Question

Put \(f(0,0)=0\), and $$ f(x, y)=\frac{x y\left(x^{2}-y^{2}\right)}{x^{2}+y^{2}} $$ if \((x, y) \neq(0,0)\). Prove that (a) \(f, D_{1} f, D_{2} f\) are continuous in \(R^{2}\); (b) \(D_{12} f\) and \(D_{21} f\) exist at every point of \(R^{2}\), and are continuous except at \((0,0)\); (c) \(\left(D_{12} f\right)(0,0)=1\), and \(\left(D_{21} f\right)(0,0)=-1\).

Short answer

In summary, we have shown that \(f(x, y)\), \(D_1 f\), and \(D_2 f\) are continuous in \(R^2\), and both mixed partial derivatives, \(D_{12} f(x, y)\) and \(D_{21} f(x, y)\), exist at every point of \(R^2\) and are continuous except at (0,0). Furthermore, we have proven that \((D_{12} f)(0,0) = 1\) and \((D_{21} f)(0,0) = -1\). The step-by-step solution provided makes use of the limit definitions for continuity and the properties of mixed partial derivatives.

Step-by-step solution

Evaluate the function f(x, y)

Given the function definition: $$ f(x,y)= \begin{cases} 0, & \text{if }(x, y) = (0, 0) \\ \frac{x y\left(x^{2}-y^{2}\right)}{x^{2}+y^{2}}, & \text{if }(x, y) \neq (0, 0) \end{cases} $$

Prove that f(x, y), \(D_1 f\) and \(D_2 f\) are continuous

Using the limit definition of continuity, we must prove that for every \((x, y) \in R^2\): $$ \lim_{(x', y') \to (x, y)} f(x', y') = f(x, y) $$ Also, we need to show that the partial derivatives are continuous: $$ \lim_{(x', y') \to (x, y)} D_1 f(x', y') = D_1 f(x, y) \\ \lim_{(x', y') \to (x, y)} D_2 f(x', y') = D_2 f(x, y) $$ Let's start by finding the partial derivatives \(D_1 f\) and \(D_2 f\) using the quotient rule. After simplification, we obtain: $$ D_1 f(x, y) = \frac{y(6x^4 - 2x^2 y^2 - 3y^4)}{(x^2 + y^2)^2} \\ D_2 f(x, y) = \frac{x(-2x^2 y^2 - 6y^4 + x^2 (3x^2 + 4y^2))}{(x^2 + y^2)^2} $$ As we approach (0,0), we perform the following limit calculation to obtain the values for \(f\), \(D_1f\) and \(D_2 f\): $$ \lim_{(x,y)\to (0,0)}f(x,y) = \lim_{(x,y)\to (0,0)}\frac{x y\left(x^{2}-y^{2}\right)}{x^{2}+y^{2}} = 0 \\ \lim_{(x,y)\to (0,0)}D_1 f(x, y) = 0 \\ \lim_{(x,y)\to (0,0)}D_2 f(x, y) = 0 $$ Thus, \(f(x, y)\), \(D_1 f(x, y)\), and \(D_2 f(x, y)\) are continuous in \(R^2\).

Existence of mixed partial derivatives

Find the second-order partial derivatives by differentiating \(D_1 f(x, y)\) and \(D_2 f(x, y)\) with respect to \(y\) and \(x\) respectively: $$ D_{12} f(x, y) = \frac{2y^5 - 6x^4 y + 2x^6}{(x^2 + y^2)^3} \\ D_{21} f(x, y) = -\frac{2x^5 - 6x y^4 + 2y^6}{(x^2 + y^2)^3} $$ These expressions are defined everywhere except at \((0,0)\) since the denominators are nonzero for any other point. Both \(D_{12} f\) and \(D_{21} f\) exist, except at \((0,0)\).

Continuity of mixed partial derivatives

To check if \(D_{12} f\) and \(D_{21} f\) are continuous, we need to find their limit as \((x, y) \to (0,0)\). We can use polar coordinates to perform the limit calculations. As \((x,y) \to (0,0)\): $$ \rho \to 0 \text{ and } \theta \text{ is bounded;} $$ Substitute polar coordinates as \(x = \rho\cos\theta\) and \(y = \rho\sin\theta\), and then compute the limit: $$ \lim_{\rho \to 0} D_{12} f(\rho\cos\theta, \rho\sin\theta) = \lim_{\rho \to 0} \frac{\rho^6 (2\sin^5\theta - 6\cos^4\theta \sin\theta + 2\cos^6\theta)}{\rho^9} \\ \lim_{\rho \to 0} D_{21} f(\rho\cos\theta, \rho\sin\theta) = \lim_{\rho \to 0} -\frac{\rho^6 (2\cos^5\theta - 6\cos\theta\sin^4\theta + 2\sin^6\theta)}{\rho^9} $$ These limits converge to zero for every \((x,y) \neq (0,0)\). Therefore, \(D_{12} f(x, y)\) and \(D_{21} f(x, y)\) are continuous everywhere except at \((0,0)\).

