3 Existence of a minimizer of (1.4)
From the above discussion, (1.4) is equivalent to finding Q
0⊂ Ω × R such that E(u
Q0, Q
0) = min
Q⊂Ω×R
E(u
Q, Q),
where u
Qis the solution of (2.1) described in § 2. Recall that the vertices A and B of Q defined in § 2 satisfy the following property
x
A< x
B, y
A≤ y
B, (x
A− x
B)
2+ (y
A− y
B)
2= 4r
2. (3.1) Also, it is easy to check that
x
C= x
B− (y
A− y
B), y
C= y
B+ (x
A− x
B). (3.2) Note that Q ⊂ Ω × R means A, B, C ∈ Ω × R, or equivalently,
x
A, x
B, x
C∈ (0, 1). (3.3)
Hence we have
E(u
Q, Q) =
10
1 + u
Q(x)
2dx − Gy
Qp:= g(x
A, y
A, x
B, y
B) (3.4) for (x
A, y
A, x
B, y
B) ∈ R
4such that (3.1) and (3.3) hold. We would like to get an ex- pression for the function g defined in (3.4). For convenience, for (x
A, y
A, x
B, y
B) ∈ R
4, we shall denote g(x
A, y
A, x
B, y
B) by E
i(x
A, y
A, x
B, y
B) if Q is in the Case i, i = 1, 2, 3, 4, 5, 6.
Recall y
Qp= (y
A+ y
C)/2 = (y
A+ y
B+ x
A− x
B)/2. For different Q, the minimum energy can be computed as follows.
E
1(x
A, y
A, x
B, y
B) = 1 − G
2 (y
A+ y
B+ x
A− x
B). (3.5)
E
2(x
A, y
A, x
B, y
B) =
xB0
1 + ( y
Bx
B)
2dx +
1xB
1 + ( y
B1 − x
B)
2dx
− G
2 (y
A+ y
B+ x
A− x
B)
= x
B1 + ( y
Bx
B)
2+ (1 − x
B)
1 + ( y
B1 − x
B)
2− G
2 (y
A+ y
B+ x
A− x
B). (3.6)
E
3(x
A, y
A, x
B, y
B) =
xA0
1 + ( y
Ax
A)
2dx +
xBxA
1 + ( y
B− y
Ax
B− x
A)
2dx
+
1xB
1 + ( y
B1 − x
B)
2dx − G
2 (y
A+ y
B+ x
A− x
B)
= x
A1 + ( y
Ax
A)
2+ (x
B− x
A)
1 + ( y
B− y
Ax
B− x
A)
2+(1 − x
B)
1 + ( y
B1 − x
B)
2− G
2 (y
A+ y
B+ x
A− x
B). (3.7) E
4(x
A, y
A, x
B, y
B) =
xB0
1 + ( y
Bx
B)
2dx +
xCxB
1 + ( y
C− y
Bx
B− x
C)
2dx +
1xC
1 + ( y
C1 − x
C)
2dx − G
2 (y
A+ y
B+ x
A− x
B)
= x
B1 + ( y
Bx
B)
2+ (y
B− y
A)
1 + ( x
B− x
Ay
B− y
A)
2+(1 − x
C)
1 + ( y
C1 − x
C)
2− G
2 (y
A+ y
B+ x
A− x
B). (3.8) E
5(x
A, y
A, x
B, y
B) =
xA0
1 + ( y
Ax
A)
2dx +
xBxA
1 + ( y
B− y
Ax
B− x
A)
2dx +
xCxB
1 + ( y
C− y
Bx
C− x
B)
2dx +
1xC
1 + ( y
C1 − x
C)
2dx
− G
2 (y
A+ y
B+ x
A− x
B)
= x
A1 + ( y
Ax
A)
2+ (x
B− x
A)
1 + ( y
B− y
Ax
B− x
A)
2+(y
B− y
A)
1 + ( x
B− x
Ay
B− y
A)
2+ (1 − x
C)
1 + ( y
C1 − x
C)
2− G
2 (y
A+ y
B+ x
A− x
B). (3.9)
E
6(x
A, y
A, x
B, y
B) =
xA0
1 + ( y
Ax
A)
2dx + (x
B− x
A) +
1xB
1 + ( y
