An algebraic approach for solving fourth-order partial differential equations

TL;DR

This paper extends the classical link between harmonic functions and holomorphic functions to fourth-order hyperbolic and elliptic PDEs using hypercomplex algebra, providing a new algebraic solution framework.

Contribution

It introduces an algebraic method for solving fourth-order PDEs by leveraging hypercomplex algebras, extending known complex analysis techniques.

Findings

Solutions derived from hypercomplex algebra components satisfy the PDEs.

The approach generalizes classical harmonic function methods to higher-order PDEs.

Provides a new algebraic perspective for solving complex PDEs.

Abstract

It is well-known that any solution of the Laplace equation is a real or imaginary part of a complex holomorphic function. In this paper, in some sense, we extend this property into four order hyperbolic and elliptic type PDEs. To be more specific, the extension is for a $c$-biwave PDE with constant coefficients, and we show that the components of a differentiable function on the associated hypercomplex algebras provide solutions for the equation.

Equations

$$\left(\frac{{\partial }^4}{\partial x^4}-2c\frac{{\partial }^4}{\partial x^2\partial y^2}+\frac{{\partial }^4}{\partial y^4}\right)u\left(x,y\right)=0,c>0.$$

$$\left(A\frac{{\partial }^4}{\partial x^4}+2B\frac{{\partial }^4}{\partial x^2\partial y^2}+C\frac{{\partial }^4}{\partial y^4}\right)u\left(x,y\right)=0,$$

$$e_1^4-2c{e}_1^2{e}_2^2+e_2^4=0.$$

$$A_c=\left\{x\mathbf{u}+y\mathbf{f}+z\mathbf{e}+v\mathbf{fe}:x,y,z,v\in \mathbb{R}\right\}$$

$$i_1=\frac{k_1}{k_1+k_2}\mathbf{u}-\frac{\mathbf{f}\sqrt{2}}{k_1+k_2}\mathbf{e},$$

$$i_2=\frac{k_2}{k_1+k_2}\mathbf{u}+\frac{\mathbf{f}\sqrt{2}}{k_1+k_2}\mathbf{e},$$

$$i_1+i_2=\mathbf{u}$$

$$i_1{i}_2=\frac{k_1{k}_2}{{\left(k_1+k_2\right)}^2}\mathbf{u}-\frac{\sqrt{2}{k}_2}{{\left(k_1+k_2\right)}^2}\mathbf{f}\mathbf{e}+\frac{\sqrt{2}{k}_1}{{\left(k_1+k_2\right)}^2}\mathbf{f}\mathbf{e}$$

$$-\frac{2}{{\left(k_1+k_2\right)}^2}\mathbf{u}+\frac{2m}{{\left(k_1+k_2\right)}^2}\mathbf{f}\mathbf{e}=0.$$

$$\mathbf{e}=\mathbf{f}\frac{k_1}{\sqrt{2}}{i}_2-\mathbf{f}\frac{k_2}{\sqrt{2}}{i}_1.$$

$$B_c=\left\{x\mathbf{u}+y\mathbf{e}|x,y\in \mathbb{R}\right\}.$$

$$\underset{\mathbb{R}\ni \epsilon \to 0}{\lim }\frac{g\left(w+\epsilon h\right)-g\left(w\right)}{\epsilon }=h{g}^{\prime }\left(w\right),$$

$$\mathbf{e}\frac{\partial }{\partial x}g\left(w\right)=\mathbf{u}\frac{\partial }{\partial y}g\left(w\right),\forall w\in B_c,$$

$$\frac{\partial u_1\left(x,y\right)}{\partial y}=\frac{\partial u_3\left(x,y\right)}{\partial x},$$

$$\frac{\partial u_2\left(x,y\right)}{\partial y}=\frac{\partial u_4\left(x,y\right)}{\partial x},$$

$$\frac{\partial u_3\left(x,y\right)}{\partial y}=\frac{\partial u_1\left(x,y\right)}{\partial x}-m\frac{\partial u_4\left(x,y\right)}{\partial x},$$

$$\frac{\partial u_4\left(x,y\right)}{\partial y}=\frac{\partial u_2\left(x,y\right)}{\partial x}-m\frac{\partial u_3\left(x,y\right)}{\partial x}.$$

