MAXIMAL DETERMINANTS OF SCHRÖDINGER OPERATORS ON BOUNDED INTERVALS

. — We consider the problem of ﬁnding extremal potentials for the functional determinant of a one-dimensional Schrödinger operator deﬁned on a bounded interval with Dirichlet boundary conditions under an L q -norm restriction ( q (cid:62) 1 ). This is done by ﬁrst extending the deﬁnition of the functional determinant to the case of L q potentials and showing the resulting problem to be equivalent to a problem in optimal control, which we believe to be of independent interest. We prove existence, uniqueness and describe some basic properties of solutions to this problem for all q (cid:62) 1 , providing a complete characterisation of extremal potentials in the case where q is one (a pulse) and two (Weierstraß’s ℘ function).


Introduction
An important quantity arising in connection with self-adjoint elliptic operators is the functional (or spectral) determinant. This has been applied in a variety of settings in mathematics and in physics, and is based on the regularisation of the spectral zeta function associated to an operator T with discrete spectrum. This zeta function is defined by where the numbers λ n , n = 1, 2, . . . denote the eigenvalues of T and, for simplicity, and without loss of generality from the perspective of this work as we will see below, we shall assume that these eigenvalues are all positive and with finite multiplicities. Under these conditions, and for many operators such as the Laplace or Schrödinger operators, the above series will be convergent on a right half-plane, and may typically be extended meromorphically to the whole of C. Furthermore, zero is not a singularity and since, formally, the regularised functional determinant is then defined by where ζ T (0) should now be understood as referring to the meromorphic extension mentioned above. This quantity appears in the mathematics and physics literature in connection to path integrals, going back at least to the early 1960's. Examples of calculations of determinants for operators with a potential in one dimension may be found in [5,12,18] and, more recently, for one-dimensional polyharmonic operators of arbitrary order [11] and the harmonic oscillator in arbitrary dimension [10]. Some of the regularising techniques for zeta functions which are needed in order to define the above determinant were studied in [19], while the actual definition (2) was given in [22]. Within such a context, it is then natural to study extremal properties of these global spectral objects and this question has indeed been addressed by several authors, mostly when the underlying setting is of a geometric nature [3,21,20,2]. In this paper, we shall consider the problem of optimising the functional determinant for a Schrödinger operator defined on a bounded interval together with Dirichlet boundary conditions. More precisely, let T V be the operator associated with the eigenvalue problem defined by where V is a potential in L q [0, 1] (q 1). For a given q and a positive constant A, we are interested in the problem of optimising the determinant given by (4) det T V −→ max, where · q denotes the norm on L q [0, 1]. For smooth bounded potentials the determinant of such operators is known in closed form and actually requires no computation of the eigenvalues themselves. In the physics literature such a formula is sometimes referred to as the Gelfand-Yaglom formula and a derivation may be found in [18], for instance-see also [5]. More precisely, for the operator T V defined by (3) we have det T V = 2y (1), where y is the solution of the initial value problem (5) −y + V y = 0 y(0) = 0, y (0) = 1.
We shall show that this expression for the determinant still holds for L q potentials and study the problem defined by (4). We then prove that (4) is well-posed and has a unique solution for all q 1 and positive A. In our first main result we consider the L 1 case where the solution is given by a piecewise constant function.
Theorem A (Maximal L 1 potential). -Let q = 1. Then for any positive number A the unique solution to problem (4) is the symmetric potential given by where χ (A) denotes the characteristic function of the interval of length centred at 1/2. The associated maximum value of the determinant is In the case of general q we are able to provide a similar result concerning existence and uniqueness, but the corresponding extremal potential is now given as the solution of a second order (nonlinear) ordinary differential equation.
The properties given in the above theorem provide a precise qualitative description of the evolution of maximal potentials as q increases from 1 to +∞. Starting from a rectangular pulse (q = 1), solutions become regular for q on (1, 2), having zero derivatives at the endpoints. In the special case of L 2 , the maximising potential can be written in terms of the Weierstraß elliptic function and has finite nonzero derivatives at the endpoints. This marks the transition to potentials with singular derivatives at the boundary for q larger than two, converging towards an optimal constant potential in the limiting L ∞ case.
