How calculate extremes of the functional?
$begingroup$
Is it also here to use the Euler-Lagrange equation? Could someone tell me how it will look like?
$${F}_{u} = int_{0}^{1} left( uu' + uu''^{2} + uu'' + u'u'' + 2u'' right) mbox{d}x$$
$$u(0) = u'(0) = u(1) = 0$$
$$u'(1) = 1$$
functional-analysis euler-lagrange-equation
asked Jan 14 at 12:33
SvsSvs
62
$endgroup$
add a comment |
$begingroup$
Is it also here to use the Euler-Lagrange equation? Could someone tell me how it will look like?
$${F}_{u} = int_{0}^{1} left( uu' + uu''^{2} + uu'' + u'u'' + 2u'' right) mbox{d}x$$
$$u(0) = u'(0) = u(1) = 0$$
$$u'(1) = 1$$
functional-analysis euler-lagrange-equation
asked Jan 14 at 12:33
SvsSvs
62
$endgroup$
add a comment |
$begingroup$
Is it also here to use the Euler-Lagrange equation? Could someone tell me how it will look like?
$${F}_{u} = int_{0}^{1} left( uu' + uu''^{2} + uu'' + u'u'' + 2u'' right) mbox{d}x$$
$$u(0) = u'(0) = u(1) = 0$$
$$u'(1) = 1$$
functional-analysis euler-lagrange-equation
asked Jan 14 at 12:33
SvsSvs
62
$endgroup$
Is it also here to use the Euler-Lagrange equation? Could someone tell me how it will look like?
$${F}_{u} = int_{0}^{1} left( uu' + uu''^{2} + uu'' + u'u'' + 2u'' right) mbox{d}x$$
$$u(0) = u'(0) = u(1) = 0$$
$$u'(1) = 1$$
functional-analysis euler-lagrange-equation
functional-analysis euler-lagrange-equation
asked Jan 14 at 12:33
SvsSvs
62
asked Jan 14 at 12:33
SvsSvs
62
asked Jan 14 at 12:33
SvsSvs
62
asked Jan 14 at 12:33
SvsSvs
62
asked Jan 14 at 12:33
SvsSvs
62
62
add a comment |
add a comment |
2 Answers
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$begingroup$
$${L} = fleft(x,u,u',u''right)$$
$$frac{ partial {L} }{ mbox{d}u } - frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) + frac{ mbox{d}^{2}}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = 0$$
But I don't know how calculate this diverates:
$$frac{ partial {L} }{ mbox{d}u } =$$
$$frac{ partial {L} }{ mbox{d}u' }= $$
$$frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) =$$
$$frac{ partial {L} }{ mbox{d}u'' } = $$
$$frac{ mbox{d}^2}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = $$
answered Jan 15 at 20:00
SvsSvs
62
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$begingroup$
Hint: The 1st, 4th & 5th term in OP's Lagrangian are total derivative terms, which don't contribute to the EL equations. The 2nd & 3rd term is $$L~=~u u^{primeprime}(u^{primeprime}+1),$$ leading to EL equation
$$0~=~frac{partial L}{partial u}+ frac{d^2}{dx^{2}}frac{partial L}{partial u^{primeprime}}~=~(3u^{primeprime}-2)u^{primeprime} +4u^{prime}u^{primeprimeprime} +2u u^{primeprimeprimeprime}.$$
answered Jan 17 at 20:58
QmechanicQmechanic
5,17711858
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2 Answers
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2 Answers
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$begingroup$
$${L} = fleft(x,u,u',u''right)$$
$$frac{ partial {L} }{ mbox{d}u } - frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) + frac{ mbox{d}^{2}}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = 0$$
But I don't know how calculate this diverates:
$$frac{ partial {L} }{ mbox{d}u } =$$
$$frac{ partial {L} }{ mbox{d}u' }= $$
$$frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) =$$
$$frac{ partial {L} }{ mbox{d}u'' } = $$
$$frac{ mbox{d}^2}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = $$
answered Jan 15 at 20:00
SvsSvs
62
$endgroup$
add a comment |
$begingroup$
$${L} = fleft(x,u,u',u''right)$$
$$frac{ partial {L} }{ mbox{d}u } - frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) + frac{ mbox{d}^{2}}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = 0$$
But I don't know how calculate this diverates:
$$frac{ partial {L} }{ mbox{d}u } =$$
