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A necessary condition for optimality in the problem of variational calculus on a conditional extremum; established by R. Clebsch [[#References|[1]]]. Suppose an extremal <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c022/c022450/c0224501.png" /> provides a conditional minimum of the functional in the [[Bolza problem|Bolza problem]]:
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<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c022/c022450/c0224502.png" /></td> </tr></table>
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{{TEX|done}}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c022/c022450/c0224503.png" /></td> </tr></table>
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A necessary condition for optimality in the problem of variational calculus on a conditional extremum; established by R. Clebsch [[#References|[1]]]. Suppose an extremal  $  x ( t): \mathbf R \rightarrow \mathbf R  ^ {n} $
 +
provides a conditional minimum of the functional in the [[Bolza problem|Bolza problem]]:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c022/c022450/c0224504.png" /></td> </tr></table>
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$$
 +
J ( x)  = \
 +
\int\limits _ { t _ {1} } ^ { {t _ 2 } }
 +
f ( t, x, \dot{x} )  dt +
 +
g ( t _ {1} , x ( t _ {1} ),\
 +
t _ {2} , x ( t _ {2} )),
 +
$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c022/c022450/c0224505.png" /></td> </tr></table>
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$$
 +
f: \mathbf R \times \mathbf R  ^ {n} \times \mathbf R  ^ {n}  \rightarrow \
 +
\mathbf R ,\  g: \mathbf R \times \mathbf R  ^ {n} \times \mathbf R \times \mathbf R  ^ {n}  \rightarrow  \mathbf R ,
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$$
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c022/c022450/c0224506.png" /></td> </tr></table>
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$$
 +
\phi ( t, x, \dot{x} )  = 0,\  \phi : \
 +
\mathbf R \times \mathbf R  ^ {n} \times \mathbf R  ^ {n}  \rightarrow  \mathbf R  ^ {m} ,\  m < n,
 +
$$
 +
 
 +
$$
 +
\psi ( t _ {1} , x ( t _ {1} ), t _ {2} , x ( t _ {2} ))  = 0,\  \psi : \mathbf R \times \mathbf R  ^ {n}
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\times \mathbf R \times \mathbf R  ^ {n}  \rightarrow  \mathbf R  ^ {p} ,
 +
$$
 +
 
 +
$$
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p  \leq  2m + 2.
 +
$$
  
 
Then, according to the multiplier method, it is an unconditional extremal of the functional
 
Then, according to the multiplier method, it is an unconditional extremal of the functional
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c022/c022450/c0224507.png" /></td> <td valign="top" style="width:5%;text-align:right;">(1)</td></tr></table>
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$$ \tag{1 }
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\int\limits _ { t _ {1} } ^ { {t _ 2 } }
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F ( t, x, \dot{x} , \lambda ) \
 +
dt + \lambda _ {0} g +
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\sum _ {\mu = 1 } ^ { p }
 +
e _  \mu  \psi _  \mu  .
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$$
  
 
Here
 
Here
  
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$$
 +
F ( t, x, \dot{x} , \lambda )  = \
 +
\lambda _ {0} f +
 +
\lambda _ {1} ( t)
 +
\phi _ {1} + \dots +
 +
\lambda _ {m} ( t)
 +
\phi _ {m} ,
 +
$$
  
 
and
 
and
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c022/c022450/c0224509.png" /></td> </tr></table>
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$$
 +
\lambda _ {0} ,\
 +
\lambda _ {i} ( t),\ \
 +
i = 1 \dots m; \ \
 +
e _  \mu  ,\ \
 +
\mu = 1 \dots p,
 +
$$
 +
 
 +
are the [[Lagrange multipliers|Lagrange multipliers]], which have to be determined together with  $  x ( t) $
 +
from the necessary conditions for an extremum of the functional (1). One such necessary condition is the Clebsch condition.
 +
 
 +
In order that  $  x ( t) $
 +
be minimal in the above problem, it is necessary that the Clebsch condition should hold. It requires that for any non-trivial set of numbers  $  \xi _ {1} $,
 +
$  i = 1 \dots n $,
 +
satisfying the equations
  
are the [[Lagrange multipliers|Lagrange multipliers]], which have to be determined together with <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c022/c022450/c02245010.png" /> from the necessary conditions for an extremum of the functional (1). One such necessary condition is the Clebsch condition.
+
$$
 +
\sum _ {i = 1 } ^ { n }
  
In order that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c022/c022450/c02245011.png" /> be minimal in the above problem, it is necessary that the Clebsch condition should hold. It requires that for any non-trivial set of numbers <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c022/c022450/c02245012.png" />, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c022/c022450/c02245013.png" />, satisfying the equations
+
\frac{\partial  \phi _ {k} ( t, x, \dot{x} ) }{\partial  \dot{x} _ {i} }
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c022/c022450/c02245014.png" /></td> </tr></table>
+
\xi _ {i}  = 0 ,\ \
 +
k = 1 \dots m,
 +
$$
  
 
the following quadratic form be non-negative:
 
the following quadratic form be non-negative:
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/c/c022/c022450/c02245015.png" /></td> </tr></table>
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$$
 +
\sum _ {i, j = 1 } ^ { n }
 +
 
 +
\frac{\partial  ^ {2} F
 +
( t, x, \dot{x} , \lambda ) }{\partial  \dot{x} _ {i} \partial  \dot{x} _ {j} }
 +
 
 +
\xi _ {i} \xi _ {j}  \geq  0.
 +
$$
  
 
The Clebsch necessary condition is directly related to the stronger necessary [[Weierstrass conditions (for a variational extremum)|Weierstrass conditions (for a variational extremum)]] and can be obtained as a corollary of the latter.
 
