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SUPPLEMENTARY MATERIAL

Electronic Appendix A. Entrance/Exit LossesOnly

In this appendix the analytical derivations to support the explanation of therole of the entrance/exit loss coefficient m on the equilibrium and stability ofdouble inlet systems are given.

A.1. Basin Tide and Inlet Velocities

Neglecting inertia and bottom friction, the linearized equations for the watermotion in the inlets are:

cmuequk = ηk − ηb, k = 1, 2, (A.16)

with, c = 4/(3πg). Assuming a pumping mode for the basin the continuityequation is:

Bdηbdt

= A1u1 +A2u2. (A.17)

The open boundary condition, which now only allows amplitude differences, is:

ηk = ηkeiωt + c.c., (A.18)

where ηk is real-valued. To arrive at an analytical solution for the basin tideand inlet velocities the following trial solutions are introduced:

ηb ∼ ηbei(ωt+ϕ) + c.c., (A.19)

uk ∼ ukei(ωt+ψk) + c.c., (A.20)

where ηb and uk are real-valued. Eliminating uk between Eqs. (A.16) and(A.17) results in an equation for the basin tide ηb. Substituting ηk and the trialsolution for ηb in this equation and separating into real and imaginary partsleads to expressions for the amplitude ηb and phase ϕ of the basin tide. Itfollows that

tanϕ =−µmueq

(A1 +A2), (A.21)

where µ = cωB. The phase ϕ is in the fourth quadrant, i.e. −π/2 < ϕ < 0.As to be expected the sea surface elevation precedes the basin surface elevation.The amplitude of the basin tide is

ηb =(A1η1 +A2η2)√

(µmueq)2 + (A1 +A2)2. (A.22)

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Substituting the trial solution for uk in Eq. (A.16) and separating into real andimaginary parts leads to expressions for the amplitude uk and phase ψk of theinlet velocity as a function of the known expressions ηb (Eq. A.22) and ϕ (Eq.A.21). Substituting these expressions it follows that

tanψk =µmueq(A1η1 +A2η2)

((µmueq)2 + (A1 +A2)2)ηk − (A1η1 +A2η2)(A1 +A2). (A.23)

When deriving the expression for sinψk from Eq. (A.23), it follows that it isalways positive and thus ψk is located in the first or second quadrant. From Eq.(A.23) it then follows that ψk is in the first quadrant when the denominator ispositive and in the second quadrant when the denominator is negative. In bothcases the velocity precedes the ocean tide. The amplitude of the inlet velocityis

uk =1

cmueq

√(µmueq ηk)2 +A2

3−k(η1 − η2)2

(µmueq)2 + (A1 +A2)2. (A.24)

A.2. Conditions for Equilibrium Cross-Sections for η1 6= η2

Eliminating ηb between Eqs. (A.16) and (A.17) results in two equations in u1

and u2. Substituting the trial solutions for uk (Eq. A.20) and taking uk = ueq,two equations are obtained with the equilibrium cross-sectional areas Aeq1 andAeq2 as unknowns:

(Aeq1 + iµmueq)ueqeiψeq

1 +Aeq2 ueqeiψeq

2 = iωBη1, (A.25)

(Aeq2 + iµmueq)ueqeiψeq

2 +Aeq1 ueqeiψeq

1 = iωBη2. (A.26)

Writing the real and imaginary parts of Eqs. (A.25) and (A.26) it can be shownthat for η1 6= η2 these equations can only be satisfied provided the phases of theinlet velocities are related as ψeq2 = π − ψeq1 . Making use of this, the followingsolution for Aeqk is obtained:

Aeqk =µm

sinψeq1

[η1 + η2

4cmueq− (−1)kueq

(1− cos2 ψeq1

2 cosψeq1

)]. (A.27)

Subtracting the imaginary parts of Eqs. (A.25) and (A.26) and substituting ψeq2

it follows that

cosψeq1 =∆η

2cmu2eq

, (A.28)

where ∆η = η1 − η2. As shown in Appendix A.1, depending on the sign of thedenominator of Eq. (A.23), ψeq1 is in the first or second quadrant. It followsfrom Eq. (A.28) that for inlets that are in equilibrium and ∆η > 0, ψeq1 is inthe first quadrant.

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With sinψeqk positive (see Appendix A.1) a requirement for Aeqk to be positiveis that the expression between square brackets in Eq. (A.27) is positive. Withthe expression for cosψeq1 given by Eq. (A.28) this leads to the condition

η1∆η

2(cm)2u4eq

≥ 1, ∆η > 0. (A.29)

A second condition for Eq. (A.27) to be a valid solution for Aeqk is that theabsolute value of cosψeq1 is equal or smaller than one, and thus from Eq. (A.28)

∆η

2cmu2eq

≤ 1, ∆η > 0. (A.30)

Similar conditions as expressed by Eqs. (A.29) and (A.30) hold for ∆η < 0. Inthat case in Eq. (A.29) η1 is replaced by η2 and in Eqs. (A.29) and (A.30) ∆ηis replaced by −∆η.