$$frac{ partial {L} }{ mbox{d}u' }= $$
$$frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) =$$
$$frac{ partial {L} }{ mbox{d}u'' } = $$
$$frac{ mbox{d}^2}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = $$
answered Jan 15 at 20:00
SvsSvs
62
$endgroup$
add a comment |
$begingroup$
$${L} = fleft(x,u,u',u''right)$$
$$frac{ partial {L} }{ mbox{d}u } - frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) + frac{ mbox{d}^{2}}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = 0$$
But I don't know how calculate this diverates:
$$frac{ partial {L} }{ mbox{d}u } =$$
$$frac{ partial {L} }{ mbox{d}u' }= $$
$$frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) =$$
$$frac{ partial {L} }{ mbox{d}u'' } = $$
$$frac{ mbox{d}^2}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = $$
answered Jan 15 at 20:00
SvsSvs
62
$endgroup$
$${L} = fleft(x,u,u',u''right)$$
$$frac{ partial {L} }{ mbox{d}u } - frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) + frac{ mbox{d}^{2}}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = 0$$
But I don't know how calculate this diverates:
$$frac{ partial {L} }{ mbox{d}u } =$$
$$frac{ partial {L} }{ mbox{d}u' }= $$
$$frac{ mbox{d}}{ mbox{d}x } left( frac{ partial {L} }{ partial u^{'} } right) =$$
$$frac{ partial {L} }{ mbox{d}u'' } = $$
$$frac{ mbox{d}^2}{ mbox{d}x^{2} }left( frac{ partial {L} }{ partial u^{''} } right) = $$
answered Jan 15 at 20:00
SvsSvs
62
answered Jan 15 at 20:00
SvsSvs
62
answered Jan 15 at 20:00
SvsSvs
62
answered Jan 15 at 20:00
SvsSvs
62
62
add a comment |
add a comment |
$begingroup$
Hint: The 1st, 4th & 5th term in OP's Lagrangian are total derivative terms, which don't contribute to the EL equations. The 2nd & 3rd term is $$L~=~u u^{primeprime}(u^{primeprime}+1),$$ leading to EL equation
$$0~=~frac{partial L}{partial u}+ frac{d^2}{dx^{2}}frac{partial L}{partial u^{primeprime}}~=~(3u^{primeprime}-2)u^{primeprime} +4u^{prime}u^{primeprimeprime} +2u u^{primeprimeprimeprime}.$$
answered Jan 17 at 20:58
QmechanicQmechanic
5,17711858
$endgroup$
add a comment |
$begingroup$
Hint: The 1st, 4th & 5th term in OP's Lagrangian are total derivative terms, which don't contribute to the EL equations. The 2nd & 3rd term is $$L~=~u u^{primeprime}(u^{primeprime}+1),$$ leading to EL equation
$$0~=~frac{partial L}{partial u}+ frac{d^2}{dx^{2}}frac{partial L}{partial u^{primeprime}}~=~(3u^{primeprime}-2)u^{primeprime} +4u^{prime}u^{primeprimeprime} +2u u^{primeprimeprimeprime}.$$
answered Jan 17 at 20:58
QmechanicQmechanic
5,17711858
$endgroup$
add a comment |
$begingroup$
Hint: The 1st, 4th & 5th term in OP's Lagrangian are total derivative terms, which don't contribute to the EL equations. The 2nd & 3rd term is $$L~=~u u^{primeprime}(u^{primeprime}+1),$$ leading to EL equation
$$0~=~frac{partial L}{partial u}+ frac{d^2}{dx^{2}}frac{partial L}{partial u^{primeprime}}~=~(3u^{primeprime}-2)u^{primeprime} +4u^{prime}u^{primeprimeprime} +2u u^{primeprimeprimeprime}.$$
answered Jan 17 at 20:58
QmechanicQmechanic
5,17711858
$endgroup$
Hint: The 1st, 4th & 5th term in OP's Lagrangian are total derivative terms, which don't contribute to the EL equations. The 2nd & 3rd term is $$L~=~u u^{primeprime}(u^{primeprime}+1),$$ leading to EL equation
$$0~=~frac{partial L}{partial u}+ frac{d^2}{dx^{2}}frac{partial L}{partial u^{primeprime}}~=~(3u^{primeprime}-2)u^{primeprime} +4u^{prime}u^{primeprimeprime} +2u u^{primeprimeprimeprime}.$$
answered Jan 17 at 20:58
QmechanicQmechanic
5,17711858
answered Jan 17 at 20:58
QmechanicQmechanic
5,17711858
answered Jan 17 at 20:58
QmechanicQmechanic
5,17711858
answered Jan 17 at 20:58
QmechanicQmechanic
5,17711858
5,17711858
add a comment |
add a comment |
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