Consider the principal part of the problem (1)
Condition 1 Assume the following conditions are satisfied:

(1)
E is a UMD space, p\in (1,\mathrm{\infty});

(2)
a\in \mathit{VMO}\cap {L}^{\mathrm{\infty}}(R), {\eta}_{1} is a VMO modulus of a, a\in S(\phi ), \mu (x)\ne 0;

(3)
Re{\omega}_{k}(x)\ne 0 and \frac{\lambda}{{\omega}_{k}}\in S(\phi ) for \lambda \in S(\phi ), 0\le \phi <\pi, k=1,2. a.e. x\in [0,1];

(4)
A(x) is a uniformly Rpositive operator in E and
A(\cdot ){A}^{1}({x}_{0})\in {L}_{\mathrm{\infty}}(0,1;L(E))\cap \mathit{VMO}(L(E)),\phantom{\rule{1em}{0ex}}{x}_{0}\in (0,1);

(5)
a(0)=a(1), A(0)=A(1) and {\eta}_{2} is a VMO modulus of A(\cdot ){A}^{1}({x}_{0}).
First, we obtain an integral representation formula for solutions.
Lemma 1 Let Condition 1 hold and f\in {L}^{p}(0,1;E). Then, for all solutions u of the problem (5) belonging to {W}^{2,p}(0,1;E(A),E), we have
where
here {B}_{ij}^{\mathrm{\prime}}(\lambda ), {D}_{ij}^{\mathrm{\prime}}(\lambda ) are uniformly bounded operators,
{U}_{\nu 0\lambda s}^{\mathrm{\prime}}(xy)=\{\begin{array}{cc}{\omega}_{1}^{i}{a}^{1}{s}^{(1\nu )/2}{A}_{\lambda}^{(1\nu )/2}{U}_{1\lambda s}(xy),\hfill & x\ge y,\hfill \\ {\omega}_{2}^{i}{a}^{1}{s}^{(1\nu )/2}{A}_{\lambda}^{(1\nu )/2}{U}_{2\lambda s}(xy),\hfill & x\le y,\hfill \end{array}
and the expression {\mathrm{\Gamma}}_{2\lambda}^{\mathrm{\prime}}(x,xy) is a scalar multiple of {\mathrm{\Gamma}}_{2\lambda}(x,xy).
Proof Consider the problem (5) for a(x)=a({x}_{0}) and A(x)=A({x}_{0}), i.e.,
Let u\in {C}^{(2)}([0,1];E(A)) be a solution of the problem (7). Taking into account the equality {L}_{0}u=({L}_{0}L)u+Lu and Theorem A_{7}, we get
Setting x={x}_{0} in above, we get (6) for u\in {C}^{(2)}([0,1];E(A)). Then the density argument, using Theorem A_{3}, gives the conclusion for
u\in {W}^{2,p}(0,1;E(A),E),\phantom{\rule{1em}{0ex}}{L}_{k}u=0.
Consider the problem (5) on (0,b), i.e.,
□
Theorem 1 Suppose Condition 1 is satisfied. Then there exists a number b\in (0,1) such that the uniform coercive estimate
\sum _{i=0}^{2}{s}^{\frac{i}{2}}\lambda {}^{1\frac{i}{2}}{\parallel {u}^{(i)}\parallel}_{{L}^{p}(0,b;E)}+{\parallel Au\parallel}_{{L}^{p}(0,b;E)}\le C{\parallel ({L}_{b}+\lambda )u\parallel}_{{L}^{p}(0,b;E)}
(9)
holds for large enough \lambda  and u\in {W}^{2,p}(0,b;E(A),E), {L}_{bk}u=0, \lambda \in S(\phi ), where C is a positive constant depending only on p, {M}_{0}, {\eta}_{1}, {\eta}_{2}.
