東吳大學數學系碩士班碩士論文
指導教授︰朱啟平 教授
一維薛丁格方程之第二特徵函數其節點位置的變化
Variation of the nodal points of the 2nd eigenfunction of one-dimensional Schrödinger equation.
研究生︰許皓強 撰
中華民國一○三年一月
誌謝 （acknowledgment）
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摘要
使用阿達瑪微分公式的手法來研究薛丁格方程第二特徵函數的節點位置與 其勢函數支撐區間位置的相對關係。
ii
Abstract
With the same idea as Hadamard differentiation formula to investigate the variation of the position of nodal points of the 2nd eigenfunction of Schrödinger equation with respect to the location of the support of its potential.
目錄 (Table of Contents)
章節 標題 頁次
誌 謝 (Acknowledgment) ……… i
摘 要 (Abstract) ……… ii
目 錄 (Table of Contents) ……… iii
第一章 介 紹 (Introduction) ……… 1
第二章 主要結果 (Main results) ………… 3
第三章 例 子 (Examples) ……… 10
參考文獻 (References) ………… 18
iii
1 Introduction
We consider one-dimensional Schr¨odinger equation with Dirichlet boundary condition :
φ00(x) − V (x)φ(x) + λφ(x) = 0, (1)
where x ∈ (−π 2,π
2), φ(−π
2) = φ(π
2) = 0, and V (x) ≥ 0 is the potential function.
Denote λ(V ) to be the 2nd Dirichlet eigenvalue of (1), φV the corresponding 2nd Dirichlet eigenfunction.
We are interested in rectangle-potentials: suppV (that is, support of V) is a closed interval and V ≡ constant on suppV . These potentials are interested in the study of the tunneling effects of the valence band problems.
Specifically, we consider the following potential in this paper: let
A = {V |suppV = [a, b] ⊆ (−π 2,π
2), V ≡ constant on [a, b], Z π2
−π2
V(x)dx = 2π}.
It is well-known that φV(x) has exactly one nodal point in (−π 2,π
2) and φ0V is absolutely continuous (see [E, Theorem 0]) for V ∈ A. Moreover, the smooth- ness of the position of nodal point with respect to the translation of support V (see [T]).
We shall use the same idea as Hadamard differentiation formula (Theorem 1) to investigate the variation of the position of nodal points with respect to the location of suppV .
In fact, we demonstrate that for V ∈ A, when the nodal point lies inside suppV, it tends to escape suppV from the left as suppV moved toward right (Corollary 1). For the nodal point lying outside suppV , as suppV moved to- ward right, the variation of the position of nodal points depends on which side are they located with respect to suppV (Theorem 2).
First we fix notations. For α ≥ 0, 0 ≤ || 1, let Vα ∈ A be the potential defined by
Vα(z) =
( α z∈ [a + , b + ] ⊆ (−π 2,π
2) 0 otherwise
with α(b − a) = 2π.
Fix α, for 0 ≤ || 1, denote φVα by u, and z() ∈ (−π 2,π
2) be the nodal point of u. Denote
λ0(0) = lim
→0
λ(Vα) − λ(V0α)
, z0(0) = lim
→0
z() − z(0)
.
Mainly we study the relation between λ0(0) and z0(0) in this article.
One purpose is, through these simple-looked V , we can see how the position of nodal points, and the eigenpairs influence each other.
The main results are stated in Section 2. The examples are given in Section 3 simulated by Matlab.
2
2 Main Results
The main tool we shall use is an improved Hadamard formula which is de- rived with the same idea as Hadamard differentiation formula with respect to the translation of suppV . We state and prove the original Hadamard formula in our case for reader’s convenience.
Theorem 1. (Hadamard formula) (ref. [H, Proposition 1.4]) Suppose (λ(), u) is the 2nd eigenpair of (1) with V ≡ Vα, Rπ2
−π2u2dx= 1 and (u0(−π2) · u00(−π2)) > 0, then λ0(0) = α[(u0)2(b) − (u0)2(a)].
Proof.
Denote V= Vα, λ() = λ(Vα) for 0 ≤ || 1.
