\magnification=\magstep1
\def\d{\displaystyle}
\def\ip#1#2{\langle #1,#2\rangle}
\def\phi{\varphi}
\def\ni{\noindent}
\centerline{\bf On constrained extrema}
\centerline{\it Thomas I.\ Vogel}
\centerline{\it Department of Mathematics}
\centerline{\it Texas A\&M University}
\centerline{\it College Station, TX}
\centerline{\it 77840}
\medskip
{\narrower\bf Assume that $I$ and $J$ are smooth functionals defined on a
Hilbert space $H$. We derive sufficient conditions for $I$ to have a
local minimum at $y$ subject to the constraint that $J$ is
constantly $J(y)$.\smallskip}


The first order necessary condition for $I$ to have a constrained
minimum at $y$ is that for some constant $\lambda$, $I'_y+\lambda
J'_y$ is identically zero. Here $I'_y$ and $J'_y$ are the
Fr\'echet derivatives of $I$ and $J$ at $y$. For the rest of the
paper, we assume that $y$ in $H$ satisfies this necessary
condition.

A common misapprehension (upon which much of the stability results
for capillary surfaces has been based) is to assume that if the
quadratic form $I''_y+\lambda J''_y$ is positive definite on the
kernel of $J'_y$ then $I$ has a local constrained minimum at $y$.
This is not correct in a Hilbert space of infinite dimension; Finn
[1] has supplied a counterexample in the unconstrained case, and
the same difficulty will occur in the constrained case. In the
unconstrained case, if (as often occurs in practice) the spectrum
of $I''_y$ is discrete and 0 is not a cluster point of the
spectrum, then $I''_y$ positive definite at a critical point $y$
implies that $I''_y$ is strongly positive, (i.e., there exists
$k>0$ such that $I''_y(x)\geq k\Vert x\Vert^2$ holds for all
$x$), and this in turn {\it does\/} imply that $y$ is a local
minimum (see [2]). However, in the constrained case, things are
not so easy. Even if $I''_y+\lambda J''_y$ has a nice spectrum (in
some sense), it is not clear that $I''_y+\lambda J''_y$ being
positive definite on the kernel of $J'_y$ implies that this
quadratic form is strongly positive on the kernel, nor that strong
positivity implies that $y$ is a local minimum.

In [3], Maddocks obtained sufficient conditions for $I''_y+\lambda
J''_y$ to be positive definite on the kernel of $J'_y$. As
Maddocks points out, this is not quite enough to say that $I$ has
a constrained minimum at $y$. Remarkably, essentially the same
conditions as Maddocks obtained  for positive definiteness do in
fact imply that $I$ has a strict local minimum at $y$ subject to
the constraint $J=J(y)$, as we shall see.

For any $h\in H$ we may say $\d J(y+h)-
J(y)=J'_y(h)+{1\over\d2}J''_y(h)+ \epsilon_1(h)\Vert h\Vert^2$,
where $\epsilon_1$ goes to zero as $\Vert h\Vert$ goes to zero. If
we consider an $h$ for which $J(y+h)=J(y)$, then of course $\d
0=J'_y(h)+{1\over\d2}J''_y(h)+\epsilon_1(h)\Vert h\Vert^2$. Now,
for that $h$ we have
$$\eqalign{
\Delta I=I(y+h)-I(y)=&I'_y(h)+{1\over2}I''_y(h)+\epsilon_2\Vert h\Vert^2\cr
=&-\lambda J'_y(h)+{1\over2}I''_y(h)+\epsilon_2\Vert h\Vert^2\cr
=&{1\over2}(I''_y+\lambda
J''_y)(h)+(\lambda\epsilon_1+\epsilon_2)\Vert h\Vert^2\cr}\eqno(1)
$$

Since $I''_y+\lambda J''_y$ is a bilinear form, there is a linear
operator $A$ defined on $H$ so that $(I''_y+\lambda
J''_y)(u,v)=\langle u, Av\rangle$. Similarly there is some element
of $H$, call it $\nabla J$, so that $J'_y$ applied to a vector $h$
is $\langle h,\nabla J\rangle$. Let $\sigma(A)$ be the spectrum of
$A$. There are three cases which often arise in practice:

{\it Theorem 1:\/}  If $\sigma(A)\cap(-\infty,c]=\emptyset$ for
some $c>0$, then $I$ has a constrained minimum at $y$.

