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Equation data
Category:4. Nonlinear Partial Differential Equations
Subcategory:4.5. Third-Order Equations
Equation(s):$\displaystyle \frac{\partial w}{\partial t} + 6w\frac{\partial w}{\partial x} + \frac{\partial^3 w}{\partial x^3} = 0,$\quad \
KdV equation.
Solution(s),
Transformation(s),
Integral(s)
:
Traveling-wave solutions:\\
$w_1(x,t)=a+\dfrac{c
   \cosh ((-6 a-1) t+x)+1}{-c \cosh ^2((-6 a-1)
   t+x)+(c-1) \cosh ((-6 a-1) t+x)+1},$\\
$w_2(x,t)=a+\dfrac{c
   \cosh ((-6 a-1) t+x)+1}{c \cosh ^2((-6 a-1)
   t+x)+(c+1) \cosh ((-6 a-1) t+x)+1},$\\   
$w_3(x,t)=a+\dfrac{c
   \cosh ((-6 a-1) t+x)+1}{-c \cosh ^2((-6 a-1)
   t+x)+(c-1) \cosh ((-6 a-1) t+x)+1},$\\
$w_4(x,t)=a+\dfrac{c
   \cosh ((-6 a-1) t+x)+1}{c \cosh ^2((-6 a-1)
   t+x)+(c+1) \cosh ((-6 a-1) t+x)+1},$\\
$w_5(x,t)=
   a+\dfrac{\dfrac{B^2}{4}+\dfrac{1}{2} \cosh ((-6
   a-1) t+x) B+\dfrac{1}{4} \sqrt{4-B^2} \sinh ((-6
   a-1) t+x) B}{\dfrac{1}{4} B^2 \cosh ^2((-6 a-1)
   t+x)+B \cosh ((-6 a-1) t+x)+1},$\\
$w_6(x,t)=
   a+\dfrac{\dfrac{B^2}{4}+\dfrac{1}{2} \cosh ((-6
   a-1) t+x) B-\dfrac{1}{4} \sqrt{4-B^2} \sinh ((-6
   a-1) t+x) B}{\dfrac{1}{4} B^2 \cosh ^2((-6 a-1)
   t+x)+B \cosh ((-6 a-1) t+x)+1},$\\
$w_7(x,t)= -\dfrac{2}{3} -\dfrac{\lambda }{6}+\dfrac{2}{\sinh ^2(\lambda
   t+x)+1},$\\
$w_8(x,t)= \dfrac{2}{3} -\dfrac{\lambda }{6}-\dfrac{2}{1-\cos ^2(x+t
   \lambda )},$\\
$w_9(x,t)= a+\dfrac{c
   \cos ((1-6 a) t+x)-1}{c \cos ^2((1-6 a)
   t+x)+(-c-1) \cos ((1-6 a) t+x)+1},$\\
$w_{10}(x,t)=a+\dfrac{c
   \cos ((1-6 a) t+x)-1}{-c \cos ^2((1-6 a)
   t+x)+(1-c) \cos ((1-6 a) t+x)+1},$\\
$w_{11}(x,t)=a+\dfrac{c
   \cos ((1-6 a) t+x)-1}{c \cos ^2((1-6 a)
   t+x)+(-c-1) \cos ((1-6 a) t+x)+1},$\\
$w_{12}(x,t)=a+\dfrac{c
   \cos ((1-6 a) t+x)-1}{-c \cos ^2((1-6 a)
   t+x)+(1-c) \cos ((1-6 a) t+x)+1},$\\
$w_{13}(x,t)= a+\dfrac{-\dfrac{B^2}{4}-\dfrac{1}{2} \cos ((1-6 a)
   t+x) B+\dfrac{1}{4} \sqrt{B^2-4} \sin ((1-6 a)
   t+x) B}{\dfrac{1}{4} B^2 \cos ^2((1-6 a) t+x)+B
   \cos ((1-6 a) t+x)+1},$\\
$w_{14}(x,t)=a+\dfrac{-\dfrac{B^2}{4}-\dfrac{1}{2} \cos ((1-6 a)
   t+x) B-\dfrac{1}{4} \sqrt{B^2-4} \sin ((1-6 a)
   t+x) B}{\dfrac{1}{4} B^2 \cos ^2((1-6 a) t+x)+B
   \cos ((1-6 a) t+x)+1},$\\
where $a$, $c$, $B$, and $\lambda$ are arbitrary constants.
Novelty:New solution(s) / integral(s)
Admin's Comment:This solutions are transformed to known solutions, see
N. A. Kudryashov, On "new travelling wave solutions" of the KdV and the KdV Burgers 
equations, {\it Communications in Nonlinear Science and Numerical Simulation}, 2009, Vol.~14, Issue~5, pp.~1891--1900.
Author/Contributor's Details
Last name:Salas
First name:Alvaro
Middle(s) name:Humberto
Country:Colombia
City:Manizales
Affiliation:Universidad Nacional de Colombia, Universidad de Caldas
Statistic information
Submission date:Mon 19 May 2008 14:21
Edits by author:0

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