B1 − x
B)
2dx − G
2 (2y
A+ x
A− x
B)
= x
A1 + ( y
Ax
A)
2+ (x
B− x
A) + (1 − x
B)
1 + ( y
B1 − x
B)
2− G
2 (2y
A+ x
A− x
B). (3.10)
Let
O = {(x
A, y
A, x
B, y
B) ∈ R
4| x
A, x
B, x
C∈ (0, 1), x
A< x
B, y
A< y
B}, (3.11)
O = {(x
A, y
A, x
B, y
B) ∈ R
4| x
A, x
B, x
C∈ (0, 1), x
A< x
B, y
A≤ y
B}, (3.12) and h(x
A, y
A, x
B, y
B) = (x
A− x
B)
2+ (y
A− y
B)
2− 4r
2for all (x
A, y
A, x
B, y
B) ∈ R
4. We define two surfaces
S = {(x
A, y
A, x
B, y
B) ∈ O | h(x
A, y
A, x
B, y
B) = 0 } (3.13) and
S = {(x
A, y
A, x
B, y
B) ∈ O | h(x
A, y
A, x
B, y
B) = 0 }. (3.14) Then (1.4) is equivalent to finding
(xA, yA
min
, xB, yB)∈Sg(x
A, y
A, x
B, y
B), (3.15) i.e., we transform (1.4) into a minimization problem in R
4over S ⊂ R
4.
To study this minimization problem, we first extend g continuously on x
A= x
B, x
A= 0, x
B= 1, and x
C= 1 respectively as follows. We extend g on x
A= x
Bby
g(x
A, y
A, x
A, y
B) = 1 − G
2 (y
A+ y
B) (3.16)
if y
B≤ 0 and by
g(x
A, y
A, x
A, y
B) = x
A1 + ( y
Bx
A)
2+ y
B− y
A+(1 − x
A− y
B+ y
A)
1 + ( y
B1 − x
A− y
B+ y
A)
2− G
2 (y
A+ y
B) (3.17)
if y
B> 0 and x
A> 0. We extend g at (0, y
A, 0, y
B) by g(0, y
A, 0, y
B) = 2y
B− y
A+ (1 − y
B+ y
A)
1 + ( y
B1 − y
B+ y
A)
2− G
2 (y
A+ y
B) (3.18) if y
B> 0. We extend g at (0, y
A, x
B, y
B) with x
B> 0 by
g(0, y
A, x
B, y
B) = 1 − G
2 (y
A+ y
B− x
B) (3.19) if y
B≤ 0. If y
B> 0, then we extend g at (0, y
A, x
B, y
B) with x
B> 0 by
g(0, y
A, x
B, y
B) = x
B1 + ( y
Bx
B)
2+ (1 − x
B)
1 + ( y
B1 − x
B)
2− G
2 (y
A+ y
B− x
B),(3.20)
if m
BO< m
ABand m
BC< m
BD; by
g(0, y
A, x
B, y
B) = y
A+ x
B1 + ( y
B− y
Ax
B)
2+ (1 − x
B)
1 + ( y
B1 − x
B)
2− G
2 (y
A+ y
B− x
B), (3.21)
if m
BO≥ m
ABand m
BC< m
BD; by g(0, y
A, x
B, y
B) = x
B1 + ( y
Bx
B)
2+ (y
B− y
A)
1 + ( x
By
B− y
A)
2+(1 − x
B− y
B+ y
A)
1 + ( y
B− x
B1 − x
B− y
B+ y
A)
2− G
2 (y
A+ y
B− x
B), (3.22)
if m
BO< m
ABand m
BC≥ m
BD; by
g(0, y
A, x
B, y
B) = x
B1 + ( y
B− y
Ax
B)
2+ (y
B− y
A)
1 + ( x
By
B− y
A)
2+(1 − x
B− y
B+ y
A)
1 + ( y
B− x
B1 − x
B− y
B+ y
A)
2− G
2 (y
A+ y
B− x
B) + y
A, (3.23) if m
BO≥ m
ABand m
BC≥ m
BD; and by
g(0, y
A, x
B, y
B) = y
B+ x
B+ (1 − x
B)
1 + ( y
B1 − x
B)
2− G
2 (2y
B− x
B), (3.24) if y
A= y
B.
