The Dequantized Schrodinger Equation and a

Complex-Valued Stationary-Action Diffusion

Representation ∗

William M. McEneaney †

August 20, 2015

Abstract

A stochastic representation for solution of the Schrodinger equationis obtained, utilizing complex-valued diffusion processes. The Maslov de-quantization is employed, where the domain is complex-valued in the spacevariable. The notion of staticization is required to relate the Hamilton-Jacobi form of the dequantized Schrodinger equation to its stochasticcontrol representation. A verification result is obtained, and existenceis reduced to a real-valued space-variable case.

Key words. Schrodinger equation, stochastic control, Hamilton-Jacobi,stationary action, staticization, complex-valued diffusion.

MSC2010. 35Q41, 60J60, 93E20, 49Lxx, 35Q93, 49L20, 30D20.

1 Introduction

We recall the Schrodinger initial value problem, given as

0 = i~ψt(s, x) + ~2

2m∆ψ(s, x)− ψ(s, x)V (x), (s, x) ∈ D, (1)

ψ(0, x) = ψ0(x), x ∈ Rn, (2)

where ~ denotes Planck’s constant, m ∈ (0,∞) denotes mass, initial conditionψ0 takes values in C, V denotes a known potential function, ∆ denotes theLaplacian with respect to the space (second) variable, D .

= (0, t) × Rn, andsubscript t will denote the derivative with respect to the time variable (the firstargument of ψ here) regardless of the symbol being used for time in the argumentlist. We also let D .

= [0, t]× Rn. We consider what is sometimes referred to asthe Maslov dequantization of the solution of the Schrodinger equation (cf., [17]),which is S : D → C given by ψ(s, x) = exp i~S(s, x). Note that ψt = i

~ψSt,

∗Research partially supported by grants from AFOSR and NSF.†University of California San Diego, La Jolla, CA 92093-0411, USA. wmceneaney@ucsd.edu

1

ψx = i~ψSx and ∆ψ = i

~ψ∆S − 1~2ψ|Sx|2c where for y ∈ Cn, |y|2c

.=∑nj=1 y

2j .

(We remark that notation | · |2c is not intended to indicate a squared norm; therange is complex.) We find that (1),(2) become

0 = −St(s, x) + i~2m∆S(s, x) +H(x, Sx(s, x)), (s, x) ∈ D, (3)

S(0, x) = φ(x), x ∈ Rn, (4)

where H : Rn × Cn → C is the Hamiltonian given by

H(x, p) = −[

12m |p|

2c + V (x)

]= statv∈Cn

v · p+ m

2 |v|2c − V (x)

, (5)

and stat will be defined in the next section. We look for solutions in the space

S .= S : D → C |S ∈ C1,2

p (D) ∩ C(D), (6)

where C1,2p denotes the space of functions which are continuously differentiable

once in time and twice in space, and which satisfy a polynomial-growth boundover the space variable. We will find it helpful to reverse the time variable,and hence we look instead, and equivalently, at the Hamilton-Jacobi partialdifferential equation (HJ PDE) problem given by

0 = St(s, x) + i~2m∆S(s, x) +H(x, Sx(s, x)), (s, x) ∈ D, (7)

S(t, x) = φ(x), x ∈ Rn. (8)

Working mainly with this last form, we will fix t ∈ (0,∞), and allow s to varyin [0, t].

Recall the semiclassical limit (cf. [1, 10, 11, 12]) where one views ~ as asmall parameter, and examines the limit as ~ ↓ 0. This leads to an HJ PDEproblem of the form

0 = St(s, x) +H(x, Sx(s, x)), (s, x) ∈ D, (9)

S(t, x) = φ(x), x ∈ Rn. (10)

In recent work (cf. [2, 18, 19, 21]) it has been shown that stationary-actionformulations of certain two-point boundary value problems (TPBVPs) for con-servative dynamical systems also take the form (9),(10). This motivates theapproach here, where we develop a stationary-action based representation forthe solution of (7),(8) (and consequently (1),(2)). Due to the complex multiplieron the Laplacian, this representation is in terms of a stationary-action stochas-tic control problem with a complex-valued diffusion coefficient. It is importantto note that the stationary-action diffusion process representation here is moreclosely related to the semiclassical limit analysis and to path-integral forms (cf.[3, 4]) than to another well-known area of research on diffusion-process represen-tations, cf. [8, 14, 22, 23, 25]. In particular, those works discuss representations(and/or lack thereof) in the spirit of the Fokker-Planck equation, while herethe representation generates S, ψ, and consequently the position probability,through the stationary-action value of the stochastic process.

It is also important to note that the results are obtained here under strongassumptions. In particular, we assume the existence of a sufficiently smooth

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potential function, defined over all of the complex space domain, which matchesV on Rn. The assumptions allow for the inclusion of the case of the quantumharmonic oscillator, when one takes the potential to be a holomorphic quadraticover the complex domain. Potentials lacking such smoothness are beyond thescope here, and it is expected that such problems will be studied with the aidof viscosity-solution techniques in a later effort.

In Section 2, we recall the definitions necessary for staticization problems.In Section 3, the underlying space domain is extended from a space over the realfield to a space over the complex field. This necessitates several other minorextensions, which are covered in the subsections. In particular, some classi-cal existence and uniqueness results for stochastic differential equations (SDEs)are trivially extended to their complex-valued counterparts. In Section 4, thestaticization-based stochastic-control value function representation for the de-quantized Schrodinger equation is obtained. More specifically, a verificationresult is obtained demonstrating that if a solution of the HJ PDE over the“complexified” domain exists, then that solution has the indicated representa-tion. Lastly, in Section 5, we indicate a result about existence of solutions ofthe HJ PDE over the complexified domain.

2 Stationarity definitions

Recall that classical systems obey the stationary action principle, where thepath taken by the system is that which is a stationary point of the actionfunctional. For this and other reasons, as in the definition of the Hamiltoniangiven in (5), we find it useful to develop additional notation and nomenclature.Specifically, in analogy with the language for minimization and maximization,we will refer to the search for stationary points as staticization, with these pointsbeing statica (in analogy with minima/maxima) and a single such point beinglabeled a staticum (in analogy with minimum/maximum). More specifically, wemake the following definitions. Suppose U is a generic normed vector space overC with G ⊆ U , and suppose F : G → C. We say v ∈ argstatF (v) | v ∈ G ifv ∈ G and either

lim supv→v,v∈G\v

|F (v)− F (v)||v − v|

= 0, (11)

or there exists δ > 0 such that G ∩ Bδ(v) = v (where Bδ(v) denotes the ballof radius δ around v). If argstatF (v) | v ∈ G 6= ∅, we define the possiblyset-valued stats operator by

statsv∈G

F (v).= statsF (v) | v ∈ G .=

F (v)

∣∣ v ∈ argstatF (v) | v ∈ G.

If argstatF (v) | v ∈ G = ∅, statsv∈G F (v) is undefined. Where applicable,we are also interested in a single-valued stat operation (note the absence ofsuperscript s). In particular, if there exists a ∈ C such that statsv∈G F (v) = a,then statv∈G F (v)

.= a; otherwise, statv∈G F (v) is undefined. At times, we may

3

abuse notation by writing v = argstatF (v) | v ∈ G in the event that theargstat is the single point v.

