Higher Order Partial Derivatives

Suppose that partial derivatives $f_{x},f_{y}$ are again partially differentiable with respect to $x,y$. Then

$$\frac{\partial}{\partial x}(\frac{\partial f}{\partial x}) = \fra... ...artial f}{\partial x}) = \frac{\partial^{2} f}{\partial y \partial x} = f_{xy}$$

$$\frac{\partial}{\partial x}(\frac{\partial f}{\partial y}) = \fra... ...(\frac{\partial f}{\partial y}) = \frac{\partial^{2} f}{\partial y^2} = f_{yy}$$

We say $f_{xx} , f_{xy}, f_{yx}, f_{yy}$ The second partial derivatives of $f$.

Evaluation

$\frac{\partial}{\partial y}(\frac{\partial f}{\partial x})$, $\frac{\partial^{2} f}{\partial y \partial x}$, To evaluate $f_{xy}$, first differentiate with respect to $x$.

If $f(x,y)$ is the class $C^{2}$ on $D$, then $f_{xy} = f_{yx}$.

Interchange the order

If a function is the class $C^2$, then it is possible to interchange the order of differentiation.

Example 4..13   Find the 2nd order derivatives of $f(x,y) = \log(x^2 + y^2)$.

SOLUTION $${f_x(x,y) = \frac{2x}{x^2 + y^2}, f_y(x,y) = \frac{2y}{x^2 + y^2}}$$,
$${f_{xx}(x,y) = \frac{\partial f_x(x,y)}{\partial x} = \frac{2(x^2 + y^2)- 2x(2x)}{(x^2 + y^2)^2} = \frac{2y^2 - 2x^2}{(x^2 + y^2)^2}}$$,
$${f_{xy}(x,y) = \frac{\partial f_x(x,y)}{\partial y} = \frac{-2x(2y)}{(x^2 + y^2)^2} = \frac{-4xy}{(x^2 + y^2)^2}}$$,
$${f_{yy}(x,y) = \frac{\partial f_y(x,y)}{\partial y} = \frac{2(x^2 ... ...{(x^2 + y^2)^2} = \frac{2x^2 - 2y^2}{(x^2 + y^2)^2}}\ensuremath{ \blacksquare}$$

Class $C^{n}$

If $f(x,y)$ has the $n$th derivatives on $D$ and they are continuous, thenwe say $f(x,y)$ is the class $C^{n}$.

Exercise 4..13   Show $${f_{xy}(0,0) \neq f_{yx}(0,0)}$$ for the following function.

$$f(x,y) = \left\{\begin{array}{cl} (xy)\frac{x^2 - y^2}{x^2+y^2} & (x,y) \neq (0,0)\\ 0 & (x,y) = (0,0) \end{array}\right.$$

SOLUTION To show $f_{xy}(0,0) \neq f_{yx}(0,0)$, we first evaluate these values.

$$f_{x}(0,y) = \lim_{h \to 0}\frac{f(h,y) - f(0,y)}{h} = \lim_{h \to 0}(hy)\frac{h^2 - y^2}{h(h^2 + y^2)} = - y$$

Thus, $f_{xy}(0,y) = -1$ and $f_{xy}(0,0) = -1$. Next

$$f_{y}(x,0) = \lim_{k \to 0}\frac{f(x,k) - f(x,0)}{k} = \lim_{k \to 0}(xk)\frac{x^2 - k^2}{k(x^2 + k^2)} = x$$

Thus $f_{yx}(x,0) = 1$ and $f_{yx}(0,0) = 1$.
Therefore, $f_{xy}(0,0) = -1 \neq f_{yx}(0,0) = 1$ $\blacksquare$

Example 4..14   If $u = \log(x^2 + y^2)$, show $u_{xx} + u_{yy} = 0$.

