Weighted Sobolev spaces for the advection operator: A variational - - PowerPoint PPT Presentation
Weighted Sobolev spaces for the advection operator: A variational method for computing shape derivatives of geometric constraints along rays. Florian Feppon Gr egoire Allaire, Charles Dapogny Julien Cortial, Felipe Bordeu S eminaire des
A variational method for computing shape derivatives of geometric constraints along rays.
Florian Feppon Gr´ egoire Allaire, Charles Dapogny Julien Cortial, Felipe Bordeu S´ eminaire des doctorants – October 17th, 2018
shape differentiation
graph spaces of the advection operator β · ∇.
shape differentiation
graph spaces of the advection operator β · ∇.
shape differentiation
graph spaces of the advection operator β · ∇.
shape differentiation
graph spaces of the advection operator β · ∇.
We rely on the method of Hadamard (figure from[1]): min
Ω J(Ω)
Ωθ = (I + θ)Ωθ with θ ∈ W 1,∞(Ω, Rd), ||θ||W 1,∞(Rd,Rd)< 1. ♥
[1]
Charles Dapogny et al. “Geometrical shape optimization in fluid mechanics using FreeFem++”. In: Structural and Multidisciplinary Optimization (2017).
We rely on the method of Hadamard (figure from[1]): min
Ω J(Ω)
Ωθ = (I + θ)Ωθ with θ ∈ W 1,∞(Ω, Rd), ||θ||W 1,∞(Rd,Rd)< 1. J(Ωθ) = J(Ω) + dJ dθ (Ω)(θ) + o(θ), where |o(θ)| ||θ||W 1,∞(Ω,Rd)
θ→0
− − − → 0, ♥
[1]
Charles Dapogny et al. “Geometrical shape optimization in fluid mechanics using FreeFem++”. In: Structural and Multidisciplinary Optimization (2017).
We rely on the method of Hadamard (figure from[1]): min
Ω J(Ω)
Ωθ = (I + θ)Ωθ with θ ∈ W 1,∞(Ω, Rd), ||θ||W 1,∞(Rd,Rd)< 1. J(Ωθ) = J(Ω) + dJ dθ (Ω)(θ) + o(θ), where |o(θ)| ||θ||W 1,∞(Ω,Rd)
θ→0
− − − → 0, In practice dJ
dθ(Ω)(θ) =
[1]
Charles Dapogny et al. “Geometrical shape optimization in fluid mechanics using FreeFem++”. In: Structural and Multidisciplinary Optimization (2017).
The signed distance function dΩ to the domain Ω ⊂ D is defined by: ∀x ∈ D, dΩ(x) = − min
y∈∂Ω ||y − x||
if x ∈ Ω, min
y∈∂Ω ||y − x||
if x ∈ D\Ω.
◮ The signed distance function dΩ(x) is differentiable at every x ∈ D such that Π∂Ω(x) = {y ∈ ∂Ω | ||x − y|| = d(x, ∂Ω)} is a singleton.
◮ The signed distance function dΩ(x) is differentiable at every x ∈ D such that Π∂Ω(x) = {y ∈ ∂Ω | ||x − y|| = d(x, ∂Ω)} is a singleton. ◮ The set Σ on which dΩ is not differentiable is called the skeleton.
∇dΩ is a unit extension of the normal constant along rays: ♥ ♥ ♥ ♥
∇dΩ is a unit extension of the normal constant along rays: ◮ ∀y ∈ ∂Ω, ∇dΩ(y) = ♥(y) is the unit outward normal to ∂Ω. ♥ ♥ ♥
∇dΩ is a unit extension of the normal constant along rays: ◮ ∀y ∈ ∂Ω, ∇dΩ(y) = ♥(y) is the unit outward normal to ∂Ω. ◮ ∀s ∈ (ζ−(y), ζ+(y)), dΩ(y + s♥(y)) = s. ♥ ♥
∇dΩ is a unit extension of the normal constant along rays: ◮ ∀y ∈ ∂Ω, ∇dΩ(y) = ♥(y) is the unit outward normal to ∂Ω. ◮ ∀s ∈ (ζ−(y), ζ+(y)), dΩ(y + s♥(y)) = s. The set ray(y) = {y + s♥(y) | s ∈ (ζ−(y), ζ+(y))} is called the ray emerging from y. ♥
∇dΩ is a unit extension of the normal constant along rays: ◮ ∀y ∈ ∂Ω, ∇dΩ(y) = ♥(y) is the unit outward normal to ∂Ω. ◮ ∀s ∈ (ζ−(y), ζ+(y)), dΩ(y + s♥(y)) = s. The set ray(y) = {y + s♥(y) | s ∈ (ζ−(y), ζ+(y))} is called the ray emerging from y. ◮ For any x ∈ D\Σ, ||∇dΩ(x)|| = 1 and ∀z ∈ ray(y), ∇dΩ(z) = ♥(y).
