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An optimal class of nondegenerate potentials for secondorder ordinary differential equations
Boundary Value Problemsvolume 2015, Article number: 189 (2015)
Abstract
By considering the Dirichlet boundary condition $x(0)=x(1)=0$, we say that $q\in L^{1}[0,1]$ is a nondegenerate potential if the ordinary differential equation $x''+q(t) x=0$ has only the trivial solution $x(t)\equiv0$ which verifies the boundary condition. Starting with a nondegenerate positive constant potential B, in this paper, we will apply the Pontryagin maximum principle (PMP) in optimal control theory to find the optimal bound $r=r(A,B)$ for any $A\in[\infty,B)$ such that any potential $q\in L^{1}[0,1]$ satisfying $A\le q \le B$ and $\int_{[0,1]} q(t)\,dt >r(A,B)$ is necessarily nondegenerate. Such a nondegeneracy problem can be considered as the dual problem in a series of papers by Li et al.
Introduction
Let us introduce the notion of nondegenerate potentials by considering the Dirichlet boundary condition.
Definition 1
Let $q\in L^{1}[0,1]$ be an integrable potential. If the following problem
has only the trivial solution $x\equiv0$, we say that $q(t)$ is a nondegenerate potential, or problem (1)(2) is nondegenerate.
If problem (1)(2) is nondegenerate, then, for any $h\in L^{1}[0,1]$, by the Fredholm alternative principle, the inhomogeneous equation
has a unique solution $x(t)$ verifying boundary condition (2). Thus one can also say that problem (1)(2) is nonresonant or invertible, as seen from different literature sources. Nondegenerate potentials are very important in many problems. For example, it is well known that nondegenerate potentials play an important role in the solvability of boundary value problems of semilinear differential equations [1–8]. The maximum and antimaximum principles for various boundary value problems are also related with nondegenerate potentials [9–12]. More importantly, in recent years, it was found that the solvability, the exact multiplicity and the stability of solutions of several interesting classes of superlinear differential equations of the LandesmanLazer type or of the AmbrosettiProdi type can be obtained from the characterization of nondegenerate potentials [1, 2, 13–16].
All of these applications are based on explicit construction of nondegenerate potentials. To describe some known classes of nondegenerate potentials, let us use
to denote all eigenvalues of
with boundary condition (2). Then the nondegeneracy can be explained using eigenvalues as follows.
Lemma 2
For $q\in L^{1}[0,1]$, problem (1)(2) is nondegenerate if and only if $\lambda_{m}(q) \ne0$ for all $m\in \mathbf{N}$.
Throughout this paper, denote
the eigenvalues of problem (3)(2) with zero potential $q=0$. Then a constant potential $q=c\in \mathbf{R}$ is nondegenerate if and only if $c \ne\mu_{m}$ for all $m\in \mathbf{N}$. Moreover, by the comparison results for eigenvalues, it is well known that if q satisfies
for some $n\in \mathbf{N}$, then q is nondegenerate. This is a trivial, but frequently used, class of nondegenerate potentials. Here $\mu_{0}$ is understood as −∞, and $q_{1} \prec q_{2}$ means that $q_{1}(t)\leq q_{2}(t)$ for all $t\in[0,1]$ and $q_{1}(t)< q_{2}(t)$ on some subset of $[0,1]$ of positive measure.
Because of Lemma 2, it is not an easy task to explicitly characterize nontrivial classes of nondegenerate potentials. However, in the study of semilinear ordinary differential equations, Li and his collaborators have applied the Pontryagin maximum principle (PMP) in the optimal control theory to construct several useful classes of nondegenerate potentials of (1) with respect to various boundary conditions like the Dirichlet, the Neumann, and the periodic boundary conditions. See [3, 4, 6, 7]. Their main results, by taking the Dirichlet boundary condition as an example, are as follows. For $\infty \le A< B\le+\infty $, denote
Starting with a nondegenerate constant potential $q\equiv A \in(\mu _{n}, \mu_{n+1})$ for some $n\in \mathbf{N}$, for any $B\in[\mu_{n+1}, +\infty)$, the optimal bound $R(A,B)$ has been found so that any $q\in \Omega_{A,B}$ satisfying
is a nondegenerate potential of problem (1)(2). In fact, the bound $R:=R(A,B)\in(A,\mu_{n+1}]$ is uniquely determined by the following equation:
Moreover, by letting
$R(A,+\infty )$ is the corresponding optimal bound so that potentials $q\in\Omega_{A,+\infty }$ satisfying
are also nondegenerate.
