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Kellerods and Kelasticas
Alain Goriely, Oxford
Part 1: Reading Joe Keller’s work on elastic rods
Part 2: Geometric stability methods for 1-D problems
Keller’s contributions to rods•About 18 publications from 1951 to 2010
•1951:Bowing of Violin Strings •2010:Ponytail motion
Keller’s contributions to rods
•Roughly divided into •Wave propagation •Contact problems •Optimal/Ideal shapes and Stability
•About 18 publications from 1951 to 2010
•1951:Bowing of Violin Strings •2010:Ponytail motion
Contact problems•Post buckling behavior of elastic tubes and rings with opposite sides in contact (1972, with Flaherty & Rubinow)
•Contact problems involving a buckled elastica (1973, with Flaherty)
•Some bubble and contact problems (1980)
Contact problems
Grotberg & Jensen 2004 (Heil)
x
y
blister
detachmentpoint
s = 0
s− = 0
s+ = 0 s+ = s̄+
s− = s̄−s = s̄
With Gaetano Napoli Tom Mullin
Stability of two rings
•Ropes in equilibrium (1987, with Maddocks)
Contact problems
Optimal/Ideal shapes•Strongest column (1960) •Tallest column (1966, with Niordson)
“To find the curve which by its revolution around an axis determines the column of greatest efficiency”
•Rope of minimum elongation (1984, with Verma)
Cox 1992
Clamped Hinged
Optimal/Ideal shapes•Strongest column (1960) •Tallest column (1966, with Niordson)•Rope of minimum elongation (1984, with Verma)
Optimal/Ideal shapes•Strongest column (1960) •Tallest column (1966, with Niordson)
McCarthy, 1999
•Rope of minimum elongation (1984, with Verma)
Optimal/Ideal shapes•Tendril shape (1984)
McMillen AG 2002
Optimal/Ideal shapes•Mobius band (1993, with Mahadevan) •Coiling of ropes (1995, with Mahadevan)
Starostin-van der Heijden, 2007
Stability methods for one-dimensional problems (work with Th. Lessinnes)
Dynamics in phase space
Dynamics in phase space
Dynamics in phase space! More generally
Dynamics in phase space! More generally
Dynamics in phase space! More generally
For initial value problems, the geometry of curves in phase space provides information on the stability of solutions.
Dynamics in phase space! More generally
For initial value problems, the geometry of curves in phase space provides information on the stability of solutions.For boundary value problems, equilibrium solutions also lie in phase space.
Dynamics in phase space! More generally
For initial value problems, the geometry of curves in phase space provides information on the stability of solutions.For boundary value problems, equilibrium solutions also lie in phase space.
Dynamics in phase space! More generally
For initial value problems, the geometry of curves in phase space provides information on the stability of solutions.For boundary value problems, equilibrium solutions also lie in phase space.
Stability?
Example: hanging rod
Take for instance: v=1.5 and M=81
Problem statementDefine the functional
with
Find a function θ which locally minimises :
or
Problem statementDefine the functional
with
Find a function θ which locally minimises :
To first order in ε,
or
Problem statementDefine the functional
with
Find a function θ which locally minimises :
To second order:
To first order in ε,
or
Problem statementDefine the functional
with
Find a function θ which locally minimises :
To second order:
To first order in ε,
or
Theorem 2 [Lessinnes&AG, 2017]: Consider a stationary solution of
and its corresponding solution in the phase plane
Similarly, the Index of S on DX([a, c]), is given by
IndexDX([a,c])
[S] = #n
s 2 (a, c] : ✓(s) 2 mo
�#n
s 2 [a, c) : ✓(s) 2 Mo
. (71)
Proof. The proof is similar to the proof of Prop. 16. We compute the Index of S on DX([c, b]).The computation on DX([a, c]) follows accordingly (e.g. after the change of variable x = �s).
We first note note that the function h(s) = ✓
0(s)
✓
0(c)
solves the IVP (??) with initial condition at cinstead of a.
