A Fast Single-Pass Method for Solving the Generalized Eikonal Equation in a Moving Medium

Abstract

We develop a fast method for approximating the solution to the generalized eikonal equation in a moving medium. Our approach consists of the following two steps. First, we convert the generalized eikonal equation in a moving medium into a Hamilton–Jacobi–Bellman equation of anisotropic eikonal type for an anisotropic minimum-time control problem. Second, we modify the Neighbor–Gradient Single-pass method (NGSPM developed by Ho et al.), so that it not only suits the converted Hamilton–Jacobi–Bellman equation but also can be faster than original NGSPM. In the case of that Mach number is not comparable than 1, we compare our method and Characteristic Fast Marching Method (CFMM developed by Dahiya) via several numerical examples to show that our method is faster and more accurate than CFMM. We also compare the numerical solutions obtained from our method with the solutions obtained using the ray theory to show that our method captures the viscosity solution accurately even when the Mach number is comparable to 1. We also apply our method to 3D example to show that our method captures the viscosity solution accurately in 3D cases.

APPENDIX

Lemma 1. Consider the function $g:\Omega \times {{S}_{1}} \to {{R}^{2}},$ $c:\Omega \times {{S}_{1}} \to {{S}_{1}}$ defined in 2.2.

Then $\left\| {g{\text{(}}{\mathbf{x}}{\mathbf{,a}}{\text{)}}} \right\| = F({\mathbf{x}})$ and for every $x \in \Omega ,$ ${{c}_{x}}: = c{\text{(}}{\mathbf{x}}{\mathbf{,a}}{\text{)}}$, is one to one mapping from ${{S}_{1}}$ to ${{S}_{1}}$ and it’s inverse mapping is

$$c_{x}^{{ - 1}}:{{S}_{1}} \to {{S}_{1}},$$

$${\mathbf{a}} \mapsto \frac{{F({\mathbf{x}}){\mathbf{a}} - {\mathbf{v}}({\mathbf{x}})}}{{\left\| {F({\mathbf{x}}){\mathbf{a}} - {\mathbf{v}}({\mathbf{x}})} \right\|}}.$$

Proof. Let us prove that $\left\| {g{\text{(}}{\mathbf{x}}{\mathbf{,a}}{\text{)}}} \right\| = F({\mathbf{x}})$.

We denote

$$l: = \left( {{\mathbf{v}}(x) \cdot {\mathbf{a}} + \sqrt {{{F}^{2}}({\mathbf{x}}) - ({{{\mathbf{v}}}^{2}}({\mathbf{x}}) - {{{({\mathbf{v}}({\mathbf{x}}) \cdot {\mathbf{a}})}}^{2}})} } \right) = \left( {\left| {{\mathbf{v}}({\mathbf{x}})} \right|\cos \alpha + \sqrt {{{F}^{2}}({\mathbf{x}}) - {{{({\kern 1pt} \left| {{\mathbf{v}}({\mathbf{x}})} \right|\sin \alpha )}}^{2}}} } \right),$$

where $\alpha$ is the angle between ${\mathbf{v}}$ and ${\mathbf{a}}$.

By the definition of $g$, we obtain that $g({\mathbf{x}}{\mathbf{,\hat {a}}}) = l{\mathbf{\hat {a}}} + {\mathbf{v}}({\mathbf{x}})$ and thus it is clear that $g({\mathbf{x}}{\mathbf{,\hat {a}}}) - {\mathbf{v}}({\mathbf{x}}) = {\mathbf{a}}{\kern 1pt} '$.

Using the second cosine formula, we can get

$$\left\| {g({\mathbf{x}},{\mathbf{\hat {a}}})} \right\| = \sqrt {{{{\left| {{\mathbf{v}}({\mathbf{x}})} \right|}}^{2}} + {{l}^{2}} - 2l\left| {{\mathbf{v}}({\mathbf{x}})} \right|\cos \alpha } = :\sqrt D .$$