$$w=x\mathbf{u}+y\mathbf{e}=\left(x-\mathbf{f}\frac{k_2}{\sqrt{2}}y\right)i_1+\left(x+\mathbf{f}\frac{k_1}{\sqrt{2}}y\right)i_2.$$

$$g\left(w\right)=\alpha \left(w_1\right)i_1+\beta \left(w_2\right)i_2,$$

$$\frac{\partial }{\partial y}\alpha \left(w_1\right)=-\mathbf{f}\frac{k_2}{\sqrt{2}}\frac{\partial }{\partial x}\alpha \left(w_1\right),$$

$$\frac{\partial }{\partial y}\beta \left(w_2\right)=\mathbf{f}\frac{k_1}{\sqrt{2}}\frac{\partial }{\partial x}\beta \left(w_2\right).$$

$$\frac{\partial }{\partial y}g\left(w\right)=\frac{\partial }{\partial y}\alpha \left(w_1\right)i_1+\frac{\partial }{\partial y}\beta \left(w_2\right)i_2$$

$$=-\mathbf{f}\frac{k_2}{\sqrt{2}}\frac{\partial }{\partial x}\alpha \left(w_1\right)i_1+\mathbf{f}\frac{k_1}{\sqrt{2}}\frac{\partial }{\partial x}\beta \left(w_2\right)i_2$$

$$\mathbf{e}\frac{\partial }{\partial x}g\left(w\right)=\left(\mathbf{f}\frac{k_1}{\sqrt{2}}{i}_2-\mathbf{f}\frac{k_2}{\sqrt{2}}{i}_1\right)\left(\frac{\partial }{\partial x}\alpha \left(w_1\right)i_1+\frac{\partial }{\partial x}\beta \left(w_2\right)i_2\right)$$

$$=-\mathbf{f}\frac{k_2}{\sqrt{2}}\frac{\partial }{\partial x}\alpha \left(w_1\right)i_1+\mathbf{f}\frac{k_1}{\sqrt{2}}\frac{\partial }{\partial x}\beta \left(w_2\right)i_2.$$

$$\mathbf{e}\frac{\partial }{\partial x}g\left(w\right)=\mathbf{u}\frac{\partial }{\partial y}g\left(w\right).$$

$$g\left(w\right)=\mathbf{u}{u}_1\left(x,y\right)+\mathbf{f}{u}_2\left(x,y\right)+\mathbf{e}{u}_3\left(x,y\right)+\mathbf{f}\mathbf{e}{u}_4\left(x,y\right)$$

$$\alpha \left(w_1\right)=\mathbf{u}\left(u_1\left(x,y\right)-\frac{k_2}{\sqrt{2}}{u}_4\left(x,y\right)\right)+\mathbf{f}\left(u_2\left(x,y\right)-\frac{k_2}{\sqrt{2}}{u}_3\left(x,y\right)\right),$$

$$\beta \left(w_2\right)=\mathbf{u}\left(u_1\left(x,y\right)+\frac{k_1}{\sqrt{2}}{u}_4\left(x,y\right)\right)+\mathbf{f}\left(u_2\left(x,y\right)+\frac{k_1}{\sqrt{2}}{u}_3\left(x,y\right)\right).$$

$$\frac{\partial }{\partial y}\alpha \left(w_1\right)=\mathbf{u}\left(\frac{\partial u_3\left(x,y\right)}{\partial x}-\frac{k_2}{\sqrt{2}}\left(\frac{\partial u_2\left(x,y\right)}{\partial x}-m\frac{\partial u_3\left(x,y\right)}{\partial x}\right)\right)$$

$$+\mathbf{f}\left(\frac{\partial u_4\left(x,y\right)}{\partial x}-\frac{k_2}{\sqrt{2}}\left(\frac{\partial u_1\left(x,y\right)}{\partial x}-m\frac{\partial u_4\left(x,y\right)}{\partial x}\right)\right)$$

Full text

1. Introduction

In this paper we are interested in finding the solution of the following equation

[TABLE]

Depending on the value of $c$ we may consider three cases. Namely, the case where $0<c<1$ and we call it as the $c$-biwave equation of the elliptic type, the case where $c>1$ and we call it as the $c$-biwave equation of hyperbolic type, and in the case where $c=1$ Eq.(1.1) is the well-known biwave equation. The biwave equation has been used in modeling of $d$-wave superconductors (see for instance [1], and references therein) or in probability theory [2, 3]. In [4] the author studied Eq.(1.1) in the case where $c<-1$ and considered its application to theory of plain orthotropy.

