## A Schrodinger singular perturbation problem

Comm. Nonlin. Sci. and Numer. Simul.

arXiv:math-ph/0511049v1 14 Nov 2005

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A Schr¨dinger singular perturbation problem o
A.G. Ramm Mathematics Department, Kansas State University, Manhattan, KS 66506-2602, USA ramm@math.ksu.edu
Abstract Consider the equation ?ε2 ?uε + q(x)uε = f (uε ) in R3 , |u(∞)| < ∞, ε = const > 0. Under what assumptions on q(x) and f (u) can one prove that the solution uε exists and limε→0 uε = u(x), where u(x) solves the limiting problem q(x)u = f (u)? These are the questions discussed in the paper.

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Let

Introduction
?ε2 ?uε + q(x)uε = f (uε) in R3 , a2 ≤ q(x), |uε (∞)| < ∞, (1.1)

ε = const > 0, f is a nonlinear smooth function, q(x) ∈ C(R3 ) is a real-valued function a = const > 0. (1.2)

We are interested in the following questions: 1)Under what assumptions does problem (1.1) have a solution? 2)When does uε converge to u as ε → 0? Here u is a solution to q(x)u = f (u). The following is an answer to the ?rst question.

(1.3)

Theorem 1.1. Assume q ∈ C(R3 ), (1.2) holds, f (0) = 0, and a is su?ciently large. More precisely, let M(R) := max|u|≤R |f (u)|, M1 (R) = max|ξ|≤R |f ′ (ξ)|, p := q(x) ? a2 , and assume that p R+M (R) ≤ R, and p +M1 (R) ≤ γ < 1, where γ > 0 is a constant and a2 a2 p := supx∈R3 |p(x)|. Then equation (1.1) has a solution uε ≡ 0, uε ∈ C(R3 ), for any ε > 0.
Math subject classi?cation: 35J60, 35B25 key words: nonlinear elliptic equations, singular perturbations

In Section 4 the potential q is allowed to grow at in?nity. An answer to the second question is: Theorem 1.2. If f (u) is a monotone, growing function on the interval [u0 , ∞), such that u limu→∞ f (u) = ∞, and f (u00 ) < a2 , where u0 > 0 is a ?xed number, then there is a solution u u uε to (1.1) such that lim uε (x) = u(x), (1.4)
ε→0

where u(x) solves (1.3). Singular perturbation problems have been discussed in the literature , , , but our results are new. In Section 2 proofs are given. In Section 3 an alternative approach is proposed. In Section 4 an extension of the results to a larger class of potentials is given.

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Proofs

Proof of Theorem 1.1. The existence of a solution to (1.1) is proved by means of the contraction mapping principle. Let g be the Green function (?ε ? + a )g = δ(x ? y) in R ,
2 2 3

g := ga (x, y, ε) ?→ 0,
|x|→∞

e? ε |x?y| g= . (2.1) 4π|x ? y|ε2
a

Let p := q ? a2 ≥ 0. Then (1.1) can be written as: uε (x) = ? gpuεdy +
R3 R3

gf (uε)dy := T (uε ).

(2.2)

Let X = C(R3 ) be the Banach space of continuous and globally bounded functions with the sup ?norm: v := supx∈R3 |v(x)|. Let BR := {v : v ≤ R}. We choose R such that T (BR ) ? BR (2.3) and T (v) ? T (w) ≤ γ v ? w , v, w ∈ BR , 0 < γ < 1. (2.4) If (2.3) and (2.4) hold, then the contraction mapping principle yields a unique solution uε ∈ BR to (2.2), and uε solves problem (1.1). The assumption f (0) = 0 guarantees that uε ≡ 0. Let us check (2.3). If v ≤ R, then T (v) ≤ v p
R3

g(x, y)dy + 3

p R + M(R) M(R) ≤ , 2 a a2

(2.5)

where M(R) := max|u|≤R |f (u)|. Here we have used the following estimate: g(x, y)dy =
R3 R3

e? ε |x?y| 1 dy = 2 . 4π|x ? y|ε2 a
a

(2.6)

