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variational methods - prescribed mean curvature equation
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\centerline{\bf Solutions to the prescribed mean curvature equation} 
\bigskip
\centerline{\bf with Dirichlet
conditions by variational methods }
\bigskip
\medskip

\centerline
{P. Amster, P. De N\'apoli and M. C. Mariani}

\centerline
{Universidad de Buenos Aires}

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\lema{Abstract:}

We apply 
variational methods in order to prove that the 
nonlinear system (1) admits at least one regular solution.

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\bigskip

\tit{Introduction}

We consider the problem of finding $u \in H^1(\Omega,\R^3)$ such that

$$\text {(1)}
\cases
\triangle u = 2 H(u) u_x \wedge u_y \text{ in } \Omega &\\
u = \gamma \text{ in } \partial \Omega
\endcases
$$
where $H$ is a given continuously 
differentiable function and $\Omega$ 
is an open subset of $\R^2$ with $C^1$ boundary.

The system of differential equations above 
is the Dirichlet problem for the
prescribed mean
curvature equation 
in isothermal coordinates, also called H-system, which
arises in the
generalized Plateau's 
problem (see [O]).

Hildebrant, Wente (see [W1,W2]) and others 
have studied the problem 
for constant
$H$. 
For this case, 
Brezis-Coron and Struwe  
have shown by variational methods
(see [BC], [S]) that 
if the boundary 
data is small and non constant,
then there are at least two weak solutions.
The problem has been also studied 
by fixed point methods (see [AMR]). 

The goal of this 
work is to show that also for non constant $H$, the
problem may be solved
by variational methods, under rather general 
conditions on $H$ and the boundary data. 

Our main result is the following:

\lema{Theorem 1}
Let
$\gamma \in H^{1/2}(\partial \Omega,\R^3) \cap L^\infty(\partial \Omega,\R^3)$, 
$R= \| \gamma -c\|_{L^\infty(\partial\Omega,\R^3)}$ for some constant $c$, 
and $B \subset \R^3$ the closed ball of radius R
centered at $c$. Then, if $H \in C^1(B)$ satisfies

$$
\| H \|_{L^\infty(B)} < \frac{1}{R}$$
then problem (1) admits a weak solution
$u \in H^1(\Omega,\R^3) \cap L^\infty(\Omega,\R^3)$, with \linebreak
$\| u-c \|_{L^\infty(B)} \le R$.

\rm
{\bf Remark:}
By a result of Bethuel ([B]) 
weak solutions of the mean curvature equation are 
$C^{2,\alpha}(\Omega)$ for $H$ bounded and 
Lipschitz continuous.
Moreover, by a result of Chanillo 
and Li 
([CL]) the solutions are 
continuous up to the boundary. 

\tit{ Variational formulation of the problem }

For simplicity, we may assume that $c=0$
and extend 
$H$ to a function of class $C^1(\R^3)$ 
such that $\| H \|_{L^\infty(\R^3)} < \frac{1}{R}$.

We define the vector field  $q:\R^3 \to \R^3$:

$$ q(x,y,z) = \frac{4}{3} (\int_0^x H(t,y,z) dt,
\int_0^y H(x,t,z) dt,\int_0^z H(x,y,t) dt) $$

It is immediate that $\div \ q = 4 H$ and
$ |q(x,y,z)| \leq  \frac{4}{3} \|H\|_{L^\infty} |(x,y,z)|$.

We'll work in the Banach space $ H^1(\Omega,\R^3) \cap 
L^\infty(\Omega,\R^3) $, with the norm:

$$ \| u \| = \| u \|_{H^1} + \| u \|_{L^\infty} $$
and consider
energy functional
$$ E(u) = D(u) + Q(u) $$
where
$D$ is the Dirichlet's integral and $Q$ is
the functional 
$$ Q(u) = \int_{\Omega} q(u) \cdot (u_x \wedge u_y)$$

We'll use the estimatives

$$ |Q(u)| \leq \int_{\Omega} |q(u)| \cdot |u_x \wedge u_y)| 
\leq \frac{2}{3}
\|H\|_{L^\infty} \|u\|_{L^\infty}\int_{\Omega} |\nabla u|^2 \tag2$$

$$
E(u) \geq \left( 1 - \frac{2}{3} \|H\|_{L^\infty} 
\|u\|_{L^\infty}\right) \int_{\Omega}
|\nabla u|^2 \tag3
$$

\tit {Technical lemmas}

In this section we present some technical facts. 

