Shoaling of Solitary Waves

Transcription

1 Shoaling of Solitary Waves by Harry Yeh & Jeffrey Knowles School of Civil & Construction Engineering Oregon State University Water Waves, ICERM, Brown U., April 2017

2 Motivation

3 The 2011 Heisei Tsunami in Japan

4 Bathymetry 200 m 1000 m 200 m 100 m 2000 m 1000 m Steep and narrow continental shelf 100 m 6500 m 200 m 1000 m tan θ = m 2000 m 7000 m 1000 m 200 m 200 m 1000 m 2000 m 100 m 200 m Very deep Japan Trench 6000 m 2000 m 1000 m 1000 m 100 km

5 Water depth 204 m GPS Wave Gage Wave period 40 ~ 50 minutes 55 cm land subsidence

6 Seabed Pressure Data and GPS Wave Gage Off Kamaishi Depth (m) GPS wave gage Pressure sensors Distance (km) h =1,600 m; x = 70 km h =1,000 m; x = 40 km h = 204 m; x = 20 km

7 Seabed Pressure Data and GPS Wave Gage Off Kamaishi Aligned at the peaks Depth (m) GPS wave gage Pressure sensors Distance (km) h =1,600 m; x = 70 km h =1,000 m; x = 40 km h = 204 m; x = 20 km

8 Seabed Pressure Data and GPS Wave Gage Off Kamaishi Aligned at the peaks Depth (m) GPS wave gage Pressure sensors Distance (km) h =1,600 m; x = 70 km h =1,000 m; x = 40 km h = 204 m; x = 20 km The temporal wave profile is very persistent.

9 Spatial Profiles The sharply peaked wave riding on the broad tsunami base appears to maintain its symmetrical waveform with increase in amplitude and narrow in wave breadth. Simple conversion (x = c t) shows that the length of the peaky wave is ~ 25 km: not too long.

10 Can this tsunami be considered as a soliton?

11 Seabed Pressure Transducers (ERI, University of Tokyo) h =1,600 m; x = 70 km. η = a sech 2 3a The breadth of the wave profile 2 λ is taken at η = 0.51 a. With this choice of length scale, the Ursell number of a solitary wave is U r = α/β = 1.0, where α = a/h; β = (h/λ) 2. ( )t ) 2 h 4 h 3 x c 0 (1+ a α = a h = β = h λ 2 = U r = α = 0.10 (The Ursell Number) β

12 Seabed Pressure Transducers (ERI, University of Tokyo) h =1,000 m; x = 40 km. The wave form becomes closer to that of soliton. η = a sech 2 3a ( )t ) 2 h 4 h 3 x c 0 (1 + a α = a h = β = h λ 2 α β = 0.33 =

13 h = 204 m; x = 20 km The Spike Riding on the Broad Tsunami resembles a soliton profile? η = a sech 2 GPS Wave Gage: 20 km off Kamaishi 3a ( )t ) 2 h 4 h 3 x c 0 (1 + a α = a h = β = h λ α β = =

14 Depth (m) GPS wave gage Pressure sensors Distance (km) Tsunami parameters: nonlinearity a Frequency dispersion β Ursell number U r Seafloor slope q. It is more or less a linear long wave with a finite seabed slope.

15 Green s Law: a Tsunami amplification (Shoaling) h ¼ : (based on linear shallow-water-wave theory) Measured runup heights onshore near Kamaishi: 15.7 m ± 6.7 m. Green s Law

16 Predicted waveform at x = 20 km using Green s law from the data at x = 70 km. η (m) 8 Does Green s law work? a h 1/4 h =1,600 m; x = 70 km h = 204 m; x = 20 km 6 4 Little amplification for the broad base wave?? x (km)

17 What We Observed from the Field Data Wave data along the east-to-west transect from Kamaishi. A unique tsunami waveform did not change much from the offshore location to the nearshore location: the waveform is comprised of a narrow spiky wave riding on the broad tsunami base at its rear portion. In spite of the persistent symmetrical waveform, the tsunami evolution is quite different from that of a soliton it is not the adiabatic evolution. As the tsunami approaches the shore, there is practically no amplification of the broad base portion of the tsunami, although the amplitude of the narrow spiky tsunami riding on the broad portion increased but not as fast as the prediction of Green s law.

