Shallow water approximations for water waves over a moving bottom. Tatsuo Iguchi Department of Mathematics, Keio University

Transcription

1 Shallow water approximations for water waves over a moving bottom Tatsuo Iguchi Department of Mathematics, Keio University 1

2 Tsunami generation and simulation Submarine earthquake occures with a seabed deformation Seabed deformation causes a deformation of water surface Tsunami propagates due to the gravity as a restorative force Engineers simulate the tsunami by solving numerically the following shallow water equations ηt + ((h + η b 1 )u ) =0, under the initial conditions u t +(u )u + g η =0 η t=0 = b 1 b 0, u t=0 =0. 2

3 x 3 Γ(t) : water surface f = ge3 Ω(t) : water region Σ(t) : seabed x =(x 1,x 2 ) Ω(t) =X =(x, x 3 ) R 3 ; b(x, t) <x 3 <h+ η(x, t)}, Γ(t) =X =(x, x 3 ) R 3 ; x 3 = h + η(x, t)}, Σ(t) =X =(x, x 3 ) R 3 ; x 3 = b(x, t)}. 3

4 Basic equations (1) (2) (3) (4) where and X Φ = 0 in Ω(t), t > 0, Φt XΦ 2 + gη =0, η t + Φ η 3 Φ=0 on Γ(t), t > 0, b t + Φ b 3 Φ=0 on Σ(t), t > 0, η(, 0) = η0 on R 2, Φ(, 0) = Φ 0 in Ω(0), Φ : velocity potential, i.e., v = X Φ : velocity = , 2 X = , 2 =( 1, 2 ), X =( 1, 2, 3 ). 4

5 Seabed deformation We assume that the seabed deforms only for time interval [0,t 0 ], so that the function b can be written in the form b(x, t) =β(x, t/t 0 ), where β(x, τ) = b0 (x) for τ 0, b 1 (x) for τ 1. A simple example of the function β is the following. b 0 (x) for τ 0, β(x, τ) = b 0 (x)+τ ( b 1 (x) b 0 (x) ) for 0 <τ<1, b 1 (x) for τ 1. 5

6 Non-dimensional form λ h δ := h λ, ε := t 0 λ/, δ,ε +0 gh x = λ x, x 3 = h x 3, t = λ gh t, Φ=λ gh Φ, η = h η, b = h b. 6

7 (1 ) δ 2 Φ+ 2 3Φ=0 in Ω(t), t > 0, (2 ) ( δ 2 Φ t Φ 2 + η ) ( 3Φ) 2 =0, δ 2( η t + Φ η ) 3 Φ=0 on Γ(t), t > 0, (3 ) 3 Φ δ 2 b Φ=(δ 2 /ε)β τ on Σ(t), t > 0, (4 ) where η(, 0) = η0 on R 2, Φ(, 0) = Φ 0 in Ω(0), Ω(t) : b(x, t) <x 3 < 1+η(x, t), Γ(t) : x 3 =1+η(x, t), Σ(t) : x 3 = b(x, t) β(x, t/ε). 7

8 Reformulation We introduce a new unknown function φ = φ(x, t) by φ(x, t) :=Φ(x, 1+η(x, t),t), which is the trace of the velocity potential on the free surface. Then, it holds that (5) φ t + η φ δ2 (1 + δ 2 η 2 ) 1( Λ DN φ + ε 1 Λ NN β τ + η φ ) 2 =0, η t Λ DN φ ε 1 Λ NN β τ = 0 for t>0, (6) η = η 0, φ = φ 0 at t =0, where Λ DN =Λ DN (η, b, δ) and Λ NN =Λ NN (η, b, δ) are linear operators depending on η, b, and δ. 8

9 The maps Λ DN and Λ NN Under suitable assumptions on η and b, for any function φ on the free surface in some class there exists a unique solution Φ of the boundary value problem (7) Φ+δ 2 2 3Φ = 0 in Ω(t), Φ=φ on Γ(t), δ 2 3 Φ b Φ=β on Σ(t). Using the solution Φ we define linear operators Λ DN =Λ DN (η, b, δ) and Λ NN =Λ NN (η, b, δ) by Namely, Λ DN (η, b, δ)φ +Λ NN (η, b, δ)β =(δ 2 3 Φ η Φ) Γ(t). Λ DN : Dirichlet data on Γ(t) Neumann data on Γ(t), Λ NN : Neumann data on Σ(t) Neumann data on Γ(t). 9

10 Letting δ 0 we obtain ηt + ((1 + η b) φ ) =0, A shallow water approximation (β τ 0) Lemma 1. M,c > 0 s >1 C, δ 1 > 0 s.t. η, b H s+3+1/2, (η, b) s+3+1/2 M, 1+η(x) b(x) c for x R 2, 0 <δ δ 1 = Λ DN (η, b, δ)φ + ((1 + η b) φ) s Cδ 2 φ s+3. Therefore, it follows from (5) that ηt + ((1 + η b) φ ) = O(δ 2 ), φ t + η φ 2 = O(δ 2 ). φ t + η φ 2 =0. Putting u := φ and taking the gradient of the second equation, we obtain ηt + ((1 + η b)u ) =0, u t +(u )u + η =0. 10