If p < ∞ and a is such that p R + M(R) ≤ R, a2 then (2.3) holds. Let us check (2.4). Assume that v, w ∈ BR , v ? w := z. Then T (v) ? T (w) ≤ p M1 (R) z + z , 2 a a2 (2.8) (2.7)

where, by the Lagrange formula, M1 (R) = max|ξ|≤R |f ′(ξ)|. If p + M1 (R) ≤ γ < 1, a2 (2.9)

then (2.4) holds. By the contraction mapping principle, (2.7) and (2.9) imply the existence and uniqueness of the solution uε (x) to (1.1) in BR for any ε > 0. Theorem 1.1 is proved. 2 Proof of Theorem 1.2. In the proof of Theorem 1.1 the parameters R and γ are independent of ε > 0. Let us denote by Tε the operator de?ned in (2.2). Then
ε→0

lim Tε (v) ? T0 (v) = 0,

(2.10)

for every v ∈ C(R3 ), where the limiting operator T0 , corresponding to the value ε = 0, is of the form: ?pv + f (v) T0 (v) = . (2.11) a2 To calculate T0 (v) we have used the following formula
ε→0

lim ga (x, y, ε) =

1 δ(x ? y), a2

(2.12)

where convergence is understood in the following sense: for every h ∈ C(R3 ) one has:
ε→0

lim

ga (x, y, ε)h(y)dy =
R3

h(x) , a2

a > 0.

(2.13)

Indeed, one can easily check that
ε→0

lim

ga (x, y, ε)dy = 0,
|x?y|≥c>0

ε→0

lim

ga (x, y, ε)dy =
|x?y|≤c

1 , a2

a > 0,

(2.14)

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where c > 0 is an arbitrary small constant. These two relations imply (2.13). We claim that if (2.10) holds for every v ∈ X, and γ in (2.4) does not depend on ε, then (1.4) holds, where u solves the limiting equation (2.2): u = T0 (u) = ?pu + f (u) . a2 (2.15)

Equation (2.15) is equivalent to (1.3). The assumptions of Theorem 1.2 imply that equation (1.3) has a unique solution. Let us now prove the above claim. Let u = Tε (u), u = uε , v = T0 (v), and limε→0 ||Tε (w) ? T0 (w)|| = 0 for all w ∈ X. Assume that Tε (v) ? Tε (w) ≤ γ v ? w , 0 < γ < 1, where the constant γ does not depend on ε, so that Tε is a contraction map. Consider the iterative process un+1 = 1 Tε (un ), u0 = v. The usual estimate for the elements un is: un ? v ≤ 1?γ Tε v ? v . Let u = limn→∞ un . This limit does exist because Tε is a contraction map. Taking n → ∞, 1 one gets u ? v ≤ 1?γ Tε (v) ? T0 (v) → 0 as ε → 0. The claim is proved. Theorem 1.2 is proved. 2 Remark 2.1. Conditions of Theorem 1.1 and of Theorem 1.2 are satis?ed if, for example, q(x) = a2 + 1 + sin(ωx), where ω = const > 0, f (u) = (u + 1)m , m > 1, or f (u) = eu . If R = 1, and f (u) = eu , then M(R) = e, M1 (R) = e, ||p|| ≤ 2, so 2+e ≤ 1 and 2+e ≤ γ < 1 a2 a2 √ provided that a > 5. For these a, the conditions of Theorem 1.1 are satis?ed and there is a solution to problem (1.1) in the ball B1 for any ε > 0.

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A di?erent approach
??y wε + a2 wε + p(εy + ξ)wε = f (wε ), |wε (∞)| < ∞, (3.1)

Let us outline a di?erent approach to problem (1.1). Set x = ξ + εy. Then

wε := uε (εy + ξ), p := q(εy + ξ) ? a2 ≥ 0. Thus wε = ? where (?? + a2 )G = δ(x ? y) in R3 , One has G(x, y)dy =
R3

G(x, y)p(εy + ξ)wε dy +
R3 R3

G(x, y)f (wε)dy, e?a|x?y| , 4π|x ? y|

(3.2)

G= 1 . a2

a > 0.

