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On a Unique Solution for the Two-Point Boundary Value Problem
Journal of Mathematical Analysis and Applications 247, 653–657 (2000) doi:10.1006/jmaa.2000.6870, available online at http://www.idealibrary.com on
NOTE On a Unique Solution for the Two-Point Boundary Value Problem Zhang Shu-Qin1 and Zhong Cheng-Kui Department of Mathematics, Lanzhou University, Lanzhou, 730000, Gansu, People’s Republic of China Submitted by George Leitmann Received March 29, 2000
An existence and uniqueness result concerned with the boundary value problem u00 + gt; ut = et; is presented.
u0 0 = u0 π = 0
© 2000 Academic Press
1. INTRODUCTION Consider the boundary value problem u00 + gt; u = et;
u0 0 = u0 π:
(1)
In [1], Huang and Shen consider the existence and uniqueness result of the solution for (1). There they consider the case that gu t; u is non-negative. Here, we consider the case that gu t; u may be negative. In this paper, we also use the following Min–Max Theorem (see [2]) to prove our main result. Min–Max Theorem (Manasevich). Let H be a real Hilbert space and let f x H → R be of class C 2 . Suppose that there exist two closed subspaces X and Y such that H = X ⊕ Y and two continuous nonincreasing functions α x 0; ∞ → 0; ∞; 1
β x 0; ∞ → 0; ∞
This work was supported by the National Natural Science Foundation of China 653 0022-247X/00 $35.00 Copyright © 2000 by Academic Press All rights of reproduction in any form reserved.
654
note
such that
Z
∞
1
αsds = ∞
Z
∞
1
βsds = ∞
2 D f x + yk; k ≥ αyk2
for all x ∈ X, y ∈ Y , and k ∈ Y , and
2 D f x + yh; h ≤ −βxh2 for all x ∈ X, y ∈ Y , and h ∈ X. Then (a)
there exists a unique v0 ∈ H such that ∇f v0 = 0;
(b)
f v0 = max x∈X miny∈Y f x + y = miny∈Y max x∈X f x + y:
In the above theorem, ∇f u and D2 f u denote the gradient and the Hessian of f at u ∈ H; respectively. In this case, f x H → H is a C 0 mapping and ∇f 0 u = D2 f u is a bounded self-adjoint linear operator on H. Let U = uutR ∈ C 2 0; π; u0 0 = u0 π = 0; ut is absolutely continπ uous and satisfies 0 u2 tdt < +∞. Then U is a real Hilbert space with the inner product Zπ u0 v0 + uv dt: u; v = 0
The norm induced in U by the inner product · will be denoted by · U . We assume that: (A)
g ∈ C 0 0; π × R; R and e ∈ C0; πy
(B)
there exists some positive integer n such that
n2 + bn + c < gu t; u < n + 12
for all u ∈ U; t ∈ 0; π;
(2)
where n2 + bn + c < 0 and b; c satisfy c>0 b + c + 1 < 0:
(3)
2. THE MAIN RESULT In this section, we consider the existence of a unique solution for Problem (1). We let gu∗ t; u = gu t; u + 2n + 12 ;
note
655
then 3n2 + 4 + bn + 2 + c < gu∗ t; u < 3n + 12 : By (3), we see that 3n2 + 4n + 2 < 3n2 + 4 + bn + 2 + c: We denote ζs = min max gu∗ t; u − 3n2 + 4n + 2
(4)
ηs = min max n + 12 − gu t; u:
(5)
uU ≤st∈0;π
uU ≤st∈0;π
Clearly, ζx 0; +∞ → 0; +∞ and η x 0; +∞ → 0; +∞ are two continuous nonincreasing functions. Define subspaces X and Y of U as X = xxt =