SOLUTION $u_x = \frac{2x}{x^2 + y^2}$.

$$u_{xx} = \frac{\partial u_x}{\partial x} = \frac{2(x^2 + y^2) - 4x^2}{(x^2 + y^2)^2} = \frac{2(-x^2 + y^2)}{(x^2 + y^2)^2}.$$

Note $u_y = \frac{2y}{x^2 + y^2}$. Then

$$u_{yy} = \frac{\partial u_y}{\partial y} =\frac{2(x^2 + y^2) - 4y^2}{(x^2 + y^2)^2} = \frac{2(x^2 - y^2)}{(x^2 + y^2)^2}.$$

Thus, $u_{xx} + u_{yy} = 0$ $\blacksquare$

Exercise 4..14   Show the following function satisfies the laplace equation. $u_{xx} + u_{yy} + u_{zz} = 0$

$$u = \frac{1}{\sqrt{x^2 + y^2 + z^2}}$$

Laplace Equation

$u_{xx} + u_{yy} = 0$ is called two dimensional Laplace equation. $u_{xx} + u_{yy} + u_{zz} = 0$ is called three dimensional Laplace equation and expressed by $\Delta u$. This represents the velocity potential of the imcompressible fluid, the potential the electrostatic field, the steady state temperature distribution of the heat conduction.

SOLUTION Let $u = (x^2 + y^2 + z^2)^{-\frac{1}{2}}$. Then

$$u_x = -\frac{1}{2}(x^2 + y^2 + z^2)^{-\frac{3}{2}}(2x) = -x(x^2 + y^2 + z^2)^{-\frac{3}{2}}$$

$$u_{xx} = \frac{\partial u_x}{\partial x}$$

$$= -(x^2 + y^2 + z^2)^{-\frac{3}{2}} -x(-\frac{3}{2})(x^2 + y^2 + z^2)^{-\frac{5}{2}}(2x)$$

$$= -(x^2 + y^2 + z^2)^{-\frac{5}{2}}\big(x^2 + y^2 + z^2 - 3x^2\big)$$

$$= -(x^2 + y^2 + z^2)^{-\frac{5}{2}}(-2x^2 + y^2 + z^2)$$

$$u_y = -\frac{1}{2}(x^2 + y^2 + z^2)^{-\frac{3}{2}}(2y) = -y(x^2 + y^2 + z^2)^{-\frac{3}{2}}$$

$$u_{yy} = \frac{\partial u_y}{\partial y}$$

$$= -(x^2 + y^2 + z^2)^{-\frac{3}{2}} -y(-\frac{3}{2})(x^2 + y^2 + z^2)^{-\frac{5}{2}}(2y)$$

$$= -(x^2 + y^2 + z^2)^{-\frac{5}{2}}\big(x^2 + y^2 + z^2 - 3y^2\big)$$

$$= -(x^2 + y^2 + z^2)^{-\frac{5}{2}}(x^2 -2y^2 + z^2)$$

$$u_z = -\frac{1}{2}(x^2 + y^2 + z^2)^{-\frac{3}{2}}(2z) = -z(x^2 + y^2 + z^2)^{-\frac{3}{2}}$$

$$u_{zz} = \frac{\partial u_z}{\partial z}$$

$$= -(x^2 + y^2 + z^2)^{-\frac{3}{2}} -z(-\frac{3}{2})(x^2 + y^2 + z^2)^{-\frac{5}{2}}(2z)$$

$$= -(x^2 + y^2 + z^2)^{-\frac{5}{2}}\big(x^2 + y^2 + z^2 - 3z^2\big)$$

$$= -(x^2 + y^2 + z^2)^{-\frac{5}{2}}(x^2 + y^2 -2z^2)$$

Thus,

$$u_{xx} + u_{yy} + u_{zz} = -(x^2 + y^2 + z^2)^{-\frac{5}{2}}(2x^2 + 2y^2 + 2z^2 - 2x^2 -2y^2 -2z^2) = 0\ensuremath{ \blacksquare}$$

Example 4..15   Show the following functions are harmonic.
1. $z = x^2 - y^2$ 2. $z = \frac{x}{x^2 + y^2}$

SOLUTION 1. $z_x = 2x, z_{xx} = 2, z_y = -2y, z_{yy} = -2$. Thus, $\Delta z = z_{xx} + z_{yy} = 2 -2 = 0 \ensuremath{ \blacksquare}$ 2.

