• Aucun résultat trouvé

# On singular perturbation problems for systems of PDEs

N/A
N/A
Protected

Partager "On singular perturbation problems for systems of PDEs"

Copied!
31
0
0

Texte intégral

(1)

## On singular perturbation problems for systems of PDEs

Ivar Ekeland and Eric Séré

Chern Institute, Nankai University, May 14th, 2018

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 1 / 31

(2)

## The problem

We want to solve a system of PDEs:

S(ε,u,Du) =f nearS(0,0,0) =0,

the linearized operatorv !Su(ε,u,Du)v+SDu(ε,u,Du)Dv has a right-inverseLε(u)for (ε,u,Du) small

kLε(u)k !∞ when ε!0

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 2 / 31

(3)

## The linear case

Consider on the 2-dimensional torus the linear di¤erential operator

ω =ω1

∂θ1

+ω2

∂θ2

where ω:= (ω1,ω2) satis…es the Diophantine condition:

ω1

ω2

p q

K

jqj2+α,α>0

Clearly ω mapsHk into Hk 1, that is, it loses one derivative. It also has an inverse:

u =

## ∑

n

unexp(ω1n1θ1+ω2n2θ2)

ωu = f () 8n 2Zd, un =fn(ω1n1+ω2n2) 1 Because of the Diophantine condition, junj K 1ω21knk1+αjfnj. So

ω1 sendsHk into Hk 1 α, that is, it loses many derivatives

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 3 / 31

(4)

## The spaces

Typically, for PDEs, both the operator and its inverse lose derivatives, so the inverse does not send you back into the initial space. For this reason one must introduce a scale of Banach spaces (Xs,k ks),s0 s <

s0 s1 s2 S =)[Xs2 Xs1 and k ks1 k ks2]

For instance,Xs =Hs(Ω)(Sobolev space) andC =\sHs with S =∞. So, in the preceding example, where Ωis the torus,

ω : Hs !Hs 1

ω1 : Hs !Hs 1 α

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 4 / 31

(5)

## Polynomials

We shall need an additional structure. Let EN,N 0 be an increasing sequence of …nite-dimensional spaces, the union being dense in all the Xs

Xs =[n 0EN

The projection ΠN :X0 !EN is assumed to satisfy:

(Polynomial growth): for all nonnegative numbers s,d satisfying s+d S, and all x 2EN, there is a constant CS such that:

kxks+d CSNdkxks k(1 ΠN)xks CSN dkxks+d

(Interpolation): for 0 t 1:

kxkts1+(1 t)s2 CSkxkts1kxk1s2 t .

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 5 / 31

(6)

## Nonlinear operators

Nonlinear operators from C into itself, such asS(u,Du) arise from maps S :Rd Rnd !Rm through some basic operations. The behaviour of the Sobolev normsHs is ruled by the so-called Hölder estimates:

(Product) Ifu andv are in L\Hs then there is a constantms depending only on s such that:

kuvks ms(kukssupjvj+kvkssupjuj)

(Composition) Letf :R!RbeC. If u 2L\Hs then f (u)2L\Hs and there is a constant `s depending on s,f,and supjuj such that:

kf (u)ks `s(1+kuks)

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 6 / 31

(7)

## Towards a statement

Let (Ys, k k0s),0 s S, be another scale of Hilbert spaces with the same properties. We consider a map F from the unit ballBs0+maxfm.`g into Y0, wheres0,m,` are given. We normalize by settingF(0) =0 and we want to solveF (u) =v . Assume F is continuous and

Gâteaux-di¤erentiable.

Let m,m0,`,`0 be given. We assume that for any S and any s S, there are constants aS andbS and a linear map L(x):Y !X satisfying, for all u with kuks0+maxfm,`g 1

F0(u)h 0s aS kuks+m+kuks0khks+m

F0(u)L(u)k = k

kL(u)kks bS kkks+`00+kkk0s0kuks+`

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 7 / 31

(8)

## A local surjection theorem

Theorem Suppose:

s >s0+maxfm,`g δ >s+`0

and S is large enough. Then, there exists some r >0 and some k >0 such that, if kvkδ <r , the equation F(u) =v can be solved with

kuks0+maxfm,`g 1 kuks kkvk0δ

For any ε>0, we can solveF(u) =v with:

v 2Yδ for anyδ =s0+maxfm,`g+`0+ε u 2Xs with s = δ (`0+ε)

The linearized operator yields ε=0, so the nonlinear operator does almost as well

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 8 / 31

(9)

## Comparison with earlier results

There is a long history of results of this type, following the seminal papers of Arnol’d, Nash and Moser, the sharpest results to date being due to Hörmander. Most of them use smoothing operators instead of projections.