In [17], the second author of this paper has used some Sobolev inequalities to characterize another important class of nonconstant nondegenerate potentials for problem (1)(2) and its pLaplacian counterpart. This class of nondegenerate potentials is used in [9, 10]. It is worth mentioning that papers [18–20] on the Lyapunovtype inequalities are also related with nondegenerate potentials.
In this paper, we study the following problem which can be considered as a dual problem to that in [3, 4, 6, 7]. Given an upper bound
In case $A\in(\mu_{n},B)$, any $q\in \Omega _{A,B}$ is always nondegenerate. See condition (4). In the following we assume that
and consider potentials in $\Omega _{A,B}$. We aim at finding the optimal value $r=r(A,B)$ such that any $q\in \Omega _{A,B}$ satisfying
is necessarily nondegenerate. The final results are as follows.
Theorem 3
Let B and A be as in (6) and (7).

For $A\in(0,\mu_{n}]$, let
$$ r:=r(A,B)\in \bigl[\mu_{n},A+{n\pi(BA)}/{\sqrt{B}} \bigr]\subset [\mu_{n},B) $$(9)be the unique solution of the following equation:
$$ {\sqrt{B}} {\cot\frac{\sqrt{B}}{2n}\frac{rA}{BA}}={\sqrt{A}} { \tan \frac{\sqrt{A}}{2n}\frac{Br}{BA}}. $$(10) 
For $A\in[\infty ,0]$, let
$$ r(A,B):=n\pi\sqrt{B}=\pi\sqrt{B}[\sqrt{B}/\pi]. $$(11)
Then, for any $q\in \Omega _{A,B}$ satisfying condition (8), we have
which implies that problem (1)(2) is nondegenerate. Moreover, the bound $r(A,B)$ is optimal in the sense that there exists $\hat{q}\in \Omega _{A,B}$ such that
and q̂ is degenerate.
One may compare equation (10) with (5). The results of Theorem 3 are obtained mainly using the PMP. However, different from the arguments in [3, 4, 6, 7], we will extensively apply the eigenvalue theory for problem (3). In fact, the nondegeneracy of potentials is a consequence of the estimates (12) on eigenvalues which are also optimal in a certain sense. Moreover, in order to prove the existence of the optimal control potentials for the optimal control problems deduced from the nondegenerate potential problems, we find that the (strong) continuous dependence of solutions and eigenvalues of (3) in potentials (with weak topology) in [21] can simplify the arguments significantly. In these senses, the present paper has given some simpler approach to the nondegeneracy problem.
The paper is organized as follows. At first, we introduce the Pontryagin maximum principle in optimal control theory and establish the connection between nondegeneracy problems and optimal control problems. Secondly, Theorem 3 is proved by solving equations in the PMP. Finally, we briefly consider nondegenerate potentials of (1) with the Neumann boundary condition and point out that the class of potentials in Theorem 3 is also nondegenerate with respect to the Neumann problem. As seen from applications of nondegenerate potentials to nonlinear differential equations mentioned above, it can be expected that the new class of nondegenerate potentials in Theorem 3 can lead to interesting applications to semilinear and superlinear differential equations.
Control systems and the Pontryagin maximum principle
For our purpose, let us introduce the following class of optimal control problems [22]. A control system consists of four elements: statecontrol trajectory, final state, set of admissible controls and cost functional.

