Question

Exencises 18-22: In each exercise, (a) Using step size \(h=01\), compute the first estimate \(y_{1}\) using Euler's method and the Runge-Kutta method (9). Let these estimates be denoted by \(y_{1}^{\&}\) and \(y_{1}^{R K}\), respectively. (b) Solve the problem analytically. (c) Compute the errors \(\left|y\left(t_{1}\right)-y_{1}\right|\) for the two estimates obtained in (a). \(y^{\prime}=y^{2}, \quad y(0)=1\)

Short answer

Question: Compute the first estimate \(y_1\) for the given ordinary differential equation \(y' = y^2\) with the initial condition \(y(0) = 1\), using Euler's method and 4th order Runge-Kutta method with a step size of 0.1. Calculate the errors for the estimates. Answer: The first estimate of the solution using: 1. Euler's method is \(y_{1}^{E} = 1.1\) with an error of 0.011111. 2. Runge-Kutta method is \(y_{1}^{R K} = 1.10545\) with an error of 0.005661.

Step-by-step solution

Identify the given information

The ODE is given by \(y^{\prime}=y^{2}\), with the initial condition \(y(0)=1\). We need to find the next value \(y_1\) using Euler's method and RK method (9) with a step size \(h=0.1\).

Euler's method

The formula for Euler's method is \(y_{n+1}=y_{n}+h \cdot y^{\prime}_{n}\). In our case, the step size \(h=0.1\), and \(y^{\prime}=y^{2}\). So for our case, the formula becomes \(y_{n+1}=y_{n}+0.1 \cdot y^{2}_n\). Applying the initial condition \(y(0)=1\): $$y_1^E = y_0 + 0.1\times y_0^2 = 1 + 0.1\times 1^2 = 1 + 0.1 = 1.1$$

Runge-Kutta method 4th order (9)

Now we compute the first estimate \(y_{1}^{R K}\) using 4th order Runge-Kutta method with the given step size \(h=0.1\). It has the following formula: $$ y_{n+1} = y_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4), $$ where \(k_1 = h \cdot f(t_n, y_n)\), \(k_2 = h\cdot f(t_n + \frac{1}{2}h, y_n + \frac{1}{2}k_1)\), \(k_3 = h\cdot f(t_n + \frac{1}{2}h, y_n + \frac{1}{2}k_2)\), \(k_4 = h \cdot f(t_n + h, y_n + k_3)\). For our problem, the function \(f(t, y) = y^2\). Using the given step size \(h=0.1\), initial condition \(y(0)=1\), and \(t_0 = 0\): \(k_1 = 0.1 \cdot f(0, 1) = 0.1 \cdot 1^2 = 0.1\), \(k_2 = 0.1 \cdot f(0+\frac{0.1}{2}, 1+\frac{0.1}{2}) = 0.1 \cdot (1.05)^2 = 0.11025\), \(k_3 = 0.1 \cdot f(0+\frac{0.1}{2}, 1+\frac{0.11025}{2}) = 0.1 \cdot (1.055)^2 = 0.111325\), \(k_4 = 0.1 \cdot f(0+0.1, 1+0.111325) = 0.1 \cdot (1.111325)^2 = 0.123442\). Now, we have all the values needed to compute \(y_1^{R K}\): \(y_{1}^{R K} = y_0 + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) = 1+\frac{1}{6}(0.1+2\times0.11025+2\times0.111325 + 0.123442) = 1.10545\).