The set {�c
2 (c, b] : h0(�c
) = 0} in (48) is the set of values of extremisers of h in (c, b]. Buth0(�) = 1
✓
0(c)
✓00(�) = �1
✓
0(c)
V✓
(✓(�)) so that �c
is a stationary point of h whenever ✓(�c
) is a stationary
point of V . Once again, the eigenvalues of S change sign when the solution ✓(s) of (10) crossesstationary points of V . The result than comes by direct application of Prop. 14 as
• If ✓c
2 m, then Sign[f(c)] = �1,
• If ✓c
2 M , then Sign[f(c)] = +1.
• For each conjugate point, if ✓(�c
) 2 m, then Sign[f(�c
)] = �1,
• while if ✓(�c
) 2 M , then Sign[f(�c
)] = +1.
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (72)
Theorem 3. Let ✓(s) be a stationary point of the functional
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (73)
and � : s 2 [a, b] !�
✓(s), ✓0(s)�
its trajectory in the phase plane. Furthermore in the phase plane,
define the set of points �M
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 < 0o
the abscissa of which are maximum points
of V (✓) and the set of points �m
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 > 0o
the abscissa of which are minimum
points of V (✓). Finally, choose c 2 (a, b) such that �(c) 2 �M
[�m
. Assuming that �(a) /2 �M
[�m
and �(b) /2 �M
[ �m
, we have the two following results.If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
= 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
= 0,(74)
then the solution ✓ is a local minimum of E .If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
> 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
> 0,(75)
then the solution ✓ is not a local minimum of E .
19
Similarly, the Index of S on DX([a, c]), is given by
IndexDX([a,c])
[S] = #n
s 2 (a, c] : ✓(s) 2 mo
�#n
s 2 [a, c) : ✓(s) 2 Mo
. (71)
Proof. The proof is similar to the proof of Prop. 16. We compute the Index of S on DX([c, b]).The computation on DX([a, c]) follows accordingly (e.g. after the change of variable x = �s).
We first note note that the function h(s) = ✓
0(s)
✓
0(c)
solves the IVP (??) with initial condition at cinstead of a.
The set {�c
2 (c, b] : h0(�c
) = 0} in (48) is the set of values of extremisers of h in (c, b]. Buth0(�) = 1
✓
0(c)
✓00(�) = �1
✓
0(c)
V✓
(✓(�)) so that �c
is a stationary point of h whenever ✓(�c
) is a stationary
point of V . Once again, the eigenvalues of S change sign when the solution ✓(s) of (10) crossesstationary points of V . The result than comes by direct application of Prop. 14 as
• If ✓c
2 m, then Sign[f(c)] = �1,
• If ✓c
2 M , then Sign[f(c)] = +1.
• For each conjugate point, if ✓(�c
) 2 m, then Sign[f(�c
)] = �1,
• while if ✓(�c
) 2 M , then Sign[f(�c
)] = +1.
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (72)
Theorem 3. Let ✓(s) be a stationary point of the functional
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (73)
and � : s 2 [a, b] !�
✓(s), ✓0(s)�
its trajectory in the phase plane. Furthermore in the phase plane,
define the set of points �M
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 < 0o
the abscissa of which are maximum points
of V (✓) and the set of points �m
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 > 0o
the abscissa of which are minimum
points of V (✓). Finally, choose c 2 (a, b) such that �(c) 2 �M
[�m
. Assuming that �(a) /2 �M
[�m
and �(b) /2 �M
[ �m
, we have the two following results.If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
= 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
= 0,(74)
then the solution ✓ is a local minimum of E .If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
> 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
> 0,(75)
then the solution ✓ is not a local minimum of E .
19
a b
Theorem 2 [Lessinnes&AG, 2017]: Consider a stationary solution of
and its corresponding solution in the phase plane
Similarly, the Index of S on DX([a, c]), is given by
IndexDX([a,c])
[S] = #n
s 2 (a, c] : ✓(s) 2 mo
�#n
s 2 [a, c) : ✓(s) 2 Mo
. (71)
Proof. The proof is similar to the proof of Prop. 16. We compute the Index of S on DX([c, b]).The computation on DX([a, c]) follows accordingly (e.g. after the change of variable x = �s).
We first note note that the function h(s) = ✓
0(s)
✓
0(c)
solves the IVP (??) with initial condition at cinstead of a.
