Dynamic optimal reinsurance and dividend-payout in finite time horizon
aa r X i v : . [ q -f i n . M F ] A ug Dynamic optimal reinsurance and dividend-payoutin finite time horizon
Chonghu Guan ∗ Zuo Quan Xu † Rui Zhou ‡ Abstract
This paper studies a dynamic optimal reinsurance and dividend-payout problem foran insurer in a finite time horizon. The goal of the insurer is to maximize its expectedcumulative discounted dividend payouts until bankruptcy or maturity which comesearlier. The insurer is allowed to dynamically choose reinsurance contracts over thewhole time horizon. This is a singular control problem and the corresponding Hamilton-Jacobi-Bellman equation is a variational inequality with fully nonlinear operator andwith gradient constraint. A comparison principle and C , smoothness for the solutionare established by penalty approximation method. We find that the surplus-time spacecan be divided into three non-overlapping regions by a ceded risk and time dependentreinsurance barrier and a time dependent dividend-payout barrier. The insurer shouldbe exposed to higher risk as surplus increases; exposed to all risk once surplus upwardcrosses the reinsurance barrier; and pay out all reserves in excess of the dividend-payoutbarrier. The localities of these regions are explicitly estimated. Keywords.
Optimal reinsurance, optimal dividend, free boundary, stochastic optimalcontrol, singular control.
Mathematics Subject Classification.
A fundamental goal of risk managers of insurance companies is to improve the solvency andstability of their companies. This goal can be reached through the choice of dividend-payout ∗ School of Mathematics, Jiaying University, Meizhou 514015, Guangdong, China. This author is partiallysupported by NNSF of China (No. 11901244), NSF of Guangdong Province of China (No. 2016A030307008).Email: . † Department of Applied Mathematics, The Hong Kong Polytechnic University, Hong Kong, China. Thisauthor is partially supported by NSFC (No. 11971409), and Hong Kong GRF (No. 15204216 and No.15202817). Email: [email protected] . ‡ Department of Applied Mathematics, The Hong Kong Polytechnic University, Hong Kong, China. Email: [email protected] . , l ]. The optimal dividend-payout strategy turns out to be an “all ornothing” policy with respect to (w.r.t.) dividend rate. That is, paying out dividends at theminimum rate 0 if the surplus is lower than a threshold, and paying out at the maximum rate ℓ otherwise. In the second case, the dividend-payout rate is unbounded. The optimal strategyis to keep the surplus under some barrier, that is, paying out all reserves in excess of thebarrier as dividends, and doing nothing under the barrier. The optimal reinsurance strategy,which clearly depends on surplus level, is shown to be that insurers should be exposed tohigher risk as surplus increases. Along this line of research, other factors including, but notlimit to, liability, regime switching, have also been taken into account in model formulation;see [29, 28]. 2nsurance market has grown furious in recent years. Insurance companies, especiallythe global insurance companies, such as AIA Group Ltd., AIG, China Life Insurance Com-pany, operate diverse businesses and offer insurance products on various term basis, such asproperty and casualty insurance, life insurance, health insurance. In this regard, it is nat-ural for insurers to consider their total risk exposure as a combination of different types ofrisk. Mathematically speaking, the total risk exposure may take any, discrete or continuous,probability distributions. Similarly, but more importantly, the reinsurance strategies shouldalso be based on different types of risk, resulting in non classical complicated reinsurancepolicies. Although the optimal reinsurance problem has been extensively investigated, mostof existing literature analyzes only for typical reinsurance policies such as proportional andexcess of loss reinsurance; see, e.g., [27, 22, 24, 18, 16, 17, 20]. This paper investigates anoptimal reinsurance and dividend-payout problem, but to bring more practical features, wedo not restrict ourself to these typical reinsurance policies. We consider a controlled diffusionsurplus process, which is a good approximation of the classical Cram´er-Lundberg process aswell-justified by Grandell [12]. A closest model to this paper is considered by Guan, Yi andChen [14], where the risk control model is relatively simple and the type of reinsurance policyis constrained to proportional ones. It turns out that the reinsurance scheme is restrictive.Tan, et al., [25] consider a similar surplus process, but their model is of infinite time horizon,so the