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# Classical exponential bounds for $p$-generalized circular and hyperbolic
functions

Yogesh J. Bagul 1, Bharti O. Fande2,a 1Department of Mathematics
K. K. M. College, Manwath
Dist: Parbhani (M. S.) - 431505, India
a Research Scholar
2Department of Mathematics
Government Vidarbha Institute of Science
and Humanities, Amravati (M. S.)-444604, India

###### Abstract.

In this article, we obtain exponential bounds for the generalized circular and
hyperbolic functions with one parameter $p$. Our results are natural
generalizations of some existing results for classical circular and hyperbolic
functions.

###### Key words and phrases:

Generalized circular functions, generalized hyperbolic functions,
eigenfunction, exponential bounds, inequalities.

###### 1991 Mathematics Subject Classification:

33B10, 33E30, 26D05.

##  1\. Introduction

For $0\leq x\leq 1,$ we know that

$\arcsin x=\int_{0}^{x}\frac{1}{(1-t^{2})^{1/2}}\,dt$

and

$\frac{\pi}{2}=\arcsin 1=\int_{0}^{1}\frac{1}{(1-t^{2})^{1/2}}\,dt.$

Hence it is easy to define the function sine on $[0,\pi/2]$ as the inverse of
arcsine and it can be then extended on $(-\infty,\infty).$ On the other hand,
an eigenfunction in the Dirichlet problem [11] for the one-dimensional
$p$-Laplacian occurs as the inverse of the function
$\phi_{p}:[0,1]\rightarrow\mathbb{R}$ given by

$\phi_{p}(x)=\int_{0}^{x}\frac{1}{(1-t^{p})^{1/p}}\,dt.$

This eigenfunction is associated with the eigenvalue containing
$\pi_{p}=\frac{2\pi}{p\sin(\pi/p)}.$ Therefore, we can denote $\phi_{p}(x)$ by
$\arcsin_{p}(x),$ i.e.,

$\arcsin_{p}(x)=\int_{0}^{x}\frac{1}{(1-t^{p})^{1/p}}\,dt$

and we evaluate

$\arcsin_{p}(1)=\frac{\pi}{p\sin(\pi/p)}=\frac{\pi_{p}}{2}.$

Thus, the inverse of $\arcsin_{p}$ on $[0,\pi_{p}/2]$ is a natural
generalization of the sine function and it is denoted by $\sin_{p}.$ For
$1<p<\infty,$ we call this function as the $p$-generalized sine function. The
$p$-generalized sine function is clearly strictly increasing on
$[0,\pi_{p}/2]$ and $\sin_{p}0=0,$ $\sin_{p}(\pi_{p}/2)=1.$ Further, we can
extend this function on $(-\infty,\infty)$ by $2\pi_{p}$ periodicity. For
instance, $\sin_{p}$ is extended on $[-\pi_{p},\pi_{p}]$ by oddness and by
defining

$\sin_{p}(x)=\sin_{p}(\pi_{p}-x);\,\,x\in[\pi_{p}/2,\pi_{p}].$

In the same way, other generalized circular and hyperbolic functions can be
defined. These generalized functions are quite similar to the classical
functions in many aspects and they have been studied extensively. For more
details, we refer the reader to [9, 16, 23, 6, 17, 18, 19, 12] and the
references thereof. In this context, many researchers have obtained the
inequalities involving generalized circular and hyperbolic functions. See, for
example [7, 8, 13, 14, 15, 21, 22, 24] and the references therein. In this
article, we establish exponential bounds for the generalized circular and
hyperbolic functions. For the corresponding classical functions, this type of
bounds can be seen in [2, 3, 4, 5, 10, 20].

##  2\. Preliminaries and Lemmas

In this section, we see the definitions of other generalized circular and
hyperbolic functions, and lemmas for proving our main results.

Let $1<p<\infty.$ Then the generalized cosine function $\cos_{p}$ is defined
by

$\cos_{p}x=\diff{\sin_{p}x}{x},\,\,x\in[0,\pi_{p}/2].$

Hence,

$\cos_{p}x=(1-\sin_{p}^{p}x)^{1/p},\,\,x\in[0,\pi_{p}/2].$

From this we get $\cos_{p}0=1$ and $\cos_{p}(\pi_{p}/2)=0,$ and

$\diff{\cos_{p}x}{x}=-\cos_{p}^{2-p}x\sin_{p}^{p-1}x,\,\,x\in[0,\pi_{p}/2].$

It is to be noted that the generalized cosine function $\cos_{p}$ is
decreasing on $[0,\pi_{p}/2].$ The generalized tangent function is defined by