The set {�c
2 (c, b] : h0(�c
) = 0} in (48) is the set of values of extremisers of h in (c, b]. Buth0(�) = 1
✓
0(c)
✓00(�) = �1
✓
0(c)
V✓
(✓(�)) so that �c
is a stationary point of h whenever ✓(�c
) is a stationary
point of V . Once again, the eigenvalues of S change sign when the solution ✓(s) of (10) crossesstationary points of V . The result than comes by direct application of Prop. 14 as
• If ✓c
2 m, then Sign[f(c)] = �1,
• If ✓c
2 M , then Sign[f(c)] = +1.
• For each conjugate point, if ✓(�c
) 2 m, then Sign[f(�c
)] = �1,
• while if ✓(�c
) 2 M , then Sign[f(�c
)] = +1.
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (72)
Theorem 3. Let ✓(s) be a stationary point of the functional
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (73)
and � : s 2 [a, b] !�
✓(s), ✓0(s)�
its trajectory in the phase plane. Furthermore in the phase plane,
define the set of points �M
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 < 0o
the abscissa of which are maximum points
of V (✓) and the set of points �m
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 > 0o
the abscissa of which are minimum
points of V (✓). Finally, choose c 2 (a, b) such that �(c) 2 �M
[�m
. Assuming that �(a) /2 �M
[�m
and �(b) /2 �M
[ �m
, we have the two following results.If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
= 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
= 0,(74)
then the solution ✓ is a local minimum of E .If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
> 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
> 0,(75)
then the solution ✓ is not a local minimum of E .
19
Similarly, the Index of S on DX([a, c]), is given by
IndexDX([a,c])
[S] = #n
s 2 (a, c] : ✓(s) 2 mo
�#n
s 2 [a, c) : ✓(s) 2 Mo
. (71)
Proof. The proof is similar to the proof of Prop. 16. We compute the Index of S on DX([c, b]).The computation on DX([a, c]) follows accordingly (e.g. after the change of variable x = �s).
We first note note that the function h(s) = ✓
0(s)
✓
0(c)
solves the IVP (??) with initial condition at cinstead of a.
The set {�c
2 (c, b] : h0(�c
) = 0} in (48) is the set of values of extremisers of h in (c, b]. Buth0(�) = 1
✓
0(c)
✓00(�) = �1
✓
0(c)
V✓
(✓(�)) so that �c
is a stationary point of h whenever ✓(�c
) is a stationary
point of V . Once again, the eigenvalues of S change sign when the solution ✓(s) of (10) crossesstationary points of V . The result than comes by direct application of Prop. 14 as
• If ✓c
2 m, then Sign[f(c)] = �1,
• If ✓c
2 M , then Sign[f(c)] = +1.
• For each conjugate point, if ✓(�c
) 2 m, then Sign[f(�c
)] = �1,
• while if ✓(�c
) 2 M , then Sign[f(�c
)] = +1.
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (72)
Theorem 3. Let ✓(s) be a stationary point of the functional
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (73)
and � : s 2 [a, b] !�
✓(s), ✓0(s)�
its trajectory in the phase plane. Furthermore in the phase plane,
define the set of points �M
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 < 0o
the abscissa of which are maximum points
of V (✓) and the set of points �m
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 > 0o
the abscissa of which are minimum
points of V (✓). Finally, choose c 2 (a, b) such that �(c) 2 �M
[�m
. Assuming that �(a) /2 �M
[�m
and �(b) /2 �M
[ �m
, we have the two following results.If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
= 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
= 0,(74)
then the solution ✓ is a local minimum of E .If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
> 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
> 0,(75)
then the solution ✓ is not a local minimum of E .
19
Define I = the number of max crossing minus the number of min crossing
= number of summits - number of valleys
= number of blue lines crossed - number of red lines crossed
a b
Theorem 2 [Lessinnes&AG, 2017]: Consider a stationary solution of
and its corresponding solution in the phase plane
Similarly, the Index of S on DX([a, c]), is given by
IndexDX([a,c])
[S] = #n
s 2 (a, c] : ✓(s) 2 mo
�#n
s 2 [a, c) : ✓(s) 2 Mo
. (71)
Proof. The proof is similar to the proof of Prop. 16. We compute the Index of S on DX([c, b]).The computation on DX([a, c]) follows accordingly (e.g. after the change of variable x = �s).
We first note note that the function h(s) = ✓
0(s)
✓
0(c)
solves the IVP (??) with initial condition at cinstead of a.