(3.3)

(3.4)

Using an argument similar to the one in the proofs of Theorem 1.1 and Theorem 1.2, one concludes that for any ε > 0 and any su?ciently large a, problem (3.1) has a unique 5

solution wε = wε (y, ξ), which tends to a limit w = w(y, ξ) as ε → 0, where w solves the limiting problem ??y w + q(ξ)w = f (w), |w(∞, ξ)| < ∞. (3.5) Problem (3.5) has a solution w = w(ξ), which is indepent of y and solves the equation q(ξ)w = f (w). (3.6)

The solution to (3.5), bounded at in?nity, is unique if a is su?ciently large. This is proved similarly to the proof of (2.9). Namely, let b2 := q(ξ). Note that b ≥ a. If there are two solutions to (3.5), say w and v, and if z := w ? v, then ||z|| ≤ b?2 M1 (R)||z|| < ||z||, provided that b?2 M1 (R) < 1. Thus z = 0, and the uniqueness of the solution to (3.5) is proved under the assumption q(ξ) > M1 (R), where M1 (R) = max|ξ|≤R |f ′ (ξ)|. Replacing ξ by x in (3.6), we obtain the solution found in Theorem 1.2.

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Extension of the results to a larger class of potentials

Here a method for a study of problem (1.1) for a larger class of potentials q(x) is given. We assume that q(x) ≥ a2 and can grow to in?nity as |x| → ∞. Note that in Sections 1 and 2 the potential was assumed to be a bounded function. Let gε be the Green function ?ε2 ?x gε + q(x)gε = δ(x ? y) in R3 , uε =
R3

As in Section 2, problem (1.1) is equivalent to

|gε (∞, y)| < ∞.

(4.1)

gε (x, y)f (uε(y))dy,

(4.2)

and this equation has a unique solution in BR if a2 is su?ciently large. The proof, similar to the one given in Section 2, requires the estimate gε (x, y)dy ≤ 1 . a2 (4.3)

R3

Let us prove inequality (4.3). Let Gj be the Green function satisfying equation (4.1) with q = qj , j = 1, 2. Estimate (4.3) follows from the inequality G1 ≤ G2 if q1 ≥ q2 .
e? ε |x?y| 4π|x?y|ε2
a

(4.4) implies (4.3).

This inequality can be derived from the maximum principle. a e? ε |x?y| If q2 = a2 , then G2 = 4π|x?y|ε2 , and the inequality gε (x, y) ≤ Let us prove the following relation:
ε→0

lim

gε (x, y)h(y)dy =
R3

h(x) q(x)

?h ∈ C∞ (R3 ),

?

(4.5)

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where C∞ is the set of C ∞ (R3 ) functions vanishing at in?nity together with their derivatives. This formula is an analog to (2.12). To prove (4.5), multiply (4.1) by h(y), integrate over R3 with respect to y, and then let ε → 0. The result is (4.5). More detailed argument is given at the end of the paper. Thus, Theorem 1.1 and Theorem 1.2 remain valid for q(x) ≥ a2 , a > 0 su?ciently large, provided that f (u) monotonically growing to in?nity and f (u00 ) < a2 for some u0 > 0. u u Under these assumptions the solution u(x) to the limiting equation (1.3) is the limit of the solution to (4.2) as ε → 0. Let us give details of the proof of (4.5). Denote the integral on the left-hand side of (4.5) by w = wε (x). From (4.1) it follows that ?ε2 ?wε + q(x)wε = h(x). (4.6)

?

Multiplying (4.6) by wε and integrating by parts yields the estimate ||wε ||L2(R3 ) ≤ c, where c > 0 is a constant independent of ε. Consequently, one may assume that wε converges weakly in L2 (R3 ) to an element w. Multiplying (4.6) by an arbitrary function ∞ φ ∈ C0 (R3 ), integrating over R3 , then integrating by parts the ?rst term twice, and then taking ε → 0, one obtains the relation: q(x)w(x)φ(x)dx =
R3 R3

h(x)φ(x)dx,

(4.7)

∞ which holds for all φ ∈ C0 (R3 ). It follows from (4.7) that qw = h. This proves formula (4.5).

References
 Berger, M., Nonlinearity and functional analysis, Acad. Press, New York, 1977.  Kantorovich, L., Akilov, G., Functional analysis, Pergamon Press, New York, 1982.  Lomov, S., Introduction into the theory of singular perturbations, AMS, Providence RI, 1992.  Ramm, A.G., Wave scattering by small bodies of arbitrary shapes, World Sci. Publishers, Singapore, 2005.  Vishik, M., Lusternik, L., Regular degeneration and boundary layer for linear di?erential equations with small parameter, Uspekhi Mat. Nauk, 12, N5 (1957), 3-122.

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