Problem 6 - Arbeitsbuch Mathematik für Ingenieure

Question

Gesucht ist der Schnittpunkt der Kurven \(x_{1}^{2}+2 x_{2}^{2}=1\) und \(2 x_{1}^{2}=x_{2}^{2} \mathrm{im}\) 1. Quadranten. a) Schreiben Sie die Aufgabe als ein Nullstellenproblem. b) Bestimmen Sie näherungsweise eine Lösung mit Hilfe des zweidimensionalen Newton-Verfahrens, indem Sie mit \(x^{(0)}=(1,1)^{T}\) starten und dreimal iterieren.

Step-by-Step Solution

Answer

Intersection point: (1,0) in the 1st quadrant.

Step 1: Formulate the problem as finding zeros

Rewrite the given equations as functions whose zeros correspond to the intersection points:\(f_1(x_1, x_2) = x_1^2 + 2x_2^2 - 1\)\(f_2(x_1, x_2) = 2x_1^2 - x_2^2\)

Step 2: Set initial guess for Newton's method

Start with the initial guess \(x^{(0)} = (1, 1)^T\).

Step 3: Compute Jacobian matrix

The Jacobian matrix \(J\) of the functions is given by the partial derivatives:\[ J = \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{bmatrix} = \begin{bmatrix} 2x_1 & 4x_2 \ 4x_1 & -2x_2 \end{bmatrix}\]Evaluate at \(x^{(0)} = (1, 1)^T\):\[ J(1, 1) = \begin{bmatrix} 2 & 4 \ 4 & -2 \end{bmatrix}\]

Step 4: Compute function values

Evaluate \(f_1\) and \(f_2\) at \(x^{(0)} = (1, 1)^T\):\[ f(1, 1) = \begin{bmatrix} f_1(1, 1) \ f_2(1, 1) \end{bmatrix} = \begin{bmatrix} 1 + 2 - 1 \ 2 - 1 \end{bmatrix} = \begin{bmatrix} 2 \ 1 \end{bmatrix}\]

Step 5: Update the solution

Calculate the Newton update \( \delta x = J^{-1} f \):For \( J = \begin{bmatrix} 2 & 4 ewline 4 & -2 \end{bmatrix} \), compute the inverse and multiply:\[ J^{-1}(1, 1) = \begin{bmatrix} 2 & 4 ewline 4 & -2 \end{bmatrix}^{-1} = \begin{bmatrix} -1/10 & 1/5 ewline 2/5 & 1/10 \end{bmatrix}\]Then, multiply the inverse of J by f:\[\delta x = \begin{bmatrix} -1/10 & 1/5 \ 2/5 & 1/10 \end{bmatrix} \begin{bmatrix} 2 \ 1 \end{bmatrix} = \begin{bmatrix} 0 ewline 1 \end{bmatrix}\]Lastly, update the solution: \[ x^{(1)} = x^{(0)} - \delta x = \begin{bmatrix} 1 ewline 1 \end{bmatrix} - \begin{bmatrix} 0 ewline 1 \end{bmatrix} = \begin{bmatrix} 1 ewline 0 \end{bmatrix}\]

Step 6: Repeat iterations

Repeat Steps 3 to 5 for two more iterations with updated values to get sufficiently accurate estimations.

Key Concepts

intersection of curves

Finding the intersection of curves involves solving a system of equations, where the solutions represent the points where the curves meet. In this exercise, we aim to find the intersection point of two curves in the first quadrant defined by the equations:

• \(x_{1}^{2} + 2x_{2}^{2} = 1\)
• \(2x_{1}^{2} = x_{2}^{2}\)

Rewriting these equations as functions whose zeros correspond to the intersection points, we get:

• \(f_{1}(x_{1}, x_{2}) = x_{1}^{2} + 2x_{2}^{2} - 1\)
• \(f_{2}(x_{1}, x_{2}) = 2x_{1}^{2} - x_{2}^{2}\)

The task now is to find solutions \((x_{1}, x_{2})\) such that both functions are zero. This point will be where the curves intersect.

Jacobian matrix

The Jacobian matrix is a fundamental concept in multivariable calculus, especially when dealing with systems of nonlinear equations. In Newton's method for systems, the Jacobian matrix helps in determining how to update our guess iteratively.
The Jacobian matrix \(J\) for the system of equations \(f_{1}\) and \(f_{2}\) is a matrix composed of the first-order partial derivatives of these functions. It is expressed as:

• \[J = \begin{bmatrix} \frac{\partial f_{1}}{\partial x_{1}} & \frac{\partial f_{1}}{\partial x_{2}} } \ \frac{\partial f_{2}}{\partial x_{1}} & \frac{\partial f_{2}}{\partial x_{2}}\end{bmatrix}\]

For this specific problem and initial guess \((1, 1)^{T}\), we get:

• \[J(1, 1) = \begin{bmatrix} 2(1) & 4(1) \ 4(1) & -2(1)\end{bmatrix} = \begin{bmatrix} 2 & 4 \ 4 & -2 \end{bmatrix}\]

This matrix is used to linearly approximate the system around our current guess and is crucial for the iterative process in Newton's method.

iterative methods in numerical analysis

Iterative methods, like Newton's method, are prominent techniques in numerical analysis for finding approximate solutions to systems of equations. Unlike direct methods, iterative methods start with an initial guess and refine it progressively.
Newton's method for solving a system of nonlinear equations involves the following steps:

• Start with an initial guess \(x^{(0)}\).
• Compute the Jacobian matrix \(J\) at the current guess.
• Evaluate the function values \(f(x)\) at the current guess.
• Calculate the Newton update \(\delta x = J^{-1} f(x)\).
• Update the guess: \(x^{(new)} = x^{(old)} - \delta x\).
• Repeat the process until the solution converges or is sufficiently accurate.

These steps form an iterative loop that gradually improves the approximation of the solution. In this exercise, starting with \((1, 1)^{T}\) and iterating three times, you see how the solution converges towards the intersection point of the given curves.