the statecontrol trajectory $(\mathbf{x},\mathbf{u})$ is characterized by a firstorder ordinary differential system
$$ \dot{\mathbf{x}}=\mathbf{f}\bigl(\mathbf{x},\mathbf{u}(t),t\bigr), \quad t\in [t_{0},t_{f}], $$(13)with a known initial state
$$ \mathbf{x}(t_{0})=\mathbf{x}_{0}, $$(14)where $\mathbf{x}, \mathbf{x}_{0}\in \mathbf{R}^{n}$, $\mathbf{u}\in \mathbf{R}^{m}$;

the final state is usually described by
$$ \mathbf{g}\bigl(\mathbf{x}(t_{f})\bigr)=0, $$(15)where $\mathbf{g}: \mathbf{R}^{n} \to \mathbf{R}^{k}$ is a known function;

the set of admissible controls is described by
$$\begin{aligned} \mathbf{U}_{[t_{0},t_{f}]} := & \bigl\{ \mathbf{u}(t): \mathbf {u}(\cdot) \mbox{ is a piecewise continuous function on }[t_{0},t_{f}] \\ &{}\mbox{such that }\mathbf{u}(t)\in U_{m}\mbox{ for all }t \in[t_{0},t_{f}] \mbox{ and} \\ &{}\mbox{problem (13)(14)(15) has solutions } \mathbf{x}(t) \bigr\} , \end{aligned}$$(16)where $U_{m}\subset \mathbf{R}^{m}$ is a known domain; and

the cost functional is a functional of $\mathbf{u}(\cdot)\in \mathbf{U}_{[t_{0},t_{f}]}$ taking the following form:
$$ J\bigl[\mathbf{u}(\cdot)\bigr]=\int_{t_{0}}^{t_{f}}L \bigl(\mathbf{x}(t),\mathbf{u}(t),t\bigr)\,dt, $$(17)where $L(\mathbf{x},\mathbf{u},t)$ is a known function.
The optimal control problem is to find an admissible control $\mathbf {u}(\cdot)\in\mathbf{U}_{[t_{0},t_{f}]}$ that maximizes the cost functional $J[\mathbf{u}(\cdot)]$. Suppose that the optimal control problem is solvable and $(\mathbf {x}^{*},\mathbf{u}^{*})$ is the optimal statecontrol trajectory . Then $\mathbf{u}^{*}$ and $\mathbf{x}^{*}$ can be characterized by the Pontryagin maximum principle [22].
Theorem 4
(Pontryagin maximum principle)
Consider the optimal control problem (13)(17), where $U_{m}\subset \mathbf{R}^{m}$ is a closed bounded set. Suppose that

$\mathbf{f}(\mathbf{x}, \mathbf{u},t)$, $\mathbf {f}_{\mathbf{x}}(\mathbf{x},\mathbf{u},t)$, $L(\mathbf{x},\mathbf {u},t)$, $\mathbf{g}(\mathbf{x})$, $\mathbf{g}_{\mathbf {x}}(\mathbf{x})$ are continuous; and

$\mathbf{f}(\mathbf{x},\mathbf{u},t)$, $\mathbf {f}_{\mathbf{x}}(\mathbf{x},\mathbf{u},t)$, $L_{\mathbf{x}}(\mathbf {x},\mathbf{u},t)$ are bounded.
Let us introduce the Hamiltonian H by
where $\psi\in \mathbf{R}^{n}$. If $(\mathbf{u}^{*},\mathbf{x}^{*})$ is the optimal solution of the problem, then there exist a vectorvalued function $\psi:[t_{0},t_{f}]\to \mathbf{R}^{n}$ and a constant vector $\mu\in \mathbf{R}^{n}$ such that $(\mathbf{x}^{*}(t),\psi(t))$ satisfies the following Hamiltonian system:
and boundary conditions
Moreover, one has
and, at any $t\in[t_{0},t_{f}]$ such that $\mathbf{u}^{*}(\cdot)$ is continuous at t, there holds
For problem (1)(2), the construction of nondegenerate potentials $q\in \Omega _{A,B}$ can be stated as the following optimal control problem. By setting $(x_{1},x_{2})=(x,\dot{x})$, equation (1) is equivalent to the following system:
the initial condition can be taken as
The final state is
The set of admissible controls is
Finally, the cost functional on $\hat{\Omega}_{A,B}$ is
In the following we consider the case that $A\in(\infty ,\mu_{n}]$ is finite.