$\tan_{p}x=\frac{\sin_{p}x}{\cos_{p}x},\,\,x\in\mathbb{R}-\left\\{k\pi_{p}+\frac{\pi_{p}}{2}:k\in\mathbb{Z}\right\\}.$

This implies

$\diff{\tan_{p}x}{x}=1+|\tan_{p}x|^{p},\,\,x\in(-\pi_{p}/2,\pi_{p}/2)$

Or

$\diff{\tan_{p}x}{x}=1+\tan_{p}^{p}x,\,\,x\in[0,\pi_{p}/2).$

The generalized secant function is denoted by $\sec_{p}$ and is defined as

$\sec_{p}x=\frac{1}{\cos_{p}x},\,\,x\in[0,\pi_{p}/2).$

Then it is easy to obtain

$\sec_{p}^{p}x=1+\tan_{p}^{p}x,\,\,x\in(0,\pi_{p}/2)$

and

$\diff{\sec_{p}x}{x}=\sec_{p}x\tan_{p}^{p-1}x,\,\,x\in[0,\pi_{p}/2).$

The generalized hyperbolic sine function $\sinh_{p}$ is the inverse of the
generalized inverse hyperbolic sine function $\operatorname{arcsinh}_{p}$
which is defined by

$\operatorname{arcsinh}_{p}x=\left\\{\begin{array}[]{cc}\int_{0}^{x}\frac{1}{(1+t^{p})^{1/p}}dt,&\quad
x\in[0,\infty),\\\ -\operatorname{arcsinh}_{p}(-x),&\quad
x\in(-\infty,0).\end{array}\right.$

The generalized hyperbolic cosine, tangent, and secant functions are denoted
by $\cosh_{p}$, $\tanh_{p},$ and $\operatorname{sech}_{p}$ respectively. They
are defined as

$\cosh_{p}x=\diff{\sinh_{p}x}{x},\,\,\tanh_{p}x=\frac{\sinh_{p}x}{\cosh_{p}x},\,\,\operatorname{sech}_{p}x=\frac{1}{\cosh_{p}x}.$

It is worth mentioning that all the above-generalized functions become
classical ones for $p=2.$ Furthermore, it is not difficult to obtain the
following formulae:

$\cosh_{p}^{p}x-\sinh_{p}^{p}=1,\,\,x>0,$
$\diff{\cosh_{p}x}{x}=\cosh_{p}^{2-p}\sinh_{p}^{p-1}x,\,\,x\geq 0,$
$\diff{\tanh_{p}x}{x}=1-\tanh_{p}^{p}x=\operatorname{sech}_{p}^{p}x,\,\,x\geq
0,$

and

$\diff{\operatorname{sech}_{p}x}{x}=-\operatorname{sech}_{p}x\tanh_{p}^{p-1}x.$

The proofs of our generalizations require the l’Hôpital’s rule of monotonicity
[1]. It is the key tool and is stated as

######  Lemma 2.1.

([1]) Let $f_{1}(x)$ and $f_{2}(x)$ be two real valued-functions which are
continuous on $[a,b]$ and derivable on $(a,b)$, where $-\infty<a<b<\infty$ and
$f_{2}^{\prime}(x)\neq 0,$ for all $x\in(a,b).$ Let,

$A(x)=\frac{f_{1}(x)-f_{1}(a)}{f_{2}(x)-f_{2}(a)},\ x\in(a,b)$

and

$B(x)=\frac{f_{1}(x)-f_{1}(b)}{f_{2}(x)-f_{2}(b)},\ x\in(a,b).$

Then, we have

  * (i)

$A(x)$ and $B(x)$ are increasing on $(a,b)$ if
$f_{1}^{\prime}(x)/f_{2}^{\prime}(x)$ is increasing on $(a,b).$

  * (ii)

$A(x)$ and $B(x)$ are decreasing on $(a,b)$ if
$f_{1}^{\prime}(x)/f_{2}^{\prime}(x)$ is decreasing on $(a,b).$

The strictness of the monotonicity of $A(x)$ and $B(x)$ depends on the
strictness of the monotonicity of $f_{1}^{\prime}(x)/f_{2}^{\prime}(x)$.

Also, we prove the following auxiliary results.

