求函数$f: \mathbb R^3 \to \mathbb R$在原点处存在方向导数的所有方向

To find all directions $\vec{v} = (a, b, c) \in \mathbb{R}^3 \setminus {0}$ where the directional derivative $\frac{\partial f}{\partial \vec{v}}(\vec{0})$ exists, we start with the definition:

$$\frac{\partial f}{\partial \vec{v}}(\vec{0}) = \lim_{t \to 0} \frac{f(t\vec{v}) - f(\vec{0})}{t} = \lim_{t \to 0} \frac{f(ta, tb, tc)}{t}$$

Since $f(\vec{0}) = 0$, we only need to analyze the limit of $\frac{f(ta, tb, tc)}{t}$ as $t \to 0$.

Step 1: Compute $f(t\vec{v})$ for $t \neq 0$ #

Substitute $x=ta$, $y=tb$, $z=tc$ into the function:
$$
f(ta, tb, tc) = \frac{(ta)^2 |tb|}{(ta)^2 |tb| + (ta - tb)^2 + (tc)^2}
$$
Simplify numerator and denominator:

• Numerator: $t^2a2 \cdot |t||b| = |t|^3a2|b|$
• Denominator: $|t|^3a2|b| + t^2(a - b)^2 + t^2c2 = t^2\left(|t|a2|b| + (a - b)^2 + c^2\right)$

Thus:
$$
\frac{f(t\vec{v})}{t} = \frac{|t|a^{2|b|}{t\left(|t|a}2|b| + (a - b)^2 + c^2\right)}
$$

Step 2: Analyze the limit by cases #

We split into cases based on the components of $\vec{v}$:

Case 1: $b = 0$

If $b=0$, the numerator becomes $|t|a^2 \cdot 0 = 0$, so $\frac{f(t\vec{v})}{t} = 0$ for all $t \neq 0$. The limit as $t \to 0$ is 0, so the directional derivative exists.

Case 2: $b \neq 0$

Here, we need to compare the right-hand ($t \to 0^+$) and left-hand ($t \to 0^-$) limits:

• For $t \to 0^+$: $|t| = t$, so the expression simplifies to $\frac{a^2|b|}{t a^2|b| + (a - b)^2 + c^2}$. As $t \to 0^+$, this tends to $\frac{a^2|b|}{(a - b)^2 + c^2}$.
• For $t \to 0^-$: $|t| = -t$, so the expression simplifies to $\frac{-a^2|b|}{-t a^2|b| + (a - b)^2 + c^2}$. As $t \to 0^-$, this tends to $\frac{-a^2|b|}{(a - b)^2 + c^2}$.

For the limit to exist, these two must be equal:
$$
\frac{a^2|b|}{(a - b)^2 + c^2} = \frac{-a^2|b|}{(a - b)^2 + c^2}
$$
Since $b \neq 0$, $|b| > 0$. The denominator $(a - b)^2 + c^2$ is non-negative:

• If $(a - b)^2 + c^2 = 0$, then $a = b$ and $c = 0$. For $\vec{v} = (b, b, 0)$ ($b \neq 0$), $\frac{f(t\vec{v})}{t} = \frac{1}{t}$, which has no limit as $t \to 0$.
• If $(a - b)^2 + c^2 > 0$, we can multiply both sides by the denominator to get $2a^2|b| = 0$. Since $|b| > 0$, this implies $a = 0$.

When $a = 0$ and $b \neq 0$, $\frac{f(t\vec{v})}{t} = 0$ for all $t \neq 0$, so the limit is 0 and the directional derivative exists.

Conclusion #

The directional derivative $\frac{\partial f}{\partial \vec{v}}(\vec{0})$ exists if and only if the direction vector $\vec{v} = (a, b, c)$ satisfies either $a = 0$ or $b = 0$ (and $\vec{v} \neq \vec{0}$, as direction vectors are non-zero by definition).

内容的提问来源于stack exchange，提问作者gbi1977