- #1
Strum
- 105
- 4
Hi everybody! First post!(atleast in years and years).
The stationary KdV equation given by
$$ 6u(x)u_{x} - u_{xxx} = 0 $$.
It has a solution given by
$$ \bar{u}(x)=-2\sech^{2}(x) + \frac{2}{3} $$
This solution obeys the indentity
$$ \int_{0}^{z}\left(\bar{u}(y) - \frac{2}{3}\right)\int_{0}^{y}\left(\bar{u}(x) - \frac{2}{3}\right)dxdy = \bar{u}(z) + \frac{4}{3} $$
Is it possible to derive this kind of identity without using the explicit form of the solution ##\bar{u}(x) ##? That is only by using the fact that ##\bar{u}(x) ## is a solution to (1). I tried differentiating (3) thrice on the left side but that generates a lot of different terms for which it is hard to use (1) to simplify. I really need some hints for this problem.
Thank you in advance :)
The stationary KdV equation given by
$$ 6u(x)u_{x} - u_{xxx} = 0 $$.
It has a solution given by
$$ \bar{u}(x)=-2\sech^{2}(x) + \frac{2}{3} $$
This solution obeys the indentity
$$ \int_{0}^{z}\left(\bar{u}(y) - \frac{2}{3}\right)\int_{0}^{y}\left(\bar{u}(x) - \frac{2}{3}\right)dxdy = \bar{u}(z) + \frac{4}{3} $$
Is it possible to derive this kind of identity without using the explicit form of the solution ##\bar{u}(x) ##? That is only by using the fact that ##\bar{u}(x) ## is a solution to (1). I tried differentiating (3) thrice on the left side but that generates a lot of different terms for which it is hard to use (1) to simplify. I really need some hints for this problem.
Thank you in advance :)