Portfolio HJB Equation

Portfolio Optimization and Stochastic Control.

Closed-Form Solutions to the HJB Equation through Lie Symmetries.

Solomon M. Antoniou

SKEMSYS

Scientific Knowledge Engineering

and Management Systems

37 Κoliatsou Street, Corinthos 20100, Greece [email protected]

Αφιερωμένο στη Γκέλλυ

Dedicated to Gelly

Abstract

We introduce a solution procedure for the Hamilton-Jacobi-Bellman (HJB)

equation associated to a portfolio optimisation problem. The HJB equation is

derived under the principle of Dynamic Programming. The solution method relies

on determining the underlying Lie point symmetries of the equation. On finding

the symmetries, we consider a linear combination of the infinitesimal generators,

and the two invariants that they determine. Using the transformations based on the

two invariants, we convert the HJB equation to an ordinary non-linear differential

equation, which can in turn be converted to an Abel differential equation. Abel’s

differential equation admits a closed form solution. What follow are twelve

families of solutions.

Keywords: Stochastic Control, Hamilton-Jacobi-Bellman equation, Portfolio

Optimization, Nonlinear PDEs, Lie Symmetries, Closed-Form Solutions.

2

1. Introduction

Stochastic Control (Bertsekas [6], [7], Evans [19], Fleming and Soner [20],

Ingersoll [28], Øksendal [42], Pham [46], Soner [49], Touzi [52]) is an

optimisation procedure used in many areas of financial applications, such as

Portfolio Optimization (Benth and Karlsen [5], Korn and Korn [31], Korn

[32], Kraft [33])

Option Pricing (Baron and Jensen [3])

Consumption-Investment Strategies (Davis, Panas and Zariphopoulou [15],

Elliott and Kopp [18], Karatzas, Lehoczky, Sethi and Shreve [29],

Karatzas, Lehoczky and Shreve [30] and Zariphopoulou [53])

Stochastic Interest Rates (Lewis [36])

Bond Pricing (Kraft [33]) and Bond Portfolio Optimization (Puhle [47])

Asset Allocation (Munk [41])

Risk (Bielecki and Pliska [8], Dmitrasinovic-Vivodic, Lari-Lavassani, Li

and Ware [16])

DSGE models (Papanicolaou [45])

Pairs Trading (Mudchanatongsuk, Primbs and Wong [40])

Pension Funds (Boulier, Trussant and Florens [13], Gerrard, Haberman and

Vigna [21])

Insurance (Hipp, [24])

Debt Crises (Stein, [50])

Various applications (Bucci et al. [14])

The basic equation used in the optimization process is the Hamilton-Jacobi-

Bellman (HJB) equation, derived under the principle of Dynamic Programming

(Bellman [4]). Despite successful predictions of the HJB equation for such

models, there is no available general solution procedure. The solutions obtained so

far are based on separation of variables. It is the purpose of this paper to introduce

3

a solution procedure (and to find some general solutions) based on Lie

Symmetries.

The Lie symmetry approach in solving differential equations was introduced by S.

Lie (Lie [37] and Lie [38]) and is best described in references Anderson and

Ibragimov [1], Bluman and Kumei [9], Dresner [17], Hereman [23], Ibragimov

[25], Ibragimov [26], Olver [43], Ovsiannikov [44] and Stephani [51]. This

technique was used for the first time in Finance by Ibragimov and Gazizov [27].

The same technique was also used by the author in solving the Bensoussan-

Crouhy-Galai equation, which models the stochastic equity volatility (Antoniou

[2]). There is however a growing list of papers in applying Lie symmetry methods

to Finance. For a (non-complete) set of references, see Bordag [10], Bordag and

Chmakova [11], Bordag and Frey [12], Goard [22], Leach, O’Hara and Sinkala

[35] and Silberberg [48].

The paper is organized as follows: In section 2 we consider a simple model, which

leads to the Hamilton-Jacobi-Bellman equation. In section 3 the Lie symmetry

analysis to this equation is performed. In other words we find the infinitesimal

generators of the Lie group. In section 4 we find general solutions to the equation,

by first considering a linear combination of the infinitesimal generators and then

by determining the invariants of the equation. The invariants allow us to convert

the equation into a nonlinear, second order ordinary differential equation. Using

another transformation, we convert this nonlinear equation to an Abel equation

possessing a closed-form solution. In section 5 we present our conclusions and the

twelve families of our solution.

