Nonlinear boundary value problems for second order

Nonlinear Studies Vol., No., pp.277-296, 2008

Nonlinear Studies c °I&S Publishers Daytona Beach, FL 2008

Nonlinear boundary value problems for second order differential equations describing concave equilibrium capillary surfaces St. Balint, L. Tanasie Department of Mathematics and Computer Science West University of Timi¸soara Bd. V. Parvan nr. 4, 300223, Timi¸soara, Romania. Abstract In this paper problems are considered: ze00 =

the

following

two

nonlinear

boundary

value

¤3/2 1 £ ¤ ρ · g · ze − pe £ 1 + (ze0 )2 − · 1 + (ze0 )2 · ze0 , r ∈ [re , Rge ], γ r

ze0 (re ) = − tan(π/2 − αg ); ze0 (Rge ) = − tan αc ; ze (Rge ) = 0, ze (r) is strictly decreasing on [re , Rge ]. zi00 =

¤3/2 1 £ ¤ ρ · g · zi − pi £ − · 1 + (zi0 )2 · zi0 , r ∈ [Rgi , ri ], 1 + (zi0 )2 γ r zi0 (Rgi ) = tan αc ; zi0 (ri ) = tan(π/2 − αg ); zi (Rgi ) = 0,

zi (r) is strictly increasing on [Rgi , ri ]. Rgi + Rge Here: 0 < Rgi < ri ≤ ≤ re < Rge ; ρ, g, γ, pe , pi , αc , αg are constants 2 having the following properties: ρ, g, γ are strictly positive and 0 < π/2 − αg < αc < π/2. Necessary or sufficient conditions are given, in terms of pe , pi for the existence of a concave solution of the above nonlinear boundary value problems (nl.b.v.p.). This kind of results can be useful in experiment planing and manufacturing technology design of single-crystal tube growth from the melt. With this aim this study was undertaken. 0 2000 AMS Subject Classification: 34B15; Keywords: nonlinear boundary value problems; existence of solutions; single-crystal tube growth

277

278

1

Introduction

For single crystal tube growth by edge-defined film-fed growth (E.F.G.) technique, in hydrostatic approximation, the free surface of the static meniscus is described by the Laplace capillary equation [1], [2]: µ γ

1 1 + R1 R2

¶ +ρ·g·z =p

(1.1)

Here γ is the melt surface tension, ρ denotes the melt density, g is the gravity acceleration, R1 , R2 denote the main radii of the free surface curvature at a point M of the free surface, z is the coordinate of M with respect to the Oz axis, directed vertically upwards, p is the pressure difference across the free surface. For the outer free surface p = pe = pm − peg − ρ · g · H and for the inner free surface p = pi = pm − pig − ρ · g · H. Here pm denotes the pressure in the meniscus melt,peg , pig denotes the gas pressure on the outer and inner free surface respectively, H denotes the melt column height between the horizontal crucible melt level and the shaper top level (see Fig. 1).

Figure 1 To calculate the outer and inner free surface shape of the static meniscus is convenient to employ the Laplace eq (1.1) in its differential form. This form of the eq(1.1) can be obtained as a necessary condition for the minimum of the free energy functional of the melt column. µ ¶ Rgi + Rge For the growth of a single crystal tube of inner radius ri ∈ Rgi , and 2 µ ¶ Rgi + Rge outer radius re ∈ , Rge the differential equation of the outer free surface 2

279 is given by: ze00 =

¤3/2 1 £ ¤ ρ · g · ze − pe £ 1 + (ze0 )2 − · 1 + (ze0 )2 · ze0 , γ r

for r ∈ [re , Rge ],

(1.2)

which is the Euler equation of the free energy functional: Rge½ Z

Ie (ze ) = re

£ ¤1/2 1 γ · 1 + (ze0 )2 + · ρ · g · ze2 − pe · ze 2

¾ · r · dr

(1.3)

ze (re ) = he > 0, ze (Rge ) = 0 and the inner free surface differential equation is given by the formula: zi00 =

¤3/2 1 £ ¤ ρ · g · zi − pi £ 1 + (zi0 )2 − · 1 + (zi0 )2 · zi0 , γ r

for r ∈ [Rgi , ri ],

(1.4)

which is the Euler equation of the free energy functional: Zri ½

¤1/2 1 £ + · ρ · g · zi2 − pi · zi γ · 1 + (zi0 )2 2

Ii (zi ) =

¾ · r · dr

(1.5)