$$z_x = \frac{x^2 + y^2 - x(2x)}{(x^2 + y^2)^2} = \frac{-x^2 + y^2}{(x^2 + y^2 )^2},$$

$$z_{xx} = \frac{-2x(x^2 + y^2)^2 - (-x^2+y^2)(2(x^2 + y^2)(2x))}{(x^2 + y^2)^4}$$

$$= \frac{(x^2 + y^2)\big(-2x(x^2 + y^2) - (4x)(-x^2 + y^2)\big)}{(x^2 + y^2)^4}$$

$$= \frac{-2x^3 -2xy^2 + 4x^3 - 4xy^2}{(x^2 + y^2)^3} = \frac{2x^3 - 6xy^2}{(x^2 + y^2)^3},$$

$$z_y = \frac{- x(2y)}{(x^2 + y^2)^2} = \frac{-2xy}{(x^2 + y^2 )^2},$$

$$z_{yy} = \frac{-2x(x^2 + y^2)^2 - (-2xy)(2(x^2 + y^2)(2y))}{(x^2 + y^2)^4}$$

$$= \frac{(x^2 + y^2)\big(-2x(x^2 + y^2) +8xy^2\big)}{(x^2 + y^2)^4}$$

$$= \frac{-2x^3 -2xy^2 + 8xy^2}{(x^2 + y^2)^3} = \frac{-2x^3 + 6xy^2}{(x^2 + y^2)^3}.$$

Thus,

$$\Delta z = z_{xx} + z_{yy} = \frac{2x^3 - 6xy^2}{(x^2 + y^2)^3} + \frac{-2x^3 + 6xy^2}{(x^2 + y^2)^3}= 0\ensuremath{ \blacksquare}$$

When we express $u_{xx} + u_{yy}, u_{xx} + u_{yy} + u_{zz}$ by $\Delta u$. Then $\Delta$ is called Laplaian and the class $C^2$ function $u$ satisfying the equation $\Delta u = 0$ is called harmonic function.

$z_x = \frac{\partial}{\partial x}\big(\frac{x}{x^2 + y^2}\big) = \frac{\frac{\p... ...^2} = \frac{x^2 + y^2 - 2x^2}{(x^2 + y^2)^2} = \frac{-x^2 + y^2}{(x^2 + y^2)^2}$

Exercise 4..15   Show the following functions are harmonic.
1. $z = e^{x}\sin{y}$ 2. $z = \tan^{-1}{\frac{y}{x}}$

Exercise4-15-2.

$z_x = \frac{\partial}{\partial x}\big(\tan^{-1}{\frac{y}{x}}\big) = \frac{\part... ...{y}{x^2}) = \frac{1}{1+ (\frac{y}{x})^2}(- \frac{y}{x^2}) = \frac{y}{x^2 + y^2}$

SOLUTION 1. $z_x = e^{x}\sin{y}$, $z_{xx} = e^{x}\sin{y}$, $z_{y} = e^{x}\cos{y}$, $z_{yy} = -e^{x}\sin{y}$.
Thus, $\Delta z = z_{xx} + z_{yy} = e^{x}\sin{y} - e^{x}\sin{y} = 0\ensuremath{ \blacksquare}$

2. $${z_x = \frac{1}{1 + (\frac{y}{x})^2} \cdot \frac{-y}{x^2} = \frac{... ...2 + y^2}, z_{xx} = -\frac{-y(2x)}{(x^2 + y^2)^2} = \frac{2xy}{(x^2 + y^2)^2}}$$
$${z_y = \frac{1}{1 + (\frac{y}{x})^2} \cdot \frac{1}{x} = \frac{x}{x^2 + y^2}, z_{yy} = -\frac{x(2y)}{(x^2 + y^2)^2} = -\frac{2xy}{(x^2 + y^2)^2}}$$. Thus,

$$\Delta z = z_{xx} + z_{yy} = \frac{2xy}{(x^2 + y^2)^2} -\frac{2xy}{(x^2 + y^2)^2} = 0\ensuremath{ \blacksquare}$$