n X a0 + am cos mt; am ∈ R; m = 0; 1; 2; · · · ; n 2 m=1 ∞ X
Y = yyt =
m=n+1
am cos mt; am ∈ R; m = n + 1; n + 2; · · ·;
(6)
(7)
P where n is as in (2) and the series yt = ∞ m=n+1 am cos mt as well as the series obtained by termwise differentiation of (7) converges uniformly on R. Clearly, we have U = X ⊕ Y and the inequalities Zπ Zπ Zπ Zπ x0 t2 dt ≤ n2 xt2 dt y 0 t2 dt ≥ n + 12 yt2 dt 0
0
0
0
for all x ∈ X; y ∈ Y . Next, we shall consider the problem u00 + g∗ t; u = 2n + 12 u + et;
u0 0 = u0 π = 0:
(8)
We note that Problem (8) is equivalent to Problem (1) by the construction of g∗ t; u. Define a function I x U → R by Z π 1 0 2 2 2 u t − Gt; ut + etut + n + 1 u dt; (9) Iu = 2 0 Ru where Gt; u = 0 g∗ t; sds; u ∈ U. Clearly, I is of class C 2 by ssumptions, A and B:
656
note
It is well known that u ∈2 o; π is a solution of problem (8) if and only if u is a critical point of I. The following is our main result. Theorem. Suppose that (A) and (B) hold, and Z∞ Z∞ ζsds = +∞; ηsds = +∞; 1
(10)
1
where ζ and η are as in (4) and (5). Then (a) there exists a unique solution u∗ of (8); (b) Iu∗ = max x∈X miny∈Y Ix + y = miny∈Y max x∈Y Ix + y, where Iu is defined in (9) and X and Y are defined in (6) and (7). Proof. First of all, it is easy to prove that a mapping ∇I x U → U is defined by Zπ u0 v0 − g∗ t; uv + etv + 2n + 12 uv dt: ∇Iu; v = 0
And for all u ∈ U; ∀v ∈ U; w ∈ U, we have Zπ 0 0
2 v w − gu∗ t; uvw + 2n + 12 wv dt: D Iuw; v = 0
For u = x + y ∈ U; x ∈ X; y ∈ Y , and h ∈ X; k ∈ Y , we have Z π 02
2 h − gu∗ t; uh2 + 2n + 12 h2 dt D Ix + yh; h = 0 Zπ n2 − gu∗ t; u + 2n + 12 h2 dt ≤ 0 Zπ = − gu∗ t; u − 3n2 + 4n + 2h2 dt 0
min
max gu∗ t; u − 3n2 + 4n + 2
Z
π
h2 dt 1 max gu∗ t; u − 3n2 + 4n + 2 min =− 2 n + 1 uU ≤xU t∈0;π Zπ Zπ h2 dt + h2 dt n2 0 0 1 min max gu∗ t; u − 3n2 + 4n + 2 ≤− 2 n + 1 uU ≤xU t∈0;π Z π Zπ 2 h0 dt + h2 dt ≤−
uU ≤xU t∈0;π
0
=−
ζ xU h2U n2 + 1
0
0
note
657
Z π 02 k − gu∗ t; uk2 + 2n + 12 k2 dt D2 Ix + yk; k = = = ≥
Z
0 π
Z
0 π
2 k0 − gu t; uk2 − 2n + 12 k2 + 2n + 12 k2 dt
2 k0 − gu t; uk2 dt
0
Z
π 0
2
k0 k0 − gu t; u n + 12 2
! dt
1 min max n + 12 − gu t; u n + 12 + 1 uU ≤yU t∈0;π Z π 1 2 k0 dt 1+ n + 12 0 1 min max n + 12 − gu t; u ≥ n + 12 + 1 uU ≤yU t∈0;π Zπ 2 0 2 k + k dt ≥
0
=
ηyU k2U n + 12 + 1
From (10) and the fact that ζs and ηs are nonincreasing, it can be proved that Z∞ Z ∞ ζs ηs ds = +∞ ds = +∞ 2 2 1 n +1 1 n + 1 + 1 and ζs/n2 + 1, ηs/n + 12 + 1 are also nonincreasing. Now, we complete the proof by the Min–Max Theorem. So Problem (1) has one unique solution by the equivalence of Problem (1) and Problem (8). REFERENCES 1. H.-W. Huang and Z.-H. Shen, On the two-point boundary value problem of Duffing equation with Neumann conditions, J. NanJing Univ. Math. Biquart. 16 No. 1 (1999), 8–11. 2. R. F. Manasevich, A min–max theorem, J. Math. Anal. Appl. 90 (1982), 64–71.
