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Chapter 5: Problem 3

Gegeben ist die nichtlineare Differentialgleichung 1. Ordnung $$ y^{\prime}=\sqrt{x+y} $$ und der Anfangswert \(y(1)=1\). Bestimmen Sie n盲herungsweise 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 Absch盲tzung 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.
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.
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.
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 \(x\).
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.

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Most popular questions from this chapter

Ein Kondensator mit der Kapazit?t \(C=5 \mu \mathrm{F}\) wird zun盲chst auf \(u_{0}=100 \mathrm{~V}\) aufgeladen und anschlieBend 眉ber einen ohmschen Widerstand von \(R=500 \Omega\) und eine Spule mit der Induktivit?t \(L=0,2 \mathrm{H}\) entladen (Bild V-69). Bestimmen Sie den zcitlichen Verlauf der Stromst盲rke \(i=i(t)\) in diesem Reihenschwingkreis. Anleitung: L枚sen Sie die Schwingungsgleichung $$ \frac{d^{2} i}{d t^{2}}+2 \delta \frac{d i}{d t}+\omega_{0}^{2} i=0 \quad\left(\delta=\frac{R}{2 L}, \omega_{0}^{2}=\frac{1}{L C}\right) $$ f眉r die Anfangswerte \(i(0)=0, u_{C}(0)=u_{0} .\) Zwischen der Kondensatorspannung \(u_{C}(t)\), der Kondensatorladung \(q(t)\) und der Stromst盲rke \(i(t)\) bestehen dabei die folgenden Zusammenh?nge: $$ C=\frac{q}{u_{C}}, \quad i=-\dot{q} \Rightarrow u_{C}(t)=\frac{1}{C} q(t)=-\frac{1}{C} \cdot \int i(t) d t $$

Ein Reihenschwingkreis enthalte den ohmschen Widerstand \(R=500 \Omega\), cinen Kondensator mit der Kapazit盲t \(C=5 \mu \mathrm{F}\) und eine Spule mit der Induk tivit盲t \(L=0.2 \mathrm{H}\). Wie lautet die station盲re L?sung der Schwingungsgleichung $$ \frac{d^{2} i}{d t^{2}}+2 \delta \frac{d i}{d t}+\omega_{0}^{2} i=\frac{1}{L} \cdot \frac{d u_{a}}{d t} \quad\left(\delta=\frac{R}{2 L}, \omega_{0}^{2}=\frac{1}{L C}\right) $$ wenn das System durch die von auBen angelegte Wechselspannung $$ u_{d}(t)=300 \mathrm{~V} \cdot \sin \left(500 \mathrm{~s}^{-1}-t\right) $$ zu erzwungenen elektrischen Schwingungen angeregt wird? Skizzieren Sie den zeitlichen Verlauf dieser Schwingung.

Zeigen Sie, daB sich die nicht-lineare Differentialgleichung 1. Ordnung $$ 4 y y^{\prime}-y^{2}=-\left(1+x^{2}\right) $$ mit Hilfe der Substitution \(u=y^{2}\) in eine lineare Differentialgleichung 1 . Ordnung 眉berf眉hren l盲Bt und bestimmen Sie die allgemeine L枚sung dieser Differentialgleichung.

Ein schwingungst?higes mechanisches Feder-Masse-System mit den Kenngr枚Ben $$ m=20 \mathrm{~kg} . \quad b=40 \mathrm{~kg} / \mathrm{s}, \quad c=100 \mathrm{~N} / \mathrm{m} $$ werde in einem Experiment durch die von au\betaen einwirkende Kraft $$ F(t)=20 \mathrm{~N} \cdot \sin (\omega t) $$ zu erzwungenen Schwingungen erregt. a) Bestimmen Sie die allgemeine L枚sung der Schwingungsgleichung. b) Wie lautet die station盲re L?sung der Schwingungsgleichung? Zeichnen Sie die Resonanzkurve \(A=A(\omega)\) sowie den Frequenzgang der Phasenverschiebung \(\varphi\) zwischen Erregerschwingung und erzwungener Schwingung. c) Bestimmen und skizzieren Sie die station盲re L?sung fur die Erregerkreisfrequenz \(\omega=1 \mathrm{~s}^{-1}\).

L?sen Sie die folgenden inhomogenen linearen Differentialgleichungssysteme 2. Ordnung durch , Aufsuchen einer partikul盲ren L枚sung": a) \(y_{1}^{\prime}=2 y_{2}+8 x\) $$ y_{2}^{\prime}=-2 y_{1} $$ $$ \text { b) } y_{1}^{\prime}=-y_{1}+y_{2}+4 \cdot \mathrm{e}^{2 x} $$ $$ y_{2}^{\prime}=-4 y_{1}+3 y_{2} $$
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