Institute of Applied Analysis and Numerical Simulations On Cavitation in Elastodynamics Jan Giesselmann IANS, University of Stuttgart joint work with A. Tzavaras (University of Crete) This work was upported by the ACMAC project - European Union FP7 14th International Conference on Hyperbolic Problems June 25–29 2012 1 / 23
Institute of Applied Analysis and Numerical Simulations Outline Introduction 1D: Concept for discontinuous solutions to second order equations 1D: Energy of solutions 3D: Solution concept and energy Summary & Prospects 2 / 23
Institute of Applied Analysis and Numerical Simulations Introduction Study fracture and cavitation in elastical solids. Once a bar breaks, or a hole forms, coherence is lost and the hypothesis of continuum implicit in the elastodynamics equations is no longer valid. Transition region from a range of loading where the model is valid to a range where it loses validity. Explore this transition at the onset of fracture. Irregular solutions to the compressible, non-linear elastodynamics equations, where a hole forms in the interior. i.e. a discontinuity of the displacement field at one point. Study the energy of the irregular solution compared to trivial solutions (which also exist). 3 / 23
Institute of Applied Analysis and Numerical Simulations Introduction Equations of elastodynamics: Search displacements Ñ R d y : B 1 ♣ 0 q ✂ r 0 , T q Ý such that det ♣ ∇ y q → 0 . satisfying the wave equation y tt ✁ div ♣ τ ♣ ∇ y qq ✏ 0 (WAVE) with stress response τ ✏ ❇ W W : R d ✂ d ❇ F : R d ✂ d Ñ R d ✂ d Ý Ñ R ; Ý . � � � The wave equation is equivalent to the following system of first order conservation laws with u ✏ ∇ y , v ✏ y t : u t ✁ ∇ v ✏ 0 (CONS) v t ✁ div ♣ τ ♣ u qq ✏ 0 . 4 / 23
Institute of Applied Analysis and Numerical Simulations Energy and admissibility Smooth solutions of (WAVE) satisfy the energy equality d ✂ 1 ✡ 2 ♣ y t q 2 � W ♣ ∇ y q ✁ div ♣ τ ♣ ∇ y q y t q ✏ 0 . dt For weak solutions (having discontinuities in ∇ y , y t ) the classical admissibility criterion is the energy inequality d ✂ 1 ✡ 2 ♣ y t q 2 � W ♣ ∇ y q ✁ div ♣ τ ♣ ∇ y q y t q ↕ 0 , dt which has to be satisfied in a weak sense. 5 / 23
✏ ⑤ ⑤ ♣ q ✏ ♣ q ✏ ❅ P ♣ q ♣ q ✏ ❅ P ❇ ♣ q ↕ ➔ → ♣ q ✏ Institute of Applied Analysis and Numerical Simulations Solutions with cavitation We study solutions of the form ✁ ✠ ⑤ x ⑤ y ♣ x , t q ✏ t ϕ x (ANSATZ) t ⑤ x ⑤ for some ϕ : r 0 , ✽q Ñ r 0 , ✽q , with ϕ ♣ 0 q → 0 , ϕ ✶ ♣ s q → 0 ❅ s → 0 . 6 / 23
♣ q ✏ ♣ q ✏ ❅ P ♣ q ♣ q ✏ ❅ P ❇ ♣ q ↕ ➔ → ♣ q ✏ Institute of Applied Analysis and Numerical Simulations Solutions with cavitation We study solutions of the form ✁ ✠ ⑤ x ⑤ y ♣ x , t q ✏ t ϕ x (ANSATZ) t ⑤ x ⑤ for some ϕ : r 0 , ✽q Ñ r 0 , ✽q , with ϕ ♣ 0 q → 0 , ϕ ✶ ♣ s q → 0 ❅ s → 0 . For solutions having the form (ANSATZ) u , v only depend on s : ✏ ⑤ x ⑤ t . 6 / 23
