<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article  PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd"><article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article"><front><journal-meta><journal-id journal-id-type="publisher-id">JAMP</journal-id><journal-title-group><journal-title>Journal of Applied Mathematics and Physics</journal-title></journal-title-group><issn pub-type="epub">2327-4352</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/jamp.2020.812220</article-id><article-id pub-id-type="publisher-id">JAMP-106017</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Physics&amp;Mathematics</subject></subj-group></article-categories><title-group><article-title>
 
 
  Hyperbolic Monge-Amp&#232;re Equation
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Fang</surname><given-names>Gao</given-names></name><xref ref-type="aff" rid="aff1"><sub>1</sub></xref><xref ref-type="corresp" rid="cor1"><sup>*</sup></xref></contrib></contrib-group><aff id="aff1"><label>1</label><addr-line>School of Mathematical Sciences, Liaocheng University, Liaocheng, China</addr-line></aff><pub-date pub-type="epub"><day>02</day><month>12</month><year>2020</year></pub-date><volume>08</volume><issue>12</issue><fpage>2971</fpage><lpage>2980</lpage><history><date date-type="received"><day>21,</day>	<month>November</month>	<year>2020</year></date><date date-type="rev-recd"><day>20,</day>	<month>December</month>	<year>2020</year>	</date><date date-type="accepted"><day>23,</day>	<month>December</month>	<year>2020</year></date></history><permissions><copyright-statement>&#169; Copyright  2014 by authors and Scientific Research Publishing Inc. </copyright-statement><copyright-year>2014</copyright-year><license><license-p>This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/</license-p></license></permissions><abstract><p>
 
 
  In this paper, based on the Lie symmetry method, the symmetry group of a hyperbolic Monge-Amp&#232;re equation is obtained first, then the one-dimensional optimal system of the obtained symmetries is given, and finally the group-invariant solutions are investigated.
 
</p></abstract><kwd-group><kwd>Hyperbolic Monge-Amp&#232;re Equation</kwd><kwd> Lie Symmetry</kwd><kwd> One-Dimensional Optimal System</kwd><kwd> Group-Invariant Solutions</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>In the 19th century, in order to study differential equations, Lie proposed Lie group theory. Due to the relatively abstract Lie group theory, it was not widely used until the 1970s. It was not until Bluman Cole wrote an intuitive and understandable book [<xref ref-type="bibr" rid="scirp.106017-ref1">1</xref>] in 1974 that Lie group theory became widely used to study and solve nonlinear partial differential equations [<xref ref-type="bibr" rid="scirp.106017-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.106017-ref3">3</xref>] [<xref ref-type="bibr" rid="scirp.106017-ref4">4</xref>] [<xref ref-type="bibr" rid="scirp.106017-ref5">5</xref>]. The basic idea of Lie group method is to simplify or solve partial differential equations by constructing group invariances as the basis of function transformation.