I Checking if a stationary point is a minimum using Lagrangian Mechanics

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To determine if a stationary point is a minimum in Lagrangian mechanics, one can use the second variation, which is analogous to the second derivative test in standard calculus. The discussion highlights a practice question where the stationary point of an integral needs to be verified as a minimum, with the initial approach involving a second derivative test on the function f(x'). The poster questions whether the condition S(a) > S_actual can be applied to I(a) in the context of Lagrangian mechanics. They express confusion about the appropriate methods for analyzing minima in this framework, indicating a need for clarification on the topic. Understanding the second variation is crucial for correctly identifying minima in Lagrangian problems.
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I'm having trouble understanding how to find out whether or not a stationary point is a minimum and I'm hoping for some clarification. In my class, we were shown that, using Euler's equation, the straight-line path:
Screenshot 2023-02-05 18.16.34.png

with constants a and b results in a stationary point of the integral:
Screenshot 2023-02-05 18.16.47.png

A certain practice question then asks to show that the stationary point corresponds to a minimum. My only attempt so far was performing a simple second derivative test on the function f(x') which turned out to be successful. However, I'm wondering if this is the only way to solve such a problem. I know that a minimum is satisfied if S(a) > S_actual, but can that same idea be mapped onto I(a), that is, is a minimum achieved if I(a) > I_actual (if that even makes sense)? I'm very new to Lagrangian mechanics and find it kind of overwhelming so forgive me if this is a silly question. It just seems that I took the calculus way of solving this when that may not be the ideal method for a class based on Lagrangian mechanics/. I appreciate any help/advice!
 
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Try googling "2nd variation in Lagrangian mechanics". (This is analog of 2nd derivatives in ordinary calculus.)
 
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