In the case where U is a Hilbert space, and G ⊆ U is an open set, F : G → Cis Frechet differentiable at v ∈ G with Riesz representation Fv(v) ∈ U if

limw→0, v+w∈G\v

|F (v + w)− F (v)− 〈Fv(v), w〉||w|

= 0. (12)

The following is immediate from the above definitions.

Lemma 1 Suppose U is a Hilbert space, with open set G ⊆ U and v ∈ G. Then,v ∈ argstatF (y) | y ∈ G if and only if Fv(v) = 0.

For a discussion of staticization for general deterministic systems, we referthe reader to [18, 20].

3 Extensions to the complex domain

Various details of extensions to the complex domain must be considered priorto the development of the representation.

3.1 Extended problem and assumptions

Although (1)–(2), (3)–(4) and (7)–(8) are typically given as HJ PDE problemsover D, we will find it convenient to change the domain to one where the spacecomponents lie over the complex field. We also extend the domain of the po-tential to Cn, where V : Cn → C, and we will abuse notation by employingthe same symbol for the extended-domain functions. Let DC

.= (0, t)× Cn and

DC = [0, t]× Cn, and define

SC.= S : DC → C : S ∈ C1,2

p (DC) ∩ C(DC) . (13)

The extended-domain form of problem (7)–(8) is

0 = St(s, x) + i~2m∆S(s, x) +H(x, Sx(s, x)), (s, x) ∈ DC, (14)

S(t, x) = φ(x), x ∈ Cn. (15)

Throughout Sections 3–4, we will assume:

(A.0) For all ~ ∈ (0, 1], there exists a solution, S = S~ ∈ SC to (14)–(15).

(A.1)V, φ : Cn → C are entire, i.e., holomorphic on all of Cn. Further,there exists C0 <∞ and q ∈ N such that |Vxx(x)|, |φxx(x)| < C0(1+|x|2q) for all x ∈ Cn.

(A.2)There exists CS < ∞ such that |Sx(r, x)| ≤ CS(1 + |x|) and|Sxx(r, x)| ≤ CS(1 + |x|2q) for all (r, x) ∈ DC.

It may be helpful to make some remarks here. First, the potential for thequantum harmonic oscillator can be extended from a quadratic function on Rn

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to a quadratic (entire function) on Cn in the obvious way. Further, g(y).=

1/|y| for y ∈ Rn can be approximated by gε ∈ C1, which exactly matchesg outside a ball of radius ε > 0 and is quadratic inside that ball, where gε

is analogous to the gravitational field generated by a uniform-density sphere.Further, one has the classical Carleman result and its extensions (cf., [13]), bywhich one may approximate a C1 function on R by an entire function such thatthe approximation errors on the real line for both the function and its derivativesgo to zero as |x| → ∞.

3.2 The underlying stochastic dynamics

We let (Ω,F , P ) be a probability triple, where Ω denotes a sample space, Fdenotes a σ-algebra on Ω, and P denotes a probability measure on (Ω,F).Let Fs | s ∈ [0, t] denote a filtration on (Ω,F , P ) (Fs0 ⊂ Fs1 ⊂ F for all0 ≤ s0 < s1 ≤ t), and let B· denote an F·-adapted Brownian motion takingvalues in Rn. For s ∈ [0, t], let

Us.= u : [s, t]× Ω→ Cn |u is F·-adapted, right-continuous and such that

E∫ ts|ur|m dr <∞ ∀m ∈ N . (16)

We supply Us with the norm

‖u‖Us.= maxm∈]1,M [

[E∫ ts|ur|m dr

]1/m,

where M ≥ 8q.We will be interested in diffusion processes given by

ξr = ξ(s,x)r = x+

∫ r

s

uρ dρ+√

~m

1+i√2

∫ r

s

dBρ.= x+

∫ r

s

uρ dρ+√

~m

1+i√2B∆r ,

(17)

where x ∈ Cn, s ∈ [0, t], u ∈ Us, and B∆r

.= Br − Bs for r ∈ [s, t]. Note

that because (1) and the dequantized versions thereof are generally defined over(0, t)×Rn, one could restrict the initial conditions to x ∈ Rn rather than x ∈ Cn,although for r ∈ (s, t], one typically has ξr ∈ Cn. We will also be interested inthe case where the control input is generated by a state-feedback. In particular,we will consider

ξ∗r = ξ∗,(s,x)r

.= x+

∫ r

s

(−1m )Sx(ρ, ξ∗,(s,x)

ρ ) dρ+√

~m

1+i√2B∆r , (18)

where presuming for now existence and uniqueness of a solution, we may definethe resulting u∗,(s,x) ∈ Us given by

u∗,(s,x)r (ω)

.= ˆu(r, ξ∗,(s,x)(ω))

.= (−1

m )Sx(r, ξ∗,(s,x)r (ω)) (19)

for all r ∈ [s, t] and ω ∈ Ω.

5

For s ∈ [0, t], we let

Xs.= ξ : [s, t]× Ω→ Cn | ξ is F·-adapted, right-continuous and such that

E supr∈[s,t]

|ξr|m <∞ ∀m ∈ N . (20)

We supply Xs with the norm

‖ξ‖Xs

.= maxm∈]1,M [

[E supr∈[s,t]

|ξr|m]1/m

.

It is natural to work with complex-valued state processes in this problemdomain. However, in order to easily apply many of the existing results regardingexistence, uniqueness and moments, we will find it handy to use a “vectorized”real-valued representation for the complex-valued state processes. We beginfrom the standard mapping of the complex plane into R2, V00 : C → R2, givenby V00(x)

.= V00(y + iz)

.= (y, z)T , where y = Re(x) and z = Im(x). This

immediately yields the mapping V0 : Cn → R2n given by V0(y+iz).= (yT , zT )T ,

where component-wise, (yj , zj)T = V00(xj) for all j ∈]1, n[ , where throughout,

for integer a ≤ b, we define ]a, b[= a, a+1, . . . b. Given control process, u ∈ Us,we define its vectorized analog by the isometric isomorphism, V : Us → Uvs ,where [V(u)]r

.= [V(v + iw)]r

.= (vTr , w

Tr )T and (vTr , w

Tr )T = V0(ur) for all

r ∈ [s, t] and ω ∈ Ω, and where

Uvs.= (v, w) : [s, t]× Ω→ R2n | (v, w) is F·-adapted, right-continuous and

such that E∫ ts|vr|m + |wr|m dr <∞ ∀m ∈ N,

(21)

‖u‖Uvs

.= maxm∈]1,M [

[E∫ ts|vr|m + |wr|m dr

]1/m. (22)

Abusing notation, we also define the isometric isomorphism, V : Xs → X vs by[V(ξ)]r

.= [V(η + iζ)]r

.= (ηTr , ζ

Tr )T for all r ∈ [s, t] and ω ∈ Ω, where

X vs.= (η, ζ) : [s, t]× Ω→ R2n | (η, ζ) is F·-adapted, right-continuous and

such that E supr∈[s,t]

[|ηr|m + |ζr|m] <∞ ∀m ∈ N ,

(23)

