Question

Reducing the Step Size These exercises examine graphically the effects of reducing step size on the accuracy of the numerical solution. A computer or programmable calculator is needed. (a) Use Euler's method to obtain numerical solutions on the specified time interval for step sizes \(h=0.1, h=0.05\), and \(h=0.025\). (b) Solve the problem analytically and plot the exact solution and the three numerical solutions on a single graph. Does the error appear to be getting smaller as \(h\) is reduced? \(y^{\prime}=y+e^{-t}, \quad y(0)=0, \quad 0 \leq t \leq 1\)

Short answer

Question: Observe and compare the numerical solutions obtained using Euler's method for step sizes \(h=0.1, h=0.05\), and \(h=0.025\) with the analytical solution for the ODE \(y^{\prime}=y+e^{-t}\), and comment on the accuracy of these numerical solutions.

Step-by-step solution

Euler's Method for given Step Sizes

In order to solve the ODE using Euler's method, we will apply the formula: \(y_{i+1} = y_i + h f(t_i, y_i)\), with \(f(t, y)=y+e^{-t}\) We will obtain the numerical solutions for each step size, \(h=0.1, h=0.05,\) and \(h=0.025\).

Solve the ODE Analytically

To find the analytical solution, we need to solve the first-order linear ODE: \(y^{\prime}-y=e^{-t}\) To solve this ODE, we first find the integrating factor: \(e^{-t}\) Multiply both sides by the integrating factor: \(e^{-t}(y^{\prime}-y)=e^{-t}e^{-t}\) Now, apply the product rule of differentiation to the left-hand side: \((ye^{-t})^{\prime} = e^{-2t}\) Integrate both sides with respect to \(t\) to find \(y\): \(\int (ye^{-t})^{\prime} dt=\int e^{-2t} dt\) \(ye^{-t}=-\frac{1}{2}e^{-2t}+C\) Now, we have to find the constant \(C\). Using the initial condition \(y(0)=0\), and replacing \(t\) and \(y\) by \(0\) and \(0\), respectively: \(0=0+C\) \(C=0\) Finally, the analytical solution of our ODE is: \(y(t)=\frac{1}{2}e^{t} - \frac{1}{2}e^{-t}\)

Compare the Numerical and Analytical Solutions

To see if the error is getting smaller as the step size \(h\) is reduced, plot the numerical solutions (obtained in Step 1) and the analytical solution (obtained in Step 2) on the same graph. If the numerical solutions get closer to the analytical solution as \(h\) decreases, then the error is reducing as the step size is reduced. By comparing the plots, the student should notice that the numerical solutions become more accurate (closer to the analytical solution) as the step size decreases. This indicates that the error in Euler's method decreases with reduced step size.

Key concepts

Numerical Solutions of ODEs

Numerical solutions of ordinary differential equations (ODEs) are approximate methods to solve differential equations when analytical solutions are either difficult or impossible to obtain. Euler's method is a fundamental numerical algorithm for solving first-order initial value problems of the form \( y^\prime = f(t, y) \) with a given initial condition \( y(t_0) = y_0 \). It operates by advancing the solution through small increments, called step sizes. Using a step size \(h \), the next value of \(y \) is estimated from the current value and the rate of change: \( y_{i+1} = y_i + h f(t_i, y_i) \).

The choice of \(h \) significantly affects the accuracy of the solution. A smaller \(h \) typically leads to more accurate results but requires more computational steps. For students working with such numerical methods, it's crucial to consider the balance between computational efficiency and the desired accuracy of the solutions.

Analytical Solutions of ODEs

In contrast to numerical methods, the analytical solutions of ODEs involve finding an exact expression that satisfies the differential equation for all values within a certain interval. Analytical techniques include methods such as separation of variables, integrating factors, and the method of undetermined coefficients, among others. These methods often require a deep understanding of calculus and related mathematical fields.

For instance, in the problem at hand, we use an integrating factor to solve the first-order linear ODE \( y' - y = e^{-t} \) analytically. The solution we obtain is \( y(t) = \frac{1}{2}e^{t} - \frac{1}{2}e^{-t} \), which represents the exact behavior of the system described by the ODE. Plotting this against numerical solutions can give students a visual understanding of how well the numerical approximation performs compared to the true solution.

Step Size Reduction

Step size reduction is a technique used in numerical methods to increase the accuracy of the solution. By decreasing the step size \(h \), we are taking smaller steps along the solution curve, which leads to a more precise approximation of the system's behavior.

When using Euler's method, reducing the step size can demonstrate a visible improvement in accuracy, as the method's error is proportionate to the step size. This is evident when comparing numerical solutions with different \(h \) values on a graph alongside the exact analytical solution. Students should notice that as \(h \) gets smaller, the numerical solution tends to converge to the analytical solution, confirming the inverse relationship between step size and error. However, too small a step size might lead to excessive computational demands, so finding a balance is key.

Initial Value Problems

Initial value problems (IVPs) are a category of differential equations that come with specific conditions at the start, or initial conditions. They are formulated by specifying the value of the unknown function at a particular point. For example, in the exercise \( y'(t) = y + e^{-t} \) with the initial condition \( y(0) = 0 \) defines an IVP where the system's behavior at \( t = 0 \) is known.

Both numerical and analytical solutions to IVPs rely on these initial conditions to determine the constant values or to start the iterative process. Teaching students how to handle IVPs with techniques like Euler's method not only enhances their problem-solving skills but also provides a practical understanding of how differential equations model real-world phenomena.