</p><p>Gao and Zhang studied a new class of dissipative hyperbolic geometric flows. By applying the Lie group method, the optimal system is obtained, and then the equation is similarly reduced and the exact solution is obtained [<xref ref-type="bibr" rid="scirp.106017-ref6">6</xref>]. They primarily study Lie symmetry analysis and exact solutions for the coupled integrable no dispersion equations, and gave the exact solution in the form of power series [<xref ref-type="bibr" rid="scirp.106017-ref7">7</xref>]. Then, by applying the classical symmetry method, Gao obtained the group invariant solution, the optimal system and the exact solution of the evolution equation of a hyperbolic curve flow [<xref ref-type="bibr" rid="scirp.106017-ref8">8</xref>]. Gao also discussed the normal hyperbolic mean curvature flow with dissipation, and obtained the symmetric optimal system and exact solutions by applying Lie symmetry method [<xref ref-type="bibr" rid="scirp.106017-ref9">9</xref>]. Ding and Wang considered symmetry group and invariant solutions of one dimensional hyperbolic inverse mean curvature flow [<xref ref-type="bibr" rid="scirp.106017-ref10">10</xref>].</p><p>In [<xref ref-type="bibr" rid="scirp.106017-ref11">11</xref>], during studying the life-span of classical solutions of hyperbolic inverse mean curvature flow, Wang deduced a hyperbolic equation with Riemann invariance that can be reduced to a hyperbolic Monge-Amp&#232; re equation, namely</p><p>u t t u x x − u x t 2 = − ( 1 + u x 2 ) 2 . (1.1)</p><p>In this paper, firstly, we investigate the symmetry group of Equation (1.1). Secondly, we discuss a one-dimensional optimal system of the obtained symmetries. Thirdly, we obtain group-invariant solutions. Finally, we draw conclusions.</p></sec><sec id="s2"><title>2. Symmetry Group</title><p>Suppose the one-parameter group of infinitesimal transformations ( x , t , u ) is given by</p><p>x * = x + ε ξ ( x , t , u ) + o ( ε 2 ) , t * = t + ε η ( x , t , u ) + o ( ε 2 ) , u * = u + ε τ ( x , t , u ) + o ( ε 2 ) , (2.1)</p><p>in which ε is a group parameter.</p><p>Let the symmetric group of Equation (1.1) be generated by the vector field in the following form:</p><p>V = ξ ( x , t , u ) ∂ ∂ x + η ( x , t , u ) ∂ ∂ t + τ ( x , t , u ) ∂ ∂ u . (2.2)</p><p>The first and second-order prolongation of V are respectively:</p><p>p r ( 1 ) ( V ) = ξ ∂ ∂ x + η ∂ ∂ t + τ ∂ ∂ u + τ x ∂ ∂ u x + τ t ∂ ∂ u t ,</p><p>p r ( 2 ) ( V ) = p r ( 1 ) ( V ) + τ x x ∂ ∂ u x x + τ x t ∂ ∂ u x t + τ t t ∂ ∂ u t t .