Then the right-hand side of the above equality is expanded as follows;

$$\begin{gathered} D = {{\left| {{\mathbf{v}}({\mathbf{x}})} \right|}^{2}} + {{l}^{2}} - 2l\left| {{\mathbf{v}}({\mathbf{x}})} \right|\cos \alpha = {{\left| {{\mathbf{v}}({\mathbf{x}})} \right|}^{2}} + {{({\kern 1pt} \left| {{\mathbf{v}}({\mathbf{x}})} \right|\cos \alpha )}^{2}} \\ + \;F{{{\mathbf{(x)}}}^{2}} - {{(\left| {{\mathbf{v}}({\mathbf{x}})} \right|\sin \alpha )}^{2}} + 2\left| {{\mathbf{v}}({\mathbf{x}})} \right|\cos \alpha \sqrt {F{{{({\mathbf{x}})}}^{2}} - {{{({\kern 1pt} \left| {{\mathbf{v}}({\mathbf{x}})} \right|\sin \alpha )}}^{2}}} - 2l\left| {{\mathbf{v}}({\mathbf{x}})} \right|\cos \alpha \\ = F{{({\mathbf{x}})}^{2}} + 2(\left| {{\mathbf{v}}({\mathbf{x}})} \right|\cos \alpha {{)}^{2}} + 2\left| {{\mathbf{v}}({\mathbf{x}})} \right|\cos \alpha \sqrt {F{{{({\mathbf{x}})}}^{2}} - {{{({\kern 1pt} \left| {{\mathbf{v}}({\mathbf{x}})} \right|\sin \alpha )}}^{2}}} - 2l\left| {{\mathbf{v}}({\mathbf{x}})} \right|\cos \alpha \\ = F{{({\mathbf{x}})}^{2}} + 2\left| {{\mathbf{v}}({\mathbf{x}})} \right|\cos \alpha ({\kern 1pt} \left| {{\mathbf{v}}({\mathbf{x}})} \right|\cos \alpha + \sqrt {F{{{({\mathbf{x}})}}^{2}} - {{{({\kern 1pt} \left| {{\mathbf{v}}({\mathbf{x}})} \right|\sin \alpha )}}^{2}}} ) - 2l\left| {{\mathbf{v}}({\mathbf{x}})} \right|\cos \alpha = F{{({\mathbf{x}})}^{2}}. \\ \end{gathered} $$

Therefore, we get $\left\| {g{\text{(}}{\mathbf{x}}{\mathbf{,a}}{\text{)}}} \right\| = F({\mathbf{x}})$.

Let us prove that ${{c}_{x}}: = c{\text{(}}{\mathbf{x}}{\mathbf{,a}}{\text{)}}$ is one to one mapping from ${{S}_{1}}$ to ${{S}_{1}}$ for every fixed $x \in \Omega$.

Assume that ${{c}_{x}}$ is not one to one mapping. Then there exists ${{{\mathbf{a}}}_{1}},\,{{{\mathbf{a}}}_{2}} \in {{S}_{1}}({{{\mathbf{a}}}_{1}} \ne {{{\mathbf{a}}}_{2}})$ and ${\mathbf{a}} \in {{S}_{1}}$ such that ${{c}_{x}}({{{\mathbf{a}}}_{1}}) = {{c}_{x}}({{{\mathbf{a}}}_{2}}) = {\mathbf{a}}$.

Otherwise, since $\left\| {g{\text{(}}{\mathbf{x}}{\mathbf{,a}}{\text{)}}} \right\| = F({\mathbf{x}})$ and $c{\text{(}}{\mathbf{x}}{\mathbf{,a}}{\text{)}} = g{\text{(}}{\mathbf{x}}{\mathbf{,a}}{\text{)/}}\left\| {g{\text{(}}{\mathbf{x}}{\mathbf{,a}}{\text{)}}} \right\|$, we get