Theorem C (Maximal L 2 potential). -Let q = 2. Then for any positive number A the unique solution to problem (4) is given by where ℘ is the Weierstraß elliptic function associated to invariants and where ω is the corresponding imaginary half-period of the rectangular lattice of periods. The corresponding (unique) value of H such that is the unique root of the polynomial Concerning possible extensions of these results to higher dimensions, the underlying space geometry mentioned above may in fact impose certain restrictions that will have to be taken into consideration when studying extremal problems for the determinant. In dimensions higher than one, different issues from those in the more standard situations analysed so far may also come into play. In particular, it is not even guaranteed that, without further restrictions, such extrema do exist -compare the results in the references above with the extremal problems related to the Laplace and Schrödinger operators considered in [6,13,14], for instance. This would thus be a first question to address when considering the problem of finding extremal potentials in domains in higher dimensions, where furthermore no formula is known for the determinant of a Schrödinger operator in general.
The paper is structured as follows. In the next section we show that the functional determinant of Schrödinger operators with Dirichlet boundary conditions on bounded intervals and with potentials in L q is well-defined and we extend the formula from [18] to this general case. The main properties of the determinant, namely boundedness and monotonicity over L q , are studied in Section 2. Having established these, we then consider the optimal control problem (4) of maximising y(1) in (5) in Sections 3 and 4, where the proofs of our main results Theorems A and B-C are given, respectively.

The determinant of one-dimensional Schrödinger operators
We consider the eigenvalue problem defined by (3) associated to the operator T V on the interval [0, 1] with potential V ∈ L q [0, 1] for q 1. Although no further restrictions need to be imposed on V at this point, for our purposes it will be sufficient to consider V to be non-negative, as we will show in Proposition 9 below. This simplifies slightly the definition of the associated zeta function given by (1) and thus also that of the determinant. Hence, in the rest of this section we assume that V is a non-negative potential.
For smooth potentials, and as was already mentioned in the Introduction, it is known that the regularised functional determinant of T V is well-defined [5,18], and this has been extended to potentials with specific singularities [16,17]. We shall now show that this is also the case for general potentials in L q [0, 1], for q 1. We first show that the zeta function associated with the operator T V as defined by (1) is analytic at the origin and has as its only singularity in the half-plane Re(s) > −1/2 a simple pole at 1/2. This is done adapting one of the approaches originally used by Riemann (see [23] and also [26, p. 21ff]). We then show that the method from [18] can be used to prove that the determinant is still given by 2y (1), where y is the solution of the initial value problem (5).
In order to show these properties of the determinant, it is useful to consider the heat trace associated with T V , defined by Our first step is to show that the behaviour of the heat trace as t approaches zero for any non-negative potential in L q (0, 1) is the same as in the case of smooth potentials.
when n goes to ∞. For a potential in L q we may thus assume the above asymptotics which imply the existence of a positive constant c such that π 2 n 2 − c λ n π 2 n 2 + c uniformly in n. For the zero potential the spectrum is given by π 2 n 2 and the heat trace associated with it becomes the Jacobi theta function defined by We are interested in the behaviour of the heat trace for potentials V for small positive t, and we will determine this behaviour by comparing it with that of ψ. For simplicity, in what follows we write We then have and, since ψ satisfies the functional equation [26, p. 22] it follows that as t → 0, it follows that Remark 2. -Note that although the heat trace for the zero potential ψ satisfies for any positive real number α, this will not be the case for general potentials, where we can only ensure that the next term in the expansion will be of order √ t.
We may now consider the extension of ζ T V to a right half-plane containing the origin.
Proposition 3. -The spectral zeta function associated with the operatorT V = −∆+V with Dirichlet boundary conditions and potential V ∈ L q [0, 1], q 1, defined by (1) may be extended to the half-plane Re(s) > −1/2 as a meromorphic function with a simple pole at s = 1/2, whose residue is given by 1/(2π).