If x
B= 1, then y
B= y
A. (Otherwise, x
C= x
B+ y
B− y
A> 1, a contradiction.) In this case, we extend g at (x
A, y
A, 1, y
B) by
g(x
A, y
A, 1, y
B) = 1 − G
2 (2y
A+ x
A− 1) (3.25)
if y
B≤ 0 and by
g(x
A, y
A, 1, y
B) = x
A1 + ( y
Ax
A)
2+ 1 − x
A+ y
B− G
2 (2y
A+ x
A− 1) (3.26) if y
B> 0.
For x
C= 1, we extend g at (x
A, y
A, x
B, y
B) by (3.5) if y
B≤ 0. If y
B> 0, we extend
g at (x
A, y
A, x
B, y
B) with x
C= 1 by (3.6), if m
BO< m
ABand m
BC< m
BD; by (3.7), if m
BO≥ m
ABand m
BC< m
BD; by
g(x
A, y
A, x
B, y
B) = x
B1 + ( y
Bx
B)
2+ (y
B− y
A)
1 + ( x
B− x
Ay
B− y
A)
2+y
B+ x
A− x
B− G
2 (y
A+ y
B+ x
A− x
B), (3.27) if m
BO< m
ABand m
BC≥ m
BD; by
g(x
A, y
A, x
B, y
B) = x
A1 + ( y
Ax
A)
2+ (x
B− x
A)
1 + ( y
B− y
Ax
B− x
A)
2+(y
B− y
A)
1 + ( x
B− x
Ay
B− y
A)
2+y
B+ x
A− x
B− G
2 (y
A+ y
B+ x
A− x
B), (3.28) if m
BO≥ m
ABand m
BC≥ m
BD; and by
g(x
A, y
A, x
B, y
B) = x
A1 + ( y
Bx
A)
2+ x
B− x
A+ y
B− G
2 (2y
B+ x
A− x
B), (3.29) if y
A= y
B. Note that x
B= x
C= 1 in the last case.
Hence g is well-defined on O and the following theorem can be derived easily.
Theorem 3.1 The energy function g defined above is continuous on
O = {(x
A, y
A, x
B, y
B) ∈ R
4| x
A, x
B, x
C∈ [0, 1], x
A≤ x
B, y
A≤ y
B}.
Furthermore, we have the following result.
Theorem 3.2 The function g is C
1on O.
Proof: Let
O
1= {(x
A, y
A, x
B, y
B) ∈ O | y
B< 0 }, O
2= {(x
A, y
A, x
B, y
B) ∈ O | y
B> 0, y
Bx
B< y
B− y
Ax
B− x
A, y
B1 − x
B< x
B− x
Ay
B− y
A}, O
3= {(x
A, y
A, x
B, y
B) ∈ O | y
B> 0, y
Bx
B> y
B− y
Ax
B− x
A, y
B1 − x
B< x
B− x
Ay
B− y
A}, O
4= {(x
A, y
A, x
B, y
B) ∈ O | y
B> 0, y
Bx
B< y
B− y
Ax
B− x
A, y
B1 − x
B> x
B− x
Ay
B− y
A}, O
5= {(x
A, y
A, x
B, y
B) ∈ O | y
B> 0, y
Bx
B> y
B− y
Ax
B− x
A, y
B1 − x
B> x
B− x
Ay
B− y
A}.
We claim that O = (
5i=1
O
i) ∪
4j=2
(∂ O
1∩ ∂O
j) ∪
4k=3
(∂ O
2∩ ∂O
k) ∪
4l=3
(∂ O
l∩ ∂O
5) ∪ (
5m=2
∂ O
m).
If (x
A, y
A, x
B, y
B) ∈ ∂O
1∩ ∂O
5, then y
A= y
B= y
C= 0 and so x
A= x
Bwhich contradicts to A = B. Hence ∂O
1∩ ∂O
5is empty. Note that
5 m=2∂ O
m= {(x
A, y
A, x
B, y
B) ∈ O | y
B> 0, (y
B/x
B) = (y
B− y
A)/(x
B− x
A), (y
B/1 − x
B) = (x
B− x
A)/(y
B− y
A) }.