$$\begin{gathered} F({\mathbf{x}})c{\text{(}}{\mathbf{x}}{\mathbf{,a}}{\text{)}} - {\mathbf{v}}({\mathbf{x}}) = g{\text{(}}{\mathbf{x}}{\mathbf{,a}}{\text{)}} - {\mathbf{v}}({\mathbf{x}}) = \left( {{\mathbf{v}}({\mathbf{x}}) \cdot {\mathbf{a}} + \sqrt {{{F}^{2}}({\mathbf{x}}) - ({{{\mathbf{v}}}^{2}}({\mathbf{x}}) - {{{({\mathbf{v}}({\mathbf{x}}) \cdot {\mathbf{a}})}}^{2}})} } \right){\mathbf{a}} \\ + \;{\mathbf{v}}({\mathbf{x}}) - {\mathbf{v}}({\mathbf{x}}) = \left( {{\mathbf{v}}({\mathbf{x}}) \cdot {\mathbf{a}} + \sqrt {{{F}^{2}}({\mathbf{x}}) - ({{{\mathbf{v}}}^{2}}({\mathbf{x}}) - {{{({\mathbf{v}}({\mathbf{x}}) \cdot {\mathbf{a}})}}^{2}})} } \right){\mathbf{a}} \\ \end{gathered} $$

(16)

and thus we get

$$F({\mathbf{x}})c({\mathbf{x}}{\mathbf{,}}{{{\mathbf{a}}}_{{\text{1}}}}) - {\mathbf{v}}({\mathbf{x}}) = g({\mathbf{x}}{\mathbf{,}}{{{\mathbf{a}}}_{{\text{1}}}}) - {\mathbf{v}}({\mathbf{x}}) = \left( {{\mathbf{v}}({\mathbf{x}}) \cdot {{{\mathbf{a}}}_{{\text{1}}}} + \sqrt {{{F}^{2}}({\mathbf{x}}) - ({{{\mathbf{v}}}^{2}}({\mathbf{x}}) - {{{({\mathbf{v}}({\mathbf{x}}) \cdot {{{\mathbf{a}}}_{{\text{1}}}})}}^{2}})} } \right){{{\mathbf{a}}}_{{\text{1}}}},$$

$$F({\mathbf{x}})c{\text{(}}{\mathbf{x}}{\mathbf{,}}{{{\mathbf{a}}}_{{\text{2}}}}{\text{)}} - {\mathbf{v}}({\mathbf{x}}) = g{\text{(}}{\mathbf{x}}{\mathbf{,}}{{{\mathbf{a}}}_{{\text{2}}}}{\text{)}} - {\mathbf{v}}({\mathbf{x}}) = \left( {{\mathbf{v}}({\mathbf{x}}) \cdot {{{\mathbf{a}}}_{{\text{2}}}} + \sqrt {{{F}^{2}}({\mathbf{x}}) - ({{{\mathbf{v}}}^{2}}({\mathbf{x}}) - {{{({\mathbf{v}}({\mathbf{x}}) \cdot {{{\mathbf{a}}}_{{\text{2}}}})}}^{2}})} } \right){{{\mathbf{a}}}_{{\text{2}}}}.$$

By using the fact that ${{c}_{x}}({{{\mathbf{a}}}_{1}}) = {{c}_{x}}({{{\mathbf{a}}}_{2}}) = {\mathbf{a}}$, we get

$$\left( {{\mathbf{v}}({\mathbf{x}}) \cdot {{{\mathbf{a}}}_{{\text{1}}}} + \sqrt {{{F}^{2}}({\mathbf{x}}) - ({{{\mathbf{v}}}^{2}}({\mathbf{x}}) - {{{({\mathbf{v}}({\mathbf{x}}) \cdot {{{\mathbf{a}}}_{{\text{1}}}})}}^{2}})} } \right){{{\mathbf{a}}}_{{\text{1}}}} = \left( {{\mathbf{v}}({\mathbf{x}}) \cdot {{{\mathbf{a}}}_{{\text{2}}}} + \sqrt {{{F}^{2}}({\mathbf{x}}) - ({{{\mathbf{v}}}^{2}}({\mathbf{x}}) - {{{({\mathbf{v}}({\mathbf{x}}) \cdot {{{\mathbf{a}}}_{{\text{2}}}})}}^{2}})} } \right){{{\mathbf{a}}}_{{\text{2}}}}.$$

Since ${{{\mathbf{a}}}_{1}},\,{{{\mathbf{a}}}_{2}} \in {{S}_{1}}$, we get ${{{\mathbf{a}}}_{1}} = {{{\mathbf{a}}}_{2}}$. This is contradictory to ${{{\mathbf{a}}}_{1}},\,{{{\mathbf{a}}}_{2}} \in {{S}_{1}}({{{\mathbf{a}}}_{1}} \ne {{{\mathbf{a}}}_{2}})$ and thus ${{c}_{x}}$ is one to one mapping.

