On some estimates for a fluid surface in a short capillary tube

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On some estimates for a fluid surface in a short capillary tube Luminit ßa Barbu Ovidius University, Department of Mathematics and Informatics, Blvd. Mamaia 124, 900527 Constant ßa, Romania article info Keywords: Capillary problem Minimum principle Critical point A priori bounds abstract In this note we investigate the problem of a fluid surface in a capillary tube of cross section X, which is assumed to be so short that the fluid rises to the top along the rim of the tube. Our main result is a minimum principle for an appropriate functional combination of the solution and its gradient. As an application of this minimum principle, we obtain some a priori estimates in terms of the curvature of @X. The proofs make use of Hopf’s maximum principles, some topological arguments regarding the local behavior of analytic functions and some computations in normal coordinates with respect to the boundary @X. Ó 2013 Elsevier Inc. All rights reserved. 1. Introduction In this note we deal with the problem of a fluid surface in a capillary tube of cross-section X, with a gravity field that is downward acting. Denoting the fluid surface by x 3 ¼ uðx 1 ; x 2 Þ, with ðx 1 ; x 2 Þ2 X R 2 , the classical Laplace–Young equation says in this case that uðx 1 ; x 2 Þ satisfies the equation (see the book of Finn [4]): div ru ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ru j j 2 q 0 B @ 1 C A ¼ cu in X: ð1:1Þ In (1.1) we assume that the physical constant c is positive and Xis a bounded strictly convex domain with smooth boundary. Also, the tube is assumed to be so short that the fluid rises to the top along the rim of the tube, which means that uðxÞ satisfies: u ¼ a on @X; with a ¼ const:> 0: ð1:2Þ According to Serrin [18], problem (1.1), (1.2) has a unique solution, provided ac 6 k min ; ð1:3Þ where k min ¼ min @X ks ðÞ and ks ðÞ is the curvature of @X. Moreover, since @X is smooth, the solution u x ðÞ is an analytic function in X(see Nirenberg [10]), a feature which will play an important role in this paper. Also, under the above assumptions on the data, the comparison principle (see Pucci and Serrin [17]) implies that u > 0 on X, while Hopf’s first maximum principle [6] implies that u x ðÞ assume its maximum value on @X and its minimum value in X. Now, let us consider the following P-function, i.e. an appropriate combination of the solution and its gradient (see the book of Sperb [19]): P x; a ð Þ :¼ 2 2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ ru j j 2 q ca 2 u 2 ; with a 2 R; ð1:4Þ 0096-3003/$ - see front matter Ó 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.amc.2013.02.016 E-mail address: [email protected] Applied Mathematics and Computation 219 (2013) 8192–8197 Contents lists available at SciVerse ScienceDirect Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc

Transcript of On some estimates for a fluid surface in a short capillary tube

Page 1: On some estimates for a fluid surface in a short capillary tube

Applied Mathematics and Computation 219 (2013) 8192–8197

Contents lists available at SciVerse ScienceDirect

Applied Mathematics and Computation

journal homepage: www.elsevier .com/ locate /amc

On some estimates for a fluid surface in a short capillary tube

0096-3003/$ - see front matter � 2013 Elsevier Inc. All rights reserved.http://dx.doi.org/10.1016/j.amc.2013.02.016

E-mail address: [email protected]

Luminit�a BarbuOvidius University, Department of Mathematics and Informatics, Blvd. Mamaia 124, 900527 Constant�a, Romania

a r t i c l e i n f o a b s t r a c t

Keywords:Capillary problemMinimum principleCritical pointA priori bounds

In this note we investigate the problem of a fluid surface in a capillary tube of cross sectionX, which is assumed to be so short that the fluid rises to the top along the rim of the tube.Our main result is a minimum principle for an appropriate functional combination of thesolution and its gradient. As an application of this minimum principle, we obtain some apriori estimates in terms of the curvature of @X. The proofs make use of Hopf’s maximumprinciples, some topological arguments regarding the local behavior of analytic functionsand some computations in normal coordinates with respect to the boundary @X.

� 2013 Elsevier Inc. All rights reserved.

