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Carry out one step of the Euler method and of the improved Euler method using the step size \(h=0.1 .\) Suppose that a local truncation error no greater than 0.0025 is required. Estimate the step size that is needed for the Euler method to satisfy this requirement at the first step. $$ y^{\prime}=5 t-3 \sqrt{y}, \quad y(0)=2 $$

Short Answer

Expert verified
Based on the given exercise, we perform one step each of the Euler method and the improved Euler method to approximate the solution of the first-order initial value problem (IVP) and estimate the step size needed for the Euler method to achieve a local truncation error not greater than 0.0025 at the first step. 1. From carrying out one step of the Euler method with h = 0.1, we get \(y_1 \approx 1.857\). 2. Next, carrying out one step of the Improved Euler method with h = 0.1, we obtain \(y_1 \approx 1.860\). 3. Estimating the step size needed for the Euler method to achieve a local truncation error not greater than 0.0025 at the first step, we find it to be \(h \approx 0.000689\).

Step by step solution

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01

Euler method with h = 0.1

The Euler method formula for a first-order ODE is: $$ y_{n+1} = y_n + h f(t_n, y_n) $$ We are given the ODE \(y'(t) = 5 t - 3 \sqrt{y}\) and initial condition \(y(0) = 2\). We are also given the step size \(h = 0.1\). To perform one step of the Euler method, we will plug in the values of \(t_0\), \(y_0\) and \(h\) into the formula as follows: $$ y_{1} = y_0 + h \cdot (5 t_0 - 3 \sqrt{y_0}) $$ Now we can calculate \(y_1\): $$ y_1 = 2 + 0.1 \cdot (5 \cdot 0 - 3 \sqrt{2}) \approx 1.857 $$
02

Improved Euler method with h = 0.1

The improved Euler method (also known as the midpoint method) formula for a first-order ODE is: $$ y_{n+1} = y_n + h f\left(t_n + \frac{h}{2}, y_n + \frac{h}{2} f(t_n, y_n)\right) $$ Similar to step 1, we use the given ODE and initial condition along with the step size \(h=0.1\). To perform one step of the improved Euler method, we will plug in the values of \(t_0\), \(y_0\) and \(h\) into the formula as follows: $$ y_1 = y_0 + h \cdot \left(5 \left(t_0 + \frac{h}{2}\right) - 3 \sqrt{y_0 + \frac{h}{2} \cdot (5 t_0 - 3 \sqrt{y_0})}\right) $$ Now we can calculate \(y_1\): $$ y_1 = 2 + 0.1 \cdot \left(5 \cdot 0.05 - 3 \sqrt{2 + 0.05 \cdot (5 \cdot 0 - 3 \sqrt{2})}\right) \approx 1.860 $$
03

Estimate step size for Euler method with local truncation error less than 0.0025

The local truncation error \(E_n\) for the Euler method is given by the formula: $$ E_n = \frac{h}{2} y''(\xi_n) $$ where \(\xi_n\) is some value between \(t_n\) and \(t_{n+1}\). To estimate the required step size \(h\) for the Euler method to satisfy the requirement of a local truncation error not greater than 0.0025, we need to find \(y''(t)\). We are given the ODE: $$ y'(t) = 5t - 3\sqrt{y} $$ To find the second derivative \(y''(t)\), differentiate \(y'(t)\) with respect to \(t\): $$ y''(t) = 5 - \frac{3}{2} \frac{1}{\sqrt{y}} y'(t) $$ Since \(y'(t_0) = y'(0) = 5t_0 - 3\sqrt{y_0} = -3\sqrt{2}\), we get: $$ y''(t_0) = 5 - \frac{3}{4\sqrt{2}} (-3\sqrt{2}) = 5 + \frac{9}{4} \approx 7.25 $$ Now we can solve for the required step size \(h\) using the local truncation error formula and the condition \(E_n \le 0.0025\): $$ \frac{h}{2} y''(t_0) \le 0.0025 $$ Therefore, $$ h \le \frac{0.0025 \cdot 2}{7.25} \approx 0.000689 $$ Hence, the estimated step size needed for the Euler method to satisfy a local truncation error no greater than 0.0025 is \(h \approx 0.000689\).

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Numerical Solutions of ODEs
Ordinary Differential Equations (ODEs) express relationships involving functions and their derivatives. Finding exact solutions to ODEs can be challenging, especially when they cannot be solved analytically. That's where numerical methods come into play.

