extremal problem-how to check istrong minima,maxima condition
The functional $I[y(x)]=int_{0}^{2}(xy^{'}+y^{'2})dx$,y(0)=1,y(2)=0
possess
a.strong minima
b.strong maxima
c.strong maxima but not weak minima
d.weak maxima but not strong minima
How do we show if the functiona is strong minima,maxima..how do we prove this?
what all we have to check?
calculus-of-variations
asked Sep 1 '14 at 10:04
amit
107312
add a comment |
The functional $I[y(x)]=int_{0}^{2}(xy^{'}+y^{'2})dx$,y(0)=1,y(2)=0
possess
a.strong minima
b.strong maxima
c.strong maxima but not weak minima
d.weak maxima but not strong minima
How do we show if the functiona is strong minima,maxima..how do we prove this?
what all we have to check?
calculus-of-variations
asked Sep 1 '14 at 10:04
amit
107312
add a comment |
The functional $I[y(x)]=int_{0}^{2}(xy^{'}+y^{'2})dx$,y(0)=1,y(2)=0
possess
a.strong minima
b.strong maxima
c.strong maxima but not weak minima
d.weak maxima but not strong minima
How do we show if the functiona is strong minima,maxima..how do we prove this?
what all we have to check?
calculus-of-variations
asked Sep 1 '14 at 10:04
amit
107312
The functional $I[y(x)]=int_{0}^{2}(xy^{'}+y^{'2})dx$,y(0)=1,y(2)=0
possess
a.strong minima
b.strong maxima
c.strong maxima but not weak minima
d.weak maxima but not strong minima
How do we show if the functiona is strong minima,maxima..how do we prove this?
what all we have to check?
calculus-of-variations
calculus-of-variations
asked Sep 1 '14 at 10:04
amit
107312
asked Sep 1 '14 at 10:04
amit
107312
asked Sep 1 '14 at 10:04
amit
107312
asked Sep 1 '14 at 10:04
amit
107312
asked Sep 1 '14 at 10:04
amit
107312
107312
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
It has Strong minima.
Legendre Condition: $F_{y'y'}$ (differentiate F w.r.t y' twice) >0 for every y then strong minima and if$ F_{y'y'}<0$ for every y' then Strong maxima.
for >0 for some y' close to p=(dy/dx, y is extremal) weak minima and for <0 for some y' close to p=(dy/dx, y is extremal) weak maxima
answered Jan 23 '15 at 9:34
user3057692
11
1
@user30567692 I tried to edit your answer but the I could not understand the second half please try to use LATEX
– happymath
Jan 23 '15 at 9:52
In your example, Fy'y' = 2>0 for every y'. Therefore it is strong minima. Let us assume thet for some problem your Fy'y' = y'+x, Now its sign is not clear and it depend on sign of p=dy/dx (where y is extremal) and range of x, So it is definitaley week ( maxima or minima will depend on sign)
– user3057692
Mar 14 '15 at 16:10
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It has Strong minima.
Legendre Condition: $F_{y'y'}$ (differentiate F w.r.t y' twice) >0 for every y then strong minima and if$ F_{y'y'}<0$ for every y' then Strong maxima.
for >0 for some y' close to p=(dy/dx, y is extremal) weak minima and for <0 for some y' close to p=(dy/dx, y is extremal) weak maxima
answered Jan 23 '15 at 9:34
user3057692
11
1
@user30567692 I tried to edit your answer but the I could not understand the second half please try to use LATEX
– happymath
Jan 23 '15 at 9:52
In your example, Fy'y' = 2>0 for every y'. Therefore it is strong minima. Let us assume thet for some problem your Fy'y' = y'+x, Now its sign is not clear and it depend on sign of p=dy/dx (where y is extremal) and range of x, So it is definitaley week ( maxima or minima will depend on sign)
– user3057692
Mar 14 '15 at 16:10
add a comment |
It has Strong minima.