Let φ, ψ ∈ C0∞[−π2,π2], then from (1) Rπ2
−π
2[u000ψ(x) − u0V0(x)ψ(x) + λ(0)u0ψ(x)]dx = 0 Integration by parts, we get
Z π2
−π2
[u00ψ0+ αχ[a,b]u0ψ]dx = λ(0) Z π2
−π2
u0ψdx (2)
Similarly, Z π2
−π
2
[u0φ0+ αχ[a+,b+]uφ]dx = λ() Z π2
−π
2
uφdx (3)
Substituting φ = u0 and ψ = uinto (2),(3) respectively, Z π2
−π
2
[u00u0+ αχ[a,b]u0u]dx = λ(0) Z π2
−π
2
u0udx (4)
Z π2
−π
2
[u0u00+ αχ[a+,b+]uu0]dx = λ() Z π2
−π
2
uu0dx (5) (5) − (4)
[λ() − λ(0)]
Z π2
−π2
uu0dx= α[
Z b+
b
uu0dx− Z a+
a
uu0dx]
= α[uu0(x1) − uu0(x2)] (6)
where x1 ∈ (b, b + ), x2∈ (a, a + ) Knowing that lim→0Rπ2
−π2 uu0dx=Rπ2
−π2 u20(x)dx = 1 ([H, proof of Proposi- tion 1.4]), we have from (6)
λ0(0) = lim
→0
λ() − λ(0)
= α[(u0)2(b) − (u0)2(a)].
In Theorem 2 we investigate the differentiation of the position of the nodal point when suppV is under a translation action. In some sense, the formula in Theorem 2 can present the ”detail” of the formulas in Theorem 1 (i.e.
(7)+(8) is the Hadamard formula in Theorem 1).
Theorem 2.
Suppose (λ(), u) is the 2nd eigenpair of (1) with V ≡ Vα, Rπ2
−π2u2dx= 1 and (u0(−π2) · u00(−π2)) > 0.
(i) Suppose z(0) ∈ [a, b], it holds
λ0(0) Z z(0)
−π2
u20dx+ αu20(a) = −z0(0)u002(z(0)) (7)
λ0(0) Z π2
z(0)
u20dx− αu20(b) = z0(0)u002(z(0)) (8) (ii) If z(0) ∈ (−π2, a), then
λ0(0) = −2π2z0(0)
[z(0) +π2]3 (9)
If z(0) ∈ (b,π2), then
λ0(0) = 2π2z0(0)
[π2− z(0)]3 (10)
From (7), (8), we see that the nodal points inside suppV tends to escape from the left of suppV to the outside as suppV moved toward right:
Corollary 1. If z(0) ∈ (a, b), then z0(0) < 0
4
Proof.
If λ0(0) ≥ 0, then (7) implies z0(0) < 0 If λ0(0) < 0, then (8) implies z0(0) < 0
We prove Theorem 2 through the following Lemmas.
Lemma 1.
If z(0) ∈ (a, b), z() ∈ (a + , b + ), then
λ0(0) Z z(0)
−π2
u20dx+ αu20(a) = −z0(0)u002(z(0))
λ0(0) Z
π 2
z(0)
u20dx− αu20(b) = z0(0)u002(z(0)) Proof.
We start by proving the first equality. Viewing u, u0 as the 1st Dirichlet eigenfunction of (1) with V ≡ 0 over [−π2, z()] and [−π2, z(0)] respectively, sim- ilar to the computation as in (4), (5). We have
Z z()
−π2
u0u00dx+ αχ[a+,z()]uu0dx= λ() Z z()
−π2
uu0dx (11) Z z(0)
−π2
u00u0dx+ αχ[a,z(0)]u0udx= λ(0) Z z(0)
−π2
u0udx (12) The right hand side of (11) − (12) is:
λ() Z z()
−π2
uu0dx− λ(0) Z z(0)
−π2
u0udx
= λ() Z z()
−π2
uu0dx− λ(0)[
Z z()
−π2
+ Z z(0)
z()
]uu0dx
= [λ() − λ(0)]
Z z()
−π
2
uu0dx− λ(0) Z z(0)
z()
uu0dx
= · {[λ() − λ(0)
]
Z z()
−π
2
uu0dx−λ(0)
Z z(0) z()
uu0dx}
= · {[λ() − λ(0)
]
Z z()
−π2
uu0dx− λ(0)z(0) − z()
uu0(x1)}, where x1 lies between z(0), z() and lim→0uu0(x1) = u20(z(0)) = 0
The left hand side of (11) − (12) is:
