1 Introduction↩︎

Let \(\tilde{M}^n\) be a complex \(n\)-dimensional Kähler manifold with complex structure \(J\) and Kähler metric \(\left<\,, \right>\). An \(n\)-dimensional submanifold \(M\) of \(\tilde{M}^n\) is called Lagrangian if \(\left<X, JY\right>=0\) for all tangent vector fields \(X\), \(Y\) on \(M\). A normal vector field \(\xi\) of a Lagrangian submanifold \(M\) is called a Hamiltonian variation if \(\xi=J\nabla f\) for some compactly supported function \(f\) on \(M\), where \(\nabla\) is the gradient on \(M\). A Lagrangian submanifold is said to be Hamiltonian stationary if it is a critical point of the volume functional for all Hamiltonian variations. As shown by Oh [1], this condition is equivalent to the divergence-free condition \[{\rm div}(JH)=0,\label{HS}\tag{1}\] where \(H\) denotes the mean curvature vector field of \(M\).

In this paper, we investigate Hamiltonian stationary Lagrangian surfaces in complex space forms. When the mean curvature is a non-zero constant, we show that the second fundamental form is parallel. Such Lagrangian surfaces have been completely classified in Theorem 7.2 of [2], Theorems A.1 and A.2 of [3]. We then turn to the case in which the mean curvature is a non-constant harmonic function. Assuming that the Gaussian curvature is constant, we prove that the ambient space must be the complex hyperbolic plane and the Gaussian curvature necessarily takes a specific negative value. This phenomenon leads to a complete classification.

2 Preliminaries↩︎

Let \(\tilde{M}^n(4\epsilon)\) be a complete and simply connected complex space form of complex dimension \(n\) and constant holomorphic sectional curvature \(4\epsilon\), that is, \(\tilde{M}^n(4\epsilon)\) is the complex Euclidean space \(\mathbb{C}^n\), the complex projective space \(\mathbb{C}P^n(4\epsilon)\) or the complex hyperbolic space \(\mathbb{C}H^n(4\epsilon)\) according as \(\epsilon=0\), \(\epsilon>0\) or \(\epsilon<0\).

Let \(M\) be a Lagrangian submanifold of \(\tilde{M}^n(4\epsilon)\). We denote the Levi-Civita connections on \(M^n\) and \(\tilde{M}^n(4\epsilon)\) by \(\nabla\) and \(\tilde{\nabla}\), respectively. The Gauss and Weingarten formulas are given respectively by \[\tilde{\nabla}_XY = \nabla_XY+h(X,Y), \quad \tilde{\nabla}_X \xi = -S_{\xi}X+D_X\xi \nonumber\] for tangent vector fields \(X\), \(Y\) and normal vector field \(\xi\), where \(h\), \(S\) and \(D\) are the second fundamental form, the shape operator and the normal connection. The mean curvature vector field \(H\) is defined by \(H=(1/n){\rm trace}\,h.\) The function \(|H|\) is called the mean curvature. We have (cf. [4]) \[\begin{align} & D_XJY=J(\nabla_XY), \tag{2}\\ & \left<h(X, Y), JZ\right>=\left<h(Y, Z), JX\right>=\left<h(Z, X), JY\right>. \tag{3} \end{align}\]

Denote by \(R\) the Riemann curvature tensor of \(\nabla\). Then the equations of Gauss and Codazzi are given respectively by \[\begin{align} \left<R(X,Y)Z,W\right>=& \left<h(Y, Z), h(X, W)\right>-\left<h(X, Z), h(Y, W)\right>\nonumber \\ &+\epsilon(\left<Y,Z\right>\left<X,W\right>-\left<X,Z\right>\left<Y,W\right>), \tag{4}\\ ({\bar\nabla}_{X}h)(Y,Z)=& ({\bar\nabla}_{Y}h)(X,Z),\tag{5} \end{align}\] where \(X,Y,Z,W\) are vectors tangent to \(M\), and \(\bar\nabla h\) is defined by \[({\bar\nabla}_{X}h)(Y,Z)= D_X h(Y,Z) - h(\nabla_X Y,Z) - h(Y,\nabla_X Z). \nonumber\]