An example: a meshed subdomain Ω ⊂ D
An example: the signed distance function dΩ:
An example: the gradient of the signed distance function ∇dΩ:
The signed distance function allows to formulate geometric constraints. ◮ Maximum thickness constraint : ∀x ∈ Ω, |dΩ(x)| ≤ dmax/2 ◮ Minimum thickness constraint: ∀y ∈ ∂Ω, |ζ−(y)| ≥ dmin/2.
In practice, one formulates geometric constraints using penalty functionals P(Ω) as follows: min
Ω J(Ω), s.t. P(Ω) ≤ 0, where P(Ω) :=
j(dΩ(x))dx. ♥ ♥
In practice, one formulates geometric constraints using penalty functionals P(Ω) as follows: min
Ω J(Ω), s.t. P(Ω) ≤ 0, where P(Ω) :=
j(dΩ(x))dx. The shape derivative of P(Ω) reads P′(Ω)(θ) =
j′(dΩ(x))d′
Ω(θ)(x)dx
with d′
Ω(θ)(y + s∇(y)) = −θ(y) · ♥(y).
|θ · ♥| θ Ω
Change of variable x = y + s♥(y) with y ∈ ∂Ω, s ∈ (ζ−(y), ζ+(y)):
P′(Ω)(θ) =
j′(dΩ(x))d′
Ω(θ)(x)dx = −
ζ+(y)
ζ−(y)
j′(s)|Dη|(y, s)ds θ(y) · ♥(y)dy
♥ ♥
Change of variable x = y + s♥(y) with y ∈ ∂Ω, s ∈ (ζ−(y), ζ+(y)):
P′(Ω)(θ) =
j′(dΩ(x))d′
Ω(θ)(x)dx = −
ζ+(y)
ζ−(y)
j′(s)|Dη|(y, s)ds θ(y) · ♥(y)dy
|Dη|(y, s) =
(1 + κi(y)s) is the Jacobian of η : (y, s) → y + s♥(y) The κi(y) are the eigenvalues of the tangential gradient ∇♥(y).
Change of variable x = y + s♥(y) with y ∈ ∂Ω, s ∈ (ζ−(y), ζ+(y)):
P′(Ω)(θ) =
j′(dΩ(x))d′
Ω(θ)(x)dx = −
ζ+(y)
ζ−(y)
j′(s)|Dη|(y, s)ds θ(y) · ♥(y)dy
|Dη|(y, s) =
(1 + κi(y)s) is the Jacobian of η : (y, s) → y + s♥(y) The κi(y) are the eigenvalues of the tangential gradient ∇♥(y). The function ∀y ∈ ∂Ω, u(y) = − ζ+(y)
ζ−(y)
j′(s)
(1 + κi(y)s)ds is “explicit” and does not involve θ.
∀y ∈ ∂Ω, u(y) = − ζ+(y)
ζ−(y)
j′(s)
(1 + κi(y)s)ds. Computing u requires:
∀y ∈ ∂Ω, u(y) = − ζ+(y)
ζ−(y)
j′(s)
(1 + κi(y)s)ds. Computing u requires:
∀y ∈ ∂Ω, u(y) = − ζ+(y)
ζ−(y)
j′(s)
(1 + κi(y)s)ds. Computing u requires:
shape differentiation
graph spaces of the advection operator β · ∇.