It is easily verified that any equation of the form

[TABLE]

where $AC>0$ and $AB<0$ can be reduced to Eq.(1.1) by changing variables. To obtain all solutions of Eq. (1.1) for $1\neq c>0$ we will use the method developed in [7]. According to such approach we need a commutative algebra with basis containing $e_{1}$, $e_{2}$ such that

[TABLE]

Then, we study monogenic functions on the subspace of this algebra containing $e_{1}$, $e_{2}$ and show that any solution of Eq. (1.1) can be obtained as a component of such monogenic functions.

2. Hyperbolic case

Firstly we study Eq. (1.1) in the case where $c>1$, which is said to be hyperbolic. Let us consider an associative commutative algebra over the real field ${\mathbb{R}}$

[TABLE]

with a basis ${\mathbf{u}}$, ${\mathbf{f}}$, ${\mathbf{e}}$, $\mathbf{fe}$, where ${\mathbf{u}}$ is the identity element of $A_{c}$ and the following Cayley table holds $\mathbf{fe}=\mathbf{ef}$, ${\mathbf{f}}^{2}={\mathbf{u}}$, ${{\mathbf{e}}}^{2}={\mathbf{u}}-m\mathbf{fe}$, where $m=\sqrt{2(c-1)}$.

The basis elements ${\mathbf{u}}$, ${\mathbf{e}}$ satisfy Eq. (1.2).

It is easily verified that for $c>1$ algebra $A_{c}$ has the following idempotents

[TABLE]

[TABLE]

where $k_{1}=\sqrt{c+1}-\sqrt{c-1}$, $k_{2}=\sqrt{c+1}+\sqrt{c-1}$.

Therefore, we have

[TABLE]

and

[TABLE]

It is easily seen that

[TABLE]

Consider a subspace $B_{c}$ of algebra $A_{c}$ of the following form

[TABLE]

defin0 2.1**.**

*A function * $g:\,B_{c}\to A_{c}$ * is called differentiable (or monogenic) on * $B_{c}$ * if for any * $B_{c}\ni w=x{\mathbf{u}}+y{\mathbf{e}}$ * there exists a unique element * $g^{\prime}\left(w\right)$ * such that for any * $h\in B_{c}$

[TABLE]

where $hg^{\prime}\left(w\right)$ is the product of $h$ and $g^{\prime}\left(w\right)$ as elements of $A_{c}$.

It follows from [7] that a function $g\left(w\right)={\mathbf{u}}\,u_{1}\left(x,y\right)+{\mathbf{f}}\,u_{2}\left(x,y\right)+{\mathbf{e}}\,u_{3}\left(x,y\right)+{\mathbf{f\,e}}\,u_{4}\left(x,y\right)$ is monogenic if and only if there exist continuous partial derivatives $\frac{\partial u_{i}\left(x,y\right)}{\partial x}$, $\frac{\partial u_{i}\left(x,y\right)}{\partial y},$ $i=1,2,3,4$ and it satisfies the following Cauchy-Riemann type conditions

[TABLE]

or

[TABLE]

[TABLE]

[TABLE]

[TABLE]

It is also proved in [7] that if $g$ is monogenic then its components $u_{i}\left(x,y\right)$ satisfies Eq. (1.1).

By passing in $B_{c}$ from the basis ${\mathbf{u}}$, ${\mathbf{e}}$ to the basis $i_{1}$, $i_{2}$, we have

[TABLE]

Lemma 2.2**.**

A function $g:\,B_{c}\to A_{c}$, where $c>1$, is differentiable if and only if it can be represented as follows

[TABLE]

where $w_{1}=x-{\mathbf{f}}\frac{k_{2}}{\sqrt{2}}y$, $w_{2}=x+{\mathbf{f}}\frac{k_{1}}{\sqrt{2}}y$ and $\alpha\left(w_{1}\right)$, $\beta\left(w_{2}\right)$ have continuous partial derivatives $\frac{\partial}{\partial x}\alpha\left(w_{1}\right),\frac{\partial}{\partial y}\alpha\left(w_{1}\right),\frac{\partial}{\partial x}\beta\left(w_{2}\right),\frac{\partial}{\partial y}\beta\left(w_{2}\right)$ satisfying

[TABLE]

[TABLE]

Proof.

