0$, which depends continuously
on $\epsilon$ in the $H^2$-norm, and is
such that $v_i(\cdot +\frac{b_i}{\epsilon})$ tends to $u_i$ in the
$H^2$-norm as $\epsilon\to 0$.
We have moreover the following asymptotic information. The solution
has the form
$$
v_i(x)=
u_i\Big(x-\frac{b_i}{\epsilon}+s_i\Big)+
\epsilon^2w_i\Big(x-\frac{b_i}{\epsilon}+s_i\Big)
$$
where $s_i$ is a vector in ${\mathbb R}^n$ and the function $w_i$ is orthogonal in $L^2$ to
the partial derivatives of $u_i$. Both $s_i$ and $w_i$ depend on
$\epsilon$ in such a way that:
{\rm(i)} $\lim_{\epsilon\to 0+} s_i=0$;
{\rm(ii)} $\lim_{\epsilon\to 0+}w_i =\eta_i$ in $H^2$;
\noindent
where $\eta_i$ is the unique solution of
$$
-\Delta \eta_i(x) +a_i\eta_i(x) -pu_i(x)^{p-1}\eta_i(x)
= -\frac12V''(b_i)(x,x)u_i(x)
$$
which is orthogonal to the partial derivatives of $u_i$.
\end{theorem}
Note that the function $\eta_i$ is in principle calculable and it is
to that extent that we consider the asymptotic form of the solution
calculable.
\subsection{First part of the proof}
In this subsection we give the proof of Theorem 1 apart from a technical lemma,
and derive much other useful information about the asymptotic form of $v$ as
$\epsilon\to 0$.
The system of equations (3) defines a non-linear operator
$$
v\mapsto F(\epsilon,v) : (H^2)^m\to (L^2)^m
$$
which depends on a parameter $\epsilon>0$. Our strategy is to solve
the equations for sufficiently small $\epsilon$ by means of a
substitution, careful consideration of the limiting problem as
$\epsilon\to 0$, and the implicit function theorem.
Introduce the
subspaces
$$
W_k= \Big\{w\in H^2: \int w D_ju_k = 0,\quad j=1,\ldots, n\Big\}
$$
and set
$$
W=W_1\times\cdots\times W_m\,.
$$
Introduce the variable vectors $s_1,\ldots,s_m$, each in ${\mathbb R}^n$, and let
$$
\xi_k= -\frac{b_k}{\epsilon}+s_k\,.
$$
Note that $\|\xi_i-\xi_j\|\to \infty$ as $\epsilon \to 0$ provided $i\not=j$.
We let $s=(s_1,\ldots,s_m)$ but we emphasize that each component of the
$m$-tuple $s$ is a vector in ${\mathbb R}^n$.
We shall now use the substitution
$$
v_i= u_i(x+\xi_i)+\epsilon^2w_i(x+\xi_i), \quad i=1,\ldots,m
$$
where $w_i\in W_i$.
The independent variables are the functions $w_i\in W_i$ and the vectors
$s_i$ implicit in $\xi_i$. We shall prove the existence of a solution
for each sufficiently small $\epsilon$, and as $\epsilon\to 0$ we shall
see that $s\to 0$ and $w_i\to \eta_i$ in the norm topology of $H^2$,
where $\eta_i$ are the functions referred to in Theorem~1.
We make the substitution and, for each $i$, translate the $i^{\rm th}$
equation, replacing $x$ by $x-\xi_i$. Divide each equation by
$\epsilon^2$. The result is a new operator equation
$f(\epsilon,s,w)=0$ involving an operator
$$
(s,w)\mapsto f(\epsilon,s,w) : ({\mathbb R}^n)^m\times W \to (L^2)^m
$$
which we proceed to describe.
To ease the exposition we split the description into three parts:
\begin{description}
\item[(A)] terms not involving $w=(w_1,\ldots,w_m)$;
\item[(B)] terms linear in $w$;
\item[(C)] terms quadratic or higher in $w$.
\end{description}
We consider each part separately with a view to taking the limit as
$\epsilon\to 0$.
\medskip
\noindent
{\bf(A)} Before division by $\epsilon^2$ the $i^{\rm th}$ component is
as follows:
$$
\displaylines{\quad
-\Delta u_i(x)+V(\epsilon(x-\xi_i))u_i(x) - G_i\Big(u_1(x+\xi_1-\xi_i),
\ldots, u_m(x+\xi_m-\xi_i)\Big)
\hfill\cr \hfill =
\Big(V(\epsilon(x-\xi_i))-a_i\Big)u_i(x) + u_i(x)^p - G_i\Big(u_1(x+\xi_1-\xi_i),
\ldots, u_m(x+\xi_m-\xi_i)\Big)
\quad\cr\hfill =
\Big(V(\epsilon(x-\xi_i))-a_i\Big)u_i(x) - g_i\Big(u_1(x+\xi_1-\xi_i),
\ldots, u_m(x+\xi_m-\xi_i)\Big)\cdot u_i(x).