Lemma 5
The optimal control problem (24)(27) associated with problem (1)(2) has an optimal control potential $q^{*}\in \Omega _{A,B}$.
Proof
As $q\equiv\mu_{n} \in\hat{\Omega }_{A,B}$, $\hat{\Omega }_{A,B}\ne \emptyset$. Define
One has $q_{k} \in\hat{\Omega }_{A,B}$ such that $\lim_{k\to \infty } J[q_{k}] =r$. It is well known that the orderinterval $\Omega _{A,B}$ is a compact subset of $L^{1}[0,1]$ when the weak topology $w_{1}$ is considered. Since $\{q_{k}\} \subset\hat{\Omega }_{A,B}\subset \Omega _{A,B}$, without loss of generality, let us assume that $q_{k} \to q^{*}$ in $(L^{1}[0,1],w_{1})$. Thus $q^{*}\in \Omega _{A,B}$.
Let $x_{k}(t)$, $k\in \mathbf{N}$, be solutions of
satisfying $(x_{k}(0),\dot{x}_{k}(0)) =(0,1)$. By the continuity dependence results of solutions on potentials in [21], as $k\to \infty $, $x_{k}$ converges in $(C[0,1],\\cdot\_{C^{0}})$ to x which is the unique solution of
satisfying $(x(0),\dot{x}(0)) =(0,1)$. Since $x_{k}(1)\equiv0$ for all k, one has $x(1)=0$. Since $x\ne0$, we see that $q^{*}\in\hat{\Omega }_{A,B}$. Finally, $q_{k} \to q^{*}$ in $(L^{1}[0,1],w_{1})$ implies that $J[q_{k}] \to J[q^{*}]$. Hence $J[q^{*}]=r$, i.e., $q^{*}$ is an optimal control. □
Lemma 6
Let $q^{*}\in\hat{\Omega}_{A,B}$ be the optimal control in Lemma 5. Define
Then any potential $q\in \Omega _{A,B}$ satisfying (8) is nondegenerate with respect to problem (1)(2).
Proof
Suppose that $q\in \Omega _{A,B}$ satisfies (8). If problem (1)(2) has nontrivial solutions, one would have $q\in\hat{\Omega }_{A,B}$ and $J[q^{*}]=\sup_{\hat{q}\in\hat{\Omega }_{A,B}} J[\hat{q}] \ge J[q]$, a contradiction with (8).
Since $q\equiv B\in \Omega _{A,B}$ is nondegenerate, one has $q^{*}\prec B$ and $r(A,B)< B$. On the other hand, as $q\equiv\mu_{n}\in \Omega _{A,B}$ is degenerate, one has $r(A,B)\ge J[\mu_{n}]=\mu_{n}$. These have given the bounds (28) for $r(A,B)$. □
Construction of nondegenerate potentials for the Dirichlet problem
In this section, we apply the PMP to find the optimal control $q^{*}$ and complete the proof of Theorem 3. To this end, we need only to take $n=2$, $m=k=1$. From the settings (24)(27), let us take
Then the Hamiltonian (18) in the PMP is
Let $q^{*}\in \Omega _{A,B}$ and $(x_{1}^{*}, x_{2}^{*})$ be the optimal solutions. Then (19)(21) in the PMP read as
where $\mu, \nu\in \mathbf{R}$ are some constants.
Note that system (31) is essentially the same as (30). In fact, both $x_{1}^{*}(t)$ and $\psi_{2}(t)$ are solutions of the following equation:
Moreover, from (32) and (33), one has
Thus there exists a constant c such that
Recall that eigenvalues $\lambda_{m}(q)$ of problem (3)(2) have the following comparison properties:
and
where $q_{1} \prec q_{2}$ means that $q_{1} \le q_{2}$ and $q_{1}(t)< q_{2}(t)$ on some subset of $[0,1]$ of positive measure. It is trivial that
for any constant γ, we see that condition $B\in(\mu_{n},\mu _{n+1})$ is equivalent to
As for the optimal potential $q^{*}$ in Lemma 5, one has the following results.