Rgi

zi (ri ) = hi > 0, zi (Rgi ) = 0 Up until 1993-1994, knowledge concerning the dependence of the shape of the meniscus free surface on the pressure difference p across the free surface for small and large Bond numbers, in the case of the growth of single crystal rods by E.F.G. technique is summarized in [1]. According to [1], for the general differential equation (1.2), (1.4) describing the free surface of the liquid meniscus, there are no complete analysis and solution. For the general equation only numerical integrations were carried out for a number of process parameter values that were of practical interest at the time. In [3], the authors investigate the pressure difference influence on the meniscus free surface shape for rods, in the case of middle-range Bond numbers (i.e., Bo=1) which most frequently occurs in practice and has been left out of the regular study in [1]. They use a numerical approach in this case to solve the meniscus surface equation written in terms of the arc length of the curve. The stability of the static meniscus free surface is analyzed by means of the Jacobi equation. The result is that a large number of static menisci having drop like shapes are unstable. The authors of [4], [5] consider automated crystal-growth processes based on weight sensors and computers. They give an expression for the weight of the meniscus, in contact with crystal and shaper of arbitrary shape, in which there are terms related to the hydrodynamic factor. Later in [6], the author shows that the hydrodynamic factor is too small to be considered in automated crystal growth. Finally in [7], [8] the authors present theoretical and numerical study of the meniscus free surface shape dependence on the pressure difference in the case of tube growth from classical semiconductor materials (i.e. 0 < αc < π/2 − αg , αg ∈ (0, π/2)) when the meniscus free surface in convex.

280 In the present paper the shape of the inner and outer free surface of the static meniscus is analyzed as function of the pressure difference p and its static stability is investigated in the case of tube growth from materials for which 0 < π/2 − αg < αc < π/2 when the free surface is concave i.e. the statistic meniscus has a drop-like shape.

2

Concave outer free surface

Let consider now the nonlinear boundary value problem (nl. b. v. p.):  ¤3/2 1 £ ¤ ρ · g · ze − pe £  1 + (ze0 )2 − · 1 + (ze0 )2 · ze0 , r ∈ [re , Rge ], ze00 =    γ r       z 0 (r ) = − tan(π/2 − α ) g e e    0   ze (Rge ) = − tan αc      ze (Rge ) = 0, ze (r) is strictly decreasing on [re , Rge ]

(2.1)

where Rgi , Rge , re , ρ, g, γ, pe , αc , αg are real numbers having the following properties: 0 < Rgi
r∗

z(r∗ + 0) =

lim z(r) r → r∗ r > r∗

(2.16)

exists and satisfies: − tan αc ≤ z 0 (r∗ + 0) ≤ − tan(π/2 − αg ),

(2.17)

[Rge − r∗ ] · tan(π/2 − αg ) ≤ z(r∗ + 0) ≤ [Rge − r∗ ] · tan αc lim z 00 (r) exists too and z 00 (r∗ + 0) ≤ 0. r → r∗ r > r∗ Due to the fact that r∗ is infimum, one of the inequalities:

The limit z 00 (r∗ + 0) =

− tan αc ≤ z 0 (r∗ + 0), z 0 (r∗ + 0) ≤ − tan(π/2 − αg ), z 00 (r∗ + 0) ≤ 0 has to be equality. The equality − tan αc = z 0 (r∗ + 0) is impossible because z 0 (r∗ + 0) > z 0 (r) ≥ − tan αc for r ∈ (r∗ , Rge ]. We show in the following that r∗ ≥ Rge /n and z 0 (r∗ + 0) = − tan(π/2 − αg ).

283 If r∗ ≥ Rge /n then we have to show only the equality z 0 (r∗ + 0) = − tan(π/2 − αg ). This can be made showing that z 00 (r∗ + 0) = 0 is impossible. Assuming that z 00 (r∗ + 0) = 0. We obtain: z(r∗ + 0) = +

pe γ n γ αc +αg − π/2 − · sin α(r∗ + 0) > · · · sin αg + g · ρ g · ρ · r∗ n−1 g·ρ Rge

n·γ n·γ n−1 · Rge · tan αc + · sin αc − · sin αc > n g · ρ · Rge g · ρ · Rge µ >

Rge −

1 Rge n

¶ · tan αc > z(Rge /n) > z(r∗ + 0)

(2.18) what is impossible. Assume now that r∗ < Rge /n and consider the difference α(Rge ) − α(Rge /n): µ α(Rge ) − α