3 Hamiltonian stationary Lagrangian surfaces with harmonic mean curvature↩︎

Let \(M\) be a Lagrangian surface in \(\tilde{M}^2(4\epsilon)\), where \(\epsilon\in\{-1, 0, 1\}\). Suppose that \(H\ne 0\) everywhere. Denote by \(K\) the Gaussian curvature of \(M\). Let \(\{e_1, e_2\}\) be a local orthonormal frame on \(M\) such that \(Je_1\) is parallel to \(H\). It follows from (3 ) that the second fundamental form takes the form \[\begin{align} \label{sf} & h(e_1, e_1)=(a-c)Je_1+bJe_2,\\ & h(e_1, e_2)=bJe_1+cJe_2, \\ & h(e_2, e_2)=cJe_1-bJe_2 \end{align}\tag{6}\] for some functions \(a\), \(b\) and \(c\).

Putting \(\omega_i^j(e_k)=\left<\nabla_{e_k}e_i, e_j\right>\), and using (2 ) and (6 ), we have \[\begin{align} (\bar\nabla_{e_1}h)(e_2, e_2)&=\{e_1c+3b\omega_1^2(e_1)\}Je_1 -\{e_1b-3c\omega_1^2(e_1)\}Je_2,\nonumber\\ (\bar\nabla_{e_2}h)(e_1, e_2)&=\{e_2b+(a-3c)\omega_1^2(e_2)\}Je_1 +\{e_2c+3b\omega_1^2(e_2)\}Je_2,\nonumber\\ (\bar\nabla_{e_1}h)(e_1, e_2)&=\{e_1b+(a-3c)\omega_1^2(e_1)\}Je_1 +\{e_1c+3b\omega_1^2(e_1)\}Je_2,\nonumber\\ (\bar\nabla_{e_2}h)(e_1, e_1)&=\{e_2(a-c)-3b\omega_1^2(e_2)\}Je_1 +\{e_2b+(a-3c)\omega_1^2(e_2)\}Je_2. \nonumber \end{align}\] Therefore, the Codazzi equation (5 ) implies \[\begin{align} & e_1c+3b\omega_1^2(e_1)=e_2b+(a-3c)\omega^2_1(e_2), \tag{7}\\ & -e_1b+3c\omega_1^2(e_1)=e_2c+3b\omega^2_1(e_2),\tag{8}\\ & e_2(a-c)-3b\omega_1^2(e_2)=e_1b+(a-3c)\omega^2_1(e_1).\tag{9} \end{align}\] Combining (8 ) and (9 ) yields \[e_2a-a\omega^2_1(e_1)=0.\label{C4}\tag{10}\]

Assume that \(M\) is Hamiltonian stationary. Then, by (1 ) we have \[e_1a+a\omega_1^2(e_2)=0. \label{C5}\tag{11}\] Using (10 ) and (11 ), we obtain \[[a^{-1}e_1, a^{-1}e_2]=0. \nonumber\] Therefore, there exists a local coordinate system \(\{u, v\}\) such that \(e_1=a\partial_u\) and \(e_2=a\partial_v\). Hence, the metric tensor is given by \[g=a^{-2}(du^2+dv^2), \label{g}\tag{12}\] which implies that \[\begin{align} &\omega_1^2(e_1)=a_v, \quad \omega_1^2(e_2)=-a_u,\tag{13} \\ &K=-(a_u)^2-(a_v)^2+a(a_{uu}+a_{vv}). \tag{14} \end{align}\] It follows from (13 ) that (7 ) and (8 ) can be rewritten as \[\begin{align} ac_u+3ba_v&=ab_v-(a-3c)a_u, \tag{15}\\ -ab_u+3ca_v&=ac_v-3ba_u.\tag{16} \end{align}\] Put \(\delta=\epsilon-K\). Then the Gauss equation (4 ) together with (6 ) yields \[\delta=2b^2-ac+2c^2. \label{G1}\tag{17}\]

