- #1

JD_PM

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- TL;DR Summary
- I want to understand what 'assuming boundary conditions' means in the context of integration by parts technique. I've stumbled upon the same issue twice and I want to understand it once and for all.

Let's present two examples

$$-\frac 1 2 \int d^3x'\big (-i \phi(x', t)\nabla^2\delta^3(x-x') \big )$$

Explicit evaluation of this integral yields

$$-\frac 1 2 \int d^3x'\big (-i \phi( \vec x', t)\nabla'^2\delta^3(\vec x-\vec x') \big ) =\frac{i}{2}\phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x') -\frac{i}{2} \int d^3 x' \nabla' \phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x')$$

$$=\frac{i}{2}\phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x') - \frac{i}{2} \nabla'\phi(\vec x', t) \delta^3(\vec x-\vec x') + \frac{i}{2} \int d^3 x' \nabla'^2 \phi(\vec x', t) \delta^3(\vec x-\vec x')$$

Now ##\frac{i}{2}\phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x'), - \frac{i}{2} \nabla'\phi(\vec x', t) \delta^3(\vec x-\vec x')## terms vanish 'assuming boundary conditions'. I do not quite get this assertion and I'd like to discuss it in further detail.

The other example is this:

$$\int_{\mathbb{R}^{n}} dx \ \partial^{m+1}f(x) \ \varphi (x) = \partial^{m}f(x) \ \varphi (x)+(-1)\int_{\mathbb{R}^{n}} dx \ \partial^{m}f(x) \ \partial \varphi (x)$$

Applying integration by parts ##m## times one gets

$$\int_{\mathbb{R}^{n}} dx \ \partial^{m+1}f(x) \ \varphi (x) =$$

$$=\partial^{m}f(x) \ \varphi (x)-\partial^{m-1}f(x) \ \partial \varphi (x) +...+(-1)^{|m+1|}\int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{m+1} \varphi (x)$$

Now ##\partial^{m}f(x) \ \varphi (x),-\partial^{m-1}f(x) \ \partial \varphi (x), +...## terms vanish due to the same reason.

This originated from #15 on here.

Thanks

$$-\frac 1 2 \int d^3x'\big (-i \phi(x', t)\nabla^2\delta^3(x-x') \big )$$

Explicit evaluation of this integral yields

$$-\frac 1 2 \int d^3x'\big (-i \phi( \vec x', t)\nabla'^2\delta^3(\vec x-\vec x') \big ) =\frac{i}{2}\phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x') -\frac{i}{2} \int d^3 x' \nabla' \phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x')$$

$$=\frac{i}{2}\phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x') - \frac{i}{2} \nabla'\phi(\vec x', t) \delta^3(\vec x-\vec x') + \frac{i}{2} \int d^3 x' \nabla'^2 \phi(\vec x', t) \delta^3(\vec x-\vec x')$$

Now ##\frac{i}{2}\phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x'), - \frac{i}{2} \nabla'\phi(\vec x', t) \delta^3(\vec x-\vec x')## terms vanish 'assuming boundary conditions'. I do not quite get this assertion and I'd like to discuss it in further detail.

The other example is this:

$$\int_{\mathbb{R}^{n}} dx \ \partial^{m+1}f(x) \ \varphi (x) = \partial^{m}f(x) \ \varphi (x)+(-1)\int_{\mathbb{R}^{n}} dx \ \partial^{m}f(x) \ \partial \varphi (x)$$

Applying integration by parts ##m## times one gets

$$\int_{\mathbb{R}^{n}} dx \ \partial^{m+1}f(x) \ \varphi (x) =$$

$$=\partial^{m}f(x) \ \varphi (x)-\partial^{m-1}f(x) \ \partial \varphi (x) +...+(-1)^{|m+1|}\int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{m+1} \varphi (x)$$

Now ##\partial^{m}f(x) \ \varphi (x),-\partial^{m-1}f(x) \ \partial \varphi (x), +...## terms vanish due to the same reason.

This originated from #15 on here.

Thanks