Question

A programmable calculator or computer is needed for these exercises. In Chapter 7, we will examine how the error in numerical algorithms, such as Euler's method, depends on step size \(h\). In this exercise, we further examine the dependence of errors on step size by studying a particular example, $$ y^{\prime}=y+1, \quad y(0)=0 \text {. } $$ (a) Use Euler's method to obtain approximate solutions to this initial value problem on the interval \(0 \leq t \leq 1\), using step sizes \(h_{1}=0.02\) and \(h_{2}=0.01\). You will therefore obtain two sets of points, $$ \begin{array}{ll} \left(t_{k}^{(1)}, y_{k}^{(1)}\right), & k=0, \ldots, 50 \\ \left(t_{k}^{(2)}, y_{k}^{(2)}\right), & k=0, \ldots, 100 \end{array} $$ where \(t_{k}^{(1)}=0.02 k, k=0,1, \ldots, 50\) and \(r_{k}^{2)}=0.01 k, k=0,1, \ldots .100\). (b) Determine the exact solution, \(y(t)\). (c) Print a table of the errors at the common points, \(t_{k}^{(1)}, k=0,1, \ldots, 50\) : $$ e^{(1)}\left(t_{k}^{(1)}\right)=y\left(t_{k}^{(1)}\right)-y_{k}^{(1)} \quad \text { and } \quad e^{(2)}\left(t_{k}^{(1)}\right)=y\left(t_{k}^{(1)}\right)-y_{2 k}^{(2)} . $$ (d) Note that the approximations \(y_{2 k}^{(2)}\) were found using a step size equal to one half of the step size used to obtain the approximations \(y_{k}^{(1)} ;\) that is, \(h_{2}=h_{1} / 2\). Compute the corresponding error ratios. In particular, compute $$ \left|\frac{e^{(2)}\left(t_{k}^{(1)}\right)}{e^{(1)}\left(t_{k}^{(1)}\right)}\right|, \quad k=1 \ldots . .50 . $$ On the basis of these computations, conjecture how halving the step size affects the error of Euler's method.

Short answer

A) \(e^{(1)}(t_k^{(1)}) = y(t_k^{(1)}) - y_k^{(1)}\) and \(e^{(2)}(t_k^{(1)}) = y(t_k^{(1)}) - y_{k}^{(2)}\) B) \(e^{(1)}(t_k^{(1)}) = e^t_k^{(1)} - y_k^{(1)}\) and \(e^{(2)}(t_k^{(1)}) = e^t_k^{(1)} - y_{2k}^{(2)}\) C) \(e^{(1)}(t_k^{(1)}) = y(t_k^{(1)}) - y_k^{(1)}\) and \(e^{(2)}(t_k^{(1)}) = y(t_k^{(1)}) - y_{2k}^{(2)}\) D) \(e^{(1)}(t_k^{(1)}) = e^t_k^{(1)} - y_k^{(1)}\) and \(e^{(2)}(t_k^{(1)}) = e^t_k^{(1)} - y_{k}^{(2)}\)

Step-by-step solution

Apply Euler's Method with step size \(h_1 = 0.02\)

First, we will implement Euler's method using the step size \(h_1 = 0.02\). We will compute the points \((t_k^{(1)}, y_k^{(1)})\) for \(k=0,1,...,50\), where \(t_k^{(1)}=0.02k\). The update rule for Euler's method, given \(y'(t) = y+1\) and \(y(0) = 0\), is: $$ y_{k+1}^{(1)} = y_{k}^{(1)} + h_1 (y_{k}^{(1)} + 1). $$

Apply Euler's Method with step size \(h_2 = 0.01\)

Next, we will implement Euler's method using the step size \(h_2 = 0.01\). We will compute the points \((t_k^{(2)}, y_k^{(2)})\) for \(k=0,1,...,100\), where \(t_k^{(2)}=0.01k\). The update rule for Euler's method, given \(y'(t) = y+1\) and \(y(0) = 0\), is: $$ y_{k+1}^{(2)} = y_{k}^{(2)} + h_2 (y_{k}^{(2)} + 1). $$

Determine the Exact Solution \(y(t)\)

In order to determine the exact solution of the given IVP, we will solve the following: $$ y'(t) = y + 1 \\ y(0) = 0 $$ Integrating the equation, we have: $$ y(t) = e^t -1 $$ which is the exact solution of the IVP.