The set {�c
2 (c, b] : h0(�c
) = 0} in (48) is the set of values of extremisers of h in (c, b]. Buth0(�) = 1
✓
0(c)
✓00(�) = �1
✓
0(c)
V✓
(✓(�)) so that �c
is a stationary point of h whenever ✓(�c
) is a stationary
point of V . Once again, the eigenvalues of S change sign when the solution ✓(s) of (10) crossesstationary points of V . The result than comes by direct application of Prop. 14 as
• If ✓c
2 m, then Sign[f(c)] = �1,
• If ✓c
2 M , then Sign[f(c)] = +1.
• For each conjugate point, if ✓(�c
) 2 m, then Sign[f(�c
)] = �1,
• while if ✓(�c
) 2 M , then Sign[f(�c
)] = +1.
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (72)
Theorem 3. Let ✓(s) be a stationary point of the functional
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (73)
and � : s 2 [a, b] !�
✓(s), ✓0(s)�
its trajectory in the phase plane. Furthermore in the phase plane,
define the set of points �M
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 < 0o
the abscissa of which are maximum points
of V (✓) and the set of points �m
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 > 0o
the abscissa of which are minimum
points of V (✓). Finally, choose c 2 (a, b) such that �(c) 2 �M
[�m
. Assuming that �(a) /2 �M
[�m
and �(b) /2 �M
[ �m
, we have the two following results.If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
= 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
= 0,(74)
then the solution ✓ is a local minimum of E .If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
> 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
> 0,(75)
then the solution ✓ is not a local minimum of E .
19
Similarly, the Index of S on DX([a, c]), is given by
IndexDX([a,c])
[S] = #n
s 2 (a, c] : ✓(s) 2 mo
�#n
s 2 [a, c) : ✓(s) 2 Mo
. (71)
Proof. The proof is similar to the proof of Prop. 16. We compute the Index of S on DX([c, b]).The computation on DX([a, c]) follows accordingly (e.g. after the change of variable x = �s).
We first note note that the function h(s) = ✓
0(s)
✓
0(c)
solves the IVP (??) with initial condition at cinstead of a.
The set {�c
2 (c, b] : h0(�c
) = 0} in (48) is the set of values of extremisers of h in (c, b]. Buth0(�) = 1
✓
0(c)
✓00(�) = �1
✓
0(c)
V✓
(✓(�)) so that �c
is a stationary point of h whenever ✓(�c
) is a stationary
point of V . Once again, the eigenvalues of S change sign when the solution ✓(s) of (10) crossesstationary points of V . The result than comes by direct application of Prop. 14 as
• If ✓c
2 m, then Sign[f(c)] = �1,
• If ✓c
2 M , then Sign[f(c)] = +1.
• For each conjugate point, if ✓(�c
) 2 m, then Sign[f(�c
)] = �1,
• while if ✓(�c
) 2 M , then Sign[f(�c
)] = +1.
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (72)
Theorem 3. Let ✓(s) be a stationary point of the functional
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (73)
and � : s 2 [a, b] !�
✓(s), ✓0(s)�
its trajectory in the phase plane. Furthermore in the phase plane,
define the set of points �M
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 < 0o
the abscissa of which are maximum points
of V (✓) and the set of points �m
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 > 0o
the abscissa of which are minimum
points of V (✓). Finally, choose c 2 (a, b) such that �(c) 2 �M
[�m
. Assuming that �(a) /2 �M
[�m
and �(b) /2 �M
[ �m
, we have the two following results.If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
= 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
= 0,(74)
then the solution ✓ is a local minimum of E .If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
> 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
> 0,(75)
then the solution ✓ is not a local minimum of E .
19
Define I = the number of max crossing minus the number of min crossing
= number of summits - number of valleys
= number of blue lines crossed - number of red lines crossed
a b
Then, if I>0, the second variation is strictly positive (local minimum=stable) if I<0, the second variation is not positive (not a minimum=unstable)
Theorem 2 [Lessinnes&AG, 2017]: Consider a stationary solution of
and its corresponding solution in the phase plane
Similarly, the Index of S on DX([a, c]), is given by
IndexDX([a,c])
[S] = #n
s 2 (a, c] : ✓(s) 2 mo
�#n
s 2 [a, c) : ✓(s) 2 Mo
. (71)
Proof. The proof is similar to the proof of Prop. 16. We compute the Index of S on DX([c, b]).The computation on DX([a, c]) follows accordingly (e.g. after the change of variable x = �s).