Solve the problem analytically

The given ODE is separable. It can be written as: $$ \frac{dy}{dt} = y^2. $$ We can solve this by separating the variables: $$ \frac{dy}{y^2} = dt. $$ Now, we integrate both sides: $$ \int \frac{dy}{y^2} = \int dt. $$ This yields: $$ -\frac{1}{y} = t + C. $$ Using the initial condition \(y(0) = 1\), we find the constant \(C\): $$ -1 = 0 + C \implies C = -1. $$ Thus, the analytical solution is: $$ -\frac{1}{y} = t - 1 \implies y(t) = \frac{1}{1-t}. $$

Compute the errors

Now, using the analytical solution, we can compute the errors at \(t_1 = t_0 + h = 0.1\): The exact value at \(t_1=0.1\) is \(y(0.1) = \frac{1}{1-0.1} = \frac{1}{0.9} = 1.111111\). The error for Euler's method is: $$ \left|y(t_{1})-y_{1}^{E}\right| = \left|1.111111 - 1.1\right| = 0.011111. $$ The error for Runge-Kutta method is: $$ \left|y(t_{1})-y_{1}^{R K}\right| = \left|1.111111 - 1.10545\right| = 0.005661. $$ In summary, we have the following results: 1. Euler's method: \(y_{1}^{E} = 1.1\) with an error of \(0.011111\). 2. Runge-Kutta method: \(y_{1}^{R K} = 1.10545\) with an error of \(0.005661\).

Key concepts

Differential Equations

Differential equations are the cornerstone of the dynamic sciences, modeling how things change over time or space. These mathematical expressions relate an unknown function and its derivatives. A common type of differential equation is an ordinary differential equation (ODE), as seen in the exercise. Here, the equation of the form \( y' = y^2 \) indicates how the rate of change of \( y \) at any point in time depends on the square of its current value. The solution to an ODE describes the behavior of the system over time, providing a function \( y(t) \) that equates to how the system evolves.

Understanding differential equations involves both finding exact analytical solutions, where possible, and numerical approximations when the equations cannot be solved explicitly. Analytical solutions give a precise formula for the function \( y(t) \), such as the solution \( y(t) = \frac{1}{1-t} \), which was derived in the problem. Though powerful, the ability to find such exact solutions is limited to ODEs that are directly integrable or can be manipulated into a form that is integrable.

Numerical Analysis

Numerical analysis is the study of algorithms that use numerical approximation for the problems of mathematical analysis. When differential equations cannot be solved analytically, numerical methods like Euler's method and the Runge-Kutta methods provide alternative approximate solutions. These methods generate numerical sequences that converge to the actual solution of the differential equations over discrete time steps.

Euler's Method: It's the most straightforward numerical method for solving ODEs, using linearity to project the function's graph forward step by step. However, its simplicity also makes it less accurate, especially for larger step sizes or over larger intervals, as illustrated in the Euler's estimate \( y_1^E = 1.1 \) for a step size \( h = 0.1 \).

Runge-Kutta Methods: These are a family of more sophisticated methods that provide greater accuracy without decreasing the step size. They do this through a weighted average of slopes calculated at several points. The fourth-order Runge-Kutta method used in the exercise computes four slopes (\( k_1, k_2, k_3, k_4 \)) at different points and averages them to estimate the next value, which gave a more precise estimate \( y_1^{RK} = 1.10545 \). By doing this, the Runge-Kutta methods typically achieve better precision than Euler's method for the same step size, which is important when modeling real-world systems where accuracy is crucial.

Error Computation

Error computation is crucial in numerical analysis to gauge the accuracy of an approximate solution. It involves calculating the difference between the approximate numerical solution and the true analytical solution, if known. There are different types of errors to consider: the local truncation error, which measures the error made in a single step, and the global error, which measures the cumulative effect of the errors over many steps.

In the context of the exercise, the local error at the first time step \( t_1 \) was computed for both Euler's and Runge-Kutta methods by comparing the estimated values \( y_1^E \) and \( y_1^{RK} \) to the exact solution \( y(0.1) \). For Euler's method, the error was found to be \( 0.011111 \), while for Runge-Kutta, it was \( 0.005661 \), demonstrating the superior accuracy of the Runge-Kutta method.Understanding the error helps in selecting the appropriate method and step size for a given problem, ensuring that results are accurate enough for the intended application. It's a balance of computational efficiency and precision, as more accurate methods typically involve more computations. Educators encourage students to both compute and reflect on these errors, fostering a deeper understanding of the strengths and limitations of numerical methods.