In the case where the mean curvature is constant, we have

From (13 ), it follows that if \(|H|=|a|/2\) is a non-zero constant, then \(\omega_1^2=0\), and hence \(K=0\). Equations (7 ) and (8 ) reduce to \[c_u-b_v=c_v+b_u=0.\label{CR}\tag{18}\] Since \(\delta=\epsilon\), differentiating (17 ) with respect to \(u\) and \(v\), and using (18 ), we have \[\begin{pmatrix} 4b & 4c-a \\ a-4c & 4b\\ \end{pmatrix} \begin{pmatrix} b_u \\ c_u \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}.\label{mt}\tag{19}\]

If \(16b^2+(a-4c)^2\ne 0\) on an open subset, then by (19 ) we obtain \(b_u=c_u=0\). Thus, it follows from (18 ) that \(b\) and \(c\) are constant. Otherwise, \(b=0\) and \(c\) is constant. In both cases, using (2 ) and (6 ), we obtain \(\bar\nabla h=0\). 0◻

Our next step is to investigate the case of non-constant mean curvature. The main result of this paper is the following.

Suppose that \(|H|\) is a harmonic function on \(M\). Then by (12 ) we have \[a_{uu}+a_{vv}=0.\label{har}\tag{20}\] Moreover, suppose that \(|H|\) is non-constant and \(K\) is constant. Then, combining (14 ) and (20 ) shows that \(K<0\) and \[a_u=\sqrt{-K}\cos\theta, \quad a_v=\sqrt{-K}\sin\theta. \label{auv}\tag{21}\] for some function \(\theta\) on \(M\). Substituting (21 ) into (20 ), we get \[-(\sin\theta)\theta_u+(\cos\theta)\theta_v=0. \label{theta1}\tag{22}\]

On the other hand, since \(a_{uv}-a_{vu}=0\) holds, by (21 ) we obtain \[(\cos\theta)\theta_u+(\sin\theta)\theta_v=0.\label{theta2}\tag{23}\] It follows from (22 ) and (23 ) that \(\theta_u=\theta_v=0\), that is, \(\theta\) is constant. Solving (21 ), we conclude that up to translations, \(a\) is given by \[a=\sqrt{-K}\{(\cos\theta)u+(\sin\theta)v\}. \label{a}\tag{24}\]

Case (i): \(b=0\) on an open subset \(\mathcal{U}\). In this case, (15 ) and (16 ) reduce respectively to \[\begin{align} & (a-3c)a_u+ac_u=0, \tag{25}\\ & 3ca_v-ac_v=0.\tag{26} \end{align}\] Differentiating (17 ) with respect to \(v\), we have \[ca_v+(a-4c)c_v=0.\label{D1}\tag{27}\] Combining (26 ) and (27 ) gives \[(a-c)c_v=0.\nonumber\]

If \(c_v\ne 0\) on an open subset in \(\mathcal{U}\), then \(a=c\). From (17 ) we see that \(a\) is constant, which contradicts our assumption. Hence we have \(c_v=0\) on \(\mathcal{U}\). Thus, by (26 ) we get \[ca_v=0.\]

Case (i.1): \(c=0\) on an open subset \(\mathcal{U}_1\subset\mathcal{U}\). In this case, from (17 ) we have \(K=\epsilon=-1\). It follows from (25 ) that \(a_u=0\) on \(\mathcal{U}_1\). Hence, using (24 ) and \(K=-1\), we obtain \(a^2=v^2\). Applying the coordinate transformation \(y=-\int v^{-1}dv\), we see that the metric tensor (12 ) becomes \[g=m^2e^{2y}du^2+dy^2\nonumber\] for some positive constant \(m\), and the second fundamental form satisfies \[h(\partial_u, \partial_u)=J\partial_u, \quad h(\partial_u, \partial_y)=h(\partial_y,\partial_y)=0.\nonumber\] We rewrite \(u\) as \(x\). According to [5], we conclude that \(\mathcal{U}_1\) is congruent to the Lagrangian surface obtained from (1).

