Ch5-Partial differentiation - Keii Chin's notes

Mathematical Methods for Physics and Engineering

1. Some properties of partial differentiation:

1.1 Clairaut’s theorem:

Let $f(x,y)$ be a function. Suppose $f$ has continuous second-order partial derivatives in a neighborhood of a point $(x_0,y_0)$. Then:

$$\frac{\partial^2 f}{\partial x\partial y} = \frac{\partial^2 f}{\partial y\partial x}$$

1.2 Total differential and total derivative:

$$df = f_x\,dx + f_y\,dy$$

(Partial derivatives exist and are continuous).

If both sides are divided by $dx$ then we find:

$$\frac{df}{dx} = f_x + f_y \frac{dy}{dx}$$

1.3 Exact total differential:

$$M(x,y)\,dx + N(x,y)\,dy$$

will be exact if

$$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$

for all pairs $(x,y)$. Well, its main purpose is to check if this total differential is the straightforward differential of any function.

1.4 The chain rule:

From 1.2 let the total differential divided by $dx$, then we have the chain rule:

$$\frac{dz}{dx} = \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} \frac{dy}{dx}$$

Example:

For plane polar coordinates which is related to Cartesian coordinates by:

$$x = r\cos\theta, \qquad y = r\sin\theta,$$

try to find expression of Laplace operator $\nabla^{2}$ written in $r$ and $\theta$. Solution is on page 160.

\nabla^2=\frac{1}{\rho}\frac{\partial}{\partial \rho}(\rho \frac{\partial}{\partial \rho})+\frac{1}{\rho^2}\frac{\partial^2}{\partial \phi^2}

2. Reciprocity relation and cyclic relation:

Start with $\left(\frac{\partial x}{\partial y}\right)_z$ and $\left(\frac{\partial y}{\partial z}\right)_x$ by substituting the latter into the former, and hold $z$ or $x$ constant respectively, we obtain: (reciprocity relation) and (cyclic relation).

Example:

$z$ is already known, try to prove:

(1) $\left(\frac{\partial x}{\partial y}\right)_z = \left(\frac{\partial x}{\partial y}\right)_z$; (2) $\left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y = -1$ (consider $x$, $y$, or $z$ constant).

Check answer p176~p177.

3. Taylor’s theorem for many-variable functions [Page 161-162]:

Two variables:

\begin{align} &f(x,y)=\sum\limits_{n=0}^{\infty}\frac{1}{n!}[(\Delta x\frac{\partial}{\partial x}+\Delta y\frac{\partial}{\partial y})^nf(x,y)]_{x_{0},y_{0}}\\ &\Delta x=x-x_0,\Delta y=y-y_0 \end{align}

or $f(x+\Delta x, y+\Delta y) = f(x,y) + f_x \Delta x + f_y \Delta y + \frac12\left(f_{xx}\Delta x^2 + 2f_{xy}\Delta x\Delta y + f_{yy}\Delta y^2\right) + \cdots$

Multi-variable:

\begin{align} &f(x_1,x_2,\dots)=\sum\limits_{n=0}^{\infty}\frac{1}{n!}[(\Delta \vec{x} \cdot\vec{\nabla})^nf(\vec{x})]_{x=x_{0}}\\ &\Delta \vec{x}=(x_1-x_0,x_2-x_0,\dots)\\ &\vec{\nabla}=(\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2},\dots) \end{align}

4. Stationary values of many-variable functions [Page 165-167]:

Two variables:

(1) Minima for $D = f_{xx}f_{yy} – f_{xy}^{2} > 0$ and $f_{xx} > 0$.

(2) Maxima for $D = f_{xx}f_{yy} – f_{xy}^{2} > 0$ and $f_{xx} < 0$.

(3) Saddle points for $D = f_{xx}f_{yy} – f_{xy}^{2} < 0$ or when $f_{xx}f_{yy} – f_{xy}^{2} < 0$.

(4) All the other situations: undetermined.

Many-variable:

The many-variable Taylor expansion stated in section 3 expanded as following:

\begin{aligned} f(x_1,x_2,\dots)&=\sum\limits_{n=0}^{\infty}\frac{1}{n!}[(\Delta \vec{x} \cdot\vec{\nabla})^nf(\vec{x})]_{x=x_{0}}\\ &=f(\vec{x_0})+\sum\limits_{i}\frac{\partial f}{\partial x_i} \mid_{\vec{x}=\vec{x_0}}\Delta x_i+\sum\limits_{i}\sum\limits_{j}\frac{\partial ^2f}{\partial x_i\partial x_j} \mid_{\vec{x}=\vec{x_0}}\Delta x_i\Delta x_j+\dots \end{aligned}

In order to find stationary points, for all $x_i$, it must have: $\frac{\partial f}{\partial x_i}=0$. So it becomes:

\begin{aligned} f(\vec{x})&=f(x_1,x_2,\dots)\\ &=f(\vec{x_0})+\sum_{i}\sum_{j}\frac{\partial^2 f}{\partial x_i \partial x_j}\Big|_{\vec{x}=\vec{x_0}}\Delta x_i \Delta x_j+\dots \end{aligned}

