Question

Let \(f(x, y)\) have continuous partial derivatives in the domain \(D\) in the \(x y\)-plane. Further let \(|\nabla f| \leq K\) in \(D\), where \(K\) is a constant. In each of the following cases, determine whether this implies that \(f\) is uniformly continuous: a) \(D\) is the domain \(x^{2}+y^{2}<1\). [Hint: If \(s\) is distance along a line segment from \(\left(x_{1}, y_{1}\right)\) to \(\left(x_{2}, y_{2}\right)\), then \(f\) has directional derivative \(d f / d s=\nabla f \cdot \mathbf{u}\) along the line segment, where \(\mathbf{u}\) is an appropriate unit vector.] b) \(D\) is the domain \(|x|<1,|y|<1\), excluding the points \((x, 0)\) for \(0 \leq x<1\).

Short answer

To summarize, we analyzed the uniform continuity of a function \(f(x, y)\) with continuous partial derivatives and \(|\nabla f| \leq K\) in two different domains: 1. In the domain \(x^2 + y^2 < 1\), the function \(f\) is uniformly continuous, as we were able to find a \(\delta\) given any \(\varepsilon > 0\) that satisfied the uniform continuity condition. 2. In the domain \(|x|<1,|y|<1\), excluding the points \((x, 0)\) for \(0 \leq x<1\), the function \(f\) is not necessarily uniformly continuous. This is because the domain is not closed, and we could find an interval around \((0,0)\) that included points where the function might not be bounded.

Step-by-step solution

Analyzing uniform continuity for domain \(x^2 + y^2 < 1\)

Consider the domain \(D\) given by \(x^2 + y^2 < 1\). It's a bounded and closed domain (a unit disc). Now we have to check if \(f\) is uniformly continuous on this domain.

Applying uniform continuity definition on given domain

Since \(|\nabla f| \leq K\), its maximum magnitude is at most \(K\) in domain \(D\). Let \(s = d((x_1, y_1), (x_2, y_2))\) be the distance along a line segment between two points in \(D\). The directional derivative of \(f\) along this line segment is given by \(df/ds = \nabla f \cdot \mathbf{u}\), where \(\mathbf{u}\) is an appropriate unit vector. By Cauchy-Schwarz inequality, we have: \(\left|\frac{df}{ds}\right| = \left|\nabla f \cdot \mathbf{u}\right| \leq |\nabla f||\mathbf{u}|\) Since \(|\nabla f| \leq K\) and \(|\mathbf{u}| = 1\), we have: \(\left|\frac{df}{ds}\right| \leq K\) Then, integrate both sides with respect to \(s\) over the interval \([0, s]\): \(\int_0^s \left|\frac{d f}{ds}\right| ds \leq \int_0^s K ds\) which gives: \(|f(x_2, y_2) - f(x_1, y_1)| \leq Ks\)

Verifying uniform continuity in the given domain

Now, given any \(\varepsilon > 0\), choose \(\delta = \varepsilon / K\). For any pair of points \((x_1, y_1)\) and \((x_2, y_2)\) in \(D\) such that \(d((x_1, y_1), (x_2, y_2)) < \delta\), we have: \(|f(x_1, y_1) - f(x_2, y_2)| \leq Ks < K\delta = \varepsilon\) This confirms that the function \(f\) is uniformly continuous in the domain \(x^2 + y^2 < 1\). #Case b)#

Analyzing uniform continuity for domain \(|x|

Now, consider the domain \(D\) given by \(|x|<1,|y|<1\), excluding the points \((x, 0)\) for \(0 \leq x<1\). We can observe that this domain is not a closed set. We can apply a similar analysis as before to check if \(f\) is uniformly continuous on this domain.

