# Integration by parts in the time derivative of a functional

I have a question regarding some calculations in Torres del Castillo's paper "Hamiltonian structures for classical fields". Let $$\phi_{(a)}$$ ($$a=1,2,\dots,n$$) be the variables that determine the state of a field and let $$x ^ i$$ ($$i=1,2,3$$) be any curvilinear coordinates. If $$F$$ is a functional given by $$F=\int\text{d}^3x\,\mathcal{F}(\phi_{(a)},\partial_i\phi_{(a)},\partial_i\partial_j\phi_{(a)},\dots,x^i,t).$$ then the time derivative of $$F$$ is $$\frac{\text{d}F}{\text{d}t}=\int\text{d}^3x\left[\frac{\partial\mathcal{F}}{\partial\phi_{(a)}}\dot{\phi}_{(a)}+\frac{\partial\mathcal{F}}{\partial(\partial_i\phi_{(a)})}\partial_i\dot{\phi}_{(a)}+\dots+\frac{\partial\mathcal{F}}{\partial{t}}\right].$$ The author states that, by performing an integration by parts and supossing that the derivatives of $$\mathcal{F}$$ are square-integrable functions, one could arrive to the expression $$\frac{\text{d}F}{\text{d}t}=\int\text{d}^3x\left[\frac{\partial\mathcal{F}}{\partial\phi_{(a)}}+\left(\frac{\text{d}}{\text{d}x^i}\right)^\dagger\frac{\partial\mathcal{F}}{\partial(\partial_i\phi_{(a)})}+\dots\right]\dot{\phi}_{(a)}+\int\text{d}^3x\,\frac{\partial\mathcal{F}}{\partial{t}},$$ where $$(\text{d}/\text{d}x^i)^\dagger$$ is the adjoint operator of $$\text{d}/\text{d}x^i$$, defined by $$\int\text{d}^3x\,f^*(\mathbf{x})\left(\frac{\text{d}}{\text{d}x^i}\right)g(\mathbf{x})=\int\text{d}^3x\,\left[\left(\frac{\text{d}}{\text{d}x^i}\right)^\dagger{f}(\mathbf{x})\right]^*g(\mathbf{x}),$$ where $$f$$ and $$g$$ are arbitrary square-integrable functions. Could someone explain to me how to get to that expression?

EDIT: Apparently, the result can be obtained from the definition of $$(\text{d}/\text{d}x^i)^\dagger$$, considering that $$\left(\frac{\text{d}}{\text{d}x^i}\right)\phi_{(a)}=\partial_i\phi_{(a)}.$$ Why should I do this consideration? When is this consideration not met?

• Please consider sharing the link to the paper. Nov 2 '18 at 23:55
– NarcosisGF
Nov 3 '18 at 5:34

The key is that, when we restrict ourselves to square-integrable functions, the adjoint operator of $$\frac{d}{dx^i}$$ is simply $$-\frac{d}{dx^i}$$. Indeed:

$$\begin{equation} \int d^3x f^*(\mathbf{x})\frac{d}{dx^i}g(\mathbf{x}) = \int d^3x g(\mathbf{x})\left(-\frac{d}{dx^i}f(\mathbf{x})\right)^* \ , \end{equation}$$

because surface terms vanish. Since the derivatives of $$\cal{F}$$ are square integrable by hypothesis, $$\frac{\partial\mathcal{F}}{\partial(\partial_i\phi_{(a)})}\rightarrow 0$$ as $$\mathbf{x}$$ goes to infinity. Given that the time derivative of a physical field cannot go to infinity, surface terms cannot survive, and we must have

$$\begin{equation} \int d^3x \partial_i\dot{\phi} \frac{\partial\mathcal{F}}{\partial(\partial_i\phi)} =\int d^3x \dot{\phi_{}}\left(-\frac{d}{dx^i} \frac{\partial\mathcal{F}}{\partial(\partial_i\phi)}\right), \end{equation}$$ from which the desired result follows.

• There is something that is not clear to me about your reasoning. Are you considering that $\partial_i\dot{\phi}_{(a)}\equiv(\text{d}/\text{d}x^i)\dot{\phi}_{(a)}$?
– NarcosisGF
Nov 3 '18 at 0:31
• I wouldn't use the $\equiv$ symbol, but yeah, since a field is a function $\phi=\phi(t,\mathbf{x})$, and the only component dependent on $x_i$ is $x_i$ itself, there is only explicit dependence, and thus the derivatives are equal in this case.
– Othin
Nov 3 '18 at 0:42