1. No category

MATHEMATICAL MODELING OF CAPILLARY ORIGAMI
Team:
Christeen Bisnath
Brendan DeCourcy
Olexsandr Iaroshenko
Emel Khan
Arnd Korn
David Nigro
Miranda Polin
Mentor: John A. Pelesko
Date: 6/14/2013
Abstract. We will investigate a liquid droplet on a specific shaped material sheet. As time elapse, the
sheet starts to fold onto the liquid. We will model this droplet on a sheet as a 2 − D hinge system; for a
single and double hinge with specified boundary conditions. First, we will formulate the energy function
as the surface area energy of the droplet and the torsion spring energy. Using variational calculus we will
optimize the functional which includes the volume constraint. We will also use numerical methods to find
the center location of the surface droplet and certain parameters.
1. Simple Hinge
• Formulation correct with pinned boundaries (nonparametric)
• Proved solutions have constant κ (curvature)
• Construct explicit solution (non-parametric)
• Formulation correct with fixed boundaries (parametric)
• Construct explicit solution (parametric)
We will consider a 2-D ridged hinge system with the
droplet fully contained within the two fixed rods on a hydrophobic sheet material. We will formulate this as an
energy minimization problem with a volume constraint
(via variational calculus) which will help us derive an ordinary differential equation representing the surface of the
droplet. We will derive algebraic equations that lead us to
find the center location of the liquid surface via numerical
methods.
The total energy of the system is comprised of the surface energy of the droplet and the elastic energy of the
spring. We will optimize the functional:
Z π+θ
Z π+θ
p
2
2
1
1
u2 +l u02 dφ + k(θ − θ0 )2 + λ
u2 dφ
(1.1) J[u(φ), θ] = γ
π−θ
π−θ
2
2
2
2
Figure 1. The surface of the
The Lagrangian includes a contribution from the voldroplet is represented by the
ume constraint (area in two dimensions) with multiplier
function u(φ) where φ ∈ [0, θ],
λ and a term for the free surface (arclength of the boundR is the radial length of the
ary) with a surface tension γ (in units energy per unit
droplet, b is the length of the
1
rods, A is the distance from the
base of the rod to the base of
the droplet, and θ is the angle
between the rods.
2
MATHEMATICAL MODELING OF CAPILLARY ORIGAMI
area) as
(1.2)
F (x, x0 , y, y 0 ) = λyx0 + γ
p
x02 + y 02 .
With it, the energy per unit length can be written as
2
Z 1
k
R
0
0
2
0
0
Jλ [t, x, x , y, y , θ] =
F (x, x , y, y )dt + (θ − θ0 ) − λ
sin θ + A0 .
2
2
0
A suitable linear combination of the Euler-Lagrange equations with respect to (x, x0 ) and (y, y 0 ) lead to
λ
(1.3)
|κ| = γ
which describes a curve of constant curvature1, such as a straight line (κ = 0), or a convex (κ > 0) or
concave (κ < 0) circle section. This result is in agreement with the derivation in polar coordinates above
and includes free surfaces beyond the opening angle of the hinge (ice cone geometry).
The other two equations come from variation of J with respect to θ and λ and read
cos θ =
Z
0
1
yx0 dt −
2k
(θ − θ0 ),
λR2
R2
sin θ = A0 .
2
These three equations are equivalent to the system of governing equations given in different variables above.
