Reconstruction of the Potential Function and its Derivatives for the Diffusion Operator

Reconstruction of the Potential Function and its Derivatives for the Diffusion Operator

Hikmet Koyunbakan and Emrah Yilmaz

Firat University, Department of Mathematics, 23119, Elazıg, Turkey Reprint requests to H. K. and E. Y.; E-mail: hkoyunbakan@gmail.com and emrah231983@gmail.com

Z. Naturforsch.63a,127 – 130 (2008); received October 24, 2007

We solve the inverse nodal problem for the diffusion operator. In particular, we obtain a reconstruc- tion of the potential function and its derivatives using only nodal data. Results are a generalization of Law’s and Yang’s works.

Key words:Diffusion Operator; Inverse Nodal Problem; Reconstruction Formula.

MSC 2000:34L40, 34A55, 34B99

1. Introduction

The inverse nodal problem was initiated by McLaughlin [1], who proved that the Sturm-Liouville problem is uniquely determined by any dense subset of the nodal points. Some numerical schemes were given by Hald and McLaughlin [2] for the reconstruc- tion of the potential. Recently Law, Yang and other authors have reconstructed the potential function and its derivatives of the Sturm-Liouville problem from the nodal points [3 – 7].

In this paper, we are concerned with the inverse nodal problem for the diffusion operator on a finite in- terval. We reconstruct the potential function and all its derivatives by using Law’s and Yang’s method [7].

The diffusion operator is written as

Ly=−y+ [q(x) +2λ^p(x)]y, (1.1) where the function q(x)∈ L²[0,π], p(x)∈L²[0,π]. Some spectral problems were extensively solved for the diffusion operator in [8 – 11].

Consider the problem

L[y] =λ^2y, (1.2)

y(0) =1, y(0) =−h, (1.3) y(π,λ) +Hy(π,λ) =0, (1.4) wherehandHare finite numbers.

Let λn be the n-th eigenvalue and 0<xⁿ₁ < ...

<xⁿ_i <π^,i=1,2,...,n−1, the nodal points of then-th

0932–0784 / 08 / 0300–0127 $ 06.00 c2008 Verlag der Zeitschrift f¨ur Naturforschung, T ¨ubingen·http://znaturforsch.com

eigenfunction. Also letI_iⁿ= [xⁿ_i,xⁿ_i+1]be thei-th nodal domain of the n-th eigenfunction and let l_iⁿ=|l_iⁿ|= xⁿ_i+1−xⁿ_i be the associated nodal length. Let j_n(x)be the largest indexjsuch that 0≤x⁽ⁿ⁾_j <x.

∆ denotes the difference operator∆a_i=a_i+1−a_i. Inductively, fork>1,∆^ka_i=∆^k−1a_i+1−∆^k−1a_i, and we introduce the difference quotient operatorδ^: δ^ai=a_i+1−ai

x_i+1−x_i=∆ai

l_i andδ^kai=δ^k−1ai+1−δ^k−1ai

l_i .

2. Main Results

Lemma 1.[12] Assume thatq∈L²[0,π]. Then, as n→∞for the problem (1.2) – (1.4),

xⁿ_i = i−¹₂

π λn − h

2λn²

+ 1 2λn²

xⁿ_i 0

(1+cos2λnt)[q(t) +2λnp(t)]dt+O

1 λn⁴

, (2.1)

l_iⁿ= π λn

+ 1 2λn²

xⁿ_i+1 xⁿ_i

(1+cos 2λnt)[q(t)

+2λnp(t)]dt+O 1

λn⁴

.

(2.2)

128 H. Koyunbakan and E. Yilmaz·Potential Function and its Derivatives Lemma 2.Suppose that f ∈L²[0,π]. Then for al-

most everyx∈[0,π], with j=j_n(x),

nlim→∞

λn

π

xⁿ_j+1 xⁿ_j

f(t)dt=f(x).

Theorem 1.[12] Suppose thatq∈L²[0,π], then

q(x) =lim

n→∞λn

2λn²lⁿ_j

π ⁻²λn−2p(x)

.

Proposition 1. Ifqis a continuous function, then (a) lim

n→∞

√λnl_i⁽ⁿ⁾=π^andli⁽ⁿ⁾=1 n+O

1 n² ; (b)l⁽ⁿ⁾_i+k/l_i+m⁽ⁿ⁾ =1+O₁

n

for any fixedk,m∈N;

(c)q_m≤λn− ^π2

(l⁽ⁿ⁾_i )² ≤q_M, whereq_m=min

[0,π]q(x)and qM=max

[0,π]q(x).

