The problem (18)–(19) is more general than the water wave problem, and therefore the solutions we have found above do not necessarily provide physical water wave solutions. In particular, the solutions are only physically relevant if the reparametrization of the free surface satisfies some non-self-intersecting condition. In other words, the solutions give rise to water waves if and only if the mapping
$$\begin{aligned} x\mapsto \left( \frac{x}{k}+({\mathcal {C}}_{kh}(v-h)(x),v(x) ):x\in {\mathbb {R}}\right) \end{aligned}$$
is injective and
$$\begin{aligned} {\mathcal {S}}=\left\{ \left( \frac{x}{k}+({\mathcal {C}}_{kh}(v-h)(x),v(x) ):x\in {\mathbb {R}}\right) \right\} \end{aligned}$$
is contained in the upper-half plane.
Periodicity
Since the solutions we have constructed in the previous section are in the function space \(C_{2\pi ,e}^{2,\alpha }({\mathbb {R}})\), they are symmetric and periodic, and it therefore suffices to study what happens in a half-period of a wave.
We will now look in more detail at the solutions along the curve \({\mathcal {K}}_{1,\pm }\). From Theorem 3.2, we know that near the bifurcation point these solutions are of the form
$$\begin{aligned} v(x)=h+s\cos (x)+o(s), \end{aligned}$$
which are monotone functions on a half-period. This added monotonicity makes these solutions easier to study than the ones for integers \(n\ge 2\), where \(\cos (nx)\) oscillates within the half-period.
However, from Lemma 4.1(i) recalled below, similar results to the ones we will provide in this section, can be shown for the curves \({\mathcal {K}}_{n,\pm }\) for all \(n\ge 2\) by working in a function space of period \(2\pi / n\).
Moreover, by denoting
$$\begin{aligned} {\mathcal {K}}_{1,\pm }={\mathcal {K}}_{\pm }^<\cup \{(m_{1,\pm }^*,Q_{1,\pm }^*,h) \}\cup {\mathcal {K}}_{\pm }^>, \end{aligned}$$
where
$$\begin{aligned} {\mathcal {K}}_{\pm }^<=\{(m(s),Q(s),v_s):s\in (-\infty ,0) \}, \end{aligned}$$
and
$$\begin{aligned} {\mathcal {K}}_{\pm }^>=\{(m(s),Q(s),v_s):s\in (0,\infty ) \}, \end{aligned}$$
(37)
from Lemma 4.1(ii) below, it suffices to study the properties of solutions along \({\mathcal {K}}_{\pm }^>\).
Lemma 4.1
Let \(h,k>0\) and the constants \(\gamma , a \in {\mathbb {R}}\) and \(b\in {\mathbb {R}}^+\) be given. For each integer \(n\ge 1\) and both choices of sign ±, denote by
$$\begin{aligned} {\mathcal {K}}_{n,\pm }=\{(m(s),Q(s),v_s):s\in {\mathbb {R}} \} \end{aligned}$$
the continuous curve of solutions of (18)–(19) in the space \({\mathbb {R}}\times {\mathbb {R}}\times C_{2\pi ,e}^{2,\alpha }\) given by the previous theorem. Then the following additional properties hold along \({\mathcal {K}}_{n,\pm }\):
-
(i)
\(v_s\) is periodic of period \(2\pi / n\) for each \(s\in {\mathbb {R}}\);
-
(ii)
\(m(-s)=m(s), \,Q(-s)=Q(s),\) and \(v_{-s}(x)=v_s(x+/ n)\) for all \(x\in {\mathbb {R}}\), for each \(s\in {\mathbb {R}}\).
We refer to (Lemma 10, [7], Sect. 4) for the proof, choosing
$$\begin{aligned} \tilde{{\mathbb {X}}}={\mathbb {R}}\times C_{2\pi / n,e}^{2,\alpha }({\mathbb {R}})\quad \text {and}\quad \tilde{{\mathbb {Y}}}=C_{2\pi / n,e}^{1,\alpha }({\mathbb {R}})\times {\mathbb {R}}. \end{aligned}$$
Injectivity
That our solutions near the bifurcation point are monotonic on \((0,\frac{\pi }{k})\) enables us to assume that along \({\mathcal {K}}_{\pm ,\text {loc}}^>\), we have \(v'(x)<0\) for all \(x\in (0,\frac{\pi }{k})\). If we can prove that this property is preserved along the entire curve \({\mathcal {K}}_{\pm }^>\), we not only rule out the possibility of the curve looping back to the bifurcation point (Theorem 3.2(b)) but also obtain that the solution must be injective between the trough and the crest. In particular, it will then be sufficient to merely check the injectivity condition along the crest line and the trough line.