Z z()
−π2
u0u00dx+ α Z z()
a+
uu0dx− ( Z z(0)
−π2
u00u0dx+ α Z z(0)
a
u0udx)
= −α[
Z a+
a
uu0dx+ Z z(0)
z()
uu0dx] − Z z(0)
z()
u0u00dx
= ·(−α)[(a + − a)(uu0)(ˆa)
+(z(0) − z())(uu0)(ˆz)
]−(z(0) − z())(u0u00)(˜z)
},
where ˆalies between a, a + , ˆz, ˜z lie between z(0) and z() and lim→0[z(0)−z()]
uu0(ˆz) = −z0(0) · u20(z(0)) = 0, Consider lim→0(11)−(12)
, we have λ0(0)Rz(0)
−π
2
u20dx= −αu20(a) − z0(0)u002(z(0))
Similarly, we prove the 2nd equality. Viewing u, u0 as the 1st Dirichlet eigenfunction over of (1) [z(),π2] and [z(0),π2] respectively, we have
Z π2
z()
u0u00+ αχ[z(),b+]uu0dx= λ() Z π2
z()
uu0dx (13) Z π2
z(0)
u00u0+ αχ[z(0),b]u0udx= λ(0) Z π2
z(0)
u0udx (14) The right hand side of (13) − (14) is:
λ() Z π2
z()
uu0dx− λ(0) Z π2
z(0)
u0udx
= λ()[
Z z(0) z()
+ Z π2
z(0)
]uu0dx− λ(0) Z π2
z(0)
uu0dx
= [λ() − λ(0)]
Z π2
z(0)
uu0dx+ λ() Z z(0)
z()
uu0dx
= · {[λ() − λ(0)]
Z π2
z(0)
uu0dx+λ()
Z z(0)
z()
uu0dx}
= · {λ() − λ(0)
Z π2
z(0)
uu0dx+ λ()z(0) − z()
uu0(x2)},
where x2 lies between z(0), z() and lim→0uu0(x2) = u20(z(0)) = 0
6
The left hand side of (13) − (14) is:
Z π2
z()
u0u00dx+ α Z b+
z()
uu0dx− ( Z π2
z(0)
u00u0dx+ α Z b
z(0)
u0udx)
= α[
Z z(0) z()
uu0dx+ Z b+
b
uu0dx] + Z z(0)
z()
u0u00dx
= ·α[(z(0) − z())(uu0)(ˆz)
+(b + − b)(uu0)(ˆb)
]+(z(0) − z())(u0u00)(˜z)
},
where ˆb lies between b, b + , ˆz, ˜z lie between z(0) and z() and lim→0[z(0)−z()]
uu0(ˆz) = z0(0) · u20(z(0)) = 0, Consider lim→0(13)−(14)
, we have λ0(0)R
π 2
z(0)u20dx= αu20(b) + z0(0)u002(z(0)).
Lemma 2.
(i) If z(0) ∈ [a, b], z() < a + , ∀, then
λ0(0) Z z(0)
−π2
u20dx= −z0(0)u002(z(0)) (ii) If z(0) ∈ [a, b], z() > b + , ∀, then
λ0(0) Z π2
z(0)
u20dx= z0(0)u002(z(0)) Proof.
(i) In this case, z(0) = a since z() → z(0) as → 0. With the same com- putation as in Lemma 1 with the 2nd term vanishes in both (11) and (12), or equivalently, plug u(a) = 0 into (7), we get the result.
(ii) In this case, z(0) = b since z() → z(0) as → 0. With the same com- putation as in Lemma 1 with the 2nd term vanishes in both (13) and (14), or equivalently, plug u(b) = 0 into (8), we get the result.
Lemma 3.
(i) If z(0) < a, z() < a + , 0 < || 1, then λ0(0) = −2π2z0(0) [z(0) +π2]3 (ii) If z(0) > b, z() > b + , 0 < || 1, then λ0(0) = 2π2z0(0)
[π2− z(0)]3 Proof.
(i) If z(0) ∈ (−π
2, a), z() < a + , then λ() = ( π
z() +π2)2, so λ0(0) = π2[−2 · (z(0) +π
2)−3· z0(0)] = −2π2z0(0) [z(0) +π2]3 (ii) If z(0) ∈ (b,π
2), z() > b + , then λ() = ( π
π
2 − z())2, so λ0(0) = π2[−2 · (π
2 − z(0))−3· (−z0(0))] = 2π2z0(0)
[π2− z(0)]3. Through Lemma 1 to Lemma 3, we prove Theorem 2 to be true.