Legendre Condition: $F_{y'y'}$ (differentiate F w.r.t y' twice) >0 for every y then strong minima and if$ F_{y'y'}<0$ for every y' then Strong maxima.
for >0 for some y' close to p=(dy/dx, y is extremal) weak minima and for <0 for some y' close to p=(dy/dx, y is extremal) weak maxima
answered Jan 23 '15 at 9:34
user3057692
11
1
@user30567692 I tried to edit your answer but the I could not understand the second half please try to use LATEX
– happymath
Jan 23 '15 at 9:52
In your example, Fy'y' = 2>0 for every y'. Therefore it is strong minima. Let us assume thet for some problem your Fy'y' = y'+x, Now its sign is not clear and it depend on sign of p=dy/dx (where y is extremal) and range of x, So it is definitaley week ( maxima or minima will depend on sign)
– user3057692
Mar 14 '15 at 16:10
add a comment |
It has Strong minima.
Legendre Condition: $F_{y'y'}$ (differentiate F w.r.t y' twice) >0 for every y then strong minima and if$ F_{y'y'}<0$ for every y' then Strong maxima.
for >0 for some y' close to p=(dy/dx, y is extremal) weak minima and for <0 for some y' close to p=(dy/dx, y is extremal) weak maxima
answered Jan 23 '15 at 9:34
user3057692
11
It has Strong minima.
Legendre Condition: $F_{y'y'}$ (differentiate F w.r.t y' twice) >0 for every y then strong minima and if$ F_{y'y'}<0$ for every y' then Strong maxima.
for >0 for some y' close to p=(dy/dx, y is extremal) weak minima and for <0 for some y' close to p=(dy/dx, y is extremal) weak maxima
answered Jan 23 '15 at 9:34
user3057692
11
edited Jan 29 '15 at 9:12
answered Jan 23 '15 at 9:34
user3057692
11
answered Jan 23 '15 at 9:34
user3057692
11
answered Jan 23 '15 at 9:34
user3057692
11
11
1
@user30567692 I tried to edit your answer but the I could not understand the second half please try to use LATEX
– happymath
Jan 23 '15 at 9:52
In your example, Fy'y' = 2>0 for every y'. Therefore it is strong minima. Let us assume thet for some problem your Fy'y' = y'+x, Now its sign is not clear and it depend on sign of p=dy/dx (where y is extremal) and range of x, So it is definitaley week ( maxima or minima will depend on sign)
– user3057692
Mar 14 '15 at 16:10
add a comment |
1
@user30567692 I tried to edit your answer but the I could not understand the second half please try to use LATEX
– happymath
Jan 23 '15 at 9:52
In your example, Fy'y' = 2>0 for every y'. Therefore it is strong minima. Let us assume thet for some problem your Fy'y' = y'+x, Now its sign is not clear and it depend on sign of p=dy/dx (where y is extremal) and range of x, So it is definitaley week ( maxima or minima will depend on sign)
– user3057692
Mar 14 '15 at 16:10
1
1
@user30567692 I tried to edit your answer but the I could not understand the second half please try to use LATEX
– happymath
Jan 23 '15 at 9:52
@user30567692 I tried to edit your answer but the I could not understand the second half please try to use LATEX
– happymath
Jan 23 '15 at 9:52
In your example, Fy'y' = 2>0 for every y'. Therefore it is strong minima. Let us assume thet for some problem your Fy'y' = y'+x, Now its sign is not clear and it depend on sign of p=dy/dx (where y is extremal) and range of x, So it is definitaley week ( maxima or minima will depend on sign)
– user3057692
Mar 14 '15 at 16:10
In your example, Fy'y' = 2>0 for every y'. Therefore it is strong minima. Let us assume thet for some problem your Fy'y' = y'+x, Now its sign is not clear and it depend on sign of p=dy/dx (where y is extremal) and range of x, So it is definitaley week ( maxima or minima will depend on sign)
– user3057692
Mar 14 '15 at 16:10
add a comment |
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