We first note note that the function h(s) = ✓
0(s)
✓
0(c)
solves the IVP (??) with initial condition at cinstead of a.
The set {�c
2 (c, b] : h0(�c
) = 0} in (48) is the set of values of extremisers of h in (c, b]. Buth0(�) = 1
✓
0(c)
✓00(�) = �1
✓
0(c)
V✓
(✓(�)) so that �c
is a stationary point of h whenever ✓(�c
) is a stationary
point of V . Once again, the eigenvalues of S change sign when the solution ✓(s) of (10) crossesstationary points of V . The result than comes by direct application of Prop. 14 as
• If ✓c
2 m, then Sign[f(c)] = �1,
• If ✓c
2 M , then Sign[f(c)] = +1.
• For each conjugate point, if ✓(�c
) 2 m, then Sign[f(�c
)] = �1,
• while if ✓(�c
) 2 M , then Sign[f(�c
)] = +1.
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (72)
Theorem 3. Let ✓(s) be a stationary point of the functional
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (73)
and � : s 2 [a, b] !�
✓(s), ✓0(s)�
its trajectory in the phase plane. Furthermore in the phase plane,
define the set of points �M
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 < 0o
the abscissa of which are maximum points
of V (✓) and the set of points �m
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 > 0o
the abscissa of which are minimum
points of V (✓). Finally, choose c 2 (a, b) such that �(c) 2 �M
[�m
. Assuming that �(a) /2 �M
[�m
and �(b) /2 �M
[ �m
, we have the two following results.If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
= 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
= 0,(74)
then the solution ✓ is a local minimum of E .If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
> 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
> 0,(75)
then the solution ✓ is not a local minimum of E .
19
Similarly, the Index of S on DX([a, c]), is given by
IndexDX([a,c])
[S] = #n
s 2 (a, c] : ✓(s) 2 mo
�#n
s 2 [a, c) : ✓(s) 2 Mo
. (71)
Proof. The proof is similar to the proof of Prop. 16. We compute the Index of S on DX([c, b]).The computation on DX([a, c]) follows accordingly (e.g. after the change of variable x = �s).
We first note note that the function h(s) = ✓
0(s)
✓
0(c)
solves the IVP (??) with initial condition at cinstead of a.
The set {�c
2 (c, b] : h0(�c
) = 0} in (48) is the set of values of extremisers of h in (c, b]. Buth0(�) = 1
✓
0(c)
✓00(�) = �1
✓
0(c)
V✓
(✓(�)) so that �c
is a stationary point of h whenever ✓(�c
) is a stationary
point of V . Once again, the eigenvalues of S change sign when the solution ✓(s) of (10) crossesstationary points of V . The result than comes by direct application of Prop. 14 as
• If ✓c
2 m, then Sign[f(c)] = �1,
• If ✓c
2 M , then Sign[f(c)] = +1.
• For each conjugate point, if ✓(�c
) 2 m, then Sign[f(�c
)] = �1,
• while if ✓(�c
) 2 M , then Sign[f(�c
)] = +1.
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (72)
Theorem 3. Let ✓(s) be a stationary point of the functional
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (73)
and � : s 2 [a, b] !�
✓(s), ✓0(s)�
its trajectory in the phase plane. Furthermore in the phase plane,
define the set of points �M
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 < 0o
the abscissa of which are maximum points
of V (✓) and the set of points �m
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 > 0o
the abscissa of which are minimum
points of V (✓). Finally, choose c 2 (a, b) such that �(c) 2 �M
[�m
. Assuming that �(a) /2 �M
[�m
and �(b) /2 �M
[ �m
, we have the two following results.If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
= 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
= 0,(74)
then the solution ✓ is a local minimum of E .If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
> 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
> 0,(75)
then the solution ✓ is not a local minimum of E .