Sufficiency can be verified directly. Indeed,

[TABLE]

[TABLE]

On the other hand, taking into account Eqs. (2.1), (2.2), we have

[TABLE]

[TABLE]

Hence,

[TABLE]

Now let us prove necessity. Suppose that a function

[TABLE]

is monogenic on $B_{c}$. Let us define

[TABLE]

[TABLE]

Thus, we have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Taking into account that

[TABLE]

we have $\frac{\partial}{\partial y}\alpha\left(w_{1}\right)=-{\mathbf{f}}\frac{k_{2}}{\sqrt{2}}\frac{\partial}{\partial x}\alpha\left(w_{1}\right)$.

Much in the same manner, it can be shown that $\frac{\partial}{\partial y}\beta\left(w_{2}\right)={\mathbf{f}}\frac{k_{1}}{\sqrt{2}}\frac{\partial}{\partial x}\beta\left(w_{2}\right)$. ∎

Remark 2.3**.**

*Considering variables $x$, $y_{1}=-\frac{k_{2}}{\sqrt{2}}y$ and $x$, $y_{2}=\frac{k_{1}}{\sqrt{2}}y,$ we have *

[TABLE]

[TABLE]

Hence, it is easily verified that if the components $\alpha_{1}$, $\alpha_{2}$ of $\alpha\left(w_{1}\right)=\alpha_{1}\left(w_{1}\right)+{\mathbf{f}}\alpha_{2}\left(w_{1}\right)$ have continuous partial derivatives $\frac{{\partial}^{2}}{\partial x^{2}}\alpha_{k}\left(w_{1}\right)$ and $\frac{{\partial}^{2}}{\partial y^{2}_{1}}\alpha_{k}\left(w_{1}\right)$, $k=1,2$ then they satisfy the wave equation

[TABLE]

Similarly, the components $\beta_{1}$, $\beta_{2}$ of $\beta\left(w_{2}\right)=\beta_{1}\left(w_{2}\right)+{\mathbf{f}}\beta_{2}\left(w_{2}\right)$ satisfy the wave equation

[TABLE]

Theorem 1**.**

$u\left(x,y\right)$* is a solution of Eq. (1.1) for $c>1$ if and only if for some $i,j\in\left\{1,2\right\}$ it can be represented as follows*

[TABLE]

where ${\alpha}_{i}\left({\omega}_{1}\right),{\beta}_{j}\left({\omega}_{2}\right)$ are four times continuous differentiable components of $\ \alpha\left({\omega}_{1}\right)$ and $\beta\left({\omega}_{2}\right)$ of monogenic function $g\left(\omega\right)$ in the decomposition (2.3) i.e.,

[TABLE]

where $\alpha\left({\omega}_{1}\right)={\alpha}_{1}\left({\omega}_{1}\right)+{\mathbf{f}}{\alpha}_{2}\left({\omega}_{1}\right)$, $\beta\left({\omega}_{2}\right)={\beta}_{1}\left({\omega}_{2}\right)+{\mathbf{f}}{\beta}_{2}\left({\omega}_{2}\right)$ satisfy Eq. (2.4).

Proof.

As it was mentioned above $u\left(x,y\right)={\alpha}_{i}\left({\omega}_{1}\right)+{\beta}_{j}\left({\omega}_{2}\right)$ is a solution for $c>1$ of Eq. (1.1).

Now suppose that $u\left(x,y\right)$ is a solution of Eq. (1.1). It is easily verified that

[TABLE]

It is easily seen that Eq. (2.5) is equivalent to the set of the following systems

[TABLE]

or

[TABLE]

Let us consider the first system. Since any solution of $\left(\frac{{\partial}^{2}}{\partial x^{2}}-\frac{{\partial}^{2}}{\partial y^{2}_{k}}\right)v_{k}\left(x,y\right)=0$ is of the form $v_{k}\left(x,y\right)=f_{1}\left(x+y_{k}\right)+f_{2}\left(x-y_{k}\right)$, where $f_{i}$, $i=1,2$ are arbitrary twice differentiable functions it follows from the second equation of the system that

[TABLE]

Thus, the first equation of the system is

[TABLE]

It is easily seen that a partial solution of Eq. (2.6) is

[TABLE]

where $F^{{}^{\prime\prime}}_{k}=f_{k}$, $k=1,2$.