11 Known results (the shallow water limit) L. V. Ovsjannikov (1974) Analytic framework, periodic boundary condition T. Kano and T. Nishida (1979) Analytic framework Y.A. Li (2006) Sobolev spaces, 2-dimension, use of conformal mapping, Green Naghdi equations ηt + ( (1 + η)u ) x =0, u t + uu x + η x = 1 3 δ2 (1 + η) 1( (1 + η) 3 (u xt + uu xx u 2 x) ) x B. A.-Samaniego and D. Lannes (2008) Sobolev spaces, 3-dimension T. I. (2009) Sobolev spaces, (n + 1)-dimension 11

12 A shallow water approximation (β τ 0) Lemma 2. M,c > 0 s >1 C, δ 1 > 0 s.t. η, b H s+3, (η, b) s+3 M, 1+η(x) b(x) c for x R 2, 0 <δ δ 1 = Λ NN (η, b, δ)β β s Cδ 2 β s+2. Therefore, it follows from (5) that η t = 1 ε β τ + 1 ε O(ε + δ2 ), φ t = 1 δ 2 2 ε 2β2 τ δ 4 ). ε 2O(ε2 By resolving these equations under the initial conditions (6), we obtain η(x, t) =η 0 (x)+β(x, t/ε) b 0 (x)+o(ε + δ 2 ), φ(x, t) =φ 0 (x)+ 1 δ 2 (8) t/ε β τ (x, τ) 2 dτ ε ε O(ε2 + δ 4 ) for the time interval 0 t ε. Particularly, we get 12 0

13 η(x, ε) =η 0 (x)+ ( b 1 (x) b 0 (x) ) + O(ε + δ 2 ), φ(x, ε) =φ 0 (x)+ 1 δ 2 1 β τ (x, τ) 2 dτ ε ε O(ε2 + δ 4 ). 0 As δ, ε +0 these data converge only in the case when δ 2 /ε also converges to some value σ. In view of this fact, we will consider the limit (9) δ, ε +0, Particularly, we can always assume that δ 2 ε σ. δ 2 ε. On the other hand, for t>εwe have ηt + ((1 + η b 1 )u ) = O(δ 2 ), u t +(u )u + η = O(δ 2 ). Therefore, taking the limit (9) of these equations we arrive at the following model problem for tsunami. 13

14 The case δ 2 ε +0 ηt + ((1 + η b 1 )u ) =0, u t +(u )u + η = 0 for t>0 η(, +0) = η0 +(b 1 b 0 ), u(, +0) = φ 0. The case δ 2 ε +0 ηt + ((1 + η b 1 )u ) =0, u t +(u )u + η = 0 for t>0 η(, +0) = η 0 +(b 1 b 0 ), ( u(, +0) = φ 0 + σ 1 β τ (,τ) 2 dτ 2 0 ). 14

15 Meiji-Sanriku earthquake Date: June 15, Seismic scale: 2 3 Duration: several minutes Magnitude: M t = The assumption of the previous tsunami model is not satisfied and it might be better to consider the limit δ +0, ε 1. 15

16 The case ε =1 In this case the full problem has the form φ t + η φ δ2 (1 + δ 2 η 2 ) 1( Λ DN φ +Λ NN b t + η φ ) 2 (10) =0, η t Λ DN φ Λ NN b t = 0 for t>0. By Lemmas 1 and 2, we have φt + η φ 2 = O(δ 2 ), η t + ((1 + η b) φ ) = b t + O(δ 2 ). Putting u = φ and letting δ 0 we obtain ηt + ((1 + η b)u ) = b t, u t +(u )u + η =0 under the zero initial condition. 16

17 A higher order approximation (Green Nagdhi equations) For simplicity, we assume that b(x, t) 0 and consider the linearized equations of (10) around the trivial flow. Since Λ DN D tanh(δ D ) (0, 0,δ)=, δ the linearized equations have the form φt + η =0, (11) D tanh(δ D ) η t δ φ =0. Using the Taylor expansion we have D tanh(δ D ) δ tanh x = x + O(x 3 ) (x 0), = D 2 + O(δ 2 )= Δ+O(δ 2 ), so that ηt +Δφ = O(δ 2 ), φ t + η =0. Letting δ 0 we obtain a linear shallow water equations. 17

