Voofie | Reducing a partial difference equation into a partial differential equation and solving for the generating function using method of characteristics

Content

1 Introduction
2 Generating Function
3 Method of Characteristics
4 To be done

Introduction

In the following section, I will try to solve the generating function of the sequence satisfying the following partial difference equation:

$\f$A_k^{n+1}=(k+1)A_{k+1}^n+(n+2-k)A_{k-1}^n\f$$ ....................................(1)

With initial condition of:

$\f$\begin{cases} A_0^0&=1\\ A_1^0&=1 \end{cases}\f$$

In general, the sequence $\f$A_k^n = 0\f$$ for k > n or k < 0, except when n = 0. The sequence arises from the derivative of sec x + tan x.

Generating Function

Define:

$\f$\displaystyle A_n(x)=\sum _{k=0}^{\infty } A_k^nx^k\f$$

Multiply (1) by $\f$x^k\f$$ and do a summation on k:

$\f$\displaystyle \sum _{k=0}^{\infty } A_k^{n+1}x^k=\sum _{k=0}^{\infty } \left((k+1)A_{k+1}^nx^k+x(n+1)A_{k-1}^nx^{k-1}-x^2(k-1)A_{k-1}^nx^{k-2}\right)\f$$

$\f$A_{n+1}(x)=x(n+1)A_n(x)+\left(1-x^2\right)A_n'(x)\f$$ ..........................(2)

Define:

$\f$\displaystyle A(x,y)=\sum _{n=0}^{\infty } A_n(x) \frac{y^n}{n!}\f$$

Multiplying (2) by $\f$\displaystyle \frac{y^n}{n!}\f$$ , and do a summation on n:

$\f$\displaystyle (n+1)\frac{A_{n+1}(x)}{(n+1)!}=x n \frac{A_n(x)}{n!}+\frac{x A_n(x)}{n!}+\frac{\left(1-x^2\right)A_n'(x)}{n!}\f$$

$\f$x A(x,y)=\left(x^2-1\right)A^{(1,0)}(x,y)+(1-x y) A^{(0,1)}(x,y)\f$$ ..................(3)

Method of Characteristics

The characteristics equation of (3) is given by:

$\f$\begin{align*} \frac{d x}{x^2-1}&=\frac{d y}{1-x y}=\frac{d z}{x z} \end{align*}\f$$ .............................................................(4)

We first let the initial values of (4) be:

$\f$x = x_0 \text{ when } y=0\f$$

First solve for the relation between x and y:

$\f$\begin{align*} \frac{d y}{d x}=\frac{1-x y}{x^2-1} \end{align*}\f$$

$\f$\begin{align*} \Rightarrow y=\frac{C-\sin ^{-1}x}{\sqrt{1-x^2}} \end{align*}\f$$

Plugging the initial condition, we have:

$\f$\begin{align*} y=\frac{\sin ^{-1}x_0-\sin ^{-1}x}{\sqrt{1-x^2}} \end{align*}\f$$ ............................................................(5)

Now, we solve for the relation between x and z using (4):

$\f$\begin{align*} \frac{d z}{x z}=\frac{d x}{x^2-1} \end{align*}\f$$

$\f$\begin{align*} \Rightarrow z=C_1\sqrt{1-x^2} \end{align*}\f$$

Since $\f$A(x,y) = z\f$$ , the constant $\f$C_1\f$$ should depend on the initial condition, i.e. the value of $\f$x_0\f$$ .

$\f$A(x,y)=F(x_0)\sqrt{1-x^2}\f$$

When y = 0,

$\f$\begin{align*} A\left(x_0,0\right)&=A_0\left(x_0\right) \\ F\left(x_0\right)\sqrt{1-x_0^2}&=1+x_0\\ F\left(x_0\right)&=\sqrt{\frac{1+x_0}{1-x_0}} \end{align*}\f$$

Therefore:

$\f$\displaystyle A(x,y)=\sqrt{\frac{1+x_0}{1-x_0}}\sqrt{1-x^2}\f$$

From (5), we knows that:

$\f$x_0=\sin \left(y\sqrt{1-x^2}+\sin ^{-1}x\right)\f$$

Finally, we have solved A:

$\f[A(x,y)=\sqrt{\frac{1+\sin \left(y\sqrt{1-x^2}+\sin ^{-1}x\right)}{1-\sin \left(y\sqrt{1-x^2}+\sin ^{-1}x\right)}}\sqrt{1-x^2}\f]$

After some simplification:

$\f$A(x,y)=\left(\sec \left(y\sqrt{1-x^2}+\sin ^{-1}x\right)+\tan \left(y\sqrt{1-x^2}+\sin ^{-1}x\right)\right)\sqrt{1-x^2}\f$$ ..............(6)

We may find an explicit formula for $\f$A_n^k\f$$ by doing power series expansion on (6), but it seems too difficult to do so.

To be done

I will try to solve it using another method, and post it up if I manage to solve it.

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Reducing a partial difference equation into a partial differential equation and solving for the generating function using method of characteristics

Introduction

Generating Function

Method of Characteristics

To be done

What is Voofie?