2. A Portfolio Selection Model and the HJB Equation

In this section we consider a selection portfolio model, optimized under the

principle of Dynamic Programming, leading to an HJB equation.

The model we consider is a Brownian model, in which the money market account

and stocks are modeled by

4

dt)t(r)t(S)t(Sd 00 , 1)0(S0 (2.1)

)]t(dWσdt)t(μ[)t(S)t(Sd , 0)0(S (2.2)

The coefficients r, μ and σ are progressively measurable processes being

uniformly bounded. Let also

)t(φ : be the number of the stocks invested at time t.

)t(S)t(φ 00 : be the money invested in the money market account at time t

)t(S)t(φ)t(S)t(φ)t(X 00φ : be the wealth of the investor

)t(X

)t(S)t(φ)t(π : be the proportion invested in stock

)t(π1 : be the proportion invested in the money market account.

The following proposition holds true (see for example Kraft [33], [34] or Øksendal

[42]).

Proposition. The wealth dynamics of a self-financing trading strategy is described

by the so-called wealth equation

)]t(dWσπdt))rμ(πr[()t(X)t(dX (2.3)

0x)0(X ( 0x0 )

We come now to the portfolio optimization one investor has with initial wealth

0x)0(X 0 and would like to have optimal final wealth )T(X* under an

optimal final strategy *φ . For a given initial wealth 0x0 , the investor

maximizes the following utility functional

))]Τ(X(U[Esup π

Aπ

(2.4)

where

π:π{A admissible and ]))Τ(X(U[E π}

and )t(Xπ is given by (2.3):

)t(dWσ)t(πdt)])t(r)t(μ()t(π)t(r[)t(X)t(dX ππ

5

0π x)0(X

The function )x(U , defined for positive real numbers, is called the utility

function, which is strictly concave, strictly increasing and continuously

differentiable, and satisfies the conditions

)x(Ulim)0(U0x

, 0)x(Ulim)(Ux

Let us define the value function

))]Τ(X(U[Esup)x,t(G πx,t

Aπ

where

]x)t(X|))Τ(X(U[E))]Τ(X(U[E ππx,t

)x(U)x,t(G

Suppose now that there is an optimal strategy *π . The corresponding HJB

equation, which maximizes the final wealth, derived under the principle of

Dynamic Programming, is given by

0)x,t(Gx2

1x])r(r[)x,t(Gsup)x,t(G xx

222xt

(2.5)

for any admissible strategy π and terminal condition )x(U)x,T(G .

We consider the case where r, μ and σ are all constants.

Pointwise maximization over π yields the first order condition

0)t(π)x,t(Gxσx)rμ()x,t(G xx22

x (2.6)

which implies that the optimal strategy is given by

)x,t(Gx

)x,t(G

σ

rμ*π

xx

x2

(2.7)

Since the derivative of (2.6) with respect to π is )x,t(Gxσ xx22

, the candidate

*π is the global maximum if 0Gxx .

Substituting *π into the HJB equation (2.5), yields a PDE for )x,t(G :

6

0)x,t(G

)x,t(Gk

2

1)x,t(Gxr)x,t(G

xx

2x2

xt (2.8)

where σ

rμk

.

The traditional way of solving equation (2.8) is by separation of variables. In fact,

considering that (Merton [39], Øksendal [42])

γ1γ )t(fxγ

1)x,t(G (2.9)

equation (2.8) yields the first order differential equation

)t(fK)t(f (2.10)

where

2k

)1(2

1r

1K (2.11)

The above differential equation (2.10) has the solution )tT(Ke)t(f , and

satisfies the condition 1)T(f . Therefore solution (2.9) is compatible with the

condition γxγ

1)x(U)x,T(G . From (2.9) and (2.7) we determine the value

function and the optimal strategy respectively. For the type of solution we have

obtained, we can check that 0Gxx for 1γ .

In section 4 we provide our own method of solution, based on Lie Symmetry

Analysis.

3. Lie Symmetry Analysis of the HJB Equation

We now consider the HJB equation (2.8) written in the equivalent form

0uk2

1uuxruu 2

x2

xxxxxt (3.1)

where we have changed the notation of the function from )x,t(G to the more

traditional notation )t,x(u .

7

We are going to perform a Lie symmetry analysis of this equation. By this we

mean that we shall determine the infinitesimal generators of the Lie symmetry

group. These would then be used to determine the invariants, needed to convert

the partial differential equation into an ordinary differential equation.