♣ q ✏ ♣ q ✏ ❅ P ♣ q ♣ q ✏ ❅ P ❇ ♣ q ↕ ➔ → ♣ q ✏ Institute of Applied Analysis and Numerical Simulations Solutions with cavitation We study solutions of the form ✁ ✠ ⑤ x ⑤ y ♣ x , t q ✏ t ϕ x (ANSATZ) t ⑤ x ⑤ for some ϕ : r 0 , ✽q Ñ r 0 , ✽q , with ϕ ♣ 0 q → 0 , ϕ ✶ ♣ s q → 0 ❅ s → 0 . For solutions having the form (ANSATZ) u , v only depend on s : ✏ ⑤ x ⑤ t . In 1D: these solutions describe discontinuous shear or fracture, in 3D: such solutions describe opening holes/cavities. 6 / 23
♣ q ✏ Institute of Applied Analysis and Numerical Simulations Solutions with cavitation We study solutions of the form ✁ ✠ ⑤ x ⑤ y ♣ x , t q ✏ t ϕ x (ANSATZ) t ⑤ x ⑤ for some ϕ : r 0 , ✽q Ñ r 0 , ✽q , with ϕ ♣ 0 q → 0 , ϕ ✶ ♣ s q → 0 ❅ s → 0 . For solutions having the form (ANSATZ) u , v only depend on s : ✏ ⑤ x ⑤ t . In 1D: these solutions describe discontinuous shear or fracture, in 3D: such solutions describe opening holes/cavities. Displacement initial boundary value problem y ♣ x , 0 q ✏ λ x , y t ♣ x , 0 q ✏ 0 ❅ x P B 1 ♣ 0 q , y ♣ x , t q ✏ λ x ❅ x P ❇ B 1 ♣ 0 q and 0 ↕ t ➔ T for some λ → 0 . 6 / 23
Institute of Applied Analysis and Numerical Simulations Solutions with cavitation We study solutions of the form ✁ ✠ ⑤ x ⑤ y ♣ x , t q ✏ t ϕ x (ANSATZ) t ⑤ x ⑤ for some ϕ : r 0 , ✽q Ñ r 0 , ✽q , with ϕ ♣ 0 q → 0 , ϕ ✶ ♣ s q → 0 ❅ s → 0 . For solutions having the form (ANSATZ) u , v only depend on s : ✏ ⑤ x ⑤ t . In 1D: these solutions describe discontinuous shear or fracture, in 3D: such solutions describe opening holes/cavities. Displacement initial boundary value problem y ♣ x , 0 q ✏ λ x , y t ♣ x , 0 q ✏ 0 ❅ x P B 1 ♣ 0 q , y ♣ x , t q ✏ λ x ❅ x P ❇ B 1 ♣ 0 q and 0 ↕ t ➔ T for some λ → 0 . Note: There is the trivial solution y ♣ x , t q ✏ λ x to this IBVP. 6 / 23
Institute of Applied Analysis and Numerical Simulations Situation in 1D Ansatz ✁ x ✠ with Y ♣✁ ξ q ✏ ✁ Y ♣ ξ q , Y ✶ ♣ ξ q → 0 ❅ ξ → 0 , y ♣ x , t q ✏ tY ξ Ñ 0 ,ξ → 0 Y ♣ ξ q → 0 . lim t Then (WAVE) amounts to ξ 2 Y ✷ ✏ ♣ τ ♣ Y ✶ qq ✶ τ W u u We impose the conditions W ✷ → 0 and W ✸ ➔ 0 . 7 / 23
Institute of Applied Analysis and Numerical Simulations Rankine Hugoniot conditions and admissibility y t ♣ x , t q ✏ Y ♣ ξ q ✁ ξ Y ✶ ♣ ξ q ✏ : V ♣ ξ q , where ξ ✏ x y x ♣ x , t q ✏ Y ✶ ♣ ξ q ✏ : U ♣ ξ q t We can write down the equations for U , V ξ U ✶ � V ✶ ✏ 0 ♣✝q ξ V ✶ � ♣ τ ♣ U qq ✶ ✏ 0 For a shock with speed σ the Rankine–Hugoniot conditions read ❞ r τ ♣ U qs ✁ σ r U s ✏ r V s and ✁ σ r V s ✏ r τ ♣ U qs ù ñ σ ✏ r U s . ❛ The eigenvalues of ♣✝q are ✟ τ ✶ ♣ U q and the Lax condition becomes U l → U r for 1 ✁ shocks and U l ➔ U r for 2 ✁ shocks . 8 / 23
