# 一維薛丁格方程第n 特徵值的震盪性

## 全文

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### 東吳大學數學系碩士班碩士論文 指導教授:朱啟平 教授

Oscillation of the n-th eigenvalue of one-dimensional Schrödinger equation

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### 誌謝(acknowledgment)

完成論文，首先誠摯的感謝我的指導教授朱啟平教授，

接下來要感謝  潘宗驛、王伶云、楊世揚助教協助我電腦 軟體的應用，讓我能夠用程式得心應手的輔助我的論文。

最後我想感謝系上所有教授以及秘書，謝謝你們這幾年 的教導及協助。

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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 n 特徵 值節點與頂點的位置與附近的震盪性。

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### Abstract

Using the method of Hadamard differential formula to investigate the position and the concussion of the nth eigenvalue node and vertex of the Schödinger equation.

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參考文獻(References)      16

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## 1 Introduction

We consider Schr¨odinger equation

φ00(x) − V (x)φ(x) + λ(x)φ(x) = 0, x ∈ (−π 2,π

2) (1)

with φ(−π2) = φ(π2) = 0, where V (x) ≥ 0 is the potential function. It is well- known (see[KZ]) that the Schr¨odinger operator H = −∆+V (x) acting on L2(Ω) with Dirichlet boundary conditions has purely discrete spectrum if Ω is a smooth bounded domain in <d. To specify the role of V (x), denote the nth Dirichlet eigenvalue of (1) by λn(V ) and let φVn(x) be the corresponding eigenfunction.

The potentials we interested are ”rectangular barriers” as following:

V := VA=

0 x ∈(−π2, A)

π

2−A x ∈[A,π2],

that is , VAis supported on [A,π2] with constant height, as well as Rπ2

π2

V = 2π.

In quantum mechanics, the rectangular barrier is a standard one-dimensional problem that demonstrates the phenomenon of wave-mechanical tunneling (i.e quantum tunneling) and wave mechanical reflection (see wiki/rectangular po- tential barrier). Since V ≥ 0, the eigenvalues are simple and λ1(V ) > 0, φVn has exactly (n-1) nodal points (see[KZ]). The eigenvalues list in ascending order are:

0 < λ1(V ) < λ2(V ) < λ3(V ) < λ4(V ) < ...

In this article, we are going to investigate how λn(VA) being varied with respect to A.

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## 2 Preliminaries

For later usage, preliminarily we consider for κ > 0, n > 0, x ∈ [−π2,π2],

F(x, κ, n) = πκ2(2x − π)2−(2x + π)2[(n − κ)2π+ π − 2x]. (2)

Lemma 1.

Fκ(x, κ, n) > 0, for x ∈ (−π2,π2), n − κ ≥ 0

Fx(x, κ, n) < 0, for x ∈ (−π2,π2) , n − κ ≥ 1

Fn(x, κ, n) < 0, for x ∈ (−π2,π2), n − κ > 0

[Proof ]

Direct computing shows that for x ∈ (−π2,π2) Fn= −2(2x + π)2(n − κ)π < 0 , if n − κ > 0

Fκ= 2πκ(2x − π)2+ 2(2x + π)2(n − κ)π > 0 , if n − κ ≥ 0.

Fx= 4πκ2(2x − π) − 4(2x + π)[(n − κ)2π+ π − 2x] + 2(2x + π)2

= 4πκ2(2x − π) − (2x + π)[4(n − κ)2π+ 4π − 8x − 2(2x + π)]

= 4πκ2(2x−π)−(2x+π)[4(n−κ)2π −12x+2π]

(3) If n − κ ≥ 1 , from (3), we have for x ∈ (−π2,π2),

Fx(x, κ) ≤ 4πκ2(2x − π) − (2x + π)[6π − 12x]

= (2x − π)[4πκ2+ 6(2x + π)] < 0 

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Lemma 2.

Fx(x, κ, n) < 0 , for x ∈ (−π2,π22nπ), 12 ≤ n − κ <1

[Proof.] from (3), we have

−Fx(x, κ, n) = (2x + π)[4(n − κ)2π −12x + 2π] + 4πκ2(π − 2x)

For fixed n ≥ 1, if (n − 1) < κ ≤ (n −12), (that is, 12 ≤ n − κ <1), then

−Fx(x, κ, n) > (2x + π)(3π − 12x) + 4π(n − 1)2(π − 2x)

= 3(2x + π)(π − 4x) + 4π(n − 1)2(π − 2x)

:= g(x)

We see that g(x) is a concave downward quadratic function, direct comput- ing shows that

g(−π2) = 8π2(n − 1)2≥0, also

g(π22nπ) = 3π2(2 − 1n)(n2 −1) + 4π2 (n−1)

2

n ,

=πn22(4n3−14n2+ 19n − 6).

