Shoaling of Solitary Waves

Harry Yeh & Jeffrey Knowles

School of Civil & Construction Engineering Oregon State University

Water Waves, ICERM, Brown U., April 2017 Motivation The 2011 Heisei Tsunami in Japan Bathymetry

200 m 200 m 1000 m 1000 m 100 m2000 m

Steep and narrow continental shelf 6500 m 100 m

200 m 200 m 2000 m 1000 m 1000 m tan θ = 0.02 7000 m

200 m 1000 m 2000 m Very deep Japan Trench 100 m 200 m 6000 m 2000 m 1000 m 200 m 1000 m

100 km GPS Wave Gage

Water depth 204 m Wave period 55 cm land subsidence 40 ~ 50 minutes Seabed Pressure Data and GPS Wave Gage Off Kamaishi

-1000 -2000

-3000

-4000

Depth (m) -5000

-6000

-7000 GPS wave gage

-8000

0 50Pressure sensors 100 150 200 Distance (km)

h =1,600 m; x = 70 km h =1,000 m; x = 40 km h = 204 m; x = 20 km Seabed Pressure Data and GPS Wave Gage Off Kamaishi

-1000 -2000

-3000

-4000

Depth (m) -5000

-6000

-7000 GPS wave gage

-8000

0 50Pressure sensors 100 150 200 Distance (km)

Aligned at the peaks

h =1,600 m; x = 70 km h =1,000 m; x = 40 km h = 204 m; x = 20 km Seabed Pressure Data and GPS Wave Gage Off Kamaishi

-1000 -2000

-3000

-4000

Depth (m) -5000

-6000

-7000 GPS wave gage -8000

0 50Pressure sensors 100 150 200 Distance (km)

Aligned at the peaks

h =1,600 m; x = 70 km The temporal wave profile is veryh =1,000 persistent. m; x = 40 km h = 204 m; x = 20 km 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.

Can this tsunami be considered as a soliton? Seabed Pressure Transducers (ERI, University of Tokyo)

2 ⎡ 3a a ⎤ h =1,600 m; x = 70 km. η = a sech 3 x − c (1+ )t ⎣⎢ 4 h ( 0 2 h )⎦⎥ 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 2 solitary wave is Ur = α/β = 1.0 , where α = a/h; β = (h/λ) .

a 5.1 α = = ≈ 0.0032 h 1600 2 2 ⎛ h ⎞ ⎛ 1600 ⎞ β = ⎜ ⎟ = ⎜ ⎟ ≈ 0.031 ⎝ λ ⎠ ⎝ 9100⎠ α U = = 0.10 (The Ursell Number) r β Seabed Pressure Transducers (ERI, University of Tokyo) h =1,000 m; x = 40 km. The wave form becomes closer to that of soliton.

2 ⎡ 3a a ⎤ η = a sech 3 x − c (1+ )t ⎣⎢ 4 h ( 0 2 h )⎦⎥

a 5.2 α = = ≈ 0.0052 h 1000 2 2 ⎛ h ⎞ ⎛ 1000 ⎞ β = ⎜ ⎟ = ⎜ ⎟ ≈ 0.016 ⎝ λ ⎠ ⎝ 7900⎠ α = 0.33 β GPS Wave Gage: 20 km off Kamaishi h = 204 m; x = 20 km The Spike Riding on the Broad Tsunami resembles a soliton profile?

2 ⎡ 3a a ⎤ η = a sech 3 x − c (1+ )t ⎣⎢ 4 h ( 0 2 h )⎦⎥

a 6.7 α = = ≈ 0.033 h 204 2 2 ⎛ h ⎞ ⎛ 204 ⎞ β = ⎜ ⎟ = ⎜ ⎟ ≈ 0.0026 ⎝ λ ⎠ ⎝ 4000⎠ α = 12.7 β 0 -1000 -2000

-3000

-4000

Depth (m) -5000 -6000

-7000 GPS wave gage -8000

0 50Pressure sensors 100 150 200 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. Tsunami amplification (Shoaling) Green’s Law: a ∝ h–¼: (based on linear shallow-water-wave theory) Measured runup heights onshore near Kamaishi: 15.7 m ± 6.7 m.

Green’s Law Does Green’s law work?

−1/4 Predicted waveform at x = 20 km using Green’s law a ∝ h from the data at x = 70 km. h =1,600 m; x = 70 km h = 204 m; x = 20 km

η (m)

6 Little amplification for the broad base wave ?? 4

x (km) 50 100 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. Laboratory Data by Pujara, Liu, & Yeh 2015 Does Green’s law work: r = − ¼ ?