We introduce the following concepts and notation (Olver [43]). Consider the base

space M, which is the Cartesian product UX of a 2-dimensional space X of

independent variables X)t,x( by a 1-dimensional space U of dependent

variables Uu . Let )1(U be the space of the first derivatives of the functions

)t,x(u with respect to x and t: )1(

tx U)u,u( . Let also )2(U be the space of the

second derivatives of the functions )t,x(u with respect to x and t:

)2(ttxtxx U)u,u,u( . Since the differential equation (3.1) is second order, we

introduce a second order jet bundle )2(M by considering the Cartesian product

)2()1()2( UUMM

The coordinates of the space )2(M are labeled by

)2(

ttxtxxtx M)u,u,u,u,u,u,t,x(z

In the space )2(M , equation (3.1) can be expressed as 0)u,t,x(Δ )2( where

2x

2xxxxxt

)2( uk2

1uuxruu)u,t,x(Δ (3.2)

Let ΔL be the solution manifold of (3.1):

)2()2(

Δ M}0Δ|Mz{L

A symmetry group ΔG of equation 0Δ is defined by

}LL:g|)M(Diffg{G ΔΔ)2(

Δ

We want to determine a subgroup of )M(Diff )2( , compatible with the structure of

ΔL . We shall first find the symmetry Lie algebra )Μ(Diff)Μ(Diff )2()2(Δ and

then use the main Lie theorem to determine ΔG . Let us denote by V

8

( )Μ(DiffV ) an element of a vector field on M (the generator of Lie

symmetries), defined by

u

)u,t,x(φt

)u,t,x(τx

)u,t,x(ξV

(3.3)

where )u,t,x(τ),u,t,x(ξ and )u,t,x(φ are smooth functions of their arguments.

The infinitesimal generators of ΔGg will have the structure of (3.3) and will

form an algebra )Μ(DiffΔ . The algebra )Μ(Diff )2(Δ will be spanned by the

vectors Vpr )2( , the second prolongation of V, defined by

tt

tt

xt

xt

xx

xx

t

t

x

x)2(

uφ

uφ

uφ

uφ

uφVVpr

The symmetries are determined by the equation (Olver [42], Theorem 2.31)

0)]u,t,x(Δ[Vpr )2()2( (3.4)

as long as

0)u,t,x(Δ )2( (3.5)

We implement next the equation 0)]u,t,x(Δ[Vpr )2()2( . We have that

0)]u,t,x(Δ[Vpr )2()2(

0]uk2

1uuxruu[Vpr 2

x2

xxxxxt)2(

or

xxt

x2

xxx

xxx uφ)ukuxr(φ)uur(ξ

0)uxru(φ xtxx (3.6)

The coefficients xφ , tφ and xxφ are calculated to be (Olver [43], Example

2.38)

txutx2xuxxux

x uuτuτuξu)ξφ(φφ

2tutxuttuxtt

t uτuuξu)τφ(uξφφ

9

2xxuuutxxxxxxuxx

xx u)ξ2φ(uτu)ξφ2(φφ

xxxut2xuu

3xuutxxu u)ξ2φ(uuτuξuuτ2

xtxuxxtuxxxuxtx uuτ2uuτuuξ3uτ2

The functions ξ,τ and φ have the following dependence

)t(ττ , )t,x(ξξ και )u,t,x(φφ

Equation (3.6) thus becomes

}u)ξφ(φ{)ukuxr()uur(ξ xxuxx2

xxxxx

txxtuxxxtt uu)τφ(u}uξφ{

2xuuxxxxuxx uφu)ξφ2(φ{

txxxu u}u)ξ2φ(

2xuuxxxxuxxx uφu)ξφ2(φ{uxr

0}u)ξ2φ( xxxu (3.7)

We now have to take into account the condition 0)u,t,x(Δ )2( . We therefore

substitute tu by xx

2x2

xu

uk

2

1uxr in the previous equation (3.7):

}u)ξφ(φ{)ukuxr()uur(ξ xxuxx2

xxxxx

xx

2x2

xxxtuxxxttu

uk

2

1uxru)(u)u(

2xuuxxxxuxx uφu)ξφ2(φ{

xx

2x2

xxxxuu

uk

2

1uxr}u)2(

2xuuxxxxuxxx uφu)ξφ2(φ{uxr

0}u)ξ2φ( xxxu (3.8)

