High-order partial derivative

2.4 High-order partial derivative

Consider $z = f (x, y)$ , several common symbols to mark the first-order derivative are $\begin{aligned} \frac{\partial f}{\partial x} (x, y) & = f_{x}^{'} (x, y) = \partial_{1} f = f_{1}^{'} \\ \frac{\partial f}{\partial y} (x, y) & = f_{y}^{'} (x, y) = \partial_{2} f = f_{2}^{'} \end{aligned}$

Mark $\begin{aligned} \frac{\partial^{2} f}{\partial y \partial x} (x, y) & ≜ \frac{\partial}{\partial y} [\frac{\partial}{\partial x} (x, y)] \\ \frac{\partial^{2} f}{\partial x \partial y} (x, y) & ≜ \frac{\partial}{\partial x} [\frac{\partial}{\partial y} (x, y)] \\ \frac{\partial^{2} f}{\partial x^{2}} (x, y) & ≜ \frac{\partial^{2} f}{\partial x \partial x} (x, y) = \frac{\partial}{\partial x} [\frac{\partial}{\partial x} (x, y)] \\ \frac{\partial^{2} f}{\partial y^{2}} (x, y) & ≜ \frac{\partial^{2} f}{\partial y \partial y} (x, y) = \frac{\partial}{\partial y} [\frac{\partial}{\partial y} (x, y)] \end{aligned}$

Given $f (x_{1}, . . ., x_{m})$ , one of the $k^{th}$ order derivatives of $f$ is

∂kf ------------ ∂xik ⋅⋅⋅∂xi1

where $i_{1}, . . ., i_{k} \in {1, . . ., m}$ . Totally there are $m^{k}$ derivatives. When $k = 2$ , mark

2 --∂-f-- ∖left(∂xj ∂xi∖right)i,j

as the Hessian matrix of $f$ at $x_{0}$ .

Theorem 2.4.1 If all of the $k^{th}$ order derivatives of $f$ are continuous (marked as $f \in C^{k}$ ), then the value of mixed partial derivative has nothing to do with the order of taking derivatives.

Proof Left as exercise. :) $◻$

Based on this theorem, if $f$ satisfies the properties above, all of the $k^{th}$ order derivatives of $f$ could be expressed uniformly as

∂kf --αm-------αi ∂xm ⋅⋅⋅∂x 1

where $α_{1} + . . . + α_{m} = k$ .

Theorem 2.4.2

1.: Assuming $f, g \in C^{k}$ , then $λ f + μ g, f \cdot g \in C^{k}$ .
2.: Assuming function $f \in C^{k}$ , mapping $G \in C^{k}$ , then $f \circ G \in C^{k}$ .

Proof Proved by mathematical induction with respect to $k$ .

1.

When $k = 1$ ,
$∂ ∂f ∂g ----[f (x )g(x)] = ---(x) g(x)+ f|(x)----(x) ∂xi ◟∂xi◝◜-◞ ◟◝◜1◞ ◟ ◝◜1◞∂◟xi◝◜-◞ C0 C C C0$
meaning $f g \in C^{1}$ .
Assuming theorem is proved for $k$ , then for $k + 1$ , $f, g \in C^{k + 1}$ , here
$∂ ∂f ∂g ----[f (x )g(x)] = ---(x) g(x)+ f (x)----(x) ∂xi ◟∂xi◝◜-◞ ◟◝k◜+◞1 ◟|◝k◜+◞1∂◟xi◝◜-◞ Ck C C Ck$
meaning $\frac{\partial}{\partial x_{i}} [f (x) g (x)] \in C^{k}$ , according to the inducting assumption $f g \in C_{k + 1}$ . Theorem is proved.

2.

Similarly. Left as exercise. :)

$◻$

Example 2.4.1

1.

In the orthogonal coordinate, assuming $u = u (x, y)$ , mark

∂2u ∂2u Δu = ---2 + --2- ∂x ∂y

where $Δ$ is the Laplace symbol. Transform the equation above into the polar coordinate system.

For any $v = v (x, y)$ , in polar coordinate system, we have $\begin{aligned} \frac{\partial v}{\partial r} & = \frac{\partial v}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial v}{\partial y} \frac{\partial y}{\partial r} = \cos θ \frac{\partial v}{\partial x} + \sin θ \frac{\partial v}{\partial y} \\ \frac{\partial v}{\partial θ} & = \frac{\partial v}{\partial x} \frac{\partial x}{\partial θ} + \frac{\partial v}{\partial y} \frac{\partial y}{\partial θ} = - r \sin θ \frac{\partial v}{\partial x} + r \cos θ \frac{\partial v}{\partial y} \end{aligned}$

Since it holds for any $v$ , therefore $\begin{aligned} \frac{\partial}{\partial r} & = \cos θ \frac{\partial}{\partial x} + \sin θ \frac{\partial}{\partial y} \\ \frac{1}{r} \frac{\partial}{\partial θ} & = - \sin θ \frac{\partial}{\partial x} + \cos θ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial x} & = \cos θ \frac{\partial}{\partial r} - \sin θ \frac{1}{r} \frac{\partial}{\partial θ} \\ \frac{\partial}{\partial y} & = \sin θ \frac{\partial}{\partial r} + \cos θ \frac{1}{r} \frac{\partial}{\partial θ} \\ \frac{\partial^{2}}{\partial x^{2}} & = (\cos θ \frac{\partial}{\partial r} - \sin θ \frac{1}{r} \frac{\partial}{\partial θ}) (\cos θ \frac{\partial}{\partial r} - \sin θ \frac{1}{r} \frac{\partial}{\partial θ}) \\ \dots \end{aligned}$

2.

Given $f : M_{n} \to M_{n}$ where $A \mapsto A^{- 1}$ . Prove that $f \in C^{\infty}$ .

Firstly, we have $\partial f (A) (B) = \partial (A^{- 1}) (B) : M_{n} \to M_{n}$ where $\partial (A^{- 1}) (B) = - A^{- 1} B A^{- 1}$ .

Notice that $A \mapsto A^{- 1}$ is derivable, meaning $- A^{- 1} B A^{- 1}$ is derivable, i.e. $f$ has second-order derivative.

Here notice that $\partial (A^{- 1})$ is derivable, meaning $f$ has third-order derivative.

Terminally, $f \in C^{\infty}$ .