Ambrosetti-Prodi type results to

fourth order fully nonlinear

differential equations

Department of Mathematics

Research Center on Mathematics and Applications (CIMA-UE)

University of Évora, PORTUGAL

 Feliz M. Minhós

THE FIFTH INTERNATIONAL CONFERENCE ON

Dynamic Systems and Applications, ATLANTA, U.S.A., MAY 30 - JUNE 2, 2007

Workshop on Topological Methods for Boundary Value Problems

u ( 1) = u’ ( 1) = u’’ ( 0) = u’’’ (1) = 0

u (4) + f (x, u, u’, u’’, u’’’) = s p(x)

f : [0,1]× R4 →R and p : [0,1] → R+ continuous, s ∈ R

Existence of solution (and location for u, u’, u’’, u’’’ ) for s ∈ R such that there are upper and lower solutions

Existence, non-existence and multiplicity depending on s.

k u (4) + b u+ = g (x)

Both endpoints are simply suported

Applications:Bending of an elastic beam

u (0) = u (1) = u’’ (0) = u’’ (1) = 0

u (4) = f (x, u, u’, u’’, u’’’)

f : [0,1]× R4 →R continuous

u (0) = u (L) = u’’ (0) = u’’ (L) = 0

(Lidstone conditions )

A. C. Lazer, P. J. McKenna, Large-amplitude periodic oscillations in suspension bridges: some new connections with nonlinear analysis. SIAM

Review, 32 (1990), 537-578.

u – displacement of a beam with lenght L

u (4) - beam bending, k - constant depending on the beam

u+ = max {0, u} – unidirectional force distended – returns to the initial state;compressed – keeps

Suspension bridges

u (0) = u (1) = u’’ (0) = u’’ (1) = 0

u (4) = f (x, u, u’, u’’, u’’’)

f : [0,1]× R4 →R continuous (Lidstone conditions )

1941

u (0) = u (1) = u’’ (0) = u’’ (1) = 0

u (4) = f (x, u, u’, u’’, u’’’)

f : [0,1]× R4 →R continuous (Lidstone conditions )

1941

u (0) = u (1) = u’’ (0) = u’’ (1) = 0

u (4) = f (x, u, u’, u’’, u’’’)

f : [0,1]× R4 →R continuous

Pedestrian Millennium Bridge

London 325m steel

Opened on 10 June 2000

80,000 to 100,000 people crossed it

Closed on 12 June 2000

(Lidstone conditions )

max sway of the deck: 70mm

movement caused by the sideways loads generate

when walking

u (0) = u (1) = u’’ (0) = u’’ (1) = 0

u (4) = f (x, u, u’, u’’, u’’’)

f : [0,1]× R4 →R continuous (Lidstone conditions )

u (0) = u (1) = u’’ (0) = u’’ (1) = 0

u (4) = f (x, u, u’, u’’, u’’’)

f : [0,1]× R4 →R continuous (Lidstone conditions )

u (4) + f (x, u, u’, u’’, u’’’) = s p(x)s ∈ R

p : [0,1] → R+ continuous,

Some works on Ambrosetti-Prodi type results:

• C. Fabry, J. Mawhin, M. Nkashama,A multiplicity result for periodic

solutions of forced nonlinear second order ordinary differential equations, Bull. London Math. Soc., 18 (1986), 173-180.

u (0) = u (π) = u’’ (0) = u’’ (π) = 0

u’’ + f (t, u, u’ ) = s , f is 2π – periodic in t• M. Senkyrik, Existence of multiple solutions for a third-order three-point regular boundary value problem, Math. Bohemica, 119 (1994) 113-121

u (4) = f (t, u ) + s sin (t)

u’ (0) = u’ (1) = u (η) = 0 , 0 ≤ η ≤ 1

• C. de Coster, L. Sanchez,Upper and lower solutions, Ambrosetti-Prodi

problem and positive solutions for a furth morder O.D.E., Riv. Mat. Pura Appl., 14 (1994), 57-82.

u (0) = u (2π) , u’ (0) = u’ (2π) u’’’ + f (t, u, u’,u’’ ) = s

u (4) + f (x, u, u’, u’’, u’’’) = s p(x)

• A. Ambrosetti, G. Prodi, On the inversion of some differential mappings with singularities between Banach spaces. Ann. Mat. Pura Appl. 93 (1972), pp. 231–246.

u ( 1) = u’ ( 1) = u’’ ( 0) = u’’’ (1) = 0

u (4) + f (x, u, u’, u’’, u’’’) = s p(x) ( Es )

(BC)

• Lower and upper solutions

Method and techniques:

