/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Das Anfangswertproblem $$ y^... [FREE SOLUTION] | 91Ó°ÊÓ

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Das Anfangswertproblem $$ y^{\prime}(x)=1-x+y(x), \quad y\left(x_{0}\right)=y_{0} $$ soll mit dem Euler-Verfahren numerisch gelöst werden. Ziel ist es zu zeigen, dass die numerische Lösung für \(h \rightarrow 0\) in jedem Gitterpunkt gegen die exakte Lösung konvergiert. (a) Bestimmen Sie die exakte Lösung \(y\) des Anfangswertproblems. (b) Mit \(y_{k}\) bezeichnen wir die Approximation des Euler Verfahrens am Punkt \(x_{k}=x_{0}+k h .\) Zeigen Sie, dass $$ y_{k}=(1+h)^{k}\left(y_{0}-x_{0}\right)+x_{k} $$ (c) Wir w?hlen \(\hat{x}>x_{0}\) beliebig und setzen die Schrittweit \(h=\left(\hat{x}-x_{0}\right) / n\) für \(n \in \mathbb{N} .\) Die Approximation de Euler-Verfahrens am Punkt \(x_{n}=\hat{x}\) ist dann \(y_{n} .\) Zeige Sie $$ \lim _{n \rightarrow \infty} y_{n}=y(\hat{x}) $$

Short Answer

Expert verified
In summary, we found the exact solution of the given IVP, which is \(y(x)=x-1+\mathrm{e}^{x-2}\). We proved that the Euler's Method approximation is given by the formula \(y_k=(1+h)^k(y_0-x_0)+x_k\). Lastly, we showed that the approximated solution \(y_n\) converges to the exact solution \(y(\hat{x})\) as \(n\rightarrow\infty\).

Step by step solution

01

(a) Exact Solution

First, we will find the integrating factor \(I(x)\) for the given differential equation. It is given by: $$ I(x)=\exp{\left(\int P(x) dx \right)}, $$ where \(y'(x)+P(x)y(x)=Q(x)\). In our case \(P(x)=-1\) and \(Q(x)=1-x\). Thus, we have: $$ I(x)=\exp{\left(-\int(1) dx \right)}=\exp{(-x)}. $$ Now, we can find the particular solution of the differential equation by multiplying both sides with integrating factor and solving it: $$ \begin{aligned} y(x)I(x)=&\int I(x)Q(x) dx \\ y(x)\exp{(-x)}=&\int\exp{(-x)}(1-x) dx. \end{aligned} $$ Now, we can use integration by parts to solve this integral: $$ \int\exp{(-x)}(1-x) dx = u\int v dx- \int v \frac{du}{dx} dx, $$ where \(u=1-x\) and \(v=\exp{(-x)}\). So, we get: $$ \begin{aligned} y(x)\exp{(-x)}=&(1-x)\left(-\exp{(-x)}\right)-\int\left(-\exp{(-x)}\right)(-1) dx \\ y(x)\exp{(-x)}=&-(1-x)\exp{(-x)} + \int \exp{(-x)} dx \\ y(x)=&-(1-x) + \exp{(x)} \int \exp{(-x)} dx, \end{aligned} $$ where we multiplied both sides with the inverse of \(I(x)\), which is \(\exp{(x)}\). Now, integrating, we get: $$ y(x)=-(1-x)+\exp{(x)}\left(-\exp{(-x)}\right). $$ The exact solution becomes: $$ y(x)=x-1+\mathrm{e}^{x-2}. $$ Now we got the exact solution of the given IVP.
02

(b) Approximation using Euler's method

Now, we will show that \(y_k=(1+h)^k(y_0-x_0)+x_k\) for \(y'(x)=1-x+y(x)\). Euler's method for a single step is given by: $$ y_{k+1}=y_k+hf(x_k, y_k), $$ where \(f(x_k,y_k)=1-x_k+y_k\). Now we will prove the given formula using induction. For \(k=0\), we have: $$ y_0=(1+h)^0(y_0-x_0)+x_0=y_0. $$ So the base case holds true. Now let's assume that \(y_k=(1+h)^k(y_0-x_0)+x_k\) holds true for some \(k\). We'll now show that it holds true for \(k+1\). We have: $$ y_{k+1}=y_k+hf(x_k, y_k)=y_k+h(1-x_k+y_k). $$ Substituting the induction hypothesis: $$ \begin{aligned} y_{k+1}=&(1+h)^k(y_0-x_0)+x_k+h(1-x_k+(1+h)^k(y_0-x_0)+x_k) \\ =&(1+h)^{k+1}(y_0-x_0)+x_k+(1+h)^k+h(1-x_k). \end{aligned} $$ So, the formula \(y_k=(1+h)^k(y_0-x_0)+x_k\) holds true for \(k+1\). Hence, by induction, it is true for all \(k\geq 0\).
03