With suitable initial conditions they form the boundary value problem.
2. Simple Hinge with gravity, free boundary
• Formulation
• Numerical Solution
2.1. Formulation. Recall the energy equation for the simple hinge without gravity from the previous section, including the volume constraint, volume = V0 :
!
Z π+θ
Z π+θ
p
2
2
1
1
2
2
(2.1)
J[u(φ), θ] = γ
u2 + u02 dφ + k(θ − θ0 ) + λ
u dφ − V0
π−θ
2
2 π−θ
2
2
Our goal now is to include the effects of gravity on the liquid in the hinge. We will be making the assumption
that the mass of the hinge is negligible in the system.
1The signed curvature of a curve in R2 given in parametric coordinates, t ∈ [0, 1] → (x(t), y(t)), is κ = (x0 y 00 − y 0 x00 )(x02 +
y 02 )−3/2 .
MATHEMATICAL MODELING OF CAPILLARY ORIGAMI
3
We consider a small fluid particle of density ρ with mass dm = ρdv, at a height h above the horizontal.
The gravitational potential of this particle is then given by
dEg = ρghdv.
By integrating this quantity over the total area of the cross section and utilizing the established notation for
the geometry, we find that the total contribution of gravity to the system energy is
1
Eg = ρg
3
(2.2)
Z
π+θ
2
u3 sin φdφ
π−θ
2
Then with the inclusion of gravity and the volume constraint, the function we wish to minimize is:
Z
(2.3)
Jg [u(φ), θ] =
π+θ
2
π−θ
2
p
1
1 2
1
3
2
02
γ u + u + ρgu sin φ + λu dφ + k(θ − θ0 )2 − λV0
3
2
2
For a more general approach to the problem, we also allow the value of the function u on the boundaries
to be a free condition.
To find the least energy state of the system, we apply the methods of the calculus of variations as before.
We require that the integrand in (2.3) satisfy the Euler-Lagrange equation, but the free boundary condition
places additional conditions on the properties of u at the endpoints. These conditions are called the natural
boundary conditions. For our system, they are
∂F ∂F =
=0
∂u0 π−θ
∂u0 π+θ
(2.4)
2
2
where F (φ, u, u0 ) is the integrand in the energy equation (2.3).
The addition of gravity to the system does not change the value of ∂F/∂u0 , so we can use our results for
the simple hinge without gravity to see that we must satisfy the boundary conditions
(2.5)
u0
π−θ
2
= u0
π+θ
2
=0
This is equivalent to the requirement that the contact angle between the fluid and the wall is π/2.
The Euler-Lagrange equation for the system reduces to the second order nonlinear differential equation
2u02
λ 1 ρg
00
u =u+
+
+
sin x (u2 + u02 )3/2
u
γu
γ
We can rearrange this equation as before to find the curvature of the surface:
λ ρg
− u sin φ
γ
γ
The second term is a perturbation from constant curvature due to gravity. As we might expect, this is a
negative value that is symmetric around φ = pi/2.
In order to find the minimal energy state of the system, we must also find ∂Jg /∂θ = 0 as before in the
zero gravity case. To simplify the resulting expression, we will use the boundary conditions (2.5) as well as
a symmetry assumption:
π−θ
π+θ
u
=u
=b
2
2
.
This results in the following relation between the system parameters:
λ 2 ρg 3
θ
kθ + γb + b + b cos
=0
2
3
2
Finally, we can completely describe our system by including the volume V0 and the boundary conditions
on u:
(2.6)
κ(φ) = −
4
(2.7)
MATHEMATICAL MODELING OF CAPILLARY ORIGAMI