In order to show injectivity, we will use the reformulation presented in the following Lemma, stated without proof (see Lemma 11 in [7]).
Lemma 4.2
If
$$\begin{aligned} V_x(x,0)\ne 0\quad \text {for all}\quad x\in (0,\pi ), \end{aligned}$$
then the injectivity condition holds if and only if
$$\begin{aligned} 0<U(x,0)<\frac{\pi }{k}\quad \text {for all}\quad x\in (0,\pi ). \end{aligned}$$
(38)
Using the definition of the periodic Hilbert transform for the strip, (38) is equivalent to
$$\begin{aligned} 0<\frac{x}{k}+({\mathcal {C}}_{kh}(v-h))(x)<\frac{\pi }{k}\quad \text {for all}\quad x\in (0,\pi ). \end{aligned}$$
Monotonicity properties
We are now ready to present a set of monotonicity properties for the pair \((m,v)\in {\mathbb {R}}\times C_{2\pi ,e}^{2,\alpha }\). These are motivated by the form of the small-amplitude solutions and will be shown to hold globally along the curve:
$$\begin{aligned}&v(x)>0\quad \text {for all}\quad x\in {\mathbb {R}}, \end{aligned}$$
(39)
$$\begin{aligned}&v\ne h, \end{aligned}$$
(40)
$$\begin{aligned}&v'(x)<0\quad \text {for all}\quad x\in (0,\pi ), \end{aligned}$$
(41)
$$\begin{aligned}&v''(0)<0\quad \text {and}\quad v''(\pi )>0, \end{aligned}$$
(42)
$$\begin{aligned}&0<\frac{x}{k}+({\mathcal {C}}_{kh}(v-h))(x)<\frac{\pi }{k}\quad \text {for all}\quad x\in (0,\pi ), \end{aligned}$$
(43)
$$\begin{aligned}&\frac{1}{k}+({\mathcal {C}}_{kh}(v'))(0)>0\quad \text {and}\quad \frac{1}{k}+({\mathcal {C}}_{kh}(v'))(\pi )>0, \end{aligned}$$
(44)
$$\begin{aligned}&\pm \left( \frac{m}{kh}-\frac{\gamma }{2kh}[v^2]-\gamma {\mathcal {C}}_{kh}(vv')-\frac{a}{6kh}[v^3]-\frac{a}{2}{\mathcal {C}}_{kh}(v^2v')\right. \nonumber \\&\left. +\left( \gamma v+\frac{a}{2}v^2\right) \left( \frac{1}{k}+{\mathcal {C}}_{kh}(v') \right) \right) >0\quad \text {for all}\quad x\in {\mathbb {R}}. \end{aligned}$$
(45)
We would like to point out that for solutions (m, Q, v) of (16), condition (45) is equivalent to
$$\begin{aligned} Q-bv(x)\ne 0\quad \text {and} \quad (v'(x))^2+\left( \frac{1}{k}+({\mathcal {C}}_{kh}(v'))(x) \right) ^2\ne 0\quad \text {for all}\quad x\in {\mathbb {R}}. \end{aligned}$$
Local properties
The question is now, to what extent are these properties satisfied along \({\mathcal {K}}_{\pm }^>\)? We introduce the following notation:
$$\begin{aligned} {\mathcal {V}}_{\pm }=\{(m,Q,v)\in {\mathbb {R}}\times {\mathbb {R}}\times C_{2\pi ,e}^{2,\alpha }({\mathbb {R}}):(39){-}(45)\,\,\text {hold} \}. \end{aligned}$$
It is not difficult to show that these properties hold for solutions on \({\mathcal {K}}_{\pm }^>\) that are near enough to the trivial one. We therefore get the following lemma. We refer the reader to Lemma 12 in [7] for a proof.