Proof By Lemma 1, for any solution u\in {W}^{2,p}(0,b;E(A),E) of the problem (8), we have
\begin{array}{rcl}{u}^{(i)}(x)& =& {\int}_{0}^{b}{\mathrm{\Gamma}}_{ib\lambda s}(x,xy)\{[a(x)a(y)]{u}^{(2)}(y)\\ +[A(x)A(y)]u(y)+f(y)\}\phantom{\rule{0.2em}{0ex}}dy+f(x),\end{array}
(10)
where
\begin{array}{rcl}{\mathrm{\Gamma}}_{ib\lambda s}(x,xy)& =& \sum _{i,j=1}^{2}\sum _{k=0}^{{m}_{1}}[{B}_{ij}^{\mathrm{\prime}}(\lambda ){\left({s}^{1}{A}_{\lambda}\right)}^{\frac{1}{2}(2+{m}_{k}ki1)}{U}_{j\lambda}(x){U}_{1\lambda}(by)]\\ +\sum _{i,j=1}^{2}\sum _{k=0}^{{m}_{2}}[{D}_{ij}^{\mathrm{\prime}}(\lambda ){\left({s}^{1}{A}_{\lambda}\right)}^{\frac{1}{2}(2+{m}_{k}ki1)}{U}_{j\lambda}(x){U}_{2\lambda}(y)]\\ +{U}_{\nu 0\lambda s}^{\mathrm{\prime}}(xy),\phantom{\rule{1em}{0ex}}\nu =0,1,2,\end{array}
(11)
here {B}_{ij}^{\mathrm{\prime}}(\lambda ), {D}_{ij}^{\mathrm{\prime}}(\lambda ) are uniformly bounded operators,
{U}_{1\lambda}(x)={e}^{{\omega}_{1}{s}^{\frac{1}{2}}{A}_{\lambda}^{\frac{1}{2}}x},\phantom{\rule{2em}{0ex}}{U}_{2\lambda}(x)={e}^{{\omega}_{2}{s}^{\frac{1}{2}}{A}_{\lambda}^{\frac{1}{2}}(bx)}
and
{U}_{\nu 0\lambda s}^{\mathrm{\prime}}(xy)=\{\begin{array}{cc}{\omega}_{1}^{i}{a}^{1}{s}^{(1\nu )/2}{A}_{\lambda}^{(1\nu )/2}{U}_{1\lambda s}(xy),\hfill & x\ge y,\hfill \\ {\omega}_{2}^{i}{a}^{1}{s}^{(1\nu )/2}{A}_{\lambda}^{(1\nu )/2}{U}_{2\lambda s}(xy),\hfill & x\le y,\hfill \end{array}\phantom{\rule{1em}{0ex}}\nu =0,1,2.
Moreover, from (10) and (11), clearly, we get
Au(x)={\int}_{0}^{b}{\mathrm{\Gamma}}_{b\lambda s}^{\mathrm{\prime}}(x,xy)\{[a(x)a(y)]{u}^{(2)}(y)+[A(x)A(y)]u(y)+f(y)\}\phantom{\rule{0.2em}{0ex}}dy,
(12)
where the expression {\mathrm{\Gamma}}_{b\lambda}^{\mathrm{\prime}}(x,xy) differs from {\mathrm{\Gamma}}_{2b\lambda}(x,xy) only by a constant.
Consider the operators
Since the operators {B}_{0\lambda} and {B}_{1\lambda} are regular on {L}^{p}(0,b;E), by using the positivity properties of A and the analyticity of semigroups {U}_{k\lambda}(x) in a similar way as in [[30], Theorem 3.1], we get
{\parallel {B}_{i\lambda}f\parallel}_{{L}^{p}(0,b;E)}\le M\lambda {}^{\frac{2i}{2}}{\parallel f\parallel}_{{L}^{p}(0,b;E)},\phantom{\rule{1em}{0ex}}i=0,1.
(13)
Since the Hilbert operator is bounded in {L}^{p}(R;E) for a UMD space E, we have
{\parallel {B}_{2\lambda}f\parallel}_{{L}^{p}(0,b;E)}\le M{\parallel f\parallel}_{{L}^{p}(0,b;E)}.