Case (i.2): \(a_v=0\) on an open subset \(\mathcal{U}_2\subset\mathcal{U}\). In this case, \(a_u\ne 0\). Differentiating (17 ) with respect to \(u\) leads to \[ca_u+(a-4c)c_u=0.\label{D2}\tag{28}\] Eliminating \(c_u\) from (25 ) and (28 ) gives \[(a-2c)(a-6c)a_u=0.\nonumber\]

If \(a\ne 2c\) on an open subset of \(\mathcal{U}_2\), then \(a=6c\), and it follows from (17 ) that \(a\) is constant, which contradicts our assumption. Hence \(a=2c\), which together with (17 ) implies \(K=\epsilon=-1\). Thus, by (24 ) we obtain \(a^2=u^2\). Applying the coordinate transformation \(x=(u+v)/2\) and \(y=(u-v)/2\), we see that the metric tensor (12 ) becomes \[g=\dfrac{2}{(x+y)^2}(dx^2+dy^2)\nonumber\] and the second fundamental form satisfies \[h(\partial_x, \partial_x)=J\partial_x, \quad h(\partial_x, \partial_y)=0, \quad h(\partial_y, \partial_y)=J\partial_y.\nonumber\] According to [6], \(\mathcal{U}_2\) is congruent to the Lagrangian surface obtained from (2) with \(m=1\).

Case (ii): \(b\ne 0\) on an open subset \(\mathcal{V}\). We put \(A=a_u\) and \(B=a_v\), which are constant. Differentiating (17 ) with respect to \(u\) and \(v\), we obtain \[\begin{align} & b_u=\frac{cA+(a-4c)c_u}{4b}, \tag{29}\\ & b_v=\frac{cB+(a-4c)c_v}{4b}.\tag{30} \end{align}\] Substituting (29 ) and (30 ) into (15 ) and (16 ) yields \[\begin{align} & a\{4bc_u-(a-4c)c_v\}=4b(3c-a)A-(12b^2-ac)B, \tag{31}\\ & a\{(a-4c)c_u+4bc_v\}=(12b^2-ac)A+12bcB. \tag{32} \end{align}\] Solving (31 ) and (32 ) for \(c_u\) and \(c_v\), we have \[\begin{align} & c_u=\frac{Af_1+Bf_2}{a(a^2+16b^2-8ac+16c^2)},\tag{33}\\ & c_v=\frac{Af_3+Bf_4}{a(a^2+16b^2-8ac+16c^2)},\tag{34} \end{align}\] where \[\begin{align} & f_1=-4ab^2-a^2c+4ac^2, \\ & f_2=-48b^3+16abc-48bc^2,\\ & f_3=4a^2b+48b^3-32abc+48bc^2, \\ & f_4=12ab^2-a^2c+4ac^2. \end{align}\] Substituting (33 ) and (34 ) into (29 ) and (30 ) gives \[\begin{align} b_u=& \frac{Ag_1+Bg_2}{a(a^2+16b^2-8ac+16c^2)},\tag{35}\\ b_v=&\frac{Ag_3+Bg_4}{a(a^2+16b^2-8ac+16c^2)},\tag{36} \end{align}\] where \[\begin{align} & g_1=-a^2 b + 8 a b c, \\ & g_2=-12ab^2+4a^2c+48b^2c -28ac^2+48c^3,\\ &g_3=a^3+12ab^2-12a^2c-48b^2c+44ac^2-48c^3,\\ &g_4=3 a^2 b - 8 a b c. \end{align}\]