19
Define I = the number of max crossing minus the number of min crossing
= number of summits - number of valleys
= number of blue lines crossed - number of red lines crossed
Then, if I>0, the second variation is strictly positive (local minimum=stable) if I<0, the second variation is not positive (not a minimum=unstable)
If I=0, Define J as the weighted difference in slopes at the end points
J= A(V’(q(b))-V’(q(a))
Theorem 2 [Lessinnes&AG, 2017]: Consider a stationary solution of
and its corresponding solution in the phase plane
Similarly, the Index of S on DX([a, c]), is given by
IndexDX([a,c])
[S] = #n
s 2 (a, c] : ✓(s) 2 mo
�#n
s 2 [a, c) : ✓(s) 2 Mo
. (71)
Proof. The proof is similar to the proof of Prop. 16. We compute the Index of S on DX([c, b]).The computation on DX([a, c]) follows accordingly (e.g. after the change of variable x = �s).
We first note note that the function h(s) = ✓
0(s)
✓
0(c)
solves the IVP (??) with initial condition at cinstead of a.
The set {�c
2 (c, b] : h0(�c
) = 0} in (48) is the set of values of extremisers of h in (c, b]. Buth0(�) = 1
✓
0(c)
✓00(�) = �1
✓
0(c)
V✓
(✓(�)) so that �c
is a stationary point of h whenever ✓(�c
) is a stationary
point of V . Once again, the eigenvalues of S change sign when the solution ✓(s) of (10) crossesstationary points of V . The result than comes by direct application of Prop. 14 as
• If ✓c
2 m, then Sign[f(c)] = �1,
• If ✓c
2 M , then Sign[f(c)] = +1.
• For each conjugate point, if ✓(�c
) 2 m, then Sign[f(�c
)] = �1,
• while if ✓(�c
) 2 M , then Sign[f(�c
)] = +1.
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (72)
Theorem 3. Let ✓(s) be a stationary point of the functional
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (73)
and � : s 2 [a, b] !�
✓(s), ✓0(s)�
its trajectory in the phase plane. Furthermore in the phase plane,
define the set of points �M
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 < 0o
the abscissa of which are maximum points
of V (✓) and the set of points �m
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 > 0o
the abscissa of which are minimum
points of V (✓). Finally, choose c 2 (a, b) such that �(c) 2 �M
[�m
. Assuming that �(a) /2 �M
[�m
and �(b) /2 �M
[ �m
, we have the two following results.If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
= 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
= 0,(74)
then the solution ✓ is a local minimum of E .If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
> 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
> 0,(75)
then the solution ✓ is not a local minimum of E .
19
Similarly, the Index of S on DX([a, c]), is given by
IndexDX([a,c])
[S] = #n
s 2 (a, c] : ✓(s) 2 mo
�#n
s 2 [a, c) : ✓(s) 2 Mo
. (71)
Proof. The proof is similar to the proof of Prop. 16. We compute the Index of S on DX([c, b]).The computation on DX([a, c]) follows accordingly (e.g. after the change of variable x = �s).
We first note note that the function h(s) = ✓
0(s)
✓
0(c)
solves the IVP (??) with initial condition at cinstead of a.
The set {�c
2 (c, b] : h0(�c
) = 0} in (48) is the set of values of extremisers of h in (c, b]. Buth0(�) = 1
✓
0(c)
✓00(�) = �1
✓
0(c)
V✓
(✓(�)) so that �c
is a stationary point of h whenever ✓(�c
) is a stationary
point of V . Once again, the eigenvalues of S change sign when the solution ✓(s) of (10) crossesstationary points of V . The result than comes by direct application of Prop. 14 as
• If ✓c
2 m, then Sign[f(c)] = �1,
• If ✓c
2 M , then Sign[f(c)] = +1.
• For each conjugate point, if ✓(�c
) 2 m, then Sign[f(�c
)] = �1,
• while if ✓(�c
) 2 M , then Sign[f(�c
)] = +1.
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (72)
Theorem 3. Let ✓(s) be a stationary point of the functional
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (73)
and � : s 2 [a, b] !�
✓(s), ✓0(s)�
its trajectory in the phase plane. Furthermore in the phase plane,
define the set of points �M
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 < 0o
the abscissa of which are maximum points
of V (✓) and the set of points �m
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 > 0o
the abscissa of which are minimum
points of V (✓). Finally, choose c 2 (a, b) such that �(c) 2 �M
[�m
. Assuming that �(a) /2 �M
[�m
and �(b) /2 �M
[ �m
, we have the two following results.If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
= 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
= 0,(74)
then the solution ✓ is a local minimum of E .If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
> 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
> 0,(75)
then the solution ✓ is not a local minimum of E .