10

We equate all the coefficients of the function u and their derivatives in the

previous equation (after multiplying by xxu ) to zero:

Coefficient of 4

x )u( :

0φk2

1uu

2 (3.9)

Coefficient of 3x )u( :

0)ξφ2(k2

1xxxu

2 (3.10)

Coefficient of 2x )u( :

0φk2

1xx

2 (3.11)

Coefficient of xxxuu :

0φk x2 (3.12)

Coefficient of xx2

x u)u( :

0)t(τk2

1 2 (3.13)

Coefficient of 2xx )u( :

0φφxr tx (3.14)

Coefficient of 2

xxx )u(u :

0))t((xrr xt (3.15)

Equations (3.9)-(3.15) are the determining equations of the Lie symmetries of

equation (3.1). We are now going to solve the system of equations (3.9)-(3.15).

From equation (3.9) we get that 0φuu , which means that φ is a linear function

with respect to u:

)t,x(βu)t,x(α)u,t,x(φ (3.16)

From equation (3.12) we get that 0φx . Therefore, using (3.16), we have

11

0βuα xx

from which we get 0αx and 0βx . Therefore the functions α and β

introduced in (3.16) are independent of x. On the other hand, since 0φx , we

get from (3.14) that 0φ t . Therefore 0βuα tt , from which we have that

0α t and 0β t , which means that functions α and β are constants. We then

obtain that

21 aua)u,t,x(φ (3.17)

Equation (3.11) then becomes an identity.

We now get from (3.13) that 0)t(τ . Therefore )t(τ is a constant:

3a)t(τ (3.18)

We get from (3.10), because of (3.17), that 0ξxx . Therefore )t,x(ξ is a linear

function with respect to x:

)t(gx)t(f)t,x(ξ (3.19)

where )t(f and )t(g are functions to be determined.

Using the previous expression, equation (3.15) becomes

0))t(g)t(gr(x)t(f

from which we get , since this equation should be true for any x, that

0)t(f and 0)t(g)t(gr

We thus have

4a)t(f (3.20)

and

tr5 ea)t(g (3.21)

Therefore the function )t,x(ξ takes the form

tr

54 eaxa)t,x(ξ (3.22)

Relations (3.22), (3.18) and (3.17) allow us to write down the generator of the

symmetries:

12

u

)aua(t

ax

)eaxa(V 213tr

54

(3.23)

Therefore we have the following

Theorem 1. The Lie algebra of the infinitesimal transformations of the HJB

equation (3.1) is spanned by the five vector fields

u

uX1

(3.24)

u

X2

(3.25)

t

X3

(3.26)

x

xX4

(3.27)

x

eX tr5

(3.28)

4. General Solutions to the HJB Equation

In order to find a solution to the HJB equation (3.1), we have to determine the

invariants of the equation. For this purpose we consider a linear combination of

the above found generators of the symmetries.

We consider the combination

21543 ΧρΧνΧμXλX

In other words we consider the generator

u

)ρuν(x

)eμxλ(t

tr

(4.1)

where the coefficients λ, μ, ν and ρ will be determined on the basis of finding

closed form solutions to the equation. The parameter appearing in (4.1) not to

be confused with appearing in the definition of k after equation (2.8).

Based on (4.1), we consider the system

13

ρuν

du

eμxλ

dx

1

dt

tr

(4.2)

In order to solve the first equation

treμxλ

dx

1

dt

(4.3)

of the above system, we have to consider two cases: rλ and rλ . Details are

given in the Appendix.

4.1. Case 1. Solution of the HJB equation in the rλ case.

If rλ , equation (4.3) has the general solution

t)λr(tλ1 e

λr

μxeC

(4.4)

Therefore one invariant of the HJB equation (3.1) is given by

t)λr(tλ eλr

μxey

(4.5)

The equation

ρuν

du

1

dt

has the general solution

)ρuνln(tνC2 .

Therefore another invariant is given by )y(υυ , where

)ρυe(ν

1u tν (4.6)

Using (4.5) and (4.6), we find that the partial derivatives of the function u

transform as

tν

yt)λr(

t eυeλr

μyλ

ν

1υu

(4.7)

yt)λr(

x υeν

1u (4.8)

14

yyt)λ2r(

xx υeν

1u (4.9)

Using (4.7)-(4.9), the original HJB equation (3.1) is transformed into

0)υ(k2

1υυy)λr(υυν 2

y2

yyyyy (4.10)

which is a nonlinear, second order ordinary differential equation.