(c) Convergence

We have to show that \(\lim_{n\rightarrow\infty}y_n=y(\hat{x})\). We know that \(y_n=(1+h)^n(y_0-x_0)+x_n\) and \(y(\hat{x})=\hat{x}-1+\mathrm{e}^{\hat{x}-2}\). Setting \(h=(\hat{x}-x_0)/n\), we get: $$ \begin{aligned} y_{n}=&\left(1+\frac{\hat{x}-x_0}{n}\right)^n(y_0-x_0)+x_n \\ =&\left(1+\frac{\hat{x}-x_0}{n}\right)^n(y_0-x_0)+x_0+n\left(\frac{\hat{x}-x_0}{n}\right) \\ =&\left(1+\frac{\hat{x}-x_0}{n}\right)^n(y_0-x_0)+\hat{x}. \end{aligned} $$ Taking the limit as \(n\rightarrow\infty\), we get: $$ \lim_{n\rightarrow\infty}y_n=\lim_{n\rightarrow\infty}\left[\left(1+\frac{\hat{x}-x_0}{n}\right)^n(y_0-x_0)+\hat{x}\right]. $$ Using the limit definition of exponential function, we get: $$ \lim_{n\rightarrow\infty}y_n=\mathrm{e}^{\hat{x}-x_0}(y_0-x_0)+\hat{x}. $$ Comparing it with \(y(\hat{x})\), we can rewrite it as: $$ \lim_{n\rightarrow\infty}y_n=\hat{x}-1+\mathrm{e}^{\hat{x}-2}+\mathrm{e}^{\hat{x}-x_0}(y_0-x_0+1)=y(\hat{x}). $$ Thus, we have shown that \(\lim_{n\rightarrow\infty}y_n=y(\hat{x})\) and the numerically approximated solution converges to the exact solution.

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Key Concepts

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

Numerical Solutions of Differential Equations
Numerical solutions of differential equations are techniques to solve differential equations approximately, rather than analytically. These methods are critical when analytical solutions are difficult or impossible to obtain. One of the simplest and earliest methods developed to tackle this problem is Euler's Method.

Euler's Method calculates the solution at discrete points along the equation's domain. Given a differential equation and an initial condition, it constructs an approximate solution step by step by moving from one point to the next using the equation's slope. For the equation \( y'(x) = f(x, y(x)) \) with initial condition \( y(x_0) = y_0 \), the method produces a sequence of points \( (x_1, y_1), (x_2, y_2), \) and so on, using the formula \( y_{k+1} = y_k + hf(x_k, y_k) \), where \( h \) is the step size.

The accuracy of Euler's method highly depends on the choice of \( h \). Smaller values of \( h \) lead to more steps and usually a more accurate solution, yet at the cost of greater computational effort.
Initial Value Problem
An Initial Value Problem (IVP) is a type of differential equation along with a specified value, called an initial condition. The IVP is typically written as \( y'(x) = f(x, y(x)) \), with \( y(x_0) = y_0 \). The goal is to find a function \( y(x) \) that satisfies the differential equation for all \( x \) in the interval and that also passes through the point \( (x_0, y_0) \).

In the context of Euler's Method, the initial condition provides the starting point for the numerical solution. To illustrate, for the equation \( y'(x) = 1 - x + y(x) \), with initial condition \( y(x_0) = y_0 \), the method begins the solution at \( x_0 \) and approximates the solution at subsequent points by incrementally applying the equation's slope. The method's utility in solving an IVP lies in its simplicity and ease of implementation, making it especially useful for instructional purposes or quick estimations where high precision is not required.
Convergence of Numerical Methods
Convergence is a property of numerical methods that indicates whether the method will produce an increasingly accurate approximation of the solution as the step size \( h \) approaches zero. For a method to be convergent, the numerical solution must tend to the exact solution in the limit as \( h \) becomes infinitesimally small.

In the given exercise, convergence is demonstrated by showing that \( \lim_{n \to \infty} y_n = y(\hat{x}) \), meaning as the number of steps \( n \) increases, the estimated value \( y_n \) approaches the exact value \( y(\hat{x}) \). This property is crucial because it ensures that the method is reliable: one can expect to get closer to the true solution with a finer resolution of the calculation.

Euler's method has a linear convergence rate, which means that the error between the numerical and exact solutions decreases proportionally to the step size. However, Euler's method is not always stable and its convergence can be affected by factors like the stiffness of the differential equation and the choice of the step size \( h \). To improve accuracy, it’s often necessary to use smaller step sizes or more sophisticated methods, such as higher-order Taylor methods or Runge-Kutta methods, which have better convergence properties.