19
Define I = the number of max crossing minus the number of min crossing
= number of summits - number of valleys
= number of blue lines crossed - number of red lines crossed
Then, if I>0, the second variation is strictly positive (local minimum=stable) if I<0, the second variation is not positive (not a minimum=unstable)
If I=0, Define J as the weighted difference in slopes at the end points
J= A(V’(q(b))-V’(q(a))
Theorem 2 [Lessinnes&AG, 2017]: Consider a stationary solution of
and its corresponding solution in the phase plane
Similarly, the Index of S on DX([a, c]), is given by
IndexDX([a,c])
[S] = #n
s 2 (a, c] : ✓(s) 2 mo
�#n
s 2 [a, c) : ✓(s) 2 Mo
. (71)
Proof. The proof is similar to the proof of Prop. 16. We compute the Index of S on DX([c, b]).The computation on DX([a, c]) follows accordingly (e.g. after the change of variable x = �s).
We first note note that the function h(s) = ✓
0(s)
✓
0(c)
solves the IVP (??) with initial condition at cinstead of a.
The set {�c
2 (c, b] : h0(�c
) = 0} in (48) is the set of values of extremisers of h in (c, b]. Buth0(�) = 1
✓
0(c)
✓00(�) = �1
✓
0(c)
V✓
(✓(�)) so that �c
is a stationary point of h whenever ✓(�c
) is a stationary
point of V . Once again, the eigenvalues of S change sign when the solution ✓(s) of (10) crossesstationary points of V . The result than comes by direct application of Prop. 14 as
• If ✓c
2 m, then Sign[f(c)] = �1,
• If ✓c
2 M , then Sign[f(c)] = +1.
• For each conjugate point, if ✓(�c
) 2 m, then Sign[f(�c
)] = �1,
• while if ✓(�c
) 2 M , then Sign[f(�c
)] = +1.
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (72)
Theorem 3. Let ✓(s) be a stationary point of the functional
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (73)
and � : s 2 [a, b] !�
✓(s), ✓0(s)�
its trajectory in the phase plane. Furthermore in the phase plane,
define the set of points �M
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 < 0o
the abscissa of which are maximum points
of V (✓) and the set of points �m
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 > 0o
the abscissa of which are minimum
points of V (✓). Finally, choose c 2 (a, b) such that �(c) 2 �M
[�m
. Assuming that �(a) /2 �M
[�m
and �(b) /2 �M
[ �m
, we have the two following results.If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
= 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
= 0,(74)
then the solution ✓ is a local minimum of E .If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
> 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
> 0,(75)
then the solution ✓ is not a local minimum of E .
19
Similarly, the Index of S on DX([a, c]), is given by
IndexDX([a,c])
[S] = #n
s 2 (a, c] : ✓(s) 2 mo
�#n
s 2 [a, c) : ✓(s) 2 Mo
. (71)
Proof. The proof is similar to the proof of Prop. 16. We compute the Index of S on DX([c, b]).The computation on DX([a, c]) follows accordingly (e.g. after the change of variable x = �s).
We first note note that the function h(s) = ✓
0(s)
✓
0(c)
solves the IVP (??) with initial condition at cinstead of a.
The set {�c
2 (c, b] : h0(�c
) = 0} in (48) is the set of values of extremisers of h in (c, b]. Buth0(�) = 1
✓
0(c)
✓00(�) = �1
✓
0(c)
V✓
(✓(�)) so that �c
is a stationary point of h whenever ✓(�c
) is a stationary
point of V . Once again, the eigenvalues of S change sign when the solution ✓(s) of (10) crossesstationary points of V . The result than comes by direct application of Prop. 14 as
• If ✓c
2 m, then Sign[f(c)] = �1,
• If ✓c
2 M , then Sign[f(c)] = +1.
• For each conjugate point, if ✓(�c
) 2 m, then Sign[f(�c
)] = �1,
• while if ✓(�c
) 2 M , then Sign[f(�c
)] = +1.