18 Laboratory Data by Pujara, Liu, & Yeh 2015 Does Green s law work: r = ¼? a 0 /h 0 a h r r r < 1/ 4

19 Solitary Wave Shoaling in the Laboratory? Laboratory data show that shoaling amplification of the solitary waves is slower than that of Green s law. As the nonlinearity increases, a h 1 10 This is a consistent trend with the field observation.

20 Background

21 Grimshaw (1970, 1971); Johnson (1973); Adiabatic: the depth variation occurs on a scale that is slower than the evolution scale of the wave, so that the wave deforms but maintains its identity of soliton. Dimensionally, the adiabatic solution can be inferred: 3a η = a 0 sech 2 0 x c 3 0 (1+ 2h)t a 4 h 0 ( ) η = a 0 h 0 h 3a sech h ah = a 0 h 0 : 1 h x ct ( ) ; c = c(a, h) a h 1 This can be formally shown with the conservation of wave action flux.

22 Synolakis and Skjelbreia (1993) Exact solution to the linear non-dispersive shallow-water wave equation with a solitarywave initial condition yields Green s law in the offshore region (Synolakis, 1991) Laboratory Observation: Two zones of gradual shoaling and rapid shoaling: a) Green s law, b) adiabatic.

23 Peregrine (1967) u t + u u x + η x = 1 3 θ 2 x 2 u xxt +θ 2 xu xt, η t + (θ x + η)u x = 0. Extension of the Boussinesq equation: x points offshore from the initial shoreline Numerical results of the solitary-wave shoaling: tan θ = 1/20. The solid line represents Green s law.

24 Preliminary Considerations

25 Preliminary Considerations When the beach slope is mild and the wave amplitude is large, i.e. L b large and L 0 small, then, it is reasonable to anticipate for the adiabatic evolution process. A problem is that the incident wave can break in the early stage of the shoaling process, because of the finite initial amplitude.

26 Preliminary Considerations When the beach slope is steep and the wave amplitude is small, i.e. L b small and L 0 large, then, the wave as a whole may not have chance to shoal due to the short shoaling distance. The wave length be too long so that only a portion of the waveform be influenced by the sloping bed. For this situation, we anticipate little shoaling of the incident wave, but the wave may amplify due to reflection.

27 Preliminary Considerations L = 4h h a0 L b = h 0 tanθ α 0 = a 0 h 0 It is important is recognize that, once we deal with a sloping bed, the propagation domain is no longer infinite, but finite. The steeper the slope, the shorter the available propagation distance. γ = L 0 /L b must be a relevant parameter to characterize the solitary wave shoaling. γ L L 0 = b 2 4tan θ 3α 0

28 Analytical Considerations

29 Variable Coefficient Korteweg-de Vries Equation The vkdv equation: η t + c η x + c x 2 η + 3c 2h ηη x + c h2 6 η xxx = 0 Here η = η(x, t) and c(x) = gh(x), in which h(x) = x tanθ + h 0. The extremum of η (x) happens when t η = 0. Hence the following equation must satisfy for the envelope of η: c η x + c x 2 η + 3c 2h ηη + c h 2 x 6 η = 0 xxx

30 Variable Coefficient Korteweg-de Vries Equation For the amplitude envelope, t η = 0. c η + c x x 2 η + 3c 2h ηη + c h 2 x 6 η = 0 xxx After normalizing the variables (ζ = η(h(x))/a 0, h = h /h 0 ), we can write: hζ '+ 1 4 ζ α 0 ζ ζ '+ 1 6 h3 tan 2 θ ζ ''' = 0 where α 0 = a 0 /h 0. Linear Non-Dispersive Case hζ '+ 1 4 ζ = 0 Therefore ζ = C 0 h 1/4. This is Green s law for linear monochromatic waves.