</p><p>The necessary and sufficient condition for Equation (1.1) to remain unchanged under an infinitesimal transformation is that the vector field should satisfy the invariance conditions of Lie symmetry:</p><p>p r ( 2 ) ( V ) ( Δ ) | Δ = 0 = 0 , (2.3)</p><p>in which Δ = u t t u x x − u x t 2 + ( 1 + u x 2 ) 2 , namely</p><p>4 τ x ( 1 + u x 2 ) u x + τ x x u t t + τ t t u x x − 2 τ x t u x t = 0 , (2.4)</p><p>in which</p><p>{ τ x = D x ( τ − ξ u x − η u t ) + ξ u x x + η u x t , τ x x = D x x ( τ − ξ u x − η u t ) + ξ u x x x + η u x x t , τ t t = D t t ( τ − ξ u x − η u t ) + ξ u x t t + η u t t t , τ x t = D x t ( τ − ξ u x − η u t ) + ξ u x x t + η u x t t , (2.5)</p><p>in which D x is total differential tox, D x x , D t t are respectively double total differential to x, t, D x t is the total differential with respect tot and then with respect tox. By substituting (2.5) into Equation (2.4), the decision equations of the original equation can be solved as follows</p><p>{ η t = η u = η x = 0 , ξ t t = ξ u u = ξ x x = 0 , τ u = ξ x , τ x = − ξ u , τ t t = 0 , ξ u t = ξ x t = ξ x u = 0 ,</p><p>the above equations can be solved as follows:</p><p>{ ξ = c 1 x + c 2 t + c 3 u + c 4 , η = c 5 , τ = c 1 u − c 3 x + c 6 t + c 7 , (2.6)</p><p>in which c 1 , c 2 , c 3 , c 4 , c 5 , c 6 , c 7 are all real constants. Substitute (2.6) into (2.2). Hence the associated seven generators for the one-parameter Lie group of infinitesimal transformations are</p><p>{ V 1 = ∂ ∂ x , V 2 = ∂ ∂ t , V 3 = ∂ ∂ u , V 4 = t ∂ ∂ x , V 5 = t ∂ ∂ u , V 6 = x ∂ ∂ x + u ∂ ∂ u , V 7 = u ∂ ∂ x − x ∂ ∂ u . (2.7)</p><p>The corresponding one-parameter transformation groups are:</p><p>{ G 1 : ( x , t , u ) → ( x + ε , t , u ) , G 2 : ( x , t , u ) → ( x , t + ε , u ) , G 3 : ( x , t , u ) → ( x , t , u + ε ) , G 4 : ( x , t , u ) → ( t ε + x , t , u ) , G 5 : ( x , t , u ) → ( x , t , t ε + u ) , G 6 : ( x , t , u ) → ( x e ε , t , u e ε ) , G 7 : ( x , t , u ) → ( u sin ε + x cos ε , t , − x sin ε + u cos ε ) .</p></sec><sec id="s3"><title>3. Optimal System</title><p>Definition 3.1: If a set { v α } α ∈ n of r-dimensional subalgebras satisfies the following conditions:</p><p>1) Any r-dimensional subalgebras are equivalent to some element in set { v α } ;</p><p>2) If β 1 ≠ β 2 , then v β 1 is not equivalent to v β 2 , then set { v α } α ∈ n is known as ther-dimensional optimal system.</p><p>Theorem 3.1: Generators in (2.7) generate an optimal system S:</p><p>{ V 1 , V 2 , V 3 , V 4 , V 5 , V 6 , V 7 , V 7 &#177; V 2 , V 7 &#177; V 6 , V 7 &#177; V 2 &#177; V 6 ,   V 6 &#177; V 2 , V 2 &#177; V 4 , V 2 &#177; V 5 , V 5 &#177; V 1 , V 4 &#177; V 3 } .</p><p>Proof: By formula [ V i , V j ] = V i V j − V j V i , we get table of Lie brackets (<xref ref-type="table" rid="table1">Table 1</xref>).</p><p>Let’s say any vector</p><p>V = l 1 V 1 + l 2 V 2 + l 3 V 3 + l 4 V 4 + l 5 V 5 + l 6 V 6 + l 7 V 7 , (3.1)</p><p>in order to set up the linear transformation of l ˜ = ( l 1 , l 2 , l 3 , l 4 , l 5 , l 6 , l 7 ) , let’s say</p><p>E i = c i j k l j ∂ l k , i = 1 , 2 , 3 , 4 , 5 , 6 , 7 , (3.2)</p><p>in which c i j k comes from [ V i , V j ] = c i j k X k . According to (3.2) and table of Lie brackets, E 1 , E 2 , E 3 , E 4 , E 5 , E 6 , E 7 can be written in</p><p>{ E 1 = l 6 ∂ l 1 − l 7 ∂ l 3 , E 2 = l 4 ∂ l 1 + l 5 ∂ l 3 , E 3 = l 7 ∂ l 1 + l 6 ∂ l 3 , E 4 = − l 2 ∂ l 1 + l 6 ∂ l 4 − l 7 ∂ l 5 , E 5 = − l 2 ∂ l 3 + l 7 ∂ l 4 + l 6 ∂ l 5 , E 6 = − l 1 ∂ l 1 − l 3 ∂ l 3 − l 4 ∂ l 4 − l 5 ∂ l 5 , E 7 = − l 3 ∂ l 1 + l 1 ∂ l 3 − l 5 ∂ l 4 + l 4 ∂ l 5 ,</p><p>for E 1 , E 2 , E 3 , E 4 , E 5 , E 6 , E 7 , Lie equation with parameter a 1 , a 2 , a 3 , a 4 , a 5 , a 6 , a 7 and initial conditions l ˜ | a i = 0 = l , i = 1 , 2 , 3 , 4 , 5 , 6 , 7 are as follows</p><disp-formula id="scirp.106017-formula3"><graphic  xlink:href="//html.scirp.org/file/21-1722175x39.png"  xlink:type="simple"/></disp-formula><p>The solution of the above equation consists of the following transformation</p><table-wrap id="table1" ><label><xref ref-type="table" rid="table1">Table 1</xref></label><caption><title> Lie brackets</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >[ V i , V j ]</th><th align="center" valign="middle" >V 1</th><th align="center" valign="middle" >V 2</th><th align="center" valign="middle" >V 3</th><th align="center" valign="middle" >V 4</th><th align="center" valign="middle" >V 5</th><th align="center" valign="middle" >V 6</th><th align="center" valign="middle" >V 7</th></tr></thead><tr><td align="center" valign="middle" >V 1</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >V 1</td><td align="center" valign="middle" >− V 3</td></tr><tr><td align="center" valign="middle" >V 2</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >V 1</td><td align="center" valign="middle" >V 3</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td></tr><tr><td align="center" valign="middle" >V 3</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >V 3</td><td align="center" valign="middle" >V 1</td></tr><tr><td align="center" valign="middle" >V 4</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >− V 1</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >V 4</td><td align="center" valign="middle" >− V 5</td></tr><tr><td align="center" valign="middle" >V 5</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >− V 3</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >V 5</td><td align="center" valign="middle" >V 4</td></tr><tr><td align="center" valign="middle" >V 6</td><td align="center" valign="middle" >− V 1</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >− V 3</td><td align="center" valign="middle" >− V 4</td><td align="center" valign="middle" >− V 5</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td></tr><tr><td align="center" valign="middle" >V 7</td><td align="center" valign="middle" >V 3</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >− V 1</td><td align="center" valign="middle" >V 5</td><td align="center" valign="middle" >− V 4</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td></tr></tbody></table></table-wrap><disp-formula id="scirp.106017-formula4"><graphic  xlink:href="//html.scirp.org/file/21-1722175x75.png"  xlink:type="simple"/></disp-formula><p>To set up an optimal system, we need to simplify the vector</p><p>l ˜ = ( l 1 , l 2 , l 3 , l 4 , l 5 , l 6 , l 7 ) , (3.3)</p><p>construct the simplest representation of (3.3) by using transforms T 1 ~ T 7 .