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 a /h 0 0 0 r -0.05 a ∝ h

-0.1 r -0.15 r < 1/ 4 -0.2

-0.25 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 1 nonlinearity increases, → a ∝ h− 10 – This is a consistent trend with the field observation. Background 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:

2 ⎡ 3a0 a ⎤ η = a0 sech ⎢ 3 (x − c0 (1+ 2h )t)⎥ ⎣ 4 h0 ⎦ ⇓

a0 h0 2 ⎡ 3a0 1 ⎤ η = sech ⎢ ()x − ct ⎥; c = c(a, h) h ⎣ 4 h h ⎦

−1 ah = a0 h0 : a ∝ h

This can be formally shown with the conservation of wave action flux. Synolakis and Skjelbreia (1993)

• Exact solution to the “linear non-dispersive” shallow-water wave equation with a solitary- wave 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. Peregrine (1967) ⎧ u u u 1 2 x2 u 2 xu , ⎪ t + x +ηx = 3θ xxt +θ xt ⎨ Extension of the Boussinesq equation: η + ⎡(θ x +η)u ⎤ = 0. ⎩⎪ t ⎣ ⎦x 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. Preliminary Considerations Preliminary Considerations

• When the beach slope is mild and the wave amplitude is large, i.e. Lb large and L0 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. Preliminary Considerations

• When the beach slope is steep and the wave amplitude is small, i.e. Lb small and L0 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. Preliminary Considerations

4h Lh= 0 003a a 0 α = 0 0 h 0 L h tan b = 0 θ

• 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 = L0/Lb must be a relevant parameter to characterize the solitary wave shoaling. L 4tan2 θ γ ≡0 = Lb 3α0 Analytical Considerations Variable Coefficient Korteweg-de Vries Equation 2 cx 3c c h The vKdV equation: ηt + c ηx + η + ηηx + ηxxx = 0 2 2h 6

Here η = η(x, t) and c(x) = gh(x), in which h(x) = x tanθ + h0.

The extremum of η (x) happens when ∂t η = 0. Hence the following equation must satisfy for the envelope of η:

2 cx 3c c h c ηx + η + ηηx + ηxxx = 0 2 2h 6 Variable Coefficient Korteweg-de Vries Equation

For the amplitude envelope, ∂t η = 0. 2 cx 3c c h c ηx + η + ηηx + ηxxx = 0 2 2h 6

After normalizing the variables (ζ = η(h(x))/a0, h = h /h0), we can write:

1 3 1 3 2 hζ '+ ζ + α 0 ζ ζ '+ h tan θ ζ ''' = 0 4 2 6 where α0 = a0/h0.

Linear Non-Dispersive Case 1 hζ '+ ζ = 0 4 −1/4 Therefore ζ = C0 h . This is Green’s law for linear monochromatic waves. Nonlinear Non-Dispersive Case

1 3 1 3 2 1 3 hζ '+ ζ + α 0 ζ ζ '+ h tan θ ζ ''' = 0 becomes hζ '+ ζ + α 0ζ ζ ' = 0 4 2 6 4 2 −1 4 θ dependency is dropped. This can be arranged as: ζ ' = −1 3 (ζ h) + 2 α 0 3 1+ 2 α 0 υ 1 Taking ζ = hυ(h) so that ζ ' = hυ '+ υ yields: dυ = − d h 5 3 υ ( 4 + 2 α 0 υ) h Integration yields: 4 lnυ + 1 ln α υ + 5 = − ln h + constant 5 5 ()0 6 4/5 1/5 ⎛ ζ ⎞ ⎛ ζ 5⎞ 1 Therefore, α + = C h− Note that this reduces to Green’s ⎝⎜ h ⎠⎟ ⎝⎜ 0 h 6⎠⎟ 0 law for α0 << 1.

This equation can be written as α ζ 5 + 5 hζ 4 − C5 = 0 0 6 0

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. Nonlinear Non-Dispersive Case

a α = 0 0 h 0

a/a0

h/h0 The wave breaking criterion, a/h = 0.78, is used here. The solution is independent of the beach slope. −r 1 a ∝ h ; r < 4 Linear Dispersive Case

1 3 1 3 2 1 3 2 1 hζ '+ ζ + α 0ζ ζ '+ h tan θ ζ ''' = 0 becomes h tan θ ζ '''+ hζ '+ ζ = 0 4 2 6 6 4 This is a third order Euler-type equation. Let C h−r . ζ = 0 Then, we find the following polynomial to satisfy the equation: tan2 θ 1 r 3 + 1 r 2 + 1 r + r − 1 = 0 ( 6 2 3 ) 4 0.25 r ~ ¼ for θ << 1, i.e. Green’s law. 0.20 r ~ 0.1 for θ ~ π/3; r ~ 0 for θ ~ π/2. 0.15 r −r 0.10 1 a ∝ h ; r < 4

0.05

0.0 0.5 1.0 1.5 θ (rad) Linear Dispersive Case

a/a0

h/h0 Very small dependency to the beach slope θ when it is ‘small’.