Lemma 3. [7] If q ∈C^N[0,π], then for k = 1,...,N,∆^kl_j=O(n^−(k+3))asn→∞and the order es- timate is independent ofj.

Lemma 4.[7] Let Φj= ∑^m

i=1φj,i with eachφj,i=

ki

p=1∏ ϕj,i,p, where each ϕj,i,p ∈U⁽ⁿ⁾_j . Suppose Φj = O(n^−v) and q is sufficiently smooth. Then δ^kΦj = O(n^−v)for allk∈N.

Lemma 5. [7] Suppose f ∈C^N[0,π] and Φj =

xj+1

x_j

f(x)dx. Then δ^kΦj = O(n⁻¹) for any k = 0,1,...,N.

Theorem 2. [7] Let Φm(x_j) =ψ1(x_j)ψ2(x_j)...

ψm(x_j), where ψi(x_j) =x_j+ki andk_i∈N∪ {0}. Ifq isC^kon[0,π], then

δ^kΦm(x_j) =





O(1), 0≤k≤m−1, m!+O(n⁻¹), k=m,

O(n⁻²), k≥m+1.



 Theorem 3. If q ∈ C^N+1[0,π], then q^(k)(x) = δ^kq(x_j)−2λnp(x)+O(n⁻¹)fork=0,1,...,N, where j= j_n(x). The order estimate is uniformly valid for compact subsets of[0,π].

Remark: Fork=1,...,N,

V_k(x_j) =







1 x_j ··· x^k_j 1 x_j+1 ··· x^k_j+1

... ... ... 1 x_j+k ··· x^k_j+k







be a(k+1)×(k+1)Vandermonde matrix. It is well known that

detV_k(x_j) =

∏

m=1

m−1

∏

m−1

∑

l_j+p

.

To prove Theorem 3, we need the following lemma.

Lemma 6.

∏k m=1l^m_j+k−m

detV_k(x_j) = 1

∏k m=1m!

+O(1

n). (2.3)

Next, we consider the following(k+1)×(k+1) matrix:

A=









1 x_j ··· x^k−1_j q(x_j) 1 x_j+1 ··· x^k−1_j+1 q(x_j+1) ... ... ... ... 1 x_j+k ··· x^k−1_j+k q(x_j+k)







.

After some operations, we obtain detA=l_jl_j+1...l_j+k−1

·det







1 x_j ··· x^k−1_j q(x_j) δ(1) δ^xj ··· · δ^q(xj)

... ... ... ...

δ(1) δ(x_j+k−1) ··· · δ^q(x_j+k−1)







= (lj)^k(l_j+1)^k−1...l_j+k−1

·det







1 xj ··· x^k−1_j q(xj) δ(1) δ^xj ··· · δ^q(x_j)

... ... ... ...

δ^k(1) δ^k(x_j) ··· · δ^kq(x_j)





.

Let B be the matrix at the right-hand side. By Lemma 4 and Theorem 2,

detB= (δ^xj)(δ^2x2j)...(δ^k−1xk−1j )(δ^kq(x_j)) +O

1 n²

=^k−1

∏

m=0(m!)δ^kq(x_j) +O 1

n

.

H. Koyunbakan and E. Yilmaz·Potential Function and its Derivatives 129 Proof of Theorem 3: Fork=1,2,...,N, let

g(x) =det







1 x ··· x^k q(x) 1 x_j ··· x^k_j q(x_j)

... ... ... ... 1 x_j+k ··· x^k_j+k q(x_j+k)





.

By Rolle’s theorem,g(x_j) =g(x_j+1) =...=g(x_j+k) = 0 implies that there is someξ1,j+i∈(x_j+i,x_j+i+1)such thatg(ξ1,j+i) =0. When we repeat the process, we can find thatξk,j∈(x_j,x_j+k)such thatg^(k)(ξk,j) =0.

In view of the definition ofg,q^(k)(ξk,j)detV_k(x_j) = k! detA. Hence

q^(k)(ξk,j) = (k!)(l_j)^k(l_j+1)^k−1...l_j+k−1 detB detV_k(xj). By Lemma 6 q^(k)(ξk,j) = δ^kq(x_j) +O(¹_n), since q ∈ C^k+1,q^(k)(x) = q^(k)(ξk,j)−2λnp(x) +O(¹_n) = δ^kq(xj)−2λnp(x) +O(¹_n).

