Does the function have total differential in [0,0]?

[twoflower]

Hi,

I have this excercise to do:

Let's have function f defined as follows:

$$f(x,y) = \frac{xy(x^2 - y^2)}{x^2 + y^2},\ \ [x,y] \neq [0,0]$$

$$f(x,y) = 0,\ \ [x,y] = [0,0]$$

Find out, whether this function has total differential in [0,0].

Well, first I observed that partial derivatives aren't continuous in [0,0], which is what I expected.

So I computed them from the definition:

$$\frac{\partial f}{\partial x}(0,0) = \lim_{t \rightarrow 0} \frac{f(t,0) - f(0,0)}{t} = 0$$

$$\frac{\partial f}{\partial y}(0,0) = \lim_{t \rightarrow 0} \frac{f(0,t) - f(0,0)}{t} = 0$$

So, if the total differential exists in [0,0], it must be zero linear transform.

Now I have to checkout, if such a transform satisfies the limit and thus it is total differential:

$$\lim_{||h|| \rightarrow 0} \frac{f((0,0) + h) - f(0,0) - L(h)}{||h||} = 0 ?$$

If I rewrite it in a slightly different way ( I put $h = (h_1,h_2)$ and $h_2 = kh_1$, I get

$$\lim_{h_1 \rightarrow 0} \frac{\frac{kh_1^2(h_1^2 - k^2h_1^2)}{h_1^2(1+k^2}}{h_1\sqrt{1+k^2}} = \frac{k(1-k^2)}{\sqrt{(1+k^2)^3}} \lim_{h_1 \rightarrow 0} h_1$$

Well, I don't know whether this goes to zero so I tried another way, ie. approaching [0,0] on the x-axis and thus putting $h_2 = 0$

The limit I got was zero, so I can't say at this moment that the total differential doesn't exist in [0,0]. Anyway, according to right results our professor gave us, total differential doesn't exist in [0,0].

How could I prove that?

Thank you very much.

 

[Tide]

Did your prof give any rationale for his or her assertion?

 

[HallsofIvy]

I don't know why you say you don't know whether
$$\lim_{h_1 \rightarrow 0} \frac{\frac{kh_1^2(h_1^2 - k^2h_1^2)}{h_1^2(1+k^2}}{h_1\sqrt{1+k^2}} = \frac{k(1-k^2)}{\sqrt{(1+k^2)^3}} \lim_{h_1 \rightarrow 0} h_1$$
goes to 0- the part involving k is finite for all k while the limit
$$\lim_{h_1 \rightarrow 0} h_1$$
is clearly 0.

Another way to handle a problem like this is to put it into polar coordinates. In polar coordinates, this reduces to
$$f(r,\theta)= r^2 sin(\theta)cos(\theta)cos(2\theta)$$
which clearly is differentiable at the origin.

 

[Tide]

Halls,

That's what I did and the trig terms further simplify to $\sin 4 \theta$ (within a factor of 2) which makes the differentiation even easier.

 

[twoflower]

I don't know why you say you don't know whether
$$\lim_{h_1 \rightarrow 0} \frac{\frac{kh_1^2(h_1^2 - k^2h_1^2)}{h_1^2(1+k^2}}{h_1\sqrt{1+k^2}} = \frac{k(1-k^2)}{\sqrt{(1+k^2)^3}} \lim_{h_1 \rightarrow 0} h_1$$
goes to 0- the part involving k is finite for all k while the limit
$$\lim_{h_1 \rightarrow 0} h_1$$
is clearly 0.
Another way to handle a problem like this is to put it into polar coordinates. In polar coordinates, this reduces to
$$f(r,\theta)= r^2 sin(\theta)cos(\theta)cos(2\theta)$$
which clearly is differentiable at the origin.

Well, I don't know polar coordinates. Maybe we sometimes touched them but didn't have them in syllabus.

So I'm trying to do it in a non-elegant way :)

So I got to the point where I need to prove the zero limit involving the total differential.

That is, I need to prove that the following goes to zero as $h = [h_1,h_2] \rightarrow [0,0]$

$$\left|\frac{h_1h_2(h_1^2 - h_2^2)}{(h_1^2 + h_2^2)\sqrt{h_1^2 + h_2^2}}\right|$$

Could you please give me some hint how to prove that? It would be sufficient to find there some part that would be bounded for all $h_1,h_2$ going to 0 and multiplying it with some zero part would give me the zero as the result.