Lemma 7
The optimal control potential $q^{*}$ in Lemma 5 must satisfy $\lambda_{n}(q^{*})=0$. Consequently, as an eigenfunction, $x_{1}^{*}(t)$ has precisely $(n+1)$ zeros in $[0,1]$, say $0=t_{0}< t_{1} < \cdots< t_{n} =1$.
Proof
Since $q^{*} \in\hat{\Omega }_{A,B}$, one has $\lambda_{m}(q^{*})=0$ for some $m\in \mathbf{N}$. As $q^{*} \le B$, by the comparison results for eigenvalues, one has $\lambda_{n+1}(q^{*}) \ge\lambda_{n+1}(B) >0$. Thus $m\le n$.
We assert that $m=n$ and hence one has $\lambda_{n}(q^{*})=0$. Otherwise, assume that $m \le n1$. Denote
One has $\lambda_{n}(q_{0})=\lambda_{n}(q^{*})>\lambda_{m}(q^{*})=0$ and $\lambda_{n}(q_{1})=\lambda_{n}(B)<0$. By the continuity of $\lambda _{n}(q_{\tau})$ in τ, there exists $\tau_{0}\in(0,1)$ such that $\lambda_{n}(q_{\tau_{0}})=0$. Since $q_{\tau_{0}}\in\hat{\Omega }_{A,B}$ and
it is a contradiction with the optimality of $q^{*}$. □
In order to deduce the optimal potential $q^{*}$, we will use properties (22) and (23) in the PMP. For simplicity, we write
which is a solution of problem (34)(2) and satisfies $(y(0),\dot{y}(0))=(0,1)$.
By (29) and (35), H takes the following form:
where $q\in[A,B]$. When t is fixed, as a function of $q\in[A,B]$, $G(t,q)$ attains its maximum at
From (23), we conclude that, at any $t\in[0,1]$ such that $q^{*}(\cdot)$ is continuous at t, there holds
If $c\le0$, we would have $q^{*}=B$, which is impossible. Hence $c>0$. Let us denote $c_{*}:= 1/\sqrt{c}>0$. Then, at any $t\in[0,1]$ such that $q^{*}(\cdot)$ is continuous at t, one has
Moreover, from (22), one has
For each $i=0,1,\ldots,n$, as $y(t_{i})=0$, one has $y(t)< c_{*}$ on some (maximal) interval $I_{i}$ containing $t_{i}$. By (37), $q^{*}(t)=B$ on $I_{i}$. Thus (38) implies that $B\dot{y}^{2}(t_{i})/c_{*}^{2}\equiv \mbox{const}$. Hence $\dot{y}(t_{i+1}) =  \dot{y}(t_{i})$. Since $\dot{y}(t_{0})$ is assumed to be 1, we conclude
Moreover, on $I_{i}$, $y(t)$ satisfies
Using conditions (39), we know that, on $I_{i}$, $y(t)$ is given by
Let us introduce
Then J is a closed set contained in $(0,1)$. Note that $q^{*}(t)=B$ on $[0,1]\backslash J$ and $q^{*}\ne B$. Thus $J\ne\emptyset$ and J consists of closed intervals
where $J_{j}=[\xi_{j},\eta_{j}]\subset(0,1)$ may shrink into a single point. However, as $q^{*}\ne B$, at least one of $J_{j}$’s is a nontrivial closed interval.
Lemma 8
The number of intervals $J_{j}$ ’s in (41), including the degenerate ones, is precisely n. Moreover, by labeling $J_{i}$ according to the order in R, one has $J_{i}\subset(t_{i1},t_{i})$ for $i=1,2,\ldots, n$, where $t_{i}$ ’s are as in Lemma 7.
Proof
Note that $t_{i}\notin J$ for all $i=0,1,\ldots,n$. Thus
Step 1. For each $i=1,2,\ldots, n$, we assert that $(t_{i1},t_{i})$ contains at most one interval $J_{j}$ from (41).