{\it Proof:} From (1) we may write $\Delta I$ as $\ip h{Ah}
+(\lambda\epsilon_1+\epsilon_2)\Vert h\Vert^2$. But $\ip h
{Ah}\geq c\Vert h\Vert^2$ (this is easily verified using the
spectral theorem, see [5]), so for $h$ sufficiently small, $\Delta
I$ is positive.

{\it Theorem 2:\/} Suppose that $\sigma(A)\cap (-\infty,\epsilon]$
consists of a single negative eigenvalue $\lambda_0$ for some
$\epsilon>0$. Let $\zeta$ solve $A\zeta=\nabla J$. ($A$ will be
invertible.) $I$ has a constrained minimum at $y$ if $J_y'(\zeta)=\ip
\zeta{A\zeta}<0$, and $I$ does not have a constrained minimum at
$y$ if $J_y'(\zeta)=\ip \zeta{A\zeta}>0$.

The proof of Theorem 2 will proceed in a series of steps.

{\it Step 1:\/} Assume that $\ip \zeta{A\zeta}<0$. Then
$I''_y+\lambda J''_y$ is strongly positive on the kernel of
$J'_y$.

{\it Proof:\/} Take $x$ in the kernel of $J'_y$. As in [4], $x$
may be written as $v+\alpha\zeta$, where $v$ is perpendicular to
$\phi_0$, the eigenfunction corresponding to $\lambda_0$. (The key
to this calculation is that $\ip \zeta{\phi_0}\neq0$. But if
$\zeta$ is orthogonal to $\phi_0$, it can be shown that $\ip
\zeta{A\zeta}>0$.) One can verify that $\ip x{Ax}=\ip v{Av}-
\alpha^2\ip \zeta{A\zeta}$, so that $\ip x{Ax} \geq \ip v{Av}$.

Let $\{E_\lambda\}$ be the spectral family associated with $A$, so
that $\d A=\int_{-\infty}^\infty\lambda dE_\lambda$. By our
assumption on $\sigma(A)$, $\d A=\lambda_0
E_{\lambda_0}+\int_\epsilon^\infty \lambda dE_\lambda$, where
$E_{\lambda_0}$ is orthogonal projection onto $\phi_0$. Therefore,
$$
\langle v,Av\rangle=\langle v,\lambda_0 E_{\lambda_0}(v)\rangle
+\int_\epsilon^\infty \lambda d\Vert E_\lambda v\Vert^2
$$
The first term vanishes, so that
$$
\ip v{Av}\geq \epsilon\int_\epsilon^\infty d\Vert E_\lambda
v\Vert^2\geq \epsilon\int_{-\infty}^\infty d\Vert E_\lambda
v\Vert^2\geq \epsilon\Vert v\Vert^2
$$
Therefore, $\ip x{Ax}\geq \epsilon\Vert v\Vert^2$.

To conclude the proof that $I''_y+\lambda J''_y$ is strongly positive on the
kernel of $J'_y$, we need to show that $\Vert v\Vert\geq k\Vert
x\Vert$ for some fixed positive constant $k$. Assume without loss
of generality that $\Vert x\Vert=1$. For any fixed $x$, $\Vert
v\Vert$ is greater than or equal to the distance from $x$ to the
line $\{c\zeta: c\in{\bf R}\}$. Consider the projection of $x$
onto $\zeta$. Its length is $\vert\ip
x{\zeta/\Vert\zeta\Vert}\vert$.
We may write $\zeta$ as $\beta\nabla J+\hat\zeta$, where
$\hat\zeta$ is perpendicular to $\nabla J$. We cannot have $\beta$
equaling 0, since by assumption, $\ip\zeta{A\zeta}=\ip
\zeta{\nabla J}<0$.

Then the projection has length at most $\Vert x\Vert\Vert
\hat\zeta\Vert/\Vert\zeta\Vert$. But
$\Vert\hat\zeta\Vert<\Vert\zeta\Vert$ (since $\beta\neq0$).
Letting $\gamma$ equal $\Vert\hat\zeta\Vert/\Vert\zeta\Vert$, we
have $\gamma<1$ and the length of the vector component of $x$
perpendicular to $\zeta$ is greater than or equal to $\sqrt{1-
\gamma^2}$. But $\Vert v\Vert$ is greater than or equal to the
length of that component, so we get our $k$ to be $\sqrt{1-
\gamma^2}$, concluding step 1.