(14)
Thus, by virtue of Theorems A_{4}, A_{6} and in view of (10)(12) for any \epsilon >0, there exists a positive number b=b(\epsilon ,{\eta}_{1},{\eta}_{2}) such that
Hence, the estimates (13)(15) imply (9). □
Theorem 2 Assume Condition 1 holds. Let u\in {W}^{2,p}(0,1;E(A),E) be a solution of (4). Then, for sufficiently large \lambda , \lambda \in S(\phi ), the following coercive uniform estimate holds:
\sum _{i=0}^{2}{s}^{\frac{i}{2}}\lambda {}^{1\frac{i}{2}}{\parallel {u}^{(i)}\parallel}_{{L}^{p}(0,1;E)}+{\parallel Au\parallel}_{{L}^{p}(0,1;E)}\le C[{\parallel (L+\lambda )u\parallel}_{{L}^{p}(0,1;E)}+{\parallel u\parallel}_{{L}^{p}(0,1;E)}].
(16)
Proof This fact is shown by a covering and flattening argument, in a similar way as in Theorem A_{8}. Particularly, by partition of unity, the problem is localized. Choosing diameters of supports for corresponding finite functions, by using Theorem 1, Theorems A_{4}, A_{6}, A_{7} and embedding Theorem A_{1} (see the same technique for DOEs with continuous coefficients [19, 20]), we obtain the assertion.
Let {Q}_{s} denote the operator in {L}^{p}(0,1;E) generated by the problem (4) for \lambda =0, i.e.,
D({Q}_{s})={W}^{2,p}(0,1;E(A),E,{L}_{k}),\phantom{\rule{2em}{0ex}}{Q}_{s}u=sa(x){u}^{(2)}+A(x)u.
□
Theorem 3 Assume Condition 1 holds. Then, for all f\in {L}^{p}(0,1;E), \lambda \in S(\phi ) and for large enough \lambda , the problem (5) has a unique solution u\in {W}^{2,p}(0,1;E(A),E). Moreover, the following coercive uniform estimate holds:
\sum _{i=0}^{2}{s}^{\frac{i}{2}}\lambda {}^{1\frac{i}{2}}{\parallel {u}^{(i)}\parallel}_{{L}^{p}(0,1;E)}+{\parallel Au\parallel}_{{L}^{p}(0,1;E)}\le C{\parallel f\parallel}_{{L}^{p}(0,1;E)}.
(17)
Proof First, let us show that the operator Q+\lambda has a left inverse. Really, it is clear that
{\parallel u\parallel}_{{L}^{p}(0,1;E)}=\frac{1}{\lambda }[{\parallel ({Q}_{s}+\lambda )u\parallel}_{{L}^{p}(0,1;E)}+{\parallel {Q}_{s}u\parallel}_{{L}^{p}(0,1;E)}].
By Theorem A_{1} for u\in {W}^{2,p}(0,1;E(A),E), we have
{\parallel {Q}_{s}u\parallel}_{{L}^{p}(0,1;E)}\le C{\parallel u\parallel}_{{W}_{s}^{2,p}(0,1;E(A),E)}.
Then, by virtue of (16) and in view of the above relations, we infer, for all u\in {W}^{2,p}(0,1;E(A),E) and sufficiently large \lambda , that there is a small ε and C(\epsilon ) such that
From the estimate (18) for u\in {W}^{2,p}(0,1;E(A),E), we get
\sum _{i=0}^{2}{s}^{\frac{i}{2}}\lambda {}^{1\frac{i}{2}}{\parallel {u}^{(i)}\parallel}_{{L}^{p}(0,1;E)}+{\parallel Au\parallel}_{{L}^{p}(0,1;E)}\le C{\parallel (Q+\lambda )u\parallel}_{{L}^{p}(0,1;E)}.