A.3. The Role of the Entrance/Exit Loss Coefficient m

From Eqs. (A.29) and (A.30) it follows that for given η1 and η2 there is only alimited range of m-values for which equilibrium cross-sectional areas exist; seeexpressions for mmin, Eq. (13), and mmax, Eq. (14). In the following the effectof varying m on Aeqk , ηeqb , ϕeq and ψeqk is investigated, where it is assumed thatη1 > η2. To show the effect on Aeqk , values of Aeqk are evaluated for m = mmin

and m = mmax. As explained in Appendix A.2, m = mmin corresponds to| cosψeq1 | = 1 and thus sinψeq1 = 0. It then follows from Eq. (A.27) that Aeqkgoes to infinity. For m = mmax and k = 2 the term between square brackets inEq. (A.27) is zero and thus Aeq2 = 0. The value of Aeq1 is positive and finite.Furthermore, it follows from Eq. (A.27) that, with cosψ1 positive, Aeq1 > Aeq2 .

To evaluate the effect of varying m on ηeqb , ϕeq and ψeqk the trial solutions,Eqs. (A.19) and (A.20), and open boundary condition, Eq. (A.18), are substi-tuted in Eq. (A.16). The real and imaginary parts of the resulting equationare

< : cmuequk cosψk = ηk − ηb cosϕ, (A.31a)

= : cmuequk sinψk = −ηb sinϕ. (A.31b)

Squaring and adding Eqs. (A.31a) and (A.31b) and substituting uk = ueqresults in

(cm)2u4eq = (ηk − ηeqb cosϕeq)

2+ (ηeqb )2 sin2 ϕeq. (A.32)

Subtracting the equations for k = 1 and k = 2 yields

η1 − ηeqb cosϕeq = ±(η2 − ηeqb cosϕeq). (A.33)

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Because of the assumption η1 > η2 the positive variant of Eq. (A.33) is notpossible. Therefore, from (A.33) it follows that

ηeqb cosϕeq =η1 + η2

2. (A.34)

Substituting for ηeqb cosϕeq in Eq. (A.32) yields

(cm)2u4eq =

(η1 − η2

2

)2

+ (ηeqb )2 sin2 ϕeq. (A.35)

With the expression for mmin (Eq. 13), it follows from Eq. (A.35) that sinϕeq =0 and, thus, with ϕeq in the fourth quadrant ϕeq = 0. It then follows from Eq.(A.34) that

ηeqb =η1 + η2

2. (A.36)

With m = mmin, the water level in the basin lies between the two ocean waterlevels. Using Eqs. (A.28) and (13), the phase angles of the velocity are ψeq1 = 0and ψeq2 = π. Summarizing, for m = mmin, ηeqb is given by Eq. (A.36), ϕeq = 0,uk = ueq, ψ

eq1 = 0 and ψeq2 = π.

Eliminating ϕeq between Eqs. (A.34) and (A.35) results in

(ηeqb )2 = (cm)2u4eq + η1η2, (A.37)

and thus ηeqb increases with increasing values of m. It then follows from Eq.(A.34) that cosϕeq decreases. With ϕeq in the fourth quadrant this impliesdecreasing values of ϕeq for increasing values of m. The basin tide increasinglylags the ocean tide. From (A.28) and with ψeq1 in the first quadrant it followsthat for increasing values of m, ψeq1 increases and, with ψeq2 = π − ψeq1 , ψeq2

decreases. Summarizing, for increasing values of m, ηeqb and ψeq1 increase andϕeq and ψeq2 decrease.

Electronic Appendix B. Entrance/Exit Lossesand Bottom Friction

This appendix explains the change in the range of m-values for which there arestable equilibriums when in addition to entrance/exit losses bottom friction isincluded in the dynamic equations. The shift in the lower limit of m can beexplained realizing that in the neighborhood of m = mmin the equilibrium cross-sectional areasAeq1 andAeq2 are large and thus bottom friction is small. From thisit is concluded that in the neighborhood of m = mmin bottom friction has littleeffect on the basin level and thus on the water level difference between oceanand basin when the system is in equilibrium. Writing Eq. (15) for equilibriumconditions results in

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c

(m+

2FL

γ√Aeqk

)u2eq = (ηk − ηb)eq. (B.38)

With the right-hand side being the same with and without bottom friction itthen follows that for a given equilibrium state adding bottom friction requiresthat the entrance/exit losses decrease and thus the value of m decreases.