Proof. -We start from where ϕ denotes the heat trace as above and the exchange between the sum and the integral is valid for Re(s) > 1/2. By Proposition 1 we may write where f (t) = O( √ t) as t approaches zero. Since, in addition, ϕ(t) = O(e −ct ) as t approaches infinity, for some c > 0, we have which is valid for Re(s) > 1 and where F is an analytic function in the half-plane Re(s) > −1/2. Due to the simple zero of 1/Γ(s) at zero we see that the expression in the right-hand side is well-defined and meromorphic in the half-plane Re(s) > −1/2, except for the simple pole at s = 1/2, showing that we may extend ζ T V (s) to this half-plane. The value of the residue is obtained by a standard computation.
Remark 4. -It is clear from the proof that the behaviour on Re(s) 1/2 will depend on the potential V . This may be seen from the simple example of a constant potential V (x) ≡ a, for which the heat trace now satisfies ϕ(t) = e −at ψ(t).
From this we see that when a is nonzero ϕ(t) has an expansion for small t with terms of the form t −(n−1)/2 , for all non-negative integers n. These terms produce simple poles at s = −(2n + 1)/2, n 0 an integer, with residues depending on a. When a vanishes we recover ζ T V (s) = ζ(2s) and there are no poles other than the simple pole at s = 1/2.
We are now ready to extend the result in [18] to the case of L q potentials with q greater than or equal to one.
Theorem 5. -The determinant of the operator T V = −∆ + V with Dirichlet boundary conditions and potential V ∈ L q [0, 1], q ∈ [1, +∞], is given by where y is the solution of the initial value problem (5).
Proof. -We shall follow along the lines of the proof in [18, Th. 1] for smooth potentials, which consists in building a one-parameter family of potentials, αV , connecting the zero potential, for which the expression for the determinant may be computed explicitly, with the potential V , and comparing the way these two quantities change.
More precisely, the main steps in this approach are as follows. For α ∈ [0, 1], we define consider the eigenvalue problem , we obtain that the eigenvalues λ k (α) and its associated (suitably normalised) eigenfunction u k (α), for any k 1, are analytic functions of α. The corresponding ζ-function is given by Although this series is only defined for Re(s) > 1/2, we know from Proposition 3 that the spectral zeta function defined by it can be extended uniquely to a meromorphic function on the half-plane Re(s) > −1/2 which is analytic at zero. We shall also define y(α, t) to be the family of solutions of the initial value problem By Proposition 7 in the next section the quantity 2y(α, 1) is well-defined for V ∈ L q [0, 1]. The idea of the proof is to show that for α ∈ [0, 1]. Since at α = 0, det T 0 = 2y(0, 1), it follows that equality of the two functions will still hold for α equal to one. The connection between the two derivatives is made through the Green's function of the operator. We shall first deal with the lefthand side of identity (10), for which we need to differentiate the series defining the spectral zeta function with respect to both α and s, and then take s = 0. We begin by differentiating the series in equation (9) term by term with respect to α to obtain where the expression for the derivative of λ j with respect to α is given by For potentials in L 1 [0, 1] we have, as we saw above, that the corresponding eigenvalues of problem (8) satisfy the asymptotics given by (6), while the corresponding eigenfunctions satisfy In a similar way, the numerator in the expression for λ j (α) satisfies where the last step follows from the Riemann-Lebesgue Lemma. Combining the two asymptotics we thus have and so the term in the series (11) is of order O(j −2s−2 ). This series is thus absolutely convergent (and uniformly convergent in α) for Re(s) > −1/2. This justifies the differentiation term by term, and also makes it possible to now differentiate it with respect to s to obtain which is uniformly convergent for s in a neighbourhood of zero and α in [0, 1]. We thus obtain ∂ ∂α Here, G α (λ = 0, t, t) is the restriction to the diagonal of the Green's function of the boundary value problem (8) at λ = 0. The exchange between the integral and the summation may be justified as above. We will now consider the right-hand side in (10).
Here we follow exactly the same computation as in [18]. For that we consider Then z(α, t) is a solution to the initial value problem: Using the variation of parameters formula, the solution of this problem is given by Therefore we obtain from which the identity (10) follows. Integrating this with respect to α yields where c is a constant independent of α. Since det T 0 = 2y(0, 1), the result follows.