The Clebsch necessary condition is directly related to the stronger necessary [[Weierstrass conditions (for a variational extremum)|Weierstrass conditions (for a variational extremum)]] and can be obtained as a corollary of the latter.

Latest revision as of 17:44, 4 June 2020


A necessary condition for optimality in the problem of variational calculus on a conditional extremum; established by R. Clebsch [1]. Suppose an extremal $ x ( t): \mathbf R \rightarrow \mathbf R ^ {n} $ provides a conditional minimum of the functional in the Bolza problem:

$$ J ( x) = \ \int\limits _ { t _ {1} } ^ { {t _ 2 } } f ( t, x, \dot{x} ) dt + g ( t _ {1} , x ( t _ {1} ),\ t _ {2} , x ( t _ {2} )), $$

$$ f: \mathbf R \times \mathbf R ^ {n} \times \mathbf R ^ {n} \rightarrow \ \mathbf R ,\ g: \mathbf R \times \mathbf R ^ {n} \times \mathbf R \times \mathbf R ^ {n} \rightarrow \mathbf R , $$

$$ \phi ( t, x, \dot{x} ) = 0,\ \phi : \ \mathbf R \times \mathbf R ^ {n} \times \mathbf R ^ {n} \rightarrow \mathbf R ^ {m} ,\ m < n, $$

$$ \psi ( t _ {1} , x ( t _ {1} ), t _ {2} , x ( t _ {2} )) = 0,\ \psi : \mathbf R \times \mathbf R ^ {n} \times \mathbf R \times \mathbf R ^ {n} \rightarrow \mathbf R ^ {p} , $$

$$ p \leq 2m + 2. $$

Then, according to the multiplier method, it is an unconditional extremal of the functional

$$ \tag{1 } \int\limits _ { t _ {1} } ^ { {t _ 2 } } F ( t, x, \dot{x} , \lambda ) \ dt + \lambda _ {0} g + \sum _ {\mu = 1 } ^ { p } e _ \mu \psi _ \mu . $$

Here

$$ F ( t, x, \dot{x} , \lambda ) = \ \lambda _ {0} f + \lambda _ {1} ( t) \phi _ {1} + \dots + \lambda _ {m} ( t) \phi _ {m} , $$

and

$$ \lambda _ {0} ,\ \lambda _ {i} ( t),\ \ i = 1 \dots m; \ \ e _ \mu ,\ \ \mu = 1 \dots p, $$

are the Lagrange multipliers, which have to be determined together with $ x ( t) $ from the necessary conditions for an extremum of the functional (1). One such necessary condition is the Clebsch condition.

In order that $ x ( t) $ be minimal in the above problem, it is necessary that the Clebsch condition should hold. It requires that for any non-trivial set of numbers $ \xi _ {1} $, $ i = 1 \dots n $, satisfying the equations

$$ \sum _ {i = 1 } ^ { n } \frac{\partial \phi _ {k} ( t, x, \dot{x} ) }{\partial \dot{x} _ {i} } \xi _ {i} = 0 ,\ \ k = 1 \dots m, $$

the following quadratic form be non-negative:

$$ \sum _ {i, j = 1 } ^ { n } \frac{\partial ^ {2} F ( t, x, \dot{x} , \lambda ) }{\partial \dot{x} _ {i} \partial \dot{x} _ {j} } \xi _ {i} \xi _ {j} \geq 0. $$

The Clebsch necessary condition is directly related to the stronger necessary Weierstrass conditions (for a variational extremum) and can be obtained as a corollary of the latter.

In problems of the calculus of variations on an unconditional extremum, in particular in the simplest problem of variational calculus, the analogue of the Clebsch condition is the Legendre condition.

In problems of optimal control, the Clebsch condition is equivalent to the non-positiveness of the second differential of the Hamilton function, which is a necessary condition in order that the Pontryagin maximum principle be fulfilled for optimal control in an open domain.

References

[1] R.F.A. Clebsch, J. für Math. , 55 (1858) pp. 254
[2] G.A. Bliss, "Lectures on the calculus of variations" , Chicago Univ. Press (1947)
How to Cite This Entry:
Clebsch condition. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Clebsch_condition&oldid=46357
This article was adapted from an original article by I.B. Vapnyarskii (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article