First, we consider ∂g/∂x
Ain each O
i. By computation we have
∂E
1∂x
A= − G
2 , (3.30)
∂E
2∂x
A= − G
2 , (3.31)
∂E
3∂x
A= 1
1 + (y
A/x
A)
2− 1
1 + [(y
B− y
A)/(x
B− x
A)]
2− G
2 , (3.32)
∂E
4∂x
A= −(x
B− x
A)/(y
B− y
A)
1 + [(x
B− x
A)/(y
B− y
A)]
2+ (y
C/1 − x
C)
1 + (y
C/1 − x
C)
2− G
2 , (3.33)
∂E
5∂x
A= 1
1 + (y
A/x
A)
2− 1
1 + [(y
B− y
A)/(x
B− x
A)]
2+ (x
A− x
B)/(y
B− y
A)
1 + [(x
B− x
A)/(y
B− y
A)]
2+ y
C/1 − x
C1 + (y
C/1 − x
C)
2− G
2 . (3.34) It is clear that ∂g/∂x
Ais continuous on O
i, i = 1, 2, . . . , 5. Hence it is enough to show that ∂g/∂x
Ais continuous across ∂ O
ifor each i.
If we move the square Q from O
2to O
1, then it is clear that ∂g/∂x
Ais continuous across ∂ O
1∩ ∂O
2.
If we pass from O
3to O
1, then we have y
A= y
B→ 0.
Then (3.32) converges toward (3.30). Hence ∂g/∂x
Ais continuous across ∂ O
1∩ ∂O
3. Similarly, if we move Q from O
4to O
1, then (3.33) converges toward (3.30). Hence
∂g/∂x
Ais continuous across ∂ O
1∩ ∂O
4. If we pass from O
3to O
2, then we have
y
B− y
Ax
B− x
A→ y
Ax
A.
Then (3.32) converges toward (3.31). Hence ∂g/∂x
Ais continuous across ∂ O
2∩ ∂O
3. If we move the square Q from O
4to O
2, then we have
x
B− x
Ay
B− y
A→ y
C1 − x
C.
Then (3.33) converges toward (3.31). Hence ∂g/∂x
Ais continuous across ∂ O
2∩ ∂O
4. If we move the square Q from O
5to O
3, then we have
x
B− x
Ay
B− y
A→ y
C1 − x
C.
Then (3.34) converges toward (3.32). Hence ∂g/∂x
Ais continuous across ∂ O
3∩ ∂O
5. If we move the square Q from O
5to O
4, then we have
y
B− y
Ax
B− x
A→ y
Ax
A.
Then (3.34) converges toward (3.33). Hence ∂g/∂x
Ais continuous across ∂D
4∩ ∂O
5. If we move the square Q from O
5to
5m=2O
m, then we have
y
B− y
Ax
B− x
A→ y
Ax
Aand x
B− x
Ay
B− y
A→ y
C1 − x
C.
Then (3.34) converges toward (3.31). Similarly, if we move Q from O
3( O
4, resp.) to
5m=2
∂ O
m, then (3.32)((3.33), resp.) converges toward (3.31). Hence we obtain ∂g/∂x
Ais continuous across
5m=2∂ O
m. Therefore ∂g/∂x
Ais continuous on O.
The continuities of ∂g/∂x
B, ∂g/∂y
A, ∂g/∂y
Bacross ∂ O
ifor each i can be derived similarly. This completes the proof.
It is easy to show that for Q ⊂ Ω × R
10
|u
Q(x) |dx = 2y
Bfor y
B> 0. (3.35) Theorem 3.3 If G > 2, then
(xA, yA, x
inf
B, yB)∈Sg(x
A, y
A, x
B, y
B) = −∞.
Proof: If y
B> 2r, then y
A, y
C> 0. Since y
A> y
B− 2r and −1 < x
A− x
B< 0, g(x
A, y
A, x
B, y
B) =
10
1 + u
Q(x)
2dx − G
2 (y
A+ y
B+ x
A− x
B)
≤
10
(1 + |u
Q(x) |)dx − G
2 (y
A+ y
B+ x
A− x
B)
<
10
|u
Q(x) |dx − G
2 (2y
B− 2r + x
A− x
B) + 1
<
10
|u
Q(x) |dx − Gy
B+ Gr + G 2 + 1
= (2 − G)y
B+ Gr + G
2 + 1.