We shall finish this section with the example of the pulse potential, of which the optimal potential in the L 1 case is a particular case.
and m > 0. A long but straightforward computation shows that the solution of is given by J.É.P. -M., 2020, tome 7 Therefore the functional determinant of the operator T = −d 2 /dx 2 + mχ S with Dirichlet boundary conditions is given by

Some properties of the determinant
Proposition 7. -The operator D is well-defined, Lipschitz on bounded sets of L q (hence continuous), and non-negative for non-negative potentials.
Proof. -For V ∈ L q [0, 1], local existence and uniqueness holds by Carathéodory for Gronwall's lemma implies that the solution is defined up to t = 1 and that Let V 1 and V 2 in L q [0, 1] be both of norm less or equal to A. Denote x 1 and x 2 the corresponding solutions defined as in (12). One has integrating one obtains By the integral version of Gronwall's inequality we obtain implying that D is Lipschitz on bounded sets. Let finally V be non-negative, and assume by contradiction that the associated y first vanishes at t > 0. As y(0) = y(t) = 0 and y (0) > 0, the function must have a positive maximum at some τ ∈ (0, t). The function y being continuously differentiable, y (τ ) = 0. Now, Then y (τ ) 1, contradicting the definition of τ .
Being Lipschitz on bounded sets, the operator D sends Cauchy sequences in L q [0, 1] to Cauchy sequences in R (Cauchy-continuity). So its restriction to the dense subset of smooth functions has a unique continuous extension to the whole space. As this restriction is equal to the halved determinant whose definition for smooth potentials is recalled in Section 1 for the operator −∆ + V with Dirichlet boundary conditions, D V is indeed the unique continuous extension of this determinant to L q [0, 1] and, in agreement with Theorem 5, We begin by proving a uniform upper bound on the maximum value of y(1) for the control problem (5), and thus for the determinant of the original Schrödinger operator given by (3).
Proof. -To prove the proposition it is enough to treat the case q = 1. The initial value problem given by equation (5) is equivalent to the integral equation We now build a standard iteration scheme defined by which converges to the solution of equation (5) -this is a classical result from the theory of ordinary differential equations which may be found, for instance, in [8], and which also follows from the computations below. We shall now prove by induction that From (14) it follows that s m+1 (m + 1) m+1 , the above may be written as Since f 0 (s) = s(t − s) t 2 /4, we have that and thus the induction hypothesis (15) holds for m = 1. Assume now that (15) holds. It follows from (14) that and, using (15), we obtain Differentiating f m (s) with respect to s and equating to zero, we obtain that this is maximal for s = (m + 1)t/(m + 2), yielding f m (s) t m+2 (m + 2) m+2 , and so On the other hand, yielding Taking t to be one finishes the proof.
We shall finally present a proof of the fact that in order to maximise the determinant it is sufficient to consider non-negative potentials. This is based on a comparison result for linear second order ordinary differential equations which we believe to be interesting in its own right, but which we could not find in the literature. Then Proof. -The proof is divided into two parts. We first show that if the potential V does not remain essentially non-negative, then u must become larger than v at some point, while never being smaller for smaller times. We then prove that once u is strictly larger than v for some time t 1 , then it must remain larger for all t greater than t 1 .
We first note that solutions of the above problems are at least in AC 2 [0, T ], and thus continuously differentiable on [0, T ]. Furthermore, u is always positive, since u(0) = 0, u (0) = 1 and u (0) 0. Let now w = u − v. While V remains essentially nonnegative, v also remains positive and w vanishes identically. Assume now that there exists a time t 0 such that for t in (0, t 0 ) the potential V is essentially non-negative and on arbitrarily small positive neighbourhoods of t 0 V takes on negative values on sets of positive measure. Then ] v close to t 0 (and zero elsewhere on these small neighbourhoods of t 0 ). Since w(t 0 ) = w (t 0 ) = 0 and w is non-negative and strictly positive on sets of positive measure contained in these neighbourhoods, w will take on positive values on arbitrarily small positive neighbourhoods of t 0 . Define now z(t) = (u 2 (t) − v 2 (t))/2. Then From the previous discussion above, we may assume the existence of a positive value t 1 such that both z and z are positive at t 1 (and thus in a small positive neighbourhood (t 1 , t 1 + δ)), while z is non-negative for all t in (0, t 1 ). Letting which is well-defined and bounded on [t 1 , t 1 + δ), we may thus write for t in [t 1 , t 1 + δ). Then z (t) − a(t)z (t) 0 and upon multiplication by a(s)ds and integration between t 1 and t we obtain a(s)ds z (t 1 ).
