Exercise 9.14

Question

Exercise 14: Define $f(0,0)=0$ and
    $$
        f(x,y)=\frac{x^3}{x^2+y^2}\qquad\hbox{if }(x,y)\ne(0,0).
    $$
    (a) Prove that $D_1f$ and $D_2f$ are bounded functions in $\R2$. (Hence $f$ is continuous.)

(b) Let $\v u$ be any unit vector in $\R2$. Show that the directional derivative $D_{\v u}f(0,0)$ exists, and that its absolute value is at most 1.

(c) Let $\gamma$ be a differentiable mapping of $\R1$ into $\R2$ (in other words, $\gamma$ is a differentiable curve in $\R2$), with $\gamma(0)=(0,0)$ and $\big|\gamma'(0)\big|>0$. Put
    $g(t)=f\big(\gamma(t)\big)$ and prove that $g$ is differentiable for every $t\in\R1$. If $\gamma\in\mathscr C'$, prove that $g\in\mathscr C'$.

(d) In spite of this, prove that $f$ is not differentiable at $(0,0)$.

analambanomenos · Accepted Answer

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This function is easier to visualize if we convert to polar coordinates, $x=r\cos\theta$, $y=r\sin\theta$, with $r\ge0$ and $0\le\theta<2\pi$. Then we have $f(r,\theta)=r\cos^3\theta$, so
    that $f$ is linear in $r$ along rays from the origin, and since $\cos^3(\theta+\pi)=-\cos^3\theta$, $f$ is a linear function along every line through the origin.

(a) Keeping in mind that along the $x$ axis $f(x,0)=x$, and along the $y$ axis $f(0,y)=0$, we have 
    \begin{align*}
        D_1f(x,y) &=
        \begin{cases}
            \displaystyle\frac{x^2(x^2+3y^2)}{(x^2+y^2)^2}, & (x,y)\ne(0,0) \
            1, & (x,y)=(0,0)
        \end{cases} \
        D_2f(x,y) &=
        \begin{cases}
            \displaystyle\frac{2x^3y}{(x^2+y^2)^2}, & (x,y)\ne(0,0) \
            0, & (x,y)=(0,0)
        \end{cases}
    \end{align*}
    Converting to polar coordinates and simplifying we have, for $r>0$ and away from the origin,
    \begin{align*}
        D_1f(r,\theta) &= \cos^2\theta(\cos^2\theta+3\sin^2\theta) \
        D_2f(r,\theta) &= -2\cos^3\theta\sin\theta.
    \end{align*}
    These do not depend on $r$ and are continuous functions of $\theta$, which attain their maximum and minimum values on the compact set $[0,2\theta]$. Excluding the origin, $D_1f$ and $D_2f$
    take these maximum and minimum values along certain rays emanating from the origin. Hence they are bounded on $\R2$.

(b) We can express $\v u$ as $\cos\theta\v e_1+\sin\theta\v e_2$ for some $0\le\theta<2\pi$. Then, from the above expression of $f$ in polar coordinates, we have $f(t\v u)=t\cos^3\theta$
    for all $t\in\R1$ so that $D_{\v u}f(0,0)=\cos^3\theta$, whose absolute value is less than 1.

