5icdsa2007 v4
TRANSCRIPT
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 ′′−−≤≤′′−+− ∞∞