Otherwise, we would have two neighboring intervals $[\xi,\eta]$ and $[\hat{\xi},\hat{\eta}]$ from $J_{j}$’s such that $t_{i1} < \xi\le\eta < \hat{\xi}\le\hat{\eta}<t_{i}$ and $y(t)< c_{*}$ on $(\eta,\hat{\xi})$. Since $y(t)$ does not change sign in $(t_{i1},t_{i})$, let us assume that $y(\eta)=y(\hat{\xi})=c_{*}$ and $0< y(t)< c_{*}$ on $(\eta,\hat{\xi})$. Then, on $(\eta,\hat{\xi})$, $\ddot{y}(t) =  q^{*}(t) y(t) =  B y(t) <0$ and $y(t)$ is strictly concave. Hence we would have $y(t) >c_{*}$ on $(\eta,\hat{\xi})$, a contradiction.
Step 2. Let $J_{j}=[\xi,\eta]$ be any interval from (41) such that $J_{j}=[\xi,\eta]\subset(t_{i1},t_{i})$. For $t\in[\eta,t_{i}]$, $y(t)$ is given by (40). Thus $c_{*}=y(\eta) \le1/\sqrt{B}$.
Step 3. For each $i=1,2,\ldots, n$, we assert that $(t_{i1},t_{i})$ contains precisely one interval $J_{j}$ from (41).
Otherwise, from Step 1, let us assume that $(t_{i1},t_{i})$ contains no $J_{j}$ from (41). Thus $y(t)< c_{*}$ on $(t_{i1},t_{i})$. On $[t_{i1},t_{i}]$, $y(t)$ is given by (40). As $y(t_{i1})=0$, we have from (40) that $t_{i}t_{i1}=\pi/\sqrt{B}$. Thus $c_{*}> y((t_{i1}+t_{i})/2)=1/\sqrt{B}$, a contradiction with the result on $c_{*}$ in Step 2. □
For $i=1,2,\ldots,n$, denote $J_{i}=[\xi_{i},\eta_{i}]\subset (t_{i1},t_{i})$. We have obtained the following formulas:
These give $q^{*}(t)$ and $y(t)$ on $[0,1]\backslash J$. Moreover, as $y(\xi_{i})=y(\eta_{i})=c_{*}$, we have from (43)
where $\tau = \arcsin(c_{*}\sqrt{B})/\sqrt{B}>0$ is a constant. Note that $\tau \leq\pi/2\sqrt{B}$.
In order to construct $q^{*}(t)$ and $y(t)$ on J, we need to distinguish two cases.
Case $0< A\le\mu_{n}$. Let us first consider a nontrivial interval from $J_{i}$’s, say $J_{1}=[\xi_{1},\eta_{1}]$, where $\xi_{1} < \eta_{1}$. Since $q^{*}\ge A>0$, equation (34) implies that $y(t)$ is strictly concave on $[t_{0},t_{1}]$. Thus $y(t) >c_{*}=y(\xi_{1})=y(\eta_{1})$ on $(\xi _{1},\eta_{1})$. By (37), one has
Since $y(t)$ satisfies
on $(\xi_{1},\eta_{1})$, $y(t)$ is symmetric with respect to $t=(\xi _{1}+\eta_{1})/2$ and
where $C\ne0$ is some constant. In order for $y(t)$ to be $C^{1}$ on $[t_{0},t_{1}]$, it is necessary and sufficient that
By formulas (43) and (46), this is
where τ is as in (44). Hence the length $J_{1}=\eta_{1}\xi_{1}$ is uniquely determined by the parameter τ.
Next, for any i, it follows from (43) and (44) that $J_{i}=[\xi_{i},\eta_{i}]$ has the same length with $J_{1}$. Hence $t_{i}t_{i1}=1/n$ for all i, i.e., $t_{i}=i/n$ for $i=0,1,\ldots,n$. By using the parameter τ in (44), one has then $J_{i}=1/n2\tau $.
Finally, let us introduce the following parameter:
Then $J_{i}=1/n2\tau = (1\alpha )/n$ and equation (47) reads as
When A and B are fixed, it is easy to verify that equation (49) has a unique solution $\alpha = \alpha (A,B)$ in the interval (48) for α. In fact, one has $0<\alpha (A,B) <n\pi/\sqrt {B}$ in this case. Moreover, $q^{*}(t)$ is determined by (42) and (45).