‖(η, ζ)‖Xvs

.= maxm∈]1,M [

[E supr∈[s,t]

(|ηr|m + |ζr|m)]1/m

. (24)

Under transformation by V, (17) becomes(ηrζr

)=

(yz

)+

∫ r

s

(vρwρ

)dρ+

√~m

1√2

(In×nIn×n

)B∆r . (25)

We may decompose S ∈ SC as

(R(r,V0(x)), T (r,V0(x)))T.= V00(S(r, x)), (26)

6

where R, T : D2.= [0, t] × R2n → R. We indicate some standard relations

between derivative components. Partial derivatives in the space variable of Sare given by

Sxj (r, x) = Ryj (r, y, z)− iRzj (r, y, z) = Tzj (r, y, z) + iTyj (r, y, z) (27)

for all (r, x) = (r, y + iz) ∈ (0, t) × Cn and all j ∈]1, n[. That is, lettinga+ ib = Sxj (r, x), one has

a = Ryj = Tzj , b = −Rzj = Tyj , (28)

evaluated at (r, y, z) for all j ∈]1, n[. Further, letting c + id.= Sxj ,xk

(r, x), onehas

c = Ryj ,yk = −Rzj ,zk = Tzj ,yk = Tyj ,zk , (29)

d = −Ryj ,zk = −Rzj ,yk = −Tzj ,zk = Tyj ,yk , (30)

evaluated at (r, y, z) for all j, k ∈]1, n[. Finally, letting e + if.= Sxj ,xk,x`

(r, x),one has

e = Ryj ,yk,y` = −Ryj ,zk,z` = −Rzj ,zk,y` = −Rzj ,yk,z`= Tzj ,yk,y` = −Tzj ,zk,z` = Tyj ,zk,y` = Tyj ,yk,z` , (31)

f = −Ryj ,yk,z` = −Ryj ,zk,y` = Rzj ,zk,z` = −Rzj ,yk,y`= −Tzj ,yk,z` = −Tzj ,zk,y` = −Tyj ,zk,z` = Tyj ,yk,y` , (32)

evaluated at (r, y, z) for all j, k, ` ∈]1, n[.One may also easily verify that for x = y + iz ∈ Cn,

V00(|x|2c) =

n∑j=1

(y2j − z2

j

2yjzj

).

More specifically, with R, T given by (26) and (yT , zT )T = V0(x),

V00(|Sx(r, x)|2c) =

(∑nj=1 R

2yj (r, y, z)− R2

zj (r, y, z)∑nj=1−2Ryj (r, y, z)Rzj (r, y, z)

)=

(∑nj=1 T

2zj (r, y, z)− T 2

yj (r, y, z)∑nj=1 2Tyj (r, y, z)Tzj (r, y, z)

). (33)

Also, letting (v0, w0).= V0(u0) for all u0 ∈ Cn, given s ∈ [0, t] and x ∈ Cn with

(yT , zT )T.= V0(x),

V0

(argstatu0∈Cn

[ n∑j=1

Sxj(r, x)u0

j + m2 |u

0|2c])

= V0

(−1m Sx(r, x)

)(34)

= 1m

(−Ry(r, y, z)Rz(r, y, z)

)= 1

m

(−Tz(r, y, z)−Ty(r, y, z)

).

7

Using the above, we see that under transformation V, (18) becomes(η∗rζ∗r

)=

(yz

)+

∫ r

s

1m

(−Ry(ρ, η∗ρ, ζ

∗ρ )

Rz(ρ, η∗ρ, ζ∗ρ )

)dρ+

√~m

1√2

(In×nIn×n

)B∆r

=

(yz

)+

∫ r

s

−1m

(Tz(ρ, η

∗ρ, ζ∗ρ )

Ty(ρ, η∗ρ, ζ∗ρ )

)dρ+

√~m

1√2

(In×nIn×n

)B∆r . (35)

Throughout, concerning both real and complex stochastic differential equations,typically given in integral form such as in (35), solution refers to a strong solu-tion.

Lemma 2 Let s ∈ [0, t), x ∈ Cn, u ∈ Us, (y, z) = V0(x) and (v, w) = V(u).There exists a unique solution, (η, ζ) ∈ X vs , to (25), and a unique solution,(η∗, ζ∗) ∈ X vs , to (35).

Proof: These are standard results. See [6], Section III.5 and Appendix D;[15], Lemma II.5.2 and Corollary II.5.12; and [16]. In the case of (35), oneshould note that by Assumption (A.2), Ry, Rz, Ty, Tz grow at most linearly iny, z.

The following is straightforward from the above definitions, cf. [24].

Lemma 3 Let s ∈ [0, t), x ∈ Cn, u ∈ Us, (yT , zT )T = V0(x) and (v, w) = V(u).ξ ∈ Xs is a solution of (17) if and only if V(ξ) ∈ X vs is a solution of (25).Similarly, ξ∗ ∈ Xs is a solution of (18) if and only if V(ξ∗) ∈ X vs is a solutionof (35).

Combining Lemmas 2 and 3, one has:

Lemma 4 Let s ∈ [0, t), x ∈ Cn and u ∈ Us. There exists a unique solution,ξ ∈ Xs, to (17), and a unique solution, ξ∗ ∈ Xs, to (18).

3.3 A relationship among the solutions

Let D2.= (0, t)× R2n and D2

.= [0, t]× R2n. By (27),

|Sx|2c =

n∑j=1

R2yj − R

2zj + 2iTyj Tzj , (36)

and by (29), (30)

∆S =

n∑j=1

Tyj ,zj − iRyj ,zj . (37)

Let (V R(V0(x)), V I(V0(x))

)T .= V00(V (x)), (38)

8

(φR(V0(x)), φI(V0(x))

)T .= V00(φ(x)), (39)

for all x ∈ Cn. Substituting (36)–(38) into (14), we have

0 =Rt + iTt − i~2m

n∑j=1

(Tyj ,zj − Ryj ,zj

)− 1

2m

n∑j=1

(R2yj − R

2zj + 2iTyj Tzj

)−(V R + iV I

)on D2, which yields

0 = Rt − ~2m

n∑j=1

Ryj ,zj − 12m

n∑j=1

(R2yj − R

2zj

)− V R ∀ (s, y, z) ∈ D2, (40)

0 = Tt − ~2m

n∑j=1

Tyj ,zj − 1m

n∑j=1

Tyj Tzj − V I ∀ (s, y, z) ∈ D2, (41)

on D2, and, of course,

R(t, y, z) = φR(y, z) ∀ (y, z) ∈ R2n, (42)

T (t, y, z) = φI(y, z) ∀ (y, z) ∈ R2n. (43)

Proposition 5 Let S ∈ SC and R, T satisfy (26) for all (r, x) ∈ DC. IfS satisfies (14)–(15), then R, T satisfy (40)–(43). Alternatively, if R, T ∈C1,2p (D2;R) ∩ C(D2;R) satisfy (40)–(43), and S ∈ SC is given by (26), then

S satisfies (14)–(15).

