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$f:\mathbb{R}^2 \to \mathbb{R}$

$f\Bigg(\begin{matrix}x\\y\end{matrix}\Bigg)=\begin{cases}\frac{xy^2}{x^2+y^2},(x,y)^T \neq(0,0)^T \\0 , (x,y)^T=(0,0)^T\end{cases}$

I need to determine all partial derivatives for $(x,y)^T \in \mathbb{R}^2$:

$f_x=y^2/(x^2+y^2)-2x^2y^2/(x^2+y^2)^2$ for $(x,y)^T \neq (0,0)$

$f_y=2xy/(x^2+y^2)-2xy^3/(x^2+y^2)^2$ for $(x,y)^T \neq (0,0)$

and $f_x=f_y=0$ for $(x,y)^T = (0,0)$.

Then I need to determine $\frac{\partial f}{\partial v}((0,0)^T)$ for all $v=(v_1,v_2)^T \in \mathbb{R}^2$.

I tried: $\frac{1}{s} (f(x+sv)-f(x))$ at $x=(0,0)^T$ is equal to $\frac{1}{s} f(sv)$=$\frac{1}{s} f\Big(\begin{matrix}sv_1\\sv_2\end{matrix}\Big)$.

Which is either equal to $0$ when the argument is $(0,0)^T$ or it is $\frac{1}{s}\frac{sv_1s^2v_2^2}{s^2v_1^2+s^2v_2^2}$ which converges to $\frac{v_1v_2^2}{v_1^2+v_2^2}$ as $s \to \infty$.

Is that correct so far?

And how do I know if $f$ is continuously partial differentiable on $\mathbb{R}^2$? According to our professor $f$ is not differentiable at $0$. How do I show that? As far as I know it has something to do with that something is not linear but I don't know what exactly. So I guess it can't be continuously partial differentiable on $\mathbb{R}^2$ as well but I am not sure about that.

Thanks for your help!

user589054
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4 Answers4

3

The partial derivatives at $(0,0)$ are

$$\partial_x f(0,0) = \lim_{h\to 0} \frac{f(h,0) - f(0,0)}h = 0$$ $$\partial_y f(0,0) = \lim_{h\to 0} \frac{f(0,h) - f(0,0)}h = 0$$

so the only candidate for the differential $Df(0,0)$ is the zero operator.

However, the limit

$$\lim_{(h_1, h_2) \to (0,0)} \frac{\|f(h_1, h_2) - f(0,0) - Df(0,0)(h_1, h_2)\|}{\|(h_1, h_2)\|} = \lim_{(h_1, h_2) \to (0,0)} \frac{\|f(h_1, h_2)\|}{\sqrt{h_1^2 + h_2^2}} = \lim_{(h_1, h_2) \to (0,0)} \frac{h_1h_2^2}{(h_1^2 + h_2^2)^{3/2}}$$ does not exist because e.g. for $h_1 = h_2$ we get $$\lim_{h_1 \to 0} \frac{h_1^3}{|h_1|^3}$$

which is $\pm 1$, depending on whether $h_1$ approaches $0$ from the left or from the right.

Therefore, $f$ is not differentiable at $(0,0)$.

mechanodroid
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1

The function $f$ is not differentiable at the origin. You can verify this fact (at least) in two ways.

1) The map $v \mapsto \frac{\partial f}{\partial v} (\mathbf{0})$ is not a linear map.

2) Using the definition of differentiability.

In this second case, you have to show that $$ \varphi(x,y) := \frac{f(x,y) - f(0,0) - \nabla f(0,0) \cdot (x,y)}{\sqrt{x^2+y^2}} = \frac{xy^2}{(x^2+y^2)^{3/2}} $$ does not go to $0$ as $(x,y) \to (0,0)$.

This fact can be easily checked since $$ \varphi(x,x) = \frac{x^3}{2^{3/2}|x|^3} $$ does not converge to $0$ as $x\to 0$.

Rigel
  • 14,434
  • Thanks! Can you provide some link to that definition? I have never seen it before – user589054 Sep 02 '18 at 14:26
  • What kind of definition of differentiability do you know? – Rigel Sep 02 '18 at 14:29
  • Let $f:D \subset \mathbb{R}^n \to X$ and $x \in D$. If there exists a $df(x)$, a $\rho>0$ with $U_\rho(x) \subset D$ and a function $\epsilon:U_\rho(0) \setminus {0}$ with $\lim_{z \to 0} \epsilon(z)=0$ so that $f(x+z)=f(x)+df(x)z+|z|\infty \epsilon(z)$ is true $\forall z \in U\rho(0) \setminus {0}$ then we call $f$ differentiable at $x$.

    ($U_\rho(0)$ denotes the open circle around $0$ with radius $\rho$)

     – user589054 Sep 02 '18 at 14:40
  • Ok, it's the same definition. In my notation, $\varphi = \epsilon$. Moreover, recall that, if $f\colon\mathbb{R^n} \to \mathbb{R}$ is differentiable at $x$, then $df(x) z = \nabla f(x) \cdot z$. – Rigel Sep 02 '18 at 14:50
  • Okay. Is my solution for $\frac{\partial f}{\partial v}((0,0)^T)$ correct? – user589054 Sep 02 '18 at 14:56
  • Yes, it seems correct. – Rigel Sep 02 '18 at 14:58
0

If a funcion is differentiable, the the directional derivative equals $\nabla_f \cdot v $. But here $\nabla_f =(0,0)$, hence all directional derivatives should be zero.

Now, setting $y=ax$ we get $$f(x,y)=f(x,ax)=\frac{ a x^3}{x^2 +a^2 x^2}=\frac{a}{1+a^2}x$$

Then, the derivative take different values (it's only zero for $a=0$... and $a=\infty$, which corresponds to the partial derivatives). Then, the function is not differentiable.

leonbloy
  • 63,430
-1

Recall that when a function is differentiable at a point $(x_0,y_0)$ the following holds

$$\frac{\partial f}{\partial v}=\nabla f(x_0,y_0)\cdot v$$

that is the directional derivative are linear function.

Therefore that is a necessary condition for the differentiabilty of $f(x,y)$ at $(x_0,y_0)$.

Since at $(0,0)$ the condition is fulfilled we need to check directly by the definition that the following holds

$$\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-\nabla f(0,0)\cdot (x,y)}{\sqrt{x^2+y^2}}=\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}}=0$$

user
  • 154,566
  • 1
    The right side of the equation above is always equal to $0$ for $(x_0,y_0)^T=(0,0)^T$, right? So I'm looking for a counterexample where the left side does not equal $0$? Can you show me how to do this? – user589054 Sep 02 '18 at 14:28
  • In that case we need to check directly differentiability by definition. I update that. – user Sep 02 '18 at 14:31