α (1) ≤ 0 , α ’ (1) ≥ 0 , α ’’ (0) ≤ 0 , α ’’’ (1) ≤ 0

(i) α is a lower-solution of (E)-(BC) if :

α (4) ≥ g ( x, α , α’, α’’, α’’’)

(ii) β is an upper-solution of (E)-(BC) if the reversed inequalities hold

Clamped beam at the right endpoint

g : [0,1]× R4 →R continuous

u (4) = g (x, u, u’, u’’, u’’’) ( E )

General existence and location theorem for (E)-(BC)

• Lower and upper solutions

Method and techniques:

• Nagumo-type growth condition in E

.)(

0

∫∞

+∞=dssh

s

E

| g ( x, y0 , y1 , y2 , y3 ) | ≤ hE (| y3 |) , in E,

with hE : [0, +∞[→R+ :

for some continuous functions γ i ≤ Гi , i = 0,1,2 ;

E = { ( x, y0 , y1 , y2 , y3 ) : γi ≤ yi ≤ Гi , i = 0, 1, 2 }

u ( 1) = u’ ( 1) = u’’ ( 0) = u’’’ (1) = 0 (BC)

u (4) = g (x, u, u’, u’’, u’’’) ( E )

A priori bound for u′′′

• ∃ γ i , Гi : [0,1]→R continuous functions : γ i ≤ Гi , i = 0,1,2 ;

Then

∃ ρ > 0 : every solution u(x) of (E) with

γ i ≤ u ( i ) ≤ Гi , i = 0,1,2, ∀x ∈ [0,1]

verifies

Lemma :

• g verifies a Nagumo-type condition in E = { ( x, y0 , y1 , y2 , y3 ) : γi ≤ yi ≤ Гi , i = 0, 1, 2 } ;

|| u′′′ ||∞ < ρ .Remark: • ρ depends only from hE , γ2, Г2 and s.

• the a priori bound does not depend on ( BC )

If

u ( 1) = u’ ( 1) = u’’ ( 0) = u’’’ (1) = 0 (BC)

u (4) = g (x, u, u’, u’’, u’’’) ( E )

• Lower and upper solutions

Method and techniques:

• Nagumo-type growth condition in E ⊂ [0,1]×R4

• Topological degree

(Invariance under homotopy)

u ( 1) = u’ ( 1) = u’’ ( 0) = u’’’ (1) = 0 (BC)

u (4) = g (x, u, u’, u’’, u’’’) ( E )

∃ u(x ) ∈ C 4 ([0,1]) solution of (E)-(BC) , ∃ N ∈ R + :

Thm 1 :

• g is continuous and verifies a Nagumo-type growth condition in

for ( x, y2 , y3 )∈[0,1]×R², α ≤ y0 ≤ β e β′ ≤ y1 ≤ α′ .

β′ (x) ≤ u′ (x) ≤ α′ (x) , -N ≤ u’’’ (x) ≤ N .

• g ( x, α, α′ , y2 , y3 ) ≥ g ( x, y0 , y1 , y2 , y3 ) ≥ g ( x, β, β′ , y2 , y3 )

α( i ) (x) ≤ u ( i ) (x) ≤ β ( i ) (x) , i = 0, 2 ,

then :

E = { ( x, y0 , y1 , y2 , y3 ) : α (i) ≤ yi ≤ β (i) , i = 0, 2, β’ ≤ y1 ≤ α’ } ;

If :

Remark: Only for s such that there are lower and upper solutions

• ∃ α, β lower and upper-solutions of (E) - (BC) : α′′ ≤ β′′ , in [0,1].

u ( 1) = u’ ( 1) = u’’ ( 0) = u’’’ (1) = 0 (BC)

u (4) = g (x, u, u’, u’’, u’’’) ( E )

u (4) + f (x, u, u’, u’’, u’’’) = s p (x) ( Es )

Existence and non-existence theorem on sThm 2:

• f : [0,1] ×R4 → R continuous function verifying Nagumo-type condition

• y0 ≥ z0 ⇒ f ( x , y0 , y1, y2 , y3 ) ≥ f ( x , z0 , z1 , z2 , z3 )

• y1 ≥ z1 , y2 ≥ z2 ⇒ f ( x , y0 , y1 , y2 , y3 ) ≤ f ( x , z0 , z1 , z 2 , z 3 )

, ,for , :0 , 1011 )(

)0 ,,1

,0

,(

)(

)0 ,0 ,0 ,0 ,(ryrysrRs

xp

ryyxf

xp

xf ≥−≤<<>∃∈∃−

•

∃ s0< s1 (possibly s0 = - ∞) :(i) for s < s0 , (Es) - (BC) has no solution

(ii) for s0< s ≤ s1 , (Es) - (BC) has a solution

then :

