t_0, y_0)) \right)$$ $$y_1 = 3 + \frac{0.1}{2} \left( \sqrt{0 + 3} + \sqrt{0.1 + 3 + 0.1 \times \sqrt{3}} \right)$$ Calculate the value: $$y_1 ≈ 3 + \frac{0.1}{2} \left( \sqrt{3} + \sqrt{3.3} \right) ≈ 3.164$$
05

Estimate the Step Size Required to Achieve the Given Local Truncation Error

We want to find the step size \(h'\) such that the local truncation error for the Euler method is no greater than 0.0025. The local truncation error for the Euler method is approximately equal to the product of the step size (\(h'\)) and the second-order derivative evaluated at the initial point \((t_0, y_0)\): $$|h' \cdot y''(t_0, y_0)| \leq 0.0025$$ First, we need to find the second-order derivative of y with respect to t: $$y' = \sqrt{t+y}$$ $$y'' = \frac{1}{2 \sqrt{t+y}} (1 + y')$$ Since \(t_0=0\) and \(y_0=3\), and we found that \(y'(t_0) = \sqrt{t_0 + y_0} = \sqrt{3}\), we can now calculate \(y''(t_0, y_0)\): $$y''(t_0, y_0) = \frac{1}{2 \sqrt{3}} (1 + \sqrt{3})$$ Now, substitute into the inequality: $$|h' \cdot \frac{1}{2 \sqrt{3}} (1 + \sqrt{3})| \leq 0.0025$$ $$h' \geq \frac{0.0025 \cdot 2 \sqrt{3}}{1 + \sqrt{3}}$$ Calculate the value: $$h' ≈ 0.0037$$ So, the step size required for the Euler method to satisfy the local truncation error requirement of no greater than 0.0025 at the first step is approximately 0.0037.

Key Concepts

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

Differential Equations
Differential equations are at the heart of much of modern science and engineering. They describe how a quantity changes in space and time, often in relation to its current state. For example, the equation
\[y' = \/(t + y), \quad y(0)=3\]
expresses a rate of change (\(y'\)) that's dependent on both the independent variable \(t\) and the unknown function \(y(t)\). This particular equation might model phenomena like population growth, chemical reactions, or motion under a force. Solving such an equation analytically can be challenging or even impossible in many cases, hence numerical methods like the Euler method are employed to find approximate solutions.
Local Truncation Error
When we approximate the solution of differential equations using numerical methods, we introduce a small error with each step we take. This is called the local truncation error, and it's a measure of how much the numerical solution deviates from the true solution at a single step, assuming that all previous steps were exact. In the context of the Euler method, the local truncation error after one step is proportional to the square of the step size (
\(h^2\)) times the second derivative of the true solution, as it is a first-order method. Reducing the step size \(h\) decreases this error, but at the cost of needing more computational steps. Our aim is usually to balance accuracy with computational efficiency.
Improved Euler Method
Also known as Heun's method, the Improved Euler Method is a simple and commonly used modification of the Euler method that improves accuracy by using an average of the slopes at the beginning and end of the interval. The formula for the Improved Euler Method is:
\[y_{n+1} = y_n + \frac{h}{2} (f(t_n, y_n) + f(t_{n+1}, y_n + h f(t_n, y_n)))\]
Notice how this incorporates an additional term compared to the standard Euler method. By taking into account the slope at the end of the interval as well as at the beginning, it can better approximate the true curve of the solution. This results in a smaller local truncation error for each step than the regular Euler method, leading to a more accurate solution over many steps.
Numerical Methods
Numerical methods are algorithms used for approximating the solutions to mathematical problems that cannot be solved exactly (analytically). They are essential in fields such as physics, engineering, and finance where precise analytical solutions are often unobtainable. Numerical methods for differential equations, such as the Euler method, the Improved Euler method, and others like the Runge-Kutta methods, transform the differential equations into algebraic equations which are solvable using standard computational techniques. One of the key goals in these methods is to minimize errors while maximizing computational efficiency.
Initial Value Problem
Our discussion centers on a specific type of differential equation problem known as an initial value problem (IVP). An IVP involves finding a function \(y(t)\) that satisfies a given differential equation and who also meets an initial condition, for example, \(y(0)=3\). The 'initial value' is the starting point, from which we use numerical methods to step through and predict subsequent values. It's essential in ensuring that the unique solution we are computing is correct for the specific scenario we're analyzing.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the initial value problem \(x^{\prime}=f(t, x, y)\) and \(y^{\prime}=g(t, x, y)\) with \(x\left(t_{0}\right)=x_{0}\) and \(y\left(t_{0}\right)=y_{0} .\) The generalization of the Adams-Moulton predictor-corrector method of Section 8.4 is $$ \begin{array}{l}{x_{n+1}=x_{n}+\frac{1}{24} h\left(55 f_{n}-59 f_{n-1}+37 f_{n-2}-9 f_{n-3}\right)} \\ {y_{n+1}=y_{n}+\frac{1}{24} h\left(55 g_{n}-59 g_{n-1}+37 g_{n-2}-9 g_{n-1}\right)}\end{array} $$ and $$ \begin{array}{l}{x_{n+1}=x_{n}+\frac{1}{24} h\left(9 f_{n+1}+19 f_{n}-5 f_{n-1}+f_{n-2}\right)} \\ {y_{n+1}=y_{n}+\frac{1}{24} h\left(9 g_{n+1}+19 g_{n}-5 g_{n-1}+g_{n-2}\right)}\end{array} $$ Determine an approximate value of the solution at \(t=0.4\) for the example initial value problem \(x^{\prime}=x-4 y, y^{\prime}=-x+y\) with \(x(0)=1, y(0)=0\). Take \(h=0.1 .\) Correct the predicted value once. For the values of \(x_{1}, \ldots, y_{3}\) use the values of the exact solution rounded to six digits: \(x_{1}=1.12883, x_{2}=1.32042, x_{3}=1.60021, y_{1}=-0.110527, y_{2}=\) \(-0.250847,\) and \(y_{3}=-0.429696\)