Lemma 9
Suppose that B is as in (6) and $A\in(0,\mu_{n}]$. Then the optimal control potential $q^{*}\in \Omega _{A,B}$ is unique. Moreover, one has $q^{*}(t) \equiv q^{*}(t(i1)/n)$ for all $t\in[(i1)/n,i/n]$, where $i=1,2,\ldots, n$, and $q^{*}_{[0,1/n]}$ is given by
where $\alpha = \alpha (A,B)\in(0, n\pi/\sqrt{B})$ is the unique solution of equation (49).
Case $A\leq0$. Denote $J_{i}=[\xi_{i},\eta_{i}]\subset(t_{i1},t_{i})$. At first, we assert that $y(t) \equiv(1)^{i+1} c_{*}$ on $J_{i}$. This is trivial when $\xi_{i}=\eta_{i}$. In the following, we assume that $\xi_{i} < \eta_{i}$. If $y(t)$ is not constant on $J_{i}$, one would have a nontrivial subinterval $(\xi,\eta)\subset[\xi_{i},\eta_{i}]$ such that
Then, on $(\xi,\eta)$, it follows from (37) that $q^{*}(t)=A\le0$ and from equation (34) that $(1)^{i+1} \ddot{y}(t)=A (1)^{i+1}y(t) \ge0$. Thus $(1)^{i+1}\dot{y}(t)$ is nondecreasing. However, (51) implies that $(1)^{i+1}\dot{y}(\xi)\ge0$ and $(1)^{i+1}\dot{y}(\eta )\le0$. Therefore $(1)^{i+1}\dot{y}(t) \equiv0$ and $y(t)$ is constant on $(\xi_{i},\eta_{i})$, a contradiction with assumption (51).
Next we assert that
In fact, for any $t\in J_{i}$, one has $(1)^{i+1}y(t) = \max_{s\in(t_{i1},t_{i})}(1)^{i+1} y(s)$. Hence one has (52).
Finally, from (43), (44) and (52), one knows that τ satisfies $\cos(\sqrt{B}\tau )=0$. Hence
Moreover, as $(1)^{i+1}y(t) \equiv c_{*}$ on $(\xi_{i},\eta_{i})$, we obtain from equation (34) that
Lemma 10
Suppose that B is as in (6) and $A\in(\infty ,0]$. Then the optimal control potential $q^{*}\in \Omega _{A,B}$ is not unique. By letting
one optimal control potential $q^{*}$ can be chosen so that $q^{*}(t) \equiv q^{*}(t(i1)/n)$ for all $t\in[(i1)/n,i/n]$, where $i=1,2,\ldots, n$, and $q^{*}_{[0,1/n]}$ is given by
Here formulas (55) and (56) are deduced from (53) and from (42), (54), respectively.
Proof of Theorem 3
Case 1: $A\in(0,\mu_{n}]$. From Lemma 9 and formula (50), one has
where $\alpha =\alpha (A,B)$ is the solution of equation (49). By letting $r=r(A,B)$, one has from (57) that $\alpha = (rA)/(BA) $ and $1\alpha = (Br)/(BA)$. Substituting into (49), we know that $r=r(A,B)$ satisfies equation (10). Moreover, one has from Lemma 6 that $r\ge \mu _{n}$ and from (48) that $r=A+\alpha (BA)< A +n\pi(BA)/\sqrt{B}$. Hence r is inside the interval (9) in this case.
Case 2: $A\in(\infty ,0]$. From (55) and (56), one has
This is (11).
In these two cases, denote
Lemma 6 asserts that any $q\in\check{\Omega }_{A,B}$ is nondegenerate. In fact, as $\check{\Omega }_{A,B}$ is a convex set containing B, for any $q\in\check{\Omega }_{A,B}$, one has $(1s) q+s B\in\check{\Omega }_{A,B}$ for $s\in[0,1]$. By Lemma 2,
As continuous functions of s, one has then
Thus $\lambda_{n}(q)$ and $\lambda_{n+1}(q)$ have the same signs as those of $\lambda_{n}(B)$ and $\lambda_{n+1}(B)$, respectively. Due to (36), we obtain (12).
Moreover, the optimality of $r(A,B)$ follows from Lemmas 5 and 6, by simply taking $\hat{q}=q^{*}$.
