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The Sine-Gordon equation is: $$u_{xx}-u_{tt}+\sin(u)=0$$ where $u=u(x,t)$. My question is: if I have the following equation: $$u_{xx}-u_{tt}+\sin(u)^k=0$$ where $k=2,3,4,\ldots,N$ is it still possible to find solitonic solutions of this equation for every $k$?

Thanks for hints or answers.

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Do you mean $\sin(u^k)$ or $(\sin u)^k$? –  enzotib Aug 18 '12 at 21:07

1 Answer 1

up vote 1 down vote accepted

In fact this belongs to a PDE of the form http://eqworld.ipmnet.ru/en/solutions/npde/npde2107.pdf.

One method to solve this PDE is by combination of variables:

Let $\xi=\dfrac{(t+C_1)^2}{4}-\dfrac{(x+C_2)^2}{4}$ , where $C_1$ and $C_2$ are arbitrary constants,

Then $\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial\xi}\dfrac{\partial\xi}{\partial t}=\dfrac{t+C_1}{2}\dfrac{\partial u}{\partial\xi}$

$\dfrac{\partial^2u}{\partial t^2}=\dfrac{\partial}{\partial t}\left(\dfrac{t+C_1}{2}\dfrac{\partial u}{\partial\xi}\right)=\dfrac{t+C_1}{2}\dfrac{\partial}{\partial t}\left(\dfrac{\partial u}{\partial\xi}\right)+\dfrac{1}{2}\dfrac{\partial u}{\partial\xi}=\dfrac{t+C_1}{2}\dfrac{\partial}{\partial\xi}\left(\dfrac{\partial u}{\partial\xi}\right)\dfrac{\partial\xi}{\partial t}+\dfrac{1}{2}\dfrac{\partial u}{\partial\xi}=\dfrac{t+C_1}{2}\dfrac{\partial^2u}{\partial\xi^2}\dfrac{t+C_1}{2}+\dfrac{1}{2}\dfrac{\partial u}{\partial\xi}=\dfrac{(t+C_1)^2}{4}\dfrac{\partial^2u}{\partial\xi^2}+\dfrac{1}{2}\dfrac{\partial u}{\partial\xi}$

$\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial\xi}\dfrac{\partial\xi}{\partial x}=-\dfrac{x+C_2}{2}\dfrac{\partial u}{\partial\xi}$

$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(-\dfrac{x+C_2}{2}\dfrac{\partial u}{\partial\xi}\right)=-\dfrac{x+C_2}{2}\dfrac{\partial}{\partial t}\left(\dfrac{\partial u}{\partial\xi}\right)-\dfrac{1}{2}\dfrac{\partial u}{\partial\xi}=-\dfrac{x+C_2}{2}\dfrac{\partial}{\partial\xi}\left(\dfrac{\partial u}{\partial\xi}\right)\dfrac{\partial\xi}{\partial t}-\dfrac{1}{2}\dfrac{\partial u}{\partial\xi}=-\dfrac{x+C_2}{2}\dfrac{\partial^2u}{\partial\xi^2}\left(-\dfrac{x+C_2}{2}\right)-\dfrac{1}{2}\dfrac{\partial u}{\partial\xi}=\dfrac{(x+C_2)^2}{4}\dfrac{\partial^2u}{\partial\xi^2}-\dfrac{1}{2}\dfrac{\partial u}{\partial\xi}$

$\therefore\dfrac{(x+C_2)^2}{4}\dfrac{\partial^2u}{\partial\xi^2}-\dfrac{1}{2}\dfrac{\partial u}{\partial\xi}-\dfrac{(t+C_1)^2}{4}\dfrac{\partial^2u}{\partial\xi^2}-\dfrac{1}{2}\dfrac{\partial u}{\partial\xi}+\sin(u)^k=0$