\lema{Lemma 2} 
$Q$ is continuous in $H^1 \cap L^\infty$.

\lema{Lemma 3}
We consider the functional

$$ I(u,h) = \int_{\Omega} H(u) \cdot (u_x \wedge u_y ) \cdot h $$
where $u \in H^1 \cap L^\infty$ and $h \in L^\infty$.

Then $I$ is continuous. 

\demost{Proof}

Lemma 2 and Lemma 3 follow from simple computations.

\lema {Lemma 4}

The functional $E$ is Gateaux differentiable in 
$H^1 \cap L^\infty$ 
for variations in $H^1_0 \cap L^\infty$, and 

$$  dE(u)(h) = 2 \int_{\Omega} \nabla u \cdot \nabla h + 
 \int_{\Omega} 4H(u) \cdot (u_x \wedge u_y) \cdot h $$
for every $u \in H^1 \cap L^\infty, h \in H^1_0 \cap L^\infty$.

\demost{Proof}

Let us first take $ u,h \in C^1 $ and define a function
$U:\Omega \times \R \to \R^3$ by:
$$ U(x,y,t) = u(x,y) + t \cdot h(x,y) $$
We'll write $u^t(x,y) = U(x,y,t)$.

We recall that the field $q$ in $\R^3$ is
associated to the 2-form $\omega$ given by:

$$ \omega(X,Y) = (X \wedge Y) \cdot q(u) $$
with $$ d\omega = \div (q) du^1 \wedge du^2 \wedge du^3 $$

Applying Stokes theorem to the cylinder 
$C = \Omega \times [0,\varepsilon ]$, we obtain:

$$ \int_{C} d(U^{*}\omega) = \int_{\partial C} U^{*}\omega $$
Moreover,
$$ d(U^{*}\omega) = 
U^{*}(d\omega) = U^{*}(\div q(u) du^1 \wedge du^2 \wedge du^3)
$$

$$ = 4H(U) \cdot \frac{\partial(U^1,U^2,U^3)}{\partial(x,y,t)} dx \wedge dy \wedge dt  
= 4H(u^t) \cdot  (u^t_x \wedge u^t_y) \cdot h $$

We have:

$$ U^{*}\omega = \omega (\frac{\partial U}{\partial x},\frac {\partial
U}{\partial y} ) dx \wedge dy = q(u^t) \cdot (u^t_x \wedge u^t_y) $$
on $\Omega \times \{\varepsilon\}$ and $\Omega \times \{ 0\}$          
and 
$$ U^{*}\omega = \omega(\frac{\partial U}{\partial \sigma},\frac{\partial
U}{\partial t}) d\sigma \wedge dt 
= q(u^t) \cdot ( \frac{ \partial U}{\partial \sigma} \wedge h) d\sigma \wedge
dt $$
on the lateral surface $\partial \Omega \times [0,\varepsilon ]$,
where $\sigma$ is the unit tangent vector on
$\partial \sigma$. 

Thus,

$$ \int_0^\varepsilon \int_{\Omega} 4H(u^t) \cdot (u^t_x \wedge u^t_y) 
\cdot h dx \wedge dy \wedge dt = $$
$$ \int_{\Omega \times \{\varepsilon\}} q(u^\varepsilon) \cdot (u^\varepsilon_x 
\wedge u^\varepsilon_y) dx \wedge dy \wedge dt -
\int_{\Omega \times \{ 0\} } q(u) \wedge (u_x \wedge u_y) \cdot h $$
$$ + \int_{\partial \Omega \times [0,\varepsilon]} q(u) \cdot 
( \frac{ \partial U}{\partial \sigma} \wedge h) d\sigma \wedge dt
$$ 

Assuming that $h=0$ on the boundary of $\Omega$, we get:

$$
\int_0^\varepsilon \int_{\Omega} 4H(u^t) \cdot (u^t_x \wedge u^t_y) \cdot h
= Q(u^\varepsilon) - Q(u).$$

By density, this formula holds for $u \in H^1 \cap L^\infty$ and
$ h \in H^1_0 \cap L^\infty $.

By dominated convergence we conclude that

$$ dQ(u)(h) = \int_{\Omega} 4H(u) \cdot  (u_x \wedge u_y) \cdot h $$
which completes the proof. 


{\bf Remark:} By lemmas 2 and 3
$dE$ is continuous, i.e. $E \in C^1(H^1_0 \cap L^\infty)$. 