We are not aware that any of these papers gets the optimal value for δ (lowest regularity of the RHS)

Every single one of them require an additional tame estimate on the second derivative:

F00(x)u,v 0s c1 kuks+p1kvkp2 +kukp1kvks+p2

+c2kxks+p3 kuks+p4 kvkp5 +kukp4kxks+p5

They then use second-order expansion to get quadratic convergence:

kxn+1 xnk 12MKkxn xn 1k2 = 1

2MKkxn xn 1k kxn xn 1k with kF00(x)k K. This is Newton’s method. However the range of convergence is extremely small: 12MKkxn xn 1k 1,so

kx1k 2(MK) 1.This is the key to the following

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 9 / 31

(10)

## The Inverse Function Theorem in Banach spaces

Theorem (Newton-Kantorovitch)

Assume F is C2 with F(0) =0, and forkxk R we have:

F0(x) 1 M and F00(x) K

For every y such that kyk min M12K,MR there is unique solution to F (x) =y in the ball kxk min MK1 ,R

Theorem ((Wazewski 1947, IE 2011))

Assume F is G-di¤erentiable with F (0) =0, and for kxk R there is a map L(x)such that

F0(x)L(x) =IY and sup

kxk RkL(x)k<M

Then for every y with kyk MR there is a solution with kxk R.

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 10 / 31

(11)

## First approach to singular perturbations

Suppose we want to solveF (ε,u) =v with F (ε,0) =0 and for ε andu small enough:

DuF (ε,u) 1 ε 1M Duu2 F (ε,u) K Then the two preceding theorems give the extimates:

(NK) kvk ε

2

M2K, kuk KMε (WE) kvk εRM, kuk R We gain one order of magnitude, but we lose uniqueness

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 11 / 31

(12)

## A singular perturbation result.

We have F(ε,u)with F (ε,0) =0. We assume

kDFε(u)hks0 aS kuks+m+kuks0khks+m

DFε(u)L(u)k = k

kL(u)kks bSε g kkks+`00+kkk0s0kuks+`

Theorem

Suppose s >s0+maxfm,`g,δ >s+`0 g0 >g , and S is large enough.

Then, for some r >0and k >0, whenever kvk0δ <rεg0 there exists for every ε>0 some uε such that:

F(ε,uε) = v kuεks0+maxfm,`g 1

kuεks kε g0kvk0δ

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 12 / 31

(13)

## A singular perturbation problem

Rauch and Métivier (2010) and Texier and Zumbrun (2013) have studied the following sysem of nonlinear coupled Schrödinger equations in

v = (u,u¯)for j,k K andn N

∂t +iλj∆ vj =

## ∑

N n=1

bjkn(v)∂vk

∂xn +cjnk (v)v¯k

∂xn vk 2 C, λj 2 R, j 6=j0 =)λj 6=λj0

The coe¢ cients bjk

n(v) andcjk

n(v) must satisfy a growth condition near the origin:

9 p 2:8a= (a1, ,aN)2Nd, ∂b

k jn

∂va + ∂c

k jn

∂va Cjajjvj(p jaj)+ and we also need some nonresonance conditions:

λj +λj0 = 0=)cjj0 cj0j =0

Imbjj = 0

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 13 / 31

(14)

## The initial condition

This system arises in optics, gives rise to several kinds of solutions, according to the initial condition:

vε(0,x) =εσ(aε(x),a¯ε(x)) There are two important classes of solutions:

aε(x) = a0 x

ε (concentration) aε(x) = a0(x)exp i

εx ξ0 (oscillation)

Here g will be an (explicit) function of σ,p andd.Applying our result, we can solve the problem for:

σ> 1 σa p+1 +d

2

p (p+1) (p 1) This is an improvement on earlier results.

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 14 / 31

(15)

## The results

We seek a solution on [0, T], whereT is independent of ε.

For σ large enough, namely:

σ> 1+N2 (oscillating case) 1 (concentrating case)

this is not a singular perturbation problem. Metivier and Rauch, using the standard inverse function theorem in Banach spaces, prove that solutions exist for all d andp.