1. Introduction

In this note we deal with the problem of a fluid surface in a capillary tube of cross-section X, with a gravity field that isdownward acting. Denoting the fluid surface by x3 ¼ uðx1; x2Þ, with ðx1; x2Þ 2 X � R2, the classical Laplace–Young equationsays in this case that uðx1; x2Þ satisfies the equation (see the book of Finn [4]):

div ruffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ruj j2

q0B@

1CA ¼ cu in X: ð1:1Þ

In (1.1) we assume that the physical constant c is positive and Xis a bounded strictly convex domain with smooth boundary.Also, the tube is assumed to be so short that the fluid rises to the top along the rim of the tube, which means that uðxÞsatisfies:

u ¼ a on @X; with a ¼ const: > 0: ð1:2Þ

According to Serrin [18], problem (1.1), (1.2) has a unique solution, provided

ac 6 kmin; ð1:3Þ

where kmin ¼min@X

k sð Þ and k sð Þ is the curvature of @X. Moreover, since @X is smooth, the solution u xð Þ is an analytic function

in X(see Nirenberg [10]), a feature which will play an important role in this paper. Also, under the above assumptions on thedata, the comparison principle (see Pucci and Serrin [17]) implies that u > 0 on X, while Hopf’s first maximum principle [6]implies that u xð Þ assume its maximum value on @X and its minimum value in X.

Now, let us consider the following P-function, i.e. an appropriate combination of the solution and its gradient (see thebook of Sperb [19]):

P x;að Þ :¼ 2� 2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1þ ruj j2

q � ca2

u2; with a 2 R; ð1:4Þ

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L. Barbu / Applied Mathematics and Computation 219 (2013) 8192–8197 8193

where u xð Þ is the solution of problem (1.1), (1.2). Payne and Philippin have proved in [13] that P x;að Þ takes its maximumvalue on @X, for a 2 0;1½ �, while P x;2ð Þ takes its maximum at a critical point of u xð Þ. As a consequence of the maximum prin-ciple for P x;1ð Þ, they also derived the following lower bound for u2

min:

u2min P a2 � 4

c1�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� ac

2kmin

� �2s2

435: ð1:5Þ

Regarding the values of a for which the auxiliary function P x;að Þ might satisfy a minimum principle, our first result states:

Theorem 1. If a 2 ð1;3=2� and c 6 2ða� 1Þkmin

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4k2

min � a2c2q

, then the auxiliary function P x;að Þ attains its minimum value onthe boundary @X.

As a consequence of Theorem 1, in the second main result of this note we derive the following upper bound for u2min :

Theorem 2. Let uðxÞ be the analytic solution of problem (1.1). If the assumptions of Theorem 1 hold, then we have the followingheight estimate:

u2min < a2 � 4

ac1�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� ð2� aÞac

2kmax

� �2s2

435; ð1:6Þ

where umin :¼minX

u xð Þ and kmax is the maximum value of the curvature k sð Þ on @X.

We note that similar minimum principles for P-functions are known only in several cases, namely for the problem of tor-sional rigidity (see [12] and [8] or [15]), the problem of torsional creep (see [14]), the problem of capillary free surfaces withoutgravity (see [9] and [14]), the soap film problem (see [1]) or for a general class of quasilinear elliptic problems, whose solutionssatisfy some convexity property (see [16]). To prove Theorem 1 and Theorem 2 we will follow the arguments recently used inBarbu and Enache [1] (see also [14]) to prove similar results for a soap film problem.

The outline of the paper is as it follows. In Section 2 we will give some preliminary results, which play an important role inthe proof of Theorem 1. Thereafter, in Section 3 we prove Theorem 1, while the proof of Theorems 2 is given in Section 4.Some final remarks will also be given in Section 5.

Finally, we note that here and in the remainder of this paper the summation convention over repeated indices (from 1 to2) is employed and the following notations are adopted: u1 :¼ @u=@x1;u2 :¼ @u=@x2 and uij :¼ @2u=ð@xi@xjÞ, for i; j 2 1;2f g.

2. Some preliminary results

In this section we formulate and prove some key lemmas, which play an important role in the proof of Theorem 1.

Lemma 3. The solution uðxÞ of problem (1.1),(1.2) has a single critical point in X.

Proof. The idea of proof is standard, being entirely based on a technique introduced by [11] to prove a similar uniquenessresult for the first eigenfunction of the Dirichlet Laplacian (see, also [14] or [1]). Therefore, we only sketch its main stepshere.