Under the transformation

υ

υ)y(w

y (4.11)

equation (4.10) takes on the form

0wy)λr(wk2

1νw]wy)λr(ν[ 322

y

(4.12)

which is an Abel differential equation.

Introducing the notation

2k2

1νγ and λrδ ( 0 ) (4.13)

equation (4.12) can be written as

wyδν

wyδwγw

32

y

(4.14)

which, in turn, under the substitution

)y(Z

1)y(w (4.15)

takes on the form

y)y(Z

y)y(ZZy

(4.16)

Introducing the function )y(V by

)y(Vy)y(Z (4.17)

we transform equation (4.16) into the equation

15

δVν

δVγVyV y

which is equivalent - after separating variables - to the equation

y

1V

V)(V

Vy2

(4.18)

4.1.I. Let 0 , i.e. 2k2

1 ( 0 ). Equation (4.18) can be written as

y

1V

VV

Vy2

(4.19)

Denoting by 4D 2 the discriminant of the trinomial VV2 , we

have the following:

4.1.Ia For 0D , i.e. 4rr or 4rr , equation (4.19) admits a

solution given implicitly by

Ayln|V|ln)(

|V|ln)(

(4.20)

where and are the two real roots of the trinomial, given by

2

D and

2

D

respectively and A is an arbitrary constant.

4.1.Ib For 0D , i.e. 4rr , equation (4.19) admits a solution given

implicitly by

Aylnp

Varctan

p)p)Vln((

2

1 22

(4.21)

where and p are two parameters defined by

2 and

2

Dp


16

4.1.Ic For 0D , i.e. 4 , equation (4.19) admits a solution given implicitly

by

Ayln|2V|ln2V

12

(4.22)

where A is an arbitrary constant.

4.1.II. Let 0 , i.e. 2k2

1 . We shall distinguish two cases: and .

4.1.IIa If , equation (4.18) becomes

y

1V

V

Vy2

In this case we have to consider two cases: and . In the former case

we may set either 2 or 2 , 0 .

We thus have the following:

4.1.IIa1 For 2 , we have to solve the equation

y

1V

V

Vy22

2

which gives upon integration

AylnV

Vln

2

1)Vln(

2

1 22

(4.23)


4.1.IIa2. For 2 , we have to solve the equation

y

1V

V

Vy22

2

which gives upon integration

AylnV

arctan)Vln(2

1 22

(4.24)


4.1.IIa3. For , we obtain the equation

17

y

1V

1V

1Vy2

or

y

1V

1V

1y

and by integration,

Ayln|1V|ln (4.25)


4.1.IIb. If , denoting by 4)(D 2 the discriminant of the trinomial

V)(V2 , we have the following:

4.1.IIb1. For 04)(D 2 , equation (4.18) admits a solution given

implicitly by

Ayln|V|ln)(

|V|ln)(

(4.26)

where and are the two real roots of the trinomial, given by

2

D and

2

D



implicitly by

Aylnp

arctanp

)p)Vln((2

1 22

(4.27)

where and p are two parameters defined by

2 and

2

Dp



implicitly by

Ayln)Vln(V

1

(4.28)

18

where is the double root of the trinomial given by

2

and A is an arbitrary constant.

The function )y(υυ is then determined integrating (compare (4.11), (4.15) and

(4.17))

)y(Vy

1y

(4.29)

where )y(V is implicitly provided by any of the (4.20)-(4.28) equations.

Therefore we find that

y

y0)(V

dexpB)y( (4.30)

where B is a constant.

4.2. Case 2. Solution of the HJB equation in the r case.

In this case the system (4.2) is given by

u

du

exr

dx

1

dttr

(4.31)

The previous system has the general solutions:

tμexC tr1

(4.32)

and

)ρuνln(tνC2 (4.33)

Therefore in this case we have the two invariants

tμexy tr (4.34)

and

)ρυe(ν

1u tν (4.35)

19

where )y(υυ .

Using (4.34) and (4.35), we find that the partial derivatives of the function u

transform as

tν

yt eυ]μ)tμy(r[ν

1υu

(4.36)

yt)rν(

x υeν

1u (4.37)

yyt)r2ν(

xx υeν

1u (4.38)

Substituting (4.36)-(4.38) into the original equation (3.1), we find that the function

)y(υυ satisfies the equation

0)υ(k2

1υυμυυν 2

y2

yyyyy (4.39)

which is a nonlinear, second order ordinary differential equation.