Rge n

¶

·

¸ pe − ρ · g · z(ξ) sin α(ξ) = α (ξ) · [Rge − Rge /n] = − γ r · 1 n−1 n αc + αg − π/2 · · · Rge > · · sin αg + cos α(ξ) n n−1 Rge 0

n−1 g·ρ n g·ρ n−1 · · Rge · tan αc + · sin αc − · · n γ Rge γ n ¸ n 1 n−1 Rge · tan αc − · sin αc · · · Rge = Rge sin αg n αc + αg − π/2 sin αg n − 1 π n · · · · Rge = αc + αg − . n−1 Rge sin αg n 2 (2.19) Hence: α(Rge /n) < π/2 − αg what is impossible. In this way it was shown that r∗ ≥ Rge /n and z 0 (r∗ + 0) = − tan(π/2 − αg ). t u Theorem 3. If for 1 < n0 < n
0 = 3/2 ∂z02 [1 + (z 0 )2 ] Hence, the Legendre condition is satisfied. The Jacobi equation: µ 2 ¶¸ · ¸ · 2 d ∂ F d ∂2F ∂ F 0 − · η − · η =0 ∂z 2 dr ∂z∂z0 dr ∂z02

(2.23)

(2.24)

in this case is given by d dr

Ã

!

r·γ 3/2

[1 + (z 0 )2 ]

· η0

−g·ρ·r·η =0

(2.25)

For the equation (2.25) the following inequalities hold: r·γ 3/2

[1 + (z 0 )2 ] Hence, the equation:

≥ re · γ · cos3 αc and

(η 0 · re · γ · cos3 αc )0 = 0

−g·ρ·r ≤0

(2.26)

(2.27)

is a Sturm type upper bound for the equation (2.25). [9]. Since every non zero solution of the equation (2.27) vanishes at most once on the interval [re , Rge ], the solution η(r) of the initial value problem:  Ã !  r · γ d  0  · η − g · ρ · r · η0 = 0  dr [1 + (z 0 )2 ]3/2 (2.28)     η(re ) = 0, η 0 (re ) = 1. has only one zero on the interval [re , Rge ] [9]. Hence the Jacobi condition is satisfied. u t

285 Theorem 5. If the solution of the i.v.p. (2.10) is convex (z 00 (r) > 0) on the interval [re , Rge ], then it is not a solution of the nl.b.v.p. (2.1). Proof. z 00 (r) > 0 on [re , Rge ] implies that z 0 (r) is strictly increasing on [re , Rge ]. Hence z 0 (re ) < z 0 (Rge ) < − tan αc < − tan(π/2 − αg ). t u Theorem 6. If n and pe satisfies: 1 m · g · ρ · Rgi m · g · ρ · Rgi

> (m · Rgi − Rgi ) · tan αc > z(m · Rgi ) > z(r∗ − 0) (3.18) what is impossible.

289 r∗

Assume now that α(m · Rgi ) − α(Rgi ):

>

m · Rgi

and

consider

the

difference

·

ρ · g · z(ξ) − pi sin α(ξ) α(m · Rgi ) − α(Rgi ) = α (ξ) · [m · Rgi − Rgi ] = − γ r

¸

0

· 1 1 αc + αg − π/2 · · (m − 1) · Rgi < − · · cos α(ξ) m−1 Rgi · sin αg − (m − 1) ·

1 g·ρ · Rgi · tan αc + · cos αg + γ m · Rgi

¸ g·ρ 1 · (m − 1) · Rgi · tan αc − · cos αg · γ m · Rgi ·

1 · (m − 1) · Rgi = −(αc + αg − π/2). sin αg

Hence: α(m · Rgi ) < π/2 − αg what is impossible. In this way it was shown that r∗ ≤ m · Rgi and z 0 (r∗ − 0) = tan(π/2 − αg ). Theorem 9. If for 1 < m0 < m
0 and z(Rgi ) = zi (Rgi ) = 0. Proof. Since eq. (3.1)1 is the Euler eq. for (3.21) it is sufficient to prove that the Legendre and Jacobi conditions are satisfied in this case. Denote by F (r, z, z 0 ) the function defined as: ¾ ½ £ ¤ 1 2 0 2 1/2 0 (3.22) · g · ρ · z − pi · z + γ · 1 + (z ) F (r, z, z ) = r · 2

290 It is easy to verify that we have: ∂2F r·γ = >0 2 3/2 ∂z0 [1 + (z 0 )2 ]

(3.23)

Hence, the Legendre condition is satisfied. The Jacobi equation: µ 2 ¶¸ · ¸ · 2 d ∂2F d ∂ F ∂ F 0 · η − =0 − · η ∂z 2 dr ∂z∂z0 dr ∂z02

(3.24)

in this case is given by d dr

Ã

!

r·γ 3/2

[1 + (z 0 )2 ]

·η

0

−g·ρ·r·η =0

(3.25)

For the equation (3.25) the following inequalities hold: r·γ [1 +

3/2 (z 0 )2 ]

≥ Rgi · γ · cos αc3 and

−g·ρ·r ≤0

(3.26)