######  Lemma 2.2.

For $p>1,$ the function $\xi(x)=\frac{x}{\tan_{p}x}$ is strictly decreasing in
$(0,\pi_{p}/2).$

######  Proof 2.3.

After differentiation, we get

$(\tan_{p}^{2}x)\xi^{\prime}(x)=\tan_{p}x-x-x\tan_{p}^{p}x:=\gamma(x)$

and

$\displaystyle\gamma^{\prime}(x)$
$\displaystyle=1+\tan_{p}^{p}x-1-\tan_{p}^{p}x-px\tan_{p}^{p-1}x$
$\displaystyle=-px\tan_{p}^{p-1}x<0$

for $p>1$ and $x\in(0,\pi_{p}/2).$ This implies that $\gamma(x)$ is strictly
decreasing in $(0,\pi_{p}/2)$ and so $\xi(x)$ is strictly decreasing in
$(0,\pi_{p}/2).$

######  Lemma 2.4.

If $p>1,$ then the function $\varsigma(x)=\frac{x}{\tanh_{p}x}$ is strictly
increasing for $x\geq 0.$

######  Proof 2.5.

Since

$(\tanh_{p}^{2}x)\varsigma^{\prime}(x)=\tanh_{p}x-x\operatorname{sech}_{p}^{p}x:=\delta(x)$

and

$\delta^{\prime}(x)=px\operatorname{sech}_{p}^{p-1}x\operatorname{sech}_{p}x\tanh_{p}^{p-1}x>0$

as $p>1$ and $x\geq 0.$ The statement follows by making the same argument as
in the proof of Lemma 2.2.

##  3\. Main results

First, we establish classical exponential bounds for the generalized circular
functions.

######  Theorem 3.1.

If $p\geq 2$ and $x\in(0,\pi_{p}/2),$ then we have

$\displaystyle e^{\alpha x^{p}}<\frac{\sin_{p}x}{x}<e^{\beta x^{p}}$ (1)

with the best possible constants
$\alpha=\frac{2^{p}\ln(2/\pi_{p})}{\pi_{p}^{p}}$ and
$\beta=-\frac{1}{p(p+1)}.$

######  Proof 3.2.

Consider the function

$f(x)=\frac{\ln\left(\frac{\sin_{p}x}{x}\right)}{x^{p}}:=\frac{f_{1}(x)}{f_{2}(x)},$

where $f_{1}(x)=\ln\left(\frac{\sin_{p}x}{x}\right)$ and $f_{2}(x)=x^{p}$ with
$f_{1}(0+)=0$ and $f_{2}(0)=0.$ Differentiation gives

$\frac{f_{1}^{\prime}(x)}{f_{2}^{\prime}(x)}=\frac{x\cos_{p}x-\sin_{p}x}{p\cdot
x^{p}\sin_{p}x}:=\frac{f_{3}(x)}{f_{4}(x)},$

where $f_{3}(x)=x\cos_{p}x-\sin_{p}x$ and $f_{4}(x)=p\cdot x^{p}\sin_{p}x$
with $f_{3}(0)=0=f_{4}(0).$ Again differentiating with respect to $x,$ we get

$\displaystyle\frac{f_{3}^{\prime}(x)}{f_{4}^{\prime}(x)}$
$\displaystyle=\frac{-1}{p}\cdot\frac{x\cos_{p}^{2-p}x\sin_{p}^{p-1}x}{x^{p}\cos_{p}x+p\cdot
x^{p-1}\sin_{p}x}$
$\displaystyle=-\frac{1}{p}\cdot\frac{1}{\left(\frac{x}{\tan_{p}x}\right)^{p-1}+p\left(\frac{x}{\tan_{p}x}\right)^{p-2}},$

which is decreasing by Lemma 2.2 as $p\geq 2.$ Applying Lemma 2.1 repeatedly
we see that $f(x)$ is strictly decreasing in $(0,\pi_{p}/2).$ Consequently, we
write

$f\left(\frac{\pi_{p}}{2}-\right)<f(x)<f(0+).$

Since $f(0+)=-\frac{1}{p(p+1)}$ and
$f(\pi_{p}/2-)=\frac{2^{p}\ln(2/\pi_{p})}{\pi_{p}^{p}},$ we obtain (1).