Thus, the general solution of the system is as follows

[TABLE]

Let us put ${\alpha}_{1}\left({\omega}_{1}\right)=\frac{k^{2}_{1}}{k^{2}_{1}-k^{2}_{2}}\left(F_{1}\left(x+y_{1}\right)+F_{2}\left(x-y_{1}\right)\right)$ and ${\beta}_{2}\left({\omega}_{2}\right)=g_{1}\left(x+y_{2}\right)+g_{2}\left(x-y_{2}\right)$.

Taking into account that $\left(\frac{{\partial}^{2}}{\partial x^{2}}-\frac{{\partial}^{2}}{\partial y^{2}_{1}}\right)\left[\frac{k^{2}_{1}}{k^{2}_{1}-k^{2}_{2}}\left(F_{1}\left(x+y_{1}\right)+F_{2}\left(x-y_{1}\right)\right)\right]=0$ and

$\left(\frac{{\partial}^{2}}{\partial x^{2}}-\frac{{\partial}^{2}}{\partial y^{2}_{2}}\right)\left[g_{1}\left(x+y_{2}\right)+g_{2}\left(x-y_{2}\right)\right]=0$ we conclude the proof for the first system.

The case of the second system can be proved similarly. ∎

3. Elliptic case

Now we consider an associative commutative algebra $A_{c}$, where $0<c<1$, over the complex field ${\mathbb{C}}$ with a basis ${\mathbf{u}}$, ${\mathbf{e}}$ and the following Cayley table ${\mathbf{u\,e}}={\mathbf{e\,u}}={\mathbf{e}}$, ${{\mathbf{e}}}^{2}={\mathbf{u}}+{\mathbf{i}}\mu{\mathbf{e}}$, where $\mu=\sqrt{2(1-c)}$. The matrix representations of ${\mathbf{u}}$ and ${\mathbf{e}}$ are

[TABLE]

Hence, we have the following traces of these representations

[TABLE]

Since

[TABLE]

then, $A_{c}$ is a semi-simple algebra [8].

By following similar steps as in Eq. (2.1) we can show that for $0<c<1$ algebra $A_{c}$ has the following idempotents

[TABLE]

where $k_{1}=\sqrt{c+1}-{\mathbf{i}}\sqrt{1-c}$, $k_{2}=\sqrt{c+1}+{\mathbf{i}}\sqrt{1-c}$.

It is also easily verified that these idempotents also satisfy

[TABLE]

and

[TABLE]

It is straightforward to see that

[TABLE]

Lemma 3.1**.**

All non-zero elements of subspace $B_{c}=\left\{x\,{\mathbf{u}}+y\,{\mathbf{e}}\mathrel{\left|\vphantom{x\,{\mathbf{u}}+y\,{\mathbf{e}}\,\,x,y\in{\mathbb{R}}}\right.\kern-1.2pt}x,y\in{\mathbb{R}}\right\}$ of algebra $A_{c}$ are invertible, that is, if $0\neq w\in B_{c}$ then there exists $w^{-1}\in B_{c}$.

Proof.

Suppose $w=s\,{\mathbf{u}}+t\,{\mathbf{e}}\in B_{c}$. Let us show that there exists $w^{-1}=x\,{\mathbf{u}}+y\,{\mathbf{e}}$, $x,y\in{\mathbb{R}}$ such that $w\,w^{-1}=1$. Indeed, the equation

[TABLE]

has a unique solution since the determinant of the system

[TABLE]

where $x,y$ are unknown, is $\Delta=s^{2}-t^{2}+{\mathbf{i}}\mu ts$ and $\Delta=0$ if and only if $s=t=0$. A function $f\left(w\right)$, $w\in B_{c}$ is said to be differentiable if it is differentiable in the common sense, i.e., for all $w\in B_{c}$ there exists the following limit

[TABLE]

It is easily seen that if $f$ is differentiable then it is monogenic and hence, it satisfies the following Cauchy-Riemann type of conditions [7]

[TABLE]

or in this case we have

[TABLE]

[TABLE]

[TABLE]

In [7] it is also proved that if a function $f\left(x,y\right)={\mathbf{u}}u_{1}\left(x,y\right)+{\mathbf{i}}\,u_{2}\left(x,y\right)+{\mathbf{e}}\,u_{3}\left(x,y\right)+{\mathbf{i}}\,{\mathbf{e}}\,u_{4}\left(x,y\right)$ is monogenic then $u_{i}\left(x,y\right)$ satisfies Eq. (1.1). We should mention that a constructive description of monogenic functions in a three-dimensional harmonic algebra was studied in [5, 6].

