Some applications of special trans function theory in physics

Abstract

Examples of applications of special trans function theory (STFT) which may be useful in obtaining closed-form analytical solutions of certain transcendental equations found in undergraduate physics is presented. The novelty of the method can be of utmost importance to the academia involved in undergraduate physics.

A. Appendix

Here, we shall show how eq. (2) is a solution to a family of transcendental equations represented by eq. (1). The proof of the same is presented in [7]. We reproduce it here so that it can be useful to the readers. According to STFT [7], eq. (1) has the analytical closed-form solution

$$\begin{aligned} \psi (\zeta )=\mathrm{trans}_{+} \left( {U(\zeta )} \right) , \end{aligned}$$

(A1)

where $\mathrm{trans}_{+} \left( {U(\zeta )} \right)$ is a new special function defined as

$$\begin{aligned} \mathrm{trans}_{+} \left( {U(\zeta )} \right) =\lim \limits _{x\rightarrow \infty } \left[ {\ln \left( {\frac{\phi (x,U(\zeta )}{\phi (x+1,U(\zeta )}} \right) } \right] \end{aligned}$$

(A2)

and $\phi \left( {x,U(\zeta )} \right)$ is defined as

$$\begin{aligned} \phi \left( {x,U(\zeta )} \right) =\sum \limits _{n=0}^{[x]} {\frac{(U(\zeta ))^{n}(x-n)^{n}}{n!}} , \end{aligned}$$

(A3)

where $\left[ x \right]$ denotes the greatest integer less than or equal to x.

Proof

The transcendental equation (1) can be identified with a partial differential equation of the following type:

$$\begin{aligned} \frac{\partial \phi (x,U(\zeta ))}{\partial x}=U(\zeta )\phi (x-1,U(\zeta )). \end{aligned}$$

(A4)

Making use of the Laplace transform, we shall solve the partial differential equation (A4). On taking Laplace transform of eq. (A4), we get

$$\begin{aligned} sF\left( {s,U(\zeta )} \right) -U(\zeta )F\left( {s,U(\zeta )} \right) \mathrm{e}^{-s}=\phi (0), \end{aligned}$$

(A5)

where

$$\begin{aligned} F\left( {s,U(\zeta )} \right) =L\left\{ {\phi \left( {x,U(\zeta )} \right) } \right\} . \end{aligned}$$

A little arrangement of eq. (A5) gives

$$\begin{aligned} F(x,U(x))= & {} \frac{\phi (0)}{s-U(\zeta )e^{-s}}\nonumber \\= & {} \frac{\phi (0)}{s\left( {1-U(\zeta )\frac{\mathrm{e}^{-s}}{s}} \right) } \end{aligned}$$

(A6)

Expansion of the denominator on the right-hand side of eq. (A6) gives

$$\begin{aligned} F\left( {x,U(\zeta )} \right) =\frac{\phi (0)}{s}\sum \limits _{n=0}^\infty {\left( {U(\zeta )} \right) ^{n}} \frac{\mathrm{e}^{-ns}}{s^{n}}. \end{aligned}$$

(A7)

The series on the right-hand side of eq. (A7) converges for $\left| {U(\zeta )\frac{\mathrm{e}^{-s}}{s}} \right| \ll 1$. Now, inverting eq. (A7) term by term, we get

$$\begin{aligned} \phi \left( {x,U(\zeta )} \right) =\sum \limits _{n=0}^\infty {\left( {U(\zeta )} \right) ^{n}\frac{\left( {x-n} \right) ^{n}}{n!}H(x-n)} , \end{aligned}$$

(A8)

where $H_({x-n})$ is the Heaviside’s unit function. The function series on the right-hand side of eq. (A8) is identical to that of eq. (A3) for ${x}>n$. Finally, we get the analytical solution to the partial differential equation (A4) by applying Laplace transform in a closed-form representation as

$$\begin{aligned} \phi \left( {x,U(\zeta )} \right) =\sum \limits _{n=0}^{[x]} {\left( {U(\zeta )} \right) ^{n}} \frac{(x-n)^{n}}{n!}. \end{aligned}$$

(A9)

According to Lerch’s theorem, eq. (A9) is the unique analytical closed-form solution to eq. (A4).

Lemma1

For any $x > 1$, the series on the right -hand side of eq. (A9) satisfies eq. (A4).