(19)
The estimate (19) implies that (4) has a unique solution and the operator {Q}_{s}+\lambda has a bounded inverse in its rank space. We need to show the rank space coincides with all the space {L}^{p}(0,1;E). It suffices to prove that there is a solution u\in {W}^{2,p}(0,1;E(A),E) for all f\in {L}^{p}(0,1;E). This fact can be derived in a standard way, approximating the equation with a similar one with smooth coefficients [19, 20]. More precisely, by virtue of [[23], Theorem 3.4], UMD spaces satisfy the multiplier condition. Moreover, by the part (2) of Lemma A_{1}, given a\in \mathit{VMO} with VMO modules \eta (r), we can find a sequence of mollifiers functions \{{a}_{h}\} converging to a in BMO as h\to 0 with a VMO modulus {\eta}_{h} such that {\eta}_{h}(r)\le \eta (r). In a similar way, it can be derived for the operator function A(x){A}^{1}({x}_{0})\in \mathit{VMO}(L(E)). □
Result 1 Theorem 3 implies that the resolvent {({Q}_{s}+\lambda )}^{1} satisfies the following sharp uniform estimate:
\sum _{i=0}^{2}{s}^{\frac{i}{2}}\lambda {}^{1\frac{i}{2}}{\parallel \frac{{d}^{i}}{d{x}^{i}}{({Q}_{s}+\lambda )}^{1}\parallel}_{L({L}^{p}(0,1;E))}+{\parallel A{({Q}_{s}+\lambda )}^{1}\parallel}_{L({L}^{p}(0,1;E))}\le C
(20)
for arg\lambda \le \phi, \phi \in (0,\pi ) and s>0.
The estimate (20) particularly implies that the operator Q is uniformly positive in {L}^{p}(0,1;E) and generates analytic semigroups for \phi \in (\frac{\pi}{2},\pi ) (see, e.g., [[22], §1.14.5]).
Remark 1 Conditions a(0)=a(1), A(0)=A(1) arise due to nonlocality of the boundary conditions (4). If the boundary conditions are local, then the conditions mentioned above are not required any more.
Consider the problem (1), where {L}_{k}u is the same boundary condition as in (4). Let {O}_{s} denote a differential operator generated by the problem (1). We will show the separability and Fredholmness of (1).
Theorem 4
Assume

(1)
Condition 1 holds;

(2)
for any \epsilon >0, there is C(\epsilon )>0 such that for a.e. x\in (0,1) and for 0<{\nu}_{0}<1, 0<{\nu}_{1}<\frac{1}{2},
Then, for all f\in {L}^{p}(0,1;E) and for large enough \lambda , \lambda \in S(\phi ), there is a unique solution u\in {W}^{2,p}(0,1;E(A),E) of the problem (1) and the following coercive uniform estimate holds:
\sum _{i=0}^{2}{s}^{\frac{i}{2}}\lambda {}^{1\frac{i}{2}}{\parallel {u}^{(i)}\parallel}_{{L}^{p}(0,1;E)}+{\parallel Au\parallel}_{{L}^{p}(0,1;E)}\le C{\parallel f\parallel}_{{L}^{p}(0,1;E)}.
(21)
Proof It is sufficient to show that the operator {O}_{s}+\lambda has a bounded inverse {({O}_{s}+\lambda )}^{1} from {L}^{p}(0,1;E) to {W}^{2,p}(0,1;E(A),E). Put {O}_{s}u={Q}_{s}u+{Q}_{0}u, where
{Q}_{0}u={s}^{\frac{1}{2}}{A}_{1}{u}^{(1)}+{A}_{0}u,u\in {W}^{2,p}(0,1;E(A),E,{L}_{k}).
