Calculo de Variacoes
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Transcript of Calculo de Variacoes
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CALCULUS OF VARIATIONS
The development of the calculus of variations begins with an integral in the following form:
I = l; dxfb,Y(z),dY/dzl. (G.1) The function in the integrand depends upon the independent variable x and on the unknown function y(x) and its derivative dy /dz . Both y(z) and its derivative are taken to be functions of the independent variable 2. The problem is to determine the function y(x) so as to minimize the value of the integral. An example of such an integral is.
This form looks like that of an integral equation because the unknown function is inside an integral operation. But it is unlike an integral equation in that the value of the integral is not known and also because the minimization of I does not uniquely determine y(z) unless a set of boundary conditions like
((3.3)
((3.4)
is specified. Here y1 and y2 are known constants. The minimization process is accomplished by introducing a parameter a so that
I ---f I ( a ) and requiring dI/dcr = 0 at Q = 0. The cr parameter is introduced by letting
where y(z) is the function that minimizes I ( Q ) with (1y = 0. The idea is that adding any function ~ ( x ) with cy # 0 causes I ( a ) to be larger than I ( a = 0).
Y(Z) + Y(Z, = Y(Z) + a d z ) ((3.5)
661
Mathematical Physics: Applied Mathematics for Scientists and Engineers Bruce R. Kusse and Erik A. Westwig
Copyright 0 2006 WILEY-VCH Verlag GmbH & Co KGaA
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662 CALCULUS OF VARIATIONS
The function ~ ( x , a) and its derivative are then introduced into the integrand of Equation G. 1.
The boundary conditions
A shorthand notation
(G.lO)
(G. 1 1 )
will be used to simplify the equations. The problem now becomes one of minimizing I ( a ) with respect to a! where
(G. 12)
(G.13)
(G. 14)
(G.15)
(G.16)
To accomplish this take d / d a of the above expression and set it equal to zero at a! = 0.
(G.17)
or
Now
so
(G.20)
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CALCULUS OF VARIATIONS 663
and
so that
((3.21)
Notice that these partial derivatives of y(x, a ) and yr(x. a ) with respect to cy are no longer a function of 0.
Substituting the5e values for the partial derivatives of y(x. 0) and g L (L. a ) into Equation G. 18 gives,
The second integral
is attacked by an integration by parts with
and
so that
((3.24)
((3.25)
((3.26)
(G.27)
The first term on the right hand side of this equation vanishes because ~ ( r ) is zero at XI and .r2.
The minimization condition then becomes
(G.29)
Since this must be true for any q(x) when (Y = 0, the condition for minimizing I (
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664 CALCULUS OF VARIATIONS
where Y = Y(X> a ) = Y(X) + Q+) (G.32)
and the function y(x) is the function we seek, the function that minimizes the original integral. In the limit (Y = 0 the above condition for minimization of the original integral becomes
and
dY (X) yx = -. dx
Equation G.33 is solved for y(x) given the boundary conditions
This is known as Eulers Equation. It can be written in equivalent form as
(G.33)
(G.34)
(G.35)
(G.36)
(G.37)
(G.38)
This equivalence can be seen by expanding the total derivative with respect to x. Equation G.33 is useful if there is no explicit y-dependence in the integrand because
it reduces to
(G.39)
Equation G.38 is useful i f f has no explicit x-dependence because it reduces to
(G.40)