Substitution of eq. (A9) into eq. (A4) yields

$$\begin{aligned}&\frac{\partial \phi \left( {x,U(\zeta )} \right) }{\partial x}-U(\zeta )\phi \left( {x-1,U(\zeta )} \right) \nonumber \\&=\frac{\partial }{\partial x}\left( {\sum \limits _{n=0}^{[x]} {\left( {U(\zeta )} \right) ^{n}\frac{(x-n)^{n}}{n!}} } \right) \nonumber \\&\quad -U(\zeta )\sum \limits _{n=0}^{[x-1]} {\left( {U(\zeta )} \right) ^{n}\frac{(x-1-n)^{n}}{n!}} \nonumber \\&=\sum \limits _{n=0}^{[x]} {\left( {U(\zeta )} \right) ^{n}\frac{(x-n)^{n-1}}{(n-1)!}}\nonumber \\&\quad -\sum \limits _{n=0}^{[x-1]} {{\left( {U(\zeta )} \right) ^{n+1}\frac{(x-(n+1))^{n}}{n!}} } \nonumber \\&\quad =\sum \limits _{n=0}^{[x-1]} {\left( {U(\zeta )} \right) ^{n+1}\frac{(x-(n+1))^{n}}{n!}} \nonumber \\&\quad -\sum \limits _{n=0}^{[x-1]} {\left( {U(\zeta )} \right) ^{n+1}\frac{(x-(n+1))^{n}}{n!}} =0 \end{aligned}$$

(A10)

for any $x \notin N$. For $x \in N$ and $x \ne$ 1, $\phi \left( {x,U(\zeta )} \right)$ is differentiable and the proof follows as well.

Remark1

It can be easily shown that the function of eq. (A9) is not differentiable at $x =$ 1, due to the term of first order. Now, from eq. (A8) we can write

$$\begin{aligned} \varphi _{n}\left( x \right) =\left( x-n \right) ^{n}H(x-n) \end{aligned}$$

and

$$\begin{aligned} \varphi _n(x)=\left\{ \begin{array}{cc} 0&{}\quad \hbox {for } x<n\\ (x-n)^{n}&{}\quad \hbox {for } x>n\\ \end{array}\right. . \end{aligned}$$

(A11)

Also,

$$\begin{aligned} \varphi '_n(x)=\left\{ \begin{array}{cc} 0&{}\quad \hbox {for } x<n\\ n(x-n)^{n-1}&{}\quad \hbox {for } x>n\\ \end{array}\right. . \end{aligned}$$

(A12)

Clearly, $\varphi _{n}(x)$ is differentiable for $n\ne 1 \quad \forall x>0.$ Also, $\varphi _{1}(x)$ for $n=1$ is differentiable $\forall x>0$ and $x\ne 1$.

Lemma2

On the particular solution to the differential eq. (A4)

By the method of separation of variables, we can write

$$\begin{aligned} \phi \left( {x,U(\zeta )} \right) =\Theta \left( {U(\zeta )} \right) \mathrm{e}^{\omega \left( {U(\zeta )} \right) x} \end{aligned}$$

(A13)

and from eqs (A4) and (A13), we have

$$\begin{aligned} \omega \left( {U(\zeta )} \right) =U(\zeta )\mathrm{e}^{-\omega \left( {U(\zeta )} \right) }. \end{aligned}$$

(A14)

From eqs (1) and (A14), we see that

$$\begin{aligned} \omega \left( {U(\zeta )} \right) =\psi (\zeta )=\psi (\zeta ;U(\zeta )). \end{aligned}$$

This means that the particular solution of the form

$$\begin{aligned}&\phi _{p} \left( {x,U(\zeta )} \right) =\Theta \left( {U(\zeta )} \right) \mathrm{e}^{\omega \left( {U(\zeta )} \right) x} \nonumber \\&\quad =\Theta \left( {U(\zeta )} \right) \mathrm{e}^{\psi (\zeta ;U(\zeta ))x} =\Theta \left( {U(\zeta )} \right) \mathrm{e}^{\psi (\zeta )x} \end{aligned}$$