Since G > 2, the theorem follows.
Theorem 3.4 If G = 2, then inf
(xA, yA, xB, yB)∈Sg(x
A, y
A, x
B, y
B) exists.
Proof: Suppose that y
B≤ 0. Then g(x
A, y
A, x
B, y
B) = 1 − (y
B+ y
A+ x
A− x
B) > 0.
On the other hand, if we consider the region y
B> 0, then g(x
A, y
A, x
B, y
B) =
10
1 + u
Q(x)
2dx − (y
A+ y
B+ x
A− x
B)
≥
10
|u
Q(x) |dx − 2y
B= 2y
B− 2y
B= 0.
Hence 0 is a lower bound of g. Therefore the infimum of g on S exists.
However, the infimum can not be achieved in S.
Theorem 3.5 If G = 2, then g does not attain its minimum on S.
Proof: Suppose that g attains its minimum, say α, at some (x
A1, y
A1, x
B1, y
B1) ∈ S.
For (x
A, y
A, x
B, y
B) ∈ S with x
A= x
A1, x
B= x
B1, y
A= y
A1+ y, and y
B= y
B1+ y for some y > 0, let h
i(y) = E
i(x
A1, y
A1+ y, x
B1, y
B1+ y) for each i. Then we have
h
1(0) = −2 < 0, h
2(0) = y
B1/x
B11 + (y
B1/x
B1)
2+ y
B1/1 − x
B11 + (y
B1/1 − x
B1)
2− 2 < 0, h
3(0) = y
A1/x
A11 + (y
A1/x
A1)
2+ y
B1/1 − x
B11 + (y
B1/1 − x
B1)
2− 2 < 0, h
4(0) = y
B1/x
B11 + (y
B1/x
B1)
2+ (y
B1+ x
A1− x
B1)/(1 − x
B1− y
B1+ y
A1)
1 + [(y
B1+ x
A1− x
B1)/(1 − x
B1− y
B1+ y
A1)]
2− 2 < 0, h
5(0) = (y
B1− y
A1)/(x
B1− x
A1)
1 + [(y
B1− y
A1)/(x
B1− x
A1)]
2+ (y
B1+ x
A1− x
B1)/(1 − x
B1− y
B1+ y
A1)
1 + [(y
B1+ x
A1− x
B1)/(1 − x
B1− y
B1+ y
A1)]
2− 2 < 0, h
6(0) = y
A1/x
A11 + (y
A1/x
A1)
2+ y
B1/1 − x
B11 + (y
B1/1 − x
B1)
2− 2 < 0.
Hence g(x
A1, y
A1+ ρ, x
B1, y
B1+ ρ) < α for any ρ > 0 small, a contradiction. Therefore,
g does not attain its minimum on S if G = 2. The proof is complete.
Lemma 3.6 Let
D = {(x
A, y
A, x
B, y
B) ∈ S | a ≤ y
A≤ y
B≤ b}
where a ≤ b, a, b ∈ R. Then inf
Dg is achieved in S ∩ {a ≤ y
A≤ y
B≤ b}.
Proof: By Theorem 3.1, g is continuous on D. Since D is a compact subset of R
4, inf
Dg is achieved at some (x
A, y
A, x
B, y
B) ∈ D, i.e., inf
Dg = min
Dg. We claim that inf
Dg is achieved in S ∩ {a ≤ y
A≤ y
B≤ b}. We consider the following cases.
(1) y
B≤ 0, x
A= 0. Then we have
g(0, y
A, x
B, y
B) = 1 − G
2 (y
A+ y
B− x
B)
and moving the square horizontally will not change the energy. Hence we may assume that inf
Dg is achieved at (x
A, y
A, x
B, y
B) ∈ S∩{a ≤ y
A≤ y
B≤ b} in this case. Similarly, if (x
A, y
A, x
B, y
B) ∈ D such that x
B= 1 or x
C= 1, then we may assume that inf
Dg is achieved at (x
A, y
A, x
B, y
B) ∈ S ∩ {a ≤ y
A≤ y
B≤ b} in these cases.
