双重sinh-Gordon方程(Double sinh-Gordon equation)是一个非线性偏微分方程。[1][2][3][4][5]. u x t = a s i n h ( u ) + b s i n h ( 2 u ) {\displaystyle u_{xt}=asinh(u)+bsinh(2u)} Remove ads行波解 v = C 5 ∗ J a c o b i C N ( C 2 + C 3 ∗ x − ( a ∗ C 5 2 − 2 ∗ b ∗ C 5 2 − 2 ∗ b − a ) ∗ t / ( C 3 ∗ ( C 5 2 − 1 ) ) , ( ( − 2 ∗ a ∗ C 5 2 + a ∗ C 5 4 + a + 2 ∗ b − 2 ∗ b ∗ C 5 4 ) ∗ ( a ∗ C 5 2 − 2 ∗ b ∗ C 5 2 − a ) ) ∗ C 5 / ( − 2 ∗ a ∗ C 5 2 + a ∗ C 5 4 + a + 2 ∗ b − 2 ∗ b ∗ C 5 4 ) ) {\displaystyle {v=_{C}5*JacobiCN(_{C}2+_{C}3*x-(a*_{C}5^{2}-2*b*_{C}5^{2}-2*b-a)*t/(_{C}3*(_{C}5^{2}-1)),{\sqrt {(}}(-2*a*_{C}5^{2}+a*_{C}5^{4}+a+2*b-2*b*_{C}5^{4})*(a*_{C}5^{2}-2*b*_{C}5^{2}-a))*_{C}5/(-2*a*_{C}5^{2}+a*_{C}5^{4}+a+2*b-2*b*_{C}5^{4}))}} v = C 5 ∗ J a c o b i D N ( C 2 + C 3 ∗ x − C 5 2 ∗ ( a ∗ C 5 2 − 2 ∗ b ∗ C 5 2 − a ) ∗ t / ( C 3 ∗ ( − 2 ∗ C 5 2 + 1 + C 5 4 ) ) , ( ( − 2 ∗ a ∗ C 5 2 + a ∗ C 5 4 + a + 2 ∗ b − 2 ∗ b ∗ C 5 4 ) ∗ ( a ∗ C 5 2 − 2 ∗ b ∗ C 5 2 − a ) ) / ( ( a ∗ C 5 2 − 2 ∗ b ∗ C 5 2 − a ) ∗ C 5 ) ) {\displaystyle {v=_{C}5*JacobiDN(_{C}2+_{C}3*x-_{C}5^{2}*(a*_{C}5^{2}-2*b*_{C}5^{2}-a)*t/(_{C}3*(-2*_{C}5^{2}+1+_{C}5^{4})),{\sqrt {(}}(-2*a*_{C}5^{2}+a*_{C}5^{4}+a+2*b-2*b*_{C}5^{4})*(a*_{C}5^{2}-2*b*_{C}5^{2}-a))/((a*_{C}5^{2}-2*b*_{C}5^{2}-a)*_{C}5))}} v = C 5 ∗ J a c o b i N C ( C 2 + C 3 ∗ x + ( a ∗ C 5 2 − 2 ∗ b ∗ C 5 2 − 2 ∗ b − a ) ∗ t / ( C 3 ∗ ( C 5 2 − 1 ) ) , ( − ( − 2 ∗ a ∗ C 5 2 + a ∗ C 5 4 + a + 2 ∗ b − 2 ∗ b ∗ C 5 4 ) ∗ ( a ∗ C 5 2 − 2 ∗ b − a ) ) / ( − 2 ∗ a ∗ C 5 2 + a ∗ C 5 4 + a + 2 ∗ b − 2 ∗ b ∗ C 5 4 ) ) {\displaystyle {v=_{C}5*JacobiNC(_{C}2+_{C}3*x+(a*_{C}5^{2}-2*b*_{C}5^{2}-2*b-a)*t/(_{C}3*(_{C}5^{2}-1)),{\sqrt {(}}-(-2*a*_{C}5^{2}+a*_{C}5^{4}+a+2*b-2*b*_{C}5^{4})*(a*_{C}5^{2}-2*b-a))/(-2*a*_{C}5^{2}+a*_{C}5^{4}+a+2*b-2*b*_{C}5^{4}))}} v = C 5 ∗ J a c o b i N D ( C 2 + C 3 ∗ x − ( a ∗ C 5 2 − 2 ∗ b − a ) ∗ t / ( C 3 ∗ ( − 2 ∗ C 5 2 + 1 + C 5 4 ) ) , ( − ( − 2 ∗ a ∗ C 5 2 + a ∗ C 5 4 + a + 2 ∗ b − 2 ∗ b ∗ C 5 4 ) ∗ ( a ∗ C 5 2 − 2 ∗ b − a ) ) / ( a ∗ C 5 2 − 2 ∗ b − a ) ) {\displaystyle {v=_{C}5*JacobiND(_{C}2+_{C}3*x-(a*_{C}5^{2}-2*b-a)*t/(_{C}3*(-2*_{C}5^{2}+1+_{C}5^{4})),{\sqrt {(}}-(-2*a*_{C}5^{2}+a*_{C}5^{4}+a+2*b-2*b*_{C}5^{4})*(a*_{C}5^{2}-2*b-a))/(a*_{C}5^{2}-2*b-a))}} v = ( a ∗ ( 2 ∗ b + a ) ) ∗ c s c ( C 1 + C 2 ∗ x − ( 2 ∗ b + a ) ∗ t / C 2 ) / a {\displaystyle {v={\sqrt {(}}a*(2*b+a))*csc(_{C}1+_{C}2*x-(2*b+a)*t/_{C}2)/a}} v = ( a ∗ ( 2 ∗ b + a ) ) ∗ c s c ( C 2 + C 3 ∗ x − ( 2 ∗ b + a ) ∗ t / C 3 ) / a {\displaystyle {v={\sqrt {(}}a*(2*b+a))*csc(_{C}2+_{C}3*x-(2*b+a)*t/_{C}3)/a}} v = ( a ∗ ( 2 ∗ b + a ) ) ∗ s e c ( C 1 + C 2 ∗ x − ( 2 ∗ b + a ) ∗ t / C 2 ) / a {\displaystyle {v={\sqrt {(}}a*(2*b+a))*sec(_{C}1+_{C}2*x-(2*b+a)*t/_{C}2)/a}} v = ( a ∗ ( 2 ∗ b + a ) ) ∗ s e c h ( C 1 + C 2 ∗ x + ( 2 ∗ b + a ) ∗ t / C 2 ) / a {\displaystyle {v={\sqrt {(}}a*(2*b+a))*sech(_{C}1+_{C}2*x+(2*b+a)*t/_{C}2)/a}} v = ( − a ∗ ( 2 ∗ b + a ) ) ∗ c s c h ( C 1 + C 2 ∗ x + ( 2 ∗ b + a ) ∗ t / C 2 ) / a {\displaystyle {v={\sqrt {(}}-a*(2*b+a))*csch(_{C}1+_{C}2*x+(2*b+a)*t/_{C}2)/a}} v = ( ( a − 2 ∗ b ) ∗ a ) ∗ c o s h ( C 2 + C 3 ∗ x − ( a − 2 ∗ b ) ∗ t / C 3 ) / ( a − 2 ∗ b ) {\displaystyle {v={\sqrt {(}}(a-2*b)*a)*cosh(_{C}2+_{C}3*x-(a-2*b)*t/_{C}3)/(a-2*b)}} v = ( ( a − 2 ∗ b ) ∗ ( 2 ∗ b + a ) ) ∗ t a n h ( C 1 + C 2 ∗ x + ( 1 / 8 ) ∗ ( a 2 − 4 ∗ b 2 ) ∗ t / ( C 2 ∗ b ) ) / ( a − 2 ∗ b ) {\displaystyle {v={\sqrt {(}}(a-2*b)*(2*b+a))*tanh(_{C}1+_{C}2*x+(1/8)*(a^{2}-4*b^{2})*t/(_{C}2*b))/(a-2*b)}} 其中 v = t a n h ( ( 1 / 2 ) ∗ u ) {\displaystyle v=tanh((1/2)*u)} Remove ads特解 u ( x , t ) = 2 a r c t a n h ( 1.5 ∗ J a c o b i C N ( 1.2 + 1.3 ∗ x + 3.2307692307692307692 ∗ t , 1.0555973258234951998 ) ) {\displaystyle u(x,t)=2arctanh(1.5*JacobiCN(1.2+1.3*x+3.2307692307692307692*t,1.0555973258234951998))} u ( x , t ) = 2 a r c t a n h ( 1.5 ∗ J a c o b i D N ( 1.2 + 1.3 ∗ x + 3.6000000000000000000 ∗ t , .94733093343134184593 ) ) {\displaystyle u(x,t)=2arctanh(1.5*JacobiDN(1.2+1.3*x+3.6000000000000000000*t,.94733093343134184593))} u ( x , t ) = 2 ∗ a r c t a n h ( 1.5 ∗ J a c o b i N C ( − 1.2 − 1.3 ∗ x + 3.2307692307692307692 ∗ t , .33806170189140663100 ∗ I ) ) {\displaystyle u(x,t)=2*arctanh(1.5*JacobiNC(-1.2-1.3*x+3.2307692307692307692*t,.33806170189140663100*I))} u ( x , t ) = 2 ∗ a r c t a n h ( 1.5 ∗ J a c o b i N D ( 1.2 + 1.3 ∗ x + .36923076923076923077 ∗ t , 2.9580398915498080213 ∗ I ) ) {\displaystyle u(x,t)=2*arctanh(1.5*JacobiND(1.2+1.3*x+.36923076923076923077*t,2.9580398915498080213*I))} u ( x , t ) = − 2 ∗ a r c t a n h ( ( 3 ) ∗ c s c ( 15.1 − 1.2 ∗ x + 2.5000000000000000000 ∗ t ) ) {\displaystyle u(x,t)=-2*arctanh({\sqrt {(}}3)*csc(15.1-1.2*x+2.5000000000000000000*t))} u ( x , t ) = − 2 ∗ a r c t a n h ( ( 3 ) ∗ c s c ( − 1.2 − 1.3 ∗ x + 2.3076923076923076923 ∗ t ) ) {\displaystyle u(x,t)=-2*arctanh({\sqrt {(}}3)*csc(-1.2-1.3*x+2.3076923076923076923*t))} u ( x , t ) = 2 ∗ a r c t a n h ( ( 3 ) ∗ s e c ( 15.1 − 1.2 ∗ x + 2.5000000000000000000 ∗ t ) ) {\displaystyle u(x,t)=2*arctanh({\sqrt {(}}3)*sec(15.1-1.2*x+2.5000000000000000000*t))} u ( x , t ) = 2 ∗ a r c t a n h ( ( 3 ) ∗ s e c h ( − 15.1 + 1.2 ∗ x + 2.5000000000000000000 ∗ t ) ) {\displaystyle u(x,t)=2*arctanh({\sqrt {(}}3)*sech(-15.1+1.2*x+2.5000000000000000000*t))} u ( x , t ) = 2 ∗ a r c t a n h ( ( 3 ) ∗ s e c h ( 1.2 + 1.3 ∗ x + 2.3076923076923076923 ∗ t ) ) {\displaystyle u(x,t)=2*arctanh({\sqrt {(}}3)*sech(1.2+1.3*x+2.3076923076923076923*t))} u ( x , t ) = 2 ∗ a r c t a n h ( ( − 3 ) ∗ c s c h ( − 15.1 + 1.2 ∗ x + 2.5000000000000000000 ∗ t ) ) {\displaystyle u(x,t)=2*arctanh({\sqrt {(}}-3)*csch(-15.1+1.2*x+2.5000000000000000000*t))} {\displaystyle } {\displaystyle } {\displaystyle } Remove ads行波图 参考文献Loading content...Loading related searches...Wikiwand - on Seamless Wikipedia browsing. 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