Deformation Near a Liquid Contact Line on an Elastic Substrate
Electronic Supplementary Material
Chung-Yuen Hui1 and Anand Jagota2,*
1
Cornell University, Department of Mechanical and Aerospace Engineering, Cornell University, Ithaca,
NY 14850
2
Lehigh University, Department of Chemical Engineering and Bioengineering Program, 111 Research
Drive, Lehigh University, Bethlehem, PA 18018
*
Corresponding author, D331 Iacocca Hall, Department of Chemical Engineering and Bioengineering
Program, Lehigh University, Bethlehem, PA 18017, anj6@lehigh.edu
1. Numerical method
To reduce the singularity of the kernel in the integral equation given by (20a), we integrate by
parts, using

 x 

v, t ( t )dt
PV 
  lim    ( t )d ln( x  t )    ( t )d ln( t  x ) 
 0
x t

x 
 


(S1a)
Using
x 
x 




(
t
)
d
ln(
x

t
)


(
x


)
ln



(
t
)
ln(
x

t
)
dt












x   ( t )d ln( t  x )   ( x   ) ln   x  ( t ) ln( t  x )dt 
(S 1b)
Combining S 1a-b, we obtain

PV

 ( t )dt

x t

   , t ( t )ln x  t dt
(S 1c)

Thus, (20a) is equivalent to
p( x ) 
1



f ( t ) ln x  t dt   ( x ) f ( x )
(S 2)

where f ( t )   , t ( t )  v, tt ( t ) . We convert the infinite domain into a finite one using the
transformation
F (v)  f  cot  v / 2  x  cot  u / 2 , t  cot  v / 2
(S 3)
1
Using this transformation, S 2 becomes:
P(u ) 
1
2


 ln sin  v / 2  sin  u / 2  (v)dv  (u ) sin  u / 2  (u )
2
sin (v  u ) / 2
2
(S 4a)
0
where
1


cot  u / 2   1

(u )  1   SL /  SV )  1  
2

 SL /  SV


cot  u / 2   1
cot  u / 2   1
(S 4b)
cot  u / 2   1
 (u)  F (u) / sin 2 (u / 2)
(S 4c)
P(u )  p  cot  u / 2    sin c
2

e

cot 2  u /2 
2
(S 4d)
The first integral in S 4a can be simplified, i.e.,
1
2

2
 ln
0
1
2
sin  (v  u ) / 2 
 (v)dv
sin  v / 2  sin  u / 2 
2
(S 5a)
2
1
 ln sin  (v  u) / 2  (v)dv  2  ln sin  v / 2  (v)dv
0
0
where we have used the fact that
2

0

  (v)dv 


f (t )
dv
dt  2  f ( t )dt  0
2
sin (v( t ) / 2) dt

(S 5b)
Using the (S 5a,b), S 4a becomes:
1
2
2
1
0 ln sin  (v  u ) / 2  (v)dv  2
2
 ln sin  v / 2  (v)dv  (u)sin  u / 2    P(u )
2
(S 6)
0
To deal with the weak logarithmic singularity when v = u, we use the following regularization:
1
2
2
 ln sin  (v  u) / 2  (v)dv
0
2
 1
sin  (v  u ) / 2 
1 

 (v)dv  
  ln
2  0
(v  u ) / 2
 2
2
(S 7)
 ln (v  u) / 2  (v)dv
0
2
For numerical purpose, it is easier to integrate in the interval  1,1 , this can be accomplished using the
change of variable,
u     1 , v     1
(S 8)
Equation S.7 becomes
1
1
 11
1
1
2
K
(



)

(

)
d


ln
(



)

(

)
d


ln sin  (  1) / 2  ( ) d




2  1
2 1 
 2 1
(S 9a)
 ( )sin 2  (  1) / 2  ( )  P( )
where K (   )  ln
sin  (   ) / 2 
.
 (   ) / 2
(S 9b)
After regularization, we can use standard quadrature techniques, where
1
n 1 i 1
1
i 1 i
 K (   ) ( )d  

n 1
K (   ) ( )d   K (ˆi   ) (ˆi )(i 1 i )
i 1
(S 10a)
ˆi  i 1  i  / 2
The logarithmic singularity of the second term in S 9a can be dealt with by exact integration, for
example,
1
n 1
i1
1
i 1
i
 ln (   )  ( )d   (ˆi )  ln (   ) d
(S 10b)
n 1
  (ˆi ) i 1    ln (i 1   )  i    ln (i   )  (i 1  i ) 
i 1
The third integral in S 9a is discretized using
1
 ln
1
2