{\it Step 2:} If $\ip \zeta{A\zeta}<0$, then $I$ has a minimum at
$y$ subject to the constraint $J=J(y)$.

{\it Proof:\/} Take an $h$ for which $J(y+h)=J(y)$. Now $h$ need
not be in the kernel of $J'_y$, but we may write $h$ as
$h_1+\alpha\zeta$, where $h_1$ is in the kernel of $J'_y$, by
taking $\alpha$ to be $\ip h{\nabla J}/\ip \zeta{\nabla J}$. (Note
that $\ip \zeta{\nabla J}=\ip\zeta{A\zeta}\neq0$.) Substituting
into equation (1),
$$
\Delta I={1\over2}\ip {h_1}{Ah_1}+\alpha\ip {h_1}{A\zeta}+
{1\over2} \alpha^2 \ip \zeta {A\zeta}+ (\lambda \epsilon_1+
\epsilon_2) \Vert h \Vert^2 \eqno{(2)}
$$
However, $\ip {h_1}{A\zeta}=\ip {h_1}{\nabla J}=0$, causing this
term to vanish. We have $0=\Delta J=J'_y(h)+\epsilon_3\Vert
h\Vert$, where $\epsilon_3$ tends to 0 as $\Vert h\Vert$ tends to
0. Thus $\alpha^2=\epsilon_3^2\Vert h\Vert^2$, and we conclude
that
$$
\Delta I={1\over2}\ip {h_1}{Ah_1}+\epsilon\Vert h\Vert^2
$$
where $\epsilon$ tends to zero as $\Vert h\Vert$ tends to 0. From
step 1, $A$ is strongly positive on the kernel of $J'_y$, so
$$
\Delta I\geq  {k\over2}\Vert h_1\Vert^2+\epsilon\Vert h\Vert^2
$$
Since $h=h_1+\alpha\zeta$, with $\alpha=-\epsilon_3\Vert h\Vert$,
it is easy to see that for $\Vert h\Vert$ sufficiently small there
holds $\d\Vert h_1\Vert\geq{1\over\d2}\Vert h\Vert$. Thus
$$
\Delta I\geq\Vert h\Vert\left({k\over8}+\epsilon\right)
$$
which must be greater than 0 for $\Vert h\Vert$ sufficiently
small. Therefore $I$ has a minimum at $y$ subject to the
constraint $J=J(y)$, concluding the proof of step 2 and the first
half of Theorem 2.

{\it Step 3:\/} Suppose that $\ip \zeta{A\zeta}>0$. Then $I$ does
not have a minimum at $y$ subject to the constraint $J=J(y)$.

{\it Proof:\/} First, $I''_y+\lambda J''_y$ is no longer positive
definite on the kernel of $J'$. Indeed, $\eta=\phi_0+c\zeta$ is in
the kernel of $J'_y$ if $\d c=-{\ip{\phi_0}{\nabla
J}\over\d\ip{\zeta}{\nabla J}}=-{\ip{\phi_0}{\nabla J}\over\d
\ip{\zeta}{A\zeta}}$, but one can verify that $\ip \eta{A\eta}<0$.

Now consider $f(r,s)=J(y+r\eta+s\nabla J)-J(y)$, a differentiable
function of $r$ and $s$. Then $\nabla f(0,0)=(0,\Vert \nabla
J\Vert^2)$, so the zero set of $f$ is tangent to the $r$ axis at
the origin. From this we conclude that there is a function $s(r)$
so that $J(y+r\eta+s(r)\nabla J)-J(y)=0$, with
$\d\lim_{r\to0}{s(r)\over\d r}=0$. From equation (1), for
$h=r\eta+s(r)\nabla J$ we have
$$
\Delta I=(I''+\lambda J'')(r\eta+s(r)\nabla
J)+(\lambda\epsilon_1+\epsilon_2) \Vert r\eta+s(r)\nabla J\Vert^2
$$
so that $\Delta I=r^2\ip\eta{A\eta}+o(r^2)$. Thus, for all $r$
sufficiently small $\Delta I<0$, indicating that we do not have a
constrained minimum, concluding the proof of Theorem 2.

