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7 Variational Principles
Corollary 7.6.1 If T is symmetric negative definite, namely (T u, u) 0, then u that
makes the functional (7.6.2) obtain the maximum is the solution of the Eq. (7.3.6).
Now to calculate the first variation of the functional corresponding to operator
Eq. (7.3.6), according to the necessary condition of the functional taking extremum,
the solution of operator Eq. (7.3.6) must satisfy
δ J [u] = δ[(T u, u) − 2(u, f )] = (T δu, u) + (T u, δu) − 2(δu, f ) = 0
Because T is the symmetric operator, (T δu, u) = (δu, T u) = (T u, δu), the
above expression can be written as
δ J [u] = 2(T u, δu) − 2(δu, f ) = 2(T u − f, δu) = 2
D
(T u − f )δud D = 0
(7.6.3)
Therefore the functional satisfied by the solution of the operator Eq. (7.3.6) can
be obtained
(T u − f, δu) =
D
(T u − f )δud D = 0
(7.6.4)
From the expression (7.3.8) or expression (7.3.9) the conclusion can be reached:
The solution of the positive definite operator Eq. (7.3.6) can be obtained by solving
the expression (7.3.8) or expression (7.3.9).
If T in the Eq. (7.3.6) is not the symmetric positive definite operator, the inner
product of expression (7.3.9) can also be made, if the inner product can be transformed
into the following form
(T u − f, δu) =
D
(T u − f )δud D = δ
D
Fd D + boundary term = 0 (7.6.5)
where, F is the function of u and D, then it is possible to find the functional
corresponding to the operator equation
J [u(D)] =
D
Fd D
(7.6.6)
The boundary condition is that the boundary term is equal to zero or the boundary
is fixed.
Example 7.6.1 Find the variational problem corresponding to the boundary value
problem of the differential equation
p(x)y
+ p
(x)y
+ q(x)y − f (x) = 0
y(x 0 ) = y 0 , y(x 1 ) = y 1