</p><p>Case 1. When l 7 ≠ 0 , Let a 1 = l 3 l 7 ( T 1 ) , a 3 = − l 1 l 7 ( T 3 ) , a 4 = l 5 l 7 ( T 4 ) , a 5 = − l 4 l 7 ( T 5 ) , make l 3 ˜ = 0 , l 1 ˜ = 0 , l 5 ˜ = 0 , l 4 ˜ = 0 , so (3.3) ⇔ ( 0 , l 2 , 0 , 0 , 0 , l 6 , l 7 ) , as a result, we obtain the following representative:</p><p>V ~ V 7 , V 7 &#177; V 2 , V 7 &#177; V 6 , V 7 &#177; V 2 &#177; V 6 .</p><p>Case 2. When l 7 = 0 , l 6 ≠ 0 , Let a 1 = − l 1 l 6 ( T 1 ) , a 4 = − l 4 l 6 ( T 4 ) , a 3 = − l 3 l 6 ( T 3 ) , a 5 = − l 5 l 6 ( T 5 ) , make l 1 ˜ = 0 , l 3 ˜ = 0 , l 4 ˜ = 0 , l 5 ˜ = 0 , so (3.3) ⇔ ( 0 , l 2 , 0 , 0 , 0 , l 6 , 0 ) , as a result, we obtain the following representative:</p><p>V ~ V 6 , V 6 &#177; V 2 .</p><p>Case 3. When l 7 = l 6 = 0 , l 2 ≠ 0 , Let a 4 = l 1 l 2 ( T 4 ) , a 5 = l 3 l 2 ( T 5 ) , make l 1 ˜ = 0 , l 3 ˜ = 0 , so (3.3) ⇔ ( 0 , l 2 , 0 , l 4 , l 5 , 0 , 0 ) , as a result, we obtain the following representative:</p><p>V ~ V 2 , V 2 &#177; V 4 , V 2 &#177; V 5 .</p><p>Case 4. When l 7 = l 6 = l 2 = 0 , l 5 ≠ 0 , Let a 2 = − l 3 l 5 ( T 2 ) , make l 3 ˜ = 0 , so (3.3) ⇔ ( l 1 , 0 , 0 , l 4 , l 5 , 0 , 0 ) .</p><p>(4.1) When l 4 ≠ 0 , Let a 2 = − l 1 l 4 ( T 2 ) , a 7 = arctan ( − l 5 l 4 ) ( T 7 ) , make l 1 ˜ = 0 , l 5 ˜ = 0 , so (3.3) ⇔ ( 0 , 0 , 0 , l 4 , 0 , 0 , 0 ) ; as a result, we obtain the following representative:</p><p>V ~ V 4 .</p><p>(4.2) When l 4 = 0 , (3.3) ⇔ ( l 1 , 0 , 0 , 0 , l 5 , 0 , 0 ) , we obtain the following representative:</p><p>V ~ V 5 , V 5 &#177; V 1 .</p><p>Case 5. When l 7 = l 6 = l 2 = l 5 = 0 , l 4 ≠ 0 , (3.3) ⇔ ( l 1 , 0 , l 3 , l 4 , 0 , 0 , 0 ) , Let a 2 = − l 1 l 4 ( T 2 ) , make l 1 ˜ = 0 , so (3.3) ⇔ ( 0 , 0 , l 3 , l 4 , 0 , 0 , 0 ) , as a result, we obtain the following representative:</p><p>V ~ V 4 , V 4 &#177; V 3 .</p><p>Case 6. When l 7 = l 6 = l 2 = l 5 = l 4 = 0 , l 3 ≠ 0 , (3.3) ⇔ ( l 1 , 0 , l 3 , 0 , 0 , 0 , 0 ) , Let a 7 = arctan ( l 1 l 3 ) ( T 7 ) , make l 1 ˜ = 0 , so (3.3) ⇔ ( 0 , 0 , l 3 , 0 , 0 , 0 , 0 ) , as a result, we obtain the following representative:</p><p>V ~ V 3 .</p><p>Case 7. When l 7 = l 6 = l 2 = l 5 = l 4 = l 3 = 0 , l 1 ≠ 0 , (3.3) ⇔ ( l 1 , 0 , 0 , 0 , 0 , 0 , 0 ) , we obtain the following representative:</p><p>V ~ V 1 .</p><p>To sum up, optimal system is</p><p>{ V 1 , V 2 , V 3 , V 4 , V 5 , V 6 , V 7 , V 7 &#177; V 2 , V 7 &#177; V 6 , V 7 &#177; V 2 &#177; V 6 ,   V 6 &#177; V 2 , V 2 &#177; V 4 , V 2 &#177; V 5 , V 5 &#177; V 1 , V 4 &#177; V 3 } .</p></sec><sec id="s4"><title>4. Group-Invariant Solutions</title><p>In this section, using the optimal system, the reduced equations and exact solutions are analyzed for Equation (1.1).</p><sec id="s4_1"><title>4.1. V = V 6 = x ∂ ∂ x + u ∂ ∂ u</title><p>The corresponding characteristic equations are</p><p>d x x = d t 0 = d u u ,</p><p>the invariances are</p><p>t , u x</p><p>the invariant solutions is</p><p>u = x f ( t ) ,</p><p>then Equation (1.1) can be reduced as</p><p>f ′ 2 − 2 f 2 − f 4 − 1 = 0. (4.1)</p><p>By solving the above equation, we can get:</p><p>f ( t ) = f ( 0 ) &#177; tan t 1 ∓ f ( 0 ) tan t ,     u = x f ( 0 ) &#177; tan t 1 ∓ f ( 0 ) tan t .