Proof: The first assertion follows by simple algebraic substitution, using(27)–(30). Now, suppose R, T ∈ C1,2

p (D2;R) ∩ C(D2;R) satisfy (40)–(43), and

let S be given by (26). We will show that R, T satisfy the Cauchy-Riemann rela-tions, and hence that S ∈ SC. After that, one may again use simple algerbrtaicsubstitutions to verify the final assertion.

Differentiating (40) with respect to yk, and (41) with respect to zk, yields

Rt,yk = ~2m

n∑j=1

Ryj ,zj ,yk + 1m

n∑j=1

(Ryj Ryj ,yk − Rzj Rzj ,yk

)+ V Ryk , (44)

Tt,zk = ~2m

n∑j=1

Tyj ,zj ,zk + 1m

n∑j=1

(Tzj Tyj ,zk + Tyj Tzj ,zk

)+ V Izk . (45)

Applying (27)–(32) in (45), one finds

Tt,zk = ~2m

n∑j=1

Ryj ,zj ,yk + 1m

n∑j=1

(Ryj Ryj ,yk − Rzj Rzj ,yk

)+ V Ryk ,

which by (44),

= Rt,yk ∀ (s, y, z) ∈ D2, ∀k ∈]1, n[. (46)

9

Also note that as φ is holomorphic,

Ryk(t, y, z) = φRyk(y, z) = φIzk(y, z) = Tzk(t, y, z) ∀ y, z ∈ Rn. (47)

By the Fundamental Theorem of Calculus,

Tzk(s, y, z) = Tzk(t, y, z)−∫ t

s

Tt,zk(σ, y, z) dσ,

which by (46),(47),

= Ryk(s, y, z) ∀ (s, y, z) ∈ D2, ∀k ∈]1, n[. (48)

Similarly, one obtains

Tyk(s, y, z) = −Rzk(s, y, z) ∀ (s, y, z) ∈ D2, ∀k ∈]1, n[. (49)

By (48),(49), the Cauchy-Riemann conditions are satisfied.

4 The verification

We will obtain a verification result demonstrating that a solution of (7),(8) isthe stationary value of the expectation of the action functional on process pathssatisfying (17).

For s ∈ (0, t) and ~ ∈ (0, 1], we define payoff J~(s, ·, ·) : Rn × Us → C by

J~(s, x, u).= E

∫ t

s

L(ξr, ur) dr + φ(ξt)

.= E

∫ t

s

m2 |ur|

2c − V (ξr) dr + φ(ξt)

, (50)

where ξ satisfies (17) with input u ∈ Us and initial state x ∈ Rn. The stationaryvalue, S~ : D → C, is given by

S~(s, x).= statu∈Us

J~(s, x, u) ∀ (s, x) ∈ D. (51)

We assume throughout Section 4 that

(A.3) argstatu∈Us J~(s, x, u) is single-valued for all (s, x) ∈ D.

This is the last assumption. We remark that one may want to weaken thisassumption to uniqueness in some prespecified subset of D, but leave that ad-ditional complication to a later effort. The main result of the section is:

Theorem 6 Let ~ ∈ (0, 1]. Suppose S ∈ SC∩C1,4(DC) satisfies (14)–(15), and

there exists CS < ∞ such that |Sxxx(r, x)|, |Stxx(r, x)|, |Sxxxx(r, x)| ≤ CS(1 +|x|2q) for all (s, x) ∈ DC. Then, S(s, x) = S~(s, x) for all (s, x) ∈ DC.

We remark that the representation is proved for general ~ ∈ (0, 1] in antici-pation of potential use in future semiclassical limit results. We begin with twolemmas.

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Lemma 7 Let s ∈ [0, t), x ∈ Cn, ~ ∈ (0, 1] and u ∈ Us. Let ξ ∈ Xs be givenby (17). Suppose S ∈ SC satisfies (14),(15). Let u∗ = u∗,(s,x), ξ∗ = ξ∗,(s,x) begiven by (18)-(19). Then,

S(s, x) = E∫ t

s

−St(r, ξr)− STx (r, ξr)ur − i~2m∆S(r, ξr) dr + φ(ξt)

and

S(s, x) = E∫ t

s

−St(r, ξ∗r )− STx (r, ξ∗r )u∗r − i~2m∆S(r, ξ∗r ) dr + φ(ξ∗t )

.

Proof: We prove only the second assertion; the proof of the first is simi-lar and somewhat simpler. Fix ~ ∈ (0, 1] and (s, x) ∈ DC. Let ξ∗ = ξ∗,(s,x)

and u∗ = u∗,(s,x) be given by (18)-(19) with S replacing S. Let (yT , zT )T =V0(x), (η∗, ζ∗) = V(ξ∗), (v∗, w∗) = V(u∗), and (R(r,V0(x)), T (r,V0(x))) =V00(S(r, x)). Note that (η∗, ζ∗) satisfy (35) with R, T replacing (R, T ). ByIto’s formula,

E[R(t, η∗t , ζ

∗t )]

= R(s, y, z) + E∫ t

s

Rt(r, η∗r , ζ∗r ) + RTy (r, η∗r , ζ

∗r )v∗r

+ RTz (r, η∗r , ζ∗r )w∗r + ~

4m

n∑j=1

[Ryj ,yj + 2Ryj ,zj + Rzj ,zj

](r, η∗r , ζ

∗r ) dr

+√

~2m

∫ t

s

[Ry(r, η∗r , ζ

∗r ) + Rz(r, η

∗r , ζ∗r )]TdBr

, (52)

where, in the interests of space we let[Ryj ,yj + 2Ryj ,zj + Rzj ,zj

](r, η∗r , ζ

∗r ) de-

note Ryj ,yj (r, η∗r , ζ∗r ) + 2Ryj ,zj (r, η∗r , ζ

∗r ) + Rzj ,zj (r, η∗r , ζ

∗r ), and use other similar

notation where helpful throughout.Now, by Assumption (A.2) and the definition of R, T , there exists CS <∞

such that

|Ry(r, y, z)|, |Rz(r, y, z)| ≤ CS(1 + |y|+ |z|) ∀ (s, y, z) ∈ D2.

Consequently, by Lemma 2 (noting that this implies (η∗, ζ∗) ∈ X vs ), there existsM1 = M1(t− s, |x|) <∞ such that

E∫ t

s

|Ry(r, η∗, ζ∗)|2 dr, E∫ t

s

|Rz(r, η∗, ζ∗)|2 dr < M1. (53)

By (53) and standard results (cf. [5], Section V.3),

E∫ t

s

[Ry(r, η∗r , ζ

∗r ) + Rz(r, η

∗r , ζ∗r )]TdBr

= 0. (54)

Combining (52) and (54) yields

R(s, y, z) = E∫ t

s

−Rt(r, η∗r , ζ∗r )− RTy (r, η∗r , ζ∗r )v∗r − RTz (r, η∗r , ζ

∗r )w∗r

11

− ~4m

n∑j=1

[Ryj ,yj + 2Ryj ,zj + Rzj ,zj

](r, η∗r , ζ

∗r ) dr + R(t, η∗t , ζ

∗t )

. (55)

Then, by (42) and (55),

R(s, y, z) = E∫ t

s

−Rt(r, η∗r , ζ∗r )− RTy (r, η∗r , ζ∗r )v∗r − RTz (r, η∗r , ζ

∗r )w∗r

− ~4m

n∑j=1

[Ryj ,yj + 2Ryj ,zj + Rzj ,zj

](r, η∗r , ζ

∗r ) dr + Re(φ(ξ∗t ))

.