######  Theorem 3.3.

Let $p\geq 1$ and $a\in(0,a),$ where $a<\frac{\pi_{p}}{2}.$ Then the
inequalities

$\displaystyle e^{\alpha_{1}x^{p}}<\cos_{p}x<e^{\beta_{1}x^{p}}$ (2)

hold with the best possible constants
$\alpha_{1}=\frac{\ln(\cos_{p}a)}{a^{p}}$ and $\beta_{1}=-\frac{1}{p}.$

######  Proof 3.4.

Suppose

$g(x)=\frac{\ln(\cos_{p}x)}{x^{p}}:=\frac{g_{1}(x)}{g_{2}(x)},\,\,x\in(0,a),$

where $g_{1}(x)=\ln(\cos_{p}x)$ and $g_{2}(x)=x^{p}$ such that
$g_{1}(0)=0=g_{2}(0).$ Then

$\displaystyle\frac{g_{1}^{\prime}(x)}{g_{2}^{\prime}(x)}$
$\displaystyle=\frac{-\cos_{p}^{1-p}x\sin_{p}^{p-1}x}{p\cdot
x^{p-1}}=-\frac{1}{p}\left(\frac{\tan_{p}x}{x}\right)^{p-1}$

which is strictly decreasing in $(0,a)$ by Lemma 2.2. By Lemma 2.1, $g(x)$ is
also decreasing in $(0,a).$ As a consequence we have

$g(a-)<g(x)<g(0+).$

The limits $g(0+)=-\frac{1}{p}$ and $g(a-)=\frac{\ln(\cos_{p}a)}{a^{p}}$ give
the desired inequalities (2).

######  Theorem 3.5.

Let $p\geq 2$ and $x\in(0,b)$ where $b<\frac{\pi_{p}}{2}.$ Then the best
possible constants $\alpha_{2}$ and $\beta_{2}$ such that the inequalities

$\displaystyle e^{\alpha_{2}x^{p}}<\frac{x}{\tan_{p}x}<e^{\beta_{2}x^{p}}$ (3)

hold true are $\frac{\ln(b/\tan_{p}b)}{b^{p}}$ and $-\frac{1}{p+1}$
respectively.

######  Proof 3.6.

Let

$h(x)=\frac{\ln\left(\frac{x}{\tan_{p}x}\right)}{x^{p}}:=\frac{h_{1}(x)}{h_{2}(x)},\,\,x\in(0,b)$

where $h_{1}(x)=\ln\left(\frac{x}{\tan_{p}x}\right)$ and $h_{2}(x)=x^{p}$
satisfying $h_{1}(0+)=0=h_{2}(0).$ After differentiating we get

$\frac{h_{1}^{\prime}(x)}{h_{2}^{\prime}(x)}=\frac{1}{p}\cdot\frac{\tan_{p}x-x\sec_{p}^{p}x}{x^{p}\tan_{p}x}:=\frac{h_{3}(x)}{h_{4}(x)}.$

Here $h_{3}(x)=\tan_{p}x-x\sec_{p}^{p}x$ and $h_{4}(x)=x^{p}\tan_{p}x$ with
$h_{3}(0)=0=h_{4}(0).$ Differentiation and elementary computations yield

$\displaystyle\frac{h_{3}^{\prime}(x)}{h_{4}^{\prime}(x)}$
$\displaystyle=\frac{-x\sec_{p}^{p}x\tan_{p}^{p-1}x}{x^{p}\sec_{p}^{p}x+px^{p-1}\tan_{p}x}$
$\displaystyle=\frac{-1}{\left(\frac{x}{\tan_{p}x}\right)^{p-1}+p\left(\frac{x}{\tan_{p}x}\right)^{p-2}\cos_{p}^{p}x}.$

Now $\frac{x}{\tan_{p}x}$ is strictly decreasing in $(0,b)$ by Lemma 2.2 and
it is positive as well in $(0,b).$ Moreover, $\cos_{p}x$ is also positive
decreasing in $(0,b).$ So we conclude that $h(x)$ is strictly decreasing in
$(0,b)$ by Lemma 2.1. The inequalities (3) follow due to the limits
$h(0+)=\frac{-1}{p+1}$ and $h(b-)=\frac{\ln(b/\tan_{p}b)}{b^{p}}.$ This
finishes the proof of Theorem 3.5.

The particular cases of which the inequalities in the above theorems are
generalizations are given below.

$\displaystyle e^{\frac{4\ln(2/\pi)}{\pi^{2}}x^{2}}<\frac{\sin
x}{x}<e^{-x^{2}/6},\,\,x\in(0,\pi/2)$ (4) $\displaystyle e^{\frac{\ln(\cos
a)}{a^{2}}x^{2}}<\cos x<e^{-x^{2}/2},\,\,x\in(0,a),\,\,a<\frac{\pi}{2}$ (5)

and

$\displaystyle e^{\frac{\ln(b/\tan b)}{b^{2}}x^{2}}<\frac{x}{\tan
x}<e^{-x^{2}/3},\,\,x\in(0,b),\,\,b<\frac{\pi}{2}.$ (6)

The inequalities (4), (5), and (6) have been obtained by putting $p=2$ in
Theorems 3.1, 3.3, and 3.5. These inequalities have already appeared in [2, 3,
10].