Remark.
When z(0) = a or z(0) = b, (7) = (9), (8) = (10) in Theorem 2, that is, (i) when z(0) = a, λ0(0) = −2π2z0(0)
(a +π2)3 = −z0(0)u002(a) Z a
−π
2
u20dx (ii) when z(0) = b, λ0(0) = 2π2z0(0)
(π2 − b)3 = z0(0)u002(b) Z
π 2
b
u20dx Proof.
(i) Since z(0) = a and suppV = [a, b], we have
u0(x) = sin[
π(x +π 2) a+π
2
] on [−π 2, a],
u00(a) = {[cos
π(x +π 2) a+π
2
] · ( π a+π
2
)}|x=a= − π (a +π
2) and
8
Z a
−π
2
u20dx = Z a
−π
2
sin2[π(x +π2) a+π2 ]dx =
Z a
−π
2
[
1 − cos2π(x +π2) a+π2
2 ]dx
= x 2 −
sin2π(x + π2) a+π2
4 π
|a−π
2 = a+π2 2
So −u002(a) Z a
−π
2
u20dx
=
− π2 (a +π
2)2 a+π
2 2
= −2π2 (a +π
2)3, and the conclusion is true.
(ii) Since z(0) = b and suppV = [a, b], we have
u0(x) = sin[
π(π 2 − x) π
2 − b ] on [b,π 2],
u00(b) = {[cos π(π
2 − x)
π
2 − b ] · ( −π π
2 − b)}|x=b= π (π
2 − b) and Z π2
b
u20dx = Z π2
b
sin2[π(π2 − x)
π
2− b ]dx = Z π2
b
[
1 − cos2π(π2 − x)
π 2 − b
2 ]dx
= x 2 +
sin2π(π2 − x)
π 2 − b
4 π
|
π 2
b =
π 2 − b
2
So u002(b) Z π2
b
u20dx
= π2 (π
2 − b)2 π 2 − b
2
= 2π2 (π
2 − b)3, and the conclusion is true.
3 Examples
We give some examples simulated by Matlab.
Example 1.
(i) V0 = 3χ[−π
4,5π12], V= 3χ[−π
4+,5π12+] with = 240π , Rπ2
−π2 V0dx=Rπ2
−π2 Vdx= 2π.
(ii) In (7), (8), we have α= 3, a = -π
4 ∼ −0.7854; b = 5π
12 ∼ 1.3090, u20(a) = u20(−π
4) = 0.5216; u20(b) = u20(5π
12) = 0.1526 (iii) Estimate z(0), z().
Observe that u0(−14π
240) = -0.0033, u0(−13π
240) = 0.0158 Let P = (−14π
240,−0.0033), Q = (−13π
240,0.0158) We shall use the slope of P Q to estimate u0(z0) and its x-intercept to estimate z0:
the slope of P Q is 0.0158 − (−0.0033) π
240
= 1.4591 and
P Qintersect x-axis at x = −14π
240 + 0.0033
1.4591 = -0.1810 Therefore z(0) ∼ -0.1810 ∈ (a, b) = (−0.7854, 1.3090), u0(z(0)) ∼ 1.4591, u002(z(0)) ∼ 2.1290
(iv) Similarly observe that u(−15π
240) = -0.0126, u(−14π
240) = 0.0066 u0(z()) ∼ 0.0066 − (−0.0126)
π 240
= 1.4668 and
z() ∼ −15π
240 + 0.0126
1.4668 = -0.1878 ∈ (a + , b + ) = (−0.7723, 1.3221) (v) λ() = 6.1388, λ(0) = 6.1538
λ0(0) ∼ λ() − λ(0)
= 6.1388 − 6.1538 π 240
= -1.1459
z0(0) ∼ z() − z(0)
= −0.1878 − (−0.1810) π
240
= -0.5195
10
(vi) Plug there approximate values into
λ0(0) Z z(0)
−π2
u20dx+ αu20(a) = −z0(0)u002(z(0)) (7) We estimate the first term:
Z z(0)
−π2
u20dx= Z −14π240
−π2
u20+ (1.45912) ·[z(0) − (−14π240)]3 3
= 0.4031000777 so the left hand side is
-1.1459 · 0.4031000777 + 3 · (0.5216) = 1.1029 the right hand side is
0.5195 · 2.1290 = 1.1061
(vii) Plug there approximate values into
λ0(0) Z π2
z(0)
u20dx− αu20(b) = z0(0)u002(z(0)) (8) Z π2
z(0)
u20dx= Z π2
−14π
240
u20+ (1.45912) · [z(0) − (−14π240)]3 3
= 0.5969000777 the left hand side is
-1.1459 · 0.5969000777 - 3 · (0.1526) = -1.1418 the right hand side is
-0.5195 · 2.1290 = -1.1061
V3= 3χ
Example 2.