</p><p>When we take f ( 0 ) = 1 , we have</p><p>u 1 = x 1 + tan t 1 − tan t ,</p><p>and</p><p>u 2 = x 1 − tan t 1 + tan t .</p><p><xref ref-type="fig" rid="fig1">Figure 1</xref> and <xref ref-type="fig" rid="fig2">Figure 2</xref> depict solutions u 1 and u 2 .</p></sec><sec id="s4_2"><title>4.2. V = V 7 = u ∂ ∂ x − x ∂ ∂ u</title><p>The corresponding characteristic equations are</p><p>d x u = d t 0 = d u − x ,</p><p>the invariances are</p><p>t , x 2 + u 2 ,</p><p>the invariant solution is given by</p><p>u = f ( t ) − x 2 ,</p><p>then Equation (1.1) can be reduced as</p><p>( f − 3 x 2 ) f ′ 2 + ( 6 x 2 f − 2 f 2 − 6 x 4 ) f ″ + 8 x 2 + 4 x 6 − 12 x 4 f + 4 = 0. (4.2)</p></sec><sec id="s4_3"><title>4.3. V = V 7 + V 2 = u ∂ ∂ x + ∂ ∂ t − x ∂ ∂ u</title><p>The corresponding characteristic equations are</p><p>d x u = d t 1 = d u − x ,</p><p>the invariance is</p><p>z = x 2 + u 2 ,</p><p>the invariant solution is given by</p><p>u = − x t + f ( z ) ,</p><p>then Equation (1.1) can be reduced as</p><p>( 2 x f ′ − t ) 4 + 2 ( 2 x f ′ − t ) 2 = 0. (4.3)</p></sec><sec id="s4_4"><title>4.4. V = V 6 + V 2 = x ∂ ∂ x + ∂ ∂ t + u ∂ ∂ u</title><p>The corresponding characteristic equations are</p><p>d x x = d t 1 = d u u ,</p><p>the invariance is</p><p>z = x e − t ,</p><p>the invariant solution is given by</p><p>u = x f ( z ) ,</p><p>then Equation (1.1) can be reduced as</p><p>− x 3 f ′ f ″ + 6 x 2 f ′ 2 f 2 e t + 4 x f 3 f ′ e 2 t + 4 x 3 f f ′ 3 + 4 x f f ′ e 2 t + ( f 4 + 2 f 2 ) = 0. (4.4)</p></sec><sec id="s4_5"><title>4.5. V = V 2 + V 4 = t ∂ ∂ x + ∂ ∂ t</title><p>The corresponding characteristic equations are</p><p>d x t = d t 1 = d u 0 ,</p><p>the invariance is</p><p>z = 2 x − t 2 ,</p><p>the invariant solution is given by</p><p>u = g ( z ) ,</p><p>then Equation (1.1) can be reduced as</p><p>8 g ′ 2 + 16 g ′ 4 − 8 g ′ g ″ + 1 = 0. (4.5)</p></sec></sec><sec id="s5"><title>5. Conclusion</title><p>This paper includes four parts: the first part is the introduction, which introduces the background knowledge of the hyperbolic Monge-Amp&#232; re equation and Lie symmetry; in the second part, symmetry group is given; in the third part, optimal system of the symmetry is discussed; in the fourth part, we obtain group-invariant solutions.</p></sec><sec id="s6"><title>Acknowledgements</title><p>I would like to thank Z G Wang teacher for drawing my attention to hyperbolic Monge-Amp&#232; re equation. I am grateful to my tutor for her suggestions, which have greatly improved this paper.</p></sec><sec id="s7"><title>Conflicts of Interest</title><p>The author declares no conflicts of interest regarding the publication of this paper.</p></sec><sec id="s8"><title>Cite this paper</title><p>Gao, F. (2020) Hyperbolic Monge-Amp&#232; re Equation. 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