Next, we establish classical exponential bounds for the generalized hyperbolic
functions.

######  Theorem 3.7.

For $x\in(0,c),$ where $c>0$ and $p\geq 2,$ we have

$\displaystyle e^{\lambda x^{p}}<\frac{x}{\sinh_{p}x}<e^{\eta x^{p}}$ (7)

with the best possible constants $\lambda=-\frac{1}{p(p+1)}$ and
$\eta=\frac{\ln(c/\sinh_{p}c)}{c^{p}}.$

######  Proof 3.8.

Let us set

$F(x)=\frac{\ln\left(\frac{x}{\sinh_{p}x}\right)}{x^{p}}:=\frac{F_{1}(x)}{F_{2}(x)},$

where $F_{1}(x)=\ln\left(\frac{x}{\sinh_{p}x}\right)$ and $F_{2}(x)=x^{p}$
with $F_{1}(0+)=0=F_{2}(0).$ Differentiation with respect to $x$ yields

$\frac{F_{1}^{\prime}(x)}{F_{2}^{\prime}(x)}=\frac{1}{p}\cdot\frac{\sinh_{p}x-x\cosh_{p}x}{x^{p}\sinh_{p}x}:=\frac{1}{p}\cdot\frac{F_{3}(x)}{F_{4}(x)}$

Here $F_{3}(0)=0=F_{4}(0)$ which forces us to differentiate again and get

$\displaystyle\frac{F_{3}^{\prime}(x)}{F_{4}^{\prime}(x)}$
$\displaystyle=\frac{-x\cosh_{p}^{2-p}x\sinh_{p}^{p-1}x}{x^{p}\cosh_{p}x+p\cdot
x^{p-1}\sinh_{p}x}$
$\displaystyle=\frac{-1}{\left(\frac{x}{\tanh_{p}x}\right)^{p-1}+p\left(\frac{x}{\tanh_{p}x}\right)^{p-2}}.$

Utilization of Lemma 2.4 implies that the function
$\frac{F_{3}^{\prime}(x)}{F_{4}^{\prime}(x)}$ is strictly increasing in
$(0,c).$ Then the repeated use of Lemma 2.1 shows that the function $F(x)$ is
also strictly increasing in $(0,c).$ Lastly, $F(0+)=-\frac{1}{p(p+1)}$ by
l’Hôpital’s rule and $F(c-)=\frac{\ln(c/\sinh_{p}c)}{c^{p}}.$

######  Theorem 3.9.

The best possible constants $\lambda_{1}$ and $\eta_{1}$ such that the double
inequality

$\displaystyle e^{\lambda_{1}x^{p}}<\cosh_{p}x<e^{\eta_{1}x^{p}}$ (8)

holds for $p\geq 1$ and $x\in(0,d),$ where $d>0,$ are
$\frac{\ln(\cosh_{p}d)}{d^{p}}$ and $\frac{1}{p}$ respectively.

######  Proof 3.10.

Consider the function

$G(x)=\frac{\ln(\cosh_{p}x)}{x^{p}}:=\frac{G_{1}(x)}{G_{2}(x)},$

where $G_{1}(x)=\ln(\cosh_{p}x)$ and $G_{2}(x)=x^{p}$ with
$G_{1}(0)=0=G_{2}(0).$ Then by elementary computations we get

$\displaystyle\frac{G_{1}^{\prime}(x)}{G_{2}^{\prime}(x)}=\frac{\cosh_{p}^{2-p}x\sinh_{p}^{p-1}x}{p\cdot
x^{p-1}\cosh_{p}x}=\frac{1}{p}\cdot\left(\frac{\tanh_{p}x}{x}\right)^{p-1}$

which is strictly decreasing in $(0,1)$ by Lemma 2.4 as $p\geq 1.$ Applying
Lemma 2.1, we conclude that $G(x)$ is strictly decreasing in $(0,d).$ Thus

$G(d-)<G(x)<G(0+).$

Theorem 3.9 follows from the limits $G(0+)=\frac{1}{p}$ and
$G(d-)=\frac{\ln(\cosh_{p}d)}{d^{p}}.$

######  Theorem 3.11.

If $p\geq 2$ and $x\in(0,m),$ where $m>0,$ then the best possible constants
$\lambda_{2}$ and $\eta_{2}$ such that the inequalities

$\displaystyle e^{\lambda_{2}x^{p}}<\frac{\tanh_{p}x}{x}<e^{\eta_{2}x^{p}}$
(9)

hold true are $-\frac{1}{p+1}$ and $\frac{\ln((\tanh_{p}m)/m)}{m^{p}}$
respectively.