By the second assumption and Theorem A_{1}, there is a small ε and C(\epsilon ) such that
\begin{array}{r}{\parallel {A}_{0}u\parallel}_{{L}^{p}(0,1;E)}\le C{\parallel {A}^{1{\nu}_{0}}u\parallel}_{{L}^{p}(0,1;E)}\\ \phantom{{\parallel {A}_{0}u\parallel}_{{L}^{p}(0,1;E)}}\le \epsilon {\parallel u\parallel}_{{W}_{s}^{2,p}(0,1;E(A),E)}+C(\epsilon ){\parallel u\parallel}_{{L}^{p}(0,1;E)},\\ {\parallel {s}^{\frac{1}{2}}{A}_{1}{u}^{(1)}\parallel}_{{L}^{p}(0,1;E)}\le C{\parallel {s}^{\frac{1}{2}}{A}^{\frac{1}{2}{\nu}_{1}}u\parallel}_{{L}^{p}(0,1;E)}\\ \phantom{{\parallel {s}^{\frac{1}{2}}{A}_{1}{u}^{(1)}\parallel}_{{L}^{p}(0,1;E)}}\le \epsilon {\parallel u\parallel}_{{W}_{s}^{2,p}(0,1;E(A),E)}+C(\epsilon ){\parallel u\parallel}_{{L}^{p}(0,1;E)}.\end{array}
(22)
In view of estimates (17) and (22), we have
\begin{array}{r}{\parallel {A}_{0}u\parallel}_{{L}^{p}(0,1;E)}<{\delta}_{1}{\parallel {Q}_{s}u\parallel}_{{L}^{p}(0,1;E)},\\ {\parallel {s}^{\frac{1}{2}}{A}_{1}{u}^{(1)}\parallel}_{{L}^{p}(0,1;E)}<{\delta}_{1}{\parallel {Q}_{s}u\parallel}_{{L}^{p}(0,1;E)}\end{array}
(23)
for u\in {W}^{2,p}(0,1;E(A),E) and {\delta}_{k}<1. By Theorem 3, the operator {Q}_{s}+\lambda has a bounded inverse {({Q}_{s}+\lambda )}^{1} from {L}^{p}(0,1;E) to {W}^{2,p}(0,1;E(A),E) for sufficiently large \lambda . So, (23) implies the following uniform estimate:
{\parallel {Q}_{0}{({Q}_{s}+\lambda )}^{1}\parallel}_{L({L}^{p}(0,1;E))}<1.
It is clear that
Then, by above relation and by virtue of Theorem 3, we get the assertion. □
Theorem 4 implies the following result.
Result 2 Suppose all the conditions of Theorem 4 are satisfied. Then the resolvent {({O}_{s}+\lambda )}^{1} of the operator {O}_{s} satisfies the following sharp uniform estimate:
\sum _{i=0}^{2}{s}^{\frac{i}{2}}\lambda {}^{1\frac{i}{2}}{\parallel \frac{{d}^{i}}{d{x}^{i}}{({O}_{s}+\lambda )}^{1}\parallel}_{L({L}^{p}(0,1;E))}+{\parallel A{({O}_{s}+\lambda )}^{1}\parallel}_{L({L}^{p}(0,1;E))}\le C
for arg\lambda \le \phi, \phi \in [0,\pi ) and s>0.
Consider the problem (1) for \lambda =0, i.e.,
Theorem 5 Assume all the conditions of Theorem 4 hold and {A}^{1}\in {\sigma}_{\mathrm{\infty}}(E). Then the problem (24) is Fredholm from {W}^{2,p}(0,1;E(A),E) into {L}^{p}(0,1;E).
Proof Theorem 4 implies that the operator {O}_{s}+\lambda has a bounded inverse {({O}_{s}+\lambda )}^{1} from {L}^{p}(0,1;E) to {W}^{2,p}(0,1;E(A),E) for large enough \lambda ; that is, the operator {O}_{s}+\lambda is Fredholm from {W}^{2,p}(0,1;E(A),E) into {L}^{p}(0,1;E). Then, by virtue of Theorem A_{2} and by perturbation theory of linear operators, we obtain the assertion. □