Problem 13 - Advanced Engineering Mathematics

Question

Consider the initial-value problem \(y^{\prime}=2 y, y(0)=1\). The analytic solution is \(y=e^{2 x}\). (a) Approximate \(y(0.1)\) using one step and Euler's method. (b) Find a bound for the local truncation error in \(y_{1}\). (c) Compare the actual error in \(y_{1}\) with your error bound. (d) Approximate \(y(0.1)\) using two steps and Euler's method. (e) Verify that the global truncation error for Euler's method is \(O(h)\) by comparing the errors in parts (a) and (d).

Step-by-Step Solution

Answer

(a) 1.2; (b) ≤0.02; (c) Actual error 0.0214, near bound; (d) 1.21; (e) Error ~O(h).

Step 1: Understanding Euler's Method for One Step

Euler's method approximates the solution of a differential equation by taking steps of a fixed size. To approximate the solution, calculate: \( y_1 = y_0 + h f(x_0, y_0) \), where \( h \) is the step size, \( f(x_0, y_0) = y' = 2y \), \( y_0 = 1 \), and \( h = 0.1 \).

Step 2: Calculate the Approximation with One Step

Using Euler's formula: \( y_1 = 1 + 0.1 \times 2 \times 1 = 1.2 \). Therefore, \( y(0.1) \) is approximately 1.2 using one step of Euler's method.

Step 3: Determine the Local Truncation Error Bound

The local truncation error in Euler's method is \( E_{ ext{local}} \leq \frac{M}{2}h^2 \), where \( M \) is a bound on \( |y''| \) on the interval. For \( y'' = 4e^{2x} \), taking \( M = 4 \) gives \( E_{ ext{local}} \leq 0.02 \).

Step 4: Compare the Actual Error with the Error Bound

The actual \( y(0.1) = e^{0.2} \approx 1.2214 \). The actual error \( |1.2214 - 1.2| = 0.0214 \) which is slightly above the error bound, demonstrating the approximation's tightness.

Step 5: Apply Euler's Method with Two Steps

Using two steps of size 0.05, calculate: 1. \( y_1 = 1 + 0.05 \times 2 \times 1 = 1.1 \)2. \( y_2 = 1.1 + 0.05 \times 2 \times 1.1 = 1.21 \).Therefore, \( y(0.1) \) is approximately 1.21 with two steps.

Step 6: Verify Global Truncation Error is O(h)

The error for one step \( |1.2214 - 1.2| \approx 0.0214 \) compared to the error for two steps \( |1.2214 - 1.21| \approx 0.0114 \) suggests the error approximately halves when the step size is halved, confirming it is \( O(h) \).

Key Concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with specific initial conditions. In simpler terms, it tells you what value a function starts with and how it changes. Imagine you have a plant that starts at 5 cm tall. The initial height (5 cm) is the initial value, and the rate at which it grows resembles the differential equation.
In our exercise, the initial value problem is given by the differential equation \( y' = 2y \) with an initial condition \( y(0) = 1 \). Here, \( y' \) represents the rate of change of \( y \), and initially, \( y \) is equal to 1 when \( x \) is zero. This sets the stage for predicting how \( y \) changes over time.

Local Truncation Error

The local truncation error is a type of error that helps us understand how much mistake we make in one step of a numerical method. Think of it as the "oops" you make each time you take a step and calculate. It's natural because when we use methods like Euler’s to approximate solutions, we tend to skip some tiny important details.
In Euler's method, for example, the local truncation error for one step is calculated using the formula \( E_{\text{local}} \leq \frac{M}{2}h^2 \). Here, \( M \) is a bound on the second derivative \(|y''|\), and \( h \) is the step size. So, in our exercise, with \( M = 4 \) and \( h = 0.1 \), the local truncation error was calculated to be \( 0.02 \). This means each step might be off by at most 0.02.

Global Truncation Error

While the local error is about mistakes in one step, the global truncation error considers the error accumulated over multiple steps. Think of it as adding up all your "oops" moments after finishing a race.
In Euler's method, the global truncation error is typically \( O(h) \), meaning it gets smaller as the step size gets smaller. This is a very important property because it tells us that the error is proportional to the size of our steps.

• In the exercise, using one step, the error was about 0.0214.
• With two steps of half the size, the error decreased to about 0.0114.

This shows that as we reduce the step size, the global error improves, supporting the theoretical error estimate of \( O(h) \).

Numerical Approximation

Numerical approximation is a powerful mathematical technique used to find approximate solutions to problems where an exact solution might be hard to get. Instead of solving complex equations exactly, we use simple, repeated calculations to get as close as possible to the real answer.
Euler's method is a basic example of numerical approximation. It uses known values to estimate unknown values of a differential equation through small steps in the desired direction. In the exercise, Euler's method was employed to approximate \( y(0.1) \).

• First approximation (one step): Resulted in \( y \approx 1.2 \).
• Second approximation (two steps): Resulted in \( y \approx 1.21 \).

These estimates show how such methods can be useful, especially when exact solutions are either difficult or impossible to find with ordinary calculations.