Ch5-Partial differentiation

Mathematical Methods for Physics and Engineering

1. Some properties of partial differentiation:

1.1 Clairaut's theorem:

Let f: \mathbb{R}^n \to \mathbb{R} be a function. Suppose f has continuous second-order partial derivatives in a neighborhood of a point (x_1,x_2,\dots,x_n) \in \mathbb{R}^n. Then: \mathbf{\frac{\partial^2f}{\partial x_i\partial x_j}=\frac{\partial^2f}{\partial x_j\partial x_i}}.

1.2 Total differential and total derivative:

\mathbf{df=\sum\limits_{i=1}^{n}\frac{\partial f}{\partial x_i}dx_i} (Partial derivatives exist and are continuous).

If both side are divided by dx_1 then we find total\ derivative\ of\ x_1: \mathbf{\frac{df}{dx_1}=\frac{\partial f}{\partial x_1}+\frac{\partial f}{\partial x_2}\frac{dx_2}{dx_1}+\cdots+\frac{\partial f}{\partial x_n}\frac{dx_n}{dx_1}}.

1.3 Exact total differential:

\mathbf{df=\sum\limits_{i=1}^{n}g_i(x_1,x_2,\dots,x_n)dx_i} will be exact if \frac{\partial g_i}{\partial x_j}=\frac{\partial g_j}{\partial x_i} for all pairs i, j. Well, its main purpose is to check if this total differential is the straightforward differential of any function.

1.4 The chain rule:

Form 1.2 let the total differential divided by du, then we have the chain rule: \mathbf{\frac{df}{du}=\sum\limits_{i=1}^{n}\frac{\partial f}{\partial x_i}\frac{dx_i}{du}}.

Example:

For plane polar coordinates which is related to Cartesian coordinates by: x=\rho cos\phi,\ y=\rho sin\phi, try to find expression of Laplace operator \Delta=\nabla^2 written in \rho and \phi. Solution is on page 160.

2. Reciprocity relation and cyclic relation:

Start with dx=(\frac{\partial x}{\partial y})_zdy+(\frac{\partial x}{\partial z})_ydz and dy=(\frac{\partial y}{\partial x})_zdx+(\frac{\partial y}{\partial z})_xdz by substituting the later into the former, and hold x or y

constant respectively, we obtain: \mathbf{(\frac{\partial x}{\partial y})_z=(\frac{\partial y}{\partial x})_z^{-1}}(reciprocity relation) and \mathbf{(\frac{\partial y}{\partial z})_x(\frac{\partial z}{\partial x})_y(\frac{\partial x}{\partial y})_z=-1}(cyclic relation).

Example:

dU=TdS-PdV is already known, try to prove:

(1) (\frac{\partial T}{\partial V})_{S}=-(\frac{\partial P}{\partial S})_{V}; (2) (\frac{\partial S}{\partial V})_{T}=(\frac{\partial P}{\partial T})_{V} (considerdS or U-ST).

Check answer p176~p177.

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

Two variables:

Multi-variable:

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

Two variables:

(1)Minima for f_{xx}>0,f_{yy}>0,f_{xy}^2<f_{xx}f_{yy}.

(2)Maxima for f_{xx}<0,f_{yy}<0,f_{xy}^2<f_{xx}f_{yy}.

(3)Saddle points for f_{xx}f_{yy}<0 or f_{xy}^2>f_{xx}f_{yy}.

(4)All the other situations: undetermined.

Many-variable:

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

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

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:

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_1,x_2,\dots) with a constrained condition g(x_1,x_2,\dots) and h(x_1,x_2,\dots), we set:

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.