λ 1 ρg
2u02
00


+
+
sin x (u2 + u02 )3/2
u =u+


u
γ
u
γ




Z π+θ

2

1


V0 =
u2 dφ


π−θ

2



2 π−θ
π+θ
0
0
u
=u
=0


2
2




π−θ
π+θ



u
=
u
=b


2
2





λ
ρg
θ


kθ + γb + b2 + b3 cos
=0
2
3
2
2.2. Numerical Solution. Our next goal is to utilize a numerical scheme to solve the system of equations
for some sensible paramater values. The general approach is to prescribe a value for the volume constraint
λ, and then use the shooting method to fit the boundary conditions.
From the equations (2.7), the system has the general form:
u00 (φ) = G(φ, u, u0 ; λ)
(2.8)
(2.9)
(2.10)
u0
π−θ
2
g(b, θ) = 0
π
= u0
=0
2
where we have used symmetry to say that u0 (π/2) = 0.
To solve the system, we make a guess for the value of u((π − θ)/2) = b, and use this value and the
constraint g(b, θ) = 0 to determine θ. Once we have the interval, we can use a numerical scheme to correct
b and θ until our boundary conditions are met.
The system to be modeled with this scheme is a centimeter scale droplet of water at room temperature
on a comparatively massless simple hinge system. The parameter values used for this model are: the
surface tension γ = 71.97 × 10−3 N/m, density ρ = 1000kg/m3 , spring constant k = 0.01N/m. Using these
parameters and varying the value of the volume constraint λ, we can find the equilibrium state for droplets
of varying volume.
MATHEMATICAL MODELING OF CAPILLARY ORIGAMI
5
As is expected from the predicted variation in curvature, the surface of the water droplet is slightly
flattened due to the effects of gravity. In order to quantify this difference we can perform a perturbation
analysis.
6
MATHEMATICAL MODELING OF CAPILLARY ORIGAMI
2.3. Perturbation analysis. Recall that if we take into acount the effect of gravity, the equation satisfied
by the free surface u is given by:
(2.11)
00
02
u u = 2u +
ρg
λ
u sin(ϕ) +
γ
γ
(u2 + u02 )3/2 + u2
r
π−θ
π+θ
γ
= u0
= 0. Let Lc =
and L0 a typical lengthscale
2
2
ρg
of the system. Then we can non-dimensionalise the previous equation to get:
with the free boundary condition u0
(2.12)
λL0
u u = 2u + u sin(ϕ) +
γ
00
02
(u2 + u02 )3/2 + u2
L20
is a non-dimensional number that represents the importance of gravity relative to capillary
L2c
effect. For small the effect of gravity is small whereas for large capillary effects are preponderant. In
order to study the effect of gravity we will study the perturbation of the solution of the problem for zero
gravity i.e. = 0 for small . Let u = u0 + u1 and expand the equation at O(1) and O(). We get the two
equations:

λL0 2

3/2

O(1) : u000 u0 = 2u02
(u0 + u02
+ u20

0 +
0 )

γ

λL0
1/2
3/2
(2.13)
(u0 u1 + u00 u01 )(u20 + u02
O() : u000 u1 + u0 u001 = 4u00 u01 + u0 sin(ϕ)(u20 + u02
+3
0 )
0 )


γ


+2u u
0 1
with =
Now if we apply the same method to the algebric equation satisfied by θ i.e. by setting θ = θ0 + θ1 and by
π θ0
π θ0
and b = + , we get the non-dimensionalised equations:
introducing a = −
2
2
2
2
(2.14)

1/2 1 2
1/2
k
1 2

O(1) :
y0 (b) + y002 (b)
+
y0 (a) + y002 (a)
θ0 +



λL
2
2

 1 λL0 0



+
y02 (b) + y02 (a) = 0


4 γ




O() : k θ1 + 1 (y02 (b) + y002 (b))−1/2 (2y0 (b)y1 (b) + 2y00 (b)y10 (b)
λL0
4
θ1
1

0


+ (2y0 (b)y0 (b) + 2y00 (b)y000 (b))) + (y02 (a) + y002 (a))−1/2 (2y0 (a)y1 (a) + 2y00 (a)y10 (a)

 2
4


θ1


− (2y0 (a)y00 (a) + 2y00 (a)y000 (a))) + y03 (b) sin(b) + y03 (a) sin(a)



2



+ 1 λL0 (2y0 (b)y1 (b) + θ1 y0 (b)y00 (b) + 2y0 (a)y1 (a) − θ1 y0 (a)y00 (a) = 0
4 γ
and then the boundary conditions become:

O(1) : u00 (b) = u00 (a) = 0
(2.15)
θ
θ
O() : 1 y000 (b) + y10 (b) = − 1 y000 (a) + y10 (a) = 0
2
2
Thus to summarise we have at leading order:

λL0 2

3/2

(u0 + u02
+ u20
u00 u0 = 2u02

0 +
0 )

γ
 0
1/2 1 2
1/2 1 λL0 2
k
1 2
(2.16)
θ0 +
y0 (b) + y002 (b)
+
y0 (a) + y002 (a)
+
y0 (b) + y02 (a) = 0


λL0
2
2
4 γ


u0 (b) = u0 (a) = 0
0
0
MATHEMATICAL MODELING OF CAPILLARY ORIGAMI
7
which is the same system as when we were neglecting gravity i.e. we have the same results as before. The
first order correction due to gravity is then given by:

