Chapter 5: Chapter 5

Problem 129

(a) Let $\mathrm{f}: \mathrm{R}^{2} \rightarrow \mathrm{R}$ be defined by $f(x, y)=2 x y\left\\{\left(x^{2}-y^{2}\right) /\left(x^{2}+y^{2}\right)\right\\}, x^{2}+y^{2} \neq 0$ and $=0, \quad \mathrm{x}=\mathrm{y}=0$. Show that $\left(\partial^{2} \mathbf{f} / \partial \mathrm{x} \partial \mathrm{y}\right) \neq\left(\partial^{2} \mathrm{f} / \partial \mathrm{x} \partial \mathrm{y}\right)$ and explain why. (b) Does there exist a function $\mathrm{f}$ with continuous second partial derivatives (i.e., an element of $\mathrm{C}^{2}$ ) such that of $/ \partial \mathrm{x}=\mathrm{x}^{2}$ and \partialf $/ \partial \mathrm{y}=\mathrm{xy}$ ?

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In summary: a) The second mixed partial derivatives of the given function are not globally equal, as they are unequal for $(x, y) \neq (0, 0)$. This is because the function is not differentiable at the origin, so it does not satisfy the necessary requirements for the equality of mixed partial derivatives. b) There does not exist a function $f$ with continuous second partial derivatives such that our given conditions $\frac{\partial f}{\partial x} = x^2$ and $\frac{\partial f}{\partial y} = xy$ are satisfied, as a function solely dependent on $y$ cannot be equal to $xy$.

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Step 1: First, we compute the first partial derivatives of $f(x,y)$ with respect to $x$ and $y$. Since the given function $f$ is defined piecewise, we will differentiate it accordingly. For $(x, y) \neq (0, 0)$, we have \[\frac{\partial f}{\partial x} = 2y\left\\{\frac{\partial}{\partial x}\left(\frac{x^2 - y^2}{x^2+y^2}\right)\right\\}\] and \[\frac{\partial f}{\partial y} = 2x\left\\{\frac{\partial}{\partial y}\left(\frac{x^2 - y^2}{x^2+y^2}\right)\right\\}\]. We need to calculate these two derivatives. #Step 2: Compute the partial derivatives of the fraction

We now compute the partial derivatives of the fraction: \[\frac{\partial}{\partial x}\left(\frac{x^2 - y^2}{x^2+y^2}\right) = \frac{4xy(x^{2}+y^{2}) - 2x(x^{2}-y^{2})(2x)}{(x^{2}+y^{2})^{2}} = \frac{4x^{2}y^{2} - 2xy((x^4 - y^4)}{(x^2 + y^2)^2}\] and \[\frac{\partial}{\partial y}\left(\frac{x^2 - y^2}{x^2+y^2}\right) = \frac{-(4xy)(x^{2}+y^{2}) - 2y(x^{2}-y^{2})(2y)}{(x^{2}+y^{2})^{2}} = \frac{-4x^{2}y^{2} - y((x^4 - y^4)}{(x^2 + y^2)^2}\] Now we can substitute these results into the expressions for our first partial derivatives. #Step 3: Write down the first partial derivatives

Step 2: The first derivatives of $f(x, y)$ with respect to $x$ and $y$ are given by the following expressions when $(x,y) \neq (0,0)$: \[\frac{\partial f}{\partial x} = 2y\left\\{\frac{4x^{2}y^{2} - 2xy(x^4 - y^4)}{(x^2 + y^2)^2}\right\\}\] and \[\frac{\partial f}{\partial y} = 2x\left\\{\frac{-4x^{2}y^{2} - y(x^4 - y^4)}{(x^2 + y^2)^2}\right\\}\]. At the origin $(x, y) = (0, 0)$, the first partials are as follows: $\frac{\partial f}{\partial x}(0, 0) = 0$, and $\frac{\partial f}{\partial y}(0, 0) = 0$. #Step 4: Compute the second mixed partial derivatives

Now we need to compute the second mixed partial derivatives from our first derivatives: \[\frac{\partial^{2} f}{\partial x \partial y} = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)\] and \[\frac{\partial^{2} f}{\partial y \partial x} = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)\]. At the origin, we have: \[\frac{\partial^2 f}{\partial x \partial y}(0, 0) = \frac{\partial}{\partial y}\left(0\right) = 0\] and \[\frac{\partial^2 f}{\partial y \partial x}(0, 0) = \frac{\partial}{\partial x}\left(0\right) = 0\]. #Step 5: Compare the second mixed partial derivatives

Step 3: When $(x, y) \neq (0, 0)$, we observe that the second mixed partial derivatives are not equal due to their different expressions: \[\frac{\partial^{2} f}{\partial x \partial y} \neq \frac{\partial^{2} f}{\partial y \partial x}\] However, at the origin $(x, y) = (0, 0)$, the second mixed partial derivatives are equal (both being equal to 0). The second mixed partial derivatives are not globally equal, since they are unequal for $(x, y) \neq (0, 0)$. This is because the function is not differentiable at the origin, so it does not satisfy the necessary requirements for the equality of mixed partial derivatives. #b)#: Given that $\frac{\partial f}{\partial x} = x^2$ and $\frac{\partial f}{\partial y} = xy$, we can attempt to find a function f satisfying these criteria. #Step 1: Integrate with respect to x

We integrate the first relation with respect to $x$: \[\int x^2 dx = \frac{1}{3} x^3 + g(y)\] where $g(y)$ is an unknown function of $y$. #Step 2: Integrate with respect to y

Step 4: Now take the partial derivative of the above expression with respect to $y$: \[\frac{\partial}{\partial y}\left(\frac{1}{3}x^3 + g(y)\right) = g'(y)\] which must also be equal to the given condition, $xy$. Therefore, we have the equation: $g'(y) = xy$. #Step 3: Verify obtained derivatives with respect to conditions

From the analysis in step 2, we see that the function $g'(y) = xy$ will not be derivable from integrating with respect to $y$, as $g'(y)$ is supposed to be solely dependent on $y$. A function solely dependent on $y$ cannot be equal to $xy$. Therefore, there does not exist a function f with continuous second partial derivatives such that our given conditions are satisfied.

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Chapter 5: Chapter 5

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