Using (33 )-(36 ), we differentiate (33 ) and (34 ) with respect to \(v\) and \(u\), respectively. A computation with a computer algebra system yields \[\begin{align} &c_{uv}=\frac{A^2h_1+ABh_2+B^2h_3}{a^2(a^2+16b^2-8ac+16c^2)^2}, \tag{37}\\ &c_{vu}=\frac{A^2h_4+ABh_5+B^2h_6}{a^2(a^2+16b^2-8ac+16c^2)^2}, \tag{38} \end{align}\] where \[\begin{align} h_1=&-12 a^4 b - 144 a^2 b^3 + 160 a^3 b c + 768 a b^3 c - 656 a^2 b c^2 + 768 a b c^3,\\ h_2=&-108 a^3 b^2 - 960 a b^4 + 18 a^4 c + 1168 a^2 b^2 c + 2304 b^4 c - 252 a^3 c^2\\ &-3840 a b^2 c^2 + 1296 a^2 c^3 + 4608 b^2 c^3 - 2880 a c^4 + 2304 c^5,\\ h_3=&-96 a^2 b^3 + 768 b^5 - 384 a b^3 c + 32 a^2 b c^2 + 1536 b^3 c^2 - 384 a b c^3 + 768 b c^4,\\ h_4=&-8 a^4 b - 96 a^2 b^3 - 768 b^5 + 128 a^3 b c + 1152 a b^3 c - 608 a^2 b c^2\\ &-1536 b^3 c^2 + 1152 a b c^3 - 768 b c^4,\\ h_5=&-92 a^3 b^2 - 192 a b^4 + 18 a^4 c + 784 a^2 b^2 c + 2304 b^4 c - 252 a^3 c^2\\ &-3072 a b^2 c^2 + 1296 a^2 c^3 + 4608 b^2 c^3 - 2880 a c^4 + 2304 c^5,\\ h_6=&-240 a^2 b^3 + 80 a^3 b c + 768 a b^3 c - 496 a^2 b c^2 + 768 a b c^3. \end{align}\] Subtracting (38 ) from (37 ), we have \[c_{uv}-c_{vu}=\frac{4b(A^2k_1+ABk_2+B^2k_3)}{a^2(a^2+16b^2-8ac+16c^2)^2},\label{integ1}\tag{39}\] where \[\begin{align} k_1=&-a^4 - 12 a^2 b^2 + 192 b^4 + 8 a^3 c - 96 a b^2 c - 12 a^2 c^2\\ & + 384 b^2 c^2 - 96 a c^3 + 192 c^4,\\ k_2=&-4 a^3 b - 192 a b^3+ 96 a^2 b c - 192 a b c^2,\\ k_3=&36 a^2 b^2 + 192 b^4 - 20 a^3 c - 288 a b^2 c\\ &+ 132 a^2 c^2 + 384 b^2 c^2 - 288 a c^3 + 192 c^4. \end{align}\] Since the compatibility condition \(c_{uv}=c_{vu}\) must hold, it follows from \((\ref{integ1})\) and the assumption \(b\ne 0\) that \[A^2k_1+ABk_2+B^2k_3=0. \label{integ2}\tag{40}\] Differentiating the left hand side of (40 ) and using (33 )-(36 ), with the aid of a computer algebra system, we get \[\begin{align} (A^2k_1+ABk_2+B^2k_3)_u&=\frac{4(A^3P_1+A^2BP_2+AB^2P_3+B^3P_4)}{a^2+16b^2-8ac+16c^2},\tag{41}\\ (A^2k_1+ABk_2+B^2k_3)_v&=\frac{8(A^3P_5+A^2BP_6+AB^2P_7+B^3P_8)}{a^2+16b^2-8ac+16c^2},\tag{42} \end{align}\] where \[\begin{align} P_1=&-a^5 - 24 a^3 b^2 - 192 a b^4 + 12 a^4 c + 168 a^2 b^2 c+ 384 b^4 c \\ & - 56 a^3 c^2 - 576 a b^2 c^2 + 168 a^2 c^3 + 768 b^2 c^3 - 384 a c^4 + 384 c^5,\\ P_2=&-2 a^4 b - 72 a^2 b^3 - 1920 b^5 + 24 a^3 b c+ 1344 a b^3 c\\ & - 264 a^2 b c^2 - 3840 b^3 c^2 + 1344 a b c^3 - 1920 b c^4,\\ P_3=&32 a^3 b^2 + 960 a b^4 - 14 a^4 c - 888 a^2 b^2 c - 2688 b^4 c + 224 a^3 c^2\\ &+ 4032 a b^2 c^2 - 1272 a^2 c^3 - 5376 b^2 c^3 + 3072 a c^4 - 2688 c^5,\\ P_4=&24 a^2 b^3 + 1152 b^5 - 8 a^3 b c - 960 a b^3 c \\ &+216 a^2 b c^2 + 2304 b^3 c^2 - 960 a b c^3 + 1152 b c^4,\\ P_5=&a^4 b + 60 a^2 b^3 + 576 b^5 - 32 a^3 b c - 672 a b^3 c \\ &+ 252 a^2 b c^2 + 1152 b^3 c^2 - 672 a b c^3 + 576 b c^4,\\ P_6=&-a^5 - 38 a^3 b^2 - 192 a b^4+ 24 a^4 c + 444 a^2 b^2 c + 1344 b^4 c - 218 a^3 c^2 \\ &- 2016 a b^2 c^2 + 924 a^2 c^3 + 2688 b^2 c^3 - 1824 a c^4 + 1344 c^5,\\ P_7=&-4 a^4 b - 180 a^2 b^3 - 960 b^5 + 96 a^3 b c+ 1248 a b^3 c\\ & - 564 a^2 b c^2 - 1920 b^3 c^2 + 1248 a b c^3 - 960 b c^4,\\ P_8=&6 a^3 b^2 - 5 a^4 c - 84 a^2 b^2 c - 192 b^4 c + 50 a^3 c^2\\ & + 288 a b^2 c^2 - 180 a^2 c^3 - 384 b^2 c^3 + 288 a c^4 - 192 c^5. \end{align}\] Thus, it follows from (40 ), (41 ) and (42 ) that \[\begin{align} & A^3P_1+A^2BP_2+AB^2P_3+B^3P_4=0, \tag{43}\\ & A^3P_5+A^2BP_6+AB^2P_7+B^3P_8=0. \tag{44} \end{align}\]