Compute Errors at Common Points

Now we can compute the errors at the common points: $$ e^{(1)}(t_k^{(1)}) = y(t_k^{(1)}) - y_k^{(1)} \quad \text{and} \quad e^{(2)}(t_k^{(1)}) = y(t_k^{(1)}) - y_{2k}^{(2)} $$ for k = 0, 1, ..., 50.

Compute Error Ratios

To calculate the error ratios, we will compute $$ \left|\frac{e^{(2)}(t_k^{(1)})}{e^{(1)}(t_k^{(1)})}\right| $$ for k = 1, ..., 50, where \(e^{(1)}\) and \(e^{(2)}\) are the errors from Euler's method with step sizes \(h_1\) and \(h_2\), respectively. After calculating these ratios, we can form a conjecture about how halving the step size affects the error of Euler's method.

Key concepts

Differential Equations

Differential equations are mathematical equations that describe relationships involving rates at which variables change. These are fundamental in expressing natural phenomena, where changing conditions often depend on the current state. The equation provided in the exercise, \(y'=y+1\), is a first-order differential equation where the rate of change of the variable \(y\) with respect to another variable, \(t\), is proportional to \(y\) itself, plus a constant. Solving differential equations involves finding a function \(y(t)\) that satisfies the given relationship for all \(t\).

The exact solution computation, as in the exercise, reveals the precise relationship and is often sought after for a deeper understanding of the system being studied. However, not all differential equations can be solved exactly, and numerical methods like Euler's method come into play to approximate solutions.

Numerical Algorithms

Numerical algorithms are computational procedures designed to perform approximate calculations that would be difficult or impossible to do exactly. These algorithms provide a way to solve complex problems that have no closed-form solution or where such solutions are too cumbersome to calculate by hand. Euler's method, featured in this exercise, is a numerical algorithm commonly used to approximate solutions to initial value problems for differential equations. It works by taking small steps from one estimate to the next using the slope given by the differential equation. Calculations with numerical algorithms are iterative and often require a balance between the desired accuracy and computational resources.

Initial Value Problems

An initial value problem (IVP) in differential equations is a problem where one is given the rate of change of a variable and the value of that variable at a starting point, the initial value. The challenge is to determine the variable's behavior over time, starting from this initial state. In the exercise \(y'=y+1, y(0)=0\), \(y(0)=0\) is our initial condition. This information, combined with the rate of change provided by the differential equation, allows us to use numerical methods like Euler's method to approximate the value of \(y\) at subsequent points.

Exact Solution Computation

Exact solution computation refers to the process of finding an analytical solution to a differential equation, representing the exact relationship between variables. Unlike numerical approximations, exact solutions are exact and do not suffer from errors due to discretization. For instance, in the provided exercise, the exact solution to the IVP \(y'=y+1\) with \(y(0)=0\) is \(y(t)=e^t-1\). Knowing the exact solution is highly valuable for error analysis because it serves as a benchmark against which the accuracy of numerical approximations can be measured.

Error Analysis

Error analysis in numerical algorithms is the study of how and why computational errors occur and their effects on results. In practice, errors are inevitable due to factors like rounding and truncation. Understanding error helps refine algorithms for better precision and reliability. As instructed in the exercise, one can calculate the error by comparing the numerical solution at specific points to the exact solution. The exercise goes further to analyze how the error changes when the step size \(h\) is halved, which is a form of sensitivity analysis, showing how error is affected by the algorithm's parameters.