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