More precisely, the shape derivative of P(Ω) reads P′(Ω)(θ) =
j′(dΩ(x))d′
Ω(θ)(x)dx =
u θ · ♥dy with d′
Ω(θ) satisfying
Ω(θ) · ∇dΩ= 0 in D\Σ
d′
Ω(θ)= −θ · ♥ on ∂Ω.
|θ · ♥| θ Ω
♥
More precisely, the shape derivative of P(Ω) reads P′(Ω)(θ) =
j′(dΩ(x))d′
Ω(θ)(x)dx =
u θ · ♥dy with d′
Ω(θ) satisfying
Ω(θ) · ∇dΩ= 0 in D\Σ
d′
Ω(θ)= −θ · ♥ on ∂Ω.
Our method: u solves the following variational problem (with ω > 0 rather arbitrary): Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(∇dΩ · ∇u)(∇dΩ · ∇v)dx = −
j′(dΩ(x))v(x)dx, ♥
More precisely, the shape derivative of P(Ω) reads P′(Ω)(θ) =
j′(dΩ(x))d′
Ω(θ)(x)dx =
u θ · ♥dy with d′
Ω(θ) satisfying
Ω(θ) · ∇dΩ= 0 in D\Σ
d′
Ω(θ)= −θ · ♥ on ∂Ω.
Our method: Take v = d′
Ω(θ):
Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(∇dΩ · ∇u)(∇dΩ · ∇v)dx = −
j′(dΩ(x))v(x)dx, ♥
More precisely, the shape derivative of P(Ω) reads P′(Ω)(θ) =
j′(dΩ(x))d′
Ω(θ)(x)dx =
u θ · ♥dy with d′
Ω(θ) satisfying
Ω(θ) · ∇dΩ= 0 in D\Σ
d′
Ω(θ)= −θ · ♥ on ∂Ω.
Our method: Take v = d′
Ω(θ):
Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(∇dΩ · ∇u)(∇dΩ · ∇v)dx = −
j′(dΩ(x))v(x)dx,
ud′
Ω(θ)ds+
ω(∇dΩ·∇u)(∇dΩ · ∇d′
Ω(θ))dx = −
j′(dΩ(x))d′
Ω(θ)(x)dx,
♥
More precisely, the shape derivative of P(Ω) reads P′(Ω)(θ) =
j′(dΩ(x))d′
Ω(θ)(x)dx =
u θ · ♥dy with d′
Ω(θ) satisfying
Ω(θ) · ∇dΩ= 0 in D\Σ
d′
Ω(θ)= −θ · ♥ on ∂Ω.
Our method: Take v = d′
Ω(θ):
Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(∇dΩ · ∇u)(∇dΩ · ∇v)dx = −
j′(dΩ(x))v(x)dx,
ud′
Ω(θ)ds+
ω(∇dΩ·∇u)(∇dΩ · ∇d′
Ω(θ))dx = −
j′(dΩ(x))d′
Ω(θ)(x)dx,
♥
More precisely, the shape derivative of P(Ω) reads P′(Ω)(θ) =
j′(dΩ(x))d′
Ω(θ)(x)dx =
u θ · ♥dy with d′
Ω(θ) satisfying
Ω(θ) · ∇dΩ= 0 in D\Σ
d′
Ω(θ)= −θ · ♥ on ∂Ω.
Our method: Take v = d′
Ω(θ):
Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(∇dΩ · ∇u)(∇dΩ · ∇v)dx = −
j′(dΩ(x))v(x)dx,
ud′
Ω(θ)ds+
ω(∇dΩ·∇u)(∇dΩ · ∇d′
Ω(θ))dx = −
j′(dΩ(x))d′
Ω(θ)(x)dx,
u (−θ · ♥)ds + 0 = −
j′(dΩ(x))d′
Ω(θ)(x)dx.
Our method: solve the variational problem
Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(∇dΩ · ∇u)(∇dΩ · ∇v)dx = −
j′(dΩ(x))v(x)dx (1)
Our method: solve the variational problem
Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(∇dΩ · ∇u)(∇dΩ · ∇v)dx = −
j′(dΩ(x))v(x)dx (1)
Under rather unrestrictive assumptions, the trace of the solution u is independent on the weight ω and is given by
∀y ∈ ∂Ω, u(y) = − ζ+(y)
ζ−(y)
j′(s)
(1 + κi(y)s)ds. (2)
Our method: solve the variational problem
Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(∇dΩ · ∇u)(∇dΩ · ∇v)dx = −
j′(dΩ(x))v(x)dx (1)
Under rather unrestrictive assumptions, the trace of the solution u is independent on the weight ω and is given by
∀y ∈ ∂Ω, u(y) = − ζ+(y)
ζ−(y)
j′(s)
(1 + κi(y)s)ds. (2)
(1) can be solved with FEM while (2) requires computing rays and curvatures!