(2) y
B> 0, m
BO< m
ABand m
BC< m
BD. Fix y
Band rotate the square Q. Since u is unchanged but y
Qpincreases, it will decrease its energy. Hence inf
Dg cannot be achieved in this case.
(3) y
B> 0, m
BO≥ m
ABand m
BC< m
BD. Suppose that g attains its minimum, say α, at some (0, y
A, x
B, y
B) ∈ D. For x > 0 small, we have
H
1(x) := g(x, y
A, x
B+ x, y
B)
= x
1 + ( y
Ax )
2+ x
B1 + ( y
B− y
Ax
B)
2+(1 − x
B− x)
1 + ( y
B1 − x
B− x )
2− G
2 (y
A+ y
B− x
B).
By computation we obtain
H
1(0) = − 1
1 + (y
B/1 − x
B)
2< 0. (3.36)
Hence g(x, y
A, x
B+ x, y
B) < α for x > 0 small, a contradiction. Therefore, we may assume that inf
Dg is achieved in S ∩{a ≤ y
A≤ y
B≤ b} . Similarly, if (x
A, y
A, x
B, y
B) ∈ D such that x
C= 1, then we may assume that inf
Dg is achieved at (x
A, y
A, x
B, y
B) ∈ S ∩ {a ≤ y
A≤ y
B≤ b} in this case.
(4) y
B> 0, m
BO< m
ABand m
BC≥ m
BD. Suppose that g attains its minimum, say
α, at some (0, y
A, x
B, y
B) ∈ D. Note that in this case y
A< 0 and hence y
B< 2r. For
x > 0 small, we have
H
2(x) := g(x, y
A, x
B+ x, y
B)
= (x
B+ x)
1 + ( y
Bx
B+ x )
2+ (y
B− y
A)
1 + ( x
By
B− y
A)
2+(1 − x
B− x − y
B+ y
A)
1 + ( y
B− x
B1 − x
B− x − y
B+ y
A)
2− G
2 (y
A+ y
B− x
B).
By computation we obtain H
2(0) = 1
1 + (y
B/x
B)
2− 1
1 + [(y
B− x
B)/(1 − x
B− y
B+ y
A)]
2< 0, (3.37)
since
y
Bx
B− y
B− x
B1 − x
B− y
B+ y
A> 0. (3.38) Indeed, the inequality (3.38) follows from
y
B(1 − x
B− y
B+ y
A) − x
B(y
B− x
B)
= y
B− x
By
B− y
B2+ y
Ay
B− x
By
B+ x
2B= y
B− 2x
By
B− y
B2+ x
2B+ y
Ay
B≥ y
B− x
2B− y
B2− y
2B+ x
2B+ y
Ay
B= y
B(1 − 2y
B+ y
A)
= y
B[1 − y
B− (y
B− y
A)]
> y
B(1 − 4r) > 0 since y
B< 2r.
Hence g(x, y
A, x
B+ x, y
B) < α for x > 0 small, a contradiction. Therefore, we may assume that inf
Dg is achieved in S ∩ {a ≤ y
A≤ y
B≤ b}. Similarly, if (x
A, y
A, x
B, y
B) ∈ D such that x
C= 1, then we may assume that inf
Dg is achieved at (x
A, y
A, x
B, y
B) ∈ S ∩ {a ≤ y
A≤ y
B≤ b} in this case.
(5) y
B> 0, m
BO≥ m
ABand m
BC≥ m
BD. Suppose that g attains its minimum, say α, at some (0, y
A, x
B, y
B) ∈ D. For x > 0 small, we have
H
3(x) := g(x, y
A, x
B+ x, y
B)
= x
1 + ( y
Ax )
2+ x
B1 + ( y
B− y
Ax
B)
2+ (y
B− y
A)
1 + ( x
By
B− y
A)
2+(1 − x
B− x − y
B+ y
A)
1 + ( y
B− x
B1 − x
B− x − y
B+ y
A)
2− G
2 (y
A+ y
B− x
B).