Let us denote vk :¼ uk; k ¼ 1;2, and write equation (1.1) in the following form:

1þ ruj j2� �

Du� uijuiuj � cu 1þ ruj j2� �3=2

¼ 0: ð2:1Þ

Differentiating (2.1) with respect to xk; k ¼ 1;2, we obtain

1þ ruj j2� �

dij � uiuj

h ivk

ij þ 2uiDu� 2uijuj � 3cuuið1þ ruj j2Þ1=2h i

vki � cvk 1þ ruj j2

� �3=2¼ 0 in X; k

¼ 1;2; ð2:2Þ

where dij is the Kronecker symbol. Since Eq. (2.2) is linear, then any linear combination of v1 and v2 of type

vðhÞ :¼ v1 cos hþ v2 sin h; with some h fixed; ð2:3Þ

also satisfies (2.2). According to maximum principle [17], if v takes a positive maximum (negative minimum) value M in X,then v � M. Also, since u ¼ a on @X, we have

vk ¼ @u@n

nk on @X; ð2:4Þ

where ðn1;n2Þ :¼ n is the outward unit normal vector on @X and @u=@n is the outward normal derivative of u xð Þ. Then

vðhÞ ¼ ð@u=@nÞn � ðcos h; sin hÞ on @X: ð2:5Þ

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On the other hand, Hopf’s second maximum principle [7] says that @u=@n > 0 on @X, so that v hð Þ vanishes at exactly twopoints on @X, there where n ? ðcos h; sin hÞ.

Now, if we assume that u xð Þ has N P 2 critical points in X, we may obtain a contradiction by studying the behavior of thenodal lines of v hð Þ exactly in the same way as in Philippin [14, p. 528]. h

Next, eventually performing a translation and/or rotation if necessary, we chose the coordinate axes such that the (single)critical point of u xð Þ is located at the origin O and u12ðOÞ ¼ u21ðOÞ ¼ 0. In the remainder of the paper this new system of coor-dinates will be considered. We have:

Lemma 4. At the origin, the solution uðxÞ of problem (1.1), (1.2) satisfies:

u11ðOÞ > 0; u22ðOÞ > 0: ð2:6Þ

Proof. Since u xð Þ attains its minimum value at the single critical point O 2 X, we clearly have u11ðOÞP 0 andu22ðOÞP 0.Therefore, to achieve the proof of the lemma, it remains to show that u11ðOÞ– 0 and u22ðOÞ– 0.

Let us assume contrariwise that u11 Oð Þ ¼ 0. We shall show that this assumption leads us to a contradiction. Indeed, if weconsider the function v1 :¼ u1, we note that it satisfies the following properties: (i) v1 is an analytic function in X(becauseu xð Þ is an analytic function in X); (ii) v1 is not identically constant (because, in such a case, u xð Þ will depend only on x2, sothat the boundary condition could not be satisfied); (iii) v1 vanishes up to a finite order m P 1 at O (becausev1 Oð Þ ¼ v1

1 Oð Þ ¼ v12 Oð Þ ¼ 0 and (ii)); (iv) v1 satisfies a differential equation with analytic coefficients and with no zero

ordered term (see (2.2)). Therefore, in some neighborhood U of the origin, the zero set of v1 consists of N :¼ mþ 1 P 2 smootharcs (nodal lines of v1), all intersecting under equal angles at the origin and dividing U in 2N P 4 sectors (see Lemma 1 andLemma 2 in [2] and [5]).

On the other hand, since v1 xð Þ vanishes at only two points on @X (see the proof of Lemma 3), it must therefore have atleast one closed nodal line in X, thing which leads us to the desired contradiction (indeed, if such a closed nodal line exists,since v1 satisfies Eq. (2.2), maximum principle [17] would imply that v1 is identically zero inside the domain bounded bythis nodal line; therefore, the analyticity of v1 will imply that v1 is in fact identically zero in X, which is impossible!).

In the same way, using the above argument with v2 instead of v1, we may also obtain that u22ðOÞ – 0 and the proof of thislemma is achieved. h

Lemma 5. If a 2 ð1;3=2� and c 6 2ða� 1Þkmin

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4k2

min � a2c2q

, then auxiliary function Pðx;aÞ attains its minimum value at the sin-gle critical point of uðxÞ or on the boundary @X.