Does it really work? Find u ∈ Vω such that ∀v ∈ Vω,
uvds+
ω(∇dΩ·∇u)(∇dΩ·∇v)dx = −
j′(dΩ(x))v(x)dx
Does it really work? Find u ∈ Vω such that ∀v ∈ Vω,
uvds+
ω(∇dΩ·∇u)(∇dΩ·∇v)dx = −
j′(dΩ(x))v(x)dx The main issue: functions of Vω are discontinuous accross the skeleton:
Does it really work? Find u ∈ Vω such that ∀v ∈ Vω,
uvds+
ω(∇dΩ·∇u)(∇dΩ·∇v)dx = −
j′(dΩ(x))v(x)dx The main issue: functions of Vω are discontinuous accross the skeleton:
Does it really work? Find u ∈ Vω such that ∀v ∈ Vω,
uvds+
ω(∇dΩ·∇u)(∇dΩ·∇v)dx = −
j′(dΩ(x))v(x)dx The main issue: functions of Vω are discontinuous accross the skeleton: We select a weight ω vanishing near the skeleton.
An analytic example...
Figure: A prescribed −j′(dΩ(x))
It works with weights vanishing near the skeleton.
(a) Mesh T ′ (skeleton manually truncated), ω = 1 (b) Mesh T , ω = 1. (c) Mesh T , ω = 2/(1 + |∆dΩ|3.5) Figure: P1 elements with ω = 1 do not allow discontinuities of test functions near the skeleton...
2 4 6 8 0.4 0.6 0.8 1.0 1.2 uAnalytic uRays uVariational
(a) Mesh T ′, ω = 1
2 4 6 8 0.4 0.6 0.8 1.0 1.2 uAnalytic uRays uVariational
(b) Mesh T , ω = 1
2 4 6 8 0.4 0.6 0.8 1.0 1.2 uAnalytic uRays uVariational
(c) Mesh T , ω = 2/(1 + |∆dΩ|3.5)
2 4 6 8 0.4 0.6 0.8 1.0 1.2 uAnalytic uRays uVariational
(d) Finer mesh T , ω = 2/(1 + |∆dΩ|3.5)
We were able to implement conveniently geometric constraints in level set based shape optimization.
We were able to implement conveniently geometric constraints in level set based shape optimization.
(a) No maximum thickness constraint (b) dmax = 0.07. Figure: Maximum thickness constraint for 2D arch.
We were able to implement conveniently geometric constraints in level set based shape optimization.
(a) No minimum thickness constraint. (b) dmin = 0.1. (c) dmin = 0.2
.
Figure: Minimum thickness constraint for 2D cantilever.
shape differentiation
graph spaces of the advection operator β · ∇.
Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(∇dΩ · ∇u)(∇dΩ · ∇v)dx =
f (x)v(x)dx ∀y ∈ ∂Ω, u(y) = ζ+(y)
ζ−(y)
f (y + s♥(y))
(1 + κi(y)s)ds ♥
Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(∇dΩ · ∇u)(∇dΩ · ∇v)dx =
f (x)v(x)dx ∀y ∈ ∂Ω, u(y) = ζ+(y)
ζ−(y)
f (y + s♥(y))
(1 + κi(y)s)ds We shall treat a more general setting: U = D\Σ, Γ = ∂Ω, β = ∇dΩ, η(y, s) = y + s♥(y). |Dη|(y, s) =
(1 + κi(y)s).
Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(β · ∇u)(β · ∇v)dx =
fvdx, ∀y ∈ Γ, u(y) = ζ+(y)
ζ−(y)
(f ◦ η |Dη|)(y, s)ds. We shall treat a more general setting: U = D\Σ, Γ = ∂Ω, β = ∇dΩ, η(y, s) = y + s♥(y). |Dη|(y, s) =
(1 + κi(y)s).