If :

u ( 1) = u’ ( 1) = u’’ ( 0) = u’’’ (1) = 0 (BC)

u (4) = g (x, u, u’, u’’, u’’’) ( E )

Step 1: ∃ s*< s1 : (Es*) - (BC) has a solution

u ( 1) = u’ ( 1) = u’’ ( 0) = u’’’ (1) = 0 (BC)

u (4) + f (x, u, u’, u’’, u’’’) = s p(x) ( Es )

Dem.:

Step 2: If ∃ σ ≤ s1 : (Eσ) - (BC) has a solution then (Es) - (BC) has a solution for σ ≤ s ≤ s1

Step 3: ∃ s0 : (i) for s < s0 , (Es) - (BC) has no solution

(ii) for s0< s ≤ s1 , (Es) - (BC) has a solution

u (4) + f (x, u, u’, u’’, u’’’) = s p (x) ( Es )

(BC)

Multiplicity result Def.:

(i) α is a strict lower-solution of (Es) - (BC) if :

α (4)+ f (x , α , α’, α’’, α’’’) > s p(x)

α ’’ ( 0) < 0 ; α’’’ (1) ≤ 0

(ii) β is a strict upper-solution of (Es) - (BC) if the reversed inequalities

hold

α ( 1) ≤ 0 ; α ’ ( 1) ≥ 0 ,

u ( 1) = u’ ( 1) = u’’ ( 0) = u’’’ (1) = 0

u (4) + f (x, u, u’, u’’, u’’’) = s p (x) ( Es )

Multiplicity result

L u = u (4)

Ns u = f (x, u, u’, u’’, u’’’) - s p(x)

Operators:

L : C 4 ([ 0,1]) × X → C ([ 0,1])

Ns : C 3([ 0,1]) × X → C ([ 0,1])

X = { y ∈C 3 ([0,1]) : y(1) = y’(1) = y’’(0) = y’’’(1) = 0 }

Lemma :

• ∃ α, β strict lower and upper-solutions of (Es) - (BC) : α′′ < β′′ , in [0,1]• f is continuous and verifies a Nagumo-type growth condition in E

then

If

Ω = { y ∈ C 3([0,1]) × X : α (i) ≤ yi ≤ β (i) , i = 0, 2, β’ ≤ y1 ≤ α’, || y′′′ ||∞ < ρ }

∃ ρ > 0 such that for

dL ( L + Ns , Ω ) = ± 1

(BC) u (1) = u’ (1) = u’’ ( 0) = u’’’ (1) = 0

)()( ,)()(for , ) ,,,,( 103210 xyxxyxyyyyxf αββα ′≤≤′≤≤↓↑

•

u (4) + f (x, u, u’, u’’, u’’’) = s p(x) ( Es )

Thm 3: If

• f : [0,1] ×R4 → R continuous function verifies Nagumo-type condition

• y0 ≥ z0 ⇒ f ( x , y0 , y1, y2 , y3 ) ≥ f (x , z0 , z1 , z2 , z3 )

• y1 ≥ z1 , y2 ≥ z2 ⇒ f ( x , y0 , y1 , y2 , y3 ) ≤ f (x , z0 , z1 , z2 , z3 )

ryrysrRstp

ryyxf

tp

xf ≥−≤<<>∃∈∃−

• 1011,for , :0 ,

)(

)0 ,,1

,0

,(

)(

)0 ,0 ,0 ,0 ,(

s0 is finite and (i) for s < s0 , (Es) - (BC) has no solution

(ii) for s = s0 , (Es) - (BC) has a solution .

then

• ∃ M ∈ R : every solution u of (Es) - (BC), for s ≤ s1 verifies u′′ < M , in [0,1]

• Assumptions of existence and non-existence theorem (Thm 2) hold

• ∃ m ∈ R : f ( x, y0 , y1 , y2 , y3 ) ≥ m p(x) in [0,1] × [- r, +∞[ 3 × R

Moreover if ∃ θ > 0, 0 ≤ η i ≤ 1 , i = 0,1 :

f ( x, y0 +η0 θ , y1 - η1 θ , y2 + θ , y3 ) ≤ f ( x, y0 , y1 , y2 , y3 ) in [0,1] × R4

(iii) for s ∈ ]s0, s1] , (Es) - (BC) has at least two solutions.then y0 ≥ z0 ⇒ f ( x , y0 , y1, y2 , y3 ) ≥ f (x , z0 , z1 , z2 , z3 )

y1 ≥ z1 , y2 ≥ z2 ⇒ f ( x , y0 , y1 , y2 , y3 ) ≤ f (x , z0 , z1 , z 2 , z 3 )

Growth speed

Multiplicity result (BC) u (1) = u’ (1) = u’’ ( 0) = u’’’ (1) = 0

Step 1: Every solution u(x) of (Es) - (BC), for s∈ ]s0, s1 ], verifies

-r < u’’(x)< M and -r < u (i) (x) < | M | , i = 0,1.

u ( 1) = u’ ( 1) = u’’ ( 0) = u’’’ (1) = 0 (BC)

u (4) + f (x, u, u’, u’’, u’’’) = s p(x) ( Es )

Dem.:

Step 2: The number s0 is finite.