This yields the positivity of z (t) (and thus of z(t)) for t greater than t 1 . Combining this with the first part of the proof shows that z is never negative for positive t.
Henceforth, we restrict the search for maximising potentials to non-negative functions. Besides, it is clear from the proof that the result may be generalised to the comparison of two potentials V 1 and V 2 where V 1 V 2 , provided V 1 is non-negative.

Maximization of the determinant over L 1 potentials
By virtue of the analysis of the previous sections, problem (4) for q = 1 can be recast as the following optimal control problem: −y + V y = 0, y(0) = 0, y (0) = 1, for fixed positive A. In order to prove Theorem A, a family of auxiliary problems is introduced: in addition to (17), the potentials are assumed essentially bounded and such that for a fixed positive B. To avoid the trivial solution V ≡ B, we suppose B > A and henceforth study the properties of problem (16)- (19). Setting x := (y, y , x 3 ), the auxiliary problem can be rewritten −x 1 (1) → min under the dynamical constraints V measurable valued in [0, B], and the boundary conditions x(0) = (0, 1, 0), free x 1 (1) and x 2 (1), x 3 (1) A.

Proposition 10. -Every auxiliary problem has a solution.
Proof. -The set of admissible controls is obviously nonempty, the control is valued in a fixed compact set, and the field of velocities is convex for any x ∈ R 3 ; according to Filippov Theorem [1], existence holds.

The dynamical system (20)-(22) is bilinear in x and
Let Φ(t) := H 1 (x(t), p(t)) be the evaluation of H 1 along the extremal (x, V, p); it is a Lipschitz function, and the maximisation condition (24) implies that V (t) = 0 when Φ(t) < 0, V (t) = B when Φ(t) > 0. If Φ vanishes identically on some interval of nonempty interior, the control is not directly determined by (24) and is termed singular. Subarcs of the trajectory corresponding to V = 0, V = B and singular control, are labelled γ 0 , γ + and γ s , respectively. Differentiating once, one obtains Φ (t) = H 01 (x(t), p(t)), where H 01 = {H 0 , H 1 } is the Poisson bracket of H 0 and H 1 ; Φ is so W 2,∞ and, differentiating again, In particular, using the definition of H, where H and p 3 are constant along any extremal. There are two possibilities for extremals depending on whether p 2 (1) = p 0 is zero or not (so-called abnormal or normal cases).
Along a singular arc, Φ ≡ 0 so (26) allows to determine the singular control provided H 101 = 0 ("order one" singular). A necessary condition for optimality (the Legendre-Clebsch condition) is that H 101 0 along such an arc; order one singular arcs such that H 101 > 0 are called hyperbolic [4]. Proof. -According to the maximum principle, the trajectory associated with an optimal potential is the concatenation of (possibly infinitely many) γ 0 , γ + and γ s subarcs. By Lemma 13, there are exactly two γ 0 arcs; if V has an infinite number of discontinuities, it is necessarily due to the presence of infinitely many switchings between γ + and γ s subarcs with H + p 3 B < 0 (this quantity must be nonpositive to ensure admissibility of the singular potential, V = −H/p 3 B, and even negative since otherwise γ + and γ s would be identical, generating no discontinuity at all). By Lemma 14, there is no γ + subarc if H + p 3 B < 0. There are thus only finitely many switchings of the potential between the constant values 0, B and −H/p 3 .
Proof. -There is necessarily some minimum t ∈ (0, 1) such that Φ(t) = 0 (otherwise V ≡ 0 which would contradict the existence of solution). There are two cases, depending on the order of the contact of the extremal with H 1 = 0.