Proof. The idea of proof is standard (see the book of Sperb [19]), being based on the construction of a second order ellipticdifferential inequality for P x;að Þ, followed by the application of Hopf’s first maximum principle [6]. More exactly, differen-tiating successively (1.4), we have

Pk ¼2uikui

ð1þ ruj j2Þ3=2 � acuuk; ð2:7Þ

Pkj ¼2 uikjui þ uikuij� �ð1þ ruj j2Þ3=2 �

6uljuluikui

ð1þ ruj j2Þ5=2 � acuukj � acujuk: ð2:8Þ

Moreover, making use of the following identity (which holds only in R2!)

uikuik ruj j2 ¼ ruj j2 Duð Þ2 þ 2uijuiukjuk � 2 Duð Þuijuiuj; ð2:9Þ

and of (2.7), (2.8), one may obtain after some reduction the following equation

DP � Pkjukuj

1þ ruj j2þWkPk ¼

c a� 2ð Þð1þ ruj j2Þ1=2 cu2ða� 1Þ � ruj j2

ð1þ ruj j2Þ1=2

" #; ð2:10Þ

where Wk is a smooth vector function which is singular at the critical point of u xð Þ (see, also [13]). Obviously, if in (2.10) wewould have

cða� 1Þu2 Pruj j2

ð1þ ruj j2Þ1=2 in X; ð2:11Þ

then the application of Hopf’s first maximum principle [6] would achieve the proof.It remains then to show that inequality (2.11) is true. For this aim, we make use of the following inequalities derived by

Payne and Philippin [13]:

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L. Barbu / Applied Mathematics and Computation 219 (2013) 8192–8197 8195

qmax 64k2

min

a2c2 � 1

!�1=2

; cu2 P ca2 � 4

ð1þ ruj j2Þ1=2 þ4

ð1þ q2maxÞ

1=2 in X; ð2:12Þ

where qmax :¼maxXruj j ¼min

@Xruj j (since, according to Payne and Philippin [13], ruj j attains its maximum value on bound-

ary). From (2.12), we see that inequality (2.11) is true if one may show that

ca2ða� 1Þ þ 4ða� 1Þð1þ q2

maxÞ1=2 P

4ða� 1Þ þ ruj j2

ð1þ ruj j2Þ1=2 in X: ð2:13Þ

Now, the right hand side of (2.13) takes its maximum value when ruj j ¼ qmax. Therefore, inequality (2.13) holds if

ca2ða� 1ÞP q2max

ð1þ q2maxÞ

1=2 : ð2:14Þ

Now, using successively (2.12) and the assumptions of the lemma, we have

q2max

ð1þ q2maxÞ

1=2 6a2c2

2kminð4k2min � a2c2Þ1=2 6 a2c a� 1ð Þ; ð2:15Þ

so that inequality (2.14) is indeed true and the proof of Lemma 5 is achieved. h

3. Proof of Theorem 1

We assume contrariwise that the minimum of P x;að Þ occurs at the origin, where u1 Oð Þ ¼ u2 Oð Þ ¼ u12 Oð Þ ¼ 0, and try toreach a contradiction.

First, evaluating (2.1), (2.7) and (2.8) at the origin and taking into account that Pðx;aÞ attains its minimum at O, we have

DuðOÞ ¼ cu Oð Þ; ð3:1Þ

respectively

P1ðO;aÞ ¼ P2ðO;aÞ ¼ 0;P11ðO;aÞ ¼ u11ðOÞð2u11ðOÞ � cau Oð ÞÞP 0;P22ðO;aÞ ¼ u22ðOÞð2u22ðOÞ � cau Oð ÞÞP 0:

ð3:2Þ

Therefore, making use of Lemma 4, (3.2) implies

2u11ðOÞ � cau Oð ÞP 0; 2u22ðOÞ � cau Oð ÞP 0; ð3:3Þ

so that, by addition, we have

2DuðOÞ � 2cau Oð ÞP 0: ð3:4Þ

Inserting now (3.1) in (3.4) and taking into account that a > 1;uðOÞ > 0, we obtain the desired contradiction and the proof isthus achieved. h

Remark 1. The minimum principle obtained in Theorem 1 is not the best possible, since no constant P would satisfy (2.10) insuch a case.