NOTE On a Unique Solution for the Two-Point Boundary Value Problem Zhang Shu-Qin1 and Zhong Cheng-Kui Department of Mathematics, Lanzhou University, Lanzhou, 730000, Gansu, People’s Republic of China Submitted by George Leitmann Received March 29, 2000
An existence and uniqueness result concerned with the boundary value problem u00 + gt; ut = et; is presented.
u0 0 = u0 π = 0
© 2000 Academic Press
1. INTRODUCTION Consider the boundary value problem u00 + gt; u = et;
u0 0 = u0 π:
(1)
In [1], Huang and Shen consider the existence and uniqueness result of the solution for (1). There they consider the case that gu t; u is non-negative. Here, we consider the case that gu t; u may be negative. In this paper, we also use the following Min–Max Theorem (see [2]) to prove our main result. Min–Max Theorem (Manasevich). Let H be a real Hilbert space and let f x H → R be of class C 2 . Suppose that there exist two closed subspaces X and Y such that H = X ⊕ Y and two continuous nonincreasing functions α x 0; ∞ → 0; ∞; 1
β x 0; ∞ → 0; ∞
This work was supported by the National Natural Science Foundation of China 653 0022-247X/00 $35.00 Copyright © 2000 by Academic Press All rights of reproduction in any form reserved.
654
note
such that
Z
∞
1
αsds = ∞
Z
∞
1
βsds = ∞
2 D f x + yk; k ≥ αyk2
for all x ∈ X, y ∈ Y , and k ∈ Y , and
2 D f x + yh; h ≤ −βxh2 for all x ∈ X, y ∈ Y , and h ∈ X. Then (a)
there exists a unique v0 ∈ H such that ∇f v0 = 0;
(b)
f v0 = max x∈X miny∈Y f x + y = miny∈Y max x∈X f x + y:
In the above theorem, ∇f u and D2 f u denote the gradient and the Hessian of f at u ∈ H; respectively. In this case, f x H → H is a C 0 mapping and ∇f 0 u = D2 f u is a bounded self-adjoint linear operator on H. Let U = uutR ∈ C 2 0; π; u0 0 = u0 π = 0; ut is absolutely continπ uous and satisfies 0 u2 tdt < +∞. Then U is a real Hilbert space with the inner product Zπ u0 v0 + uv dt: u; v = 0
The norm induced in U by the inner product · will be denoted by · U . We assume that: (A)
g ∈ C 0 0; π × R; R and e ∈ C0; πy
(B)
there exists some positive integer n such that
n2 + bn + c < gu t; u < n + 12
for all u ∈ U; t ∈ 0; π;
(2)
where n2 + bn + c < 0 and b; c satisfy c>0 b + c + 1 < 0:
(3)
2. THE MAIN RESULT In this section, we consider the existence of a unique solution for Problem (1). We let gu∗ t; u = gu t; u + 2n + 12 ;
note
655
then 3n2 + 4 + bn + 2 + c < gu∗ t; u < 3n + 12 : By (3), we see that 3n2 + 4n + 2 < 3n2 + 4 + bn + 2 + c: We denote ζs = min max gu∗ t; u − 3n2 + 4n + 2
(4)
ηs = min max n + 12 − gu t; u:
(5)
uU ≤st∈0;π
uU ≤st∈0;π
Clearly, ζx 0; +∞ → 0; +∞ and η x 0; +∞ → 0; +∞ are two continuous nonincreasing functions. Define subspaces X and Y of U as X = xxt =
n X a0 + am cos mt; am ∈ R; m = 0; 1; 2; · · · ; n 2 m=1 ∞ X
Y = yyt =