From (16),

$$\frac{{F({\mathbf{x}})c{\text{(}}{\mathbf{x}}{\mathbf{,a}}{\text{)}} - {\mathbf{v}}({\mathbf{x}})}}{{\left\| {F({\mathbf{x}})c{\text{(}}{\mathbf{x}}{\mathbf{,a}}{\text{)}} - {\mathbf{v}}({\mathbf{x}})} \right\|}} = \frac{{\left( {{\mathbf{v}}({\mathbf{x}}) \cdot {{{\mathbf{a}}}_{{\text{2}}}} + \sqrt {{{F}^{2}}({\mathbf{x}}) - ({{{\mathbf{v}}}^{2}}({\mathbf{x}}) - {{{({\mathbf{v}}({\mathbf{x}}) \cdot {{{\mathbf{a}}}_{{\text{2}}}})}}^{2}})} } \right){\mathbf{a}}}}{{\left( {{\mathbf{v}}({\mathbf{x}}) \cdot {{{\mathbf{a}}}_{{\text{2}}}} + \sqrt {{{F}^{2}}({\mathbf{x}}) - ({{{\mathbf{v}}}^{2}}({\mathbf{x}}) - {{{({\mathbf{v}}({\mathbf{x}}) \cdot {{{\mathbf{a}}}_{{\text{2}}}})}}^{2}})} } \right)}} = {\mathbf{a}}$$

and thus we see that ${{c}_{x}}$ is and it’s inverse mapping is

$$c_{x}^{{ - 1}}:{{S}_{1}} \to {{S}_{1}},$$

$${\mathbf{a}} \mapsto \frac{{F({\mathbf{x}}){\mathbf{a}} - {\mathbf{v}}({\mathbf{x}})}}{{\left\| {F({\mathbf{x}}){\mathbf{a}} - {\mathbf{v}}({\mathbf{x}})} \right\|}}.$$

We also present a detailed proof of Lemma 2.3.

Proof of Lemma 2.3.

By the definition of $\tilde {f}{\text{(}}{\mathbf{x}}{\mathbf{,a}}{\text{)}}: = {\mathbf{v}}(x) \cdot {\mathbf{a}} + \sqrt {{{F}^{2}}({\mathbf{x}}) - ({{{\mathbf{v}}}^{2}}({\mathbf{x}}) - {{{({\mathbf{v}}({\mathbf{x}}) \cdot {\mathbf{a}})}}^{2}})}$, we get

$$\begin{gathered} \tilde {f}({\mathbf{x}}{\mathbf{,a}}) = {\mathbf{v}}({\mathbf{x}}) \cdot {\mathbf{a}} + \sqrt {{{F}^{2}}({\mathbf{x}}) - ({\text{(}}{{{\mathbf{v}}}^{{\mathbf{2}}}}({\mathbf{x}}) - {{{{\text{(}}{\mathbf{v}}({\mathbf{x}}) \cdot {\mathbf{a}}{\text{)}}}}^{{\mathbf{2}}}})} = {{{v}}_{1}}({\mathbf{x}}){{a}_{1}} + {{{v}}_{2}}({\mathbf{x}}){{a}_{2}} \\ + \;\sqrt {{{F}^{2}}({\mathbf{x}}) - ({{{v}}_{1}}({\mathbf{x}})(1 - a_{1}^{2}) + {{{v}}_{2}}({\mathbf{x}})(1 - a_{2}^{2}) - 2{{{v}}_{1}}({\mathbf{x}}){{{v}}_{2}}({\mathbf{x}}){{a}_{1}}{{a}_{2}})} . \\ \end{gathered} $$

