Revisit Euler Buckling problem joint with Atia Afroz Toshizumi - PowerPoint PPT Presentation

Revisit Euler Buckling problem joint with Atia Afroz Toshizumi Fukui (Saitama University) 15 September 2018, fjv 2018 (Nha Trang)

Buckling of rod (with pinned ends)

Buckling of rod (with pinned ends) λ

Buckling of rod (with pinned ends) λ λ

Buckling of rod (with pinned ends) λ λ λ

Buckling of rod (with pinned ends) λ λ λ λ

Euler buckling When λ < λ ∗ , nothing happens.

Euler buckling When λ < λ ∗ , nothing happens. When λ > λ ∗ , we have buckling.

Euler buckling When λ < λ ∗ , nothing happens. When λ > λ ∗ , we have buckling. where λ ∗ is Euler’s critical load.

Euler buckling When λ < λ ∗ , nothing happens. When λ > λ ∗ , we have buckling. where λ ∗ is Euler’s critical load. θ ( x ( s ) , u ( s )) s λ s : arc length parameter l : the length of the rod θ : angle κ : curvature

Energy formulation θ ( x ( s ) , u ( s )) s λ Minimize energy E = S + λ T where ∫ l S = 1 κ 2 ds strain energy , 2 0 T = x ( s ) potential energy

Variational problem ▶ Minimize the energy E = S + λ T

Variational problem ▶ Minimize the energy E = S + λ T ▶ Investigate zero of X → X ∗ , u �→ [ ϕ �→ D φ E ] where X is some Banach space, X ∗ dual space of X , and 1 ( D φ E ) u = lim t ( E | u + t φ − E | u ) t → 0

Variational problem ▶ Minimize the energy E = S + λ T ▶ Investigate zero of X → X ∗ , u �→ [ ϕ �→ D φ E ] where X is some Banach space, X ∗ dual space of X , and 1 ( D φ E ) u = lim t ( E | u + t φ − E | u ) t → 0 ▶ We would like to know the bifurcation of the zero set above.

Let us start compute θ ( x ( s ) , u ( s )) s λ ▶ s : arc length parameter, x ′ = cos θ , u ′ = sin θ

Let us start compute θ ( x ( s ) , u ( s )) s λ ▶ s : arc length parameter, x ′ = cos θ , u ′ = sin θ ▶ κ : curvature u ′′ κ = d θ ds = d ds sin − 1 u ′ = (1 − ( u ′ ) 2 ) 1 / 2 ▶ T : potential energy ∫ l ∫ l ∫ l (1 − ( u ′ ) 2 ) 1 / 2 ds T = x ( s ) = x ′ ds = cos θ ds = 0 0 0

Variational formulation ▶ Minimize E ∫ l ∫ l E = 1 u ′′ (1 − ( u ′ ) 2 ) 1 / 2 ds (1 − ( u ′ ) 2 ) 1 / 2 ds + λ 2 0 0

Variational formulation ▶ Minimize E ∫ l ∫ l E = 1 u ′′ (1 − ( u ′ ) 2 ) 1 / 2 ds (1 − ( u ′ ) 2 ) 1 / 2 ds + λ 2 0 0 ▶ on X X = { u ∈ H 2 [0 , l ] : u (0) = u ( l ) = 0 }

Variational formulation ▶ Minimize E ∫ l ∫ l E = 1 u ′′ (1 − ( u ′ ) 2 ) 1 / 2 ds (1 − ( u ′ ) 2 ) 1 / 2 ds + λ 2 0 0 ▶ on X X = { u ∈ H 2 [0 , l ] : u (0) = u ( l ) = 0 } ▶ where H 2 [0 , l ] is Sobolev space on [0 , l ]

Variational formulation ▶ Minimize E ∫ l ∫ l E = 1 u ′′ (1 − ( u ′ ) 2 ) 1 / 2 ds (1 − ( u ′ ) 2 ) 1 / 2 ds + λ 2 0 0 ▶ on X X = { u ∈ H 2 [0 , l ] : u (0) = u ( l ) = 0 } ▶ where H 2 [0 , l ] is Sobolev space on [0 , l ] ▶ First variation formula: ∫ l u ′ ( u ′′ ) 2 u ′′ ϕ ′′ λ u ′ [ ( ) ϕ ′ ] ( D φ E ) u = 1 − ( u ′ ) 2 + (1 − ( u ′ ) 2 ) 2 − ds (1 − ( u ′ ) 2 ) 1 / 2 0

