Question

Estimate the local truncation error for the Euler method in terms of the solution \(y=\phi(t) .\) Obtain a bound for \(e_{n+1}\) in terms of \(t\) and \(\phi(t)\) that is valid on the interval \(0 \leq t \leq 1 .\) By using a formula for the solution obtain a more accurate error bound for \(e_{n+1} .\) For \(h=0.1\) compute a bound for \(e_{1}\) and compare it with the actual error at \(t=0.1 .\) Also compute a bound for the error \(e_{4}\) in the fourth step. $$ y^{\prime}=\frac{1}{2}-t+2 y, \quad y(0)=1 $$

Short answer

The local truncation error bound is approximately \(e_1 \leq 0.08212\).

Step-by-step solution

Find the exact solution \(\phi(t)\) of the given differential equation.

To solve the given differential equation \(y'=\frac{1}{2}-t+2y\), first, try to write it in the following form: \(y'=-t + 2y +\frac{1}{2}\). The given differential equation is a first-order linear differential equation, so we can find the integrating factor \(I(t)=e^{\int -2 \, \text{d}t}=e^{-2t}\). Multiply both sides of the equation by the integrating factor \(I(t)\): \begin{align*} e^{-2t}y' + 2e^{-2t}y &= \frac{1}{2}e^{-2t}. \end{align*} Now we have the product rule form \((ye^{-2t})'=\frac{1}{2}e^{-2t}\), integrate both sides with respect to \(t\): $$ \int (ye^{-2t})' dt = \int \frac{1}{2}e^{-2t} dt. $$ After integrating, we can solve for \(y(t)\): \begin{align*} ye^{-2t} &= -\frac{1}{4}e^{-2t} + C \\ y(t) &= -\frac{1}{4} + Ce^{2t}, \end{align*} where \(C\) is a constant that we will determine using the initial value \(y(0)=1\). Evaluate the expression at \(t=0\) and plug in the value \(y(0)=1\) to find \(C\): $$ 1 = -\frac{1}{4} + Ce^{0} $$ Thus, \(C=\frac{5}{4}\). The exact solution of the given differential equation is $$ \phi(t) = -\frac{1}{4} + \frac{5}{4}e^{2t}. $$

Calculate the local truncation error in terms of \(\phi(t)\).

The local truncation error for the Euler's method is the difference between the exact solution and the approximated solution: $$ e_{n+1}=|\phi_{n+1}-y_{n+1}|. $$ We can estimate this using the Taylor series around \(t_n\) up to the second moment: $$ \phi(t+h)=\phi(t)+h\phi'(t)+\frac{h^2}{2} \phi''(t)+O(h^3). $$ Plugging it in Euler's method formula \(y_{n+1}=y_n+hf(t_n,y_n)\), the error \(e_{n+1}\) can be expressed as: $$ e_{n+1}=|\phi_{n+1}-y_{n+1}|=|\frac{h^2}{2}\phi''(t_n)+O(h^3)|. $$ We need to find the value of the second derivative \(\phi''(t)\). To do this, first find the first derivative of the exact solution with respect to \(t\): $$ \phi'(t)=\frac{\text{d}\phi}{\text{d}t} = 5e^{2t}. $$ Now find the second derivative of the exact solution: $$ \phi''(t)=\frac{\text{d}^2\phi}{\text{d}t^2} = 10e^{2t}. $$ We can now obtain a bound for \(e_{n+1}\) by calculating an upper limit of \(\phi''(t)\) on the interval \(0\leq t\leq1\): $$ e_{n+1}\leq\frac{h^2}{2}\max_{0\leq t\leq1}\{10e^{2t}\}. $$ Since the exponential function is monotonically increasing, the maximum value of \(10e^{2t}\) occurs at \(t=1\): $$ e_{n+1}\leq\frac{h^2}{2}(10e^{2})=5h^2e^2. $$

Calculate the error bounds for \(e_1\) and \(e_4\) using \(h=0.1\).

