證明 Second (Partial) Derivative Test

Posted on May 18, 2021

Second (Partial) Derivative Test

Let $f : D \rightarrow \mathbb{R}$ be a function in two variables $x$ and $y$. Suppose that $f, f_{x}, f_{y}, f_{xx}, f_{yy}, f_{xy}, f_{yx}$ are all continuous in an open disk centered at $(a, b)$ and $f_{x}(a, b) = f_{y}(a, b) = 0$. (Then $f_{xy}(a, b) = f_{yx}(a, b)$)

Let

\[D(x, y) = f_{xx}(x, y) f_{yy}(x, y) - \left(f_{xy}(x, y)\right)^{2}\]

Then

  1. if $D(a, b) > 0$ and $f_{xx}(a, b) > 0$, then $f(a, b)$ is a local minimum of $f$,
  2. if $D(a, b) > 0$ and $f_{xx}(a, b) < 0$, then $f(a, b)$ is a local maximum of $f$,
  3. if $D(a, b) < 0$, then $(a, b)$ is saddle point of $f$,
  4. if $D(a, b) = 0$, no conclusion.

$D(x, y)$ is called the Discriminant or Hessian of $f$.

Proof of Second (Partial) Derivative Test

Let $\vec{u} = <u_{1}, u_{2}>$ be any unit vector in $\mathbb{R}^{2}$. Let $\delta$ be a positive number such that $f, f_{x}, f_{y}, f_{xx}, f_{yy}, f_{xy}, f_{yx}$ are all continuous in $U_{\delta}(a, b)$.

Graph of $U_{\delta}(a, b)$, $\phi_{\vec{u}}(\delta)$ and $\phi_{\vec{u}}(-\delta)$

Define

\(\phi_{\vec{u}}(t) = f_{x}(a+tu_{1}, b+tu_{2})\), where $t \in (-\delta, \delta)$.

By the Chain Rule, one has

\[\phi'_{\vec{u}}(t) = f_{x}(a+tu_{1}, b+tu_{2})\cdot u_{1} + f_{y}(a+tu_{1}, b+tu_{2}) \cdot u_{2}\]

So

\[\phi'_{\vec{u}}(0) = f_{x}(a, b)\cdot u_{1} + f_{y}(a, b) \cdot u_{2}=0\]

Hence $(a, b)$ is a critical point of $f$, therefore $f_{x}(a,b)=f_{y}(a,b)=0$.

Use The Chain Rule again, one has

\[\phi''_{\vec{u}}(t) = f_{xx}(a+tu_{1}, b+tu_{2}) \cdot u_{1}^2 + f_{yy}(a+tu_{1}, b+tu_{2}) \cdot u_{2}^{2} \\ + 2f_{xy}(a+tu_{1}, b+tu_{2}) \cdot u_1 u_2\]

So

\[\phi''_{\vec{u}}(0) = f_{xx}(a, b) \cdot u_{1}^2 + 2f_{xy}(a, b) \cdot u_1 u_2 + f_{yy}(a, b) \cdot u_{2}^{2} = 0\]

Observe that $-4D(a, b) = 4f^2_{xy}(a, b) - 4 f_{xx}(a, b) f_{yy}(a, b)$ is the discriminant of the quadratic polynomial $Q(X) = f_{xx}(a, b) X^2 + 2f_{xy}(a, b) X + f_{yy}(a, b)$.

And we know that

\[\frac{\phi''_{\vec{u}}(0)}{u_2^{2}} = f_{xx}(a, b) \cdot \left(\frac{u_{1}}{u_2}\right)^2 + 2f_{xy}(a, b) \cdot \left(\frac{u_1}{u_2}\right) + f_{yy}(a, b)\]

which is equal to $Q(\frac{u_1}{u_2})$ (if $u_2 = 0$, we can divide $u_1$ instead, because $\vec{u}$ is a unit vector, $u_1u_2 \neq 0$).

So if the discriminant of $Q$, $-4D(a, b) < 0$, it indicates $Q$ have no root. Then we can tell $Q$ is always positive or negative by its leading coefficient $f_{xx}(a, b)$. If $f_{xx}(a, b) > 0$, then $Q$ is always postive. If$f_{xx}(a, b) < 0$, then $Q$ is always negative.

In conclusion, we have

  1. Suppose $D(a, b) > 0$ and $f_{xx}(a, b) > 0$. Then one has $\phi''_{\vec{u}}(0) > 0$ for all $\vec{u}$. Therefore, f(a, b) is a local minimum.
  2. Suppose $D(a, b) > 0$ and $f_{xx}(a, b) < 0$. Then one has $\phi''_{\vec{u}}(0) < 0$ for all $\vec{u}$. Therefore, f(a, b) is a local maximum.
  3. Suppose $D(a, b) < 0$. Then one has \(\phi^{\prime\prime}_{\vec{u}}(0) > 0\) for some $\vec{u}$ and has $\phi^{\prime\prime}_{\vec{u}}(0) < 0$ for some $\vec{u}$. Therefore, $(a, b)$ is saddle point of $f$.

Credit

  • 中央大學微積分講義