Examples of such more general settings:
Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(β · ∇u)(β · ∇v)dx =
fvdx, (3) ◮ The space Vω considered is the so-called graph space Vω =
ωv 2dx < +∞ and
ω|β · ∇v|2dx < +∞
Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(β · ∇u)(β · ∇v)dx =
fvdx, (3) ◮ The space Vω considered is the so-called graph space Vω =
ωv 2dx < +∞ and
ω|β · ∇v|2dx < +∞
(u, v) →
uvds +
ω(β · ∇u)(β · ∇v)dx is a coercive bilinear form on Vω.
Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(β · ∇u)(β · ∇v)dx =
fvdx, (3) ◮ The space Vω considered is the so-called graph space Vω =
ωv 2dx < +∞ and
ω|β · ∇v|2dx < +∞
(u, v) →
uvds +
ω(β · ∇u)(β · ∇v)dx is a coercive bilinear form on Vω. ◮ Ingredients required:
Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(β · ∇u)(β · ∇v)dx =
fvdx, (3) ◮ The space Vω considered is the so-called graph space Vω =
ωv 2dx < +∞ and
ω|β · ∇v|2dx < +∞
(u, v) →
uvds +
ω(β · ∇u)(β · ∇v)dx is a coercive bilinear form on Vω. ◮ Ingredients required:
◮ Existence of traces on Γ for functions v ∈ Vω
Find u ∈ Vω such that ∀v ∈ Vω,
uvds +
ω(β · ∇u)(β · ∇v)dx =
fvdx, (3) ◮ The space Vω considered is the so-called graph space Vω =
ωv 2dx < +∞ and
ω|β · ∇v|2dx < +∞
(u, v) →
uvds +
ω(β · ∇u)(β · ∇v)dx is a coercive bilinear form on Vω. ◮ Ingredients required:
◮ Existence of traces on Γ for functions v ∈ Vω ◮ Poincar´ e inequality ∀v ∈ Vω,
ωv 2dx ≤ C
v 2dy +
ω|β · ∇v|2dx
Vω =
ωv 2dx < +∞ and
ω|β · ∇v|2dx < +∞
◮ Existence of traces on Γ for functions v ∈ Vω ◮ Poincar´ e inequality ∀v ∈ Vω,
ωv 2dx ≤ C
v 2dy +
ω|β · ∇v|2dx
Vω =
ωv 2dx < +∞ and
ω|β · ∇v|2dx < +∞
◮ Existence of traces on Γ for functions v ∈ Vω ◮ Poincar´ e inequality ∀v ∈ Vω,
ωv 2dx ≤ C
v 2dy +
ω|β · ∇v|2dx
Vω =
ωv 2dx < +∞ and
ω|β · ∇v|2dx < +∞
◮ Existence of traces on Γ for functions v ∈ Vω ◮ Poincar´ e inequality ∀v ∈ Vω,
ωv 2dx ≤ C
v 2dy +
ω|β · ∇v|2dx
Here β = ∇dΩ and div(∇dΩ) / ∈ L∞(U). We need a different analysis.
The key idea: use the flow η(y, s) of β ∈ C1(U) to do a change of variables x = η(y, s). d ds η(y(s), s) = β(η(y(s), s)), η(0, y) = y, ∀y ∈ Γ
Examples of such more general settings:
The key idea: use the flow η(y, s) of β ∈ C1(U) to do a change of variables x = η(y, s). d ds η(y(s), s) = β(η(y(s), s)), η(0, y) = y, ∀y ∈ Γ
The key idea: use the flow η(y, s) of β ∈ C1(U) to do a change of variables x = η(y, s). d ds η(y(s), s) = β(η(y(s), s)), η(0, y) = y, ∀y ∈ Γ Replace the space
Vω =
ωv 2dx < +∞ and
ω|β · ∇v|2dx < +∞
The key idea: use the flow η(y, s) of β ∈ C1(U) to do a change of variables x = η(y, s). d ds η(y(s), s) = β(η(y(s), s)), η(0, y) = y, ∀y ∈ Γ Replace the space
Vω =
ωv 2dx < +∞ and
ω|β · ∇v|2dx < +∞
v |
ζ+(y)
ζ−(y)
α˜ v 2dsdy < +∞ and
ζ+(y)
ζ−(y)
α|∂s ˜ v|2dsdy < +∞
with α = ω ◦ η|Dη|.
