I need to find an extremal of the functional $I[u] = \int_{0}^{1}(u'')^2dx - (u(0) - u(1))^2 \\ $, without any boundary conditions. Here's my solution:
$ \\ \frac{d I[u+th]}{dt}|_{t=0}=2u''h'|^{1}_{0} -2u'''h|^{1}_{0}+2(u(1)-u(0))h|^{1}_{0}+2\int_{0}^{1}u^{(4)}hdx.$ When$\ \ h\in \mathrm{C}^{\infty }[0, 1], h(0)=h(1). \\ $ By the equation, we get $2\int_{0}^{1}u^{(4)}hdx=0 \Rightarrow u(x)=\mathrm{C}_{1}\frac{x^3}{6}+\mathrm{C}_{2}\frac{x^2}{2}+\mathrm{C}_{3}x+\mathrm{C}_{4}. \\ $
And if $ h(0)=h(1) \Rightarrow u(1)-u^{(3)}(1)-u(0)+u^{(3)}=0\Rightarrow \mathrm{C}_{1} = 0, \mathrm{C}_{2}=2\mathrm{C}_{3}.\\ $
So the answer is $ u(x)=x^2+\mathrm{C}_{3}x+\mathrm{C}_{4} \\ $. Is it Correect?
Here is one more question, how to find extremal of functional $ \int_{0}^{1}udx$, using previous functional with Lagrange multiplier? Thanks in advance!
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stboy
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How did you get $C_2 = 2$? – A rural reader Apr 27 '23 at 14:27
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@Aruralreader $u^{(3)}$ is constant so we get $u(0)=u(1) $ and by $u''h'|^{1}{0} = 0$ we have $ u''(0) = u''(1)$. Unfortunately I don't have proof of $h'(0) = h'(1)$. So, $ \mathrm{C}{2} = 2\mathrm{C}_{3}$ – stboy Apr 27 '23 at 15:10
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@Gonçalo Seems to be right but how can it be proven or where I have made a mistake? – stboy Apr 27 '23 at 21:24
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@stboy: Goncalo's examples are instructive. The usual argument is that, if $u$ is an extremum or stationary point of the functional, it will satisfy some condition like an ODE. The catch is that it requires an extremum or stationary point to exist in the first place. – A rural reader Apr 27 '23 at 22:04
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@Gonçalo It's clear, thanks very much! – stboy Apr 29 '23 at 12:47