Variational formulation ▶ Minimize E ∫ l ∫ l E = 1 u ′′ (1 − ( u ′ ) 2 ) 1 / 2 ds (1 − ( u ′ ) 2 ) 1 / 2 ds + λ 2 0 0 ▶ on X X = { u ∈ H 2 [0 , l ] : u (0) = u ( l ) = 0 } ▶ where H 2 [0 , l ] is Sobolev space on [0 , l ] ▶ First variation formula: ∫ l u ′ ( u ′′ ) 2 u ′′ ϕ ′′ λ u ′ [ ( ) ϕ ′ ] ( D φ E ) u = 1 − ( u ′ ) 2 + (1 − ( u ′ ) 2 ) 2 − ds (1 − ( u ′ ) 2 ) 1 / 2 0 ▶ Investigate the bifurcation of zero of Φ : X × R → X ∗ , Φ( u , λ ) = [ ϕ �→ ( D φ E ) u ]

Sobolev space W k , p [0 , l ] F [0 , l ] = { f : [0 , l ] → R } / ∼ a.e. . f ∼ a.e. g means f and g coincide except measure zero set. W k , p [0 , l ] = { u ∈ F [0 , l ] : ∥ u ∥ k , p < ∞} ( k ) 1 ( k ) 2 , ∑ ∥ D i u ∥ 2 ∥ u ∥ k , p = Sobolev norm p i i =0 {(∫ l ) 1 / p , 0 | u | p ds 1 ≤ p < ∞ , ∥ u ∥ p = sup {| u ( s ) | : s ∈ [0 , l ] } , p = ∞ , where D i u denote the i th order distributional derivatives of u . H k [0 , l ] = W k , 2 [0 , l ] which is a Hilbert space.

Let us see the bifurcation from trivial solution ▶ u = 0 is a solution Φ(0 , λ ) = 0

Let us see the bifurcation from trivial solution ▶ u = 0 is a solution Φ(0 , λ ) = 0 ▶ Set u = ∑ ∞ m =1 y m u m , y m ∈ R , 1 sin m π s u m = , m = 1 , 2 , 3 , . . . √ l l / 2

Let us see the bifurcation from trivial solution ▶ u = 0 is a solution Φ(0 , λ ) = 0 ▶ Set u = ∑ ∞ m =1 y m u m , y m ∈ R , 1 sin m π s u m = , m = 1 , 2 , 3 , . . . √ l l / 2 ▶ Consider ∞ ∑ Φ( y 1 , y 2 , . . . , λ ) = Φ( y m u m , λ ) m =1

▶ By computation, ( π 2 m 2 ) π 2 m 2 ∂ Φ u ∗ (0 , λ ) = − λ m ∂ y m l 2 l 2 Setting Φ = Φ 1 u ∗ 1 + Φ 2 u ∗ 2 + · · · , ∂ Φ1   ∂ y 1 ∂ Φ2   ∂ y 2 J Φ =   ∂ Φ3   ∂ y 3   ...

▶ By computation, ( π 2 m 2 ) π 2 m 2 ∂ Φ u ∗ (0 , λ ) = − λ m ∂ y m l 2 l 2 Setting Φ = Φ 1 u ∗ 1 + Φ 2 u ∗ 2 + · · · , ∂ Φ1   ∂ y 1 ∂ Φ2   ∂ y 2 J Φ =   ∂ Φ3   ∂ y 3   ... ▶ Inverse mapping theorem implies that u = 0 is only solution near 0 when λ ̸ = π 2 m 2 l 2

▶ By computation, ( π 2 m 2 ) π 2 m 2 ∂ Φ u ∗ (0 , λ ) = − λ m ∂ y m l 2 l 2 Setting Φ = Φ 1 u ∗ 1 + Φ 2 u ∗ 2 + · · · , ∂ Φ1   ∂ y 1 ∂ Φ2   ∂ y 2 J Φ =   ∂ Φ3   ∂ y 3   ... ▶ Inverse mapping theorem implies that u = 0 is only solution near 0 when λ ̸ = π 2 m 2 l 2 ▶ When λ = π 2 m 2 l 2 , we have pitchfolk bifurcation at u = 0.

Lyapunov-Schmidt reduction ▶ When λ = λ ∗ = π 2 n 2 l 2 , the matrix ∂ Φ1   ∂ y 1 ∂ Φ2   ∂ y 2 J Φ =   ∂ Φ3   ∂ y 3   ... is of corank 1.