Therefore, we get

$$\begin{gathered} \left\| {\tilde {f}{\text{(}}{\mathbf{x}}{\mathbf{,a}}{\text{)}} - \tilde {f}{\text{(}}{\mathbf{y}}{\mathbf{,a}}{\text{)}}} \right\| = \left| {{{{v}}_{1}}({\mathbf{x}}){{a}_{1}} - {{{v}}_{1}}({\mathbf{y}}){{a}_{1}} + {{{v}}_{2}}({\mathbf{x}}){{a}_{2}}\mathop - \limits_{_{{}}} {{{v}}_{2}}({\mathbf{y}}){{a}_{2}}} \right. \\ + \;\sqrt {{{F}^{2}}({\mathbf{x}}) - ({{{v}}_{1}}({\mathbf{x}})(1 - a_{1}^{2}) + {{{v}}_{2}}({\mathbf{x}})(1 - a_{2}^{2}) - 2{{{v}}_{1}}({\mathbf{x}}){{{v}}_{2}}({\mathbf{x}}){{a}_{1}}{{a}_{2}})} \\ \left. { - \;\sqrt {{{F}^{2}}({\mathbf{y}}) - ({{{v}}_{1}}({\mathbf{y}})(1 - a_{1}^{2}) + {{{v}}_{2}}({\mathbf{y}})(1 - a_{2}^{2}) - 2{{{v}}_{1}}({\mathbf{y}}){{{v}}_{2}}({\mathbf{y}}){{a}_{1}}{{a}_{2}})} } \right| \leqslant \left| {{{{v}}_{1}}({\mathbf{x}}){{a}_{1}} - {{{v}}_{1}}({\mathbf{y}}){{a}_{1}}} \right| \\ \end{gathered} $$

$$\begin{gathered} + \;\left| {{{{v}}_{2}}({\mathbf{x}}){{a}_{2}} - {{{v}}_{2}}({\mathbf{y}}){{a}_{2}}} \right| + \left| {\sqrt {{{F}^{2}}({\mathbf{x}}) - ({{{v}}_{1}}({\mathbf{x}})(1 - a_{1}^{2}) + {{{v}}_{2}}({\mathbf{x}})(1 - a_{2}^{2}) - 2{{{v}}_{1}}({\mathbf{x}}){{{v}}_{2}}({\mathbf{x}}){{a}_{1}}{{a}_{2}})} } \right. \\ \left. { - \;\sqrt {{{F}^{2}}({\mathbf{y}}) - ({{{v}}_{1}}({\mathbf{y}})(1 - a_{1}^{2}) + {{{v}}_{2}}({\mathbf{y}})(1 - a_{2}^{2}) - 2{{{v}}_{1}}({\mathbf{y}}){{{v}}_{2}}({\mathbf{y}}){{a}_{1}}{{a}_{2}})} } \right| \leqslant \left| {{{{v}}_{1}}({\mathbf{x}}) - {{{v}}_{1}}({\mathbf{y}})} \right| \\ \end{gathered} $$

$$\begin{gathered} + \;\left| {{{{v}}_{2}}({\mathbf{x}}) - {{{v}}_{2}}({\mathbf{y}})} \right| + \left| {\sqrt {{{F}^{2}}({\mathbf{x}}) - ({{{v}}_{1}}({\mathbf{x}})(1 - a_{1}^{2}) + {{{v}}_{2}}({\mathbf{x}})(1 - a_{2}^{2}) - 2{{{v}}_{1}}({\mathbf{x}}){{{v}}_{2}}({\mathbf{x}}){{a}_{1}}{{a}_{2}})} } \right. \\ \left. { - \;\sqrt {{{F}^{2}}({\mathbf{y}}) - ({{{v}}_{1}}({\mathbf{y}})(1 - a_{1}^{2}) + {{{v}}_{2}}({\mathbf{y}})(1 - a_{2}^{2}) - 2{{{v}}_{1}}({\mathbf{y}}){{v}_{2}}({\mathbf{y}}){{a}_{1}}{{a}_{2}})} } \right|. \\ \end{gathered} $$