Lemma 4.3
For either choice of sign ±, let (37) be the curve of solutions of (18)–(19). Then there exists \(\epsilon >0\) sufficently small such that
$$\begin{aligned} \{(m(s),Q(s),v_s):s\in (0,\epsilon ) \} \subset {\mathcal {V}}_{\pm }. \end{aligned}$$
Main result
We now state and prove the main result of this section.
Theorem 4.4
For either choice of sign ±, let (37) be the curve of solutions of (18)–(19). Then one of the following alternatives occurs:
-
(A1)
\({\mathcal {K}}_{\pm }^>\subset {\mathcal {V}}_{\pm }\), in which case alternative (a) in Theorem 3.2 occurs;
-
(A2)
there exists some \(s^*\in (0,\infty )\) such that \(\{(m(s),Q(s),v_s):s\in (0,s^*) \}\subset {\mathcal {V}}_{\pm }\), while \((m(s^*),Q(s^*),v_{s^*})\) satisfies (39)–(42), (44) and (45), and instead of (43), it satisfies
$$\begin{aligned}&0<\frac{x}{k}+({\mathcal {C}}_{kh}(v-h) )(x)\le \frac{\pi }{k}\quad \text {for all}\quad x\in (0,\pi ), \\&\frac{x_0}{k}+({\mathcal {C}}_{kh}(v-h) )(x_0)=\frac{\pi }{k}\quad \text {for some}\quad x_0\in (0,\pi ). \end{aligned}$$
In other words, the first alternative (A1) means that all solutions on \({\mathcal {K}}_{\pm }^>\) correspond to physical water waves with the qualitative properties described above, whereas alternative (A2) means that such solutions exist until \(s=s^*\), at which point the wave profile self-intersects on the line strictly above the trough.
Since the qualitative properties hold in a neighborhood of the bifurcation point, the idea is to use a continuation argument: we need to check to what extend these properties are satisfied along the global curve of solutions. We are therefore faced with two possible options: either the properties hold along the entire curve, in which case (39)–(45) hold with strict inequalities. Or, there exists a first point at which at least one of the properties fails, and thus that particular property must then have a non-strict inequality.
Therefore, we rewrite (39)–(45), replacing all the strict inequalities with non-strict ones and argue by contradiction. We work with the harmonic functions U and V from the conformal map \(U+iV\) described in the beginning of the paper, evaluated at the surface of \({\mathcal {R}}_{kh}\). Notice that from the evenness of v, we can deduce the function V(x, y) is even in the x variable and the function U(x, y), being the harmonic conjugate of \(-V\) up to a constant, is odd in the x variable.
Exploiting the harmonicity and parity of U and V and using the Cauchy–Riemann equations, we then use a series of strong maximum principle arguments, along with the Hopf and Serrin inequalites at the boundary, to show that the inequalites in (39)–(42), (44)–(45) and the first inequality in (43) must be strict.
The proof of this theorem presents very strong similarities to the proof of the analoguous theorem in [7]. However, for the sake of readability, we will provide most of the details: the maximum principle arguments are rather delicate and it is essential to carefully verify that the added stratification term does not cause too many complications. Indeed, in past literature, one major difficulty of proving nodal properties for stratified waves lies in the fact that the zeroth order term generally has the wrong sign (for example, see [1, 3]).
Proof of Theorem 4.4
We begin by fixing the choice of sign ±: we choose the \(+\) sign but all arguments follow identically for the − sign.
First off, we define the set
$$\begin{aligned} I=\{s\in (0,\infty ):(m(s),Q(s),v_s)\in {\mathcal {V}}_+ \}. \end{aligned}$$
Notice that \({\mathcal {V}}_+\) is an open set in \({\mathbb {R}}\times {\mathbb {R}}\times C_{2\pi ,e}^{2,\alpha }({\mathbb {R}})\). This implies that I is an open subinterval of \((0,\infty )\). We therefore have two options. The first one is \(I=(0,\infty )\), in which case \({\mathcal {K}}_+^>\subset {\mathcal {V}}_+\). This would imply that the alternative (b) in Theorem 3.2 is excluded and (a) must be valid. In this case, there is nothing more to show.