corresponding Hamilton-Jacobi-Bellman (HJB) equation is an ordinary differentialequation (ODE), which is much easier to handle than that of a partial differential equation(PDE) in our case.In our model, both drift and volatility of the controlled surplus process depend on thereinsurance policy. The target of the insurer is to maximize the expected cumulative dis-counted dividend payouts until bankruptcy or a given maturity time which comes earlier.The model has the following features: First, we do not confine the reinsurance contracts tobe some particular ones such as proportional reinsurance or excess of loss reinsurance. Theinsurance company can chose its reinsurance policy freely at its will subject to the expectedvalue premium principle. Second, the reinsurance contracts are chosen dynamically depend-ing on the surplus level. It turns out that the optimal reinsurance policy is a feedback onethat depends on surplus, time and ceded risk. Third, the insurance claims can admit anyprobability distribution having a finite third moment. For bounded claims, we have a betterunderstanding of the optimal reinsurance strategy. Finally, we consider a finite time horizonproblem, making the HJB equation being an extremely challenging fully nonlinear one withgradient constraint.Like many existing studies, we adopt a PDE approach to study the problem. Becausethe problem is a singular control problem with respective to (w.r.t.) dividend payout, the3orresponding HJB equation turns out be a variational inequality problem with gradientconstraint. As an unavoidable consequence of the choice of the time-dependent reinsurancepolicy, there is a functional optimization problem appearing in the operator, making theproblem become a fully nonlinear one. Using the method in [8, 9, 7], we show that thegradient function itself satisfies an obstacle problem. Finally, a comparison principle, anoptimal control and necessary properties (such as C , smoothness and uniqueness) for thefull nonlinear problem are derived by penalty approximation method.Our model provides numerous economic insights. We show that there is a smooth, time-dependent, dividend-payout barrier that divides the surplus-time space into a non-dividend-payout region and a dividend-payout region. The insurer should pay out all reserves in excessof the dividend-payout barrier (that is, all reserves in the dividend-payout region). Further-more, we find a ceded risk and time dependent smooth reinsurance barrier that divides thenon-dividend-payout region into a reinsurance region and a non-reinsurance region, in anincreasing order of surplus. As ceded risk increases, the non-reinsurance region is shrink-ing, while, the reinsurance region is expanding. The former never vanishes, but the latterdisappears when the ceded risk is smaller than an explicitly given constant. The insurancecompany should be exposed to higher risk as surplus increases in the reinsurance region;exposed to all risk once its surplus goes into the non-reinsurance region. We also provideaccurate explicitly estimations for the localities of these regions. This makes the relevantnumerical solution scheme can be easily established. When the claims are bounded randomvariables, we show there is a uniform non-action region, in which the insurer should beexposed to all risk and not to pay out dividends.The reminder of the paper is organised as follows. In Section 2, we formulate the optimalreinsurance and dividend-payout problem. Section 3 poses the corresponding HJB equationand gives a verification theorem. The properties such as the existence and uniqueness of aclassical solution to the HJB problem are also provided. Section 4 is devoted to the studyof the reinsurance, non-reinsurance and dividend-payout regions. The proofs of some resultsare given in the appendices. This paper investigates an optimal reinsurance and dividend-payout problem. We first needto model the cash reserves for an insurance company (the insurer), let us start from theclassical Cram´er-Lundberg model.In the classical Cram´er-Lundberg model, there are two different components that affectthe cash reserve (also called surplus) dynamics. The first one is the receiving payments of4remium from the policyholders at a constant rate p continuously. The other one is theoutgoing payments for claims. If we denote the total number of claims received until time t by N t and the size of the i th claim by Z i , then R t , the company’s surplus at time t , is givenby R t = R + pt − N t X i =1 Z i . (2.1)where N t is a Poisson process with intensity 1, and all claims are independent and identicallydistributed random variables independent of N t .Sometimes, the