Theorem 4. Suppose that q in (1.1) isC^N+1 on [0,π](N ≥1), and let j = j_n(x) for eachx∈[0,π]. Then, asn→∞,

q(x) =λn

2λn²lⁿ_j

π ⁻²λn−2p(x)

+O 1

n

,

and, for allk=1,2,...,N,

q^(k)(x)=2λn^3/2δ^klj

π ⁻²λnδ^kp(xj)−2λnp^(k)(x)+O(1).

Proof: The uniform approximation for q is evi- dent. Suppose thatqis continuously differentiable on [0,π]. Apply the intermediate value theorem on Propo- sition 1c, then there is someξ⁽ⁿ⁾_j ∈(x⁽ⁿ⁾_j ,x⁽ⁿ⁾_j+1)such that

λn²l⁽ⁿ⁾_j

π ⁼

1−q(ξj⁽ⁿ⁾) λn

_−1/2

=1+q(ξj⁽ⁿ⁾) 2λn +O

1 λn²

.

Hence 2λn

λn²lⁿ_j π ⁻¹

−q(ξ⁽ⁿ⁾j ) =O 1

n²

,

2λn

λn²lⁿ_j π ^−p(x)

+2λnp(x)−q(ξj⁽ⁿ⁾) =O 1

n²

.

Applying the mean value theorem, when n is suffi- ciently large, then

q(x) =q(ξ⁽ⁿ⁾j )−2λnp(x) +O 1

n

.

Then we employ a modified Pr¨ufer substitution due to Ashbaugh-Benguria [13] to solve the boundary condi- tionshandH,x=0 andx=π, respectively:

y=r(x)sin√ λθ(x), y=√

λ^r(x)cos√ λθ(x),

so that

θ=cos²√

λθ(x)−q(x)sin²√ λθ(x) λ

−2p(x)sin²√ λθ(x).

(2.4)

Integrating (2.4) fromx_jtox_j+1,

√π λn

=

x_j+1

xj

cos²√

λθ(x)dx

−1 λ

x_j+1

xj

q(x)sin²√

λθ(x)dx

−2

xj+1

xj

p(x)sin²√

λθ(x)dx,

(2.5)

it results by Lemma 5 from (2.5) that

√π λn

=−q(xj) 2λn

l_j+c_j+O 1

n^N+4

−p(x_j)l_j+d_j+O 1

n^N+4

+O 1

n

, (2.6)

where cj= 1

2λn

∑

q^(k)(x_j)

(k+1)!l^k+1_j =O 1

n⁴

,

d_j=

∑

k=1

p^(k)(x_j)

(k+1)!l^k+1_j =O 1

n²

.

Summarizing from (2.6), q(x_j) =−2π^√λn

l_j −2λnp(x_j)+2λn(c_j+d_j)+O(1).

130 H. Koyunbakan and E. Yilmaz·Potential Function and its Derivatives Therefore

δ^q(xj) =−2πλn ∆l_j

l²_jl_j+1−2λnδ^p(xj) +O(1), and so, fork=1,2,...,N,

δ^kq(x_j) =−2πλnδ^k−1 ∆lj

l²_jl_j+1

−2λnδ^kp(xj) +O(1).

(2.7)

If we use the results of Theorem 3 and Theorem 5, we get

q^(k)(x) =δ^kq(x_j)−2λnp(x) +O(n⁻¹),

q^(k)(x) =2λn^3/2δ^klj

π ⁻²λnδ^kp(x_j)−2λnp(x)+O(1).

Theorem 5. Assume thatqisC^N+1on[0,π]. Then, fork=1,2,...,N,

δ^kq(xj) =2λn^3/2δ^klj

π ⁻²λnδ^kp(xj) +O(1).

The estimate is independent ofj.

Proof:In view of derivations and the fact thatδ^lj=

∆lj

lj =^O(n_O(n⁻⁴₋1⁾) =O(n⁻³), it suffices to show that δ^k−1

π²∆l_j l²_jl_j+1

=−δ^k−1 λn∆l_j

l_j

+O 1

n³

. (2.8)

But δ

λn∆l_j

l_j −π²∆l_j l²_jl_j+1

=δ

λn− π² l_jl_j+1

∆l_j l_j

=π^{2(∆lj+∆lj+1)} l²_jl_j+1l_j+2 +

λn− π² l_jl_j+1

δ^lj

=O 1

n³

.

Thus, (2.8) follows by Lemma 4. If we write (2.8) in (2.7), then

δ^kq(x_j) =−2πλn

−δ^k−1λn∆l_j π^2lj +O

1 n³

−2λnδ^kp(x_j) +O(1), (2.9)

δ^kq(x_j) =2λn^3/2δ^klj

π ⁻²λnδ^kp(x_j) +O(1).

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