(i) V0 = 4χ[−π6,π3], V= 4χ[−π6+,π3+] with = 240π , Rπ2
−π2 V0dx=Rπ2
−π2 Vdx= 2π.
(ii) In (7), (8), we have α= 4, a = -π
6 ∼ −0.5236; b = π
3 ∼ 1.0472, u20(−π
6) = 0.0870; u20(π
3) = 0.6509 (iii) Estimate z(0), z().
Observe that u0(−23π
240) = -0.0044, u0(−22π
240) = 0.0131 Let P = (−23π
240,−0.0044), Q = (−22π
240,0.0131) We shall use the slope of P Q to estimate u0(z0) and its x-intercept to estimate z0:
the slope of P Q is 0.0131 − (−0.0044) π
240
= 1.3369 and
P Qintersect x-axis at x = −23π
240 + 0.0044
1.3369 = -0.2978 Therefore z(0) ∼ -0.2978 ∈ (a, b) = (−0.5236, 1.0472), u0(z(0)) ∼ 1.3369, u002(z(0)) ∼ 1.7873
(iv) Similarly observe that u(−24π
240) = -0.0140, u(−23π
240) = 0.0036 u0(z()) ∼ 0.0036 − (−0.0140)
π 240
= 1.3445 and
z() ∼ −24π
240 + 0.0140
1.3445 = -0.3037 ∈ (a + , b + ) = (−0.5105, 1.0603) (v) λ() = 6.2599, λ(0) = 6.2312
λ0(0) ∼ λ() − λ(0)
= 6.2599 − 6.2312 π 240
= 2.1925
z0(0) ∼ z() − z(0)
= −0.3037 − (−0.2978) π
240
= -0.4507
12
(vi) Plug there approximate values into
λ0(0) Z z(0)
−π2
u20dx+ αu20(a) = −z0(0)u002(z(0)) (7) Z z(0)
−π2
u20dx= Z −23π240
−π2
u20+ (1.33692) ·[z(0) − (−23π240)]3 3
= 0.2163000214 the left hand side is
2.1925 · 0.2163000214 + 4 · (0.0870) = 0.8222 the right hand side is
0.4507 · 1.7873 = 0.8055
(vii) Plug there approximate values into
λ0(0) Z π2
z(0)
u20dx− αu20(b) = z0(0)u002(z(0)) (8) Z π2
z(0)
u20dx= Z π2
−23π
240
u20+ (1.33692) · [z(0) − (−23π240)]3 3
= 0.7837000214 the left hand side is
2.1925 · 0.7837000214 - 4 · (0.6509) = -0.8853 the right hand side is
-0.4507 · 1.7873 = -0.8055
V04= 4χ[−π6,π3]
Example 3.
(i) V0 = 5χ[−π4,3π20], V= 5χ[−π4+,3π20+] with = 240π , Rπ2
−π2 V0dx=Rπ2
−π2 Vdx= 2π.
(ii) In (7), (8), we have α= 5, a = −π
4 ∼ −0.7854; b = 3π
20 ∼ 0.4712, u20(−π
4) = 0.9436; u20(3π
20) = 0.0634 (iii) Estimate z(0), z().