(56)

Similarly,

T (s, y, z) = E∫ t

s

−Tt(r, η∗r , ζ∗r )− TTy (r, η∗r , ζ∗r )v∗r − TTz (r, η∗r , ζ

∗r )w∗r

− ~4m

n∑j=1

[Tyj ,yj + 2Tyj ,zj + Tzj ,zj

](r, η∗r , ζ

∗r ) dr + Im(φ(ξ∗t ))

.

(57)

Now, applying V−100 to (56),(57), one has

S(s, x) = R(s, y, z) + iT (s, y, z)

= E∫ t

s

−St(r, ξ∗r )− [Ry(r, η∗r , ζ∗r ) + iTy(r, η∗r , ζ

∗r )]T v∗r

− [Tz(r, η∗r , ζ∗r )− iRz(r, η∗r , ζ∗r )]T iw∗r

− ~4m

n∑j=1

[Ryj ,yj+ 2Ryj ,zj+ Rzj ,zj+ i(Tyj ,yj+ 2Tyj ,zj+ Tzj ,zj )

](r, η∗r , ζ

∗r ) dr

+ S(t, ξ∗t )

,

and using (27), this is

= E∫ t

s

−St(r, ξ∗r )− STx (r, ξ∗)u∗r

− ~4m

n∑j=1

[Ryj ,yj+ 2Ryj ,zj+ Rzj ,zj+ i(Tyj ,yj+ 2Tyj ,zj+ Tzj ,zj )

](r, η∗r , ζ

∗r ) dr

+ S(t, ξ∗t )

. (58)

Also, by (29), (30),

[Ryj ,yj + 2Ryj ,zj + Rzj ,zj ](r, η∗r , ζ∗r ) = −2Im(Sxj ,xj

),

i[Tyj ,yj + 2Tyj ,zj + Tzj ,zj ](r, η∗r , ζ∗r ) = 2iRe(Sxj ,xj

),

12

Applying these identities in (58) yields

S(s, x) = E∫ t

s

−St(r, ξ∗r )− STx (r, ξ∗)u∗r − i~2m∆S(r, ξ∗) dr + S(t, ξ∗t )

.

Lemma 8 Let ~ ∈ (0, 1], and suppose that S ∈ SC satisfies (14)–(15). Then,S(s, x) = J~(s, x, u∗,(s,x)) for all (s, x) ∈ DC, where u∗,(s,x) is given by (18)-(19)with S in place of S.

Proof: Fix ~ ∈ (0, 1] and (s, x) ∈ DC. Let ξ∗ = ξ∗,(s,x) and u∗ = u∗,(s,x) begiven by (18)-(19) with S replacing S. From the second assertion of Lemma 7,we have

S(s, x) = E∫ t

s

−St(r, ξ∗r )− STx (r, ξ∗)u∗r − i~2m∆S(r, ξ∗) dr + φ(ξ∗t )

,

which by (19),

= E∫ t

s

−St(r, ξ∗r ) + 1m |Sx(r, ξ∗r )|2c − i~

2m∆S(r, ξ∗) dr + φ(ξ∗t )

= E

∫ t

s

−St(r, ξ∗r ) + 12m |Sx(r, ξ∗r )|2c + V (ξ∗r )− i~

2m∆S(r, ξ∗) dr

+

∫ t

s

m2 |(−1m )Sx(r, ξ∗r )|2c − V (ξ∗r ) dr + φ(ξ∗t )

,

which by (5) and (14),

= E∫ t

s

m2 |(−1m )Sx(r, ξ∗r )|2c − V (ξ∗r ) dr + φ(ξ∗t )

,

which by the choice of u∗ = u∗,(s,x) and (50),

= J~(s, x, u∗,(s,x)).

Proof: [proof of Theorem 6.] Fix (s, x) ∈ DC. Let L(x, v).= m

2 |v|2c−V (x) for

all x, v ∈ Cn. For compactness of nontation, let ξ∗ = ξ∗,(s,x) and u∗ = u∗,(s,x).By Lemma 8,

S(s, x) = E∫ t

s

L(ξ∗r , u∗r) dr + φ(ξ∗t )

= J~(s, x, u∗). (59)

It remains to be shown the u∗ is the argstat over Us of J~(s, x, ·).Let u ∈ Us and δ

.= u − u∗ ∈ Us. Let ξ ∈ Xs be the trajectory generated

by u, i.e., the solution of (17), and let ∆.= ξ − ξ∗ ∈ Xs, where we note that

∆r =∫ rsδρ dρ for all (r, ω) ∈ [s, t]× Ω. By (59),

J~(s, x, u∗) = S(s, x) = ES(t, ξt)+[S(s, x)− ES(t, ξt)

],

which by Lemma 7 and (15),

13

= ES(t, ξt)+ E∫ t

s

−St(r, ξr)− STx (r, ξr)ur − i~2m∆S(r, ξr) dr

= Eφ(ξt)+ E

∫ t

s

L(ξr, ur) dr

+ E

∫ t

s

−L(ξr, ur) dr − St(r, ξr)− STx (r, ξr)ur − i~2m∆S(r, ξr) dr

. (60)

Now, by (5), (14),

0 = St(r, x) + statv∈Cn

STx (r, ξ∗)v + m

2 |v|2c − V (x)

+ i~

2m∆S(r, x).

Taking x = ξ∗r in this, and using (27), (34) and the definition of u∗, we have

0 = St(r, ξ∗r ) + STx (r, ξ∗)u∗r + m

2 |u∗r |2c − V (ξ∗r ) + i~

2m∆S(r, ξ∗r ) (61)

for all (r, ω) ∈ (s, t)× Ω. Combining (60) and (61), one has

J~(s, x, u∗,(s,x)) = E∫ t

s

L(ξr, ur) dr + φ(ξt)

+ E

∫ t

s

L(ξ∗r , u∗r)− L(ξr, ur) + St(r, ξ

∗r )− St(r, ξr)

+ STx (r, ξ∗r )u∗r − STx (r, ξr)ur + i~2m [∆S(r, ξ∗r )−∆S(r, ξr)] dr

= J~(s, x, u) + E

∫ t

s

L(ξ∗r , u∗r)− L(ξr, ur) + St(r, ξ

∗r )− St(r, ξr)

+ STx (r, ξ∗r )u∗r − STx (r, ξr)ur + i~2m [∆S(r, ξ∗r )−∆S(r, ξr)] dr

.

