Problem 12 - A First Course in Real Analysis

Question

A vector function \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) is given. Verify that the Inverse function theorem is applicable and find the inverse function \(g\). \(y_{1}=x_{1} \cos \left(\pi x_{2} / 2\right), y_{2}=x_{1} \sin \left(\pi x_{2} / 2\right), x_{1}>0,-1

Step-by-Step Solution

Answer

The Inverse function theorem is applicable as the determinant of the Jacobian is always positive for the given domain. The inverse function is given by \( g(y_{1}, y_{2}) = (\sqrt{y_{1}^2 + y_{2}^2}, \frac{2}{\pi} \arctan2(y_{2}, y_{1})) \).

Step 1: Review the Inverse Function Theorem conditions

The Inverse function theorem states that a function, f, is locally invertible if it is continuously differentiable (i.e., it is a C1 function) and its Jacobian matrix, J(f), is non-singular (its determinant is not zero) at a point in its domain. In this context, the function f must have a non-zero determinant of its Jacobian matrix to apply the theorem.

Step 2: Calculate the Jacobian Matrix

The Jacobian matrix J(f) of the vector function f is a 2x2 matrix composed of the partial derivatives of the functions with respect to x1 and x2. Calculating these partial derivatives we get: \[ J(f) = \begin{pmatrix} \frac{\partial y_{1}}{\partial x_{1}} & \frac{\partial y_{1}}{\partial x_{2}} \ \frac{\partial y_{2}}{\partial x_{1}} & \frac{\partial y_{2}}{\partial x_{2}} \end{pmatrix} = \begin{pmatrix} \cos(\frac{\pi x_{2}}{2}) & -\frac{\pi x_{1}}{2}\sin(\frac{\pi x_{2}}{2}) \ \sin(\frac{\pi x_{2}}{2}) & \frac{\pi x_{1}}{2}\cos(\frac{\pi x_{2}}{2}) \end{pmatrix}. \]

Step 3: Verify the conditions of the Inverse Function Theorem

To verify that the Inverse function theorem is applicable, compute the determinant of J(f): \[ \text{det}(J(f)) = \text{det}\begin{pmatrix} \cos(\frac{\pi x_{2}}{2}) & -\frac{\pi x_{1}}{2}\sin(\frac{\pi x_{2}}{2}) \ \sin(\frac{\pi x_{2}}{2}) & \frac{\pi x_{1}}{2}\cos(\frac{\pi x_{2}}{2}) \end{pmatrix} = \cos^2(\frac{\pi x_{2}}{2}) + \frac{\pi^2 x_{1}^2}{4}\sin^2(\frac{\pi x_{2}}{2}). \] Since \( x_{1} > 0 \) and \( -1 < x_{2} < 1 \), both \( \cos(\frac{\pi x_{2}}{2}) \) and \( \sin(\frac{\pi x_{2}}{2}) \) are non-zero, the determinant is always positive; therefore, J(f) is non-singular, and the function f is locally invertible.

Step 4: Compute the Inverse Function

To compute the inverse function g, we need to solve the given equations for \( x_1 \) and \( x_2 \). From the given equations: \[ x_{1} = \sqrt{y_{1}^2 + y_{2}^2} \] and \[ x_{2} = \frac{2}{\pi} \arctan2(y_{2}, y_{1}), \] This yields the inverse function \( g: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} \), where \( g(y_{1}, y_{2}) = (\sqrt{y_{1}^2 + y_{2}^2}, \frac{2}{\pi} \arctan2(y_{2}, y_{1})) \). The two-argument arctangent function, \( \arctan2(y, x) \), computes \( \text{the angle} \) whose tangent is \( y/x \), taking into account the sign of both arguments to find the correct quadrant.

Key Concepts

Jacobian Matrix

Understanding the Jacobian matrix is crucial when studying multivariable functions and their transformations. It's constructed from the first-order partial derivatives of a vector function. For a function from \( \mathbb{R}^{n} \rightarrow \mathbb{R}^{m} \), the Jacobian matrix is an \(m \times n\) matrix. Each element of this matrix represents how a component of the output vector changes with a slight variation in a component of the input vector.

In the exercise, the vector function \( f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} \) has its Jacobian matrix constructed by taking the partial derivatives of each output component with respect to each input component. The calculation of the matrix's elements is straightforward yet critical for determining if the function \( f \) is invertible near a given point by checking the Jacobian determinant.

Locally Invertible Function

A function is called locally invertible at a point if it can be inverted in a region around that point. The Inverse Function Theorem provides us with a way to verify the local invertibility of \( C^1 \) functions—that is, functions which are continuously differentiable.

In our case, the function in question needs to meet specific criteria to be locally invertible. Firstly, it should be continuously differentiable; this ensures that the function behaves 'nicely' enough to consider local inversion. Secondly, the Jacobian determinant at the point of interest must not be zero because a zero determinant would imply that the function is 'flattening' the space in a way that local inversion is not possible. By examining the determinant of the Jacobian matrix in the exercise, we can conclude that the function \( f \) is locally invertible around any point where \( x_{1} > 0 \) and \( -1 < x_{2} < 1 \) due to the determinant being positive.

Continuous Differentiability

Continuous differentiability, denoted as \( C^1 \) differentiability, refers to functions that have continuous first derivatives. This is an essential condition in many mathematical theorems because it implies a certain smoothness of the function. A \( C^1 \) function does not have any sharp 'corners' or 'edges'—the graph of the function has a smooth curve.

For functions from \( \mathbb{R}^{n} \) to \( \mathbb{R}^{m} \) to be \( C^1 \), all the first-order partial derivatives must exist and be continuous. This property ensures local predictability and stability of the function's behavior, which are necessary for the application of the Inverse Function Theorem as shown in the original exercise.

Determinant of a Matrix

The determinant of a matrix is a special value that can be computed from its elements and provides important properties about the matrix and the linear transformation it represents. In two dimensions, the determinant gives information about the area distortion that occurs when a region in \( \mathbb{R}^{2} \) is transformed by the matrix.

The determinant being non-zero is synonymous with the matrix being invertible, meaning that the transformation it represents is not compressing the space into a lower dimension. In the given exercise, the determinant of the Jacobian matrix, which is always positive under the stated conditions, proves that the function does not 'squash' the space into a lower dimension, and hence invertibility is attained locally. This is because the positive determinant signifies a preservation of orientation and guarantees that near each point, the function is a bijection.