(c) Since $f$ is differentiable away from the origin, we only need to show that $g'(0)$ exists. Since $\gamma(t)$ is differentiable at $t=0$, there are functions $\delta_1(t)$ and 
    $\delta_2(t)$ such that, for $j=1,2$, $\gamma_j(t)=t\gamma_j'(0)+t\delta_j(t)$ and $\lim\delta_j(t)=0$ as $t\rightarrow0$. Hence
    \begin{align*}
        g'(0) &= \lim_{t\rightarrow0} \frac{g(t)}{t} \
        &= \lim_{t\rightarrow0} \frac{f\big(\gamma_1(t),\gamma_2(t)\big)}{t} \
        &= \lim_{t\rightarrow0} \frac{f\big(t\gamma_1'(0)+t\delta_1(t),t\gamma_2'(0)+t\delta_2(t)\big)}{t} \
        &= \lim_{t\rightarrow0} \frac{1}{t}\cdot\frac{t^3\big(\gamma_1'(0)+\delta_1(t)\big)^3}{t^2\Big(\big(\gamma_1'(0)+\delta_1(t)\big)^2+\big(\gamma_2'(0)+\delta_2(t)\big)^2\Big)} \
        &= \frac{\gamma_1'(0)^3}{\big|\gamma'(0)\big|^2}
    \end{align*}
    Also, if $\gamma_r(t)$ and $\gamma_\theta(t)$ are the polar coordinates of $\gamma(t)$ for $t\ne 0$, then we have
    \begin{align*}
        \lim_{t\rightarrow0}\cos\gamma_\theta(t) &= \lim_{t\rightarrow0}\frac{\gamma_1(t)}{\big|\gamma(t)\big|} \
        &= \lim_{t\rightarrow0}\frac{t\big(\gamma'(0)+\delta_1(t)\big)}{t\sqrt{\big(\gamma_1'(0)+\delta_1(0)\big)^2+\big(\gamma_2'(0)+\delta_2(0)\big)^2}} \
        &= \frac{\gamma_1'(0)}{\big|\gamma'(0)\big|} \
        \lim_{t\rightarrow0}\sin\gamma_\theta(t) &= \lim_{t\rightarrow0}\frac{\gamma_2(t)}{\big|\gamma(t)\big|} \
        &= \lim_{t\rightarrow0}\frac{t\big(\gamma'(0)+\delta_2(t)\big)}{t\sqrt{\big(\gamma_1'(0)+\delta_1(0)\big)^2+\big(\gamma_2'(0)+\delta_2(0)\big)^2}} \
        &= \frac{\gamma_2'(0)}{\big|\gamma'(0)\big|}.
    \end{align*} 
    For $t$ where $\gamma(t)\ne\v 0$, we can apply the special case of the chain rule, Theorem 9.15, worked out in Example 9.18 to get
    \begin{align*}
        g'(t) &= D_1f\big(\gamma_1(t),\gamma_2(t)\big)\gamma_1'(t)+D_2f\big(\gamma_1(t),\gamma_2(t)\big)\gamma_2'(t) \
        &= \cos^2\gamma_\theta(t)\big(\cos^2\gamma_\theta(t)+3\sin^2\gamma_\theta(t)\big)\gamma_1'(t)-2\big(\cos^3\gamma_\theta(t)\sin\gamma_\theta(t)\big)\gamma_2'(t)
    \end{align*}
    Now assume that $\gamma\in\mathscr C'$. Applying the above limits we get
    \begin{align*}
        \lim_{t\rightarrow0}g'(t) &= \frac{\gamma_1'(0)^2}{\big|\gamma'(0)\big|^2}\Bigg(\frac{\gamma_1'(0)^2}{\big|\gamma'(0)\big|^2}+3\frac{\gamma_2'(0)^2}{\big|\gamma'(0)\big|^2}\Bigg)\gamma_1'(0)-
             2\Bigg(\frac{\gamma_1'(0)^3}{\big|\gamma'(0)|^3}\frac{\gamma_2'(0)}{\big|\gamma'(0)\big|}\Bigg)\gamma_2'(0) \
        &= \frac{\gamma_1'(0)^3\big(\gamma_1'(0)^2+\gamma_2'(0)^2\big)}{\big|\gamma'(0)\big|^4} \
        &= \frac{\gamma_1'(0)^3}{\big|\gamma'(0)\big|^2} \
        &= g'(0).
    \end{align*}
    Hence $g$ is continuous at 0, and so $g\in\mathscr C'$.

(d) Let $\v u=\big(\cos(\pi/4),\sin(\pi/4)\big)$, a unit vector, and suppose $f$ were differentiable at $(0,0)$. Then by formula (40) in Example 9.18 in the text, the directional derivative
    would be
    $$
        D_{\v u}f(0,0)=D_1(0,0)\cos(\pi/4)+D_2(0,0)\sin(\pi/4)=\cos(\pi/4)=\sqrt{2}/2.
    $$
    However, by part (b), $D_{\v u}f(0,0)=\cos^3(\pi/4)=\sqrt{2}/4$. Hence $f$ cannot be differentiable at $(0,0)$.

Exercise 9.14

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Exercise 9.14

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