Show that the modified Euler formula of Problem 22 is identical to the improved Euler formula of \(\mathrm{Eq} .(5)\) for \(y^{\prime}=f(t, y)\) if \(f\) is linear in both \(t\) and \(y .\)

Determine an approximate value of the solution at \(t=0.4\) and \(t=0.5\) using the specified method. For starting values use the values given by the Runge- Kutta method; see Problems 1 through 6 of Section 8.3 . Compare the results of the various methods with each other and with the actual solution (if available). $$ \begin{array}{l}{\text { (a) Use the fourth order predictor-corrector method with } h=0.1 . \text { Use the corrector }} \\ {\text { formula once at each step. }} \\ {\text { (b) Use the fourth order Adams-Moulton method with } h=0.1} \\ {\text { (c) Use the fourth order backward differentiation method with } h=0.1 .}\end{array} $$ $$ y^{\prime}=\left(t^{2}-y^{2}\right) \sin y, \quad y(0)=-1 $$

The modified Euler formula for the initial value problem \(y^{\prime}=f(t, y), y\left(t_{0}\right)=y_{0}\) is given by $$ y_{n+1}=y_{n}+h f\left[t_{n}+\frac{1}{2} h, y_{n}+\frac{1}{2} h f\left(t_{n}, y_{n}\right)\right] $$ Following the procedure outlined in Problem \(14,\) show that the local truncation error in the modified Euler formula is proportional to \(h^{3}\).

Consider the initial value problem $$ y^{\prime}=3 t^{2} /\left(3 y^{2}-4\right), \quad y(0)=0 $$ (a) Draw a direction field for this equation. (b) Estimate how far the solution can extended to the right. Let \(t_{1 / 4}\) be the right endpoint of the interval of existence of this solution. What happens at \(t_{M}\) to prevent the solution from continuing farther? (c) Use the Runge-Kutta method with various step sizes to determine an approximate value of \(t_{M}\). (d) If you continue the computation beyond \(t_{\mathrm{H}}\), you can continue to generate values of \(y .\) What significance, if any, do these values have? (e) Suppose that the initial condition is changed to \(y(0)=1 .\) Repeat parts (b) and (c) for this problem.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free