But how do we decide a minima or a maxima of a many-variable function? It’s simple. For a N-dimension function, given point $\vec{x}=\vec{x_0}$, no matter how to change the variable $\vec{x}=(x_1,x_2,x_3,\dots,x_N)$ at an arbitrary rate $\Delta x=\sum\limits_r a_r\vec{e_r}$ around $\vec{x}=\vec{x_0}$, it always has $\Delta f=f(\vec{x})-f(\vec{x_0})>0$ or $\Delta f=f(\vec{x})-f(\vec{x_0})<0$, then this point is a minima or a maxima. We then rewrite $\Delta$ f as:

\begin{align} \Delta f&=\sum\limits_{i}\sum\limits_{j}\frac{\partial ^2f}{\partial x_i\partial x_j} \mid_{\vec{x}=\vec{x_0}}\Delta x_i\Delta x_j\\ &=\frac{1}{2}\Delta x^TH\Delta x=\sum_{r}\frac{1}{2}\lambda_{r}a_{r}^{2} \end{align}

where $M_{ij}=\frac{\partial ^2f}{\partial x_i\partial x_j}\mid_{\vec{x}=\vec{x_0}}$ called Hessian matrix, and $M\vec{e_r}=\lambda_r\vec{e_r}$, $\vec{e_r}\vec{e_s}=\delta_{rs}$. We require $\Delta f>0$ or $\Delta f<0$ for all $a_r$(change at an arbitrary rate $\Delta x=\sum\limits_r a_r\vec{e_r}$ around $\vec{x}=\vec{x_0}$), then we conclude that this point $\vec{x}=\vec{x_0}$ is a minima when all eigenvalues($\lambda_r$) of M are positive, a maxima when all eigenvalues($\lambda_r$) of M are negative. See an example on page 167.

5. Method of Lagrange undetermined multipliers [Page 168]:

When need to find stationary points of $f(x,y,\ldots)$ with a constrained condition $g(x,y,\ldots)=0$ and $h(x,y,\ldots)=0$, we set:

d(f+\lambda g+\mu h)=\sum\limits_{i}\frac{\partial(f+\lambda g+\mu h)}{\partial x_i}dx_i=0

This requires $\frac{\partial(f+\lambda g+\mu h)}{\partial x_i}=0$ for all $x_i$. You can expand to more conditions. See an example about Boltzmann distribution($n_k = Ce^{\mu E_k}$) on page 172.

6. Envelopes [Page 175]:

The definition of envelope: $f(x,y,t)=0,\ \frac{\partial f(x,y,t)}{\partial t}=0$. See an example:

Find the envelope of a length L ladder leaning on a vertical wall. The answer is given on page 175.

Notes:

You need to find the simplest equation of $f$ before you continue to $\frac{\partial f(x,y,t)}{\partial t}=0$. Here’s why.

The textbook skipped the most important part about $\frac{\partial f(x,y,t)}{\partial t}=0$ which may mislead reader. The Ladder is $\frac{x}{a}+\frac{y}{b}=1$,$a^2+b^2=L^2$, Then you take partial differentiation of $f(x,y,a)=\frac{x}{a}+\frac{y}{L^2-a^2}-1=0$ where you get $a=\frac{Lx^{1/3}}{(x^{2/3}+y^{2/3})^{1/3}}$ which leads to the envelope equation $x^{2/3}+y^{2/3}=L^{2/3}$. It seems perfect, before you have to find a simplest equation of this ladder.

If you write the ladder as $f(x,y,a)=\sqrt{L^2-a^2}x+ay-a\sqrt{L^2-a^2}=0$, then you’ll find it is difficult to find the envelope although there’s no procedural problems. Even in this book, its solution is complicated that it’s really difficult for you to get the answer $a=\frac{Lx^{1/3}}{(x^{2/3}+y^{2/3})^{1/3}}$ from $\frac{\partial f(x,y,a)}{\partial a}=-\frac{x}{a^2}+\frac{ay}{(L^2-a^2)^{3/2}}=0$. You’d better write it as $f(x,y,a)=x+\frac{a}{b}y-a=x+\frac{a}{\sqrt{L^2-a^2}}y-a=0$ where $\frac{\partial f}{\partial a}=\frac{L^2-a^2+a^2}{(L^2-a^2)^{3/2}}y-1=0\rightarrow y=\frac{(L^2-a^2)^{3/2}}{L^2}$, because of symmetry, and $x=\frac{(L^2-b^2)^{3/2}}{L^2}$. Finally, add them together it’s $x^{2/3}+y^{2/3}=L^{2/3}$.

7. Differentiation of integrals [Page 179]:

If $I(x)=\int^{v(x)}_{u(x)}f(x,t)dt$, then $$\frac{\partial I}{\partial x}=f(x,v(x))\frac{dv}{dx}-f(x,u(x))\frac{du}{dx}+\int^{v(x)}_{u(x)}\frac{\partial f(x,t)}{\partial x}dt$$.