######  Proof 3.12.

Let us introduce the function

$H(x)=\frac{\ln\left(\frac{\tanh_{p}x}{x}\right)}{x^{p}}:=\frac{H_{1}(x)}{H_{2}(x)},$

where $H_{1}(x)=\ln\left(\frac{\tanh_{p}x}{x}\right)$ and $H_{2}(x)=x^{p}$
with $H_{1}(0)=0=H_{2}(0).$ By differentiation we get

$\frac{H_{1}^{\prime}(x)}{H_{2}^{\prime}(x)}=\frac{1}{p}\cdot\frac{x\operatorname{sech}_{p}^{p}x-\tanh_{p}x}{x^{p}\tanh_{p}x}:=\frac{H_{3}(x)}{H_{4}(x)},$

where $H_{3}(x)=x\operatorname{sech}_{p}^{p}x-\tanh_{p}x$ and
$H_{4}(x)=x^{p}\tanh_{p}x$ satisfying $H_{3}(0)=0=H_{4}(0).$ Then

$\displaystyle\frac{H_{3}^{\prime}(x)}{H_{4}^{\prime}(x)}$
$\displaystyle=\frac{-p\cdot
x\operatorname{sech}_{p}^{p}x\tanh_{p}^{p-1}x}{p\cdot
x^{p-1}\tanh_{p}x+x^{p}\operatorname{sech}_{p}^{p}x}$
$\displaystyle=\frac{-p}{p\cdot\cosh_{p}^{p}x\left(\frac{x}{\tanh_{p}x}\right)^{p-2}+\left(\frac{x}{\tanh_{p}x}\right)^{p-1}}.$

Now $\frac{x}{\tanh_{p}x}$ is clearly positive in $(0,m)$ and it is also
increasing by Lemma 2.4. This implies that
$\frac{H_{3}^{\prime}(x)}{H_{4}^{\prime}(x)}$ is strictly increasing, fortiori
$H(x)$ is strictly increasing in $(0,m)$ by Lemma 2.1. Therefore

$H(0+)=-\frac{1}{p+1}<H(x)<\frac{\ln((\tanh_{p}m)/m)}{m^{p}}=H(m-).$

This completes the proof of Theorem 3.11.

An immediate consequence of Theorem 3.5 and Theorem 6 is the following
corollary.

######  Corollary 3.13.

Let $p\geq 2$ and $b<\frac{\pi_{p}}{2}.$ Then it is true that

$\displaystyle\frac{x}{\tan_{p}x}<\frac{\tanh_{p}x}{x},\,\,x\in(0,b).$ (10)

Indeed, the inequality (10) is valid in $(0,\pi_{p}/2).$

In the end, we remark that by putting $p=2$ in Theorems 3.7, 3.9, and 3.11, we
achieve the following existing inequalities.

$\displaystyle e^{-x^{2}/6}<\frac{x}{\sinh x}<e^{\frac{\ln(c/\sinh
c)}{c^{2}}x^{2}},\,\,x\in(0,c),\,\,c>0,$ (11) $\displaystyle
e^{\frac{\ln(\cosh d)}{d^{2}}x^{2}}<\cosh
x<e^{x^{2}/2},\,\,x\in(0,d),\,\,d>0,$ (12)

and

$\displaystyle e^{-x^{2}/3}<\frac{\tanh x}{x}<e^{\frac{\ln((\tanh
m)/m)}{m^{2}}x^{2}},\,\,x\in(0,m),\,\,m>0.$ (13)

The above inequalities (11), (12), and (13) were established and discussed in
[2, 4, 5, 10].

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††thanks: These authors contributed equally to this work††thanks: These
authors contributed equally to this work

# Surface acoustic wave stimulated Brillouin scattering in thin-film lithium
niobate waveguides

Kaixuan Ye  Nonlinear Nanophotonics Group, MESA+ Institute of Nanotechnology,
University of Twente, Enschede, Netherlands  Hanke Feng  Department of
Electrical Engineering $\&$ State Key Laboratory of Terahertz
and Millimeter Waves, City University of Hong Kong, Hong Kong, China  Yvan
Klaver  Nonlinear Nanophotonics Group, MESA+ Institute of Nanotechnology,
University of Twente, Enschede, Netherlands  Akshay Keloth  Nonlinear
Nanophotonics Group, MESA+ Institute of Nanotechnology,
University of Twente, Enschede, Netherlands  Akhileshwar Mishra  Nonlinear
Nanophotonics Group, MESA+ Institute of Nanotechnology,
University of Twente, Enschede, Netherlands  Cheng Wang<EMAIL_ADDRESS>Department of Electrical Engineering $\&$ State Key Laboratory of Terahertz
and Millimeter Waves, City University of Hong Kong, Hong Kong, China  David
Marpaung<EMAIL_ADDRESS>Nonlinear Nanophotonics Group, MESA+
Institute of Nanotechnology,
University of Twente, Enschede, Netherlands