31 Nonlinear Non-Dispersive Case hζ '+ 1 4 ζ α 0 ζ ζ '+ 1 6 h3 tan 2 θ ζ ''' = 0 becomes hζ '+ 1 4 ζ α 0 ζ ζ ' = This can be arranged as: ζ ' = (ζ h) α 0 Taking ζ = hυ(h) so that ζ ' = hυ '+ υ yields: θ dependency is dropped α 0 υ υ ( α 0 υ) dυ = 1 h d h Integration yields: Therefore, ζ h 4/5 4 lnυ + 1 ln ( α υ + 5 ) = ln h + constant ζ α 0 h /5 = C 0 h 1 This equation can be written as α 0 ζ hζ 4 C 0 5 = 0 Note that this reduces to Green s law for α 0 << 1. There are five roots, two of which are complex, another two which are negative, and one that is positive. It must therefore be that the positive real root represents the physical amplitude.

32 Nonlinear Non-Dispersive Case α 0 = a 0 h 0 a/a 0 h/h 0 The wave breaking criterion, a/h = 0.78, is used here. The solution is independent of the beach slope. a h r ; r < 1 4

33 Linear Dispersive Case hζ '+ 1 4 ζ α 0 ζ ζ '+ 1 6 h3 tan 2 θ ζ ''' = 0 becomes 1 6 h3 tan 2 θ ζ '''+ hζ '+ 1 4 ζ = 0 This is a third order Euler-type equation. Let ζ = C 0 h r. Then, we find the following polynomial to satisfy the equation: tan 2 θ ( 1 r r r ) + r 1 = r r ~ ¼ for θ << 1, i.e. Green s law. r ~ 0.1 for θ ~ π/3; r ~ 0 for θ ~ π/2. a h r ; r < θ (rad)

34 Linear Dispersive Case a/a 0 h/h 0 Very small dependency to the beach slope θ when it is small. a h r ; r < 1 4

35 Numerical Approach

36 The Euler Code Higher-Order Pseudo-Spectral Method Dommermuth and Yue (1987); Tanaka (1993); Jia (2014)!φ!x!x +! φ!z!z = 0 in h 0 +! ζ (!x,!t )!z!η(!x,!t ) ζ! t! + ζ!!!x φ!x φ!!z = 0 on!z = h 0 + ζ! (!x,!t )!η! t +! φ!x!η!x! φ!z = 0 on!z =!η(!x,!t )!φ! t + g!η (!φ 2!x + φ! ) 2!z = 0 on!z =!η(!x,!t ) (!x,!z) = (λ 0 x, λ 0 z);!t = (λ 0 c 0 ) t;! φ = λ 0 c 0 φ;!η = λ 0 η;! ζ = λ0 ζ The kinematic and dynamic boundary conditions at the free surface, z = η (x, t): η t = (1+ η x 2 )φ z φ x s η x and φ t s = 1 2 where φ s (x, t) = φ(x, η(x,t), t) ( 1+ η 2 ) φ x z ( ) φ x s ( ) 2 η

37 The Euler Code: (Dommermuth and Yue,1987) Taking a perturbation expansion of velocity potential ϕ together with the Taylor expansion about z = 0: φ s (x, t) = M M m m=1 k=0 η k ε m k! k φ m z k z = 0 Introduce a linear combination of basis functions which also satisfy Laplace s equation: φ m (x, z, t) = A m (x, z, t) + B m (x, z, t) Modeling the vertical velocity: A m (x, z, t) = n=0 A m n (t) cosh(k (z + h)) n cosh k n h e i k n x k A m z k B m (x, z, t) = B m 0 (t)z + B m n (t) sinh k z n cosh k n h ei kn x k B m z k Therefore: φ s (x, t) = M M m m=1 k=0 n=1 ε m η k k! k z k ( A m + B m ) For given ϕ s, we expand for A mn and B mn. z = 0 = 0 at z = h when k is odd, = 0 at z = 0 when k is even.

38 At the bottom surface, z = h + ζ (x) The Euler Code To determine B mn, we need the bottom boundary condition: φ (x, z = h + ζ, t) = M M m m=1 k=0 ζ k ε m k! k z k (A m + B m ) z = h Substitute them to the bottom boundary condition at z = h + ζ (x): ζ x φ x φ z = 0, on z = h + ζ (x) becomes M M m ζ k k+1 ζ x φ x ε m k! z (A + B ) k+1 m m m=1 k=0 z = h = 0, on z = h + ζ (x) For given ζ x and ϕ s, we successively determine B mn and A mn with the use of FFT. Then, we express z φ (x,η, t) = M m=1 M m k=0 η k ε m k! k+1 ( A m + B m ) z k+1 z =0

