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--- basic_tools:variational_calculus:the_variational_problem [2018/03/14 15:39]
iiqof
+++ basic_tools:variational_calculus:the_variational_problem [2018/03/14 16:25] (current)
iiqof Clarified things
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-\int_a^b  \frac{\partial F}{\partial q}\phi + \frac{\partial F}{\partial \dot q}\dot\phi dt = \frac{d}{dt}\left[\frac{\partial F}{\partial \dot q}\phi\right]^b_a + \int_a^b  \left(\frac{\partial F}{\partial q} - \frac{d}{dt}\frac{\partial F}{\partial \dot q}\right)\phi dt
+\int_a^b  \frac{\partial F}{\partial q}\phi + \frac{\partial F}{\partial \dot q}\dot\phi dt = \left[\frac{\partial F}{\partial \dot q}\phi\right]^b_a + \int_a^b  \left(\frac{\partial F}{\partial q} - \frac{d}{dt}\frac{\partial F}{\partial \dot q}\right)\phi dt
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-Saying that, now the [[basic_tools:variational_calculus:fundamental_lemma|fundamental lemma of variational calculus]] enters. The variation, $\phi$, is arbitrary, this means by the interior of the parenthesis is equal to zero, giving:
+Saying that, now the [[basic_tools:variational_calculus:fundamental_lemma|fundamental lemma of variational calculus]] enters. The variation, $\phi$, is arbitrary but  with $\phi(a)=\phi(b)=0$. The later assertion, makes the term outside the integral zero. On the other hand, by the fundamental lemma, the interior parentheisi is zero, giving:
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