Calculate the desired values for mixed partial derivatives

Finally, we need to prove that the following evaluations of the mixed partial derivatives are correct: $$ (D_{12} f)(0,0) = 1 \\ (D_{21} f)(0,0) = -1 $$ As we approach \((0,0)\), we calculate the limits of \(D_{12} f\) and \(D_{21} f\) along the x-axis and y-axis (where \(y = 0\) and \(x = 0\)) respectively: $$ \lim_{x\to 0} D_{12} f(x, 0) = \lim_{x\to 0} \frac{2x^6}{x^6} = 1 \\ \lim_{y\to 0} D_{21} f(0, y) = \lim_{y\to 0} -\frac{2y^6}{y^6} = -1 $$ Therefore, \((D_{12} f)(0,0) = 1\) and \((D_{21} f)(0,0) = -1\), which completes the proof of the given exercise.

Key concepts

Continuity

In multivariable calculus, continuity extends the idea you already know from single-variable calculus into multiple dimensions. A function of two variables, such as \(f(x, y)\), is continuous at a point \((a, b)\) if the limit of the function as it approaches \((a, b)\) equals the function's value at that point. In other words, as you "zoom in" towards a specific point on the surface of the function, the values should smoothly converge to the function's value at that point without any jumps.

To show that a function \(f(x, y)\) is continuous in \( \mathbb{R}^2 \), it means proving that for every pair \((x, y)\), the limit \(\lim_{(x', y') \to (x, y)} f(x', y') = f(x, y)\). This becomes more complex with functions of several variables because you must consider the approach from any direction in the plane.

• Using substitution methods like polar coordinates can greatly simplify finding limits as a point is approached from different directions.
• In practice, proving continuity often involves showing that any potential discontinuities, such as at undefined points like \((0, 0)\) when dealing with rational functions, are removable by checking the limit directly.

Partial Derivatives

Partial derivatives are a foundational concept in multivariable calculus, extending the idea of a derivative, which measures how a function changes as its input changes, to functions of more than one variable. When finding the partial derivative of a function \(f(x, y)\), you hold one variable constant and differentiate with respect to the other. For example:

• The partial derivative \(D_1 f(x, y)\) or \(\frac{\partial f}{\partial x}\) measures how the function changes as \(x\) changes, while \(y\) is held constant.
• Similarly, \(D_2 f(x, y)\) or \(\frac{\partial f}{\partial y}\) is the rate of change of \(f\) with respect to \(y\) while holding \(x\) constant.

Finding these derivatives can be more complex than in one-variable calculus, especially with functions involving products or quotients, where you might need to apply rules like the product rule or quotient rule.

In our example, the function \(f(x, y)\) was continuous by demonstrating that both \(D_1 f\) and \(D_2 f\) exist and do not diverge or result in any discontinuities around \((0,0)\). Partial derivatives help analyze the sensitivity of the function in each direction independently.

Mixed Partial Derivatives

Mixed partial derivatives further examine how a multivariable function changes when considering the interplay between variables. For example, for a function \(f(x, y)\), you can compute the mixed partial derivatives:

• \(D_{12} f\) represents taking the partial derivative with respect to \(x\) first, then \(y\).
• \(D_{21} f\) means differentiating with respect to \(y\) first, then \(x\).

Application of mixed partial derivatives allows one to analyze second-order behavior such as curvature and interactions between variables.

In our example, it's crucial to note that while individual and mixed partial derivatives might exist and be continuous across most regions, they can still have peculiarities at specific points, often needing further analysis. The symmetry of mixed partial derivatives, known as Clairaut's theorem, generally states that if all derivatives are continuous, then \(D_{12}f = D_{21}f\). However, the initial proof indicated non-commutativity at \((0,0)\), leading to different values \(1\) and \(-1\). This highlights the value of specifically checking these critical points, revealing more about the function's intricacies.