m=n+1
am cos mt; am ∈ R; m = n + 1; n + 2; · · ·;
(6)
(7)
P where n is as in (2) and the series yt = ∞ m=n+1 am cos mt as well as the series obtained by termwise differentiation of (7) converges uniformly on R. Clearly, we have U = X ⊕ Y and the inequalities Zπ Zπ Zπ Zπ x0 t2 dt ≤ n2 xt2 dt y 0 t2 dt ≥ n + 12 yt2 dt 0
0
0
0
for all x ∈ X; y ∈ Y . Next, we shall consider the problem u00 + g∗ t; u = 2n + 12 u + et;
u0 0 = u0 π = 0:
(8)
We note that Problem (8) is equivalent to Problem (1) by the construction of g∗ t; u. Define a function I x U → R by Z π 1 0 2 2 2 u t − Gt; ut + etut + n + 1 u dt; (9) Iu = 2 0 Ru where Gt; u = 0 g∗ t; sds; u ∈ U. Clearly, I is of class C 2 by ssumptions, A and B:
656
note
It is well known that u ∈2 o; π is a solution of problem (8) if and only if u is a critical point of I. The following is our main result. Theorem. Suppose that (A) and (B) hold, and Z∞ Z∞ ζsds = +∞; ηsds = +∞; 1
(10)
1
where ζ and η are as in (4) and (5). Then (a) there exists a unique solution u∗ of (8); (b) Iu∗ = max x∈X miny∈Y Ix + y = miny∈Y max x∈Y Ix + y, where Iu is defined in (9) and X and Y are defined in (6) and (7). Proof. First of all, it is easy to prove that a mapping ∇I x U → U is defined by Zπ u0 v0 − g∗ t; uv + etv + 2n + 12 uv dt: ∇Iu; v = 0
And for all u ∈ U; ∀v ∈ U; w ∈ U, we have Zπ 0 0
2 v w − gu∗ t; uvw + 2n + 12 wv dt: D Iuw; v = 0
For u = x + y ∈ U; x ∈ X; y ∈ Y , and h ∈ X; k ∈ Y , we have Z π 02
2 h − gu∗ t; uh2 + 2n + 12 h2 dt D Ix + yh; h = 0 Zπ n2 − gu∗ t; u + 2n + 12 h2 dt ≤ 0 Zπ = − gu∗ t; u − 3n2 + 4n + 2h2 dt 0
min
max gu∗ t; u − 3n2 + 4n + 2
Z
π
h2 dt 1 max gu∗ t; u − 3n2 + 4n + 2 min =− 2 n + 1 uU ≤xU t∈0;π Zπ Zπ h2 dt + h2 dt n2 0 0 1 min max gu∗ t; u − 3n2 + 4n + 2 ≤− 2 n + 1 uU ≤xU t∈0;π Z π Zπ 2 h0 dt + h2 dt ≤−
uU ≤xU t∈0;π
0
=−
ζ xU h2U n2 + 1
0
0
note
657
Z π 02 k − gu∗ t; uk2 + 2n + 12 k2 dt D2 Ix + yk; k = = = ≥
Z
0 π
Z
0 π
2 k0 − gu t; uk2 − 2n + 12 k2 + 2n + 12 k2 dt
2 k0 − gu t; uk2 dt
0
Z
π 0
2
k0 k0 − gu t; u n + 12 2
! dt
1 min max n + 12 − gu t; u n + 12 + 1 uU ≤yU t∈0;π Z π 1 2 k0 dt 1+ n + 12 0 1 min max n + 12 − gu t; u ≥ n + 12 + 1 uU ≤yU t∈0;π Zπ 2 0 2 k + k dt ≥
0
=
ηyU k2U n + 12 + 1
From (10) and the fact that ζs and ηs are nonincreasing, it can be proved that Z∞ Z ∞ ζs ηs ds = +∞ ds = +∞ 2 2 1 n +1 1 n + 1 + 1 and ζs/n2 + 1, ηs/n + 12 + 1 are also nonincreasing. Now, we complete the proof by the Min–Max Theorem. So Problem (1) has one unique solution by the equivalence of Problem (1) and Problem (8). REFERENCES 1. H.-W. Huang and Z.-H. Shen, On the two-point boundary value problem of Duffing equation with Neumann conditions, J. NanJing Univ. Math. Biquart. 16 No. 1 (1999), 8–11. 2. R. F. Manasevich, A min–max theorem, J. Math. Anal. Appl. 90 (1982), 64–71.