Homework Help: How do you know a function is well defined?

Tags:
1. Oct 11, 2014

minor_embedding

1. The problem statement, all variables and given/known data
I was given the following function

f(x,y) =
\begin{cases}
\frac{x^2y}{x^4+y^2} & (x,y) \neq 0 \\
0 & (x,y) = 0
\end{cases}

Which of the following are true?
(I) f is not continuous at (0, 0).
(II) f is differentiable everywhere
(III) f as a well defined partial derivatives everywhere (i.e. $\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$ are both defined)
(IV) f is continuous at (0, 0) but not differentiable at (0, 0).

3. The attempt at a solution

I know 1 is true. Since 1 is true 2 is not true and IV is definitely not true. But the answer states that 3 is also true.

I don't think I understand what it means to be a well defined partial derivative.

2. Oct 12, 2014

haruspex

Are you sure (I) is true?

3. Oct 12, 2014

Staff: Mentor

Seems like it's true to me (i.e., that f is discontinuous at (0, 0)). Along the path y = x2 $\lim_{(x, y) \to (0, 0)} f(x, y) \neq f(0, 0)$.

4. Oct 12, 2014

minor_embedding

So 1 is definitely true, but what does it mean to be a well defined partial derivative?

5. Oct 12, 2014

Simon Bridge

The problem statement tells you what this means ... namely that both partials are defined everywhere.

6. Oct 12, 2014

HallsofIvy

In polar coordinates, $x= r cos(\theta)$, $y= r sin(\theta)$ so that
$$\frac{x^2y}{x^4+ y^2}= \frac{r^3 cos^2(\theta)sin(\theta)}{r^4 sin^4(\theta)+ r^2 cos^2(\theta)}= r\frac{cos^2(\theta)sin(\theta)}{r^2 sin^2(\theta)+ cos^2(\theta)}$$. As (x, y) goes to (0, 0), along any path, r goes to 0 so that fraction goes to 0. This function is continuous at (0, 0).

7. Oct 12, 2014

Orodruin

Staff Emeritus
Did you try the path $\theta(r) = r$? (Also, your denominator mixes up sin and cos for x and y.) Mark44 already showed that this path leads to a non-zero limit in #3.

8. Oct 12, 2014

pasmith

Let $0 < |C| \leq \frac12$ and $\alpha = \frac{1}{2C}(1 + \sqrt{1 - 4C^2}) \neq 0$ so that $C\alpha^2 - \alpha + C = 0$. Then on the continuous path $(x(t),y(t)) = (t, \alpha t^2)$ we have $$C(x^4 + y^2) - x^2 y = (C\alpha^2 - \alpha + C)t^4 = 0$$ so that $$\lim_{t \to 0} f(x(t),y(t)) = C \neq 0$$ and $f$ is not continuous at the origin.

Last edited: Oct 12, 2014
9. Oct 12, 2014

minor_embedding

I realize that my question probably sounds pretty idiotic with or without the problem statement. But what I'm confused about is if it's not continuous at the origin, then it can't be differentiable at the origin. So then how is it defined everywhere? Does the second part of the piecewise take care of that?

10. Oct 12, 2014

gopher_p

I think the gist of the "well-defined" part of (II) is that you can use easy derivative rules and formulas to find the partials away from the origin, but you need to appeal to the definition of the partial derivative to determine whether $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exist at the origin.

11. Oct 12, 2014

haruspex

It fails to be continuous because there exist paths to the origin which produce different limits. But a partial derivative means keeping one of the variables constant, so the 'paths' for these are parallel to the axes only. On those paths, the function has a limit of zero at the origin, consistent with the value there.

12. Oct 12, 2014

vela

Staff Emeritus
You seem to be assuming if all the partial derivatives of a function exist at a point, the function is differentiable at that point. This isn't the case.

13. Oct 12, 2014

Simon Bridge

The phrasing tells you what you are supposed to be looking for - you are not asked if the function is differentiable everywhere, only if it has "a well defined derivative" everywhere - then the statement tells you what it means by "well defined derivative".

So - if you want to say that it does not have well defined derivatives everywhere, in the sense stated, then you should prove that by showing that one or other partial does not exist somewhere. Have you tried that?

Note: the two statements ("f is differentiable everywhere", and "f'x, f'y exist everywhere") may be equivalent - but since you are unsure you should check it out.