\newpage

\lema {Corollary 5}

If $u$ is a critical point of functional $E$ for variations in  $H^1_0
\cap L^\infty$ then $u$ is a weak solution of
(1).

\rm
\tit{Proof of Theorem 1}

We fix $R^{\prime} > R$ such that  $ \| H \|_{L^\infty} R^{\prime} < 1 $
and consider the nonempty set:

$$ K = \{ u \in H^1(\Omega,\R^3): u = \gamma \text{ in } \partial \Omega ,
 \| u \| \leq R^{\prime} \} $$

Then we obtain

\lema{Lemma 6}

$E$ achieves a minimum in $K$.

\demost{Proof}

By (3), the functional 
$E$ is bounded from below in $K$. Let 
$ E_0 = \inf_{v \in K} E(v) $ and 
consider a minimizing sequence $(u^n)$ 

By (3) and Poincar\'e's inequality $u^n$ is
bounded in $H^1$, and then we may suppose that

$$ u^n \to \su \text{ in } H^1 \text{weakly} $$
and
$$ u^n \to \su \text{ a.e. } $$

By the continuity of $q$ we conclude that:

$$ q(u^n) \to q(\su) \text{ a.e. } $$
and since $|q(u^n)| \leq \frac{4}{3} \|H \|_{L^\infty} $, 
we may also assume that:

$$ q(u^n) \to q(\su) \text{ in } L^\infty \text{ weak* }$$

Set $ \theta^n = u^n - u \in H^1_0$. Then  
$ \| \theta^n \|_{L^\infty} \leq 2R^{\prime} $, and

$$ E(u^n) = \int_{\Omega} |\nabla \su|^2  + 
2 \int_{\Omega} \nabla u \cdot \nabla \theta^n$$ 
$$ + \int_{\Omega} |\nabla \theta^n|^2 
+ \int_{\Omega} q(u^n) \cdot (\su_x + \theta^n_x) \wedge (\su_y +
\theta^n_y) $$

Thus,
$$\int_{\Omega} \nabla u \cdot \nabla \theta^n \to 0 $$

Moreover, being
$$\int_{\Omega} q(u^n) \cdot (\su_x \wedge \theta^n_y) =
- \int_{\Omega} \theta^n_y \cdot (q(u^n) \wedge \su_x) $$
then
$$\int_{\Omega} q(u^n) \cdot (\su_x \wedge \theta^n_y)  \to 0,$$
since
$ \theta^n_y \to 0 $ weakly in  $L^2$, and 

$$ |q(u^n) \wedge \su_x|^2 \leq 
   \left( \frac{4}{3} |u^n|\right )^2 \cdot |\su_x|^2 
\leq \left( \frac{4}{3} R^{\prime} \right) |\su_x|^2.$$

In the same way, we conclude that 

$$ \int_{\Omega} q(u^n) \cdot (\theta^n_x \wedge \su_y) \to 0 $$

We also have the estimative:

$$ |\int_{\Omega} q(u^n) \cdot  \theta^n_x \wedge \theta^n_y| 
\leq \int_{\Omega} 4 \|H\|_{L^\infty} \|u^n)\|_{L^\infty} 
|\theta^n_x \wedge \theta^n_y|
$$
$$ 
\leq \frac{2}{3} \|H\|_{L^\infty} R^{\prime} \int_{\Omega} |\nabla \theta^n|^2 
\leq \frac{2}{3} \int_{\Omega} |\nabla \theta^n|^2 
$$

Finally as $ q(u^n) \to q(\su) $ weak* in $L^\infty$:

$$ \int_{\Omega} q(u^n) \cdot (\su_x \wedge \su_y) =
\int_{\Omega} q(\su) \cdot (\su_x \wedge \su_y) + o(1) $$

Then,
$$ E(u^n) \geq E(u) + o(1) + \frac{1}{3} \int_{\Omega} |\nabla \theta^n|^2 $$
and we can conclude that
$E(u) = E_0 $ and $ \int_{\Omega} |\nabla \theta^n|^2 \to 0$.

\lema{Lemma 7}

Let $u \in K $ satisfy $E(u) = \inf_{v \in K} E(v)$.
Then $u$ is a weak solution of (1).

\demost{Proof}

Let us fix a nonnegative test function 
$ \eta \in 
{\cal D}(\Omega)$.
Then $ u(1-\varepsilon \eta) \in K $ for $\varepsilon \geq 0 $
small enough.

