One-dimensional wrinkling of thin membranes
Bertrand Desmons
Aspirant FNRS
Institute of Mathematics, University of Mons (Belgium)
We assume that there is no variation other than those in the comPhysical experiments [3]
( x(s), y(s))
u(s)
pression direction ⇒ one-dimensional parametrization.
A film, initially flat, is laid on
Minimization of the energy due to :
a substrate such as water and
folds (which is measured by its curvature);
L−δ
compressed horizontally at both 0
potential energy due to the displacement of the substrate underedges.
It first starts to wrinneath.
kle, taking a sinusoidal form, but,
⇒ We are seeking for minima of the functional
for larger compressions, seems to
Z L
Z L
concentrate at the centre.
|u0 |2 ds + 21 K
E : X → R : u 7→ 12
yu (s))2 cos u(s) ds
(1)
0
Steps leading to the limit problem δ → 0:
1.
2.
3.
4.
0
Rs
where yu (s) = 0 sin u(t) dt, K is a constant relative to the substrate and X describes the space of admissible
Boundedness of {E (uδ )}δ for any (uδ )δ >0, minimizers. functions:
Inequality kuδ k2X ≤ 2E (uδ ) + KL3 max{1, kuδ k2∞ }, imZ L
Z L
n
o
RL
1
X = u ∈ H0 (]0, L[; R) cos u(s) ds = L − δ and
sin u(s) ds = 0 .
(2)
plies uδ * u0 but, as 0 1 − cos uδ = δ, u0 = 0.
0
0
√
The sequence (uδ / δ ) is also bounded in H01; so
√
√
uδ / δ * u∗ with ku∗ k L2 = 2.
Going to the limit δ → 0. . .
This gives then the eigenvalue problem
Let us call αδ and βδ the two Lagrange multipliers ap- Euler-Lagrange equation of the problem:
pearing in the Euler-Lagrange equation (see (3)). By a

iv
00
Z L
Z L

good choice of “test functions” v, we can deduce estiw
−
αw
+ Kw = 0

∂E (uδ ) · v + αδ
sin uδ · v + βδ
cos uδ · v = 0 (3)
mates on these constants, permitting us to divide this
w(0) = w( L) = 0
(5)
√
0
0