Proof of Theorem A. -The proof is done in three steps: first, existence and uniqueness are obtained for each auxiliary problem (17)- (19) with bound V B on the potential, B large enough; then existence for the original problem (16)- (19) with unbounded control is proved. Finally, uniqueness is obtained.
where shc(z) := (1/z) sh z. The function to be maximised is continuous on the compact triangle defining the constraints on (s, ), so one retrieves existence. As we know that an optimal arc must start and end with γ 0 arcs, the solution cannot belong to the part of the boundary corresponding to = 2 min{s, 1 − s}, extremities included; as a result, ∂y/∂s must be zero at a solution, so s = 1/2 (potential symmetric with respect to t = 1/2). Since one has y(s = 1/2, ) = (1 + A/4) + A 2 /24 + o( ), too small a cannot maximise the function; so, for B large enough, the point on the boundary = A/B cannot be solution, and one also has that Assume that there exists an increasing sequence (C k ) k of subsets of C such that Proof. -Let x be the limit of the stationary sequence (x k ) k ; assume, by contradiction, that f := f (x) > inf C f (including the case inf C f = −∞). Then there is y ∈ C such that f (y) < f . By density, there is k ∈ N and y k ∈ C k such that y − y k ε where, by continuity of f at y, ε > 0 is chosen such that then f (y k ) f (y) + d/2 < f , which is contradictory as k can be taken large enough in order that x k = x. So f (x) = inf C f , whence existence.

End of proof of Theorem
The Hamiltonian is unchanged, but the constraint on the potential is now just V 0; as a consequence, Φ = p 2 x 1 + p 3 must be nonpositive because of the maximisation condition. Normality is proved as in Lemma 11 and p 2 (1) can be set to −1. Regarding strong activation of the L 1 constraint (p 3 < 0) the argument at the beginning of the proof of Lemma 12 immediately implies that p 3 must be negative as Φ cannot be positive, now. So Φ(0) = Φ(1) = p 3 < 0 and V is equal to 0 in the neighbourhood of t = 0+ and t = 1−. Because existence holds, Φ must first vanish at some t ∈ (0, 1) (otherwise V ≡ 0 which is obviously not optimal); necessarily, Φ (t) = 0. One verifies as before that there cannot be interior γ 0 subarcs, which prevents accumulation of γ 0 γ s or γ s γ 0 switchings. In particular, V must be piecewise constant and the right limit Φ (t+) exists. The same kind of reasoning as in the proof of Corollary 17 rules out the fold case Φ (t+) nonzero; so Φ (t+) = 0 and V switches from 0 to singular at t. The impossibility of interior γ 0 subarcs implies a γ 0 γ s γ 0 structure, which is the solution obtained before.
Remark 19. -This proves in particular that the generalised "impulsive" potential equal to A times the Dirac mass at t = 1/2 is not optimal, whatever A > 0, the combination γ 0 γ s γ 0 giving a better cost. More precisely, the optimal value of the determinant is 1 + A/4 + A 3 /192 + o(A 3 ), so it is asymptotic when A tends to zero to the value obtained for the Dirac mass (equal to 1 + A/4, as is clear from (i) in the proof above).

Maximization of the determinant over L q potentials, q > 1
We begin the section by proving that maximising the determinant over L 1 potentials is estimated by maximising the determinant over L q , letting q tend to one. To this end, we first establish the following existence result. (Note that the proof is completely different from the existence proof in Theorem A.) Let A > 0, and consider the following family of problems for q in [1, ∞]: Proposition 20. -Existence holds for q ∈ [1, ∞].
Proof. -The case q = ∞ is obvious, while Theorem A deals with q = 1. Let then q belong to (1, ∞). By using Gronwall lemma as in (13), it is clear that D is bounded on the closed ball of radius A of L q [0, 1]. So the value of the problem is finite. Let (V k ) k be a maximising sequence. As V k q A for any k, up to taking a subsequence one can assume the sequence to be weakly- * converging in L q (L r ) * (1/q + 1/r = 1) towards some V . Clearly, V q A. Let x k be associated with V k according to (12). The sequence (x k ) k so defined is bounded and equicontinuous as, for any t s in [0, 1], by Hölder inequality. Using Ascoli's Theorem (and taking a subsequence), (x k ) k converges uniformly towards some x. As x(1) = lim k x k (1) is equal to the value of the problem, it suffices to check that x = C(V )x to conclude. Being a bounded sequence, (C(V k )) k is equicontinuous in L q (L r ) * , and (x k · χ |[0,t] ) k converges towards x · χ |[0,t] in L r for any t in [0, 1] (χ |[0,t] denoting the characteristic function of the interval [0, t]). So which concludes the proof.
