A variational principle for tangent vector fields

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

The gradient of a Neumann eigenfunction is a vector field tangent to the boundary. Rohleder discovered that these gradients are exactly the minimizers of a variational problem for tangent vector fields, with the same minimum value \(\mu\) [Roh21, Theorem 1.1]. This lesson proves that principle on our class of domains, with its equality case, and an identity that rewrites the divergence–curl energy as a Dirichlet energy plus a boundary curvature term. The next lesson transfers this energy to the boundary circle.

Notation is that of Neumann eigenfunctions and the hot-spots problem. In particular \(\Omega\) is simply connected, \(\mu<\lambda_D\) (Proposition 4.1 there), and Green's formulas (2.1)–(2.2) and the Poincaré inequalities of Lemma 2.3 there are available.

1. The vector variational principle

A vector field \(X\in H^1(\Omega;\mathbb R^2)\) is tangent if \(\operatorname{tr}X\cdot\nu=0\) almost everywhere on \(\partial\Omega\). For such fields put \[ \begin{gathered} \mathcal Q(X)=\int_\Omega\bigl(|\operatorname{div}X|^2+|\operatorname{curl}X|^2\bigr),\\ \operatorname{curl}X=\partial_1X_2-\partial_2X_1, \end{gathered} \] and write \(\nabla^\perp j=(-\partial_2j,\partial_1j)\) for a scalar \(j\).

Proposition 1.1 (Rohleder's principle; [OpenAI-HS, Proposition 3.1]). For every tangent \(X\in H^1(\Omega;\mathbb R^2)\), \[ \mathcal Q(X)\ge\mu\int_\Omega|X|^2, \tag{1.1} \] with equality if and only if \(X=\nabla v\) for some \(v\in V\).

Proof. Step 1: a Neumann potential. By Lemma 2.3(2) of the first lesson, \(\int\nabla v\cdot\nabla\phi\) is an inner product on the mean-zero subspace of \(H^1(\Omega)\), equivalent to the \(H^1\) inner product. The Riesz–Fréchet theorem gives a mean-zero \(v\in H^1(\Omega)\) with \(\int\nabla v\cdot\nabla\phi=-\int(\operatorname{div}X)\phi\) for mean-zero \(\phi\). For \(\phi=1\) both sides vanish, the right side because Green's formula (2.1) of the first lesson and the tangent condition give \(\int\operatorname{div}X=\int_{\partial\Omega}\operatorname{tr}X\cdot\nu\,ds=0\). So the identity holds for all \(\phi\in H^1(\Omega)\). By (ER N) with \(k=0\), \(v\in H^2(\Omega)\), and as in the proof of Proposition 3.1(3) of the first lesson, \(\Delta v=\operatorname{div}X\) almost everywhere and \(\partial_\nu v=0\).

Step 2: a Dirichlet potential. By (B10) and (B29) of Boundary energy, local inverses, and harmonic data, there is \(j\in H^1_0(\Omega)\cap H^2(\Omega)\) with \(\Delta j=\operatorname{curl}X\) almost everywhere.

Step 3: \(\nabla^\perp j\) is tangent. For \(F\in C^2(\overline\Omega)\), \(\nabla^\perp F\cdot\nu=-\nabla F\cdot\tau=-\partial_\tau F\), and (SOB 1) applied to the divergence-free field \(\nabla^\perp F\) gives \(\int_{\partial\Omega}\partial_\tau F\,ds=0\). Let \(j_n\in C^\infty(\overline\Omega)\) converge to \(j\) in \(H^2(\Omega)\) (SOB 2), and let \(\psi\in C^\infty(\overline\Omega)\). Applying this to \(F=j_n\psi\), \[ \begin{aligned} &\int_{\partial\Omega}(\nabla^\perp j_n\cdot\nu)\psi\,ds\\ &\quad=-\int_{\partial\Omega}(\partial_\tau j_n)\psi\,ds\\ &\quad=\int_{\partial\Omega}j_n\,\partial_\tau\psi\,ds . \end{aligned} \] As \(n\to\infty\), the traces converge in \(L^2(\partial\Omega)\), and the right side tends to \(\int\operatorname{tr}j\,\partial_\tau\psi\,ds=0\). Since such \(\psi\) have dense traces, \(\operatorname{tr}(\nabla^\perp j)\cdot\nu=0\).