Let h(n) = 4n3−14n2+ 19n − 6, then h0(n) = 12n2−28n + 19 > 0, ∀n.

So h(1) = 3 > 0 reveals that h(n) > 0, accordingly g(π22nπ) > 0 for all n ≥ 1.

We see that g(x) > 0, that is Fx<0 ,∀x ∈ (−π2,π22nπ). 

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## 3 Main results

In the following , first we shall study the variation of λn(VA) w.r.t A, specifically around those A’s that coincide with the zeros or peaks of φVnA(x).

kn

VnA

### (x)

fig 1.

Suppose A is the k-th zero of φVnA(x), n ≥ 2, 1 ≤ k ≤ n − 1. For convenience we denote φVnA , λVnA by φkn , λkn in this subsection .Then φkn is the k-th Dirichlet eigenfunction of φ00(x) + λφ = 0 in (−π2, A), as well as the (n − k)th Dirichlet eigenfunction of φ00(x) + (λ − VA)φ(x) = 0 in (A,π2). That is,

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







φkn00(x) + λknφkn(x) = 0, x ∈ (−π2, A),

φkn(−π2) = φkn(A) = 0,

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where λkn is the k-th eigenvalue of (4);

and









φkn00(x) + (λknπ

2−Akn(x) = 0, x ∈ (A,π2), φkn(A) = φkn(π2) = 0,

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where λknπ

2−A is the (n − k)th eigenvalue of (5).

From (4) we know that λkn= (kπ)

2

(π2+A)2,

from (5) we know that λkn= [(n−k)π](π 2 2−A)2 + π

2−A. Hence (π(kπ)2

2+A)2 = [(n−k)π]

2+2π(π2−A)

(π2−A)2 , that is, A must satisfy

π2k2(π − 2A)2= (π + 2A)2[(n − k)2π2+ π2−2Aπ]

Proposition 1 For n ≥ 2, 1 ≤ k ≤ n − 1,

(i)Suppose Akn = A is the kth-zero of φVnA(x), x ∈ [−π2,π2], then F (Akn, k, n) = 0 (ii)For fixed n,k, there is exactly one root of F (x, k, n) = 0 in (−π2,π2), and Akn is well-defined.

[Proof ]

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kn12

n

### (x)

fig 2.

Suppose A is the k-th peak of φVnA(x) n ≥ 1, 1 ≤ k ≤ n, Denote φVnA, λn(VA) by ψnk, ηknfor convenience in this subsection, then ψnk is k-th Dirichlet-Neumann eigenfunction of φ00+ λφ = 0 in (−π2, A) as well as the (n − k + 1)th Neumann- Dirichlet eigenfunction of φ00(x) + (λ − VA)φ(x) = 0 in (A,π2). That is,

ψkn00(x) + ηknψnk(x) = 0 , x ∈(−π2, A)

ψkn(−π2) = 0, ψkn0(A) = 0 (6)

where ηnk is the kth eigenvalue of (6);

ψkn00+ (ηknπ

2−Akn(x) = 0, x ∈ (A,π2)

ψkn0(A) = 0, ψkn(π2) = 0. (7)

where ηnkπ

2−A is the (n − k + 1)th eigenvalue of (7).

From (6), we know that ηkn= (k−

1 2)2π2 (A+π2)2 ,

from (7), we know that ηnk = (n−k+

1 2)2π2 (π2−A)2 + π

2−A. Just replace k by k − 12 in proposition 1, we have

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As in proposition 1, Ak−

1

n 2 is welldefined , and we have Proposition 2.