By passing from the basis ${\mathbf{u}}$, ${\mathbf{e}}$ to the basis $I_{-}$, $I_{+}$ we have

[TABLE]

∎

Lemma 3.2**.**

A function $f:\,B_{c}\to A_{c}$, $0<c<1$, is differentiable if and only if it can be represented as follows

[TABLE]

where $w_{1}=x_{1}+{\mathbf{i}}\,y_{1}$, $x_{1}=x$, $y_{1}=-{\mathbf{i}}\,\frac{k_{2}}{\sqrt{2}}y$, $w_{2}=x_{2}+{\mathbf{i}}\,y_{2}$, $x_{2}=x$, $y_{2}={\mathbf{i}}\,\frac{k_{1}}{\sqrt{2}}y$ and $\alpha\left(w_{1}\right)$, $\beta\left(w_{2}\right)$ are analytical functions of variables $w_{1}$, $w_{2},$ respectively, as follows

[TABLE]

Proof.

The sufficiency can be verified directly. Indeed,

[TABLE]

[TABLE]

[TABLE]

On the other hand, taking into account Eqs. (2.4), (3.1), we have

[TABLE]

[TABLE]

Hence,

[TABLE]

Now let us prove necessity. Suppose that a function

[TABLE]

is monogenic on $B_{c}$, i.e., ${\mathbf{e}}\,\frac{\partial}{\partial x}f\left(w\right)=\frac{\partial}{\partial y}f\left(w\right)$.

By using Eq. (3.1) we can represent $f\left(w\right)$ in the following manner

[TABLE]

where

[TABLE]

[TABLE]

Consider

[TABLE]

[TABLE]

Then, taking into account Eq. (3.1), we have

[TABLE]

[TABLE]

Therefore,

[TABLE]

[TABLE]

Suppose ${\alpha}\left({\omega}_{1}\right)={\alpha}_{1}\left({\omega}_{1}\right)+{\mathbf{i}}\,{\alpha}_{2}\left({\omega}_{1}\right)$ and ${\beta}\left({\omega}_{2}\right)={\beta}_{1}\left({\omega}_{2}\right)+{\mathbf{i}}\,{\beta}_{2}\left({\omega}_{2}\right)$. It follows from the proof of Lemma 3.2 that ${\alpha}_{i}\left({\omega}_{1}\right)+{\beta}_{j}\left({\omega}_{2}\right)$, $i,j\in\left\{1,2\right\}$ are solutions of Eq. (1.1) for $0<c<1$. ∎

Theorem 2**.**

$u\left(x,y\right)$* is a solution of Eq. (1.1) for $0<c<1$ if and only if for some $i,j\in\left\{1,2\right\}$ it can be represented as follows*

[TABLE]

where ${\alpha}_{i}\left({\omega}_{1}\right),{\beta}_{j}\left({\omega}_{2}\right)$ are components of $\ \alpha\left({\omega}_{1}\right)$ and $\beta\left({\omega}_{2}\right)$ of monogenic function $g\left(\omega\right)$ in the decomposition (2.3) i.e.,

[TABLE]

where $\alpha\left({\omega}_{1}\right)$, $\beta\left({\omega}_{2}\right)$ are complex analytical functions of respective variables.

Proof.

As mentioned above ${\alpha}_{i}\left({\omega}_{1}\right)+{\beta}_{j}\left({\omega}_{2}\right)$, $i,j\in\left\{1,2\right\}$ are solutions of Eq. (1.1) for $0<c<1$.

If $u\left(x,y\right)$ is a solution of Eq. (1.1) for $0<c<1$ much in the same way as in proving Theorem 1 we can show that $u\left(x,y\right)={\alpha}_{i}\left({\omega}_{1}\right)+{\beta}_{j}\left({\omega}_{2}\right)$. ∎