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I am trying to find Hamilton's equations for a general Hamiltonian given by $$H[u]=\int_\mathbf{R} \phi(u,u_x)dx$$

Suppose $$\frac{\delta f[u]}{\delta u(x)}\equiv \frac{\partial f}{\partial u}-\frac{\partial }{\partial x}\frac{\partial f}{\partial u_x}+\left(\frac{\partial }{\partial x}\right)^2\frac{\partial f}{\partial u_{xx}}+...$$ then Hamilton's equations are given by $$\frac{\partial u}{\partial t}=\{u,H[u]\}=\frac{\partial }{\partial x}\frac{\delta H[u]}{\delta u(x)}$$

But I don't know how to evaluate the last quantity. Would someone mind teaching me, please?

My guess would be

$$\frac{\partial }{\partial x}\int_\mathbf{R} \frac{\partial \phi}{\partial u}-\frac{\partial }{\partial x}\frac{\partial \phi}{\partial u_x}dx$$

Then $$\left[\frac{\partial \phi}{\partial u}-\frac{\partial }{\partial x}\frac{\partial \phi}{\partial u_x}\right]_{-\infty}^{\infty}$$ Would that be right?

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up vote 0 down vote accepted

What I suspect is stopping OP is different notations for the functional/variational derivative. Let there be given a functional

$$\tag{1} F[u]~=~\int \! dx ~f(x), $$

where the integrand notation $f(x)$ is a short-hand notation for the following function

$$\tag{2} f(x)~=~f(u(x), u^{\prime}(x),u^{\prime\prime}(x),\ldots;x) .$$

If the underlying variational problem (with given boundary conditions) is well-posed, then the functional derivative exists, and is usually denoted $$\tag{3} \frac{\delta F}{\delta u(x)}. $$

It is given by the Euler-Lagrange formula

$$\tag{4} \frac{\delta F}{\delta u(x)} ~=~ \frac{\partial f(x)}{\partial u(x)}-\frac{d}{dx}\left( \frac{\partial f(x)}{\partial u^{\prime}(x)}\right)+ \frac{d^2}{dx^2}\left( \frac{\partial f(x)}{\partial u^{\prime\prime}(x)}\right)+\ldots, $$

Unfortunately, another notation for the functional derivative is also often used

$$\tag{5} \frac{\delta f(x)}{\delta u(x)} ~:=~ \frac{\partial f(x)}{\partial u(x)}-\frac{d}{dx}\left( \frac{\partial f(x)}{\partial u^{\prime}(x)}\right)+ \frac{d^2}{dx^2}\left( \frac{\partial f(x)}{\partial u^{\prime\prime}(x)}\right)+\ldots. $$

Comparing eqs. (4) and (5), it becomes easy to confuse the integrand $f(x)$ with the integral $F$.

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Aha! Thanks, Qmechanic!! – valerie v. May 18 '13 at 10:19

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