Step 4: the remainder is a gradient. Put \(Z=X-\nabla v-\nabla^\perp j\in H^1(\Omega;\mathbb R^2)\). It is tangent, and since \(\operatorname{div}\nabla^\perp j=0\), \(\operatorname{curl}\nabla v=0\) and \(\operatorname{curl}\nabla^\perp j=\Delta j\), \[ \operatorname{div}Z=0,\qquad\operatorname{curl}Z=0 . \] In the sense of distributions \(\Delta Z_1=\partial_1\operatorname{div}Z-\partial_2\operatorname{curl}Z=0\) and \(\Delta Z_2=\partial_2\operatorname{div}Z+\partial_1\operatorname{curl}Z=0\), so \(Z_1,Z_2\) are real analytic in \(\Omega\) (Anisotropic derivative classes and analyticity, Theorem 3.1). The Cauchy–Riemann equations for \(Z_1-iZ_2\) are exactly \(\operatorname{div}Z=0\) and \(\operatorname{curl}Z=0\), so \(Z_1-iZ_2\) is holomorphic, and it has a primitive \(G\) on the simply connected domain \(\Omega\) (Lemma A.2 of Dirichlet series and Euler products). The function \(r=\operatorname{Re}G\) satisfies \(\nabla r=(\operatorname{Re}G',-\operatorname{Im}G')=Z\) in \(\Omega\).

We show \(r\in H^1(\Omega)\). For \(M>0\) let \(r_M=\max(-M,\min(r,M))\). Approximating the truncation by smooth functions with derivative in \([0,1]\), as in Lemma 3.2 of the first lesson, shows that \(r_M\) has weak gradient \(\mathbf 1_{\{|r|<M\}}Z\); hence \(r_M\in H^1(\Omega)\) with \(|\nabla r_M|\le|Z|\). Fix a disc \(B\) with \(\overline B\subseteq\Omega\). By Lemma 2.3(2) of the first lesson, \[ \begin{aligned} \|r_M\|&\le C_B\bigl(\|\nabla r_M\|+|(r_M)_B|\bigr)\\ &\le C_B\Bigl(\|Z\|+\max_{\overline B}|r|\Bigr). \end{aligned} \] Letting \(M\to\infty\), Fatou's lemma gives \(r\in L^2(\Omega)\). Then \(r_M\to r\) and \(\nabla r_M\to Z\) in \(L^2\) by dominated convergence, so \(r\in H^1(\Omega)\) with \(\nabla r=Z\). Green's formula (2.1) of the first lesson, with the tangent condition and \(\operatorname{div}Z=0\), gives \[ \begin{aligned} \int_\Omega|Z|^2&=\int_\Omega Z\cdot\nabla r=-\int_\Omega(\operatorname{div}Z)r\\ &\quad+\int_{\partial\Omega}(\operatorname{tr}Z\cdot\nu)\operatorname{tr}r\,ds=0 . \end{aligned} \] Hence \[ X=\nabla v+\nabla^\perp j . \tag{1.2} \]

Step 5: the energy. By (2.1) of the first lesson with the tangent field \(\nabla^\perp j\), \(\int\nabla v\cdot\nabla^\perp j=-\int v\operatorname{div}\nabla^\perp j=0\). So the two summands in (1.2) are orthogonal in \(L^2\), and \(\|X\|^2=\|\nabla v\|^2+\|\nabla j\|^2\). Also \(\operatorname{div}X=\Delta v\) and \(\operatorname{curl}X=\Delta j\), so \[ \mathcal Q(X)=\|\Delta v\|^2+\|\Delta j\|^2 . \] Green's formula (2.2) with \(\partial_\nu v=0\), the Cauchy–Schwarz inequality and the definition of \(\mu\) (\(v\) has mean zero) give \[ \begin{aligned} &\|\nabla v\|^2=-\int v\Delta v\\ &\quad\le\|v\|\|\Delta v\|\\ &\quad\le\mu^{-1/2}\|\nabla v\|\|\Delta v\|, \end{aligned} \tag{1.3} \] so \(\|\Delta v\|^2\ge\mu\|\nabla v\|^2\). Since \(\operatorname{tr}j=0\), the same argument with \(\lambda_D\) gives \(\|\Delta j\|^2\ge\lambda_D\|\nabla j\|^2\). Therefore \[ \begin{aligned} \mathcal Q(X)&\ge\mu\|\nabla v\|^2+\lambda_D\|\nabla j\|^2\\ &=\mu\|X\|^2+(\lambda_D-\mu)\|\nabla j\|^2, \end{aligned} \] and (1.1) follows from \(\mu<\lambda_D\).

