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PostPosted: Mon, 9 Apr 2012 12:23:34 UTC 
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Joined: Thu, 22 Jun 2006 01:08:00 UTC
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hi, i am confused with this integral, the problem reads:

given the D.E. \partial _t u+\partial^3 _x u+u \partial_x u =0 (KdV equation), denote the soliton solution by S(x-\lambda t) where S represents the shape of the soliton and \lambda the speed. Write the equation for S by substituting in the soliton equation, integrate once and show that the constant of integration vanishes using the fact that the S has to decay at infinity.

what i have done
taking v=(x-\lambda t)
\partial _t S=-\lambda \partial _v S dt
\partial _x S=\partial _v S dx
\partial^3 _x S=\partial^3 _v S dx

so he soliton equation reads -\lambda \partial _v S dt+\partial^3 _v S dx+S \partial _v S dx=0
And then I don't know what to do, any idea?

thanks!


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PostPosted: Mon, 9 Apr 2012 12:28:10 UTC 
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
ender wrote:
hi, i am confused with this integral, the problem reads:

given the D.E. \partial _t u+\partial^3 _x u+u \partial_x u =0 (KdV equation), denote the soliton solution by S(x-\lambda t) where S represents the shape of the soliton and \lambda the speed. Write the equation for S by substituting in the soliton equation, integrate once and show that the constant of integration vanishes using the fact that the S has to decay at infinity.

what i have done
taking v=(x-\lambda t)
\partial _t S=-\lambda \partial _v S dt
\partial _x S=\partial _v S dx
\partial^3 _x S=\partial^3 _v S dx

so he soliton equation reads -\lambda \partial _v S dt+\partial^3 _v S dx+S \partial _v S dx=0
And then I don't know what to do, any idea?

thanks!


What are the dt,dx doing here? Once you correct that, follow the instruction
Quote:
integrate once and ...

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Mon, 9 Apr 2012 12:46:43 UTC 
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Joined: Thu, 22 Jun 2006 01:08:00 UTC
Posts: 135
well yes here dx=1 and dt=1 but still i get then the equation

-\lambda \partial_v S+\partial^3_v S+S \partial S=0

integrating we have
-\lambda S+\partial^2_v S+\frac{S^2}{s}=0
right?
but i think the solution i should get is

-\frac{1}{2}u^2-u_{xx}


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PostPosted: Mon, 9 Apr 2012 12:49:29 UTC 
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6635
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
ender wrote:
well yes here dx=1 and dt=1 but still i get then the equation

-\lambda \partial_v S+\partial^3_v S+S \partial S=0

integrating we have
-\lambda S+\partial^2_v S+\frac{S^2}{s}=0
right?
but i think the solution i should get is

-\frac{1}{2}u^2-u_{xx}


Huh? What is this mysterious s you have in the denominator here? And you haven't shown why the RHS is zero...

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Mon, 9 Apr 2012 13:08:14 UTC 
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Joined: Thu, 22 Jun 2006 01:08:00 UTC
Posts: 135
sorry the denominator is not an S is a 2, (typo). The RHS is 0 because it the profile decays at x=infinity and this also holds fot t=0, it follows from this.


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