# Derivation of an integral identity from the kdv equation.

1. May 22, 2014

### 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 with out 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.

2. May 22, 2014

### pasmith

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

3. May 22, 2014

### Strum

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

4. May 22, 2014

### pasmith

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 $u'' \to 0$, $u' \to 0$ and $u \to u_0$ as $|x| \to \infty$ then integrating once yields $$3u^2 = u'' + 3u_0^2.$$ Multiplying by $u'$ and integrating once more yields $$\frac12 u'^2 = u^3 - 3u_0^2u + 2u_0^3 = (u - u_0)^2(u + 2u_0).$$ A further messy integration yields $$u(x) = u_0 - 3u_0 \mathrm{sech}^2 \left(\sqrt{\frac{3u_0}{2}} x\right).$$

Last edited: May 22, 2014
5. May 22, 2014

### pasmith

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

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

6. May 22, 2014

### 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 :)