insurance company needs to protect itself by sharing its risk with a cededcompany (the reinsurer). The insurer buys reinsurance contracts from the reinsurer. Givena reinsurance contract I ( · ), the reinsurer should compensate I ( z ) to the insurer when aclaim amount z for the ceded risk is received by the insurer. This function I ( · ) is knownas the ceded loss function, and H ( z ) := z − I ( z ) is known as the retained loss function.A reinsurance policy (or strategy) consists of a series of reinsurance contracts { I t } t > overtime, where I t denotes the reinsurance contract signed at time t . Note that the reinsurancecontracts are dynamically chosen by the insurer, usually time and surplus dependent.The presence of reinsurance modifies the risk exposure of the insurer. It distorts theincoming and outgoing cash flow of the insurer’s surplus process (2.1). Under big portfolios,as well-justified by Grandell [12], the surplus process R t can be approximated by the followingdiffusion processd R t = ( p − p ( I t ) − E R t − [ Z − I t ( Z )])d t + q E R t − [( Z − I t ( Z )) ] d W t , (2.2)where { W t } t > is a standard Brownian motion independent of the random claim Z , E R t − [ · ] = E [ · | R t − ], and p ( I t ) denotes the reinsurance premium corresponding to the reinsurance con-tract I t . Remark that the contract I t may depend on the status of the surplus process up to t , so it is stochastic.In this paper we consider the expected value premium principle for both insurance andreinsurance contracts, given by p = (1 + δ ) E [ Z ] , p ( I t ) = (1 + ρ ) E R t − [ I t ( Z )] , where δ , ρ > R t = ( − γ + ρ E R t − [ Z − I t ( Z )])d t + q E R t − [( Z − I t ( Z )) ] d W t , γ = ( ρ − δ ) E [ Z ]. We impose that ρ > δ , i.e., γ >
0, to ensure that reinsurance isnon-cheap. If reinsurance is too cheap, ρ < δ , then the insurer can simply eliminate allits risk exposure by ceding all arising claims to the reinsurer, reaping the certain profit of δ − ρ > ρ = 1 from now on.In this paper, we assume that the insurer will pay out part of its surplus as dividends toits shareholders. Let L t be the cumulative dividend extracted from the surplus process until t , which is a non-decreasing c`adl`ag (right continuous with left limits) process. It is chosen bythe insurer according to its surplus status. Then the new surplus process { R s } s > t beginningat time t with an initial value x satisfies, recalling that H ( z ) = z − I ( z ), d R s = (cid:16) − γ + E R s − [ H s ( Z )] (cid:17) d s + p E R s − [ H s ( Z )] d W s − d L s , s > t,R t − = x > , (2.3)This is also a c`adl`ag process. It jumps at the same time as L with the same jump size butnegative sign, namely R s − R s − = − ( L s − L s − ) for any s > t . Define the ruin time of theinsurer as τ = inf (cid:8) s > t | R s (cid:9) . (2.4)The insurer is not allowed to pay out dividends more than the existing surplus, so L s − L s − R s − at any time s . As a consequence, R s = R s − − ( L s − L s − ) >
0. So the surplus of theinsurance company is R τ = 0 at the ruin time.The objective of our optimal reinsurance and dividend-payout model is to find a re-tained loss policy H t = { H s } s > t − (or equivalently, a reinsurance policy I t = { I s } s > t − ) and adividend-payout policy L t = { L s } s > t − to maximize the expectation of discounted cumulativedividend payouts until bankruptcy or a given maturity T > V ( x, t ) = sup H t , L t E " Z T ∧ τt − e − c ( s − t ) d L s (cid:12)(cid:12)(cid:12)(cid:12) R t − = x , x > , t T, (2.5)where c is a positive discount factor. This is a singular control problem.In the rest of this paper, we will investigate the value function and provide the optimalreinsurance and dividend-payout strategies. We now study Problem (2.5) by dynamic programming principle.6he claims are allowed to be discrete or continuous distributed. Let F ( z ) denote theircommon cumulative distribution function. Because the claims are nonnegative, F (0 − ) = 0.For technical reason, we assume the third moment is finite and denote µ k := Z ∞ z k d F ( z ) < ∞ , for k = 1 , , . Also the retained loss function (or equivalently, the ceded function) is subject to the con-straint 0 H ( Z ) Z. We introduce the following variational inequality min (cid:26) v t − sup H ∈H (cid:18) v xx Z ∞ H ( z ) d F ( z ) + v x Z ∞ H ( z )d F ( z ) (cid:19) + γv x + cv, v x − (cid:27) = 0in Q T := (0 , + ∞ ) × (0 , T ] ,v (0 , t ) = 0 , < t T,v ( x,
0) = x, x > , (3.1)where H := (cid:8) H : [0 , ∞ ) → [0 , ∞ ) | H ( z ) z (cid:9) . This variational inequality is linked to the optimal reinsurance and dividend-payout problem(2.5) by the following verification theorem.
