On singular perturbation problems for systems of PDEs

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On singular perturbation problems for systems of PDEs

Ivar Ekeland and Eric Séré

CEREMADE, Université Paris-Dauphine

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

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The problem

We want to solve a system of PDEs:

S(ε,u,Du) =f nearS(0,0,0) =0,

the linearized operatorv !^S^u(ε,u,Du)v+S_Du(ε,u,Du)Dv has a right-inverseL_ε(u)for (ε,u,Du) small

k^Lε(u)k !∞ when ε!⁰

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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

j^qj²⁺^α^,^α>0

Clearly ∂_ω mapsH^k into H^k ¹, that is, it loses one derivative. It also has an inverse:

u =

∑

n

u_nexp(ω1n₁θ1+ω2n₂θ2)

∂_ωu = f () 8ⁿ 2Z^d, un =fn(ω1n1+ω2n2) ¹ Because of the Diophantine condition, j^uⁿj ^K ¹^ω2¹kⁿk¹⁺^αj^fⁿj^{. So}

∂_ω¹ sendsH^k into H^k ¹ ^α, that is, it loses many derivatives

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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 (X_s,k k^s),s₀ s < _∞

s₀ s₁ s₂ S =)[X_s₂ X_s₁ and k k^s1 k k^s2]

For instance,X_s =H^s(Ω)(Sobolev space) andC^∞ =\^s^H^s ^with ^S =∞. So, in the preceding example, where Ωis the torus,

∂ω : H^s !^H^s ¹

∂_ω¹ : H^s !^H^s ¹ ^α

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Polynomials

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

X_s =[ⁿ ⁰^E^N

The projection ΠN :X₀ !^E^N is assumed to satisfy:

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

k^xk^s⁺^d ^C^S^N^dk^xk^s k(1 ΠN)xk^s ^CSN ^dk^xks+d

(Interpolation): for 0 t 1:

k^xkts1+(1 t)s2 C_Sk^xk^ts1k^xk¹s2 ^t .

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Nonlinear operators

Nonlinear operators from C^∞ into itself, such asS(u,Du) arise from maps S :R^d R^nd !^R^m through some basic operations. The behaviour of the Sobolev normsH^s is ruled by the so-called Hölder estimates:

(Product) Ifu andv are in L^∞\^H^s then there is a constantm_s depending only on s such that:

kûvks m_s(kûkssupj^vj+k^vkssupjûj)

(Composition) Letf :R!^R^be^C^∞^{. If} ^u 2^L^∞\^H^s ^then f (u)2^L^∞\^H^s and there is a constant `s depending on s,f,and supj^uj ^{such that:}

k^f (u)ks `s(1+k^uks)

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Towards a statement

Let (Y_s, k k⁰s),0 s S, be another scale of Hilbert spaces with the same properties. We consider a map F from the unit ballB_s₀₊_max_f_m_._`_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 a_S andb_S and a linear map L(x):Y !^X satisfying, for all u with k^uks0+maxf^m,`g 1

F⁰(u)h ⁰_s a_S k^uks+m+k^uks0k^hks+m

F⁰(u)L(u)k = k

k^L(u)kks bS k^kks+`⁰0+k^kk⁰s0k^uks+`

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A local surjection theorem

Theorem Suppose:

s >s₀+maxf^m,`g δ >s+`⁰

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

k^uks0+maxf^m,`g 1 k^uks kk^vk⁰δ

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

v 2^Y^δ ^{for any}^δ =s₀+maxf^m,`g+`⁰+ε u 2^X^s ^with ^s = δ (`⁰+ε)

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

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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 ⁰_s c1 k^uks+p1k^vkp2 +k^ukp1k^vks+p2

+c₂k^xks+p3 k^uks+p4 k^vkp5 +k^ukp4k^xks+p5

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

k^xⁿ⁺¹ ^xⁿk ¹₂^MKk^xⁿ ^xⁿ ¹k² = ¹

2MKk^xⁿ ^xⁿ ¹k k^xⁿ ^xⁿ ¹k with k^F⁰⁰(x)k ^K. This is Newton’s method. However the range of convergence is extremely small: ¹₂MKk^xⁿ ^xⁿ ¹k ^1,^so

k^x¹k ²(MK) ¹.This is the key to the following

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The Inverse Function Theorem in Banach spaces

Theorem (Newton-Kantorovitch)

Assume F is C² with F(0) =0, and fork^xk ^{R we have:}

F⁰(x) ¹ M and F⁰⁰(x) K

For every y such that k^yk ^min M¹²K,_M^R there is unique solution to F (x) =y in the ball k^xk ^min MK¹ ,R

Theorem ((Wazewski 1947, IE 2011))

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

F⁰(x)L(x) =I_Y and sup

k^xk ^Rk^L(x)k<M

Then for every y with k^yk M^R there is a solution with k^xk ^R.

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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 D_uu² F (ε,u) K Then the two preceding theorems give the extimates:

(NK) k^vk ^ε

2

M²K, k^uk _KM^ε (WE) k^vk ^εR_M^, k^uk ^R We gain one order of magnitude, but we lose uniqueness

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A singular perturbation result.