(i) For n ≥ 1, 1 ≤ k ≤ n, suppose Ak−

1 2

n = A is the kth peak of φVnA(x), x ∈(−π2,π2), then F (Ak−

1 2

n , k −12, n) = 0

(ii) An−

1

n 2 < π2π

2n

[proof ]

We need only to prove (ii). Since φ

V

An−1 n 2

n is the 1st Neumann-Dirichlet eigen- function of φ00+ (λ − V

An−

1 2 n

)φ in [An−

1

n 2,π2] , to get the nontrivial solution φ, we see that λ−V

An−

1 n 2

>0, and there is only one ”41wave” of φ

V

An−1 n 2

n

on [An−

1 2

n ,π2]. On the other hand, φ

V

An−1 n 2

n is the nth Dirichlet-Neumann eigen- function of φ00+ λφ = 0 in [−π2, An−

1 2

n ], there are totally (2n-1) ”14 waves” of φ

V

An−1 n 2

n in (−π2, An−

1

n 2) . Since 0 < λ − V

An−

1 n 2

< λ, the length of ”14 wave” in [An−

1

n 2,π2] is larger than that in [−π2, An−

1

n 2]. That is

π 2 − An−

1

n 2 >An−

1

n 2 +π2 (2n − 1)

So

(2n − 1)(π

2 − An−12) > An−

1

n 2 +π 2

and

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π 2 − π

2n= π(n − 1) 2n > An−

1

n 2

and we complete the proof of (ii) 

Theorem 1. For n ≥ 1, km= m2, m= 1, 2, 3..., (2n − 1),

Aknm is increasing with respect to m and is decreasing w.r.t n. That is,

A

1

n2 < A1n < A

3

n2 < A2n < ... as well as A

1 2

1 > A

1 2

2 > ... > A

1

n2 > ...etc.

[Proof.]

From propositions 1,2, for fixed n, we know that Aknm is the root of

F(x, km, n) = 0 with n − km12, n ≥1, m = 1, 2, ..., (2n − 1)and F (x, κ, n) is strictly decreasing in (−π2,π2), therefore for x ∈ (−π2,π2).

F(x,κ,n)=0 can be represented as F (x(κ), κ, n) = 0 for fixed n and

F(x(n), κ, n) = 0, for fixed κ. From Lemmas 1,2 and implicit differentiation, for fixed n, we have dx = −FFκx >0 , if n − κ ≥ 1 ,x ∈ (−π2,π2), or if

1

2 ≤ n − κ <1, x ∈ (−π2,π22nπ). Observe that Aknm ∈(−π2,π2),

n − km ≥ 1, if m = 1, 2, ..., (2n − 2); also from proposition 2(ii), Aknm ∈ (−π2,π22nπ), n − km=12, if m= 2n − 1, therefore Aknm is increasing w.r.t. m . On the other hand , for fixed κ, we have dxdn= −FFnx <0, so Aκnm is decreasing w.r.t. n. 

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n

A

κn

### , κ = k or k −

12

Below we shall use Hadamard differentiation formula, we formulate them ac- cording to our situation in the following.

Theorem A. [ref. (H, Theorem 2.5.11)] Suppose V, V are rectangular barriers with supp(V )=[A , L] where V has height τ > 0 on supp V ; supp(V)=[A,L]

V has height τ () := τL−A(L−A) on supp V.

Suppose τ ()is smooth with respect to , andRL

−Lφ2n,Vdx= 1, then

λ0n(0) := lim

→0

λn(V) − λn(V )

 = τ0(0)(L − A)[

RL

Aφ2n,V(x)dx

L − A − φ2n,V(A)] (8)

Theorem 2.

λn(VA) is increasing w.r.t. A around A = Aκn, k = 1, 2, ..., n − 1

λn(VA) is decreasing w.r.t. A around A = Aκ−

1

n 2, k = 1, 2, ..., n

[Proof ]

Assume A > A for  > 0 , then τ = π

2−A > π

2−A = τ (0) for  > 0, so τ0(0) > 0 in (8). When V = VAkn , A = Akn is a zero of φn,VAk

n, from (8), we have λ0n>0 at V = VAkn. When V = V

Ak−

1 2 n

, A= Ak−

1

n 2 is a peak of φn,V

Ak−1/2 n

, from (8) , we have λ0n <0 at V = VAk−1/2

n . That is , the nth Dirichlet eigen- value of (1) will increase w.r.t. A around A = Akn , and will decrease around A= AAk−1/2

n . 

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## 4 Examples

### 4.1 Example for Theorem 1.

Recall F (x, κ, n) = πκ2(2x − π)2−(2x + π)2[(n − κ)2π+ π − 2x] , x ∈[−π2,π2], κ > 0 , n > 0

(1) A

1 2

1 is the root of F (x,12,1) = π4(2x − π)2−(2x + π)2[54π −2x] = 0

A

1 2

1 ≈ −0.69907; from equation (6), we also know that

λ

1 2

1 = (

1 2π)2 (π2+A

1 2 1)2

≈3.24698

(2) A

1 2

2 is the root of F (x,12,2) = π4(2x − π)2−(2x + π)2[134π −2x] = 0

A

1 2

2 ≈ −0.932392; from equation (6), we also know that

λ

1 2

2 = (

3 2π)2 (π2+A

1 2 2)2

≈6.05408

(3) A

1 2

3 is the root of F (x,12,3) = π4(2x − π)2−(2x + π)2[(294)π − 2x] = 0

A

1 2

3 ≈ −1.09757; from equation (6), we also know that

λ

1 2

3 = (

5 2π)2 (π2+A

1 2 3)2

≈11.018

We see : A

1 2

1 > A

1 2

2 > A

1 2

3 (see figure 3.)

