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The variational problem is following:

$$ J[u,x] = \int _{ t _{ 0 } } ^{ t _{ 1 } } u \left( x \left( t \right) , t \right) {\text{d}t}$$

So how to calculate the $\delta J$ ?


I am new to this subject, and one immature thought is:

$$ \begin{aligned}\delta J & = \int _{ t _{ 0 } } ^{ t _{ 1 } } \left( u + \delta u \right) \left[ x + \delta x , t \right] - u \left( x , t \right) {\text{d}t} \\ & = \int _{ t _{ 0 } } ^{ t _{ 1 } } \left( \frac{ \partial u }{ \partial x } \delta x + \delta u \right) {\text{d}t}\end{aligned} $$

I don't know how to handle the $\delta u$

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Your reasoning is correct, you just need to unpack the notation a bit. You have shown that $$ \delta J[u,x](\delta u, \delta x) = \int_{t_0}^{t_1}u'(x(t),t)\delta x(t) + \delta u(x(t),t)~dt. $$ You can think of this as a directional derivative in the direction $(\delta u,\delta x)$. Since $J$ depends on two functions, it makes sense that the directional derivative will contain terms relating to the perturbation in each input, both $u$ and $x$.

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  • $\begingroup$ Thank you! I will try to solve my problem in this way :) $\endgroup$
    – randolf
    May 18 at 6:31

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