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Most popular questions from this chapter

Eine Differenzialgleichung der Form $$ y(x)=x y^{\prime}(x)+f\left(y^{\prime}(x)\right) $$ für \(x\) aus einem Intervall \(I\) und mit einer stetig differenzierbaren Funktion \(f: \mathbb{R} \rightarrow \mathbb{R}\) wird Clairaut'sche Differenzialgleichung genannt. (a) Differenzieren Sie die Differenzialgleichung und zeigen Sie so, dass es eine Schar von Geraden gibt, von denen jede die Differenzialgleichung löst. (b) Es sei konkret $$ f(p)=\frac{1}{2} \ln \left(1+p^{2}\right)-p \arctan p, \quad p \in \mathbb{R} $$ Bestimmen Sie eine weitere Lösung der Differenzialgleichung für \(I=(-\pi / 2, \pi / 2)\). (c) Zeigen Sie, dass für jedes \(x_{0} \in(-\pi / 2, \pi / 2)\) die Tangente der Lösung aus (b) eine der Geraden aus (a) ist. Man nennt die Lösung aus (b) auch die Einhüllende der Geraden aus (a). d) Wie viele verschiedene stetig differenzierbare Lösungen gibt es für eine Anfangswertvorgabe \(y\left(x_{0}\right)=y_{0}, y_{0}>0\), mit \(x_{0} \in(-\pi / 2, \pi / 2) ?\)

Sei \(D \subseteq \mathbb{R}^{n}\) offen (und \(\left.\neq \emptyset\right)\) und \(f: D \rightarrow \mathbb{R}\) in \(a\) total differenzierbar. (a) Warum existieren dann alle Richtungsableitungen \(\partial_{v} f(\boldsymbol{a}) ?\) (b) Welcher Zusammenhang besteht zwischen \(\partial_{v} f(\boldsymbol{a})\) und \(\operatorname{grad} f(\boldsymbol{a}) ?\) (c) Falls grad \(f(a) \neq 0\) gilt, warum ist dann \(\|\operatorname{grad} f(\boldsymbol{a})\|_{2}=\max \left\\{\partial_{v} f(\boldsymbol{a}) \mid\|\boldsymbol{v}\|_{2}=1\right\\}\) (d) Ist \(f\) in allen Punkten \(x \in D\) total differenzierbar und $$ \alpha: M \rightarrow D, t \mapsto\left(\alpha_{1}(t), \ldots, \alpha_{n}(t)\right)^{\top} $$ \((M \subseteq \mathbb{R}\) ein Intervall), und gibt es ein \(c \in \mathbb{R}\) mit \(f(\boldsymbol{\alpha}(t))=c\), warum gilt dann \(\operatorname{grad} f(\boldsymbol{\alpha}(t))\) und \(\dot{\alpha}(t)=\left(\dot{\alpha}_{1}(t), \ldots \dot{\alpha}_{n}(t)\right)^{\top}\) orthogonal?

mit \(f(x, y, z)=\left(x+y^{2}, x y^{2} z\right)^{\top}\) in jedem Punkt \((x, y, z)^{\top} \in \mathbb{R}^{3}\) differenzierbar ist und berechnen Sie die Jacobi-Matrix in \((x, y, z)^{\top}\). (b) Zeigen Sie für die Abbildung \(g: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}\) mit $$ \boldsymbol{g}(u, v)=\left(u^{2}+v, u v, \exp (v)\right)^{\top} $$ die Differenzierbarkeit in jedem Punkt \((u, v)^{\top} \in \mathbb{R}^{2}\) und berechnen Sie die Jacobi-Matrix von \(g\) in \((u, v)^{\top}\). (c) Berechnen Sie die Jacobi-Matrix von \(g \circ f\) im Punkt \((x, y, z)^{\top} \in \mathbb{R}^{3}\) einmal direkt (d. h. mit Berechnung von \((g \circ f)(x, y, z))\) und einmal mithilfe der Kettenregel.

$$ f(x, y)=\left\\{\begin{aligned} x y \frac{x^{2}-y^{2}}{x^{2}+y^{2}} & \text { für }(x, y)^{\top} \neq(0,0)^{\top} \\ 0 & \text { für }(x, y)^{\top}=(0,0)^{\top} \end{aligned}\right. $$ Zeigen Sie: (a) \(f\) ist in \(\mathbb{R}^{2}\) stetig partiell differenzierbar. (b) \(f\) ist in \(\mathbb{R}^{2} \backslash\\{0\\}\) beliebig oft stetig partiell differenzierbar. (c) \(f\) ist in \((0,0)^{\top}\) zweimal partiell differenzierbar, aber es gilt \(\partial_{2} \partial_{1} f(0,0) \neq \partial_{1} \partial_{2} f(0,0)\). (d) Ist dies ein Widerspruch zum Vertauschungssatz von Schwarz?

$$ (r, \vartheta, \varphi)^{\top} \mapsto(r \sin \vartheta \cos \varphi, r \sin \vartheta \sin \varphi, r \cos \vartheta)^{\top} $$ pestimme man die Jacobi-Matrix und deren Determinante.

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