The other option is that I is not equal to the entire interval \((0,\infty )\) and this is the case we will consider. Choosing the \(\epsilon \) from the local properties lemma, we have \((0,\epsilon )\subset I\). We now denote by \(s^*\) the upper end-point of the largest interval containing \((0,\epsilon )\) and contained in I. In other words, we have \((0,s^*)\subset I\) but \(s^*\notin I\). We will investigate the properties of the solution \((m(s^*),Q(s^*),v_{s^*})\).
Claim 1:
\(v_{s^*}\ne h\)
The proof of this claim is quite technical and completely identical to the one in [7], so we omit it here.
For notational simplicity, for the remainder of the proof, we will denote \((m(s^*), Q(s^*),v_{s^*})\) by \((m(s),Q(s),v_s)\). Note that this is the limit of the solutions satisfying the monotonicity properties.
From the definition of I, we therefore have the following inequalities:
$$\begin{aligned}&v(x)\ge 0\quad \text {for all}\quad x\in {\mathbb {R}},\quad \text {and } v\text { is even and of period}\,2\pi , \end{aligned}$$
(46)
$$\begin{aligned}&v'(x)\le 0\quad \text {for all}\quad x\in [0,\pi ],\quad \text {so that}\quad v'\ge 0\quad \text {in}\quad [\pi ,2\pi ], \end{aligned}$$
(47)
$$\begin{aligned}&0\le \frac{x}{k}+({\mathcal {C}}_{kh}(v-h))(x)\le \frac{\pi }{k}\quad \text {for all}\quad x\in [0,\pi ], \end{aligned}$$
(48)
$$\begin{aligned}&\frac{m}{kh}-\frac{\gamma }{2kh}[v^2]-\gamma {\mathcal {C}}_{kh}(vv')-\frac{a}{6kh}[v^3]-\frac{a}{2}{\mathcal {C}}_{kh}(v^2v')\nonumber \\&\qquad +\left( \gamma v+\frac{a}{2}v^2\right) \left( \frac{1}{k}+{\mathcal {C}}_{kh}(v') \right) \ge 0\quad \text {for all}\quad x\in {\mathbb {R}}. \end{aligned}$$
(49)
In other words, either (39)–(45) are satisfied and the inequalities are strict, or they aren’t, in which case we have equality signs. Using sharp forms of maximum principles, we now prove that all the above inequalities, except for the second one in (48) are strict. To this end, we define the rectangular domain
$$\begin{aligned} {\mathcal {R}}:=\{(x,y):0<x<\pi ,\quad \text {and}\quad -kh<y<0 \} \end{aligned}$$
with boundary \(\partial {\mathcal {R}}\) divided up into the following segments:
$$\begin{aligned} \begin{alignedat}{2}&\partial {\mathcal {R}}_t=\{(x,0):0\le x\le \pi \}&\qquad&\partial {\mathcal {R}}_b=\{(x,-kh):0\le x\le \pi \}\\&\partial {\mathcal {R}}_l=\{(0,y):-kh\le y\le 0 \}&\qquad&\partial {\mathcal {R}}_r=\{(\pi ,y):-kh\le y\le 0 \}. \end{alignedat} \end{aligned}$$
Claim 2: \(v(x)>0\) for all \(x\in {\mathbb {R}}\).
Assume that the claim does not hold. Then (46) and (47) imply that \(v(\pi )=0\). Since V is harmonic in \({\mathcal {R}}_{kh}\), it would therefore have a global minimum in \(\overline{{\mathcal {R}}_{kh}}\) at \((\pi ,0)\). Using the Hopf boundary point lemma, this would imply that \(V_y(\pi ,0)<0\). However, since (48) is equivalent to
$$\begin{aligned} 0=U(0,0)\le U(x,0)\le U(\pi ,0)=\frac{\pi }{k},\quad \forall x\in [0,\pi ], \end{aligned}$$
(50)
using the Cauchy–Riemann equations, we get
$$\begin{aligned} V_y(\pi ,0)=U_x(\pi ,0)\ge 0. \end{aligned}$$
This is a contradiction, and thus the claim has been proven.
Claim 3: (45) with the \(+\) sign holds.