Step 6: equality. If equality holds, then \(\nabla j=0\), so \(j=0\) by Lemma 2.3(1) of the first lesson, and \(\|\Delta v\|^2=\mu\|\nabla v\|^2\). If \(\Delta v=0\), then (1.3) gives \(\nabla v=0\) and \(X=0=\nabla0\). Otherwise the inequalities in (1.3) are equalities, so \(\|\nabla v\|^2=\mu\|v\|^2\), and \(v\in V\) by Proposition 3.1(2) of the first lesson. Conversely, for \(v\in V\) the field \(\nabla v\) is tangent, \(\operatorname{div}\nabla v=-\mu v\) and \(\operatorname{curl}\nabla v=0\), so \(\mathcal Q(\nabla v)=\mu^2\|v\|^2=\mu\|\nabla v\|^2\). \(\square\)

For tangent \(X,Y\in H^1(\Omega;\mathbb R^2)\) put \[ \begin{aligned} &\mathcal E(X,Y)=\int_\Omega\operatorname{div}X\operatorname{div}Y\\ &\quad+\int_\Omega\operatorname{curl}X\operatorname{curl}Y-\mu\int_\Omega X\cdot Y, \end{aligned} \] and \(\mathcal E(X)=\mathcal E(X,X)\).

Corollary 1.2. \(\mathcal E(X)\ge0\) for every tangent \(X\in H^1(\Omega;\mathbb R^2)\). If \(\mathcal E(X)=0\), then \(X=\nabla v\) for some \(v\in V\), and \(\mathcal E(X,Y)=0\) for every tangent \(Y\in H^1(\Omega;\mathbb R^2)\).

Proof. The first two assertions are Proposition 1.1. For the third, \(0\le\mathcal E(X+tY)=2t\,\mathcal E(X,Y)+t^2\mathcal E(Y)\) for all real \(t\), which forces \(\mathcal E(X,Y)=0\). \(\square\)

Rohleder proved the principle for bounded simply connected Lipschitz domains with piecewise smooth boundary and convex corners, in the form of a minimum over tangent fields of the Dirichlet energy plus a curvature term [Roh21, Theorem 1.1]; the identity of Section 2 connects the two forms. His curvature is negative where the domain is convex; the curvature \(\kappa\) below is positive there.

2. The curvature identity

Let \(\kappa=\det(\tau,\partial_\tau\tau)\) on \(\partial\Omega\), where \(\partial_\tau\) is differentiation with respect to arclength in the direction \(\tau\). On the unit circle, \(\tau=(-\sin\theta,\cos\theta)\), \(\partial_\tau\tau=(-\cos\theta,-\sin\theta)\) and \(\kappa=1\).

Lemma 2.1 ([OpenAI-HS, Lemma 3.2]). For every tangent \(X\in C^2(\overline\Omega;\mathbb R^2)\), \[ \begin{aligned} \mathcal Q(X)&=\int_\Omega|\nabla X|^2\\ &\quad+\int_{\partial\Omega}\kappa|X|^2\,ds, \end{aligned} \tag{2.1} \] where \(|\nabla X|^2=\sum_{i,k}(\partial_kX_i)^2\).