is decreasing and v q tends to v 1 when q → 1+. 1], V q V r so the radius A ball of L r is included in the radius A ball of L q . As a result of the inclusion of the admissible potentials, v q v r . Because of monotonicity, the limit v := lim q→1+ v q exists, and v v 1 . Moreover, as proved in Theorem A, the unique maximising potential V 1 for q = 1 is actually essentially bounded so for any q 1. By continuity of D on L 1 , the left-hand side of the previous inequality tends to D V1 = v 1 when q → 1+, and one conversely gets that v 1 v.
Fix q > 1 and A > 0. As in Section 3, we set x := (y, y , x 3 ) to take into account the L q constraint. Then problem (29) can be rewritten −x 1 (1) → min under the dynamical constraints and the boundary conditions x(0) = (0, 1, 0), free x 1 (1) and x 2 (1), x 3 (1) A q . The potential is measurable and can be assumed non-negative thanks to the comparison result from Proposition 9. For such an L q optimal potential V , the Pontryagin maximum principle holds for the same reason as in the proof of Theorem A (see step (iii)). So there exists a nontrivial pair (p 0 , p) = (0, 0), p 0 0 a constant and p : [0, 1] → (R 3 ) * a Lipschitz covector function such that, a.e. on [0, 1], and where the Hamiltonian H is now equal to (With f (x, V ) denoting the dynamics (30)-(32) in compact form.) In addition to the boundary conditions on x, the following transversality conditions hold (note that p 3 is again a constant): p 1 (1) = −p 0 , p 2 (1) = 0, and p 3 0 with complementarity Although the system is not bilinear in (x, V ) anymore, the adjoint equation holds unchanged, and one proves normality and strong activation of the L q constraint similarly to the case q = 1.
Lemma 23. -The cost multiplier p 0 is negative.
Proof. -By contradiction: if p 0 = 0, one has p 2 (and p 1 ) identically zero by (33), so p 3 cannot also be zero and must be negative. Then, H = p 3 V q and the maximisation condition implies V = 0 a.e. since V is non-negative. This is contradictory as the zero control is admissible but clearly not optimal.
We will not set p 0 = −1 but will use the fact that p 3 is also negative to use a different normalisation instead.
Proof. -Observe that, as in Lemma 12, p 2 x 1 is positive on (0, 1). Now, assume by contradiction that p 3 = 0: then H = p 1 x 2 +p 2 x 1 V , which would prevent maximisation of H on a nonzero measure subset.
Proof. -Because Ψ is non-negative, it is clear from the maximisation condition that for all t ∈ (0, 1). Since Ψ is an absolutely continuous function, we can differentiate once to get and iterate to obtain Using the fact that the Hamiltonian H is constant along an extremal, and substituting p 1 x 2 by H −V Ψ−p 3 V q and V by its expression, the following second order differential equation is obtained for Ψ: We can normalise p 3 in order that −p 3 = B q−1 q , which gives the desired differential equation for Ψ, as well as the desired expression for V .
According to what has just been proved, t → (Ψ(t), Ψ (t)) parametrises the curve y 2 = f (x) where f (that depends on H, A and q) is Since Ψ (0) = H − c(A, q) is positive, H > c(A, q) > 0 and f has a local minimum (resp. maximum) at As g(c(A, q)) < 0 and g(H) → ∞ when H → ∞ (note that 1 + 1/α < 2), g has one zero in (c(A, q), ∞), and only one in this interval as is clear inspecting g . Proof. -The derivative of ψ is strictly positive for t in (0, 1/2) as is clear from Figure 1, case (i). As a consequence, ψ is strictly increasing on [0, 1/2], and so is V by virtue of (34).
Recalling Proposition 25, one eventually gets that the maximal potential for q = 2 is for the unique H in (9A 2 /2, h * (A)) such that Ψ(1) = 0, that is This proves Theorem C.