\quad\cr }
$$
The second term in the last line consists of sums of monomials. In
each we have the unshifted factor $u_i(x)$ together with at least one
other {\em shifted\/} factor of the form $u_k(x+\xi_k-\xi_i)$ for
which $k\not=i$. Such a monomial tends to 0 in the $L^2$-norm as
$\epsilon\to 0$. In fact the convergence to 0 is faster than that of
any power of $\epsilon$ because of the exponentially fast decrease of
the function $\phi$ at infinity.
Division by $\epsilon^2$ leads therefore to the limit
$$
\frac12 V''(b_i)(x-s_i,x-s_i) u_i(x),
$$
where the second derivative $V''(b_i)$ is regarded as a
symmetric, bilinear form. Note that the limit is attained in the
$L^2$-norm thanks to the boundedness of the second derivatives of $V$.
But in fact, owing to the rapid decrease of $u_i(x)$ we get the same
result if the second derivatives of $V$ have polynomial growth.
The expression (A) defines a mapping
$$
f_0:{\mathbb R}_+ \times ({\mathbb R}^n)^m \to (L^2)^m
$$
where ${\mathbb R}_+$ denotes the interval $[0,\infty[$ and the $i^{\rm th}$ component
of $f_0$ is given by
$$
\displaylines{\quad
\Big(f_0(\epsilon, s)\Big)_i=
\epsilon^{-2}\Big(V(\epsilon(x-\xi_i))-a_i\Big)u_i(x)
\hfill\cr\hfill
{} - \epsilon^{-2}g_i\Big(u_1(x+\xi_1-\xi_i),
\ldots, u_m(x+\xi_m-\xi_i)\Big)\cdot u_i(x)
\quad\cr}
$$
if $\epsilon>0$, and
$$
\Big(f_0(0, s)\Big)_i =
\frac12 V''(b_i)(x-s_i,x-s_i) u_i(x)\,.
$$
Note that the derivative of $f_0(\epsilon, s)$ with respect to $s$
converges to the corresponding derivative of $f_0(0, s)$ as
$\epsilon\to 0$. Convergence occurs in the uniform operator topology
(the operator norm) thanks to the boundedness of the second
derivatives of $V$ (and, as before, polynomial growth would suffice).
\medskip
\noindent
{\bf (B)} After division by $\epsilon^2$ the $i^{\rm th}$ component comprises
the following terms:
$$
\displaylines{\quad
-\Delta w_i(x) +V(\epsilon(x-\xi_i))w_i(x) -pu_i(x)^{p-1}w_i(x)
\hfill\cr\hfill
{}-g_i\Big(u_1(x+\xi_1-\xi_i),\ldots, u_m(x+\xi_m-\xi_i)\Big)w_i(x)
\hfill\cr\hfill
-\sum_{k=1}^m\Big(D_{v_k}g_i\Big)\Big(u_1(x+\xi_1-\xi_i),\ldots, u_m(x+\xi_m-\xi_i)\Big)
u_i(x)w_k(x+\xi_k-\xi_i).
\quad\cr}
$$
(Note that $D_{v_k}g_i$ denotes the partial derivative of $g_i$ with respect to the variable $v_k$.)
This expression may be thought of as
$$
f_1(\epsilon, s) w
$$
where $ f_1(\epsilon, s)$ is a linear mapping from
$W$ to $(L^2)^m$ for each $\epsilon>0$ and $s$.
The main difficulty we have to face is the fact that $f_1(\epsilon,
s)$ does not behave well in the operator norm as $\epsilon\to 0$, as
we now proceed to see.
The last two terms in the expression consist of sums of monomials of the
form
$$
u_{k_1}(x+\xi_{k_1}-\xi_i)\cdots u_{k_{p-1}}(x+\xi_{k_{p-1}}-\xi_i)w_i(x)
$$
and of the form
$$
u_{k_1}(x+\xi_{k_1}-\xi_i)\cdots u_{k_{p-2}}(x+\xi_{k_{p-2}}-\xi_i)
u_i(x)w_k(x+\xi_k-\xi_i)
$$
Let us consider these as linear maps acting on the functions $w_i$.
