# A Two-Dimensional mKdV Linear Map and Its Application in Digital Image Cryptography
## 2 Formulation of the Problem
### Traveling Wave Solution for P\(\Delta\)E mKdV (O\(\Delta\)E mKdV)  
In [13; 25], a four-parameter family of mappings derived from the generalized P\(\Delta\)E mKdV in Equation (4) was studied. Note that a system of ordinary difference equations, O\(\Delta\)E mKdV, can be derived from Equation (4) via the restriction of traveling wave solution with  
\[V_{l,m}=V_{n},n=z_{1}l+z_{2}m, \tag{7}\]  
where \(z_{1}\) and \(z_{2}\) are relatively prime integers (see [26; 27] for other applications). Substituting Equation (7) into Equation (4), the following discrete mapping can be obtained:  
\[\theta_{1}V_{n}V_{n+z_{2}}-\theta_{2}V_{n+z_{1}}V_{n+z_{1}+z_{2}}-\theta_{3}V_ {n}V_{n+z_{1}}+\theta_{4}V_{n+z_{2}}V_{n+z_{1}+z_{2}}=0 \tag{8}\]The map in Equation (8) represents an infinite hierarchy of mappings labeled with \(z_{1}\) and \(z_{2}\). For fixed \(z_{1}\) and \(z_{2}\), Equation (8) is a mapping \(\mathbb{R}^{z_{1}+z_{2}}\ \rightarrow\ \mathbb{R}^{z_{1}+z_{2}}\). In [24], for \(z_{1}=1\) and \(z_{2}=2\), the map in Equation (8) has been found to possess several properties--for example, the (anti)measure-preserving property. Consider the discrete mapping in Equation (8). Let \(z_{1}=1\) and \(z_{2}=2\). This gives the following relation:  
\[V_{n+3}=\frac{V_{n}(\theta_{3}V_{n+2}-\theta_{1}V_{n+1})}{(\theta_{4}V_{n+1}- \theta_{2}V_{n+2})}. \tag{9}\]  
The discrete Equation (9) can be written as follows:  
\[ \begin{array}{l}V_{n+2}^{\prime}=\frac{V_{n}(\theta_{3}V_{n+2}-\theta_{1}V_ {n+1})}{V_{n}^{\prime}(\theta_{4}V_{n+1}-\theta_{2}V_{n+2})}\\ V_{n+1}^{\prime}=V_{n+2}\\ V_{n}^{\prime}=V_{n+1}\end{array}  \tag{10}\]  
where the prime denotes the upshift. If \(\zeta_{n}^{0}=\frac{V_{n+2}}{V_{n+1}}\) and \(\zeta_{n}^{1}=\frac{V_{n+1}}{V_{n}}\), then Equation (10) can be written as  
\[\zeta_{n+1}=\mathbf{h}_{\theta}(\zeta_{n}) \tag{11}\]  
where  
\[ \begin{array}{cccc}\mathbf{h}_{\theta}:&\mathbb{R}^{2}&\rightarrow&\mathbb{ R}^{2}\\ &(x,y)&\mapsto&\Big{(}\frac{-1}{xy}\frac{(\theta_{3}V_{-}\theta_{1})}{(\theta_{ 2}x-\theta_{4})},x\Big{)}.\end{array} \]  
Let us assume that \(\theta_{2}\) is not equal to zero. If \(\alpha=\frac{\theta_{1}}{\theta_{3}}\), \(\beta=\frac{\theta_{3}}{\theta_{2}}=1\), and \(\lambda=\frac{\theta_{4}}{\theta_{2}}=1\), then the following 2D map can be derived from Equation (11):  
\[ \begin{array}{cccc}\mathbf{h}_{\alpha,1,1}:&\mathbb{R}^{2}&\rightarrow& \mathbb{R}^{2}\\ &(x,y)&\mapsto&\Big{(}\frac{(\alpha-x)}{xy(x-1)},x\Big{)}.\end{array}  \tag{12}\]  
Furthermore, its integral normal form is (see [13] for other integral normal forms):  
\[H(x,y:\alpha)=\ \alpha\bigg{(}\frac{1}{x}+\frac{1}{y}\bigg{)}\ +\ (x+y)\ -\bigg{(}\frac{1}{xy}\bigg{)}\ -\ (xy) \tag{13}\]  
where \(\alpha\in\mathbb{R}\). Thus, for all \(n\in\mathbb{N}\), the solution for Equation (12) lies in a level set of \(H(x,y:\alpha)\). The mapping Equation (12) has several properties:  
* Integrable: The mapping \(\mathbf{h}_{\alpha,1,1}\) has an integral Equation (13), \(H(x,y:\alpha)=H(\frac{(\alpha-x)}{xy(x-1)},\)\(x:\alpha)\). In other words, \(H(x,y:\alpha)\) is evidently a constant of the mapping's motion (the orbits of all points in the plane lie on the level sets given by \(H(x,y:\alpha)=C\) where for any orbit is determined from the initial condition, i.e., \(C=H(x_{0},y_{0}:\alpha)\). * Measure-preserving: The mapping \(\mathbf{h}_{\alpha,1,1}\) is measure-preserving, i.e., \[|D\mathbf{h}_{\alpha,1,1}|=\frac{\rho(x,y)}{\rho\Big{(}\frac{(\alpha-x)}{xy(x -1)},x\Big{)}}\] where \[\rho(x,y)=\frac{(x+y)}{(xy)^{2}}[\partial_{\alpha}H(x,y:\alpha)]^{-1}=\frac{ 1}{x\,y}\]
* Reversible: There is exists a reversing symmetry \(G(x,y)=(y,x)\) such that \(G^{\circ}\mathbf{h}_{\alpha,1,1}{}^{\circ}G^{-1}=\mathbf{h}_{\alpha,1,1}^{-1}\). It's means that \(\mathbf{h}_{\alpha,1,1}\) is reversible (\(\mathbf{h}_{\alpha,1,1}{}^{\circ}G^{\circ}\mathbf{h}_{\alpha,1,1}=G\)). Notethat the mapping \(G\) is an involution, i.e., \(G\circ G=\mathbf{Id}\), and \(\mathbf{Id}\) is identity mapping.