In explaining the downward shift of the maximum m-value the amountof bottom friction that is added is assumed to be small compared to the en-trance/exit losses. In that case it is reasonable to assume that in the neighbor-hood of m = mmax the qualitative behavior of ηeqb , Aeq1 and Aeq2 as a functionof m is the same as in the absence of bottom friction, i.e. ηeqb increases and Aeq2decreases with increasing values of m and Aeq1 � Aeq2 .

When including bottom friction the governing equations for inlet 1 and 2are respectively

cueq

(m+ 2FL

γ√A1

)u1 = η1 − ηb, (B.39)

cueq

(m+ 2FL

γ√A2

)u2 = η2 − ηb. (B.40)

Substituting the trial solutions for ηb and uk, respectively Eqs. (A.19) and(A.20), separating into real and imaginary parts and, subsequently, squaringand adding these parts results in

(cueq)2

(m+

2FL

γ√A1

)2

u21 = η2

1 + η2b − 2η1ηb cosϕ, (B.41)

(cueq)2

(m+

2FL

γ√A2

)2

u22 = η2

2 + η2b − 2η2ηb cosϕ. (B.42)

Writing Eqs. (B.41) and (B.42) for equilibrium conditions yields

c2u4eq

(m+

2FL

γ√Aeq1

)2

= η21 + (ηeqb )2 − 2η1η

eqb cosϕeq, (B.43)

c2u4eq

(m+

2FL

γ√Aeq2

)2

= η22 + (ηeqb )2 − 2η2η

eqb cosϕeq. (B.44)

Eliminating ηeqb cosϕeq between Eqs. (B.43) and (B.44) the following expressionfor ηeqb as a function of m, Aeq1 and Aeq2 is derived

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(ηeqb )2 =(cm)2u4eq + η1η2 + c2u4

eq

(η2

η2 − η1

) 4mFL

γ√Aeq1

+

(2FL

γ√Aeq1

)2

− c2u4eq

(η1

η2 − η1

) 4mFL

γ√Aeq2

+

(2FL

γ√Aeq2

)2 .

(B.45)

Making use of Aeq1 � Aeq2 , Eq. (B.45) is simplified to

(ηeqb )2 = (cm)2u4eq+η1η2−c2u4

eq

(η1

η2 − η1

) 4mFL

γ√Aeq2

+

(2FL

γ√Aeq2

)2 . (B.46)

This equation relates m, ηeqb and Aeq2 in the neighborhood of m = mmax whenthe system is in equilibrium. The system is perturbed by increasing the value ofm with an amount of ∆m (with ∆m positive) and keeping all other parametersconstant. As a result the system is assumed to go to a new equilibrium. Theresulting changes in ηeqb and Aeq2 are respectively ∆ηb and ∆A2. Here, it isassumed that the perturbations are small compared to the equilibrium values.In Eq. (B.46) replacing m by meq+∆m, Aeq2 by Aeq2 +∆A2 and ηeqb by ηeqb +∆ηbresults in an equation relating meq, ηeqb and Aeq2 for a new equilibrium. This

equation can be further simplified by expanding 1/√Aeq2 in a Taylor series and

only considering terms of perturbed order, i.e. ∆m, ∆ηb and ∆A2. The resultis

∆m · 2c2u4eq

[meq

(η1 − η2

η1

)+

2FL

γ√Aeq2

]=

∆ηb · 2ηeqb

[η1 − η2

η1

]+

∆A2 · 2c2u4eq

[meqFL

γ(Aeq2 )3/2+ 2

(FL

γAeq2

)2].

(B.47)

In this equation, the expressions in square brackets are all positive (η1 > η2).Because ηeqb increases and Aeq2 decreases with increasing values of m, ∆ηb ispositive and ∆A2 is negative. It then follows from Eq. (B.47) that for ∆m tobe positive the term multiplying ∆ηb has to be larger than the absolute valueof the term multiplying ∆A2, which is only possible for Aeq2 > 0. In turn, thisimplies that in the presence of bottom friction the maximum value of m forwhich equilibrium is possible is smaller than mmax.