The case when σ is smaller than these bounds has been investigated by Texier and Zumbrun using a Nash-Moser theorem. We will compare our results to theirs

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 15 / 31

(16)

## Comparison: two space dimensions

Concentrating case

TZ requirep 4 and 2(p+1)9 <σ<1.

ES gives anyp 3, and (p+1)(p2p 1 1)<σ<1 Oscillating case

TZ requirep 3 and p+15 <σ<2

ES gives anyp 2 and (p+1)(p3p 2 1) <σ<2

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 16 / 31

(17)

## Comparison: three space dimensions

Concentrating case

TZ requiresp 4 and p+14 <σ<1.

ES gives anyp 3, and 2(p+1)(p4p 1 1) <σ<1 Oscillating case

TZ gives anyp 2 and 2(p+1)11 <σ< 52 ES gives anyp 2 and 2(p+1)(p7p 4 1) <σ< 52

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 17 / 31

(18)

## Proof of the Theorem

We introduce positive numbers α>1,θ <1,σ<S and βwith αβ<σ.

We choose an integerN0 2 and we de…ne two in…nite sequences of integres Nn andMn byNn = N0αn and Mn = Nnθ , so that Nn ! very fast andNn 1 Mn Nn with dim(Mn)<<dim(Nn). We then construct by induction a sequence un 2ENn such that:

ΠM0 nF (un) =Π0Mn 1v for n 1

8n 1, kun un+1ks0 CNnαβ σ+s0kvk0d

8n 1, kun un+1kσ CNnαβkvkd0

Since β σ/α<0 the sequence xn is Cauchy in Hs0. So it converges to some u2 Hs0 and by continuityF(u) =v.Using the interpolation inequality, we see that un will also converge in all norms s <σ αβ

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 18 / 31

(19)

## Initialization

Introduce a norm N1 onEN1 and EN0

1 :

N1(z) =kzkδ+N1 θα(σ δ)kzkσ

Set B1 :=fz 2EN1 j N1(z) 1g and de…ne a functionf1 :B1 !EM0 1

by:

f1(u) =Π0M1F (u) Lemma

If (1 θ) (σ δ)>θm+maxf`,θ`0g+g/η, then Df1(u)has a right inverse L1(u)and we have:

kL(u)kN1 Cε g M1`0+N1`

We then apply the local surjection theorem to solvef1(u1) =ΠM0v, provided:

kvk0δ Cεg M1`0+N1` 1

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 19 / 31

(20)

## The induction

Introduce a norm Nn onENn :

Nn(z) =kzks0 +Nnσ1+skzkσ

Set Bn := nz 2ENn j Nn(z) ε gNnαβ1σ+s0kvk0δ

o

and de…ne a functionfn :Bn !ENn by:

fn(z) =ΠM0 n(F(un 1+z) F(un 1))

Taking into account the induction assumption ΠN0n 1F (un 1) =Π0Nn 2v, this can be rewritten as:

fn(z) =en+hn

hn = Π0Mn 1 Π0Mn 2 v 2EM0 n 1

en = Π0Mn 1 Π0Mn F (un 1)2EM0 n Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 20 / 31

(21)

## The induction

Lemma Assume:

(1+α θα) (σ s0) > αβ+α(m+`) +`0+ g αη (1 θ) (σ s0) > m+θ`0+ g

αη Then Dfn(z)has a right-inverse Ln(z)on the ball Nn(z) ε gNnαβ1σ+s0kvk0δ and :

kLn(z)kNn ε g

Nnβ+`Nn iσ+s0+`0 M1`0+N1` +Mn`0

!

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 21 / 31

(22)

## The induction

We have to solve fn(z) =en+hn with Nn(z) ε gNnαβ1σ+s0kvk0δ.By the local surjection theorem, this is possible provided

kLn(z)kNn kenkNn+khnkNn ε gNnαβ1σ+s0kvk0δ.