A monomial of the first kind defines a linear map that tends to 0 with
$\epsilon$ in the operator norm provided at least two distinct shifts
are present to cause the function multiplying $w_i(x)$ to converge uniformly
to 0. This occurs unless $k_1=\cdots=k_{p-1}$. Similarly a
monomial of the second kind defines a linear map that tends to 0 with
$\epsilon$ in the operator norm unless $k_1=\cdots=k_{p-2}=i$.
Throwing out terms that tend to 0 in the operator norm leaves
$$\displaylines{\quad
-\Delta w_i(x) +V(\epsilon(x-\xi_i))w_i(x) -pu_i(x)^{p-1}w_i(x)
\hfill\cr\hfill
-\sum_{k=1,k\not=i}^m\lambda_{ik}u_k(x+\xi_k-\xi_i)^{p-1}w_i(x)
\hfill\cr\hfill
-\sum_{k=1,k\not=i}^m\gamma_{ik}u_i(x)^{p-1}w_k(x+\xi_k-\xi_i)
\quad\cr}
$$
for certain constants $\gamma_{ik}$. This defines a linear map acting
on the functions $w_i$ but it is plain that it does not attain a limit in the
norm topology, but only in the strong operator topology, as $\epsilon\to
0$.
In fact the strong operator limit is the linear mapping $f_1(0,s)$
given by
$$
f_1(0,s)w= -\Delta w_i(x) +a_iw_i(x) -pu_i(x)^{p-1}w_i(x).
$$
Note that it is independent of $s$.
\medskip
\noindent
{\bf(C)} This may be written as $\epsilon^2f_2(\epsilon,s,w)$
and tends to zero, along with any derivatives it possesses, as
$\epsilon\to 0$. The convergence is uniform for $s$ and $w$ in bounded sets.
\medskip
The limiting problem is the following.
$$\displaylines{\quad
\frac12 V''(b_i)(x-s_i,x-s_i) u_i(x)
-\Delta w_i(x) +a_iw_i(x) -pu_i(x)^{p-1}w_i(x)=0,
\hfill\cr\hfill
i=1,\ldots,m.\quad\cr}
$$
It has a non-degenerate solution $s_i=0$, $w_i=\eta_i$, $i=1,\ldots, m$,
where $\eta_i(x)$ is the unique solution in $W_i$ of
$$
-\Delta \eta_i(x) +a_i\eta_i(x) -pu_i(x)^{p-1}\eta_i(x)
= -\frac12V''(b_i)(x,x)u_i(x)
$$
(see section 6.2).
For $\epsilon>0$ our problem takes the form
$$
f(\epsilon,s,w)=
f_0(\epsilon, s)
+f_1(\epsilon,s)w
+\epsilon^2f_2(\epsilon,s,w)=0\,. \eqno{(4)}
$$
At this point we would like to apply the implicit function
theorem to derive a solution for all sufficiently small
$\epsilon>0$. But this requires that the derivative w.r.t.~$(s,w)$
converges in the operator-norm as $\epsilon\to 0$. This fails for
terms (B). However we can still use the implicit function theorem
via a modification which is discussed in the appendix (see Theorem 4
in section 6).
For convenience let us denote the space $({\mathbb R}^n)^m\times W$ by $E$
and the space $(L^2)^m$ by $F$. Define an operator-valued function
$A:{\mathbb R}_+\to L(E,F)$ given by
$$
A(\epsilon)(\sigma,z)= D_sf_0(0,0)\sigma +f_1(\epsilon,0)z. \eqno{(5)}
$$
To apply Theorem 4 we have to
check the following properties of $A$.
\begin{description}
\item{(1)} $A$ is continuous for $\epsilon>0$ w.r.t.~the strong
operator-topology. (This is needed to ensure that the solution depends
continuously on $\epsilon$; see Theorem~4.)
\item{(2)} The limit $\lim_{\epsilon\to
0}A(\epsilon)=D_{(s,w)}f(0,0,\eta)$ is attained in the strong
operator topology. (The importance of $s=0$, $w=\eta$ is that it is
the solution of the limiting problem. Compare condition (c) of
Theorem 4.)
\item{(3)} The limit $\lim_{\epsilon\to 0,s\to
0,w\to\eta}\Big(A(\epsilon)-D_{(s,w)}f(\epsilon,s,w)\Big)=0$ is attained in
the operator-norm topology. (Compare condition (d) of Theorem 4.)
\item{(4)} There exist $M>0$ and $\epsilon_0>0$ such that
$A(\epsilon)$ is invertible for $0\le\epsilon<\epsilon_0$ and
$\|A(\epsilon)^{-1}\|