(A15)

satisfies the differential equation (A4) under the condition that $\psi (\zeta )$ satisfies eq. (1). Consequently, eq. (A15) is an asymptotic function to eq. (A4) and hence

$$\begin{aligned} \lim \limits _{x\rightarrow \infty } \left( {\frac{\phi (x,U(\zeta ))}{\phi _{p} (x,U(\zeta ))}} \right) =1 \end{aligned}$$

(A16)

based on the functional theory. Now, by the principle of unique solution and functional theory, we have

$$\begin{aligned}&\lim \limits _{x\rightarrow \infty } \left( {\frac{\phi (x+1,U(\zeta ))}{\phi (x,U(\zeta ))}} \right) =\frac{\phi _{p} (x+1,U(\zeta ))}{\phi _{p} (x,U(\zeta ))}\nonumber \\&\quad =\frac{\Theta \left( {U(\zeta )} \right) \mathrm{e}^{\omega (U(\zeta ))(x+1)}}{\Theta \left( {U(\zeta )} \right) \mathrm{e}^{\omega (U(\zeta ))x}}=\mathrm{e}^{\omega (U(\zeta ))}\nonumber \\&\quad =\mathrm{e}^{\psi (\zeta )} \end{aligned}$$

(A17)

and

$$\begin{aligned} \begin{array}{ll} \psi (\zeta )&{}=\mathrm{trans}_{+} \left( {U(\zeta )} \right) \\ &{}=\lim \limits _{x\rightarrow \infty } \left[ \ln \left( \dfrac{\phi (x+1,U(\zeta ))}{\phi (x,U(\zeta ))} \right) \right] \\ \end{array}. \end{aligned}$$

(A18)

As $\psi (\zeta )=U(\zeta )\mathrm{e}^{-\psi (\zeta )}$, eq. (A18) can be written as

$$\begin{aligned} \psi (\zeta )=\lim \limits _{x\rightarrow \infty } \left( {U(\zeta )\frac{\phi (x,U(\zeta ))}{\phi (x+1,U(\zeta ))}} \right) . \end{aligned}$$

(A19)

More explicitly,

$$\begin{aligned} \begin{array}{l} \psi (\zeta )=\mathrm{trans}_{+} \left( {U(\zeta )} \right) \\ \quad =\lim \limits _{x\rightarrow \infty } \left( {U(\zeta )\frac{\sum \nolimits _{n=0}^{[x]} {\frac{(U(\zeta ))^{n}(x-n)^{n}}{n!}} }{\sum \nolimits _{n=0}^{[x+1]} {\frac{(U(\zeta ))^{n}(x+1-n)^{n}}{n!}} }} \right) \\ \end{array}. \end{aligned}$$

(A20)

Equation (A20) defines a new special function namely, $\mathrm{trans}_{+} (U(\zeta ))$. The essential part of STFT is the existence of eq. (A16). It is clear that by applying unique solution principle, we have eq. (A16). Thus, it is proved that eq. (A1) is the solution to eq. (1).

To illustrate the application of STFT, let us consider the following transcendental equation as a mathematical example:

$$\begin{aligned} x=\ln (y)+y. \end{aligned}$$

(A21)

Taking exponential on both sides of eq. (A21), we have

$$\begin{aligned} \mathrm{e}^{x}=y\mathrm{e}^{y}. \end{aligned}$$

(A22)

We can rewrite eq. (A22) as

$$\begin{aligned} y=\mathrm{e}^{x}\mathrm{e}^{-y}. \end{aligned}$$

(A23)

Equation (A23) takes the form of eq. (1) and hence by the application of STFT, for $\mathrm{e}^{x}\in R$, we have the analytical closed-form solution of eq. (A21) as

$$\begin{aligned} y= & {} \mathrm{trans}_{+} (\mathrm{e}^{x}) \nonumber \\= & {} \lim \limits _{M\rightarrow \infty } \left( {\mathrm{e}^{x}\left( {\frac{\sum \nolimits _{n=0}^{[M]} {(\mathrm{e}^{x})^{n}\frac{(M-n)^{n}}{n!}} }{\sum \nolimits _{n=0}^{[M+1]} {(\mathrm{e}^{x})^{n}\frac{(M+1-n)^{n}}{n!}} }} \right) } \right) , \end{aligned}$$

(A24)

where [M] is the greatest integer less than or equal to M. Equation (A24) is a new special function that gives explicit analytical solution to eq. (A21).