3
$\begingroup$

Exercise :

Show that the integral $\int (xu-3tu^2) \mathrm{d}x$ remains unchanged (is a constant of motion) for the equation KdV.

Attempt :

Let $u$ be a solution of the KdV equation, thus satisfying :

$$u_t + u_{xxx} + 6uu_x = 0$$

Let also $I = \int_{-\infty}^\infty (xu-3tu^2) \mathrm{d}x$ and then the derivative with respect to time, will be :

$$\frac{\mathrm{d}I}{\mathrm{d}t}=\int_{-\infty}^\infty\frac{\partial}{\partial t}(xu-3tu^2)\mathrm{d}x=\int_{-\infty}^\infty(xu_t-3u^2-6tuu_t)\mathrm{d}x$$ $$=$$ $$\int_{-\infty}^\infty (-xu_{xxx} - 6xuu_x - 3u^2 + 6tuu_{xxx}+36tu^2u_x)\mathrm{d}x$$

How would one proceed now to show that $\frac{\mathrm{d}I}{\mathrm{d}x} = 0$, thus the expressions above are equal to $0$ ?

$\endgroup$
  • $\begingroup$ Are $u$ and $t$ functions of $x$? $\endgroup$ – Aqua May 31 '18 at 9:50
  • $\begingroup$ @ChristianF The function $u$ is a solution of the KdV equation as $u(x,t)$. $t$ and $x$ are variables. I meant differentiation by $t$ by the way, thus $\frac{\mathrm{d}I}{\mathrm{d}t}$. I corrected the title expression, body and calculations are okay and correct. Sorry for that ! $\endgroup$ – Rebellos May 31 '18 at 9:52
3
$\begingroup$

Starting from $$ \int_{-\infty}^{\infty} (xu_t - 3u^2-6tuu_t) \, dx, $$ the second term can be integrated by parts: $$ \int_{-\infty}^{\infty} -3u^2 \, dx = [-3xu^2]_{-\infty}^{\infty} + \int_{-\infty}^{\infty} 6xuu_x \, dx = \int_{-\infty}^{\infty} 6xuu_x \, dx, $$ since the boundary terms must vanish for $\int xu$ to be finite. We can apply KdV to replace the first two terms of the whole integral: $$ \int_{-\infty}^{\infty} (xu_t +6xuu_x -6tuu_t) \, dx = \int_{-\infty}^{\infty} (-xu_{xxx}-6tuu_t) \, dx = \int_{-\infty}^{\infty} (u_{xx}-6tuu_t) \, dx , $$ where for the last equality we have integrated by parts again. The first term is the $x$-derivative of $u_x$, so vanishes when integrated, leaving us with $$ -6t\int_{-\infty}^{\infty} uu_t \, dx $$ Lastly, $uu_t = -uu_{xxx}-3u^2u_x$. The second term of this is obviously the derivative of $-u^3$, so vanishes when integrated. The other term can be integrated by parts, $$ \int -uu_{xxx} \, dx = 0 + \int u_x u_{xx} \, dx = \int \frac{1}{2}(u_x^2)_x \, dx, $$ and so is also a derivative. Hence all the terms vanish.


Alternatively, we can say: $$ \int_{-\infty}^{\infty} (xu_t - 3u^2-6tuu_t) \, dx = \int_{-\infty}^{\infty} ((x-6tu)(-u_{xxx}-6uu_x) - 3u^2 ) \, dx \\ = \int_{-\infty}^{\infty} (-(x-6tu)u_{xxx}-6xuu_x + 36tu^2u_x - 3u^2 ) \, dx \\ = \int_{-\infty}^{\infty} ( (1-6tu_x)u_{xx}-((x-6tu)u_{xx})_x - 3(xu^2)_x + 3u^2+ 12t(u^3)_x - 3u^2 ) \, dx \\ = \int_{-\infty}^{\infty} (u_x-3tu_x^2 -(x-6tu)u_{xx} - 3xu^2 +12tu^3 )_x \, dx \\ = [u_x-3tu_x^2 -(x-6tu)u_{xx} - 3xu^2 +12tu^3 ]_{-\infty}^{\infty} = 0, $$ where the first equality uses KdV, the second regroups terms, the third uses the product rule in the form $f'g=(fg)'-fg'$ on the first three terms, the fourth writes everything possible as derivatives, the fifth from integrating and the final one comes from the decay at $\infty$.

$\endgroup$
  • $\begingroup$ Can you explain more thoroughly or present your thoughts and the calculations ? I seem to get lost without seeing the compact integration by parts calculation and I also got lost by trying to figure out what "term" I must follow each time. Much appreciated ! $\endgroup$ – Rebellos May 31 '18 at 12:45
  • $\begingroup$ I've updated my answer. $\endgroup$ – Chappers May 31 '18 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.