Note that this is equivalent to
$$\begin{aligned} \begin{aligned}&\zeta _y+\gamma VV_y+\frac{a}{2}V^2V_y\\&\quad =\frac{m}{kh}-\frac{\gamma }{2kh}[v^2]-\gamma {\mathcal {C}}_{kh}(vv')-\frac{a}{6kh}[v^3]\\&\qquad -\frac{a}{2}{\mathcal {C}}_{kh}(v^2v')+\left( \gamma v+\frac{a}{2}v^2\right) \left( \frac{1}{k}+{\mathcal {C}}_{kh}(v') \right) \ge 0\quad \text {for all}\quad x\in {\mathbb {R}}. \end{aligned} \end{aligned}$$
We know from Theorem 3.2 that
$$\begin{aligned} Q-2bv(x)>0\qquad \forall x\in {\mathbb {R}}. \end{aligned}$$
(51)
Therefore, since we have
$$\begin{aligned} \left( \zeta _y+\gamma VV_y+\frac{a}{2}V^2V_y\right) ^2=(Q-2bV)(V_x^2+V_y^2)\qquad \text {on}\quad y=0 \end{aligned}$$
(52)
the proof of the claim boils down to verifying that
$$\begin{aligned} \left( \zeta _y+\gamma VV_y+\frac{a}{2}V^2V_y\right) ^2=(Q-2bV)(V_x^2+V_y^2)>0\qquad \text {on}\quad y=0. \end{aligned}$$
This is equivalent to
$$\begin{aligned} v'(x)^2+\left( \frac{1}{k}+ {\mathcal {C}}_{kh}(v')(x)\right) ^2>0,\quad \forall x\in {\mathbb {R}} \end{aligned}$$
(53)
or
$$\begin{aligned} V_x^2(x,0)+V_y^2(x,0)>0,\quad \forall x\in {\mathbb {R}}. \end{aligned}$$
We argue by contradiction.
We assume that there exists a point \(x_0\in {\mathbb {R}}\) such that \(V_x^2(x_0,0)+V_y^2(x_0,0)=0\). This would imply that
$$\begin{aligned} \zeta _y(x_0,0)+\gamma V(x_0,0)V_y(x_0,0)+\frac{a}{2}V^2(x_0,0)V_y(x_0,0)=0. \end{aligned}$$
(54)
Notice that (49) is equivalent to
$$\begin{aligned} \zeta _y(x,0)+\gamma V(x,0)V_y(x,0)+\frac{a}{2}V^2(x,0)V_y(x,0)\ge 0\qquad \text {for all}\quad x\in {\mathbb {R}}. \end{aligned}$$
(55)
Combining (54) with (55) ensures that \(\zeta _y+\gamma VV_y+\frac{a}{2}V^2V_y=O((x-x_0)^2)\) as \(x\rightarrow x_0\) at \(y=0\). From (52), we thus have that \(V_x^2(x,0)+V_y^2(x,0)=O((x-x_0)^4)\) as \(x\rightarrow x_0\). Moreover, from the second derivative of the right-hand side of (52), we see that
$$\begin{aligned} V_{xx}(x_0,0)=V_{xy}(x_0,0)=0. \end{aligned}$$
(56)
From the evenness and periodicity of v, we can assume, without loss of generality that \(0\le x_0\le \pi \). We now examine the three different possibilities for \(x_0\).
(i) \(x_0=0\) : this would imply, using (47), that V has a global maximum in \(\overline{{\mathcal {R}}_{kh}}\). By the Hopf boundary point lemma, we therefore get \(V_y(0,0)>0\). However this would mean that
$$\begin{aligned} \frac{1}{k}+{\mathcal {C}}_{kh}(v')(0)=V_y(0,0)>0, \end{aligned}$$
which is a contradiction and so \(x_0\ne 0\).
-
(ii)
\(0<x_0<\pi \) : this would imply that \(V_x\) has its global maximum in \(\overline{{\mathcal {R}}}\) at \((x_0,0)\). Indeed, \(V_x=0\) on the vertical sides by oddness and on the bottom by the boundary condition, whereas (47) indicates that \(V_x\le 0\) on the top. By the Hopf boundary point lemma, we get \(V_{xy}(x_0,0)>0\), a contradiction to (56).