Institute of Applied Analysis and Numerical Simulations One class of possible solutions Thus, we investigate ✩ Y ♣ 0 q � αξ : 0 ➔ ξ ➔ σ ✫ Y ♣ ξ q : ✏ ✁ Y ♣ 0 q � αξ : ✁ σ ➔ ξ ➔ 0 (1D-ANSATZ) λξ : ⑤ ξ ⑤ → σ ✪ Y λ t x = − σt x = σt α − tY (0) + αx tY (0) + αx Y (0) ξ α λx λx x λ 9 / 23
Institute of Applied Analysis and Numerical Simulations One class of possible solutions U ✏ 2 Y ♣ 0 q δ ξ ✏ 0 � αχ t⑤ ξ ⑤➔ σ ✉ � λχ t⑤ ξ ⑤→ σ ✉ V ✏ Y ♣ 0 q χ t 0 ➔ ξ ➔ σ ✉ ✁ Y ♣ 0 q χ t✁ σ ➔ ξ ➔ 0 ✉ V U Y (0) λ λ α α − σ ξ σ − Y (0) ξ − σ σ 10 / 23
Institute of Applied Analysis and Numerical Simulations One class of possible solutions U ✏ 2 Y ♣ 0 q δ ξ ✏ 0 � αχ t⑤ ξ ⑤➔ σ ✉ � λχ t⑤ ξ ⑤→ σ ✉ V ✏ Y ♣ 0 q χ t 0 ➔ ξ ➔ σ ✉ ✁ Y ♣ 0 q χ t✁ σ ➔ ξ ➔ 0 ✉ V U Y (0) λ λ α α − σ ξ σ − Y (0) ξ − σ σ What is the meaning of τ ♣ U q in this case? 10 / 23
Institute of Applied Analysis and Numerical Simulations slic–solutions We propose the following notion of solution Definition We call y P C ♣r 0 , T q , L p ♣ B 1 ♣ 0 qq a s ingular l imiting i nduced from c ontinuum (slic)–solution provided for all ψ P C ✽ 0 ♣ R q such that ➺ 1 supp ♣ ψ q ⑨ r✁ 1 , 1 s , ψ ♣ x q dx ✏ 1 , and ψ ♣ x q ✏ ψ ♣✁ x q ✁ 1 the following holds in D ✶ n Ñ✽ r y n tt ✁ ♣ τ ♣ y n x qq x s ✏ 0 lim where y n ♣ x , t q ✏ y ✝ x n ψ ♣ n ☎q . 11 / 23
Institute of Applied Analysis and Numerical Simulations δ -shocks This problem is conected to the notion of δ -shocks for hyperbolic conservation laws. There is a vast literature on this field, see the work of Danilov, Shelkovic. In contrast, our notion of solution is based on the underlying structure of the second order problem. For the p-system in fluid mechanics we have lim U Ñ✽ p ♣ U q ✏ 0 such that ♣ p ♣ δ x ✏ 0 qq♣ 0 q ✏ 0 seems reasonable. See chapter 9.7 in Dafermos’ book. 12 / 23
Institute of Applied Analysis and Numerical Simulations Are there slic–solutions? Lemma Let y P H 1 ♣r 0 , T s , L 2 ♣ B 1 ♣ 0 qqq ❳ L 2 ♣r 0 , T s , H 1 ♣ B 1 ♣ 0 qqq with essinf y x → 0 satisfy ➺ T ➺ ❅ ϕ P C 1 y t ϕ t ✁ τ ♣ y x q ϕ x dxdt ✏ 0 0 ♣♣ 0 , T q ✂ B 1 ♣ 0 qq , 0 B 1 ♣ 0 qq then y is a slic–solution. Thus, slic–solutions generalize standard weak solutions. Lemma A function y given by (1D-ANSATZ) with Y ♣ 0 q � ασ ✏ λσ is a slic–solution if and only if Y ♣ 0 q ✏ τ ♣ λ q ✁ τ ♣ α q τ ♣ u q and lim ✏ 0 . σ u u Ñ✽ 13 / 23
Institute of Applied Analysis and Numerical Simulations Energy of slic–solutions We define the energy at time t of a slic–solution y to be ➺ 1 t ♣ x , t qq 2 � W ♣ y n 2 ♣ y n E ♣ y , t q : ✏ lim x ♣ x , t qq dx . n Ñ✽ B 1 ♣ 0 q Because of l’Hospitals Theorem it holds W ♣ u q ✏ lim u Ñ✽ τ ♣ u q . lim u Ñ✽ u Lemma In case lim u Ñ✽ τ ♣ u q ✏ ✽ a slic–solution y given by (1D-ANSATZ) satisfies E ♣ y , t q ✏ ✽ for all t → 0 and E ♣ y , 0 q ➔ ✽ . 14 / 23
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