Let $f=f(q,p)$, $g=g(q,p)$ and Possion bracket $$\{f,g\}=\frac{\partial f}{\partial q}\frac{\partial g}{\partial p}-\frac{\partial f}{\partial p}\frac{\partial g}{\partial q}. \tag{1}$$ Then Hamilton's equations for Hamiltonian $H=H(q,p)$ are $$\dot{q}=\{q,H(q,p)\}, \tag{2}$$ $$\dot{p}=\{p,H(q,p)\}. \tag{3}$$ Now let $\mathcal{F}(\mathbb{R})$ be a space of function on $\mathbb{R}$ so that for $u\in\mathcal{F}$ we have $u$ and its derivatives decay at $\pm\infty$. I am trying to construct Hamilton's equations for Hamiltonian $$H=\int_{-\infty}^{\infty}\left(\frac{1}{2}u_x^2-u^3\right)\,dx \tag{4}$$ using Poisson bracket $$\{F,G\}(u)=\int_{-\infty}^{\infty}\frac{\delta F}{\delta u}\frac{d}{dx}\left(\frac{\delta G}{\delta u}\right)\,dx. \tag{5}$$ Is it correct if I take $H=H(u,v)$ with $u=u$ and $v=u_x$? If so then should I leave the Hamilton's equations as $$\dot{u}=\{u,H\}=-\int_{-\infty}^\infty\left(6uu_x+u_{xxx}\right)\,dx=-\int_{-\infty}^\infty\left(6uv+v_{xx}\right)\,dx ,\tag{6}$$ $$\dot{v}=\{v,H\}=\int_{-\infty}^\infty u_{xx}\,dx=\int_{-\infty}^\infty v_{x}\,dx ,\tag{7}$$ or evaluate the integrals?


1 Answer 1

  1. Hint: Eq. (6) in its current form (v4) is meaningless since the lhs. depends on $x$, while the rhs. is integrated over $x$. The functional derivative $$\tag{A}\frac{\delta F}{\delta u(x^{\prime})}~\stackrel{(B)}{=}~\frac{\delta u(x)}{\delta u(x^{\prime})}~=~\delta(x\!-\!x^{\prime})$$ in eq. (6) for the functional $$\tag{B} F[u]~:=~u(x)~=~\int \! dx^{\prime}~u(x^{\prime})~\delta(x^{\prime}\!-\!x)$$ includes a delta function, which removes the integral on the rhs. of eq. (6). The next eq. (7) is similar.

  2. OP asks the following in a comment.

    I can't take $\delta u/\delta u$ as 1?

    OP is apparently thinking of a 'same-$x$' functional derivative $\frac{\delta u(x)}{\delta u(x)}=1$, which is also discussed in e.g. this and this Phys.SE posts. However, the functional derivatives in eq. (5) are not 'same-$x$' functional derivatives, so the answer to the comment is No.

  • $\begingroup$ I can't take $\delta u/\delta u$ as 1? $\endgroup$
    – Sukan
    Apr 9, 2016 at 23:54
  • $\begingroup$ $\uparrow$ @Sukan: No, not in the above formulation. I updated the answer. $\endgroup$
    – Qmechanic
    Apr 9, 2016 at 23:56

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