Observe that u0(18π
240) = -0.0009, u0(19π
240) = 0.0132 Let P = (18π
240,−0.0009), Q = (19π
240,0.0132)
We shall use the slope of P Q to estimate u0(z0) and its x-intercept to estimate z0:
the slope of P Q is 0.0132 − (−0.0009) π
240
= 1.0772 and
P Qintersect x-axis at x = 18π
240 + 0.0009
1.0772 = 0.2365 Therefore z(0) ∼ 0.2365 ∈ (a, b) = (−0.7854, 0.4712), u0(z(0)) ∼ 1.0772, u002(z(0)) ∼ 1.1604
(iv) Similarly observe that u(17π
240) = -0.0011, u(18π
240) = 0.0130 u0(z()) ∼ 0.0130 − (−0.0011)
π 240
= 1.0772 and
z() ∼ 17π
240 + 0.0011
1.0772 = 0.2236 ∈ (a + , b + ) = (−0.7723, 0.4843) (v) λ() = 5.6550, λ(0) = 5.7119
λ0(0) ∼ λ() − λ(0)
= 5.6550 − 5.7119 π 240
= -4.3468
z0(0) ∼ z() − z(0)
= 0.2236 − (−0.2365) π
240
= -0.9855
14
(vi) Plug there approximate values into
λ0(0) Z z(0)
−π2
u20dx+ αu20(a) = −z0(0)u002(z(0)) (7) Rz(0)
−π
2
u20dx= Z 18π240
−π2
u20+ (1.07722) · [z(0) −18π240]3 3
= 0.8271000003 the left hand side is
-4.3468 · 0.8271000003 + 5 · (0.9436) = 1.1228 the right hand side is
0.9855 · 1.1604 = 1.1436
(vii) Plug there approximate values into
λ0(0) Z π2
z(0)
u20dx− αu20(b) = z0(0)u002(z(0)) (8) Rπ2
z(0)u20dx= Z π2
18π 240
u20+ (1.07722) ·[z(0) −18π240]3 3
=0.1729000003 the left hand side is
-4.3468 · 0.1729000003 - 5 · (0.0634) = -1.0686 the right hand side is
-0.9855 · 1.1604 = -1.1436
V5 = 5χ
Example 4.
(i) V0 = 8χ[0,π4], V= 8χ[0+,π4+] with = 240π , Rπ2
−π2 V0dx=Rπ2
−π2 Vdx= 2π.
(ii) In (9), we have α= 8, a = 0; b = π
4 ∼ 0.7854, u20(0) = 0.0954; u20(π
4) = 0.9791 (iii) Estimate z(0), z().
Observe that u0(−28π
240) = -0.0082, u0(−27π
240) = 0.0048 Let P = (−28π
240,−0.0082), Q = (−27π
240,0.0048) We shall use the slope of P Q to estimate u0(z0) and its x-intercept to estimate z0:
the slope of P Q is 0.0048 − (−0.0082) π
240
= 0.9931 and
Therefore P Q intersect x-axis at x = −28π
240 + 0.0082
0.9931 = -0.3583 z(0) ∼ −0.3583 < a = 0 , u0(z(0)) ∼ 0.9931
(iv) Similarly observe that u(−29π
240) = -0.0136, u(−28π
240) = 0.0001 u0(z()) ∼ 0.0001 − (−0.0136)
π 240
= 1.0446 and
z() ∼ −29π
240 + 0.0136
1.0446= −0.3666 < a + = 0.0131 (v) λ() = 6.8060, λ(0) = 6.7124
λ0(0) ∼ λ() − λ(0)
= 6.8060 − 6.7124 π 240
= 7.1505
z0(0) ∼ z() − z(0)
= −0.3666 − (−0.3583) π
240
= -0.6341
16
Plug there approximate values into λ0(0) = −2π2z0(0)
[z(0) +π2]3 (9)
the left hand side is 7.1505
the right hand side is
−2π2(−0.6341) [−0.3583 +π
2]3 = 12.5166
1.7826 = 7.0215
V08 = 8χ[0,π4]
References
[E] Everitt, W. N. , Kwong, M. K. and Zettl, A. , Oscillation of eigenfunc- tions of weighted regular Sturm-Liouville problem. J. London Math. Soc.
27(1983), 106-120.
[H] Harrell E.M. II, Kr¨oger, P, Kurata, K. , On the placement of an obsta- cle or a well so as to optimize the fundamental eigenvalue, SIAM J. Math.
Anal. Vol.33, No.1, p.240-259.
[T] Trynin, A. Yu, Differential properties of zeros of eigenfunctions of the Sturm-Liouville Problem, Ufa Mathematical Journal, vol.3, no.4, (2011), p.130-140.
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