-
(iii)
\(x_0=\pi \) : Since the mapping \(x\mapsto V(x,y)\) is even about \(\pi \), for every \(y\in [-kh,0]\), we have
$$\begin{aligned} V_{xxx}(\pi ,0)=V_{xyy}(\pi ,0)=0. \end{aligned}$$
(57)
As above, \(V_x\) has its global maximum in \(\overline{{\mathcal {R}}}\) at \((\pi ,0)\). Taking into account (56) and (57), the Serrin edge-point lemma gives us that \(V_{xxy}(\pi ,0)<0\). The Cauchy–Riemann equations thus imply that \(U_{xxx}(\pi ,0)<0\). Similarly, we can deduce that \(U_x(\pi ,0)=U_{xx}(\pi ,0)\) since \(V_y(\pi ,0)=0\) by assumption and (56) holds. As a result, we obtain
$$\begin{aligned} U(x,0)-U(\pi ,0)=\frac{1}{6}U_{xxx}(\pi ,0)(x-\pi )^3+O((x-\pi )^4)>0\quad \text {as}\quad x\nearrow \pi . \end{aligned}$$
This is a contradiction to (50).
From points (i)-(iii), we can deduce that \(x_0\) doesn’t exist and we have shown that
$$\begin{aligned} V_x^2(x,0)+V_y^2(x,0)>0\quad \text {for all}\quad x\in {\mathbb {R}}. \end{aligned}$$
Since we have \(V_x(\pi ,0)=0\), we necessarily have \(V_y(\pi ,0)\ne 0\). From (50) and the Cauchy–Riemann equations, we have \(V_y(\pi ,0)=U_x(\pi ,0)\ge 0\). As a result, we obtain
$$\begin{aligned} \frac{1}{k}+({\mathcal {C}}_{kh}(v'))(\pi )=V_y(\pi ,0)>0. \end{aligned}$$
This concludes the proof of the claim. \(\quad \square \)
Remark 2
Notice that we have also proved (44) and (53).
Remark 3
For the next part of the proof, we will need \((m,Q,v)\in {\mathbb {R}}\times {\mathbb {R}}\times C_{2\pi ,e}^3({\mathbb {R}})\). However, notice that by bootstrapping the improvement of regularity that is based on Theorem 3.1 in the proof of Theorem 3.2, it is not difficult to show that for any solution (m, Q, v) of (18)–(19), the function v is necessarily of class \(C^{\infty }\) on \({\mathbb {R}}\).
In addition, we will need the fact that
$$\begin{aligned} V_x^2+V_y^2>0\qquad \text {in}\quad \overline{{\mathcal {R}}_{kh}}. \end{aligned}$$
(58)
We omit the proof here as it is slightly technical, and is completely identical to the one in [7], located directly before Lemma 15.
Remark 4
Notice that from the dynamic boundary condition, we have
$$\begin{aligned} \zeta _y=-\gamma VV_y-\frac{a}{2}V^2V_y\pm (Q-2bV)^{1/2}(V_x^2+V_y^2)^{1/2} \qquad \text {at}\, (x,0), \forall x\in {\mathbb {R}}. \end{aligned}$$
(59)
In what follows, we will be applying maximum-principle arguments to the function \(f:\overline{{\mathcal {R}}_{kh}}\rightarrow {\mathbb {R}}\) given by
$$\begin{aligned} f:=\frac{V_x\zeta _y-V_y\zeta _x}{V_x^2+V_y^2}, \end{aligned}$$
which is well-defined thanks to (58). Notice that f is in fact the (pseudo) horizontal velocity of the fluid, \(-\psi _{X}\). Moreover, f is a harmonic function in \({\mathcal {R}}_{kh}\) as it is the imaginary part of
$$\begin{aligned} \frac{-(\zeta _y+i\zeta _x)}{V_y+iV_x}, \end{aligned}$$
a quotient of two holomorphic functions. In addition, we clearly have \(f=0\) on \(\partial {\mathcal {R}}_l\cup \partial {\mathcal {R}}_b\cup \partial {\mathcal {R}}_r\). On the top, \(\partial {\mathcal {R}}_t\) we have
$$\begin{aligned} f=\frac{V_x(\zeta _y+\gamma VV_y+\frac{a}{2}V^2V_y)}{V_x^2+V_y^2}=\pm \frac{V_x(Q-2bV)^{1/2}}{(V_x^2+V_y^2)^{1/2}}, \end{aligned}$$
(60)
where we used (20) to get the left-hand side and (59) to obtain the right-hand side.
