Derivation of an integral identity from the kdv equation.

[Strum]

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 :)

 

[pasmith]

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
$$ \int_{0}^{z}\left(\bar{u}(y) - \frac{2}{3}\right)\int_{0}^{y}\left(\bar{u}(x) - \frac{2}{3}\right)dxdy = \bar{u}(z) + \frac{4}{3} $$

Are you sure? The left hand side is, using the express solution, <br /> 4 \int_0^z \mathrm{sech}^2 (y) \int_0^y \mathrm{sech}^2 (x) \,dx\,dy<br /> = 4 \int_0^z \mathrm{sech}^2 (y) \tanh (y)\,dy = 2\tanh^2 (z) = 2(1 + \mathrm{sech}^2 (z)) but the right hand side is
2(1 - \mathrm{sech}^2(z)).

 

[Strum]

$\tanh(x)^{2} = 1- \sech(x)^{2}$

 

[pasmith]

$\tanh(x)^{2} = 1- \sech(x)^{2}$

Oops

On further analysis, I suspect that the integral identity is a consequence of \bar u - \frac23 being a multiple of \mathrm{sech}^2(x), not of \bar u being a solution of the ODE.

If you look for solutions of the ODE which satisfy u&#039;&#039; \to 0, u&#039; \to 0 and u \to u_0 as |x| \to \infty then integrating once yields <br /> 3u^2 = u&#039;&#039; + 3u_0^2. Multiplying by u&#039; and integrating once more yields <br /> \frac12 u&#039;^2 = u^3 - 3u_0^2u + 2u_0^3 = (u - u_0)^2(u + 2u_0). A further messy integration yields u(x) = u_0 - 3u_0 \mathrm{sech}^2 \left(\sqrt{\frac{3u_0}{2}} x\right).

 

[pasmith]

I have worked out how to get the identity. We start from <br /> \frac12 u&#039;^2 = (u - u_0)^2(u + 2u_0) and make two assumptions: firstly that -2u_0 \leq u(x) &lt; u_0 and secondly that u(x) attains its minimum at x = 0. Thus u&#039;(x) &gt; 0 in x &gt; 0 and u(0) = -2u_0.

Starting from <br /> \frac12 u&#039;^2 = (u - u_0)^2(u + 2u_0) we take the square root and divide by u + 2u_0 to obtain <br /> \frac{u&#039;}{\sqrt 2 \sqrt{u + 2u_0}} = u_0 - u. Integrating once then yields <br /> \left[\sqrt{2}\sqrt{u(x) + 2u_0}\right]_0^y = -\int_0^y u(x) - u_0\,dx. Since by assumption u(0) + 2u_0 = 0 we have <br /> \sqrt{2}\sqrt{u(y) + 2u_0} = -\int_0^y u(x) - u_0\,dx. Multiplying by u_0 - u(y) then gives <br /> \sqrt{2}\sqrt{u(y) + 2u_0}(u_0 - u(y)) = u&#039;(y) = (u(y)- u_0) \int_0^y u(x) - u_0\,dx. A further integration then yields
<br /> u(z) - u(0) = \int_0^z (u(y)- u_0) \int_0^y u(x) - u_0\,dx\,dy<br /> and since u(0) = -2u_0 we obtain finally <br /> \int_0^z (u(y)- u_0) \int_0^y u(x) - u_0\,dx\,dy = u(z) + 2u_0 as required.

 

[Strum]

Thank you very much! I've just this minute completed my own similar solution from the simplicfications you did earlier.
Start from
$$ \frac{u'^{2}}{(u-u_{0})^{2}} = 2(u + 2u_{0}) $$
Then use the identity $$\partial_{x}\left( \frac{u'}{u-u_{0}}\right) = \frac{u''}{u-u_{0}} - \frac{u'^{2}}{(u-u_{0})^{2}} $$ and
$$u'' = 3(u^{2}-u_{0}^{2}) $$
to find $$\partial_{x}\left( \frac{u'}{u-u_{0}}\right) = u-u_{0}$$
Integrate twice and we get
$$ u = \int_{0}^{z}(u(y) - u_{0})\left( \int_{0}^{y}(u(x) - u_{0} )dx+ u'(0)\right) dy + u(0) $$

So all in all the conclusion must be: yes, you can derive the identity without using the explicit form of $\bar{u}$ as long as we specify some boundary conditions. This was quite a nice problem :)