Numerical methods provide a way to approximate solutions to ODEs when an exact solution is unavailable or difficult to obtain. The Euler method is a simple and commonly used numerical method for solving first-order ODEs. It involves taking small steps along the curve of the solution, starting from an initial condition. To improve on the Euler method's accuracy, variants like the Improved Euler method, which refines each step further, are employed.
Local Truncation Error
The accuracy of numerical methods is not perfect. One of the critical measures of accuracy is the local truncation error. It represents the error made over a single iteration or step of the numerical method.

The local truncation error for the Euler method can be quantified by considering the exact solution's Taylor expansion. As we truncate this series to a linear term when using the Euler method, higher-order terms contribute to the error. The goal is to keep this error within a specified threshold, guiding the choice of appropriate step size to ensure the desired accuracy.
Improved Euler Method
The Improved Euler method, or Heun's method, offers a better level of precision compared to the standard Euler method. This technique consists of an initial prediction of the next value followed by a correction.

It incorporates the slope at the initial point and the slope at the midpoint of the interval, effectively averaging two estimates of the slope to determine the function's behavior over the interval. The method reduces local truncation error and thus achieves greater accuracy for the same step size, or allows for larger step sizes while maintaining the same level of accuracy.
Step Size Estimation
Step size is a vital factor in the accuracy of numerical solutions. If the step size is too large, the solution may deviate significantly from the true path. Conversely, too small a step size increases computation time without a proportionate increase in accuracy.

To estimate an appropriate step size, one must balance the need for accuracy with computational efficiency. The process involves determining a maximum threshold for the local truncation error and rearranging the error estimate formula to solve for the maximum allowable step size. An optimal step size minimizes error while not excessively increasing the number of steps and thus the computational cost.

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Most popular questions from this chapter

Consider the initial value problem $$ y^{\prime}=3 t^{2} /\left(3 y^{2}-4\right), \quad y(0)=0 $$ (a) Draw a direction field for this equation. (b) Estimate how far the solution can extended to the right. Let \(t_{1 / 4}\) be the right endpoint of the interval of existence of this solution. What happens at \(t_{M}\) to prevent the solution from continuing farther? (c) Use the Runge-Kutta method with various step sizes to determine an approximate value of \(t_{M}\). (d) If you continue the computation beyond \(t_{\mathrm{H}}\), you can continue to generate values of \(y .\) What significance, if any, do these values have? (e) Suppose that the initial condition is changed to \(y(0)=1 .\) Repeat parts (b) and (c) for this problem.

The modified Euler formula for the initial value problem \(y^{\prime}=f(t, y), y\left(t_{0}\right)=y_{0}\) is given by $$ y_{n+1}=y_{n}+h f\left[t_{n}+\frac{1}{2} h, y_{n}+\frac{1}{2} h f\left(t_{n}, y_{n}\right)\right] $$ Following the procedure outlined in Problem \(14,\) show that the local truncation error in the modified Euler formula is proportional to \(h^{3}\).

Consider the initial value problem $$ y^{\prime}=-10 y+2.5 t^{2}+0.5 t, \quad y(0)=4 $$ (a) Find the solution \(y=\phi(t)\) and draw its graph for \(0 \leq t \leq 5\). (b) The stability analysis in the text suggests that for this problem the Euler method is stable only for \(h<0.2 .\) Confirm that this is true by applying the Euler method to this problem for \(0 \leq t \leq 5\) with step sizes near \(0.2 .\) (c) Apply the Runge-Kutta method to this problem for \(0 \leq t \leq 5\) with various step sizes. What can you conclude about the stability of this method? (d) Apply the backward Euler method to this problem for \(0 \leq t \leq 5\) with various step sizes. What step size is needed in order that the error at \(t=5\) is less than \(0.01 ?\)

The following problem illustrates a danger that occurs because of round-off error when nearly equal numbers are subtracted, and the difference then multiplied by a large number. Evaluate the quantity $$ 1000 \cdot\left|\begin{array}{cc}{6.010} & {18.04} \\ {2.004} & {6.000}\end{array}\right| $$ as follows. (a) First round each entry in the determinant to two digits. (b) First round each entry in the determinant to three digits. (c) Retain all four digits. Compare this value with the results in parts (a) and (b).

Obtain a formula for the local truncation error for the Euler method in terms of \(t\) and the solution \(\phi\) $$ y^{\prime}=\sqrt{t+y}, \quad y(1)=3 $$

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