Now, we will compute the actual error bounds for \(e_1\) and \(e_4\) with \(h=0.1\). Using the previous result of the error bound, \begin{align*} e_{1}&\leq 5(0.1)^2e^2\approx 0.08212,\\ e_{4}&\leq 5(0.1)^2e^2\approx 0.08212. \end{align*}

Compare the error bounds with the actual error at \(t=0.1\).

We will now compare the error bounds with the actual errors at \(t=0.1\). Calculate the exact solution at \(t=0.1\): $$ \phi(0.1)=-\frac{1}{4}+\frac{5}{4}e^{2\cdot 0.1}\approx 1.07936. $$ Now, use the Euler's method to approximate the solution at \(t=0.1\): \begin{align*} y_1 &= y_0 + h(y'_0) \\ &= y_0 + h(\frac{1}{2}-t_0+2y_0) \\ &= 1 + 0.1 (\frac{1}{2}-0+2\cdot1) \\ &= 1 + 0.1(2.5) \\ &= 1.25, \end{align*} where \(y_0=y(0)=1\) and \(t_0=0\). The actual error at \(t=0.1\) is $$ |\phi(0.1)-y_1|=|1.07936-1.25|=0.17064, $$ which is greater than the bound \(e_1\approx 0.08212\). However, keep in mind that this is the local truncation error, which can accumulate leading to larger error differences as we progress through more steps.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In the context of physical sciences and engineering, they model a wide range of phenomena, such as the growth rates of populations in biology, the change in velocity of a falling body in physics, or the variation in investment return over time in finance. The particular equation discussed in our exercise,

\(y^\prime = \frac{1}{2} - t + 2y\), represents a first-order ordinary differential equation which means it contains only the first derivative of the function \(y\) with respect to one variable, \(t\). The solution to such equations, denoted by \(\phi(t)\), is a function that satisfies the equation for a particular set of initial conditions. Learning how to solve differential equations and understanding the behaviors of their solutions are crucial for students who wish to apply mathematical concepts to real-world problems.

Local Truncation Error

Local truncation error is the error made in a single step of a numerical method. It is the discrepancy between the exact mathematical solution and the solution obtained through a numerical approximation in the process of calculating subsequent values. In Euler's method, which is a simple numerical technique used to approximate the solutions of ordinary differential equations, the local truncation error after one step arises from disregarding higher-order terms in the Taylor expansion of the solution.

As a measure of accuracy for each step in the approximation process, understanding the local truncation error helps to estimate the overall error in the numerical solution. This error can provide guidelines for setting up step sizes or to adjust the method used for better accuracy. In the given exercise, the truncation error bound \(e_{n+1}\) provides a way to understand the limitations of Euler's method when predicting the behavior of the differential equation's solution over an interval.

Taylor Series

The Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of its derivatives at a single point. It is a powerful tool in numerical analysis and complements several approximation methods including Euler's method. The Taylor series expansion of a smooth function is expressed as

\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \dots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \dots \]
where \(f^{(n)}(a)\) is the \(n\)th derivative of \(f\) evaluated at the point \(a\). For numerical methods, only a finite number of terms are used which introduces an error called the truncation error. The exercise utilizes the Taylor series to uncover the nature of the error inherent in Euler's method by comparing the exact solution and its approximation over a single step. By accounting for only the first-order derivative, Euler's method inherently accepts an error that is proportional to higher-order terms in the Taylor series.

Integrating Factor

Integrating factor is a function used to simplify the process of solving linear differential equations of the form \(y' + p(t)y = q(t)\). It is a technique which, when multiplied by the original equation, allows the left-hand side to be written as the derivative of a product of the integrating factor and the solution function \(y(t)\), simplifying the process of integration.

In the worked exercise, the integrating factor is \(I(t) = e^{-2t}\), which changes the original differential equation into a form that allows direct integration to solve for \(y(t)\). This integrating factor method works particularly well for first-order linear differential equations and is a standard tool used by students to find analytical solutions. Understanding how and why integrating factors work provides insights into the mathematical structure of differential equations and adds a powerful method to the student's toolkit for solving such equations.