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

k^DFε(u)hks⁰ a_S k^uks+m+k^uks0k^hks+m

DF_ε(u)L(u)k = k

k^L(u)kks b_Sε ^g k^kks+`⁰0+k^kk⁰s0k^uks+`

Theorem

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

Then, for some r >0and k >0, whenever k^vk⁰δ <rε^g⁰ there exists for every ε>0 some u_ε such that:

F(ε,u_ε) = v k^uεks0+maxf^m^,`g 1

k^u^εks kε ^g⁰k^vk⁰δ

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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∆ v_j =

∑

N n=1

b_j^k_n(v)^∂v^k

∂x_n +c_jn^k (v)^∂^v^¯^k

∂x_n v_k 2 ^C^, ^λ^j 2 ^R^, ^j 6=j⁰ =)^λ^j 6=λj⁰

The coe¢ cients b_j^k

n(v) andc_j^k

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

9 ^p ²^:8^a= (a₁, ,a_N)2^N^d^, ^∂b

k jn

∂v^a + ^∂c

k jn

∂v^a C_j_a_jj^vj⁽^p ^j^a^j⁾⁺ and we also need some nonresonance conditions:

λj +λj⁰ = 0=)^c^jj⁰ ^c^j⁰^j =0

Imbjj = 0

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The initial condition

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

v_ε(0,x) =ε^σ(a_ε(x),a¯_ε(x)) There are two important classes of solutions:

a_ε(x) = a⁰ x

ε (concentration) a_ε(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 ^σ^a p+1 +^d

2

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

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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

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Comparison: two space dimensions

Concentrating case

TZ requirep 4 and _2(p+1)⁹ <σ<1.

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

TZ requirep 3 and _p₊₁⁵ <σ<2

ES gives anyp 2 and _(p+1)(p^3p ² ₁₎ <σ<2

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Comparison: three space dimensions

Concentrating case

TZ requiresp 4 and _p+1⁴ <σ<1.

ES gives anyp 3, and _2(p_+1)(p^4p 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)(p^7p ⁴ ₁₎ <σ< ⁵₂

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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 = N₀^αⁿ and Mn = N_n^θ , so that Nn !^∞ very fast andN_n ₁ M_n N_n with dim(M_n)<<dim(N_n). We then construct by induction a sequence u_n 2^E^Nn such that:

ΠM⁰ nF (un) =_Π⁰_M_n ₁v for n 1

8ⁿ ^1, k^uⁿ ^uⁿ⁺¹ks0 CNn^αβ ^σ⁺^s⁰k^vk⁰d

8ⁿ ^1, k^uⁿ ^uⁿ⁺¹kσ CNn^αβk^vkd⁰

Since β σ/α<0 the sequence xn is Cauchy in Hs0. So it converges to some u2 ^H^s0 and by continuityF(u) =v.Using the interpolation inequality, we see that u_n will also converge in all norms s <σ αβ

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Initialization

Introduce a norm N¹ ^on^E^N1 and E_N⁰

1 :

N¹(z) =k^zkδ+N₁ ^θ^α⁽^σ ^δ⁾k^zkσ

Set B¹ ^:=f^z 2^E^N1 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:

k^L(u)k_N₁ ^C^ε ^g ^M1^`⁰+N₁^`

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

k^vk⁰δ Cε^g M₁^`⁰+N₁^` ¹

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The induction

Introduce a norm Nⁿ ^on^E^Nn :

Nⁿ(z) =k^zks0 +N_n^σ₁⁺^sk^zkσ

Set Bⁿ ^:= ⁿz 2^E^Nn j Nⁿ(z) ε ^gN_n^αβ₁^σ⁺^s⁰k^vk⁰δ

o

and de…ne a functionf_n :Bⁿ !^E^Nn by:

fn(z) =ΠM⁰ n(F(un 1+z) F(un 1))

Taking into account the induction assumption ΠN⁰n 1F (u_n ₁) =_Π⁰_N_n ₂v, this can be rewritten as:

fn(z) =en+hn

hn = _Π⁰_M_n ₁ _Π⁰_M_n ₂ v 2^EM⁰ n 1

e_n = _Π⁰_M_n ₁ _Π⁰_M_n F (u_n ₁)2^EM⁰ n Ivar Ekeland and Eric Séré (Institute) On singular perturbation problems for systems of PDEs

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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 ^gN_n^αβ₁^σ⁺^s⁰k^vk⁰δ and :

k^Lⁿ(z)k_Nn ε ^g

Nn^β^+`N_{n i}^σ⁺^s⁰^+`⁰ M₁^`⁰+N₁^` +M_n^`⁰

!

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The induction

We have to solve fn(z) =en+hn with Nⁿ(z) ε ^gN_n^αβ₁^σ⁺^s⁰k^vk⁰δ.By the local surjection theorem, this is possible provided

k^Lⁿ(z)k_Nn k^eⁿk_N_n+k^hⁿk_N_n ^ε ^g^Nn^αβ1^σ⁺^s⁰k^vk⁰δ.

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

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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

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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)2^B Y there exists some ε>0 with:

DF x⁰ L(x)v v ak^vk

Assume moreover that supfk^L(x)k j^x 2^Bg<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 =I_Y

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.

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THE INVERSE FUNCTION THEOREM IN BANACH SPACES

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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) j^a ^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¯) 8^x

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

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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}

8^ξ 2^X^, ^lim_t ¹_t [f (x+tξ) f (x)] =h^Df (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), k^x^¯ ^x⁰k ^R

k^Df (x¯)k ^f (x₀) inff R Corollary

There is a sequence xn such that:

f (x_n)!^inf^f k^Df (x_n)k !⁰ Proof.

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

ε and let ε= ¹_n

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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 tk^uk 8(t,u)

t!lim+0

1

t (f (x¯ +tu) f (x¯)) ^f (0)

R k^uk 8^u h^Df (x),ui ^f (0)

R k^uk 8^u^{, or} k^Df (x)k ^f (0) R

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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:

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

k^y^¯k _M^R there is some x such that:¯

k^x^¯k ^Mk^y^¯k F (x¯) =y¯

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