Step 3: For s ∈ ]s0, s1 ] there is a second solution of (Es) - (BC).

Step 4: For s = s0 , (Es) - (BC) has a solution.

( Bs )

(BC)

Mechanical properties of the human spine byApplication :

y1 ( L /2) = 0 ; y1’ ( L /2) = 0

y1’’ ( - L /2) = 0 y1’’’ ( L /2) = 0;

a continuous cantilever beam model

Goal: analyze the overall deformation of the spine under various loading conditions (aircraft ejections, vehicle crash situations),

certain form of scoliosis

G. Noone and W.T.Ang, The inferior boundary condition of a continuouscantilever beam model of the human spine, Australian Physical & Engineering

Sciences in Medicine, Vol. 19 no 1 (1996) 26-30.

A. Patwardhan, W. Bunch, K. Meade, R. Vandeby and G. Knight, A biomechanical analog of curve progression and orthotic stabilization in

idiopathic scoliosis, J. Biomechanics, Vol 19, no 2 (1986) 103-117.

EI y1 (4)(x) - P y1’’(x) = s + Q ( y1’’’(x) ) + P y0’’ (x)

( Bs )

(BC)

Q : transverse load

y(x) : total lateral displacement of the beam-column, with length L

y0(x) : initial lateral displacement (known)

y1(x) : lateral displacement due to the axial and transverse loads

(unknown)

EI : is the flexural rigidity of the beam-column,

P : axial load

y(x) = y0(x) + y1(x)

EI y1 (4)(x) - P y1’’(x) = s + Q ( y1’’’(x) ) + P y0’’(x)

y1 ( L /2) = 0 ; y1’ ( L /2) = 0

y1’’ ( - L /2) = 0 ; y1’’’ ( L /2) = 0

( Bs )

(BC)

(a) Frontal X-ray picture of the flexible waist part

Th7-Th12, L1-L5 of the spine of a patient with

lumbar scoliosis, caused by a difference in leg length of

0.5 cmand standing in an upright muscle-relaxed position.

(b) Extracted contour picture of(a)

EI y1 (4)(x) - P y1’’(x) = s + Q ( y1’’’(x) ) + P y0’’(x)

y1 ( L /2) = 0 ; y1’ ( L /2) = 0

y1’’ ( - L /2) = 0 ; y1’’’ ( L /2) = 0

( Bs )

(BC)

f ( x , z0 , z1, z2, z3 ) = - ( P / EI ) ( z2 + y0(x) ) – Q(z3 ) / EI p(x) ≡ 1 /EI

Let a and b positive such that

0. 2 8

3)( 2 ≤+

++−

∞QPLEIab

Then

++−−−= 432

234

384

231

2624

1 )( LxLx

Lx

Lxbxβ

are, respectively, lower and upper solutions of ( Bs )-(BC) for s such that

( ) ( ) , )( )( )( )( 00 yPxPxQEIasyPxPxQEIb ′′−′′−′′′−≤≤′′−′′−′′′−− ααββ

and 384

153

2624

1 )( 432

234

−+−−= LxLx

Lx

Lxaxα

EI y1 (4)(x) - P y1’’(x) = s + Q ( y1’’’(x) ) + P y0’’(x)

y1 ( L /2) = 0 ; y1’ ( L /2) = 0

y1’’ ( - L /2) = 0 ; y1’’’ ( L /2) = 0

x ∈ [-L/2, L/2] .

( Bs )

(BC)

α (x) ≤ y1 (x) ≤ β (x)

For L = EI = P = 1,

α

β

∃ y1 (x) solution of ( Bs )-(BC) :

EI y1 (4)(x) - P y1’’(x) = s + Q ( y1’’’(x) ) + P y0’’(x)

y1 ( L /2) = 0 ; y1’ ( L /2) = 0

y1’’ ( - L /2) = 0 ; y1’’’ ( L /2) = 0

−+−−= 432

234

384

153

2624

1 )( LxLx

Lx

Lxaxα

++−−−= 432

234

384

231

2624

1 )( LxLx

Lx

Lxbxβ

00 |||| 8

11 ||||

8

11 yQasyQb ′′−−≤≤′′−+− ∞∞