Lyapunov-Schmidt reduction ▶ When λ = λ ∗ = π 2 n 2 l 2 , the matrix ∂ Φ1   ∂ y 1 ∂ Φ2   ∂ y 2 J Φ =   ∂ Φ3   ∂ y 3   ... is of corank 1. ▶ So Φ( y 1 , y 2 , . . . , λ ) = 0 defines y m ( m ̸ = n ) as functions of y n and λ by implicit function theorem.

Lyapunov-Schmidt reduction ▶ When λ = λ ∗ = π 2 n 2 l 2 , the matrix ∂ Φ1   ∂ y 1 ∂ Φ2   ∂ y 2 J Φ =   ∂ Φ3   ∂ y 3   ... is of corank 1. ▶ So Φ( y 1 , y 2 , . . . , λ ) = 0 defines y m ( m ̸ = n ) as functions of y n and λ by implicit function theorem. ▶ Now problem is of finite dimension and we conclude pitchfolk bifurcation at u = 0.

Pitchfolk bifurcarion F ( x , λ ) = x 3 − λ x = 0 ✲ λ F (0 , 0) = F x (0 , 0) = F xx (0 , 0) = F λ (0 , 0) = 0, F xxx (0 , 0) ̸ = 0, F x λ (0 , 0) ̸ = 0

Bifurcation set B and hysteresis set H F ( x , λ, α ) = 0 B = { α : ∃ ( x , λ ) F ( x , λ, α ) = 0 , F x ( x , λ, α ) = F λ ( x , λ, α ) = 0 } , H = { α : ∃ ( x , λ ) F ( x , λ, α ) = 0 , F x ( x , λ, α ) = F xx ( x , λ, α ) = 0 } α ∈ B α ∈ B transcritical pitchfolk bifurcation bifurcation ✲ λ ✲ λ ✲ λ α ̸∈ H α ∈ H α ̸∈ H

Modified problem (due to Golubitsky and Schaefer) α 2 θ ( x ( s ) , u ( s )) s λ ▶ Minimize E = S + λ T + α 2 u where ∫ l S = 1 ( κ − α 1 ) 2 ds 2 0

Modified problem (due to Golubitsky and Schaefer) α 2 θ ( x ( s ) , u ( s )) s λ ▶ Minimize E = S + λ T + α 2 u where ∫ l S = 1 ( κ − α 1 ) 2 ds 2 0 ▶ Φ( u , λ, α 1 , α 2 ) = [ ϕ �→ ( D φ E ) ( u ,λ,α ) ]

Modified problem (due to Golubitsky and Schaefer) α 2 θ ( x ( s ) , u ( s )) s λ ▶ Minimize E = S + λ T + α 2 u where ∫ l S = 1 ( κ − α 1 ) 2 ds 2 0 ▶ Φ( u , λ, α 1 , α 2 ) = [ ϕ �→ ( D φ E ) ( u ,λ,α ) ] ▶ Apply Lyapunov-Schmidt reduction, and find W so that ⟨ Φ( xu 1 + W ( x , λ, α )) , u m ⟩ = 0 , m = 2 , 3 , . . .

Modified problem (due to Golubitsky and Schaefer) α 2 θ ( x ( s ) , u ( s )) s λ ▶ Minimize E = S + λ T + α 2 u where ∫ l S = 1 ( κ − α 1 ) 2 ds 2 0 ▶ Φ( u , λ, α 1 , α 2 ) = [ ϕ �→ ( D φ E ) ( u ,λ,α ) ] ▶ Apply Lyapunov-Schmidt reduction, and find W so that ⟨ Φ( xu 1 + W ( x , λ, α )) , u m ⟩ = 0 , m = 2 , 3 , . . . ▶ F ( x , λ, α ) = ⟨ Φ( xu 1 + W , λ, α ) , u 1 ⟩ is p- K -versal unfolding of pitchfolk bifurcation.

Example of p- K -versal unfolding F ( x , λ, α 1 , α 2 ) = x 3 − λ x + α 1 x 2 + α 2 H = { α 3 B = { α 2 = 0 } , 1 = 27 α 2 } The bifurcation diagrams of the zeros of f α ( x , λ ) = F ( x , λ, α ) are shown as follows: α 2 ✻ H ❄ ✲ ■ ✻ α 1 B

We consider Minimize E = S + λ T + α 2 u on X = H 2 [0 . l ] where ∫ l S = 1 ( κ − α 1 κ ) 2 ds 2 0 [ ] 1 i =1 a i cos 2 i π s √ a 0 + ∑ ∞ where κ = . l l / 2