{\it Theorem 3:\/} If $\sigma(A)\cap (-\infty,0)$ consists of more
than one point, $I$ does not have a constrained minimum at $y$.

{\it Proof:\/} Suppose that $\nu$ and $\mu$ are in $\sigma(A)\cap
(-\infty,0)$, with $\nu<\mu$. Let $E_\lambda$ be the spectral
decomposition of $A$, so that $E_\lambda$ is not constant in any
neighborhood of $\nu$ nor in any neighborhood containing $\mu$.
Take an $\epsilon>0$ so that the two $\epsilon$ neighborhoods
around $\nu$ and $\mu$ are disjoint and contained in $(-
\infty,0)$. Then $E_{\nu+\epsilon}-E_{\nu-\epsilon}$ is nonzero,
i.e., is a nontrivial projection. Therefore there is some
$\phi_0\neq 0$ so that $\left(E_{\nu+\epsilon}-E_{\nu-
\epsilon}\right)\phi_0=\phi_0$. I claim that $\ip
{\phi_0}{A\phi_0}<0$.

Indeed, $\ip {\phi_0}{A\phi_0}=\ip {\phi_0}{\int_{-
\infty}^\infty\lambda dE_\lambda(\phi_0)}$, which is $\int_{-
\infty}^\infty \lambda d\ip {E_\lambda(\phi_0)}{\phi_0}$, where
the latter just a Stieljes integral. But beyond $\nu+\epsilon$,
$E_\lambda(\phi_0)=\phi_0$, so we only get a negative
contribution. It is certainly strictly negative, since for
$\lambda<\nu-\epsilon$, $E_\lambda(\phi_0)=0$.

Now find a $\phi_1$ for $\mu$ in the same fashion. We need to show
that $\ip {\phi_0}{A\phi_1}=0$. But
$\d\ip{\phi_0}{A\phi_1}=\int_{-\infty}^\infty\lambda
d\ip{\phi_0}{E_\lambda \phi_1}$, and it is routine to show that
$\ip {\phi_0}{E_\lambda\phi_1}=0$ for all $\lambda$.

We may take $c_0$ and $c_1$, not both zero, so that
$c_0\phi_0+c_1\phi_1$ is perpendicular to $\nabla J$. Then
$\ip{c_0\phi_0+c_1\phi_1}{Ac_0\phi_0+Ac_1\phi_1}=c_0^2\ip
{\phi_0}{A\phi_0}+c_1^2\ip{\phi_1}{A\phi_1}<0$. The proof now
proceeds as in step 3 of Theorem 2.

{\it Note:\/} It often occurs in practice that the spectrum of $A$
is discrete and may be written as $\lambda_0<\lambda_1 \leq
\lambda_2 \leq\ldots$, with 0 not a cluster point of $\sigma(A)$.
In this special case, the parts of the hypotheses of the above
theorems which relate to $\sigma(A)$ are as follows. In Theorem 1
we require that $0<\lambda_0$, in Theorem 2 we require that
$\lambda_0<0<\lambda_1$ (in addition to the hypotheses on
$\zeta$), and in Theorem 3 we require that $\lambda_0<\lambda_1<0$.


\bigskip
\centerline{\bf REFERENCES}
\medskip
\ni[1] Finn, R., Editorial comment on {\it Stability of a
Catenoidal Liquid Bridge}\/, by Lianmin Zhou, to appear in Pac.\
J.\ Math.
\smallskip
\ni[2] Gelfand, I.M., and Fomin, S.V., {\bf Calculus of
Variations}, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1963.
\smallskip
\ni[3] Maddocks, J.H., {\it Stability and Folds\/} Arch.\ Rat.\
Mech.\ Anal., vol.\ 99, 301-328, 1987.
\smallskip
\ni[4] Maddocks, J.H., Stability of nonlinear elastic rods, Arch.\
Rat.\ Mech.\ Anal., vol.\ 85, pp.\ 311-354, 1984.
\smallskip
\ni[5] Schechter, M., {\bf Spectra of Partial Differential
Operators}, North-Holland Publishing Co., Amsterdam, 1971.
\bye