4. Proof of Theorem 2

From Theorem 1 we know that P x;að Þ takes its minimum value at some point Q 2 @X. This fact implies that

2� 2

ð1þ q2minÞ

1=2 �ac2

a26 2� 2

ð1þ ruj j2Þ1=2 �ac2

u2; ð4:1Þ

where qmin is the minimum value of ruj j on @X. Evaluating (4.1) at the unique minimal point of u xð Þ, we obtain the followinginequality

u2min 6 a2 þ 4

ac1

ð1þ q2minÞ

1=2 � 1

!; ð4:2Þ

where umin is the minimum value of u xð Þ on X.Next, we construct an upper bound for 1= 1þ q2

min

� �. Since P x;að Þ cannot be constant on X and takes its minimum value at

the point Q 2 @X, Hopf’s second maximum principle [7] implies that @P Q ;að Þ=@n < 0, or

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8196 L. Barbu / Applied Mathematics and Computation 219 (2013) 8192–8197

unn

ð1þ u2nÞ

3=2 <aac

2at Q ; ð4:3Þ

where un and unn are the first and second outward normal derivatives of u xð Þ on @X (we have used that un > 0 on @X). On theother hand, since the boundary @X is smooth, equation (1.1) may be evaluated on @X, in normal coordinates with respect to@X, as it follows

unn

1þ u2n

� �3=2 þkun

1þ u2n

� �1=2 ¼ ac on @X: ð4:4Þ

Now, the insertion of (4.4) in (4.3) leads us to

qmin

ð1þ q2minÞ

1=2 >ac 2� að Þ

2kmax; ð4:5Þ

so that

11þ q2

min

< 1� ð2� aÞac2kmax

� �2

: ð4:6Þ

Finally, insertion of (4.6) in (4.2) leads to

u2min < a2 � 4

ac1�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� acð2� aÞ

2kmax

� �2s2

435; ð4:7Þ

and the proof of Theorem 2 is achieved. h

5. Some remarks for fluid surfaces in long capillary tubes

When the capillary tube is assumed to be so long that the fluid does not reach the top of the tube at any point along itsperiphery, the surface solution uðxÞ satisfies the following boundary condition

1þ jruj2� ��1=2 @u

@n¼ cos h; on @X; ð5:1Þ

where 0 < h < p2 is the contact angle between the fluid surface and the capillary wall. According to Ural’ceva [22] (see, also

[20] or [21]), problem (1.1), (5.1) has a classical solution uðxÞ, so that we can consider again the auxiliary function P x;að Þdefined in (5.1), but with uðxÞ being now solution to problem (1.1), (5.1). According to Payne and Philippin [13], P x;1ð Þ takesits maximum value on @X and we have

u2 Að Þ � u2 Cð Þ 6 4c

1� sin hð Þ; ð5:2Þ

where A 2 @X is a point corresponding to a minimum boundary value of u xð Þ and C 2 X is the unique minimal point of u xð Þ(see [3]). Moreover, in the same paper [13], Payne and Philippin also proved that P x;2ð Þ takes its maximum value at the un-ique minimal point of u xð Þ. Therefore, from (2.10) we conclude that P x;2ð Þ attains its minimum value on @X, unless P x;2ð Þ isconstant. We assume contrariwise that P x;2ð Þ might be constant on X and show that this assumption leads to acontradiction.

Indeed, if P x;2ð Þ is constant on X , then at point A 2 @X, where the tangential derivative satisfies us ¼ 0, we have@P=@n ¼ 0, or

unn ¼ cuð1þ u2nÞ

3=2; ð5:3Þ

since un > 0 on @X. On the other hand, evaluating Eq. (1.1) at the same point A 2 @X, in normal coordinates with respect tothe boundary @X, we have

unn ¼ cu 1þ u2n

� �3=2 � uss þ kunð Þ 1þ u2n

� �: ð5:4Þ

Combining now (5.3) and (5.4), we get

�kun ¼ uss; at point A 2 @X; ð5:5Þ

where the left hand side of (5.5) is strictly negative (since k sð Þ > 0 and un > 0), while the right hand side is non-negative(since A is the point corresponding to a minimum boundary value of u xð Þ). Therefore, we have reached a contradiction, sothat P x;2ð Þ attains its minimum value at a point B 2 @X, where u xð Þ itself is a maximum, and we may derive the followinginequality

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L. Barbu / Applied Mathematics and Computation 219 (2013) 8192–8197 8197

u2 Bð Þ � u2 Cð Þ > 2c

1� sin hð Þ: ð5:6Þ

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