Hence, the equation: (η 0 · Rgi · γ · cos3 αc )0 = 0

(3.27)

is a Sturm type upper bound for the equation (3.25). [9]. Since every non zero solution of the equation (3.27) vanishes at most once on the interval [Rgi , ri ], the solution η(r) of the initial value problem:  Ã !  r · γ d  0  · η − g · ρ · η0 = 0  dr [1 + (z 0 )2 ]3/2 (3.28)     η(Rgi ) = 0, η 0 (Rgi ) = 1. has only one zero on the interval [Rgi , ri ][9]. Hence the Jacobi condition is satisfied. u t Theorem 11. If the solution of the i.v.p. (3.1) is convex (z 00 (r) > 0) on the interval [Rgi , ri ], then it is not a solution of the nl.b.v.p. (3.1). Proof. z 00 (r) > 0 on [Rgi , ri ] implies that z 0 (r) is strictly increasing on [Rgi , ri ]. Hence z 0 (Rgi ) = tan αc < z 0 (ri ). t u Theorem 12. If m and pi satisfies: 1 366.62[P a]. Numerical integration of the initial value problem (2.10) illustrate that for pe = 370, 460[P a] there exists re ∈ (0.0047, 0.0048)[m] such that the solution of

293 the initial value problem is a concave solution of the nl.b.v.p. (2.1) on [re , Rge ], and shows also that for pe = 320[P a] there exists re ∈ [0.0047, 0.0048)[m] such that the solution of the initial value problem is a concave solution of the nl.b.v.p. (2.1) on [re , Rge ]. (Figs 5, 6)

Figure 5 - ze versus r for pe = 320, 370, 460[P a]

Figure 6 - αe versus r for pe = 320, 370, 460[P a] Consequently, condition (2.9) is not a necessary condition. B. Inner free surface Inequality (3.2) is a necessary condition for the existence of a concave solution of the nl.b.v.p. (3.1). Is this condition also sufficient? Computation shows that for the considered numerical data and for m = 1.03 (ri = Rgi · m = 0.004326[m]), the inequality (3.2) becomes: 53.597[P a] ≤ pe ≤ 134.902[P a]. Numerical integration of the initial value problem (3.10) illustrate that for pi = 90[P a] that the solution of the initial value problem is a concave solution of the nl.b.v.p. (3.1) on [Rgi , m · Rgi ] and shows that for pi = 54, 134[P a] the solution of the initial value problem is not anymore a concave solution of the nl.b.v.p. (3.1) on [Rgi , m · Rgi ]. (Figs 7, 8)

294

Figure 7 - zi versus r for pi = 54, 90, 134[P a]

Figure 8 - αi versus r for pi = 54, 90, 134[P a] Consequently, condition (3.2) is a necessary one, but it is not a sufficient. Inequality (3.8) · is a sufficient ¸ condition which assure that there is no ri in the Rgi + Rge closed interval Rgi , for which the nl.b.v.p. (3.1) possessis a concave 2 solution. Computation illustrate that for the considered numerical data inequality (3.8) becomes: pi < −31.99507[P a]. Numerical integration of the · initial value problem ¸ Rgi + Rge such (2.10) illustrate that for pi = −35, −70[P a] there is no ri ∈ Rgi , 2 that the solution of the initial value problem is a concave solution of the · nl.b.v.p. (3.1) ¸ Rgi + Rge on [Rgi , ri ], and shows also that for pi = −10[P a] there is no ri ∈ Rgi , 2 such that the solution of the initial value problem is a concave solution of the nl.b.v.p. (3.1) on [Rgi , ri ]. (Fig 9)

295

Figure 9 - zi and αi , respectively versus r for pi = −10, −35, −70[P a] Consequently, condition (3.2) is a sufficient one, but it is not a necessary condition. Inequality (3.9) is a sufficient condition for the existence of a point ri ∈ [Rgi , m · Rgi ] such that the solution of the initial value problem (3.10) is a concave solution of the nl.b.v.p. (3.1) on the interval [Rgi , ri ]. Is this condition also a necessary condition? Computation shows that for the considered numerical data and for m = 1.04 the inequality (3.9) becomes: pi > 93.50[P a]. Numerical integration of the initial value problem (3.10) illustrate that for pi = 120[P a] there exists ri ∈ (0.0042, 0.0.004368)[m] such that the solution of the initial value problem is a concave solution of the nl.b.v.p. (3.1) on (0.0042, 0.0.004368)[m] and show also that also for pi = 90[P a] there exists ri ∈ (0.0042, 0.0.004368)[m] such that the solution of the initial value problem is a concave solution of the nl.b.v.p. (3.1) on [Rgi , ri ]. (Figs 10, 11).

Figure 10 - zi versus r for pi = 90, 120, 170[P a]

296

Figure 11 - αi versus r for pi = 90, 120, 170[P a] Consequently, condition (3.9) is not a necessary condition.

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