Case 3: $A=\infty $. In this case, one has $\Omega _{\infty ,B} \supset \Omega _{A,B}$ for any $A\in(\infty ,0]$. By the meaning of $r(\infty ,B)$, one has $r(\infty ,B) \ge r(0,B) = n\pi\sqrt{B}$.
On the other hand, suppose that $q\in \Omega _{\infty ,B}$ satisfies
Since $q\le B$, one has
For any $A\in(\infty ,0)$, define
Then $q_{A}\ge q$ and $q_{A}\in \Omega _{A,B}$. Let $0<\varepsilon < \elln\pi\sqrt {B}$ be a constant. See (58). Then $q_{A}\varepsilon \in \Omega _{A\varepsilon ,B}$ and
Thus $q_{A}\varepsilon \in\check{\Omega }_{A\varepsilon ,B}$. From inequality (12), for $q_{A}\varepsilon $, we know that
When $A\downarrow \infty $, the measure of $E_{A}$ tends to 0 and $\q_{A}q\ _{1}\to0$. Therefore
See (60). Combining with (59), we know that q also satisfies (12). Thus $r(\infty ,B)\le n\pi\sqrt{B}$. □
Remark 11
Let B be as in (6). For the case $A=\infty $, Theorem 3 means that any potential in
is nondegenerate. Note that these nondegenerate potentials $q(t)$ may be unbounded from below.
Let us fix B as in (6) and consider $r(A,B)$ as a function of $A\in(\infty , \mu_{n}]$. From equation (10), it is easy to prove that $r=r(A,B)$ is continuous and nonincreasing in A. Moreover, one has $r(\mu_{n},B) = \mu_{n} =(n\pi)^{2}$. As a function of A, the graph of $r(A,B)$ is as in Figures 1 and 2.
Nondegenerate potentials for the Neumann problem
For equation (1) with the Neumann boundary condition
the nondegenerate potentials can be defined similarly as in Definition 1. Given a potential $q\in L^{1}[0,1]$. The eigenvalues of problem (3)(61) are
Note that
One has $\hat{\mu}_{m}=\mu_{m}$ for $m\in \mathbf{N}$.
By the approach in the preceding sections, for the Neumann problem, the results are as follows.
Theorem 12
Let B and A be as in (6) and (7), where $n\in \mathbf{N}$. By letting $r(A,B)$ and $\check{\Omega }_{A,B}$ be as those for the Dirichlet problem, any $q\in\check{\Omega }_{A,B}$ is also nondegenerate with respect to problem (1)(61).
For the Neumann problem, one has the zeroth eigenvalue. See (62) and (63). This leads to the following problem. Suppose that B and A are such that
What is the optimal lower bound $r(A,B)$ for $\int_{0}^{1} q(t)\,dt$ so that any $q\in \Omega _{A,B}$ is nondegenerate?
Theorem 13
Let B and A be as in (64). Then $r(A,B)\equiv0$. That is, any $q\in \Omega _{A,B}$ satisfying
is a nondegenerate potential of problem (1)(61). Moreover, the lower bound 0 in (65) is optimal.
Proof
This theorem is simply a restatement of some classical results. In fact, for any $q\in L^{1}[0,1]$, it is well known that
See [17]. Under assumption (65), one has $\hat{\lambda}_{0}(q) <0$. On the other hand, for any $q\in \Omega _{A,B}$, as $q\le B$, one has $\hat{\lambda}_{1}(q) \ge\hat{\lambda}_{1}(B)= \mu_{1}B >0$. See assumption (64). Thus q is a nondegenerate potential of problem (1)(61).
As for the optimality, one needs only to notice that the zero potential $q=0 \in \Omega _{A,B}$ is degenerate with respect to problem (1)(61). □
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Acknowledgements
The authors would like to thank Zhiyuan Wen for helpful discussions. This work is supported by the National Natural Science Foundation of China (Grant No. 11231001 and No. 11371213) and the National 111 Project of China (Station No. 111201).
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Keywords
 nondegenerate potential
 eigenvalue
 optimal control
 Pontryagin maximum principle
 boundary value problem