###### Abstract

We report the first-ever experimental observation of backward stimulated
Brillouin scattering (SBS) in thin-film lithium niobate (TFLN) waveguides. The
peak Brillouin gain coefficient of the z-cut LN waveguide with a crystal
rotation angle of 20$\degree$ is as high as 84.9 m-1W-1, facilitated by
surface acoustic waves (SAW) at 8.06 GHz.

Stimulated Brillouin scattering (SBS), a nonlinear optomechanical interaction
involving light and GHz acoustic waves, has spurred groundbreaking
applications Eggleton2019 , from sub-Hertz level laser Gundavarapu2018 ,
ultra-high selectivity filter Marpaung2015Low-powerSelectivity , to optical
gyroscope Lai2020 . Exploiting the SBS effect in chip-scale photonic devices
for Brillouin-based applications requires a large Brillouin gain coefficient
in a scalable photonic integrated platform. Remarkably, strong SBS effect with
Brillouin gain coefficient over 100 m-1W-1 has been observed in chalcogenide
Pant2011  and suspended-silicon Kittlaus2016  waveguides. While promising,
these platforms are either volatile or mechanically unstable. On-chip SBS has
also been explored in low-loss and scalable silicon nitride platforms
Botter2022 ; Gundavarapu2018 . Nevertheless, Brillouin gain coefficients in
these platforms are limited to below 1 m-1W-1.

Very recently, theoretical works predict a strong SBS effect in thin-film
lithium niobate (TFLN) waveguides Rodrigues2023 . TFLN waveguides, recognized
for their low loss, scalability, and versatility Boes2023 , have enabled
unprecedented performances and functions in modulators Wang2018 , optical
frequency combs Zhang2019 , and quantum optics McKenna2020 . While the
optomechanical effect has been investigated in TFLN waveguides, previous
experiments rely on electric interdigital transducers to externally excite the
acoustic waves Sarabalis2021 ; Shao2019 . The SBS effect in TFLN waveguides
has never been observed.

Here, we report, to the best of our knowledge, the first-ever experimental
observation of backward SBS signals from TFLN waveguides. Our investigation
encompasses both z-cut and x-cut TFLN waveguides. Notably, we observed the SBS
effect in air-cladded z-cut TFLN waveguides with various crystal rotation
angles ($\theta$), confirming the crystal orientation dependence of the SBS
strength. Furthermore, we experimentally demonstrated the enhancement of the
SBS effect by surfaced acoustic waves by comparing x-cut TFLN waveguides with
both silica and air cladding. The massive Brillouin gain coefficient (84.9
m-1W-1 in z-cut TFLN waveguides with $\theta$=20$\degree$) in the mature TFLN
platform makes it an ideal candidate for integrated Brillouin-based
applications.

The air-cladded z-cut TFLN waveguides under test with $\theta$ = 0$\degree$,
20$\degree$, and 40$\degree$ are illustrated in Fig.1 (a). We fabricate the
waveguides using a commercial z-cut LN wafer (NANOLN) with a 400 nm-thick TFLN
layer. The etch depth of the z-cut waveguide is 200 nm. All z-cut TFLN
waveguides are 1-cm long. For waveguides with $\theta$ = 20$\degree$ and
40$\degree$, the angled section is 0.5 cm. All z-cut LN waveguides have a
coupling loss of 5 dB per facet and a propagation loss of 0.8 dB/cm.

Fig.1 (b) and (c) shows the simulated electric field and corresponding
acoustic response of the z-cut TFLN waveguides with $\theta$ = 20$\degree$ in
COMSOL. The waveguide supports surface acoustic waves at 7.95 GHz, resulting
in a large overlap between the optical and acoustic fields for the SBS
interaction.