Using a computer algebra system, we find that the resultant of the left-hand sides of (43 ) and (44 ) with respect to \(A\) is, up to a non-zero constant factor, \[B^9(2b^2-ac+2c^2)^3(a^2 + 16 b^2 - 8 a c + 16 c^2)^7 (a^2 + 48 b^2 - 24 a c + 48 c^2)^2Q,\label{R1}\tag{45}\] where \[\begin{align} Q=&5 a^6 + 120 a^4 b^2 + 720 a^2 b^4 + 2304 b^6 - 60 a^5 c - 720 a^3 b^2 c \\ &- 3456 a b^4 c + 300 a^4 c^2 + 3168 a^2 b^2 c^2 + 6912 b^4 c^2 - 1008 a^3 c^3 \\ &- 6912 a b^2 c^3 + 2448 a^2 c^4 + 6912 b^2 c^4 - 3456 a c^5 + 2304 c^6. \end{align}\] Combining (17 ) and (45 ), we find that (45 ) simplifies to \[B^9 \delta^3 (a^2 + 8 \delta)^7 (a^2 + 24 \delta)^2 (5 a^6 + 60 a^4 \delta + 180 a^2 \delta^2 + 288 \delta^3),\label{R2}\nonumber\tag{46}\] which must vanish on \(\mathcal{V}\). Taking into account that \(\nabla a\ne 0\), we have \[B\delta=0.\nonumber\]

If \(\delta\ne 0\), then \(B=0\), which implies that (40 ) and (44 ) reduce respectively to \[\begin{align} &a^4 + 12 a^2 b^2 - 192 b^4 - 8 a^3 c + 96 a b^2 c + 12 a^2 c^2 \nonumber\\ &- 384 b^2 c^2 + 96 a c^3 - 192 c^4=0,\tag{47}\\ &a^4 + 60 a^2 b^2 + 576 b^4 - 32 a^3 c - 672 a b^2 c \nonumber \\ &+252 a^2 c^2 + 1152 b^2 c^2 - 672 a c^3 + 576 c^4=0.\tag{48} \end{align}\] Subtracting (47 ) from (48 ), we derive \[24 (2 b^2 - a c + 2 c^2) (a^2 + 16 b^2 - 8 a c + 16 c^2)=0. \label{abc3}\tag{49}\] Combining (17 ) and (49 ) implies \[a^2+8\delta=0,\nonumber\] which contradicts \(\nabla a\ne 0\). Hence \(\delta=0\), that is, \(K=\epsilon=-1\).

