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Does the pair of equations \(x=u^{2}-v^{2}, y=2 u v\), have an inverse on a neighborhood of \((0,0)\) ?

Short Answer

Expert verified
No, the mapping does not have an inverse near \((0,0)\) because the Jacobian determinant is zero.

Step by step solution

01

Understand the Problem

We need to determine if the mapping given by \((u, v) \to (x, y) = (u^2 - v^2, 2uv)\) is invertible around the point \((0,0)\). For the mapping to be invertible, it must be one-to-one and onto in a neighborhood of \((0,0)\). A common approach is to check if the Jacobian determinant at this point is non-zero.
02

Find the Jacobian Matrix

The Jacobian matrix \(J\) for the transformation is derived from the partial derivatives of \(x = u^2 - v^2\) and \(y = 2uv\):\[J = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} = \begin{bmatrix} 2u & -2v \ 2v & 2u \end{bmatrix}.\]
03

Evaluate the Jacobian Determinant

The determinant of the Jacobian \(J\) is calculated as follows:\[\det(J) = (2u)(2u) - (-2v)(2v) = 4u^2 + 4v^2 = 4(u^2 + v^2).\]At the point \((u, v) = (0, 0)\), substitute in the values to find:\[\det(J) = 4(0^2 + 0^2) = 0.\]
04

Analyze the Result

Since the Jacobian determinant at \((0,0)\) is zero, the linear approximation of the system is singular there. This means the mapping is not locally invertible in any neighborhood of \((0,0)\). Therefore, there is no inverse function around \((0, 0)\).

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Key Concepts

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

Jacobian determinant
The Jacobian determinant plays a crucial role in examining the invertibility of a function at a specific point. When we have a transformation defined by equations like \( x = u^2 - v^2 \) and \( y = 2uv \), the first step in analyzing invertibility is to construct the Jacobian matrix. This matrix is formed by taking the partial derivatives of each component of the transformation with respect to each variable.
For our example, the Jacobian matrix \( J \) is:
\[J = \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} = \begin{bmatrix} 2u & -2v \ 2v & 2u \end{bmatrix}.\]
Next, we calculate the determinant of the Jacobian matrix from this step. The determinant value helps identify if the mapping is locally invertible. In this case, the determinant is \( 4(u^2 + v^2) \), and as we substitute \((u,v) = (0,0)\), we find it equals zero.
  • When the determinant is non-zero, it indicates that the mapping may be invertible in a neighborhood around the point.
  • If the determinant is zero, as with our exercise, it suggests that the mapping is not invertible at that point.
The zero determinant here indicates a singularity, meaning the transformation loses its unique inverse characteristic in any neighborhood around this critical point.
invertibility
Invertibility is a key concept when studying functions and transformations in mathematics. A function is said to be invertible around a point if we can reverse the function and find a unique pre-image for every image in some neighborhood of that point. For our particular mapping given by \((u, v) \rightarrow (x, y) = (u^2 - v^2, 2uv)\), determining invertibility involves analyzing the properties of the Jacobian determinant.
The crux of the matter lies in whether the function is both one-to-one and onto around a specific point. If it is, then an inverse function exists locally.
  • **One-to-One:** Each unique input produces a unique output, meaning the function does not "fold over" itself in the neighborhood.
  • **Onto:** Every possible output in the neighborhood is covered by the function.
In our exercise, we found that the Jacobian determinant equals zero at \((0,0)\), indicating that the function lacks local invertibility. It cannot be reversed around this point, missing both one-to-one and onto properties due to the singular nature of the transformation. Hence, there is no inverse function in the immediate vicinity of \((0, 0)\).
neighborhood mapping
Understanding neighborhood mapping is crucial when we want to explore the local behavior of functions around specific points. Think of a neighborhood as a small area around a particular point, like the vicinity of a house. In mathematics, it refers to an interval or vicinity where certain properties, such as invertibility, can be studied.
For a function like \((u, v) \rightarrow (x, y) = (u^2 - v^2, 2uv)\), neighborhood mapping involves observing how points in one small area map to another through the transformation. In this small region, checking for properties like invertibility becomes feasible by looking at things like the Jacobian determinant.
  • **Local Behavior:** Examine how inputs close to a point, such as \((0,0)\), alter or transform to outputs. This behavior gives insights into whether the function holds certain properties in that neighborhood.
  • **Jacobian's Role:** By checking the Jacobian determinant, we can determine if the function remains stable enough for an inverse mapping, that is, converting output back to input in the neighborhood.
In this context, the zero value of the Jacobian determinant at \((0,0)\) tells us that the function does not map uniquely from input to output at this point, rendering it non-invertible in any immediate neighborhood. Therefore, there is no way to consistently reverse the mapping in the local region around \((0, 0)\).

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Most popular questions from this chapter

(Double limits and iterated limits, revisited) A bit of care is needed with the statement "If a double limit exists, so do the two iterated limits, and the two iterated limits equal the double limit." Let $$ f(x, y)=\left\\{\begin{array}{ll} y+x \sin \frac{1}{y} & \text { if } y \neq 0 \\ 0 & \text { if } y=0 \end{array}\right. $$ (a) Show that \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)=0\) but \(\lim _{x \rightarrow 0} \lim _{y \rightarrow 0} f(x, y)\) does not exist.

Let \(f\) and \(g\) have continuous first partial derivatives on an open set \(D \subset \mathbb{R}^{2}\) and let \(T: D \rightarrow \mathbb{R}^{2}\) be defined by $$ T(u, v)=(f(u, v), g(u, v)) $$ Prove that if \(T\) is one-to-one on \(D\), then the set \(T(D)\) is open.

Suppose that the dependencies of the variables can be described informally as " \(w\) depends on all three of \(x, y\), and \(z\) and each in turn depends on both \(s\) and \(t . "\) Express this schematically and write a chain rule for it.

Is it still true that the gradient is in the direction of greatest change for the function if \(f: \mathbb{R}^{n} \rightarrow \mathbb{R}\) where \(n=1 ?\)

(Cauchy-Riemann Equations) The two equations introduced in Exercise 12.2.2(c) $$ \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \text { and } \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} $$ are called the Cauchy-Riemann equations and are fundamental in complex analysis. This is so because a necessary and sufficient condition that a continuous function of a complex variable defined in a neighborhood \(N\) of a point in the complex plane be analytic is that its real and imaginary parts satisfy the Cauchy-Riemann equations. This means that if $$ f(x, y)=(u(x, y), v(x, y)) $$ then both equations in (3) are valid for all \((x, y) \in N\). [In complex notation we write \(z=x+i y\) and \(f(z)=u+i v\).] Show that each of the following functions \(f=u+i v\) is analytic in a neighborhood of the origin. (a) \(u=x^{2}-y^{2}\) and \(v=2 x y\) (b) \(u=e^{x} \cos y\) and \(v=e^{x} \sin y\) (c) \(u=3 x+y\) and \(v=3 y-x\)

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