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (72)
Theorem 3. Let ✓(s) be a stationary point of the functional
E [✓] =
Z
b
a
(✓0(s)�A)2
2� V
�
✓(s)�
ds (73)
and � : s 2 [a, b] !�
✓(s), ✓0(s)�
its trajectory in the phase plane. Furthermore in the phase plane,
define the set of points �M
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 < 0o
the abscissa of which are maximum points
of V (✓) and the set of points �m
=n
(✓, ✓0)�
�
dVd✓ = 0, d2V
d✓2 > 0o
the abscissa of which are minimum
points of V (✓). Finally, choose c 2 (a, b) such that �(c) 2 �M
[�m
. Assuming that �(a) /2 �M
[�m
and �(b) /2 �M
[ �m
, we have the two following results.If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
= 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
= 0,(74)
then the solution ✓ is a local minimum of E .If both
8
<
:
#n
s 2 [c, b] : �(s) 2 �m
o
�#n
s 2 (c, b] : �(s) 2 �M
o
> 0,
#n
s 2 [a, c] : �(s) 2 �m
o
�#n
s 2 [a, c) : �(s) 2 �M
o
> 0,(75)
then the solution ✓ is not a local minimum of E .
19
Define I = the number of max crossing minus the number of min crossing
= number of summits - number of valleys
= number of blue lines crossed - number of red lines crossed
Then, if I>0, the second variation is strictly positive (local minimum=stable) if I<0, the second variation is not positive (not a minimum=unstable)
If I=0, Define J as the weighted difference in slopes at the end points
J= A(V’(q(b))-V’(q(a))
Then, if J#0, unstable if J>0, ? numerical
One max, no min ⇒ I>0 ⇒ Stable
Stability: Count the number of max-number of min
Example: hanging rod
I=-1Unstable
K=2Dirichletunstable
I=1Stable
I=-1Unstable
K=2Dirichletunstable
K=0Dirichlet stable
I=1Stable
I=-1Unstable
K=0Dirichlet stable
K=2Dirichletunstable
K=0Dirichlet stable
I=0but J=0Unstable
I=1Stable
I=-1Unstable
K=0Dirichlet stable
2 more stable cases and 2 more unstable ones (not shown)
K=2Dirichletunstable
K=0Dirichlet stable
I=0but J=0Unstable
Conclusions
! I wish I were as creative as Keller
Thomas Lessinnes
Geometric Stability (Nonlinearity- 2017) TL, AGStability of birods (SIAM J. Appl Math- 2016), TL, AG
Bi-rods
Growing rods
Conclusions
! I wish I were as creative as Keller
! If you can draw a solution, you know its stability
Thomas Lessinnes
Geometric Stability (Nonlinearity- 2017) TL, AGStability of birods (SIAM J. Appl Math- 2016), TL, AG
Bi-rods
Growing rods
The second variation
The second variation
The second variation
The second variation
Sturm-Liouville problem.
Stability iff λ>0
The second variation
Sturm-Liouville problem.
Stability iff λ>0
First auxiliary problem
The second variation
Sturm-Liouville problem.
Stability iff λ>0
First auxiliary problem
The second variation
Sturm-Liouville problem.
Stability iff λ>0
First auxiliary problem
The second variation
Sturm-Liouville problem.
Stability iff λ>0
First auxiliary problem
The second variation
Conjugate point to a
Sturm-Liouville problem.
Stability iff λ>0
First auxiliary problem
Conjugate point to a
Sturm-Liouville problem.
Stability iff λ>0
First auxiliary problem
Conjugate point to a
Sturm-Liouville problem.
Stability iff λ>0
Second auxiliary problem
First auxiliary problem
Conjugate point to a
Sturm-Liouville problem.
Stability iff λ>0
Second auxiliary problem
⇒ The roots of h’ are the conjugate points
First auxiliary problem
Conjugate point to a
Sturm-Liouville problem.
Stability iff λ>0
Second auxiliary problem
⇒ The roots of h’ are the conjugate points
⟺ The roots of q’’ are the conjugate points
First auxiliary problem
Conjugate point to a
Sturm-Liouville problem.
Stability iff λ>0
Second auxiliary problem
⇒ The roots of h’ are the conjugate points
⟺ The roots of q’’ are the conjugate points
⟺ The extrema of V give the conjugate points