N soliton solution KdV equation
I find many references to the "N-soliton" solution to the KdV equation
$$
u_{t} - uu_{x} + u_{xxx} = 0
$$
of the form $u(x, 0) = N(N+1) sech^{2}(x)$. However, I require the $N$ soliton solution for
$$
u_{t} + uu_{x} + u_{xxx} = 0
$$
instead. Does anyone the answer and know a paper where this is found?
Many thanks
pde
edited Dec 9 at 18:57
lcv
717410
asked Dec 9 at 17:38
Hello
113
add a comment |
I find many references to the "N-soliton" solution to the KdV equation
$$
u_{t} - uu_{x} + u_{xxx} = 0
$$
of the form $u(x, 0) = N(N+1) sech^{2}(x)$. However, I require the $N$ soliton solution for
$$
u_{t} + uu_{x} + u_{xxx} = 0
$$
instead. Does anyone the answer and know a paper where this is found?
Many thanks
pde
edited Dec 9 at 18:57
lcv
717410
asked Dec 9 at 17:38
Hello
113
Because that second term is 'quadratic', if $u$ satisfies $u_t-uu_x+u_{xxx}=0$ then $v=-u$ satisfies $v_t+vv_x+v_{xxx}=0$, so the equations are equivalent.
– Steven Stadnicki
Dec 9 at 18:59
Thank you very much
– Hello
Dec 9 at 19:01
add a comment |
I find many references to the "N-soliton" solution to the KdV equation
$$
u_{t} - uu_{x} + u_{xxx} = 0
$$
of the form $u(x, 0) = N(N+1) sech^{2}(x)$. However, I require the $N$ soliton solution for
$$
u_{t} + uu_{x} + u_{xxx} = 0
$$
instead. Does anyone the answer and know a paper where this is found?
Many thanks
pde
edited Dec 9 at 18:57
lcv
717410
asked Dec 9 at 17:38
Hello
113
I find many references to the "N-soliton" solution to the KdV equation
$$
u_{t} - uu_{x} + u_{xxx} = 0
$$
of the form $u(x, 0) = N(N+1) sech^{2}(x)$. However, I require the $N$ soliton solution for
$$
u_{t} + uu_{x} + u_{xxx} = 0
$$
instead. Does anyone the answer and know a paper where this is found?
Many thanks
pde
pde
edited Dec 9 at 18:57
lcv
717410
asked Dec 9 at 17:38
Hello
113
edited Dec 9 at 18:57
lcv
717410
asked Dec 9 at 17:38
Hello
113
edited Dec 9 at 18:57
lcv
717410
edited Dec 9 at 18:57
lcv
717410
edited Dec 9 at 18:57
lcv
717410
717410
asked Dec 9 at 17:38
Hello
113
asked Dec 9 at 17:38
Hello
113
asked Dec 9 at 17:38
Hello
113
113
Because that second term is 'quadratic', if $u$ satisfies $u_t-uu_x+u_{xxx}=0$ then $v=-u$ satisfies $v_t+vv_x+v_{xxx}=0$, so the equations are equivalent.
– Steven Stadnicki
Dec 9 at 18:59
Thank you very much
– Hello
Dec 9 at 19:01
add a comment |
Because that second term is 'quadratic', if $u$ satisfies $u_t-uu_x+u_{xxx}=0$ then $v=-u$ satisfies $v_t+vv_x+v_{xxx}=0$, so the equations are equivalent.
– Steven Stadnicki
Dec 9 at 18:59
Thank you very much
– Hello
Dec 9 at 19:01
Because that second term is 'quadratic', if $u$ satisfies $u_t-uu_x+u_{xxx}=0$ then $v=-u$ satisfies $v_t+vv_x+v_{xxx}=0$, so the equations are equivalent.
– Steven Stadnicki
Dec 9 at 18:59
Because that second term is 'quadratic', if $u$ satisfies $u_t-uu_x+u_{xxx}=0$ then $v=-u$ satisfies $v_t+vv_x+v_{xxx}=0$, so the equations are equivalent.
– Steven Stadnicki
Dec 9 at 18:59
Thank you very much
– Hello
Dec 9 at 19:01
Thank you very much
– Hello
Dec 9 at 19:01
add a comment |
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Because that second term is 'quadratic', if $u$ satisfies $u_t-uu_x+u_{xxx}=0$ then $v=-u$ satisfies $v_t+vv_x+v_{xxx}=0$, so the equations are equivalent.
– Steven Stadnicki
Dec 9 at 18:59
Thank you very much
– Hello
Dec 9 at 19:01