Moreover, notice that on \(\partial {\mathcal {R}}_t\), f does not vanish identically and has a constant sign (which depends on the sign ±). By the maximum principle, f therefore has a strict sign in \({\mathcal {R}}\).
Claim 4: \(v'(x)<0\) for all \(x\in (0,\pi )\).
Arguing by contradiction, we assume that for some \(x_0\in (0,\pi )\), we have \(v'(x_0)=0\). Since, as a result, \(x_0\) is a global maximum for \(v'\) on \((0,\pi )\), we have \(v''(x_0)=0\). We can clearly see that we must then also have \(f(x_0,0)=0\) and therefore f has at \((x_0,0)\) a global extremum in \({\mathcal {R}}\cup \partial {\mathcal {R}}\). Applying the Hopf boundary point lemma, we get \(f_y(x_0,0)\ne 0\). The idea is now to show that we actually have \(f_y(x_0,0)=0\), which would lead to a contradiction.
For simplicity, we write
$$\begin{aligned} f=\frac{p}{q}\qquad \text {where}\quad p=V_x\zeta _y-V_y\zeta _x\qquad \text {and}\quad q=V_x^2+V_y^2. \end{aligned}$$
Clearly, for every point in \(\overline{{\mathcal {R}}_{kh}}\), we have
$$\begin{aligned} f_y=\frac{p_yq-pq_y}{q^2}. \end{aligned}$$
In particular, since at \((x_0,0)\) we have \(p=V_x=V_{xx}=0\), we obtain
$$\begin{aligned} f_y=\frac{p_y}{q}=\frac{V_{xy}\zeta _y-V_y\zeta _{xy}}{V_y^2}. \end{aligned}$$
(61)
Differentiating (59) with respect to x, we get:
$$\begin{aligned} \begin{aligned} \zeta _{xy}=&-\gamma (V_xV_y+VV_{xy})-\frac{a}{2}(2VV_xV_y+V^2V_{xy})\\ {}&\pm \left( \frac{-bV_x(V_x^2+V_y^2)^{1/2}}{(Q-2bV)^{1/2}}+\frac{(Q-2bV)^{1/2}(V_xV_{xx}+V_yV_{xy})}{(V_x^2+V_y^2)^{1/2}} \right) \end{aligned} \end{aligned}$$
at (x, 0) for all \(x\in {\mathbb {R}}\). Therefore, at the point \((x_0,0)\), where we already know that \(V_x=V_{xx}=0\), we have
$$\begin{aligned}&\zeta _y=-\gamma VV_y-\frac{a}{2}V^2V_y\pm (Q-2bV)^{1/2}|V_y|, \end{aligned}$$
(62)
$$\begin{aligned}&\zeta _{xy}=-\gamma VV_{xy}-\frac{a}{2}V^2V_{xy}\pm \frac{(Q-2bV)^{1/2}V_yV_{xy}}{|V_y|}. \end{aligned}$$
(63)
We can now clearly see that at the point \((x_0,0)\), we have
$$\begin{aligned} V_{xy}\zeta _y=V_y\zeta _{xy}. \end{aligned}$$
Consequently, using (61) we get that \(f_y(x_0,0)=0\), thus obtaining the desired contradiction which proves the claim.
Claim 5: \(v''(0)<0\) and \(v''(\pi )>0\).
Since \(v'(0)=v'(\pi )=0\), and \(v'(0)\le 0\) on \([0,\pi ]\) we get
$$\begin{aligned} v''(0)\le 0\qquad \text {and}\qquad v''(\pi )\ge 0. \end{aligned}$$
We therefore only need to prove that these two inequalities are strict. Again, we argue by contradiction.