v |
ζ+(y)
ζ−(y)
α˜ v 2dsdy < +∞ and
ζ+(y)
ζ−(y)
α|∂s ˜ v|2dsdy < +∞
v |
ζ+(y)
ζ−(y)
α˜ v 2dsdy < +∞ and
ζ+(y)
ζ−(y)
α|∂s ˜ v|2dsdy < +∞
Existence of traces and Poincar´ e inequality in Vα are obtained for quite general weights α satisfying the following assumptions:
v |
ζ+(y)
ζ−(y)
α˜ v 2dsdy < +∞ and
ζ+(y)
ζ−(y)
α|∂s ˜ v|2dsdy < +∞
Existence of traces and Poincar´ e inequality in Vα are obtained for quite general weights α satisfying the following assumptions:
v |
ζ+(y)
ζ−(y)
α˜ v 2dsdy < +∞ and
ζ+(y)
ζ−(y)
α|∂s ˜ v|2dsdy < +∞
Existence of traces and Poincar´ e inequality in Vα are obtained for quite general weights α satisfying the following assumptions:
gα : y − → ζ+(y)
ζ−(y)
α(y, s)ds is uniformly bounded, i.e. gα ∈ L∞(Γ).
v |
ζ+(y)
ζ−(y)
α˜ v 2dsdy < +∞ and
ζ+(y)
ζ−(y)
α|∂s ˜ v|2dsdy < +∞
Existence of traces and Poincar´ e inequality in Vα are obtained for quite general weights α satisfying the following assumptions:
gα : y − → ζ+(y)
ζ−(y)
α(y, s)ds is uniformly bounded, i.e. gα ∈ L∞(Γ).
hα : y − → ζ+(y)
ζ−(y)
s α−1(y, t)dt
is uniformly bounded: hα ∈ L∞(Γ).
An example: the circle Ω = {x ∈ R2 | ||x||2 < 1}. ◮ U = Ω\{0} ❡
An example: the circle Ω = {x ∈ R2 | ||x||2 < 1}. ◮ U = Ω\{0} ◮ dΩ(x) = r − 1, ∇dΩ = ❡r, div(∇dΩ) = 1
r , α = r.
An example: the circle Ω = {x ∈ R2 | ||x||2 < 1}. ◮ U = Ω\{0} ◮ dΩ(x) = r − 1, ∇dΩ = ❡r, div(∇dΩ) = 1
r , α = r.
◮ The third assumption is satisfied because 1 r s
1
1 t dtdr = 1 s log(s)ds < +∞
An example: the circle Ω = {x ∈ R2 | ||x||2 < 1}, U = Ω \ {0}. Poincar´ e inequality to prove (weight ω = 1) is
f 2dx ≤
|❡r · ∇f |2dx +
f 2dy With polar coordinates for a radial function f : 1 f (r)2rdr ≤ C
1 f ′(r)2rdr
To prove for f ∈ C1([0, 1]): 1 f (r)2rdr ≤ C
1 f ′(r)2rdr
To prove for f ∈ C1([0, 1]): 1 f (r)2rdr ≤ C
1 f ′(r)2rdr
r
1 f ′(t)dt (*)
To prove for f ∈ C1([0, 1]): 1 f (r)2rdr ≤ C
1 f ′(r)2rdr
r
1 f ′(t)dt (*)
◮ Use Cauchy-Schwartz inequality:
1
f ′(t)dt
r
1
|f ′(t)|t1/2t−1/2dt ≤ 1 f ′(r)2rdr 1/2 r
1
1 t dt 1/2
To prove for f ∈ C1([0, 1]): 1 f (r)2rdr ≤ C
1 f ′(r)2rdr
r
1 f ′(t)dt (*)
◮ Use Cauchy-Schwartz inequality:
1
f ′(t)dt
r
1
|f ′(t)|t1/2t−1/2dt ≤ 1 f ′(r)2rdr 1/2 r
1
1 t dt 1/2 ◮ Use Young’s inequality (a + b)2 ≤ 2a2 + 2b2, multiply (*) by r and integrate: 1 f (r)2rdr ≤ 2f (1)2 1 rdr + 2 1 f ′(r)2rdr 1 r r
1
1 t dtdr
Much more details in the preprint. Feppon, F., Allaire, and Dapogny, C. A variational formulation for computing shape derivatives of geometric constraints along rays. HAL preprint hal-01879571. (2018)