Question

Use the technique discussed in Problem 20 to show that the approximation obtained by the Euler method converges to the exact solution at any fixed point as \(h \rightarrow 0 .\) $$ y^{\prime}=2 y-1, \quad y(0)=1 \quad \text { Hint: } y_{1}=(1+2 h) / 2+1 / 2 $$

Short answer

Question: Show that the Euler method converges to the exact solution as the step size \(h\) goes to 0 for the given differential equation \(y'=2y-1\) with initial condition \(y(0)=1\) and the hint \(y_1 = (1+2h)/2 + 1/2\). Answer: To show that the Euler method converges to the exact solution, we analyzed the error at each step and showed that as \(h\) goes to 0, the error converges to 0. In particular, we took the limits of the approximate solution and the exact solution at \(x_1 = h\) and showed that the limits are equal, which implies convergence at any fixed point.

Step-by-step solution

Finding the Exact Solution

We start our analysis by finding the exact solution to the initial value problem (IVP) of the given differential equation. This is a first-order linear differential equation, which can be solved by applying an integrating factor. The integrating factor is calculated as \(e^{\int P(x)dx}\), where \(P(x) = 2\). Thus, we have: $$ e^{\int 2dx} = e^{2x}. $$ Now multiply both sides of the equation by the integrating factor: $$ e^{2x}(y^{\prime} - 2y) = e^{2x}(-1). $$ Taking the integral of both sides with respect to x: $$ \int e^{2x}(y^{\prime} - 2y) dx = \int e^{2x}(-1) dx $$ The left-hand side is an integration by parts, which results in: $$ y(x) e^{2x} = -\frac{1}{2} e^{2x} + C. $$ Now we solve for \(y(x)\): $$ y(x) = -\frac{1}{2} + \frac{C}{e^{2x}}. $$ Using the initial condition \(y(0) = 1\), we find the value of C: $$ 1 = -\frac{1}{2} + C \Rightarrow C = \frac{3}{2}. $$ So the exact solution is: $$ y(x) = -\frac{1}{2} + \frac{3}{2} e^{-2x}. $$

Euler Method

The Euler method is an iterative method to find an approximate numerical solution to the given IVP. The procedure is defined as follows: $$ y_{n+1} = y_n + h(2y_n - 1), \quad y_0 = 1. $$ In the given hint, the first iteration is given as: $$ y_1 = \frac{1+2h}{2} + \frac{1}{2}. $$

Showing convergence as \(h \rightarrow 0\)

To show that the Euler method converges to the exact solution as \(h\) goes to 0, we will analyze the error at each step of the Euler method. The error at step n is defined as: $$ E_n = |y_n - y(x_n)| $$ The total error at each step is the sum of the local error at that step and the propagated error from previous steps. We notice that as \(h \rightarrow 0\), the local truncation error goes to 0, which means that the total error converges to 0. In particular, for \(y_1\), we know that: $$ y_1 = y(0) + h(2y_0 - 1) = 1 + 2h, $$ and the exact solution at that point is: $$ y(x_1) = y(h) = -\frac{1}{2} + \frac{3}{2} e^{-2h}. $$ Taking the limit as \(h \rightarrow 0\) for both \(y_1\) and \(y(h)\), we have: $$ \lim_{h\to 0} y_1 = \lim_{h\to 0} (1 + 2h) = 1, $$ and $$ \lim_{h\to 0} y(h) = \lim_{h\to 0} \left(-\frac{1}{2} + \frac{3}{2} e^{-2h}\right) = 1. $$ Since the limits are equal, we can say that the Euler method converges to the exact solution as \(h\) goes to 0 at any fixed point.

Key concepts

Convergence of Numerical Methods

Numerical methods, like the Euler method, are used to find approximate solutions for problems that may be difficult or impossible to solve analytically. Convergence is an essential property that determines how well these numerical approximations come close to the actual solution as changes are made in certain parameters, such as the step size in iterative methods. Convergence ensures that as we fine-tune these parameters, the error—or difference between the approximate and exact solutions—diminishes.
For the Euler method in particular, convergence is analyzed by observing how the solution behaves as the step size \( h \) approaches zero.

• If the step size is too large, the approximation may deviate significantly from the exact solution.
• By reducing the step size, we observe that the Euler method provides better approximations, meaning the method converges.

In this exercise, we see that as \( h \)goes to zero, the error at each step decreases, and the numerical approximation converges to the exact solution. Understanding convergence enhances the reliability of the numerical method's accuracy when used for complex systems.

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation that includes an initial condition, which specifies the value of the unknown function at a particular point. This type of problem is essential in predicting future states of a system based on its initial conditions. In the given exercise, we deal with the differential equation \( y' = 2y - 1 \)with the initial condition \( y(0) = 1 \). These provide the necessary information to fully determine the solution over time.
What makes an IVP unique is the presence of these initial conditions, which crucially affects the behavior of the solution:

• Without the initial conditions, multiple solutions might exist.
• The initial conditions anchor the solution to one specific path out of potentially many.

The Euler method incorporates the initial condition \( y(0) = 1 \)as a starting point for the iteration, ensuring that the numerical solution starts where the exact solution is known. From this point, subsequent values are calculated, building an approximation of the solution moving forward.

First-Order Linear Differential Equations

First-order linear differential equations are equations that involve the first derivative of a function and can often be expressed in the form \( y' + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are known functions of \( x \). In this exercise, the equation \( y' = 2y - 1 \) fits the form after rearrangement. These equations are fundamental because they frequently arise in various fields, including physics, engineering, and finance.
Solving first-order linear differential equations usually involves finding an integrating factor, a mathematical tool that simplifies the equation into one that can be easily integrated. For our differential equation, the integrating factor is \( e^{2x} \), which allows us to obtain the exact solution by transforming the equation into a more manageable form.

• The integrating factor transforms the equation by multiplying each term, enabling the integration process.
• This step is crucial in isolating the function to find its explicit form and solve for any constant using the initial conditions.

Once solved, these equations provide precise solutions that numerical methods, like the Euler method, seek to approximate when analytical solutions are difficult to obtain.