 w0 (0) = w0 ( L) = 0
equation by δ, giving the limit case (4).
As δ → 0, we obtain:
Z L
Z LZ ·
Z ·
Z L
(where w0 = u∗) with the additional
constraint
Z L
The equation (5) admits nontrivial solutions only if α <
√
u∗0 v0 + K
u∗
v +α
u∗ v + β
v = 0 (4)
−2 K; the characteristic polynomial admits then two pairs
0
0
0
0
0
0
√
Z L
of purely imaginary complex roots. For convenience rea- where α = lim αδ and β = lim βδ / δ (up to subsequences).
02
w
=
2.
(6)
π
π
δ
→
0
δ
→
0
sons, let us denote their moduli by L µ and L ν with, w.l.o.g.
0
µ > ν.
By successively imposing the
ν
Numerical experiments: continuation algorithm applied to the “limit solutions”
n 1
boundary conditions on a genn 2
π4
2
n 3
Plot of the energies for the first two solutions for K = k l 4 with, respectively k = 0.01,
eral
solution
w,
we
get
the
two
n 4
n 5
n 6
2, 3. Below are the curves u obtained by continuation, followed by the resulting curves
equations
in
µ
and
ν
for
which
n 7
n 8
( x, y). (Length L is fixed to 10 and δ varies from 0.05 to 3.5 by step 0.05.)
a part of the solutions is drawn.
◦
◦
◦
◦
◦
◦
◦
◦
1D or 2D space of solutions for
equation (5)?
µ
2D
space
⇔
(
µ,
ν
)
in
intersec2
3
4
5
6
7
8
9
10
π
π
π
π
tion ⇔ (µ, ν ) ∈ N × N with
µ sin 2 µ cos 2 ν − ν sin 2 ν cos 2 µ = 0
µ + ν even.
µ cos π2 µ sin π2 ν − ν cos π2 ν sin π2 µ = 0
1D space ⇔ (µ, ν ) in (only)
one curve.
70
160
60
140
300
250
120
50
200
100
40
150
80
30
60
20
100
40
10
50
20
0
0
0 0.5 1 1.5 2 2.5 3 3.5
0
0 0.5 1 1.5 2 2.5 3 3.5
0 0.5 1 1.5 2 2.5 3 3.5
π2 2
(µ
L2
+ ν 2 ). Thus, for fixed K, the
What about eigenvalues? We get −α =
lowest eigenvalue −α is found for (µ, ν ) lying on the curve closest to the
diagonal (depending on K, it is curve n◦1 or n◦2).
2
2
2
1
1
1
0
0
0
−1
−1
−1
−2
−2
−2
Possible degeneracy of eigenvalues: e.g. for the first one, for all k ∈ N,
K=
π4
L4
k 2 (k + 2)2
0
1
2
will give a 2D first eigenspace.
Some properties on the “limit solutions”
s
1
Form of the solutions: We have w(s) = ŵ L − 2 with, in case 1D space,
3
40
250
200
150
100
50
2
3
40
0
4
0
3
3
−1
2
2
−2
1
1
−3
0
0
−4
ŵ(t) = A cos( π2 µ ) cos(πνt) − cos( π2 ν ) cos(πµt)
ŵ(t) = A sin π2 µ sin(πνt) − sin π2 ν sin(πµt)
0
1
2
3
40
2
4
6
8
10
1
0
1
2
3
40
2
4
6
8
10
0
2
2
1
1
0
0
−1
−1
−2
−2
1
1
2
2
3
3
40
40
250
200
150
100
50
2
4
6
8
10
0
−1
−2
0
1
2
3
40
250
200
150
100
50
2
nth
1
250
200
150
100
50
4
where A is determined (up to sign) by (6).
In case of 2D space, ŵ can be any linear combination of two functions like
above, with adjusted coefficients; fixing constraint (6) selects an ellipse in
the space of solutions.
Symmetry: if w(s) is a solution of (5) then w( L − s) is also solution; it is
±w(s) in case 1D-space and can be described from w(s) in case 2D-space.
Parity of all solutions in case 1D-space: oddness on red curves, evenness
on blue ones.
Number of roots depending on µ and ν: in case 1D-space,
0
– At least n − 1 inner roots when being on the
curve.
– Increases by 2 after a crossing for the curve going below. It seems that
0
each newly created root comes from an edge.
1
– All roots are simple.
This behaviour is found (at least numerically) also for solutions living in
the 2D spaces.
References
0
1
2
3
40
250
200
150
100
50
0
4
4
3
3
2
2
1
1
1
2
3
40
250
200
150
100
50
1
0
−1
−2
2
3
40
2
4
6
8
10
0
0
−1
−1
0
1
2
3
40
2
4
6
8
10
0
1
2
3
40
2
4
6
8
10
[1] H.-G. Grunau. Positivity, change of sign and buckling eigenvalues in a one-dimensional fourth
order model problem. Advances in differential equations, 7(2):177–196, 2002.
Future work
[2] M. A. Peletier. Sequential buckling: a variational analysis. SIAM J. Math. Anal., 32(5):1142–
Evolution of the “degeneracy points” (2D space) when δ grows.
1168, September 2000.
Study of the equation as K → 0 (related to the Euler-Bernoulli elastica prob- [3] L. Pocivavsek, R. Dellsy, A. Kern, S. Johnson, B. Lin, K. Y. C. Lee, and E. Cerda. Stress and fold
localization in thin elastic membranes. Science, 320:912–916, May 2008.
lem). Continuation algorithm applied to elastica curves.
[4] J. A. Zasadzinski, J. Ding, H. E. Warriner, A. J. Bringezu, and A. J. Waring. The physics and
Study of the equation for large values of K.
physiology of lung surfactants. Current Opinion in Colloid and Interface Science, 6(5):506–513,
Variable coefficients in equation (5).
November 2001.