$\left(\dfrac{(x+C_2)^2}{4}-\dfrac{(t+C_1)^2}{4}\right)\dfrac{\partial^2u}{\partial\xi^2}-\dfrac{\partial u}{\partial\xi}+\sin(u)^k=0$

$-\xi\dfrac{\partial^2u}{\partial\xi^2}-\dfrac{\partial u}{\partial\xi}+\sin(u)^k=0$

$\xi\dfrac{\partial^2u}{\partial\xi^2}+\dfrac{\partial u}{\partial\xi}-\sin(u)^k=0$

Which reduce the problem to an ODE. As the general solution of an ODE only have arbitrary constants as the arbitrary parts, but the general solution of a PDE should contain arbitrary functions as the arbitrary parts, so the above method only find a class of particular solutions rather than general solutions of the PDE.

Another approach to solve this PDE is by change of variables using in the conventional wave equation:

Let $\mu=x+t$ , $\eta=x-t$ ,

Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial\mu}\dfrac{\partial\mu}{\partial x}+\dfrac{\partial u}{\partial\eta}\dfrac{\partial\eta}{\partial x}=\dfrac{\partial u}{\partial\mu}+\dfrac{\partial u}{\partial\eta}$

$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial\mu}+\dfrac{\partial u}{\partial\eta}\right)=\dfrac{\partial}{\partial\mu}\left(\dfrac{\partial u}{\partial\mu}+\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\mu}{\partial x}+\dfrac{\partial}{\partial\eta}\left(\dfrac{\partial u}{\partial\mu}+\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\eta}{\partial x}=\dfrac{\partial^2u}{\partial\mu^2}+\dfrac{\partial^2u}{\partial\mu\eta}+\dfrac{\partial^2u}{\partial\mu\eta}+\dfrac{\partial^2u}{\partial\eta^2}=\dfrac{\partial^2u}{\partial\mu^2}+2\dfrac{\partial^2u}{\partial\mu\eta}+\dfrac{\partial^2u}{\partial\eta^2}$

$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial\mu}\dfrac{\partial\mu}{\partial t}+\dfrac{\partial u}{\partial\eta}\dfrac{\partial\eta}{\partial t}=\dfrac{\partial u}{\partial\mu}-\dfrac{\partial u}{\partial\eta}$

$\dfrac{\partial^2u}{\partial t^2}=\dfrac{\partial}{\partial t}\left(\dfrac{\partial u}{\partial\mu}-\dfrac{\partial u}{\partial\eta}\right)=\dfrac{\partial}{\partial\mu}\left(\dfrac{\partial u}{\partial\mu}-\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\mu}{\partial t}+\dfrac{\partial}{\partial\eta}\left(\dfrac{\partial u}{\partial\mu}-\dfrac{\partial u}{\partial\eta}\right)\dfrac{\partial\eta}{\partial t}=\dfrac{\partial^2u}{\partial\mu^2}-\dfrac{\partial^2u}{\partial\mu\eta}-\dfrac{\partial^2u}{\partial\mu\eta}+\dfrac{\partial^2u}{\partial\eta^2}=\dfrac{\partial^2u}{\partial\mu^2}-2\dfrac{\partial^2u}{\partial\mu\eta}+\dfrac{\partial^2u}{\partial\eta^2}$

$\therefore\dfrac{\partial^2u}{\partial\mu^2}+2\dfrac{\partial^2u}{\partial\mu\eta}+\dfrac{\partial^2u}{\partial\eta^2}-\dfrac{\partial^2u}{\partial\mu^2}+2\dfrac{\partial^2u}{\partial\mu\eta}-\dfrac{\partial^2u}{\partial\eta^2}+\sin(u)^k=0$

$4\dfrac{\partial^2u}{\partial\mu\eta}+\sin(u)^k=0$

Which is still no hope to generally solve this PDE. According to http://eqworld.ipmnet.ru/en/solutions/npde/npde2103.pdf, note that when $\sin(u)^k$ changes to $be^{\beta u}$ , the above approach can find the general solution.

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