This implies∣∣J~(s, x, u∗,(s,x))− J~(s, x, u)∣∣

≤ E∫ t

s

∣∣L(ξ∗r , u∗r)− L(ξr, ur) + St(r, ξ

∗r )− St(r, ξr)

+ STx (r, ξ∗r )u∗r − STx (r, ξr)ur + i~2m [∆S(r, ξ∗r )−∆S(r, ξr)]

∣∣ dr.= E

∫ t

s

∣∣Ξr(ω)∣∣ dr. (62)

Note that by Taylor’s Theorem, for x, x, v, v ∈ Cn and r ∈ (s, t), and usingthe assumed bounds on derivatives and the definition of L,∣∣∣L(x, v)− L(x, v) + St(r, x)− St(r, x) + Sx(r, x)v − Sx(r, x)v

+ i~2m [∆S(r, x)−∆S(r, x)]

∣∣∣14

≤∣∣∣− Vx(x)(x− x) +mv(v − v) + Stx(r, x)(x− x)

+ Sxx(r, x)v(x− x) + Sx(r, x)(v − v)

+[Sx(r, x)v − Sx(r, x)v − Sxx(r, x)v(x− x)− Sx(r, x)(v − v)

](63)

+ i~2m (∆S)x(r, x)(x− x)

∣∣∣+K1

(1 + |x|2q + |x|2q

)|x− x|2 +m|v − v|2,

for appropriate K1 = K1(C0, CS , ~,m) <∞. This implies∣∣Ξr∣∣ ≤∣∣∣− Vx(ξ∗r )∆r +mu∗rδr + Stx(r, ξ∗r )∆r

+ Sxx(r, ξ∗r )u∗r∆r + Sx(r, ξ∗r )δr +[Sx(r, ξr)ur

− Sx(r, ξ∗r )u∗r − Sxx(r, ξ∗r )u∗r∆r − Sx(r, ξ∗r )δr]

+ i~2m (∆S)x(r, ξ∗r )∆r

∣∣∣+K1

(1 + |ξr|2q + |ξ∗r |2q

)|∆r|2 +m|δr|2 ∀ (r, ω) ∈ (s, t)× Ω. (64)

Now recall that

u∗ = −1m Sx(r, ξ∗r ) = argstat

v∈Cn

[m2 |v|

2c + Sx(r, ξ∗r )v

],

and consequently, by Lemma 1,

mu∗r + Sx(r, ξ∗) = 0. (65)

Substituting (65) into (64) yields∣∣Ξr∣∣ ≤∣∣∣− Vx(ξ∗r )∆r + Stx(r, ξ∗r )∆r + Sxx(r, ξ∗r )u∗r∆r +[Sx(r, ξr)ur

− Sx(r, ξ∗r )u∗r − Sxx(r, ξ∗r )u∗r∆r − Sx(r, ξ∗r )δr]

+ i~2m (∆S)x(r, ξ∗r )∆r

∣∣∣+K1

(1 + |ξr|2q + |ξ∗r |2q

)|∆r|2 +m|δr|2 ∀ (r, ω) ∈ (s, t)× Ω. (66)

Also, more generally, recalling definitions (19),

ˆu(r, x) = −1m Sx(r, x) = argstat

v∈Cn

[L(x, v) + Sx(r, x)v

]∀ (r, x) ∈ DC. (67)

Note that

− Vx(ξ∗r ) + Stx(r, ξ∗) + Sxx(r, ξ∗r )u∗r + i~2m (∆S)x(r, ξ∗)

=∂

∂x

[− V (x) + St(r, x) + Sx(r, x)v + i~

2m Sxx(r, x)]∣∣∣∣x=ξ∗x, v=ˆu(r,ξ∗r )

where the partial derivative notation indicates that the derivative is taken only

over explicitly appearing arguments, and this is

=d

dx

[L(x, ˆu(r, x)) + St(r, x) + Sx(r, x)ˆu(r, x) + i~

2m Sxx(r, x)]∣∣∣∣x=ξ∗r

− ∂

∂v

[L(x, v) + Sx(r, x)v

]∣∣∣∣x=ξ∗r , v=ˆu(r,ξ∗r )

d

dxˆu(r, x)

∣∣∣∣x=ξ∗r

,

15

which by (67),

=d

dx

[L(x, ˆu(r, x)) + St(r, x) + Sx(r, x)ˆu(r, x) + i~

2m Sxx(r, x)]∣∣∣∣x=ξ∗r

,

which by (14) and (67),

=d

dx

[0]

= 0. (68)

Substituting (68) into (66), we have∣∣Ξr∣∣ ≤ ∣∣∣Sx(r, ξr)ur − Sx(r, ξ∗r )u∗r − Sxx(r, ξ∗r )u∗r∆r − Sx(r, ξ∗r )δr

∣∣∣+K1

(1 + |ξr|2q + |ξ∗r |2q

)|∆r|2 +m|δr|2 ∀ (r, ω) ∈ (s, t)× Ω,

which implies

E∫ t

s

∣∣Ξr∣∣ dr ≤ m‖δ‖2Us +K1E∫ t

s

(1 + |ξr|2q + |ξ∗r |2q

)|∆r|2 dr

+ E∫ t

s

∣∣∣Sx(r, ξr)ur − Sx(r, ξ∗r )u∗r − Sxx(r, ξ∗r )u∗r∆r − Sx(r, ξ∗r )δr

∣∣∣ dr. (69)

Now,

E∫ t

s

∣∣∣Sx(r, ξr)ur − Sx(r, ξ∗r )u∗r − Sxx(r, ξ∗r )u∗r∆r − Sx(r, ξ∗r )δr

∣∣∣ dr(70)

= E∫ t

s

∣∣∣[Sx(r, ξr)− Sx(r, ξ∗r )]δr+

[Sx(r, ξr)− Sx(r, ξ∗r )− Sxx(r, ξ∗r )∆r

]u∗r

∣∣∣ dr.By Taylor’s Theorem and the assumptions,∣∣∣[Sx(r, x)− Sx(r, x)

](v − v)

∣∣∣ ≤ Cs(1 + |x|2q + |x|2q)|x− x| |v − v|

and ∣∣∣Sx(r, x)− Sx(r, x)− Sxx(r, x)(x− x)∣∣∣ ≤ CS

2

(1 + |x|2q + |x|2q

)|x− x|2

for all x, x, v, v ∈ Cn and r ∈ (s, t). Applying these inequalities in (70), we have

E∫ t

s

∣∣∣Sx(r, ξr)ur − Sx(r, ξ∗r )u∗r − Sxx(r, ξ∗r )u∗r∆r − Sx(r, ξ∗r )δr

∣∣∣ dr≤ E

∫ t

s

Cs(1 + |ξr|2q + |ξ∗r |2q

)|∆r| |δr| dr

+ E∫ t

s

Cs

2

(1 + |ξr|2q + |ξ∗r |2q

)|∆r|2 |u∗r | dr

≤ CS

2 E∫ t

s

(1 + |ξr|2q + |ξ∗r |2q

)|δr|2 dr + CS

2 E∫ t

s

(1 + |ξr|2q + |ξ∗r |2q

)|∆r|2 dr

16

+ CS

2 E∫ t

s

(1 + |ξr|2q + |ξ∗r |2q

)|u∗r | |∆r|2 dr. (71)