Let us assume that for some \(x_0\in \{0,\pi \}\), we have \(v''(x_0)=0\). Then we clearly have \(V_x=V_{xx}=0\) at \((x_0,0)\). Since f admits a global extremum at \((x_0,0)\) in \({\mathcal {R}}\cup \partial {\mathcal {R}}\), the Serrin edge-point lemma ensures that not all first and second derivatives of f can be equal to zero at the point \((x_0,0)\). However, we will now show that they all do in fact vanish, which will be a contradiction.
Since \(f=0\) on \(\partial {\mathcal {R}}_l\cup \partial {\mathcal {R}}_r\) and f is harmonic, we must therefore have \(f_y=f_{yy}=f_{xx}=0\) on \(\partial {\mathcal {R}}_l\cup \partial {\mathcal {R}}_r\). From the definition (60) of f on \(\partial {\mathcal {R}}_t\), we can see that \(f_x(x_0,0)\) contains a factor of \(V_x\) or \(V_{xx}\) for every term, and therefore, \(f_x(x_0,0)=0\).
We now check the mixed derivative \(f_{xy}(x_0,0)\). Using the same notation as for the previous claim, we get
$$\begin{aligned} f_{xy}=f_{yx}=\frac{(p_{xy}q+p_yq_x-p_xq_y-pq_{xy})q^2-(p_yq-pq_y)2qq_x}{q^4}. \end{aligned}$$
Since at the point \((x_0,0)\) we have
$$\begin{aligned} \zeta _x=V_x=\zeta _{xy}=V_{xy}=V_{xx}=V_{yy}=p=p_x=p_y=0, \end{aligned}$$
we get
$$\begin{aligned} f_{xy}=\frac{p_{xy}}{q}. \end{aligned}$$
Differentiating p and evaluating it at \((x_0,0)\), we get
$$\begin{aligned} p_{xy}=V_{xxy}\zeta _y-V_y\zeta _{xxy}. \end{aligned}$$
We calculate \(\zeta _{xxy}(x_0,0)\) by taking the derivative with respect to x of (63). Taking into account that \(V_x=V_{xx}=V_{xy}=0\) at \((x_0,0)\), we are left with
$$\begin{aligned} \zeta _{xxy}=-\gamma VV_{xxy}-\frac{a}{2}V^2V_{xxy}\pm \frac{(Q-2bV)^{1/2}V_yV_{xxy}}{|V_y|}. \end{aligned}$$
Since (62) is also valid at \((x_0,0)\) we clearly have
$$\begin{aligned} V_{xxy}\zeta _y=V_y\zeta _{xxy} \end{aligned}$$
and therefore, \(f_{xy}(x_0,0)=0\). As a result, all first and second derivatives of f vanish at the point \((x_0,0)\) and we have obtained the desired contradiction and proven the claim.
Claim 6: condition (A2) in Theorem 4.4 holds.
Recall that (48) is valid. We first assume that the first part of (A2) fails. This means, there exists an \(x_0\in (0,\pi )\) such that
$$\begin{aligned} 0=\frac{x_0}{k}+{\mathcal {C}}_{kh}(v-h)(x_0)=U(x_0,0). \end{aligned}$$
This would imply that the harmonic function U in \({\mathcal {R}}\) has at \((x_0,0)\) a global minimum in \(\overline{{\mathcal {R}}}\). Therefore, by the Hopf boundary point lemma, we get \(U_y(x_0,0)<0\). However, from the Cauchy–Riemann equations, we would then have \(V_x(x_0,0)>0\), a contradiction to (47). As a result, the first part of (A2):
$$\begin{aligned} 0<\frac{x}{k}+({\mathcal {C}}_{kh}(v-h) ) (x)\le \frac{\pi }{k}\quad \text {for all}\quad x\in (0,\pi ), \end{aligned}$$
is true.
The second part of (A2):
$$\begin{aligned} \frac{x_0}{k}+({\mathcal {C}}_{kh}(v-h) )(x_0)=\frac{\pi }{k}\quad \text {for some}\quad x_0\in (0,\pi ), \end{aligned}$$
must also necessarily be true. Otherwise, taking into account everything we have proved thus far, we would have \((m(s^*),Q(s^*),v_{s^*})\in {\mathcal {V}}_+\), which would be a contradiction to the definition of \(s^*\). This completes the proof of the claim and of the theorem.