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There's an integral ${\int\limits_{t_1}^{t_2}}(\frac{\partial{L}}{\partial{q}}{\delta}q+\frac{\partial{L}}{\partial{v}}{\delta}v)dt=0$. [1.]

$ {\delta}v={\frac{d{\delta}q}{dt}}$ [2.]

I should get $ [\frac{\partial{L}}{\partial{v}}{\delta}q]_{t_1}^{t_2}+{\int\limits_{t_1}^{t_2}}(\frac{\partial{L}}{\partial{q}}-\frac{d}{dt}\frac{\partial{L}}{\partial{v}}){\delta}q dt = 0$ [3.] from [1.] using integration by parts and [2.], but I don't know how exactly should I calculate it. This is taken from Landau's and Lifshitz's "Mechanics" more precisely Chapter I, ยง2.

Qmechanic
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Qwedfsf
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2 Answers2

1

Given $$ \int_{t_1}^{t_2}\textrm{d}t\,\left(\frac{\partial L}{\partial q}\,\delta q + \frac{\partial L}{\partial v}\,\delta v \right)= \int_{t_1}^{t_2}\textrm{d}t\,\left(\frac{\partial L}{\partial q}\,\delta q\right) + \int_{t_1}^{t_2}\textrm{d}t\,\left(\frac{\partial L}{\partial v}\,\frac{d}{dt}\delta q \right) $$ then the second contribution on the right hand side can be re-written as $$ \int_{t_1}^{t_2}\textrm{d}t\,\left(\frac{\partial L}{\partial v}\,\frac{d}{dt}\delta q \right) = \int_{t_1}^{t_2}\textrm{d}t\,\left(\frac{d}{dt}\left(\frac{\partial L}{\partial v} \delta q\right)-\delta q\frac{d}{dt}\left(\frac{\partial L}{\partial v} \right) \right). $$ The first piece can be integrated out, being a total derivative, and the second one can be grouped together with the first overall contribution.

P. S. My personal experience is to never read Landau & Lifshitz's to learn from.

hft
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gented
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Partial integration is employed only for the second term in (1): $$\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}\frac{\partial L}{\partial v}\delta v=\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}\frac{d}{dt}\left(\frac{\partial L}{\partial v}\delta q\right)dt-\underset{{t}_{1}}{\overset{{t}_{2}}{\int }}\left(\frac{d}{dt}\frac{\partial L}{\partial v}\right)\delta qdt$$ The first term comes out as $$\frac{\partial L}{\partial v}\delta {q|}_{{t}_{1}}^{{t}_{2}},$$ the second one is left under the integral.

Soba noodles
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