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I'm working on the following problem:

Consider the $f:\mathbb{R}^2 \mapsto \mathbb{R}$ defined by $$f(x,y) = \frac{xy(x^2-y^2)}{x^2+y^2}$$ for everywhere outside the origin and $f(0,0)=0$ at the origin is. Show that the second derivative of $f$ exists but the mixed partials are not equal at the origin.

My work:

I'm getting for $\nabla f = <\frac{yx^4+4x^2y^3-y^5}{(x^2+y^2)^2},\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}>$. My analysis show that the partials are continuous at zero and take the value $0$ at the origin. I'm having trouble analyzing the second derivative:

$$ A = \begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix} = \begin{bmatrix}\frac{-4xy^3(x^2-3y^2)}{(x^2+y^2)^3}&\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}\\\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}&\frac{4yx^3(y^2-3x^2)}{(x^2+y^2)^3}\end{bmatrix} $$

$$\lim_{\|h\|\rightarrow 0} \frac{\nabla f((0,0)+h)-\nabla f((0,0))- A \cdot h\|}{\|h\|} \\ = \lim_{\|h\|\rightarrow 0} \frac{\|\nabla f(h)-\langle f_{xx}h_x+f_{xy}h_y,f_{yx}h_x+f_{yy}h_y\rangle\|}{\|h\|} \\= \lim_{\|h\|\rightarrow 0} \frac{\|\nabla f(h)-\langle\frac{-4xy^3(x^2-3y^2)}{(x^2+y^2)^3}h_x+\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}h_y,\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}h_x+\frac{4yx^3(y^2-3x^2)}{(x^2+y^2)^3}h_y\rangle\|}{\|h\|}$$

When I do a polar conversion on this thing and send $r \rightarrow 0$ (uniform for all $\theta$), I get all my $r$'s canceling out -- leaving a limit dependent on theta. But according to the problem, this ratio should go to zero though. Does anyone know what's going on?

I've seen other problems like this that show partials are not continuous at the the origin. That's different though from just establishing existance of the total derivative, right?

yoshi
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  • You have exactly the same expressions for $f_{xy}$ and $f_{yx}$ so if your calculations are correct the mixed partials are equal everywhere, in contradiction of what you are supposed to prove. How confident are you in your calculations? – saulspatz Apr 10 '18 at 05:22
  • pretty confident, I did them on wolfram alpha. also I think I saw the same expression computed in another question - I would have linked it except I have too many windows open. – yoshi Apr 10 '18 at 05:30
  • I just did it in sympy and I agree with you. The mixed partials must be discontinuous though, or they'd be equal. I guess you have to compute them at the origin from the definition. – saulspatz Apr 10 '18 at 05:43

1 Answers1

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To show that the second derivative exists, all you have to show is that the second partials exist at $(0,0)$. For example $$ f_{xy}(0,0)=\lim_{y\to0}{\frac{f_x(0,y)-f_x(0,0)}{y}}=\lim_{y\to0}{\frac{-y^5}{y^5}}=-1 $$ I also got $$f_{yx}(0,0)=+1,\text{ and }f_{xx}(0,0)=f_{yy}(0,0)=0$$ but I'm getting sleepy, so no guarantees.

EDIT $$f_{xx}(0,0)=\lim_{x\to 0}\frac{f_x(x,0)-f_x(0,0)}{x-0}=\lim_{x\to0}\frac{0}{x\cdot x^4}=0\\ f_{yy}(0,0)=\lim_{y\to 0}\frac{f_y(0,y)-f_y(0,0)}{y-0}=\lim_{y\to 0}\frac{0}{y\cdot y^4}=0 $$

EDIT

To round this off, I verified that the second derivative doesn't exist at the origin, and I might as well share my calculations. We know that if the derivative exists it must be $$\pmatrix{0&1\\-1&0}$$ so that the linear approximation to $\nabla f(x,y)$ at $(0,0$) will be $\pmatrix{y\\-x}$. So we need to show that $$ \lim_{(x,y)\to(0,0)}\frac{\left\lvert\left\lvert\nabla f(x,y)-(y,-x)^{\top}\right\rvert\right\rvert}{||(x,y)||} $$ is not equal to $0$. We will show the limit goes to $\infty$ as $(x,y)\to(0,0)$ along the curve $x=y^2$. It suffices to show the numerator goes to $\infty.$ Also, since $||(a,b)||<|a|,$ it is enough to show this for the first component. Substituting $x=y^2$ we have$$ \lim_{y\to0}\frac{y^9+4y^7-y^5}{(y^4+y^2)^2}-y=\lim_{y\to0}\frac{2y^7-2y^5}{y^8+2y^6+y^4}=\infty $$

saulspatz
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  • Okay -- but don't we need to evaluate the actual ratio as I did to obtain whether the second derivative exists at the origin. Computing the entries in $A$ basically saying, "when the second derivative exists, the value is $A(x,y)$" -- but we haven't established existence yet – yoshi Apr 10 '18 at 11:21
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    Yes, of course. I was getting too sleepy when I wrote that. I should have said that we know what the derivative has to be, so we can just use that in evaluating the limit of the ratio. Sorry. – saulspatz Apr 10 '18 at 12:31
  • @saulpatz how did you get $f_{xx}$ and $f_{yy}$ to equal $0$. I get that they don't exist. For $f_{xx}$ I obtained this by taking $\lim_{x \rightarrow 0} f_x$ – yoshi Apr 12 '18 at 03:15
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    I added my computations. They use $$\lim_{(x,y)\to(0,0)}f_x(x,y)=\lim_{(x,y)\to(0,0)}f_y(x,y)=0$$ which you said you verified. I haven't checked this. Are you saying it's incorrect? – saulspatz Apr 12 '18 at 15:49
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    I just confirmed $f_x\to0, f_y\to0\text{ as }(x,y)\to(0,0)$ so I think the calculations are all correct. – saulspatz Apr 12 '18 at 16:02
  • Ah! While I was computing $f_{xx}$, I was computing $f_x(x,y)$, not $f_x(x,0)$ and was obtaining some value that was $y$ dependant. But indeed, $y$ should be set to zero if we are evaluating at zero. Thank you!! – yoshi Apr 13 '18 at 00:45