As the function
$ \varphi(\varepsilon) = E(u(1-\varepsilon \cdot \eta))$
has a minimum in $ \varepsilon = 0 $, then
$ \varphi^{\prime}(0) = dE(u)(-u \cdot \eta) \geq 0 $, i.e.

$$
 2 \int_{\Omega} \nabla u \cdot \nabla (-\eta \cdot u) 
+  \int_{\Omega} 4H(u) (u_x \wedge u_y) \cdot (-\eta u) \geq 0 \; 
\forall \; \eta \in {\cal D}(\Omega)
$$

By Green's formula,

$$ \int_{\Omega} \triangle(|u|^2) \cdot \eta = 
- \int_{\Omega} \nabla(|u|^2) \nabla(\eta)
= -\int_{\Omega} 2  ( \nabla u \cdot  u ) \cdot \nabla \eta $$ 

It follows 
that 

$$ -\frac{1}{2} \triangle |u|^2 + |\nabla u |^2 + 2H(u) u \cdot u_x \wedge 
u_y \leq 0 \text { in } {\cal D}^{\prime}(\Omega) $$

Then, as 

$$ |2H(u) u \cdot ( u_x \wedge u_y)| \leq \| H \|_{L^\infty} 
R^{\prime} |\nabla u|^2 $$
it follows that

$$ -\triangle |u|^2 \leq 0 \text { in } { \cal D}^{\prime}(\Omega) $$
and then by Stampacchia's maximum principle (see
[GT]) we get:

$$ sup_{\Omega} |u| = sup_{\partial \Omega } |u| = R $$

Since $ R < R^{\prime} $, $u$ is interior in $K$. 
It follows that 
$dE(u)(v) = 0$ for any $v \in H^1_0 \cap L^\infty $, 
and by corollary 5 $u$ 
is a weak solution of (1).

\bigskip

\tit{References}

[AMR] Amster P. Mariani, M.C, Rial, D.F: Existence and uniqueness
of H-System's 
solutions with Dirichlet 
conditions. 
To appear in Nonlinear Analysis, Theory, Methods, and Applications.

[B] Bethuel, F:  Un r\'esultat de regularit\'e pour les
solutions de l'equation des surfaces \`a courboure moyenne prescrite. 
R.Acad. Sci. Paris S\'er. I Math 314 (1992) $n^{o}$ 13, 1003-100.

[B-C] Brezis, H., Coron, J., Multiple Solutions of H-Systems and 
Rellich's conjecture. Comm.Pure Appl. Math 37, 1984, 149-187.

[CL] Chanillo S., Li Y.: Continuity of Solutions 
of Uniformly Elliptic Equations in $\R^2$. Manuscripta math. 
77,415-433 (1992)

[GT] Gilbarg, D., Trudinger N.: Elliptic Partial 
Differential Equations of Second Order.
Springer-Verlag 1977
 
[O] Osserman, R: A Survey of Minimal Surfaces. Van
Nostrand Reinhold Company, 1969

[S] Struwe, M.: Variational Methods, Applications to Nonlinear 
Partial Differential Equations and Hamiltonian Systems, 
Springer-Verlag, p. 180.

[W1] Wente, H.: An Existence Theorem for Surfaces 
of Constant Mean Curvature. Journal of Mathematical Analysis and
Applications 26, p. 318-344 (1969)

[W2] Wente, H.: The differential equation 
$\Delta X = 2H (X_u \wedge X_v)$ 
with vanishing boundary values.
Proceedings of the American Mathematical Society 50 (1975), 131-7. 

\bigskip

{\bf P.Amster$^*$, P. De N\'apoli and M. C. Mariani$^*$}

\noi Dpto. de Matem\'atica, 
Fac. de Cs. Exactas y Naturales, Universidad de Buenos Aires.

\noi $^*$ Consejo Nacional de Investigaciones Cient\'\i ficas y T\'ecnicas 
(CONICET)

\bigskip

{\bf Address for correspondence:} 

\noi Prof. P. Amster and M. C. Mariani,

\noi Dpto. de Matem\'atica - Fac. de Cs. Exactas y Naturales, UBA

\noi Pab. I, Ciudad Universitaria (1428) 

\noi Buenos Aires, Argentina

{\bf E-mail:} pamster\@dm.uba.ar - mcmarian\@dm.uba.ar



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