λL0

1/2
3/2

(u0 u1 + u00 u01 )(u20 + u02
u000 u1 + u0 u001 = 4u00 u01 + u0 sin(ϕ)(u20 + u02
+3

0 )
0 )

γ




+2u0 u1




k
1



θ1 + (y02 (b) + y002 (b))−1/2 (2y0 (b)y1 (b) + 2y00 (b)y10 (b)


λL
4

 θ10
1
+ (2y0 (b)y00 (b) + 2y00 (b)y000 (b))) + (y02 (a) + y002 (a))−1/2 (2y0 (a)y1 (a) + 2y00 (a)y10 (a)
(2.17)
2
4


 θ1
0
0
00


(2y
(a)y
(a)
+
2y
(a)y
(a)))
+
y03 (b) sin(b) + y03 (a) sin(a)
−
0

0
0
0
 2


1 λL0



+
(2y0 (b)y1 (b) + θ1 y0 (b)y00 (b) + 2y0 (a)y1 (a) − θ1 y0 (a)y00 (a) = 0


4
γ



θ
θ

 1 y000 (b) + y10 (b) = − 1 y000 (a) + y10 (a) = 0
2
2
which can be solved knowing the leading order solution using a shooting method for example.
3. Two Hinged Problem
For a slightly more complex model, we can derive the energy equations for a system with two hinges. In
this model, we will ignore gravity. The surface energy will be unchanged, but the spring energy will need to
be adjusted to account for the second hinge.
The geometry will be described as in the following figure., which has similiar measurments as the one
hinge system.
u (φ)
a
.
θ=
π
2
+
b
c
α
α
2
a
.
c
Since we are assuming symmetry, the energy for each spring will be identical. As a result, the spring
energy to be added to our model is:
(3.1)
Es = k
π
2
+
2
α
− θ0
2
The volume of the fluid region will also change:
(3.2)
V0 =
1
2
Z
0
α
u2 dφ −
a2
sin α
8 cos2 α/2
The complete energy equation including the volume constraint is
Z
(3.3)
J[u(φ), α] =
0
α
p
π α
2
1 2
a2
2
02
γ u + u + λu dφ + k
+ − θ0 − λ
sin α + V0
2
2
2
8 cos2 α/2
As one can see, the integral has not changed from the single hinge case, so the Euler-Lagrange equation
will be the same as before. Therefore the curvature will be the same constant value derived earlier in (1.3).
To complete the description of the model, we include the fact that our minimized energy state will satisfy
∂J/∂α = 0:
(3.4)
π α
p
λ
a2
∂J
=k
+ − θ0 + γ u(α)2 + u0 (α)2 + u(α)2 +
=0
∂α
2
2
2
8 sin2 α/2
8
MATHEMATICAL MODELING OF CAPILLARY ORIGAMI
3.1. Non-dimensionalisation.
2
Z α
Z αp
1
a2
π−α
2
2
02
u dφ −
Jλ [u, α] = γ
− θ0 + λ
− 2A0
u + u dφ + k
2
2
4 tan α2
0
0
Introduce a length scale L such that u = Lu∗ and divide the functional by k to derive the nondimensional
parameter groups
γL
λL2
λa2
λA0
A1 =
,
A2 =
,
A3 =
,
A4 =
,
k
k
k
k
each of which represents balancing of different contributions to the energy, for example from elastic and
capillary effects.
4. Conclusion
In this project we attempted to mathematically justify the results of capillary origami experiments. We
used simple two dimensional models of massless single and double hinged materials interacting with a fluid
drop, with and without gravitational effects. By considering the systems without gravity, we were able to
show that the fluid surface in its equilibrium state had constant curvature. As expected, the inclusion of
gravity resulted in a symmetric negative contribution to the surface curvature, but at certain scales this
contribution becomes negligible.
For the single hinge zero gravity model, we were able to derive equations which can be solved numerically
to find the complete geometry of the system using the length of the material wall and the volume constraint
constant.
To consider more general surfaces, we derived parametric equations for the energy density, and similar
analysis resulted in the same constant curvature conclusion.
Future work in this area would consider the effects of electromagnetic energy on the system.