Here,

$$\begin{gathered} \left| {\sqrt {{{F}^{2}}({\mathbf{x}}) - ({{{v}}_{1}}({\mathbf{x}})(1 - a_{1}^{2}) + {{{v}}_{2}}({\mathbf{x}})(1 - a_{2}^{2}) - 2{{{v}}_{1}}({\mathbf{x}}){{{v}}_{2}}({\mathbf{x}}){{a}_{1}}{{a}_{2}})} } \right. \\ \left. { - \;\sqrt {{{F}^{2}}({\mathbf{y}}) - ({{{v}}_{1}}({\mathbf{y}})(1 - a_{1}^{2}) + {{{v}}_{2}}({\mathbf{y}})(1 - a_{2}^{2}) - 2{{{v}}_{1}}({\mathbf{y}}){{{v}}_{2}}({\mathbf{y}}){{a}_{1}}{{a}_{2}})} } \right| \\ \end{gathered} $$

$$ \leqslant \,\frac{{\left| {{{F}^{2}}({\mathbf{x}})\, - \,({{{v}}_{1}}({\mathbf{x}}){\kern 1pt} (1\, - \,a_{1}^{2})\, + \,{{{v}}_{2}}({\mathbf{x}}){\kern 1pt} (1\, - \,a_{2}^{2})\, - \,2{{{v}}_{1}}({\mathbf{x}}){{{v}}_{2}}({\mathbf{x}}){{a}_{1}}{{a}_{2}})\, - \,{{F}^{2}}({\mathbf{y}})\, - \,({{{v}}_{1}}({\mathbf{y}}){\kern 1pt} (1 - a_{1}^{2})\, + \,{{{v}}_{2}}({\mathbf{y}}){\kern 1pt} (1\, - \,a_{2}^{2})\, - \,2{{{v}}_{1}}({\mathbf{y}}){{{v}}_{2}}({\mathbf{y}}){{a}_{1}}{{a}_{2}})} \right|}}{{\left| {\sqrt {{{F}^{2}}({\mathbf{x}}) - {{{\left\| {{\mathbf{v}}({\mathbf{x}})} \right\|}}^{2}}\mathop {\sin }\nolimits^2 {{\alpha }_{1}}} + \sqrt {{{F}^{2}}({\mathbf{y}}) - {{{\left\| {{\mathbf{v}}({\mathbf{y}})} \right\|}}^{2}}\mathop {\sin }\nolimits^2 {{\alpha }_{1}}} } \right|}}$$

$$ \leqslant \frac{{\left| {{{F}^{2}}({\mathbf{x}}) - {{F}^{2}}({\mathbf{y}})} \right| + \left| {{v}_{1}^{2}({\mathbf{x}}) - {v}_{1}^{2}({\mathbf{y}})} \right|(1 - a_{1}^{2}) + \left| {{v}_{2}^{2}({\mathbf{x}}) - {v}_{2}^{2}({\mathbf{y}})} \right|(1 - a_{2}^{2}) + 2\left| {{{{v}}_{1}}({\mathbf{x}}){{{v}}_{2}}({\mathbf{x}}) - {{{v}}_{1}}({\mathbf{y}}){{{v}}_{2}}({\mathbf{y}})} \right|{{a}_{1}}{{a}_{2}}}}{{2\sqrt {{{F}^{2}}({\mathbf{x}}) - {{{\left\| {{v}({\mathbf{x}})} \right\|}}^{2}}} \sqrt {{{F}^{2}}({\mathbf{x}}) - {{{\left\| {{v}({\mathbf{x}})} \right\|}}^{2}}} }}$$

$$ \leqslant \frac{{\left| {{{F}^{2}}({\mathbf{x}}) - {{F}^{2}}({\mathbf{y}})} \right| + \left| {{v}_{1}^{2}({\mathbf{x}}) - {v}_{1}^{2}({\mathbf{y}})} \right| + \left| {{v}_{2}^{2}({\mathbf{x}}) - {v}_{2}^{2}({\mathbf{y}})} \right| + 2\left| {{{{v}}_{1}}({\mathbf{x}}){{{v}}_{2}}({\mathbf{x}}) - {{{v}}_{1}}({\mathbf{x}}){{v}_{2}}({\mathbf{y}}) + {{{v}}_{1}}({\mathbf{x}}){{v}_{2}}({\mathbf{y}}) - {{{v}}_{1}}({\mathbf{y}}){{{v}}_{2}}({\mathbf{y}})} \right|}}{{2g_{1}^{2}}}$$

