Proving the Second Derivative Test using directional derivatives

This is a proof which only uses directional derivatives.

We want to prove the Second Derivative Test.

Suppose the second partial derivatives of $f$ are continuous on a disk with center (a, b), and suppose that $f_{x} (a, b) = f_{y} (a, b) = 0$ . Let
$H = | \begin{matrix} f_{x x} & f_{x y} \\ f_{y x} & f_{y y} \end{matrix} | = f_{x x} (a, b) f_{y y} (a, b) - [f_{x y} (a, b)]^{2}$
(a) If $H > 0, f_{x x} (a, b) > 0$ , then $f (a, b)$ is a local minimum.

(b) If $H > 0, f_{x x} (a, b) < 0$ , then $f (a, b)$ is a local maximum.

(c) If $H < 0$ , then $f (a, b)$ is a saddle point.

(d) If $H = 0$ , inconclusive.

Let $u = (h, k)$ , which is a unit vector.

\begin{aligned} D_{u}^{2} f (x, y) & = D_{u} [D_{u} f (x, y)] \\ = D_{u} [\nabla f \cdot u] \\ = (f_{x x} h + f_{y x} k) h + (f_{x y} h + f_{y y} k) k \end{aligned}

Since the second partial derivatives are continuous, $f_{x y} = f_{y x}$

\begin{aligned} D_{u}^{2} f (x, y) & = f_{x x} h^{2} + 2 f_{x y} h k + f_{y y} k^{2} \\ = k^{2} (f_{x x} (\frac{h}{k})^{2} + 2 f_{x y} \frac{h}{k} + f_{y y}) \end{aligned}

Let $z = \frac{h}{k}$ , $g (z) = f_{x x} z^{2} + 2 f_{x y} z + f_{y y}$

Use the quadratic function above to find the range of solutions of $D_{u}^{2} f (x, y)$ .

Let $d = (2 f_{x y})^{2} - 4 f_{x x} f_{y y} = 4 (f_{x y}^{2} - f_{x x} f_{y y}) = - 4 H$

(1) If $d < 0$ , means that $D_{u}^{2} f (x, y)$ can only be either positive or negative.

We can know that the point $(a, b)$ must be a local minimum or a local maximum.

\begin{aligned} D_{u}^{2} f (x, y) & = f_{x x} h^{2} + 2 f_{x y} h k + f_{y y} k^{2} \\ = f_{x x} (h^{2} + 2 \frac{f_{x y}}{f_{x x}} h k + \frac{f_{y y}}{f_{x x}} k^{2}) \\ = f_{x x} [(h + \frac{f_{x y}}{f_{x x}} k)^{2} + (f_{x x} f_{y y} - f_{x y}^{2}) k^{2}] \\ = f_{x x} [(h + \frac{f_{x y}}{f_{x x}} k)^{2} + H k^{2}] \end{aligned}

We know that $H > 0$ , so,

if $f_{x x} > 0$ , the equation above will always be positive, which means a local minimum at $(a, b)$ .

Otherwise if $f_{x x} < 0$ , it will always be negative, which means a local maximum at $(a, b)$ .

(2) If $d > 0$ , means that $D_{u}^{2} f (x, y)$ can be both positive, zero or negative, so

$H < 0$ tells us that the function has a saddle point at $(a, b)$ .

(3) If $d > 0$ , means that $D_{u}^{2} f (x, y)$ can only be either positive and zero or negative and zero, so

$H = 0$ cannot tell that whether $(a, b)$ is a local minimum, local maximum or a saddle point.

Note: Posted at The “second derivative test” for $f(x,y)$.

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