39 The Euler Code Substituting into the kinematic and dynamic boundary conditions at z = 0, φ t s M M m η ( ) k ε m = η x 2 m=1 k=0 k! k+1 k+1 z k+1 ( A m + B m ) M M m 2 η k η t = (1 + η x ) ε m s ( A k! z k+1 m + B m ) φ x η x m=1 k=0 z = 0 z = s ( φ x ) 2 η The above equations are no longer function of z, and we solve with the spectral method. Horizontal spatial derivatives in wavenumber space. We use M = 5 based on our sensitivity analysis for satisfying the no-flux boundary condition on the sloping bed. The 4 th order Runge-Kutta method for time stepping.

40 Validation Our treatment of the bottom boundary condition is adapted from Dommermuth and Yue (1987), although they did not demonstrate the scheme. Hence, we validate it by numerically observing the no-flux condition on the sloping bed. Velocity normal to the bed boundary vanishes except at x = 0 (beach toe). velocity (m/s) x (m)

41 Validations of the Numerical Results Comparison of numerically predicted shoaling (solid circles) with the laboratory data by Grilli (1994) Comparisons of numerical values (solid marks) with the large-scale laboratory experiments (hollow marks) tan θ =1/34.7; a 0 = 0.044m, h 0 = 0.44m. Composite beach: tan θ 1 =1/12 and tan θ 2 =1/24; h 0 = 1.888m. circles: a 0 = m, triangles: a 0 = m, squares: a 0 = m, diamonds: a 0 = m.

42 Results

43 Shoaling on a Very Mild Slope: tan θ = 1/400 α 0 = 0.1 Adiabatic evolution? γ L 0 L b = 4tan2 θ 3α 0 = Note the trailing tail formation that is not a soliton.

44 Shoaling on a Very Mild Slope: tan θ = 1/400 α 0 = 0.1 Adiabatic evolution process γ L 0 L b = 0.009

45 Shoaling of Wave for tan θ = 1/20 α 0 = 0.1 γ L 0 L b = 4tan2 θ 3α 0 = 0.18

46 Shoaling of Wave for tan θ = 1/20 α 0 = 0.1; γ L 0 L b = 4tan2 θ 3α 0 = 0.18 Comparison with the results of the vkdv theory.

47 Momentum Flux and Wave Reflection a 0 = 0.1m; h 0 = 1.0 m; tan θ = 0.02 Wave profile at t = 23.6 s Amplification in space η (m) a (m) x (m) Momentum flux at x = 10m Max flux = m 3 /s 2 x (m) Velocity field at t = 23.6 s h u 2 (m 3 /s 2 ) u (m/s) Max flux = m 3 /s 2 t (s) x (m)

48 Evolution of Energy Flux: α 0 = 0.10 γ L 0 L b = 4 tan2 θ 3α 0 ; L 0 = 4 h 0 3a 0 h 0 ; a 2 h 1/2 a2 h 1/2 a 0 2 h 0 1/2 Amplifies toward the beach toe. The amplification growth is slower than Green s law, when γ is large. When γ is small, the growth rate can exceed that of Green s law near the shore: approach to the adiabatic evolution process but it breaks early.

49 Evolution of Energy Flux: α 0 = 0.10 γ L 0 L b = 4 tan2 θ 3α 0 ; L 0 = 4 h 0 3a 0 h 0 ; a 2 h 1/2 a2 h 1/2 a 0 2 h 0 1/2 α 0 = 0.1; γ = α 0 = 0.1; γ = 0.18 Amplifies toward the beach toe. The amplification growth is slower than Green s law, when γ is large. When γ is small, the growth rate can exceed that of Green s law near the shore: approach to the adiabatic evolution process but it breaks early.

50 Evolution of Energy Flux: α 0 = 0.05 γ L 0 L b = 4 tan2 θ 3α 0 ; L 0 = 4 h 0 3a 0 h 0 ; a 2 h 1/2 a2 h 1/2 a 0 2 h 0 1/2 Amplifies toward the beach toe. The amplification growth is slower than Green s law. When γ is small, the growth rate can exceed that of Green s law near the shore.