Proof. Expanding the squares, \[ \begin{aligned} &|\operatorname{div}X|^2+|\operatorname{curl}X|^2\\ &\quad=|\nabla X|^2+2\det\nabla X, \end{aligned} \] with \(\det\nabla X=\partial_1X_1\partial_2X_2-\partial_2X_1\partial_1X_2\). Since \(\operatorname{div}\nabla^\perp F=0\), one has \(\operatorname{div}(X_1\nabla^\perp X_2)=\nabla X_1\cdot\nabla^\perp X_2\), which is \(-\det\nabla X\), and \(\operatorname{div}(X_2\nabla^\perp X_1)=\det\nabla X\). By (SOB 1) and \(\nabla^\perp F\cdot\nu=-\partial_\tau F\), \[ \begin{aligned} &2\int_\Omega\det\nabla X\\ &\quad=\int_{\partial\Omega}(X_2\nabla^\perp X_1-X_1\nabla^\perp X_2)\cdot\nu\,ds\\ &\quad=\int_{\partial\Omega}(X_1\partial_\tau X_2-X_2\partial_\tau X_1)\,ds . \end{aligned} \] On \(\partial\Omega\) write \(X=a\tau\) with \(a=X\cdot\tau\in C^1(\partial\Omega)\). Then \[ \begin{aligned} &X_1\partial_\tau X_2-X_2\partial_\tau X_1\\ &\quad=a^2(\tau_1\partial_\tau\tau_2-\tau_2\partial_\tau\tau_1)\\ &\qquad+a\,\partial_\tau a\,(\tau_1\tau_2-\tau_2\tau_1)\\ &\quad=\kappa a^2=\kappa|X|^2 .\qquad\square \end{aligned} \]

The next lesson uses (2.1) for fields that are smooth up to the boundary, and passes to Lipschitz boundary data by an approximation argument.

3. Exercises

3.1. On the unit disc let \(X=(-x_2,x_1)\). Show that \(X\) is tangent, and check (2.1) by computing both sides.

3.2. For the field of Exercise 3.1, find \(v\) and \(j\) in (1.2), and show \(\mathcal E(X)>0\).

3.3. On the unit disc let \(X=(1,0)\). Show \(\mathcal Q(X)=0<\mu\int|X|^2\), and explain why this does not contradict Proposition 1.1.

3.4. In a boundary chart at \(x_0\) (Section 1 of the first lesson), show that \(\kappa(x_0)=\gamma''(0)\). Deduce that \(\kappa\ge0\) when \(\Omega\) is convex.

3.5. Show that the tangent condition is used in Step 1 of the proof of Proposition 1.1, and give an example of a nontangent field \(X\) on the unit disc for which no \(v\in H^1\) satisfies \(\Delta v=\operatorname{div}X\), \(\partial_\nu v=0\).

4. Solutions

3.1. On the circle \(X=\tau\), so \(X\cdot\nu=0\). Here \(\operatorname{div}X=0\) and \(\operatorname{curl}X=2\), so \(\mathcal Q(X)=4\pi\). The derivatives are \(\partial_2X_1=-1\), \(\partial_1X_2=1\) and \(0\) otherwise, so \(\int|\nabla X|^2=2\pi\); with \(\kappa=1\) and \(|X|=1\) on the circle, the boundary term is \(2\pi\).

3.2. \(\operatorname{div}X=0\) gives \(v=0\). The function \(j=(|x|^2-1)/2\) vanishes on the circle and \(\nabla^\perp j=(-x_2,x_1)=X\), so \(X=\nabla^\perp j\). Step 5 of the proof gives \(\mathcal E(X)\ge(\lambda_D-\mu)\|\nabla j\|^2>0\), because \(\nabla j=x\ne0\).

3.3. \(\operatorname{div}X=\operatorname{curl}X=0\), while \(\int|X|^2=\pi\). On the circle \(X\cdot\nu=x_1\), which is not zero, so \(X\) is not tangent.

3.4. With \(\Omega\) above the graph, \(\nu=(\gamma',-1)/\sqrt{1+\gamma'^2}\) and \(\tau=i\nu=(1,\gamma')/\sqrt{1+\gamma'^2}\). At \(x_1=0\), \(\gamma'=0\), \(ds=dx_1\), \(\tau=(1,0)\) and \(\partial_\tau\tau=(0,\gamma''(0))\), so \(\kappa=\gamma''(0)\). If \(\Omega\) is convex, its intersection with the chart rectangle lies above every tangent line of the graph near \(0\), so \(\gamma\) is convex near \(0\) and \(\gamma''(0)\ge0\).

3.5. The compatibility \(\int\operatorname{div}X=0\) is needed for \(\phi=1\) in Step 1. For \(X=(x_1,x_2)\) on the unit disc, \(\operatorname{div}X=2\); a solution would satisfy \(\int\Delta v=\int_{\partial\mathbb D}\partial_\nu v\,ds=0\), but \(\int\operatorname{div}X=2\pi\).

References