Lemma Assume:

δ >s0+ α

ϑ(σ s0 αβ+`") (α 1)β>(1 ϑ) (σ s0) +ϑm+`"+g

η

`"=max (α 1)`+`0,ϑ`0

Then the local surjection theorem obtains, and un can be found

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 22 / 31

(23)

## Showing that the values of the parameters are compatible

α,β,ϑ andσ must satisfy, with `"=maxf(α 1)`+`0,ϑ`0g 1

α <ϑ<1

(1 ϑ) (σ δ)>ϑm+max `,ϑ`0 +g η β>max `,ϑ`0 + ϑ

α(σ δ) σ >αβ+s

(1+α ϑα) (σ s0)>αβ+α(m+`) +`0+ g η (1 ϑ) (σ s0)>m+ϑ`0+ g

αη δ >s0+ α

ϑ(σ s0 αβ+`") (α 1)β>(1 ϑ) (σ s0) +ϑm+`"+g

η

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 23 / 31

(24)

## Continuous sections

Theorem (Eric Séré)

Let X and Y be Banach spaces, B X the unit ball, and F :B !Y with F (0) =0.Suppose F is continuous and Hadamard-di¤erentiable on X . Assume there are a map L:B ! L(X,Y)and a constant a<1 such that, for every (x,v)2B Y there exists some ε>0 with:

DF x0 L(x)v v akvk

Assume moreover that supfkL(x)k jx 2Bg<M.Denote by B0 Y the ball of radius (1 a)rM 1. Then there exists a continuous map G :B0 !B such that F G =IY

So, there is no uniqueness in solving F(ε,x) =y, but we can ask for continuous dependence on the right-hand side. Our plan is to extend this to the singular perturbation problem with loss of derivative and show continuous dependence on v andε.

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 24 / 31

(25)

THE INVERSE FUNCTION THEOREM IN BANACH SPACES

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 25 / 31

(26)

## A variational principle

Theorem (Ekeland, 1972)

Let (X,d) be a complete metric space, and f :X !R[ f+g be a lower semi-continuous map, bounded from below:

f(x,a) ja f (x)g is closed in X R inff > ∞

Suppose f (x0)<∞. Then for every R >0, there exists somex such that:¯ f (x¯) f (x0)

d(x,¯ x0) R

f (x) f (x¯) f (x0) inff

R d(x,x¯) 8x

The proof relies on the Baire theorem: if Fn,n2N, is a sequence of closed bounded subsets of a complete metric space, with Fn+1 Fn and diam Fn !0, then their intersection is a singleton.

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 26 / 31

(27)

## First-order version

De…nition

We shall say that f is Gâteaux-di¤erentiableat x if there exists a continous linear map Df (x):X !X such that

8ξ 2X, limt 1t [f (x+tξ) f (x)] =hDf (x), ξi In other words, the restriction of f to every line is di¤erentiable (for instance, partial derivatives exist). Note that a G-di¤erentiable function need not be continuous.

Theorem

Let (X,d) be a complete metric space, and f :X !R[ f+∞g be a lower semi-continuous map, bounded from below. Then for every x02 X and R >0, there exists some x such that:¯

f (x¯) f (x0), kx¯ x0k R

kDf (x¯)k f (x0) inff R Corollary

There is a sequence xn such that:

f (xn)!inff kDf (xn)k !0 Proof.

Just take x0 such that f (x0) inff ε and R=p

ε and let ε= 1n

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 27 / 31

(28)

## Proof

For simplicity, take x0 =0 and inff =0. Apply EVP to x =x¯ +tu and let u !0. We get:

f (x¯ +tu) f (x¯) f (0)

R tkuk 8(t,u)

t!lim+0

1

t (f (x¯ +tu) f (x¯)) f (0)

R kuk 8u hDf (x),ui f (0)

R kuk 8u, or kDf (x)k f (0) R

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 28 / 31

(29)

## A non-smooth inverse function theorem

Theorem

Let X and Y be Banach spaces. Let F :X !Y be continuous and Gâteaux-di¤erentiable, with F (0) =0. Assume that the derivative DF(x) has a right-inverse L(x), uniformly bounded in a neighbourhood of0:

8v 2Y, DF(x)L(x)v =v supfkL(x)k j kxk Rg<M Then, for every y such that¯

ky¯k MR there is some x such that:¯

kx¯k Mky¯k F (x¯) =y¯

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 29 / 31

(30)

## Proof

Take any y¯ with ky¯k MR. Consider the functionf :X !R de…ned by:

f (x) =kF (x) y¯k

It is continuous and bounded from below, so that we can apply EVP. We can …nd x¯ with:

f (x¯) f (0) =ky¯k MR kx¯k Mky¯k R 8x, f (x) f (x¯) f (0)