Now, using Holder’s inequality,

E∫ t

s

(1 + |ξr|2q + |ξ∗r |2q

)|δr|2 dr

≤

1 +[2E∫ t

s

|ξ∗r |4q + |ξ∗r + ∆r|4q dr]1/2

‖δ‖2Us

≤ C1

1 + ‖ξ∗‖2qXs

+[E∫ t

s

|∆r|4q dr]1/2

‖δ‖2Us , (72)

for appropriate C1 = C1(q) <∞. Substituting (72) and (71) into (69) yields

E∫ t

s

∣∣Ξr∣∣ dr ≤ m+ C1CS

2

[1 + ‖ξ∗‖2qXs

+[E∫ t

s

|∆r|4q dr]1/2]

‖δ‖2Us

+ (K1 + CS

2 )E∫ t

s

(1 + |ξr|2q + |ξ∗r |2q

)|∆r|2 dr

+ CS

2 E∫ t

s

(1 + |ξr|2q + |ξ∗r |2q

)|u∗r | |∆r|2 dr,

which by the definition of u∗ and Assumption (A.2),

≤m+ C1CS

2

[1 + ‖ξ∗‖2qXs

+[E∫ t

s

|∆r|4q dr]1/2]

‖δ‖2Us

+ C2E∫ t

s

(1 + |ξ∗r |4q + |ξ∗r + ∆r|4q

)|∆r|2 dr,

for appropriate C2 = C2(C0, Cs, CS , q) <∞, and this is

≤m+ C1CS

2

[1 + ‖ξ∗‖2qXs

+[E∫ t

s

|∆r|4q dr]1/2]

‖δ‖2Us

+ C3E∫ t

s

(|∆r|2 + |ξ∗r |4q|∆r|2 + |∆r|4q+2

)dr, (73)

for appropriate C3 = C3(C0, CS , CS , q) <∞.Next one should note some simple estimates. First, using Holder’s inequality,

E∫ t

s

|∆r|2 dr = E∫ t

s

∣∣∣ ∫ r

s

δρ dρ∣∣∣2 dr ≤ (t− s)E

∫ t

s

|δρ|2 dρ.= (t− s)‖δ‖22

≤ (t− s)‖δ‖2Us . (74)

Similarly, with Holder’s inequality and some simple calculations,

E∫ t

s

|∆r|4q+2 dr = E∫ t

s

∣∣∣ ∫ r

s

δρ dρ∣∣∣4q+2

dr ≤ (t− s)4q+2E∫ t

s

|δρ|4q+2 dρ

.= (t− s)4q+2‖δ‖4q+2

4q+2 ≤ (t− s)4q+2‖δ‖4q+2Us , (75)

17

and

E∫ t

s

|∆r|4q dr ≤ (t− s)4q‖δ‖4qUs . (76)

Next,

E∫ t

s

|ξ∗r |4q|∆r|2 dr ≤[E∫ t

s

|ξ∗r |8q]1/2[

E∫ t

s

∣∣∣ ∫ t

s

δρ dρ∣∣∣4 dr]1/2

≤ (t− s) ‖ξ∗‖4qXs

[E∣∣∣ ∫ t

s

δρ dρ∣∣∣4]1/2≤ (t− s)5/2 ‖ξ∗‖4qXs

[E∫ t

s

|δρ|4 dρ]1/2

≤ (t− s)5/3 ‖ξ∗‖4qXs‖δ‖2Us . (77)

Substituting (74)–(77) into (73), one finds

E∫ t

s

∣∣Ξr∣∣ dr ≤ m+ C1CS

2

[1 + ‖ξ∗‖2qXs

+ t2q‖δ‖2qUs]‖δ‖2Us

+ C3

[‖δ‖2Us + t5/3‖ξ∗‖4qXs

‖δ‖2Us + t4q+2‖δ‖4q+2Us

]≤ C4

[1 + t4q+2 +

(1 + t5/3

)‖ξ∗‖4qXs

][‖δ‖2Us + ‖δ‖4q+2

Us

], (78)

for appropriate choice of C4 = C4(C0, CS , CS , q) < ∞. Substituting (78) into(62) yields ∣∣J~(s, x, u∗,(s,x))− J~(s, x, u)

∣∣ ≤ C‖δ‖2Us ,for ‖δ‖Us ≤ 1 and appropriate choice of C = C(t, x, C0, Cs, Cs, q) < ∞. Bydefinition, this implies that u∗ = argstatu∈Us J

~(s, x, u), where uniqueness ofthe argstat is guaranteed by Assumption (A.3).

It may be worth noting the following, which reflects the uniqueness impliedby the above representation.

Corollary 9 In addition to (A.0)–(A.3), assume the conditions of Theorem 6.There exists a unique solution S ∈ SC to (14),(15), where S = S~. There alsoexists a solution, S ∈ S, to (7),(8), given by S(r, y) = S~(r,V−1

0 ((yT , 0)T )) forall (r, y) ∈ D. Lastly, any other solution in S to (7),(8) cannot be extendedholomorphically to a solution of (14),(15) in SC.

Proof: The existence of S follows from Assumption (A.0), and the uniquenessfollows from Theorem 6. Let S be as given in the corollary statement. Notethat

St(r, y) = St(r, y + i0) ∀ (r, y) ∈ D. (79)

Also, by (27), for (r, y) ∈ D, and the fact that S agrees with S on D,

|Sx(r, y + i0)|2c =

n∑j=1

[R2yj − T

2yj + 2iRyj Tyj

]=

n∑j=1

[R2yj − T

2yj + 2iRyj Tyj

]

18

=

n∑j=1

(Syj (r, y))2 = |Sy(r, y)|2c , (80)

where we take (R(r,V0(x)), T (r,V0(x)))T = V00(S(r, x)) for all (r, x) ∈ DC and(R(r, y), T (r, y))T = V00(S(r, y)) for all (r, y) ∈ D. Similarly, using (29), (30),

∆S(r, y + i0) =

n∑j=1

Sxj ,xj(r, y + i0)

=

n∑j=1

[Ryj ,yj (r, y + i0) + iTyj ,yj (r, y + i0)

]=

n∑j=1

[Ryj ,yj (r, y + i0) + iTyj ,yj (r, y + i0)

]= ∆S(r, y) (81)

for all (r, y) ∈ D. Then, by (14) and (79)–(81),

St(r, y) + i~2m∆S(r, y) +H(y, Sy(r, y))

= St(r, y + i0) + i~2m∆S(r, y + i0) +H(y + i0, Sx(r, y + i0)) = 0 (82)

for all (r, y) ∈ D. That S also satisfies the terminal condition is obvious, andwe see that S is a solution of (7),(8).

Regarding the last assertion, recall that if two holomorphic functions on Cnagree on x = y + iz ∈ Cn | z = 0, then they agree on all of Cn, where thisfollows in the standard way from power series representations (cf. [9]). Notingthe uniqueness of S = S~ yields the assertion.

Remark 10 The results concerning S in Corollary 9 also extend to (1),(2) and(3),(4) in the obvious ways.

5 Existence

The results of Sections 3–4 were conditioned on an assumption of existence ofa solution to (14),(15). However, for this class of systems, one can use simplecomplex-analysis equivalences to reduce the question of existence of a solutionof the complex HJ PDE problem to that of existence of a solution of a realHJ PDE problem. The latter belongs to a class of problems for which thereis a wide variety of existence results under varying assumptions, which are notrepeated here. In this section, we drop the earlier assumptions.