Under the transformation

υ

υ)y(w

y (4.40)

equation (4.39) takes the form

32y www)w( (4.41)

which is an Abel differential equation. This equation may also be treated as an

equation with separable variables. However it will here be considered as an Abel

differential equation, since in that case we obtain simpler expressions as solutions.

In solving (4.38), we shall consider two cases: 0μ and 0μ .

4.2.I. If 0μ , we find that the general solution to the equation (4.41) is given by

Ayγ

ν)y(w

(4.42)

where A is an arbitrary constant. Integrating further υ

υ)y(w

y , we find that

20

γ

ν

)Ayγ(B)y(υ (4.43)

where B is an arbitrary constant.

4.2.II. If 0μ , equation (4.37) takes the form

wμν

wμwγw

32

y

(4.44)

4.2.IIa. If 0γ , under the substitution

)y(Y)y(w (4.45)

equation (4.44) takes the form

)νγ(μY)γν(

YγΥ

2

y

( νγ )

and by integration we obtain

AyγYln)νγ(μY)γν( 2 (4.46)


Therefore )y(υ is the solution to the equation

)y(

y (4.47)

where )y(Υ is given implicitly by (4.46). Therefore we find that

y

y0)(

dexpB)y( (4.48)

where B is an arbitrary constant.

4.2.IIb. For 0γ (i.e. 2k

2

1ν ), equation (4.41) becomes

wμν

wμw

3

y

(4.49)

Under the substitution

21

)y(X

1w

the above equation (4.49) takes the form

XXy

which has the general solution

Ay222 (4.50)

where A is a constant.

Therefore )y(υ is the solution to the equation

)y(X

1

υ

υy (4.51)

where )y(X is given implicitly by (4.50). Hence, we find that

y

y0

)ξ(X

ξdexpB)y(υ (4.52)

where B is a constant.

5. Conclusions and Discussion.

Gathering the results of the previous section, we arrive at the following

Theorem 2. The HJB equation (3.1) admits the following twelve families of

general solutions:

Solution I.

C)(V

dexpeB)t,x(u

y

y

t

0

(5.1)

t)λr(tλ e

λr

μxey

( 0μ ) , r (5.2)

where )y(V is a function defined implicitly by

Ayln|)y(V|ln)(

|)y(V|ln)(

(5.3)

22

The parameters and are defined by

2

D and

2

D (5.4)

respectively, where λrδ ( 0 ) and 4D 2. The parameter is such

that 2k2

1 ( 0 ) and the parameter satisfies either one of the inequalities

4rr or 4rr .

Solution II

C)(V

dexpeB)t,x(u

y

y

t

0

(5.5)

t)λr(tλ eλr

μxey

( 0μ ) , r (5.6)


Aylnp

)y(Varctan

p)p))y(Vln((

2

1 22

(5.7)

The parameters and p are defined by

2 and

2

Dp (5.8)

respectively, where λrδ ( 0 ) and 4D 2.

The parameter is such that 2k2

1 ( 0 ) and the parameter takes values

within the range 4rr .

Solution III.

C)ξ(Vξ

ξdexpeB)t,x(u

y

y

tν

0

(5.9)

t)λr(tλ e

λr

μxey

( 0μ ) , r (5.10)

23


Ayln|2)y(V|ln2)y(V

12

(5.11)

The parameter is defined by λrδ ( 0 ) and satisfies the equation

4 . The parameter is such that 2k2

1 ( 0 ).

Solution IV

C)ξ(Vξ

ξdexpeB)t,x(u

y

y

tν

0

(5.12)

t)λr(tλ eλr

μxey

( 0μ ) , 0 , (5.13)


Ayln)y(V

)y(Vln

2

1))y(Vln(

2

1 22

(5.14)

where 2ωνδ , λrδ ( 0 ) and 2k2

1 .

Solution V

C)(V

dexpeB)t,x(u

y

y

t

0

(5.15)

t)λr(tλ eλr

μxey

0μ , 0 , (5.16)


Ayln)y(V

arctan))y(Vln(2

1 22

(5.17)

The parameter is defined through 2 where λrδ ( 0 ) and

2k2

1

24

Solution VI

C)(V

dexpeB)t,x(u

y

y

t

0

(5.18)

t)λr(tλ eλr

μxey

0μ , 0 , (5.19)


Ayln|1)y(V|ln (5.20)

The parameters satisfy , λrδ ( 0 ) and 2k2

1 .