$$ \leqslant \frac{{2{{F}_{{\max }}}\left| {F({\mathbf{x}}) - F({\mathbf{y}})} \right| + \left| {{v}_{1}^{2}({\mathbf{x}}) - {v}_{1}^{2}({\mathbf{y}})} \right| + \left| {{v}_{2}^{2}({\mathbf{x}}) - {v}_{2}^{2}({\mathbf{y}})} \right| + 2\left| {{{{v}}_{1}}({\mathbf{x}})} \right|\left| {{{{v}}_{2}}({\mathbf{x}}) - {{{v}}_{2}}({\mathbf{y}})} \right| + 2\left| {{{{v}}_{2}}({\mathbf{y}})} \right|\left| {{{{v}}_{1}}({\mathbf{x}}) - {{{v}}_{1}}({\mathbf{y}})} \right|}}{{2g_{1}^{2}}}$$

$$ \leqslant \,\frac{{2{{F}_{{\max }}}{\kern 1pt} \left| {F({\mathbf{x}}) - F({\mathbf{y}})} \right| + 2{{V}_{{\max }}}{\kern 1pt} \left| {{{{v}}_{1}}({\mathbf{x}}) - {{{v}}_{1}}({\mathbf{y}})} \right| + 2{{V}_{{\max }}}{\kern 1pt} \left| {{{{v}}_{2}}({\mathbf{x}}) - {{{v}}_{2}}({\mathbf{y}})} \right| + 2{{V}_{{\max }}}{\kern 1pt} \left| {{{{v}}_{1}}({\mathbf{x}}) - {{{v}}_{1}}({\mathbf{y}})} \right| + 2{{V}_{{\max }}}\left| {{{{v}}_{2}}({\mathbf{x}}) - {{{v}}_{2}}({\mathbf{y}})} \right|}}{{2g_{1}^{2}}}$$

$$ \leqslant \frac{{{{F}_{{\max }}}}}{{g_{1}^{2}}}\left| {F({\mathbf{x}}) - F({\mathbf{y}})} \right| + \frac{{2{{V}_{{\max }}}}}{{g_{1}^{2}}}\left| {{{{v}}_{1}}({\mathbf{x}}) - {{{v}}_{1}}({\mathbf{y}})} \right| + \frac{{2{{V}_{{\max }}}}}{{g_{1}^{2}}}{v}\left| {{{v}_{2}}({\mathbf{x}}) - {{{v}}_{2}}({\mathbf{y}})} \right|$$

and from the assumption (2.2), we get

$$\begin{gathered} \left| {f({\mathbf{x}}{\mathbf{,a}}) - f({\mathbf{y}}{\mathbf{,a}})} \right| \leqslant \left| {{{{v}}_{1}}({\mathbf{x}}) - {{{v}}_{1}}({\mathbf{y}})} \right| + \left| {{{{v}}_{2}}({\mathbf{x}}) - {{{v}}_{2}}({\mathbf{y}})} \right| \\ + \;\frac{{{{F}_{{\max }}}}}{{g_{1}^{2}}}\left| {F({\mathbf{x}}) - F({\mathbf{y}})} \right| + \frac{{2{{V}_{{\max }}}}}{{g_{1}^{2}}}\left| {{{{v}}_{1}}({\mathbf{x}}) - {{{v}}_{1}}({\mathbf{y}})} \right| + \frac{{2{{V}_{{\max }}}}}{{g_{1}^{2}}}\left| {{{{v}}_{2}}({\mathbf{x}}) - {{{v}}_{2}}({\mathbf{y}})} \right| \\ \leqslant \;\left( {{{L}_{{{{{v}}_{1}}}}} + {{L}_{{{{{v}}_{2}}}}} + \frac{{{{F}_{{\max }}}{{L}_{F}} + 2{{V}_{{\max }}}{{L}_{{{{{v}}_{1}}}}} + 2{{V}_{{\max }}}{{L}_{{{{{v}}_{2}}}}}}}{{g_{1}^{2}}}} \right)\left\| {{\mathbf{x}} - {\mathbf{y}}} \right\| \\ \end{gathered} $$

and this completes the proof of Lemma 2.3.