Undergrad Assuming boundary conditions when integrating by parts

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SUMMARY

The discussion focuses on the application of boundary conditions in integration by parts, specifically in the context of evaluating integrals involving delta functions and derivatives. Two examples illustrate how terms vanish under certain boundary conditions, which is crucial for simplifying expressions. The first example involves the integral of a product of a function and the Laplacian of a delta function, while the second example demonstrates the iterative application of integration by parts. Both examples highlight the importance of boundary conditions in determining the validity of the resulting expressions.

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  • Understanding of integration by parts in calculus
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  • Knowledge of partial derivatives and their applications
  • Basic concepts of boundary conditions in mathematical analysis
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JD_PM
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TL;DR
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 :smile:
 
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It means exactly what it says. Here is an example.

##\int_a^bf(x)g'(x)dx=f(x)g(x)|_a^b-\int_a^bf'(x)g(x)dx=f(b)g(b)-f(a)g(a)-\int_a^bf'(x)g(x)dx##

if at least on of the functions vanishes on the boundary, the first two terms in the last expression will be zero.
 
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