Theorem 11 Suppose V, φ are holomorphic on Cn. Suppose also that R ∈C1,3(D2) ∩ C(D2) satisfies (40),(42). Then, there exists T ∈ C1,3(D2) ∩ C(D2)satisfying (41),(43) such that for each s ∈ [0, t], T (s, ·, ·) is a harmonic conju-gate of R(s, ·, ·). Further, letting S(s, x) = V−1

00 (R(s,V0(x)), T (s,V0(x))) for all(s, x) ∈ DC, S satisfies (14),(15).

19

Proof: Let T (t, ·, ·) = φI(·, ·). Then, by (42), T (t, ·, ·) is a complex conjugateof R(t, ·, ·), and satisfies (43). Now we begin the construction of T for s ∈ (0, t).Recalling that there is a free constant in the harmonic conjugate, for each s ∈(0, t), we let

T (s, 0, 0).= T (t, 0, 0)−

∫ t

s

− 1m

n∑j=1

Ryj (ρ, 0, 0)Rzj (ρ, 0, 0)

+ ~2m

n∑j=1

Ryj ,yj (ρ, 0, 0) + V I(0, 0) dρ,

.= T (t, 0, 0)−

∫ t

s

[− 1

m

n∑j=1

RyjRzj + ~2m

n∑j=1

Ryj ,yj + V I](ρ, 0, 0) dρ

(83)

which implies that for all s ∈ (0, t),

Tt(s, 0, 0) =[− 1

m

n∑j=1

RyjRzj + ~2m

n∑j=1

Ryj ,yj + V I](s, 0, 0), (84)

which by (27)–(30),

=[

1m

n∑j=1

TyjTzj + ~2m

n∑j=1

Tyj ,zj + V I](s, 0, 0), (85)

which implies that (41) is satisfied at (s, 0, 0) for all s ∈ (0, t).By (27) and (44), for all (s, y, z) ∈ D2 and all k ∈]1, n[,

Tt,zk = Rt,yk = ~2m

n∑j=1

Ryj ,zj ,yk + 1m

n∑j=1

[RyjRyj ,yk −RzjRzj ,yk ] + V Ryk , (86)

which by (27)–(32),

= ~2m

n∑j=1

Tyj ,zj ,zk + 1m

n∑j=1

[TzjTyj ,zk + TyjTzj ,zk ] + V Izk

=d

dzk

~

2m

n∑j=1

Tyj ,zj + 1m

n∑j=1

TyjTzj + V I. (87)

For z1 ∈ R, let z1(z1).= (z1, 0, 0 . . . 0)T ∈ Rn, and note that for s ∈ (0, t)

Tt(s, 0, z1(z1)) = Tt(s, 0, 0) +

∫ z1

0

Tt,z1(s, 0, z1(ζ)) dζ, (88)

which by (87),

= Tt(s, 0, 0) +

∫ z1

0

d

dz1

[~

2m

n∑j=1

Tyj ,zj + 1m

n∑j=1

TyjTzj + V I](s, 0, z1(ζ))

dζ

20

= Tt(s, 0, 0) +[

~2m

n∑j=1

Tyj ,zj + 1m

n∑j=1

TyjTzj + V I](s, 0, z1(z1))

−[

~2m

n∑j=1

Tyj ,zj + 1m

n∑j=1

TyjTzj + V I](s, 0, 0),

which by (85),

=[

~2m

n∑j=1

Tyj ,zj + 1m

n∑j=1

TyjTzj + V I](s, 0, z1(z1)),

which implies that (41) is satisfied at (s, 0, z1(z1)) for all s ∈ (0, t) and z1 ∈ R.Proceeding from here similarly, first for z2 and then z3 and so on, yields finally

Tt(s, 0, z) =[

~2m

n∑j=1

Tyj ,zj + 1m

n∑j=1

TyjTzj + V I](s, 0, z) ∀ s ∈ (0, t), z ∈ Rn.

We now proceed to integrate along the y-components. First, for k ∈]1, n[,differentiating (40) with respect to zk, we have

Rt,zk = ~2m

n∑j=1

Ryj ,zj ,zk + 1m

n∑j=1

[RyjRyj ,zk −RzjRzj ,zk ] + V Rzk , (89)

which by (27)–(32),

= − ~2m

n∑j=1

Tyj ,zj ,yk − 1m

n∑j=1

[TzjTyj ,yk + TyjTzj ,yk ]− V Iyk

= − d

dyk

~

2m

n∑j=1

Tyj ,zj + 1m

n∑j=1

TyjTzj + V I. (90)

By (27) and (90), we have

Tt,yk = −Rt,zk =d

dyk

~

2m

n∑j=1

Tyj ,zj + 1m

n∑j=1

TyjTzj + V I. (91)

For y1 ∈ R, let y1(y1).= (y1, 0, 0 . . . 0)T ∈ Rn, and note that for s ∈ (0, t) and

z ∈ Rn,

Tt(s, y1(y1), z) = Tt(s, 0, z) +

∫ y1

0

Tt,y1(s, y1(η), z) dη,

which by (91),

= Tt(s, 0, z) +

∫ y1

0

d

dy1

[~

2m

n∑j=1

Tyj ,zj + 1m

n∑j=1

TyjTzj + V I](s, y1(η), z)

dη

= Tt(s, 0, z) +[

~2m

n∑j=1

Tyj ,zj + 1m

n∑j=1

TyjTzj + V I](s, y1(y1), z)

21

−[

~2m

n∑j=1

Tyj ,zj + 1m

n∑j=1

TyjTzj + V I](s, 0, z),

which by (41),

=[

~2m

n∑j=1

Tyj ,zj + 1m

n∑j=1

TyjTzj + V I](s, y1(y1), z),

which implies that (41) is satisfied at (s, y1(y1), z) for all s ∈ (0, t), y1 ∈ R andz ∈ Rn. Proceeding from this along each component of y, one finally obtains

Tt(s, y, z) =[

~2m

n∑j=1

Tyj ,zj + 1m

n∑j=1

TyjTzj + V I](s, y, z) ∀ (s, y, z) ∈ D2, (92)

which is (41). By construction, T has the indicated smoothness. Lastly, byProposition 5, one obtains the assertions concerning S.

Remark 12 As S obtained in Theorem 11 is holomorphic in x for each s ∈ (0, t](and hence C∞ in the space variable), noting that R, T are related to S by (26),one immediately sees that R, T are C1,∞(D2) ∩ C(D2).

It should be remarked that the analogous result to Theorem 11, where onesupposes existence of the solution to (41),(43) rather than (40),(42) is obtainedin an analogous manner, and the redundant proof is omitted. The result is:

Theorem 13 Suppose V, φ are holomorphic on Cn. Suppose also that T ∈C1,3(D2) ∩C(D2) satisfies (41),(43). Then, there exists R ∈ C1,3(D2) ∩C(D2)satisfying (40),(42) such that for each s ∈ [0, t], T (s, ·, ·) is a harmonic conju-gate of R(s, ·, ·). Further, letting S(s, x) = V−1

00 (R(s,V0(x)), T (s,V0(x))) for all(s, x) ∈ DC, S satisfies (14),(15).

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