18 To obtain a higher oder approximation, we use the Taylor expansion Then, we have tanh x = x 1 3 x3 + O(x 5 ) (x 0). D tanh(δ D ) δ = D δ2 D 4 + O(δ 4 ) = Δ 1 3 δ2 Δ 2 + O(δ 4 ). Neglecting the term O(δ 4 ), we obtain ηt +Δφ δ2 Δ 2 φ =0, (12) φ t + η =0. This system has non-trivial solution of the form η(x, t) =η 0 e i(ξ x ωt), if ξ R 2 and ω C satisfy φ(x, t) =φ 0 e i(ξ x ωt) ω 2 ( δ2 ξ 2 ) ξ 2 = 0 (Dispersion relation for (12)). Therefore, the initial value problem for (12) is in general ill-posed, and (12) is not good approximation for the linearized equations (11). 18

19 On the other hand, in view of the relation D tanh(δ D ) δ φ = Δ ( φ δ2 Δφ)+O(δ 4 ), let us introduce a new variable ψ satisfying the relation φ δ2 Δφ = ψ + O(δ 4 ). This implies that φ = ψ + O(δ 2 ), so that φ = ψ 1 3 δ2 Δφ + O(δ 4 )=(1 1 3 δ2 Δ)ψ + O(δ 4 ). This motivates us to introduce a new variable ψ by ψ = ( δ2 Δ ) 1 φ. Then, it follows from (11) that ηt +Δψ = O(δ 4 ), ψ t + η = 1 3 δ2 Δψ t. Putting u = ψ and neglecting the term O(δ 4 ), we obtain ηt + u =0, (13) u t + η = 1 3 δ2 Δu t. Corresponding nonlinear equations are called the Green Nagdhi equations. 19

20 The dispersion relation for (13) is ( δ2 ξ 2) ω 2 ξ 2 =0, so that the initial value problem for (11) is well-posed. Therefore, we can expect that the solution for (11) can be approximated by the solution (13) up to order O(δ 4 ). Remark. If we do not use the Taylor expansion tanh x = x 1 3 x3 + O(x 5 ) (x 0) but the Padé approximation x tanh x = O(x5 ) (x 0), 3 x2 we can directly obtain a linearized Green Nagdhi equations (13) from the linearized water wave equations (11). 20

21 A higher order approximation (nonlinear version) Lemma 3. Let n =2. M,c > 0 s >1 C, δ 1 > 0 s.t. η, b H s+5+1/2, (η, b) s+5+1/2 M, 1+η(x) b(x) c for x R 2, 0 <δ δ 1 = Λ DN (η, b, δ)φ + ((1 + η b) φ ) + δ (T 2 (η, b) φ ) s Cδ 4 φ s+5+1/2, Λ NN (η, b, δ)β β δ (1 2 + η b) ( 1 2 (1 + η b) β +( η)β)} s Cδ 4 β s+4, where T = T (η, b) is a differential operator acting on vector fields defined by T (η, b)u = 1 3 ( (1 + η b) 3 u ) ( (1 + η b) 2 b u ) 1 2 (1 + η b)2 ( b)( u) + (1 + η b)( b)( b u). 21

22 Introducing new variables u by ( δ 2 ) φ = 1+ T (η, b) u + δ 2( b t η η b 2 (1 + η b) b ) t and neglecting the terms of O(δ 4 ), we can derive from the full water wave equations (5) the following Green Nagdhi equations (14) η t + ((1 + η b)u ) = b t, ( (1 + η b)+δ 2 T (η, b) ) u t +(1+η b) η +(1+η b)(u )u +δ 2 ( 1 3 (1 + η b)3( ( u) (u ) ) ( u) ) ( (1 + η b) 2 (u ) 2 b ) (1 + η b)2( ( u) (u ) ) ( u) b +(1 + η b) ( (u ) 2 b ) b +(1 + η b) 2 (u b t ) + 2(1 + η b)(u b t ) η (1 + η b)2 b tt +(1+η b)b tt η } =0. 22

23 Theorem 1. Let M 0,c 0 > 0, and s>3. Then, there exist a time T>0and constants C, δ 0 > 0 such that for any δ (0,δ 0 ], η 0 H s+7+1/2, φ 0 H s+7, b C 3 ([0,T]; H s+8 ) satisfying η 0 s+7+1/2 + φ 0 s+7 M 0, b(t),b t (t),b tt (t),b ttt (t) s+8 M 0 for t [0,T], 1+η 0 (x) b 0 (x) c 0 > 0 for x R 2, the initial value problem (10) and (6) has a unique solution (η, φ) =(η δ,φ δ )on the time interval [0,T] satisfying sup η δ (t) ζ δ (t) s C 0 δ 4, 0 t T where (ζ δ,v δ ) is a unique solution of the Green Nagdhi equations (14) with the initial conditions (ζ δ,v δ )=(η 0,u δ 0) at t =0. Here, u δ 0 is defined by ( δ 2 ) φ 0 = 1+ 1+η 0 b(, 0) T (η 0,b(, 0)) u δ 0 + δ 2( b t (, 0) η (1 + η 0 b(, 0)) b t (, 0) ). 23

24 Thank you very much for your attention! 24