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My Question: Find first and second variation of following functional $$J(u)=\int_{t_0}^{t_1}(1+u'^2)\sqrt{u}dt$$

My Answer: I found first variation of the functional as $$\delta J=\int_{t_0}^{t_1}\big[\frac{(1-u'^2)}{2\sqrt{u}}- 2u''\sqrt{u} \big]\delta u dt$$ Is it right?

But I can' t find second variation of the functional. Could you help me?

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1 Answer

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Your expression for the first variation is right. The integrand for the second variation is given by

$$ \frac14\frac{(\frac{d}{dt}u(t))^2-4(\frac{d^2}{dt^2}u(t))u(t)-1}{u(t)^{3/2}} $$

This result is obtained using the Euler-Lagrange equation with the form

$$\frac{\partial L}{\partial u}-\frac{d}{dt}\left(\frac{\partial L}{\partial u'}\right)+\frac{d^2}{dt}\left(\frac{\partial L}{\partial u''}\right)=0$$

where $L$ is given by the integrand in the first variation

$${\color{blue}L}=\frac{(1-u'^2)}{2\sqrt u}-2u''\sqrt{u}$$

More formally we write

$$\delta^2 J=\int_{t_0}^{t_1}\big[\frac14\frac{(\frac{d}{dt}u(t))^2-4(\frac{d^2}{dt^2}u(t))u(t)-1}{u(t)^{3/2}}]\delta^2 u dt$$

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