Applying uniform continuity definition on given domain

As in Case a), consider the directional derivative of \(f\) along a line segment and using Cauchy-Schwarz inequality we have: \(\left|\frac{df}{ds}\right| \leq K\) Integrating with respect to \(s\) over the interval \([0, s]\): \(|f(x_2, y_2) - f(x_1, y_1)| \leq Ks\)

Verifying uniform continuity in the given domain

Consider a sequence of points \((x_n, y_n) = (\frac{1}{n+1}, \frac{1}{n+1})\) in the domain. The sequence converges to point \((0,0)\). Now consider the sequence \((x'_n, y'_n) = (\frac{1}{n+1}, 0)\). The sequence \((x'_n, y'_n)\) also converges to point \((0,0)\), but it's not in the domain \(D\). Any neighborhood of \((0,0)\) will contain some points \((x'_n, y'_n)\). Observe that we can find an interval around \((0,0)\) which includes points where the function might not be bounded, due to the reason mentioned above. In conclusion, we cannot ensure uniform continuity of \(f\) on the domain \(|x|<1,|y|<1\), excluding the points \((x, 0)\) for \(0 \leq x<1\).

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with the differentiation and integration of vector fields. These are functions with vectors as outputs that vary through an area or space.
In essence, we use it to understand how vector fields such as fluid flow, electric, or gravitational fields behave in physical space.
In vector calculus, a key concept is the gradient, represented as \(abla f\), which combines partial derivatives of a function. - **Gradient**: The gradient of a function \(f(x, y)\) is a vector pointing in the direction of the greatest rate of increase of \(f\). It consists of the partial derivatives \(\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\).
- **Applications**: These are often used in finding tangent planes, Lagrange multipliers, or analyzing motions within fields.
By understanding vector calculus, we analyze functions operating in multi-dimensional spaces, providing a foundational skill for physics, engineering, and data science.

Directional Derivative

In vector calculus, the directional derivative provides the rate of change of a function in a specific direction. This tool helps understand how quickly a function value changes when you move in a particular direction in the input space.
Imagine you're hiking up a mountain: the steepest path uphill represents the direction of the gradient. However, if you want to move in any other direction, the directional derivative helps estimate the change in elevation relative to that path.Here’s how it works:- **Definition**: The directional derivative of a function \(f(x, y)\) in the direction of a unit vector \(\mathbf{u} = (u_1, u_2)\) is given by: \[ D_\mathbf{u}f(x, y) = abla f \cdot \mathbf{u} \] where \(abla f\) is the gradient of the function.- **Calculation**: It combines the gradients' components with \(\mathbf{u}\)'s components as \(abla f \cdot \mathbf{u} = \frac{\partial f}{\partial x}u_1 + \frac{\partial f}{\partial y}u_2\).Directional derivatives thus allow detailed analysis of multi-variable functions beyond the main gradient direction.

Cauchy-Schwarz Inequality

The Cauchy-Schwarz Inequality is a fundamental principle in linear algebra and analysis. It provides a bound on the dot product of two vectors. This inequality is essential when working with directions and magnitudes, especially in optimization and computational solutions.
This principle asserts:- For any two vectors \(\mathbf{a}\) and \(\mathbf{b}\) in an inner product space, the Cauchy-Schwarz Inequality states: \[ |\mathbf{a} \cdot \mathbf{b}| \leq ||\mathbf{a}|| \cdot ||\mathbf{b}|| \]- Here, \(|\mathbf{a} \cdot \mathbf{b}|\) represents the dot product, and \(||\mathbf{a}||\) and \(||\mathbf{b}||\) are the magnitudes (or norms) of \(\mathbf{a}\) and \(\mathbf{b}\), respectively.This inequality ensures that the dot product of vectors does not exceed the product of their magnitudes.
In the context of directional derivatives, it can be used to bound the rate of change of the function by limiting the influence of direction through the vector's size, as seen in linking \(|abla f \cdot \mathbf{u}|\) with \(|abla f||\mathbf{u}|\).
Understanding and applying this inequality is crucial for solving problems involving vectors and ensuring results fall within expected limits.