We experimentally characterized the Brillouin gain profile of the z-cut TFLN
waveguides with a pump-probe modulated lock-in amplifier setup Gyger2020 ;
Botter2022 ; Botter2023 . Fig.1 (d) shows the measured SBS peaks from the TFLN
waveguides with different $\theta$. We calculated the Brillouin gain
coefficients by comparing the SBS peak from fibers in the setup (total length
5 m) and the waveguide peaks. Specifically, the z-cut TFLN waveguide at
$\theta$= 0$\degree$ exhibits a Brillouin gain coefficient of 9.45 m-1W-1 at
8.34 GHz, while at $\theta$= 20$\degree$, the Brillouin gain coefficient
reaches as high as 84.9 m-1W-1 at 8.06 GHz. For $\theta$= 40$\degree$, the
Brillouin gain coefficient is reduced to 25.6 m-1W-1 at 7.88 GHz. The crystal
orientation dependence of the SBS strength matches well with our simulation in
Fig.1 (e). Note that the simulated Brillouin gain coefficients are around 50%
lower than the measured values, which could be potentially due to the
underestimated acoustic quality factor in our simulation model (Qac=1000).
Additionally, the SBS signal is further validated through vector network
analyzer (VNA) measurement Botter2022 ; Botter2023 , as shown in Fig.1 (f).

Fig. 1: (a) Artistic representation of the air-cladded z-cut TFLN waveguides
with crystal orientation angle $\theta$ of 0$\degree$, 20$\degree$, and
40$\degree$. (b) Simulated electric field and (c) the corresponding acoustic
response of the air-cladded z-cut TFLN waveguides at the Brillouin frequency
shift ($\theta$ = 20$\degree$). (d) Lock-in amplifier measurement results of
the Brillouin gain profile of TFLN waveguides with different rotation angles
$\theta$. The Brillouin gain coefficients are calculated by comparing
waveguide peaks with fiber peaks. (e) Simulated Brillouin gain density plot as
a function of $\theta$. Stars indicate data points we measured. (f) SBS peak
from the z-cut TFLN waveguides ($\theta$= 20$\degree$) from the VNA
measurement. VNA: vector network analyzer. Fig. 2: (a)Lock-in amplifier
measurement results of the x-cut TFLN waveguides with and without silica
cladding, displayed together with SBS peaks from 5 m of fibers in the setup.
Measured (b) Stokes and (c) anti-Stokes signal from the x-cut TFLN waveguides
without silica cladding.

We also characterized Brillouin gain profiles of x-cut TFLN waveguides with an
LN layer thickness of 500 nm and an etching depth of 250 nm. As shown in Fig.2
(a), the measured Brillouin gain coefficient of the x-cut TFLN waveguide (2.8
cm long) with $\theta$ = 0$\degree$ is 26.1 m-1W-1 at 8.70 GHz. For further
verification, we have also characterized the sample with the VNA measurement,
as shown in Fig.2 (b) and (c). Notably, the measured Stokes signal revealed a
narrow linewidth of 11.9 MHz. To verify the role of surface acoustic waves in
facilitating the SBS process, we also characterized samples with silica
cladding on top of the x-cut TFLN waveguides (1 cm long). The Brillouin gain
coefficients of the silica-cladded sample were reduced by nearly fourfold,
primarily due to the diminished confinement of the acoustic wave compared to
the air-cladded counterparts.

In conclusion, we have observed the first-ever SBS signal from TFLN waveguides
in both z- and x-cut. The substantial Brillouin gain coefficients highlight
the immense potential for leveraging SBS in the mature TFLN platform. This
opens avenues for intersecting integrated Brillouin lasers, as well as
Brillouin-based microwave photonics signal processing with existing
functionalities such as modulators and optical frequency combs in the TFLN
platform with unparalleled performances.

## Author Contribution

D.M. and K.Y. developed the concept and proposed the physical system. K.Y. and
H.F. designed the photonic circuits. H.F. fabricated the photonic circuits.
K.Y. and Y.K. developed and performed numerical simulations. K.Y. and Y.K.
performed the experiments with input from A.K., and A.M.. K.Y. and D.M. wrote
the manuscript with input from everyone. D.M. and C.W. led and supervised the
entire project.

###### Acknowledgements.

The authors acknowledge funding from the European Research Council
Consolidator Grant (101043229 TRIFFIC), Nederlandse Organisatie voor
Wetenschappelijk Onderzoek (NWO) Start Up (740.018.021), the Research Grants
Council, University Grants Committee (N_CityU11320, CityU11204022), and
Croucher Foundation (9509005).

## Disclosures

The authors declare no conflicts of interest. During the preparation of this
manuscript, we notice a similar work by Caique C. Rodrigues et al.
rodrigues2023onchip  also reports the SBS process in the TFLN platform.

## Data Availability

The data of this study are available from the corresponding authors upon
reasonable request.

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