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? \(y^{\prime}=y^{-1}, \quad y(0)=1, \quad 0 \leq t \leq 1\)

Short answer

#Answer# The effect of reducing the step size in Euler's method is that it increases the accuracy of the numerical solution for the given IVP. As the step size decreases, the numerical solution becomes closer to the exact solution.

Step-by-step solution

Applying Euler's method

To apply Euler's method with different step sizes, we first need to set up the equation for Euler's method as follows:$$y_{n+1} = y_n + hf(t_n, y_n)$$Where \(f(t_n, y_n) = y_n^{-1}\). Let's calculate the numerical solutions for the different step sizes \(h=0.1, h=0.05,\) and \(h=0.025\).

Solving the problem analytically

To solve the given IVP, we need to find an analytical solution:$$y^{\prime} = y^{-1}$$With initial condition \(y(0) = 1\). We can rewrite the given differential equation as a separable equation:$$ydy = dt$$Integrate both sides to find the exact solution:$$\int y dy = \int dt$$This gives us the exact solution:$$\frac{1}{2} y^2 = t + C$$Plug the initial condition \(y(0) = 1\) into the equation to find the constant C:$$\frac{1}{2} (1)^2 = 0 + C$$So, \( C= \frac{1}{2}\). Therefore, the exact solution is:$$y(t) = \sqrt{2t + 1}$$

Sketch the graph for numerical solutions and exact solution

Now, let's sketch the graph for each numerical solution and the exact solution. You will need to use a calculator or computer to plot the results of the Euler's method calculations along with the exact solution, which is \(y(t) = \sqrt{2t + 1}\).

Comparing the errors for the numerical solutions

Observe the plotted graph and compare the errors for the different step sizes. The error should appear to be getting smaller as \(h\) gets reduced. This is because a smaller step size results in more accurate approximations for the derivative at each time step. The numerical solution will be closer to the exact solution as the step size decreases.

Key concepts

Numerical Solutions Using Euler's Method

Euler's method provides a way to approximate solutions to differential equations, especially when an exact solution is difficult or impossible to find. This numerical technique helps predict the behavior of a function over small intervals. This method is especially useful in situations where we can only gather discrete data points.
In our exercise, Euler's method is applied using different step sizes. The advancement of the solution from one small step to the next is given by the formula:

• Initial Condition: Start at a given point with a known value of the function, here, it is \( y(0) = 1 \).
• Recursive Formula: \( y_{n+1} = y_n + hf(t_n, y_n) \), where \( f(t_n, y_n) = y_n^{-1} \).
• Iterate over several steps: Depending on your step size \( h \), repeat this formula to get approximations \( y_1, y_2, ... \).

This technique uses successive approximations and is particularly useful for understanding the behavior of solutions, even though it results in a set of values rather than a continuous function.

Impact of Step Size Reduction on Accuracy

The size of the step \( h \) plays a crucial role in the accuracy of Euler's method. The smaller the step size, the closer the numerical approximation gets to the true solution of the differential equation. Reducing the step size comes with both advantages and some cautionary aspects.

• Increased Accuracy: Smaller steps result in less error over each iteration. Errors accumulate over time steps, so reducing \( h \) minimizes this accumulation.
• Computational Load: Reducing the step size increases the number of calculations necessary. With \( h = 0.05 \) vs. \( h = 0.1 \), we double the calculations to cover the same interval.
• Analysis: Plotting results with various \( h \) values reveals visibly that approximations converge toward the exact solution, showing reduced deviation.

Overall, reducing the step size is a trade-off between increased accuracy and computational demand, perfectly illustrated in this task.

Exact Solution Comparison and Error Observation

When you solve an initial value problem (IVP) analytically, like in this exercise, you aim to find an exact solution. Here, the exact solution to the differential equation \( y' = y^{-1} \) is \( y(t) = \sqrt{2t + 1} \). This offers a precise curve against which we can compare our numerical estimates.
To compare the accuracy:

• Graphical Comparison: Plot both the numerical solutions at different step sizes and the exact solution on the same graph.
• Error Evaluation: As the step size \( h \) is reduced (from 0.1 to 0.025), it's observable that the difference between the Euler's calculations and \( y(t) = \sqrt{2t + 1} \) diminishes.
• Interpretation: The decrease in error with a decreasing step size visually shows how refined our approximation becomes.

This graphical method effectively showcases how smaller steps inch our numerical results closer to the mathematically exact solution. Such comparisons provide insights into optimizing numerical methods for practical applications.