# A Two-Dimensional mKdV Linear Map and Its Application in Digital Image Cryptography
## 2 Formulation of the Problem
### Two-Dimensional Linear Map Derived from a Nonlinear \(O\Delta E\) mKdV Map  
An essential element in the construction of the linear system in Equation (12) is a fixed point. A fixed point can be obtained by finding the critical point of the integral function. The critical points of the integral in Equation (12) are the solutions of  
\[\frac{1+x^{2}y-x^{2}y^{2}-y\alpha}{x^{2}y}=0 \tag{26}\]  
and  
\[\frac{1+xy^{2}-x^{2}y^{2}-x\alpha}{xy^{2}}=0 \tag{27}\]  
After taking the difference between the two equations, we have a line \(y=x\). By substituting these solutions into one of the equations, we have the following critical points that depend on \(\alpha\):  
\[(x^{*},y^{*})=\{(x_{i},y_{i}),\,i=1,2,3,4,5,6\} \tag{28}\]  
where \(x_{1}=y_{1}=\frac{1-a^{2}-\sqrt{1-2a^{2}+4a^{3}+a^{4}}}{2a}\), \(x_{2}=y_{2}=\frac{1-a^{2}+\sqrt{1-2a^{2}+4a^{3}+a^{4}}}{2a}\), \(x_{3}=y_{3}=\frac{1}{4}-\frac{1}{2}\sqrt{\frac{1}{4}-\frac{\sqrt{2}(-4+a)}{ \tau_{1}}-\frac{\tau_{1}}{3}}-\frac{1}{3}\sqrt{\frac{1}{2}}-\frac{1}{2}\sqrt{ \frac{1}{2}+\frac{\sqrt{2}(-4+a)}{\tau_{1}}+\frac{\tau_{1}}{3}\sqrt{2}}-\frac {1-8a}{4\sqrt{\frac{1}{4}-\frac{\sqrt{2}(-4+a)}{\tau_{1}}-\frac{\tau_{1}}{3} \sqrt{2}}}\), \(x_{4}=y_{4}=\frac{1}{4}-\frac{1}{2}\sqrt{\frac{1}{4}-\frac{\sqrt{2}(-4+a)}{ \tau_{1}}+\frac{\tau_{1}}{3}\sqrt{2}}\). \[ \begin{array}{l}\frac{1}{4}-\frac{1}{2}\sqrt{\frac{1}{4}-\frac{\sqrt[3]{2}(-4+ \alpha)}{\tau_{1}}-\frac{\tau_{1}}{3}\sqrt[3]{2}}+\frac{1}{2}\sqrt{\frac{1}{2}+ \frac{\sqrt[3]{2}(-4+\alpha)}{\tau_{1}}+\frac{\tau_{1}}{3}\sqrt[3]{2}}-\frac{1 -8\alpha}{4\sqrt[3]{\frac{1}{4}-\frac{\sqrt[3]{2}(-4+\alpha)}{\tau_{1}}-\frac {\tau_{1}}{3}\sqrt[3]{2}}},\,x_{5}=y_{5}=\\ \frac{1}{4}+\frac{1}{2}\sqrt{\frac{1}{4}-\frac{\sqrt[3]{2}(-4+\alpha)}{\tau_{1 }}-\frac{\tau_{1}}{3}\sqrt[3]{2}}-\frac{1}{2}\sqrt{\frac{1}{2}+\frac{\sqrt[3]{2 }(-4+\alpha)}{\tau_{1}}+\frac{\tau_{1}}{3}\sqrt[3]{2}}-\frac{1-8\alpha}{4 \sqrt[3]{\frac{1}{4}-\frac{\sqrt[3]{2}(-4+\alpha)}{\tau_{1}}-\frac{\tau_{1}}{3 }\sqrt[3]{2}}},\,x_{6}=y_{6}=\\ \frac{1}{4}+\frac{1}{2}\sqrt{\frac{1}{4}-\frac{\sqrt[3]{2}(-4+\alpha)}{\tau_{1 }}-\frac{\tau_{1}}{3}\sqrt[3]{2}}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{\sqrt[3]{2 }(-4+\alpha)}{\tau_{1}}+\frac{\tau_{1}}{3}\sqrt[3]{2}}-\frac{1-8\alpha}{4 \sqrt[3]{\frac{1}{4}-\frac{\sqrt[3]{2}(-4+\alpha)}{\tau_{1}}-\frac{\tau_{1}}{3 }\sqrt[3]{2}}},\,\text{where }\tau_{1}=\\ \left(-9+9\alpha^{2}+\sqrt{3}\sqrt{283-192\alpha-6\alpha^{2}-4\alpha^{3}+27 \alpha^{4}}\right)^{1/3}.\end{array} \]  
Let \(I\) be an interval in \(\mathbb{R}\) (finite). Assume that \(\alpha\in I\). For \(\alpha=\{9.79,-4.79,0.1249\}\), we then have some fixed point values, as seen in Table 2.  
To construct a linear map derived from the map in Equation (12) around a fixed point \((x^{*},y^{*})\), we can use the values of \(\alpha\) in Table 2. Thus, a linear map can be constructed based on the Jacobian matrix \(A_{mKdV}\) as follows:  
\[\vartheta=-\frac{(-9.79-x)\left(-\frac{-9.79-x}{-(1+x)x^{2}y}-\frac{-9.79-x}{ (-1+x)^{2}xy}-\frac{1}{(-1+x)xy}\right)}{(-1+x)xy^{2}} \tag{29}\]  
From Equation (12) and Equation (29), for \((x^{*})y^{*}(x^{*}-1)\
eq 0\), we have the following linear map:  
\[\left( \begin{array}{c}x^{\prime}\\ y^{\prime}\end{array} \right)=\left( \begin{array}{cc}\frac{1+x*(ax*-2)}{(x*)^ {2}y*(x*-1)^{2}}&\frac{(ax*-1)}{(y*)^{2}x*(x*-1)}\\ 1&0\end{array} \right)\left( \begin{array}{c}x\\ y\end{array} \right) \tag{30}\]  
For example, at \(\alpha=-9.79\) and the point \((x_{6},x_{6})=(2.5571,2.5571)\), we have a linear map--namely,  
\[\left( \begin{array}{c}x^{\prime}\\ y^{\prime}\end{array} \right)=\left( \begin{array}{cc}-1.68064&-1\\ 1&0\end{array} \right)\left( \begin{array}{c}x\\ y\end{array} \right) \tag{31}\]  
Once more, based on the encryption algorithm in Figure 2(a), we used 2D linear mKdV map in Equation (31) to hide the plain image in Figure 2. Applying the linear map in Equation (31), we obtained the cipher image shown in Figure 9.  
\begin{table}
\begin{tabular}{c c c} \hline
**No. ** & \(\alpha\) & **Fix Point** & **Fixed Point Positions** \\ \hline
1.