−r 1 a ∝ h ; r < 4 Numerical Approach The Euler Code Higher-Order Pseudo-Spectral Method Dommermuth and Yue (1987); Tanaka (1993); Jia (2014)

! ! 0 in h !(x,t!) z !(x,t!) φx!x! + φz!z! = − 0 +ζ ! ≤ ! ≤ η ! ! ! ! ! 0 on z h !(x,t!) ζ t! +ζ x! φx! − φz! = ! = − 0 +ζ ! ! ! ! ! 0 on z !(x,t!) ηt! +φx! ηx! − φz! = ! = η ! φ! + gη! + 1 φ! 2 + φ! 2 = 0 on z! = η!(x!,t!) t! 2 ()x! z! (x!,z!) = (λ x, λ z); t! = (λ c ) t; φ! = λ c φ; η! = λ η; ζ! = λ ζ 0 0 0 0 0 0 0 0

The kinematic and dynamic boundary conditions at the free surface, z = η (x, t):

2 2 2 s s 1 2 1 s ηt = (1+ ηx )φz − φx ηx and φt = 2 ()1+ ηx ()φz − 2 ()φx − η where φ s (x, t) = φ(x, η(x,t), t) The Euler Code: (Dommermuth and Yue,1987) Taking a perturbation expansion of velocity potential ϕ together with the Taylor expansion about z = 0: M M −m η k ∂k φ φ s (x, t) = ε m m ∑∑ k k! z m=1 k=0 ∂ z = 0 Introduce a linear combination of basis functions which also satisfy Laplace’s equation: φm (x, z, t) = Am (x, z, t) + Bm (x, z, t) Modeling the vertical velocity:

∞ k cosh(k (z + h)) i k x ∂ A (x, z, t) (t) n e n m 0 at z h when k is odd, Am = ∑ Am n ↔ k = = − n=0 cosh knh ∂z ∞ k sinh k z i k x ∂ B (x, z, t) B (t)z (t) n e n m 0 at z 0 when k is even. Bm = m 0 + ∑ Bm n ↔ k = = n=1 cosh knh ∂ z Therefore: M M −m η k ∂k φ s (x, t) = ε m A + B ∑∑ k ()m m k! z m=1 k=0 ∂ z = 0

s For given ϕ , we expand for Amn and Bmn. The Euler Code

To determine Bmn, we need the bottom boundary condition: At the bottom surface, z = − h + ζ (x)

M M −m ζ k ∂k φ (x, z = −h + ζ , t) = ε m (A + B ) ∑∑ k m m k! z m=1 k=0 ∂ z = −h Substitute them to the bottom boundary condition at z = − h + ζ (x):

ζ x φx − φz = 0, on z = −h +ζ (x) becomes M M −m ζ k ∂k+1 ζ φ − ε m (A + B ) = 0, on z = −h +ζ (x) x x ∑ ∑ k! z k+1 m m m=1 k=0 ∂ z = −h

s For given ζx and ϕ , we successively determine Bmn and Amn with the use of FFT.

M M −m k k+1 η ∂ ( A m + B m ) Then, we express (x, , t) m ∂zφ η = ∑ ∑ ε k+1 m=1 k=0 k! ∂z z =0 The Euler Code Substituting into the kinematic and dynamic boundary conditions at z = 0,

2 ⎧ M M −m k k+1 ⎛ η ∂ ⎞ 2 φ s = 1 1 + η 2 ε m A + B − 1 φ s − η ⎪ t 2 ()x ∑∑ k+1 ()m m 2 ()x ⎜ k! z ⎟ ⎪ ⎝ m=1 k=0 ∂ z = 0 ⎠ ⎨ M M −m k k+1 ⎪ 2 m η ∂ s η = (1 + η ) ε A + B − φ η ⎪ t x ∑∑ k+1 ()m m x x m 1 k 0 k! ∂ z ⎩ = = z = 0

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 4th order Runge-Kutta method for time stepping. 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) Validations of the Numerical Results

Comparison of numerically predicted Comparisons of numerical values shoaling (solid circles) with the (solid marks) with the large-scale laboratory data by Grilli (1994) laboratory experiments (hollow marks)

tan θ =1/34.7; a0 = 0.044m, h0 = 0.44m. Composite beach: tan θ1 =1/12 and tan θ2 =1/24; h0 = 1.888m. circles: a0 = 1.038 m, triangles: a0 = 0.755 m, squares: a0 = 0.472 m, diamonds: a0 = 0.189 m. Results Shoaling on a Very Mild Slope: tan θ = 1/400

α0 = 0.1 Adiabatic evolution?

L 4tan2 θ γ ≡ 0 = = 0.009 L 3α b 0

Note the trailing tail formation that is not a soliton. Shoaling on a Very Mild Slope: tan θ = 1/400

L α = 0.1 γ ≡ 0 = 0.009 0 L b Adiabatic evolution process Shoaling of Wave for tan θ = 1/20

α0 = 0.1

L 4tan2 θ γ ≡ 0 = = 0.18 L 3α b 0 Shoaling of Wave for tan θ = 1/20 L 4tan2 θ α = 0.1; γ ≡ 0 = = 0.18 0 L 3α b 0 Comparison with the results of the vKdV theory. Momentum Flux and Wave Reflection

a0 = 0.1m; h0 = 1.0 m; tan θ = 0.02

Wave profile at t = 23.6 s Amplification in space

η (m) a (m)

x (m) x (m)

Momentum flux at x = −10m Velocity field at t = 23.6 s Max flux = 4.63 × 10-5 m3/s2

2 h u u 3 2 (m /s ) (m/s)

Max flux = 0.1146 m3/s2 t (s) x (m) Evolution of Energy Flux: α0 = 0.10