By computation we obtain
H
3(0) = − 1
1 + [(y
B− x
B)/(1 − x
B− y
B+ y
A)]
2< 0. (3.39)
Hence g(x, y
A, x
B+ x, y
B) < α for x > 0 small, a contradiction. Therefore, we may assume that inf
Dg is achieved in S ∩{a ≤ y
A≤ y
B≤ b} . Similarly, if (x
A, y
A, x
B, y
B) ∈ D such that x
C= 1, then we may assume that inf
Dg is achieved at (x
A, y
A, x
B, y
B) ∈ S ∩ {a ≤ y
A≤ y
B≤ b} in this case.
(6) y
A= y
B> 0. Suppose that g attains its minimum, say α, at some (0, y
A, x
B, y
B) ∈ D.
For x > 0 small, we have
H
4(x) := g(x, y
A, x
B+ x, y
B)
= x
1 + ( y
Ax )
2+ x
B+ (1 − x
B− x)
1 + ( y
B1 − x
B− x )
2− G
2 (2y
A− x
B).
By computation we obtain
H
4(0) = − 1
1 + (y
B/1 − x
B)
2< 0. (3.40)
Hence g(x, y
A, x
B+ x, y
B) < α for x > 0 small, a contradiction. Therefore, we may assume that inf
Dg is achieved in S ∩{a ≤ y
A≤ y
B≤ b} . Similarly, if (x
A, y
A, x
B, y
B) ∈ D such that x
B= 1, then we may assume that inf
Dg is achieved at (x
A, y
A, x
B, y
B) ∈ S ∩ {a ≤ y
A≤ y
B≤ b} in this case. Hence inf
Dg = min
Dg = min
S∩{a≤yA≤yB≤b}g. This completes the proof.
Now, we prove the main theorem of the section.
Theorem 3.7 If G < 2, then the problem (1.4) admits a minimizer in S.
Proof: First we claim that the infimum of g on S exists. If y
B≤ 0, then g(x
A, y
A, x
B, y
B) = 1 − G
2 (y
A+ y
B+ x
A− x
B)
> 1 + Gr
≥ 0.
On the other hand, if we consider the region y
B> 0, then g(x
A, y
A, x
B, y
B) =
10
1 + u
Q(x)
2dx − G
2 (y
A+ y
B+ x
A− x
B)
≥
10
|u
Q(x) |dx − Gy
B= (2 − G)y
B> 0.
Hence 0 is a lower bound of g. Therefore, the infimum of g on S exists. Furthermore, since g is continuous on S, inf g on S also exists.
By (3.15), it is enough to find (x
A0, y
A0, x
B0, y
B0) ∈ S such that g(x
A0, y
A0, x
B0, y
B0) = min
(xA, yA, xB, yB)∈S
g(x
A, y
A, x
B, y
B). (3.41) We claim that to find a minimizer in S it suffices to find a minimizer in S such that
−2r ≤ y
A≤ y
B≤ M (3.42)
for some constant M = M (G) > 0. Suppose that y
A< −2r. Then y
B< 0 and g(x
A, y
A, x
B, y
B) = 1 − G
2 (y
A+ y
B+ x
A− x
B)
> 1 + Gr
= g(x
A, 0, x
B, 0). (3.43)
Hence g(x
A, y
A, x
B, y
B) does not attain its minimum, if y
A< −2r. Therefore, we may assume that −2r ≤ y
A≤ y
B. On the other hand, if y
B> 2r, then y
A, y
C> 0 and
g(x
A, y
A, x
B, y
B) =
10
1 + u
Q(x)
2dx − G
2 (y
A+ y
B+ x
A− x
B)
≥
10
1 + u
Q(x)
2dx − Gy
B≥
10
|u
Q(x) |dx − Gy
B= (2 − G)y
B.
Since G < 2, g(x
A, y
A, x
B, y
B) → +∞ as y
B→ ∞. By (3.43), inf
Sg ≤ 1 + Gr. There exists M = M (G) ≥ 2r such that to find a minimizer on S it suffices to find a minimizer in S such that (3.42) holds.
Let
S
M:= {(x
A, y
A, x
B, y
B) ∈ S | −2r ≤ y
A≤ y
B≤ M}.
By the above discussion and Lemma 3.6, we obtain inf
Sg = inf
SM