From (17 ) and \(b\ne 0\), we see that \(c\ne0\). Changing the sign of \(e_1\) if necessary, we may assume that \(c>0\). Put \[\begin{align} f=\frac{\sqrt{b^2+c^2-b\sqrt{b^2+c^2}}}{\sqrt{2}(b^2+c^2)},\quad k=\frac{\sqrt{b^2+c^2+b\sqrt{b^2+c^2}}}{\sqrt{2}(b^2+c^2)}, \label{fk} \end{align}\tag{50}\] which are non-zero and unequal everywhere. It follows from (17 ) and (50 ) that \[\begin{align} a=\frac{\sqrt{f^2+k^2}}{fk},\quad b=\frac{k^2-f^2}{(f^2+k^2)^{\frac{3}{2}}},\quad c=\frac{2fk}{(f^2+k^2)^{\frac{3}{2}}}.\label{abcfk} \end{align}\tag{51}\] We make the following change of basis: \[\begin{align} \tilde{e}_1=\frac{k}{\sqrt{f^2+k^2}}e_1+\frac{f}{\sqrt{f^2+k^2}}e_2, \quad \tilde{e}_2=\frac{f}{\sqrt{f^2+k^2}}e_1-\frac{k}{\sqrt{f^2+k^2}}e_2. \label{change} \end{align}\tag{52}\] Then, using (6 ), (51 ) and (52 ), we find \[\begin{align} h(\tilde{e}_1, \tilde{e}_1)=f^{-1}J\tilde{e}_1,\quad h(\tilde{e}_1, \tilde{e}_2)=0, \quad h(\tilde{e}_2, \tilde{e}_2)=k^{-1}J\tilde{e}_2. \end{align}\]

According to [6], there exists a local coordinate system \(\{x, y\}\) such that \(\partial_x=f\tilde{e}_1\) and \(\partial_y=k\tilde{e}_2\), where \(f\) and \(k\) satisfy \[\begin{align} \frac{f_y}{k}=\frac{k_x}{f}, \quad \Bigl(\frac{f_y}{k}\Bigr)_y+\Bigl(\frac{k_x}{f}\Bigr)_x=fk.\label{od1} \end{align}\tag{53}\] The Hamiltonian stationary condition (1 ) is equivalent to (cf. [7]) \[\Bigl(\frac{k}{f}\Bigr)_x+\Bigl(\frac{f}{k}\Bigr)_y=0.\label{od2}\tag{54}\]

By Theorem 3.1 of [8], up to translations and sign, the exact solutions of the over-determined PDE system (53 )-(54 ) are given by \[\begin{align} f=\lambda m\,{\rm csch}\Bigl(\frac{\lambda(m^2x+y)}{\sqrt{1+m^2}}\Bigr), \quad &k=\lambda\,{\rm csch}\Bigl(\frac{\lambda(m^2x+y)}{\sqrt{1+m^2}}\Bigr); \tag{55}\\ f=\lambda m\,\sec\Bigl(\frac{\lambda(m^2x+y)}{\sqrt{1+m^2}}\Bigr), \quad &k=\lambda\,\sec\Bigl(\frac{\lambda(m^2x+y)}{\sqrt{1+m^2}}\Bigr);\tag{56}\\ f=\frac{m\sqrt{1+m^2}}{m^2x+y}, \quad &k=\frac{\sqrt{1+m^2}}{m^2x+y},\tag{57} \end{align}\] where \(\lambda\) and \(m\) are positive real numbers. We note that \(f=mk\). By the first equation of (51 ) we have \[a=\frac{\sqrt{1+m^2}}{mk}.\nonumber\] Furthermore, it follows from (52 ) that \[\begin{align} \partial_u=\frac{1}{1+m^2}(\partial_x+m^2\partial_y),\quad \partial_v=\frac{m}{1+m^2}(\partial_x-\partial_y). \end{align}\]

Since \(a_u\) and \(a_v\) are constant, the solutions (55 ) and (56 ) are excluded. Considering also that \(a_u^2+a_v^2=1\) and \(f\ne k\), we see that \(f\) and \(g\) are given by (57 ) with \(m\ne 1\). In [9], it is proved that the corresponding surface is congruent to the Lagrangian surface obtained from case (2) with \(m\ne 1\). 0◻

References↩︎