Really Stuck on Partial derivatives question

Question

Ok so im really stuck on a question. It goes: Consider $$u(x,y) = xy \frac {x^2-y^2}{x^2+y^2} $$ for $(x,y)$ $ \neq $ $(0,0)$ and $u(0,0) = 0$. calculate $\frac{\partial u} {\partial x} (x,y)$ and $\frac{\partial u} {\partial y} (x,y)$ for all $ (x,y) \in \Bbb R^2. $ show that $ \frac {\partial^2 u} {\partial x \partial y} (0,0) \neq \frac {\partial^2 u} {\partial y \partial x} (0,0) $. Check, using polar coordinates, that $ \frac {\partial u}{\partial x} and \frac {\partial u}{\partial y} $ are continuous at $(0,0)$

Any help really appreciated. Cheers

Answer 1

We are given:

$$u(x, y)=\begin{cases} xy \frac {x^2-y^2}{x^2+y^2}, ~(x, y) \ne (0,0)\\\\ ~~~0, ~~~~~~~~~~~~~~~(x, y) = (0,0)\;. \end{cases}$$

I am going to multiply out the numerator for ease in calculations, so we have:

$$\tag 1 u(x, y)=\begin{cases} \frac {x^3y - xy^3}{x^2+y^2}, ~(x, y) \ne (0,0)\\\\ ~~~0, ~~~~~~~~~~~~~~(x, y) = (0,0)\;. \end{cases}$$

We are asked to:

• (a) Find: $\displaystyle \frac{\partial u} {\partial x} (x,y) ~\forall x~ \in \Bbb R^2$

• (b) Find: $\displaystyle \frac{\partial u} {\partial y} (x,y) ~\forall x~ \in \Bbb R^2$

• (c) Show that $ \displaystyle \frac {\partial^2 u} {\partial x \partial y} (0,0) \neq \frac {\partial^2 u} {\partial y \partial x} (0,0)$

• (d) Check, using polar coordinates, that $\displaystyle \frac {\partial u}{\partial x} \text{and} \frac {\partial u}{\partial y} $ are continuous at $(0,0)$

Using $(1)$, for part $(a)$, we get:

$\tag 2 \displaystyle \frac{\partial u} {\partial x} (x,y) = \frac{(3x^2y- y^3)(x^2+y^2) - 2x(x^3y - xy^3)}{(x^2 + y^2)^2} = \frac{x^4y + 4x^2y^3-y^5}{(x^2+y^2)^2}$

Using $(1)$, for part $(b)$, we get:

$\tag 3 \displaystyle \frac{\partial u} {\partial y} (x,y) = \frac{(x^3-3xy^2)(x^2+y^2) - 2y(x^3y - xy^3)}{(x^2+y^2)^2} = \frac{x^5 - 4x^3y^2-xy^4}{(x^2+y^2)^2}$

Next, we need mixed partials, so using $(2)$, we have:

$ \tag 4 \displaystyle \frac {\partial^2 u} {\partial x \partial y} (x, y) = \frac{(x^4+ 12x^2y^2-5y^4)(x^2+y^2)^2 - 2(x^2+y^2)(2y)(x^4y + 4x^2y^3-y^5)}{(x^2+y^2)^4} = \frac{x^6 + 9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3} = \frac {\partial^2 u} {\partial y \partial x} (x, y)$

Thus, we get:

$$\tag 5 \displaystyle \frac {\partial^2 u} {\partial x \partial y} (x, y) = \frac {\partial^2 u} {\partial y \partial x} (x, y) = \begin{cases} \frac{x^6 + 9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}, ~(x, y) \ne (0,0)\\\\ ~~~~~~~~~~~~~~~~~~0, ~~~~~~~~~~~~~~~~~~~~~~~~(x, y) = (0,0)\;. \end{cases}$$

Now, for part $(c)$, we want to show that $ \frac {\partial^2 u} {\partial x \partial y} (0,0) \neq \frac {\partial^2 u} {\partial y \partial x} (0,0)$, so we need to find the limits of each mixed partial.

We have:

$ \tag 6 \displaystyle \frac{\partial^2 u} {\partial x \partial y} (0,0) = \lim\limits_{h \to 0} \frac{\frac{\partial u}{\partial x} (0, h) - \frac{\partial u}{\partial x} (0, 0)}{h} = \lim\limits_{h \to 0} \frac{-h^5/h^4}{h} = \lim\limits_{h \to 0} \frac{-h}{h} = -1$, and

$ \tag 7 \displaystyle \frac{\partial^2 u} {\partial y \partial x} (0,0) = \lim\limits_{h \to 0} \frac{\frac{\partial u}{\partial y} (h, 0) - \frac{\partial u}{\partial y} (0, 0)}{h} = \lim\limits_{h \to 0} \frac{h^5/h^4}{h} = \lim\limits_{h \to 0} \frac{h}{h} = +1$

$\therefore$, for part $(c)$, we have shown that:

$$\frac {\partial^2 u} {\partial x \partial y} (0,0) \neq \frac {\partial^2 u} {\partial y \partial x} (0,0)$$

as desired.

Can you handle part $(d)$?

Regards