Exercise 9.27

Question

Exercise 27: Put $f(0,0)=0$, and
    $$
        f(x,y)=\frac{xy(x^2-y^2)}{x^2+y^2}\qquad\hbox{if }(x,y)\ne(0,0).
    $$
    Prove that

(a) $f$, $D_1f$, $D_2f$ are continuous in $\R2$;

(b) $D_{12}f$ and $D_{21}f$ exist at every point of $\R2$, and are continuous except at $(0,0)$;

(c) $D_{12}f(0,0)=1$, and $D_{21}f(0,0)=-1$.

analambanomenos · Accepted Answer

(a) Converting to polar coordinates for $(x,y)\ne(0,0)$, we have
    \begin{align*}
        f(r\cos\theta,r\sin\theta) &= \frac{r^4\cos\theta\sin\theta(\cos^2\theta-\sin^2\theta)}{r^2} \
        &= \frac{r^2\sin2\theta\cos2\theta}{2} \
        &= \frac{r^2\sin4\theta}{4} \
        \big|f(x,y)\big| &\le\frac{r^2}{4}=\frac{x^2+y^2}{4}
    \end{align*}
    which converges to $0=f(0,0)$ as $(x,y)\rightarrow(0,0)$.

We have
    $$
        D_1f(0,0) = \lim_{h\rightarrow 0}\frac{f(h,0)}{h} = \lim_{h\rightarrow 0}\frac{0}{h}=0.
    $$
    For $(x,y)\ne(0,0)$, we have 
    \begin{align*}
        D_1f(x,y) &= \frac{(x^2+y^2)(3x^2y-y^3)-2x(x^3y-xy^3)}{(x^2+y^2)^2} \
        &= \frac{x^4y+4x^2y^3-y^5}{(x^2+y^2)^2} \
        D_1f(r\cos\theta,r\sin\theta) &= \frac{r^5(\cos^4\theta\sin\theta+4\cos^2\theta\sin^3\theta-\sin^5\theta)}{r^4} \
        \big|D_1f(x,y)\big| &\le 6r = 6\sqrt{x^2+y^2}
    \end{align*}
    which converges to $0=D_1f(0,0)$ as $(x,y)\rightarrow(0,0)$.
    
    We have
    $$
        D_2f(0,0) = \lim_{h\rightarrow 0}\frac{f(0,h)}{h} = \lim_{h\rightarrow 0}\frac{0}{h}=0.
    $$
    For $(x,y)\ne(0,0)$, we have 
    \begin{align*}
        D_2f(x,y) &= \frac{(x^2+y^2)(x^3-3xy^2)-2y(x^3y-xy^3)}{(x^2+y^2)^2} \
        &= \frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2} \
        D_2f(r\cos\theta,r\sin\theta) &= \frac{r^5(\cos^5-4\cos^3\theta\sin^2\theta-\cos\theta\sin^4\theta)}{r^4} \
        \big|D_1f(x,y)\big| &\le 6r = 6\sqrt{x^2+y^2}
    \end{align*}
    which converges to $0=D_2f(0,0)$ as $(x,y)\rightarrow(0,0)$.

(b) For $(x,y)\ne(0,0)$, we have
    \begin{align*}
        D_{12}f(x,y) &= \frac{(x^2+y^2)(5x^4-12x^2y^2-y^4)-4x(x^5-4x^3y^2-xy^4)}{(x^2+y^2)^3} \
        &= \frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3} \
        D_{12}f(r\cos\theta,r\sin\theta) &= \frac{r^6(\cos^6\theta+9\cos^4\theta\sin^2\theta-\sin^6\theta)}{r^6} \
        &= \cos^6\theta+9\cos^4\theta\sin^2\theta-\sin^6\theta
    \end{align*}
    So $D_{12}f$ has a constant value along the rays emanating from the origin. Since this value is not a constant function of $\theta$, we see that $D_{12}f(x,y)$ does not converge to a limit as
    $(x,y)\rightarrow(0,0)$, and so $D_{12}f(x,y)$ cannot be continuous at the origin. Also, for $(x,y)\ne(0,0)$ we can apply Theorem 9.41 and conclude that $D_{21}f(x,y)=D_{12}f(x,y)$ is also not
    continuous at the origin.

(c) We have
    \begin{gather*}
        D_{12}f(0,0)=\lim_{h\rightarrow 0}\frac{D_2f(h,0)}{h}=\lim_{h\rightarrow 0}\frac{h^5/h^4}{h}=\lim_{h\rightarrow 0}1=1 \
        D_{21}f(0,0)=\lim_{h\rightarrow 0}\frac{D_1f(0,h)}{h}=\lim_{h\rightarrow 0}\frac{-h^5/h^4}{h}=\lim_{h\rightarrow 0}-1=-1
    \end{gather*}

Exercise 9.27

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Exercise 9.27

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