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figure 3.

On the other hand,

(4) A

1 2

3 is the root of F (x,12,3) = π4(2x − π)2−(2x + π)2[(294)π − 2x] = 0

A

1 2

3 ≈ −1.09757; from equation (6), we also know that

1

= (

1π)2

11.018

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(5) A13 is the root of F (x, 1, 3) = π(2x − π)2−(2x + π)2[5π − 2x] = 0

A13≈ −0.625503; from equation (4), we also know that

λ13= (π π2

2+A13)2 ≈11.045 (6) A

3 2

3 is the root of F (x,32,3) = 4(2x − π)2−(2x + π)2[134π −2x] = 0

A

3 2

3 ≈ −0.155689; from equation (6), we also know that

λ

3 2

3 = (

3 2π)2 (π2+A

3 2 3)2

≈11.0893

(7) A23 is the root of F (x, 2, 3) = 4π(2x − π)2−(2x + π)2[2π − 2x] = 0

A23≈0.308752; from (4), we also know that

λ23= (π(2π)2

2+A23)2 ≈11.1751 (8) A

5 2

3 is the root of F (x,52,3) = 254π(2x − π)2−(2x + π)2[54π −2x] = 0

A

5 2

3 ≈0.754791; from equation (6), we also know that

λ

5 2

3 = (

5 2π)2 (π2+A

5 2 3)2

≈11.4055

We see : A

1 2

3 < A13< A

3 2

3 < A23< A

5 2

3

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### 4.2 Examples for Theorem2

(1) A = A

1 2

3 = −1.09757,

λ

1 2

3 := λ

1 2

3(VA) = (

1 2π)2

(π2+A)2 ≈11.018

Numerically:(by matlab)

Al= −1.09955 < A

1 2

3, λ3(VA) = 11.0320 > λ

1 2

3

Ar= −1.07337 > A

1 2

3, λ3(VA) = 11.0096 < λ

1 2

3,

we see λ3is decreasing around A = A

1 2

3

(2) A = A13= −0.625503,

λ13:= λ13(VA) =(ππ2

2+A)2 ≈11.045 Numerically:(by matlab)

Al= −0.62831 < A13, λ3(VA) = 11.0322 < λ13

Ar= −0.60213 > A13, λ3(VA) = 11.0622 > λ13,

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(3) A = A

3 2

3 = −0.155689,

λ

3 2

3 := λ

3 2

3(VA) = (

3 2π)2

(π2+A)2 ≈11.0893

Numerically:(by matlab)

Al= −0.15707 < A

3 2

3, λ3(VA) = 11.1127 > λ

3 2

3

Ar= −0.13089 > A

3 2

3, λ3(VA) = 11.0847 < λ

3 2

3,

we see λ3is decreasing around A = A

3 2

3

(4) A = A23= 0.308752,

λ23:= λ23(VA) =(π(2π)2

2+A)2 ≈11.1751 Numerically:(by matlab)

Al= 0.28797 < A23, λ3(VA) = 11.1228 < λ23

Ar= 0.31415 > A23, λ3(VA) = 11.1824 > λ13,

we see λ3is increasing around A = A23

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(5) A = A

5 2

3 = 0.754791,

λ

5 2

3 := λ

5 2

3(VA) = (

5π 2)2

(π2+A)2 ≈11.4055

Numerically:(by matlab)

Al= 0.73303 < A

5 2

3, λ3(VA) = 11.5118 > λ

5 2

3

Ar= 0.78539 > A

3 2

3, λ3(VA) = 11.3845 < λ

5 2

3

we see λ3is decreasing around A = A

5 2

3

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References

[KZ] Kong,Q. and Zettl,A. , Eigenfunctions of regular Stum-Liouville problems, J.D.E. —3—(1996), 1-19

[H] Henrot,A., Extreme problems for eigenvalues of elliptic operators, Front.Math., Birh¨auser Verlag, Basel, 2006

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