利用fexp方法求對稱正則長波方程的精確解畢業(yè)論文-文庫吧資料
2025-07-01 15:16本頁面
【正文】 a?? 33014146hbeeFa????1(0)?2 414041bab???32,?023396ha? (6)0220226AehF???(6)220223Ah???，當(dāng) 時， 可化為30()ab?2306ahb??F032cos()F???3 414043ahbab????2023,13026,4hhb?,當(dāng) 時,22314aeb??(0)?13a??可化為F351cosh()??4 341404aab???133,?,220606beaea????紅河學(xué)院本科畢業(yè)論文（設(shè)計）5202,416hba??20()ab??5 3404140a??2323,ba?0122,4h?,當(dāng)107()eFb??????1(0)ab?時, 可化為10? F81(coshin)ab??6 340440aab???1321,b?012,ah? 00911()eaFe?????, 當(dāng) 時,0ab?0b??可化為F011(sinh)cos()2a???7 34413ahab????20220,220,4ha??,202212()4hbaee????2??8 3 32ba???011320,1,4hba??01021,ba?? 2202224(hbaebF?????2104(ae???, 2022)(e???01)??9 04403h????21213,ab03421,6h? 430131()abeF????301aeb??03,?第二章 對稱正則長波方程的精確解610 344041ahaba????2021,b?1123,46ah?? 3202241330(),6(,).aehbeFab???????11 340404ab??21,b??320,4ah, (2)150aeFb???02)??12 34414ba???1202a?3,9h, (3)016aeFb???0)?13 340441aab????2130,b?3021,9ah? 2317()1(abeF????213aeb???320,??14 3041404bab???1221a?3,9h, (7)4183aeFb???430)??15 4041404bab???1221a?3,6h, (6)319aeFb???30)??16 404041ab????2213,ab 211320 11(2)3(),0,)aebFab???????紅河學(xué)院本科畢業(yè)論文（設(shè)計）71236,abh??當(dāng) 時, 可化成31b??F2123aeb???17 304041a???1a?213,b42215,h? 2334422113( ))eee??????當(dāng) 時, 可化成3b??F241(sinh2)cos()si)/????18 340404ahab???2132,b?01120,a? 20222324()abeaeF???????203(,)?第二章 對稱正則長波方程的精確解819 3410413ahbab????23,2044,16ah?? (4)4025016hbeaeFa????,當(dāng)40(,)?240?時, 可化成F4260sinh()ab?20 34104ahbab???203,2324,1ah???,342734()aee????430(,)??21 3120bab??130,?h202143()/,a???43hbab 23214428()baeFe??????430)a??22 30104132ha????? 21,?4032/6ab ,(3)(4)109ebF?????310(,)??23 34041hab???2??1,41302/,9ab, (3)30aeFb???0)?24 34104ha????132,?012344,ba?,2314103()abebeF??????410(,)?紅河學(xué)院本科畢業(yè)論文（設(shè)計）925 3404104ahba????132?,6 ,6234aeFb???24(0)??26 3404141ahba????22b?3,6 ,403aeFb???04()?27 404140ahba???132021?2,9 ,734aeFb???34(0)?28 3404101ahba????22b?3,6 ,8435aeFb???4(0)?29 404103ahba????22,141/,5?,234413631())aebF??????44(0,)??30 30413ahba????22,6,?314040//?,34410370()abeFb??????44(,)??31 12ahaba??30213,b???2401,/, 0381aFbe?????01(,)b??當(dāng) 時, 可化為1? 03912coshaF??第二章 對稱正則長波方程的精確解1032 304041ahaba????213,b2h, 234024aebFh???3(0)a??33 3041401aha????23,b?402,16h??,24416aebFh???,當(dāng) 時, 24(0)b?2441?可化為F242cosh()a??34 3041404ahbab???23,241213,6hb??, ,當(dāng)1343226ebha???130)??時, 可為3421??F143cosh(2)a???35 34421302022,/,/?????,當(dāng) 時,0145()abe???1b可化為F046tanh()2?36 3403202aba????41325,h?2441015,aabhb???417()abe????，當(dāng) 時,41(0,??14b??可化為F485tanh()2?37 314132? ??340,/,ahba?204 04/8hb, , 當(dāng)4049()abe????40(,)b?時, 可化為4F450tanh(2)??紅河學(xué)院本科畢業(yè)論文（設(shè)計）1138 340420aab???21,h22333140,aabb?????,231351()abeF????,當(dāng) 時, 可化31(0,)???31b??F為 352tanhb???39 341412a??03213,bb??2022299,ahh?,3053()eF???,當(dāng) 時, 03(,)ab??30b??可化為 054tanh()2?40 3414041bab????220201,ah314 221,8bah???,315()abeF?????, 當(dāng) 時, 13(0,)??31b??可化為 156ta
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