418
7 Variational Principles
By similar to the above methods, the higher order derivatives of the functional
can be deduced, and the conclusion can be reached: The second order and higher
than second order derivatives of the simplest functional are all operators.
7.6 Variational Principles of Operator Equations
The variational principle of the symmetric positive definite operator equation is given
below.
Theorem 7.6.1 Let T be a symmetric positive definite operator, the domain of
definition is D, the range is T (D), u ∈ D, f ∈ T (D), if the operator equation
T u = f
(7.6.1)
exists the solution u = u 0 , the necessary and sufficient conditions satisfying the
u 0 are that the functional
J [u] = (T u, u) − 2(u, f )
(7.6.2)
obtains a minimum.
In Eq. (7.3.6), the free term f is the known function of an independent variable,
it is called the driving function or forcing function.
Proof The necessity. If u = u 0 is the solution of Eq. (7.3.6), then T u 0 = f . Let
u = u 0 + η, obviously u ∈ D. The functional (7.6.2) may be written as
J [u] = [T (u 0 + η), u 0 + η] − 2(u 0 + η, f )
Making use of the symmetry of the operator T (T u, v) = (u, T v), and according
to the property of the inner product (u, v) = (v, u), expanding the above expression,
we have
J [u] = (T u 0 , u 0 ) + (T u 0 , η) + (T η, u 0 ) + (T η, η) − 2(u 0 , f ) − 2(η, f )
= (T u 0 , u 0 ) + (T u 0 , η) + (T u 0 , η) + (T η, η) − 2(u 0 , f ) − 2( f, η)
= J [u 0 ] + 2(T u 0 − f, η) + (T η, η)
But T u 0 − f = 0, and according to the definition of positive definite operator,
(T η, η) > 0, there is
J [u] > J [u 0 ]
This shows that J [u] reaches the minimum at u = u 0 .
7 Variational Principles
By similar to the above methods, the higher order derivatives of the functional
can be deduced, and the conclusion can be reached: The second order and higher
than second order derivatives of the simplest functional are all operators.
7.6 Variational Principles of Operator Equations
The variational principle of the symmetric positive definite operator equation is given
below.
Theorem 7.6.1 Let T be a symmetric positive definite operator, the domain of
definition is D, the range is T (D), u ∈ D, f ∈ T (D), if the operator equation
T u = f
(7.6.1)
exists the solution u = u 0 , the necessary and sufficient conditions satisfying the
u 0 are that the functional
J [u] = (T u, u) − 2(u, f )
(7.6.2)
obtains a minimum.
In Eq. (7.3.6), the free term f is the known function of an independent variable,
it is called the driving function or forcing function.
Proof The necessity. If u = u 0 is the solution of Eq. (7.3.6), then T u 0 = f . Let
u = u 0 + η, obviously u ∈ D. The functional (7.6.2) may be written as
J [u] = [T (u 0 + η), u 0 + η] − 2(u 0 + η, f )
Making use of the symmetry of the operator T (T u, v) = (u, T v), and according
to the property of the inner product (u, v) = (v, u), expanding the above expression,
we have
J [u] = (T u 0 , u 0 ) + (T u 0 , η) + (T η, u 0 ) + (T η, η) − 2(u 0 , f ) − 2(η, f )
= (T u 0 , u 0 ) + (T u 0 , η) + (T u 0 , η) + (T η, η) − 2(u 0 , f ) − 2( f, η)
= J [u 0 ] + 2(T u 0 − f, η) + (T η, η)
But T u 0 − f = 0, and according to the definition of positive definite operator,
(T η, η) > 0, there is
J [u] > J [u 0 ]
This shows that J [u] reaches the minimum at u = u 0 .