n
sin  (  1) / 2  ( )d   (ˆi ) ln
i 1
sin  (ˆi  1) / 2 
(i 1 i )
 (ˆi  1) / 2
(S 10c)
n
  (ˆi ) (i 1  1) ln(i 1  1)  (i  1) ln(i  1)  i 1  i  
i 1
Combining S 10a,b,c, we obtained the discretization of the integral operators on the LHS of S 9a.
Evaluating both sides of the discretized S 9 at    j ,1  j  n resulting in a system of n linear equations
 
with n unknowns  j    j which we solve using Matlab.
3
When the solid surface tensions are different, the solution procedure for vertical displacements has to
be modified and we also need separately to solve for horizontal displacements. The principal change in
the calculation of vertical displacements is the loss of symmetry around the origin which previously
allowed us to solve for v,xx and to integrate numerically out from the origin with the condition that v,x =
0 at x=0. When solid surface tensions are different, we first integrate the solution as in the symmetric
case. We then integrate it out sufficiently far from the origin such that the elastic stresses dominate,
and here the solution is symmetric. By matching the solution at large positive and negative distances,
we find the value of v,x at x=0, hence completing the solution.
Because of incompressibility, the horizontal and vertical deformation problems are decoupled.
The horizontal displacement problem is simpler. In particular, the shear traction is given by
 1
 lv cos  c   sl   sv 

q   2
 0

x 1
(S11)
x 1
For this case we can directly apply the results from Johnson, equations 2.30b, 2.30d, and 2.32a[1]. After
normalization, we get
 /    1  x  1
du
1
   cos  c  sl sv
ln
 lv /  sv   x  1
dx

1
u 
2

 /    1 
2
2
 cos  c  sl sv
  x  1 ln  x  1   x  1 ln  x  1



/

lv
sv




(S12)
2. Solution including horizontal displacements
Within small-strain, linear, elasticity our formulation can handle situations such as  SV   SL and/or
sin  c   1
, either of which result in horizontal as well as vertical surface displacements. Thus, one
expects a non-symmetrical deformed surface when  SV   SL even if  c   / 2 . Figure S1 illustrates
this by an example where we have assigned  c   / 2 as a fixed value and have chosen a moderately
large value of  =10. For the case of  lv /  sv  1 (shown already in Figure 4a) elastic forces are small in
the region near x  1 , the shape is symmetric, and approximately given by Neumann’s triangle. As the
parameter  lv /  sv decreases from unity, the shape of the solid surface rotates to satisfy equilibrium
(primarily due to surface tensions). Note, however, that even for moderate difference in solid surface
tensions, the calculated shape near the contact line is unphysical – this is an indication of the limitation
of the small-strain, linear theory in handling the large rotations. That is, in the limit of large  , the
condition  SV   SL causes significant rotations of the surface in the vicinity of the contact line,
invalidating the linear theory used. For this reason, we do not here pursue further cases where there is
4
significant horizontal displacement – those will likely require a more general numerical treatment, for
example using finite element methods with surface tension elements [2-4]. Combining force and
configurational energy balance for the problem of a liquid drop on an elastic half-space will likely lead
also to a range of contact angles permissible for both conditions to be satisfied simultaneously, and will
also require more involved numerical analysis.
2
1.6
 = 10.0
 = /2
c
lv/sv=1
1.4
 /
Normalized distance y/
1.8
1.2
1
0.8
sl
sv
1.00
0.90
0.80
0.70
0.6
0.4
0.2
0
-5
0
Normalized distance x/
5
Figure S1. Deformed surface shapes for a surface with both vertical and horizontal surface
displacements.
1.
2.
3.
4.
Johnson, K.L., Contact Mechanics. 1985, Cambridge: Cambridge University Press.
Xu, X., et al., Gravity and Surface Tension Effects on Shape Change of Soft Materials. Langmuir,
2013.
Jagota, A., C. Argento, and S. Mazur, Growth of adhesive contacts for Maxwell viscoelastic
spheres. Journal of Applied Physics, 1998. 83(1): p. 250-259.
Hui, C.-Y., et al., Constraints on Micro-Contact Printing Imposed by Stamp Deformation”, .
Langmuir, 2002. 18: p. 1394-1404.
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