354
5 Variational Problems of Conditional Extrema
Proof According to the expression (1.3.24) and expression (1.3.33) in Chap. 1, the
original functional can be written in the following form
J =
¨
S
∂u
∂ N
dS =
¨
S
∇u · ndS =
¨
S
∇u · dS =
¨
S
∂u
∂ x
dydz +
∂u
∂ y
dzdx +
∂u
∂z
dxdy
The variation of the functional is
δ J =
˜
S δ
∂u
∂ N
dS =
˜
S
δ
∂u
∂ x
dydz + δ
∂u
∂ y
dzdx + δ
∂u
∂z
dxdy
=
˜
S
∂δu
∂ x
dydz +
∂δu
∂ y
dzdx +
∂δu
∂z
dxdy
where, The last integral uses the property that the variational and derivative can be
exchanged the order.
For the integral of
∂δu
∂ x
dydz, since the variation δu of u is only the function of y
and z, it has nothing to do with x, that is
∂δu
∂ x
= δ
∂u
∂ x
= 0, in other words, on the plane
Oyz, the variation with respect to u x is zero. Similarly, there are δ
∂u
∂ y
=
∂δu
∂ y
= 0,
δ
∂u
∂z
=
∂δu
∂z
= 0, therefore the integral of the above expression is zero, namely
δ J =
˜
S δ
∂u
∂ N
dS = 0. Quod erat demonstrandum.
Lemma 5.4.1 can also be proved in another way.
Proof According to the Gauss formula, the original functional can be written as
J =
¨
S
∂u
∂ N
dS =
¨
S
∇u · ndS =
˚
V
udV =
˚
V
(u xx + u yy + u zz )dV
(5.1)
Let F = u = u xx + u yy + u zz , according to the proof process of theorem 2.9.1,
the variation of the functional can be written as
δ J =
B +
˚
V
∂
2
∂ x 2 F u xx +
∂
2
∂ y 2 F u yy +
∂
2
∂z 2 F u zz
dV
(5.2)
where,
B are the terms related to the boundary integral, since the integral boundary
is fixed, there is
B = 0. Furthermore, F u xx = F u yy = F u zz = 1 of the integral
term are a constant, the partial derivatives
∂
2
∂ x 2 F u xx =
∂
2
∂ y 2 F u yy =
∂
2
∂z 2 F u zz = 0, thus,
no matter whether the functional obtains extremum, there is δ J = 0. Quod erat
demonstrandum.
Corollary 5.4.1 Let the function u(x, y) be continuously differential the function
of second order in the planar domain D(x, y), Γ is the curve boundary of D,
∂u
∂ N
is the normal derivative of u on the curve Γ , then the variation of the functional
J =
Γ
∂u
∂ N
dΓ is zero.
Consider two-dimensional boundary value problem. Let the mixed type functional
5 Variational Problems of Conditional Extrema
Proof According to the expression (1.3.24) and expression (1.3.33) in Chap. 1, the
original functional can be written in the following form
J =
¨
S
∂u
∂ N
dS =
¨
S
∇u · ndS =
¨
S
∇u · dS =
¨
S
∂u
∂ x
dydz +
∂u
∂ y
dzdx +
∂u
∂z
dxdy
The variation of the functional is
δ J =
˜
S δ
∂u
∂ N
dS =
˜
S
δ
∂u
∂ x
dydz + δ
∂u
∂ y
dzdx + δ
∂u
∂z
dxdy
=
˜
S
∂δu
∂ x
dydz +
∂δu
∂ y
dzdx +
∂δu
∂z
dxdy
where, The last integral uses the property that the variational and derivative can be
exchanged the order.
For the integral of
∂δu
∂ x
dydz, since the variation δu of u is only the function of y
and z, it has nothing to do with x, that is
∂δu
∂ x
= δ
∂u
∂ x
= 0, in other words, on the plane
Oyz, the variation with respect to u x is zero. Similarly, there are δ
∂u
∂ y
=
∂δu
∂ y
= 0,
δ
∂u
∂z
=
∂δu
∂z
= 0, therefore the integral of the above expression is zero, namely
δ J =
˜
S δ
∂u
∂ N
dS = 0. Quod erat demonstrandum.
Lemma 5.4.1 can also be proved in another way.
Proof According to the Gauss formula, the original functional can be written as
J =
¨
S
∂u
∂ N
dS =
¨
S
∇u · ndS =
˚
V
udV =
˚
V
(u xx + u yy + u zz )dV
(5.1)
Let F = u = u xx + u yy + u zz , according to the proof process of theorem 2.9.1,
the variation of the functional can be written as
δ J =
B +
˚
V
∂
2
∂ x 2 F u xx +
∂
2
∂ y 2 F u yy +
∂
2
∂z 2 F u zz
dV
(5.2)
where,
B are the terms related to the boundary integral, since the integral boundary
is fixed, there is
B = 0. Furthermore, F u xx = F u yy = F u zz = 1 of the integral
term are a constant, the partial derivatives
∂
2
∂ x 2 F u xx =
∂
2
∂ y 2 F u yy =
∂
2
∂z 2 F u zz = 0, thus,
no matter whether the functional obtains extremum, there is δ J = 0. Quod erat
demonstrandum.
Corollary 5.4.1 Let the function u(x, y) be continuously differential the function
of second order in the planar domain D(x, y), Γ is the curve boundary of D,
∂u
∂ N
is the normal derivative of u on the curve Γ , then the variation of the functional
J =
Γ
∂u
∂ N
dΓ is zero.
Consider two-dimensional boundary value problem. Let the mixed type functional