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Electronic Appendix C. Linear Stability En-trance/Exit Losses Only

In this appendix the linear stability of the equilibrium for entrance/exit lossesonly, as summarized in Section 3, is discussed in detail. Linear stability isinvestigated by considering Eq. (1), which determines the rate of change of theequilibrium after a perturbation:

Ak = Aeqk + ∆Ak. (C.48)

Here, it is assumed that the equilibrium value is much larger than the pertur-bation, i.e. Aeqk � ∆Ak. Next, the system of Eq. (1) for k = 1, 2 is linearizedaround the equilibrium solution (i.e. only terms linear in the perturbation areretained). Therefore, the amplitude of the inlet velocity uk , Eq. (7), is lin-earized. The first order Taylor expansion for uk around the equilibrium solutionhas the form

uk(A1, A2) = ueqk + ∆uk

= uk(Aeq1 , Aeq2 ) + ∆A1

∂uk∂A1

∣∣∣(A1,A2)

+ ∆A2∂uk∂A2

∣∣∣(A1,A2)

,(C.49)

where

∂uk∂A1

∣∣∣∣(A1,A2)

=− 1

cmueq

√(µmueq ηk)2 + (A2)2(η1 − η2)2

(µmueq)2 + (A1 +A2)2·(

A1 +A2

(µmueq)2 + (A1 +A2)2

),

(C.50)

and

∂uk∂A2

∣∣∣∣(A1,A2)

=1

cmueq

(1√

(µmueq ηk)2+A22(η1−η2)2

(µmueq)2+(A1+A2)2

·

[A2(η1 − η2)2

(µmueq)2 + (A1 +A2)2

]−√

(µmueq ηk)2 +A22(η1 − η2)2

(µmueq)2 + (A1 +A2)2·

[A1 +A2

(µmueq)2 + (A1 +A2)2

]).

(C.51)

Here, c = 4/(3πg) and µ = cωB. Using Eq. (C.48) and considering only termsof perturbed order ∆Ak, it follows that:

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∆uk =1

(cm)2u3eq

(Aeq3−k(η1 − η2)2

(µmueq)2 + (Aeq1 +Aeq2 )2

)∆A3−k−

ueq

(Aeq1 +Aeq2

(µmueq)2 + (Aeq1 +Aeq2 )2

)(∆A1 + ∆A2).

(C.52)

With uk = ueqk + ∆uk the expression for dAk(uk)/dt (Eq. 1) can be written as

dAkdt

(uk) =M

L

(1 + 3

∆ukueq

+O(∆u2k)

)(C.53)

Considering only terms of perturbed order ∆uk and substituting Eq. (C.52) inEq. (C.53) result in:

dAkdt

∣∣∣∣∆Ak

=3M

(cm)2u4eqL

(Aeq3−k(η1 − η2)2∆A3−k

(µmueq)2 + (Aeq1 +Aeq2 )2−

(cm)2u4eq(A

eq1 +Aeq2 )(∆A1 + ∆A2)

(µmueq)2 + (Aeq1 +Aeq2 )2

).

(C.54)

With ∆Ak = Akeλt the eigenvalue problem for Ak can now be written as

λ

[A1

A2

]=

[−ζξ ζ (σ1 − ξ)

ζ (σ2 − ξ) −ζξ

] [∆A1

∆A2

], (C.55)

with

ζ =3M

(cm)2u4eqL

(1

(µmueq)2 + (Aeq1 +Aeq2 )2

),

ξ = (cm)2u4eq(A

eq1 +Aeq2 ),

σk = Aeqk (η1 − η2)2.

If Re(λ) > 0, the equilibrium is unstable as the amplitude of the perturbation in-creases in time. Re(λ) < 0 indicates that the equilibrium is stable. The stabilityof the equilibrium can be deduced from the matrix in (C.55) by investigating itstrace and the determinant (i.e. without explicitly calculating the eigenvalues),as the trace is equal to the sum of the eigenvalues and the determinant to theirproduct. Hence, the conditions for a stable equilibrium are a negative trace anda positive determinant. From the matrix in (C.55) it follows that the trace isalways negative, since ζ and ξ are always positive. In case of similar forcingamplitudes (σ1 = σ2 = 0), the determinant is zero. Hence, one eigenvalue iszero and the other negative. It can thus be concluded that for similar forcingamplitudes there are an infinite number of equilibriums on a straight line whichare neutrally stable. If the system is perturbed, it returns to equilibrium, butnot necessarily the same equilibrium.

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For different forcing amplitudes (σ1 6= σ2) the determinant of the matrix in(C.55) is positive provided

Aeq1Aeq2

+Aeq2Aeq1

>(η1 − η2)2

(cm)2u4eq

− 2. (C.56)

It can be shown that the left-hand side of (C.56) is minimal, and equal to 2,if both equilibrium cross-sections are equal. Hence, if η1 − η2 < 2cmu2

eq, thedeterminant is always positive. From (11) it follows that if an equilibrium exists,η1 − η2 is always smaller than 2cmu2

eq and, therefore, the determinant of thematrix in (C.55) is always positive. As the trace is negative and the determinantpositive, the equilibriums given by (C.56) are always stable. Summarizing,for different forcing amplitudes and provided that an equilibrium exists, thisequilibrium is always stable.

38