51 Evolution of Energy Flux: α 0 = 0.01 γ L 0 L b = 4 tan2 θ 3α 0 ; L 0 = 4 h 0 3a 0 h 0 ; a 2 h 1/2 a2 h 1/2 a 0 2 h 0 1/2 Amplifies toward the beach toe, but approximately follows Green s law thereafter: a 2 h 1/2 ~ constant. The steeper the beach slope, the greater the amplification prior to reaching the beach toe.

52 Evolution of Energy Flux: α 0 = γ L 0 L b = 4 tan2 θ 3α 0 ; L 0 = 4 h 0 3a 0 h 0 ; a 2 h 1/2 a2 h 1/2 a 0 2 h 0 1/2 See the wave amplification starts far offshore due to reflection from the steep beach when γ is large. This because the wavelength is so long in comparison with the (steep) beach length.

53 Evolution of Energy Flux: α 0 = γ L 0 L b = 4 tan2 θ 3α 0 ; L 0 = 4 h 0 3a 0 h 0 ; a 2 h 1/2 a2 h 1/2 a 0 2 h 0 1/2 Follows Green s law when γ is small, say γ < 1.0. Then, the amplification becomes a h r ; r < 1 4 for γ > 1.0.

54 Evolution of Energy Flux: α 0 = γ L 0 L b = 4 tan2 θ 3α 0 ; L 0 = 4 h 0 3a 0 h 0 ; a 2 h 1/2 a2 h 1/2 a 0 2 h 0 1/2 See the wave amplification starts far offshore due to reflection from the steep beach when γ is large. This because the wavelength is so long in comparison with the (steep) beach length.

55 Conclusion

56 Summary Solitary wave amplifies while it approaches the beach toe. Shoaling of a solitary wave can follow Green s law (a h ¼ ), when the nonlinearity parameter α 0 is small (< 0.01) and the beach slope parameter γ = L 0 /L b is smaller than O(1). Shoaling of a solitary wave can follow the adiabatic evolution (a h 1 ), when the nonlinearity parameter α 0 is large (~ 0.1) and the beach slope parameter γ = L 0 /L b is very small (< O(0.01)). Shoaling of a solitary wave takes place at the slower rate (a h r ; r < ¼) than Green s law, when the nonlinearity parameter α 0 ~ O(0.1) and the beach slope parameter γ = L 0 /L b is also small (~ O(0.1)). This is the vkdv limit. The findings are qualitatively consistent with the numerical results provided by Peregrine (1967).

57 Peregrine (1967) u t + u u x + η x = 1 3 θ 2 x 2 u xxt +θ 2 xu xt, η t + (θ x + η)u x = 0. Extension of the Boussinesq equation: x points offshore from the initial shoreline Numerical results of the solitary-wave shoaling: θ = 1/20. The solid line represents Green s law. Also note the different rate of amplification with α 0. α 0 = 0.05, r ¼; α 0 = 0.1 & 0.15, r < ¼; α 0 = 0.2, r > ¼.

58 Summary There must be a factor(s) other than the parameter γ = L 0 /L b that influence the shoaling. Possibly, 1. Wave reflection at the beach toe. 2. Wave runup onto the dry shore. 3. Development of the wave skewness. For real co-seismic tsunamis, α 0 = O(10 3 ) and γ = O(10 1 ) ~ O(1), the wave should shoal as the rate less than Green s law. r ¼. To realize the adiabatic evolution (r = 1), γ < O(10-2 ) and α 0 O(10-1 ). It is possible to happen for a landslide generated tsunami. But it would likely radiate out because of a small source area for a landslide.

59 Shoaling of Tsunamis Event α 0 = a 0 /h 0 tan θ ϒ = L 0 /L b 2011 Heisei East Japan Indian Ocean, Thailand Indian Ocean, India Crescent City, California Tsunami Source in Alaska Off New York Submarine Landslide

60 Shoaling on a Very Mild Slope: tan θ = 1/400 α 0 = 0.1 γ L 0 L b = For example, h 0 = 200 m, a 0 = 20 m!! Possible offshore landslide generated tsunamis off New York. But it would likely radiate out because of a small source area. Adiabatic shoaling process unlikely occurs for tsunamis.