Solution VII

C)(V

dexpeB)t,x(u

y

y

t

0

(5.21)

t)λr(tλ e

λr

μxey

( 0μ ) (5.22)


Ayln|)y(V|ln)(

|)y(V|ln)(

(5.23)

The parameters are given by

2

D,

2

D and 04)(D 2 ,

and 0 , where 2k

2

1νγ and λrδ ( 0 )

Solution VIII

C)(V

dexpeB)t,x(u

y

y

t

0

(5.24)

t)λr(tλ e

λr

μxey

( 0μ ) (5.25)

25


Aylnp

)y(Varctan

p)p))y(Vln((

2

1 22

(5.26)


2,

2

Dp , 04)(D 2 , and 0

where 2k2


Solution IX

C)(V

dexpeB)t,x(u

y

y

t

0

(5.27)

t)λr(tλ eλr

μxey

( 0μ ) (5.28)


Ayln))y(Vln()y(V

1

(5.29)


2, 04)(D 2 , and 0

where 2k

2


Solution X

C)Ayγ(eB)t,x(u γ

ν

tν (5.30)

where

trexy (5.31)

26

A, B, C are arbitrary constants and ν is a free parameter with 0ν , and

2k2

1νγ with 0γ .

Solution XI

C)(

dexpeB)t,x(u

y

y

t

0

(5.32)

where )y(Y is a function defined implicitly by

Ayγ)y(Yln)νγ(μ)y(Y)γν( 2 ( 0γ ) (5.33)

and

tμexy tr ( 0μ ) (5.34)

A, B, C are arbitrary constants and μ, ν are free parameters with 0μ , 0ν , and

2k2

1νγ with 0γ .

Solution XII

C)ξ(X

ξdexpeB)t,x(u

y

y

tν

0

(5.35)

where )y(X is a function defined implicitly by

Ay2)y(2)y(2 (5.36)

and

tμexy tr ( 0μ ) (5.37)

provided that 2k

2

1ν ( 0γ ). A, B, C are arbitrary constants and μ, ν are free

parameters with 0μ , 0ν .■

Only one solution that has been dealt with in the literature so far. It is the solution

(5.30)-(5.31). The equation (2.9) is a member of this family of solutions when we

27

choose to place 0CA and 1B , and an obvious adjustment (and renaming)

of the parameters, using the same boundary condition

γxγ

1)x(U)T,x(u

The other solutions appear for the first time in this paper. It is obvious that the

parameters and the various constants need to be adjusted to match the appropriate

boundary conditions. It remains to be seen if the general solutions found in this

article are to have any practical interest in Finance and Stochastic Control.

In any case, the Hamilton-Jacobi-Bellman equation associated to an optimization

problem seems to have a rich mathematical structure and thus deserves further

investigation.

Appendix. Solution of equation (4.3).

Equation (4.3) can be written as

0dt)ex(xd tr (A.1)

Using standard methods, we find that equation (A.1) admits te as an integrating

factor. Therefore we have to find the general solution of the complete first order

differential equation

0dt)eex(dxe t)r(tt (A.2)

We suppose that (A.2) admits a solution of the form

C)t,x(U (A.3)

Since

0dxx

Udt

t

UdU

(A.4)

and on comparing this equation with (A.2), we obtain the following system of

differential equations

t)r(t eex

t

U

(A.5)

28

tex

U

(A.6)

Integrating (A.5) with respect to t, we need to distinguish two cases:

r and r .

A1. If r , we obtain from equation (A.5) the solution

)x(er

exU t)r(t

(A.7)

where )x( is a function to be next determined. From (A.7)

)x(ex

U t

(A.8)

(comparing (A.8) to (A.6)) we find that 0)x( , i.e. 1C)t( . Therefore, using

(A.7) and (A.3), we obtain the following solution of (4.3) when r :

Cer

ex t)r(t

(A.9)

which is equation (4.4).

A2. If r , we get from (A.5)

text

U (A.10)

which upon integration with respect to t gives us

)x(texU t (A.11)

where )x( is a function to be determined next.

From (A.11)

)x(ex

U t

(A.12)

(comparing (A.12) to (A.6)), we get 0)x( , i.e. 2C)t( . Therefore using

(A.11) and (A.3), we get the following solution of (4.3) when r :

2tr Ctex (A.13)

which is equation (4.34).

29

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