R kx x¯k=f (x¯) 1

M kx x¯k I claim F(x¯) =y.¯

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 30 / 31

(31)

## Proof (ct’d)

Assume F(x¯)6=y¯. The last equation can be rewritten:

8t 0, 8u 2X, f (x¯ +tu) f (x¯) t

1 M kuk Simplify matters by assuming X is Hilbert. Then:

F(x¯) y¯

kF(x¯) y¯k,DF(x¯)u =hDf (x¯),ui M1 kuk

We now take u = L(x¯) (F(x¯) y¯), so thatDF(x¯)u = (F (x¯) y¯). We get a contradiction:

kF (x¯) y¯k kL(x¯)k

M kF (x¯) y¯k<kF(x¯) y¯k

Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

Chern Institute, Nankai University, May 14th, 2018 31 / 31

Références

Documents relatifs

The ACCESS procedure provides access permission checking on the server, the FSSTAT procedure returns dynamic information about a file system, the FSINFO procedure

If the gateway is authorized to send multiple successive Notify commands, it will proceed as follows. When the gateway exits the &#34;notification state&#34;, it resets the

These packages, when used in conjunction with the packages currently defined in RFC 2705 (Media Gateway Control Protocol Version 1.0) [1], allow an MGCP Call Agent to

1) The Call Agent asks the first gateway to &#34;create a connection&#34; on the first endpoint. The gateway allocates resources to that connection, and responds to

The base Media Gateway Control Protocol (MGCP) includes audit commands that only allow a Call Agent to audit endpoint and/or connection state one endpoint at a time..

Default time-out values may be over-ridden by the Call Agent for any Time-Out event defined in this document (with the exception of those that have a default value

However, as long as the gateway is in service and able to respond to MGCP commands, it can apply the endpoint configuration command to endpoints specified by the

If an AIGP attribute is received and its first AIGP TLV contains the maximum value 0xffffffffffffffff, the attribute SHOULD be considered to be malformed and SHOULD be

A corpus of 10 professional audio descriptions in English (WHW-EN-Pr), including both the text and the audiovisual file: 6,799 words... The corpus includes audio descriptions

Other interesting fields related to AVT are automatic sign language translation and signing avatars (Wolfe et al. Still, very often this type of image-related research is

This framework has been used to illuminate the roles of the field and cultural domains in the generation of advertising ideas’ (Vanden Bergh and Stuhlfaut 2006). In this paper

Prior research suggests a lack of gender diversity in creative departments since subjectivity of creativity has a negative impact on women (Mallia 2009), however our data

investigation focused on creativity in PR although authors in the PR literature cite it as part of the professional competence of a PR practitioner (see Wilcox et al., 2007,

It is, however, not very probable that similar citizen programmes will be set up for flora and insects in the near future, even though national structures with the means to

9 As outlined in these brief summaries, the role of science in Professional Foul and Hapgood has much in common with Stoppard’s use of chaos theory in Arcadia, since catastrophe

Finally, he was able to avoid the vexing issue of ‘popular’ religion and its nagging oppositions between rich/poor, emotional/rational,

The same piece of music might frame the beginning and end of a performance (as a Moby track does Tender Offer), or a particular question might punctuate a video, as the question

6 (1989) 95) on the existence of minimal (in the sense of Giaquinta and Guisti) heteroclinic solutions of a nonlinear elliptic PDE.. Bangert’s work is based on an earlier paper of

We relied on the display of those values to design a set of interactions that offer new services for exploratory design and structure-based interaction: query and selection

We then obtain a two-to-one correspondence, via the logarithmic derivative, between these functions and a subclass of hyperbolically monotone functions of finite type1.

subsequent Indo-Pacific populations likely derive from Philippine populations, that all colonies in the invasive range were polygyne (i.e. with more than one queen), and that

Expression of the two enzymes in tomato up-regulated most intrinsic carotenogenic genes, and efficiently directed carbon flux into carotenoids, leading to massive accumulation of

εἰ δὴ τὸ ὂν καὶ τὸ ἓν ταὐτὸν καὶ μία φύσις τῷ ἀκολουθεῖν ἀλλή- λοις ὥσπερ ἀρχὴ καὶ αἴτιον, ἀλλ’ οὐχ ὡς ἑνὶ λόγῳ δηλού- μενα (δι- αφέρει δὲ οὐθὲν οὐδ’ ἂν ὁμοίως