Theorem 3.1 (Verification Theorem)
If there exits a solution v ∈ C , (cid:0) Q T \{ (0 , } (cid:1) T C (cid:0) Q T (cid:1) to Problem (3.1) such that it is increasing and concave w.r.t. x . Then v ( x, T − t ) = V ( x, t ) , (3.2) where V , defined by (2.5) , is the value function of optimal reinsurance and dividend-payoutproblem. The proof is given in Appendix A. Clearly, this result tells us that Problem (3.1) admits atmost one classical solution. For simplicity we call (3.1) the HJB equation for our problem(2.5), although it is a time-changed equation of the standard definition (see [31]).If γ > µ , then it is easy to check that v ≡ x is the solution to Problem (3.1). In this caseone can see that the drift of the surplus process in (2.3) is either negative if E [ I t ( Z )] > x for Problem (2.5). The problem is boring in this case. So inthe rest of this paper, we assume 0 < γ < µ .To tackle Problem (3.1), one hopes to solve the supremum in the PDE, namelysup H ∈H (cid:16) v xx Z ∞ H ( z ) d F ( z ) + v x Z ∞ H ( z )d F ( z ) (cid:17) , or sup H ∈H Z ∞ (cid:16) H ( z ) v xx + H ( z ) v x (cid:17) d F ( z ) , which is obviously equivalent to Z ∞ sup h z (cid:16) h v xx + hv x (cid:17) d F ( z ) . (3.3)We now rewrite (3.3) by an explicit function of v x and v xx . Define h ∗ ( z, y ) := argmax h z (cid:18) − h y + h (cid:19) = y , if y > z ; z, if y z . (3.4)Notice v x > v xx Z ∞ (cid:18) h ∗ (cid:16) z, − v xx v x (cid:17)(cid:19) d F ( z ) + v x Z ∞ h ∗ (cid:16) z, − v xx v x (cid:17) d F ( z ) , that is, A (cid:18) − v xx v x (cid:19) v xx + B (cid:18) − v xx v x (cid:19) v x , where functions A ( y ) and B ( y ) are defined by A ( y ) = R /y z (1 − F ( z ))d z, if 0 < y < + ∞ ; R ∞ z (1 − F ( z ))d z = µ , if y , and B ( y ) = R /y (1 − F ( z ))d z, if 0 < y < + ∞ ; R ∞ (1 − F ( z ))d z = µ , if y . We note that A ( y ) and B ( y ) are both decreasing functions in C ( R ), which satisfy0 < A ( y )
12 min { y − , µ } , < B ( y ) min { y − , µ } , for y >
0; (3.5) y A ′ ( y ) = y B ′ ( y ) = F ( y − ) − , for y >
0; (3.6) A ′ ( y ) = B ′ ( y ) = 0 , for y . (3.7)8efine an operator L v := A (cid:18) − v xx v x (cid:19) v xx + B (cid:18) − v xx v x (cid:19) v x − γv x − cv. Then Problem (3.1) can be rewritten as min { v t − L v, v x − } = 0 in Q T ,v (0 , t ) = 0 , < t T,v ( x,
0) = x, x > . (3.8)This is a variational inequality problem for a fully nonlinear elliptic operator with gradientconstraint. A usual approach to study this kind of problem is to transform it into a variationalinequality problem for its gradient. Then the gradient constraint becomes value constraintand the new variational inequality becomes a well-studied obstacle problem; see [8, 9, 14, 15].In this paper, we adopt the same approach.Below we use intuitive argument to find a variational inequality for the gradient of v . Wefirst note that ∂ x (cid:20) A (cid:18) − v xx v x (cid:19) v xx + B (cid:18) − v xx v x (cid:19) v x (cid:21) = A (cid:18) − v xx v x (cid:19) v xxx + B (cid:18) − v xx v x (cid:19) v xx + (cid:20) A ′ (cid:18) − v xx v x (cid:19) v xx + B ′ (cid:18) − v xx v x (cid:19) v x (cid:21) ∂ x (cid:18) − v xx v x (cid:19) = A (cid:18) − v xx v x (cid:19) v xxx + B (cid:18) − v xx v x (cid:19) v xx , where we used (3.6) and (3.7) to get the last equation. By this, one can easily deduce that ∂ x ( L v ) = T v x , where the operator T is defined as T u := A (cid:16) − u x u (cid:17) u xx + B (cid:16) − u x u (cid:17) u x − γu x − cu. Next, we deduce a boundary condition for v x . Define a continuous function f ( y ) := − yA ( y ) + B ( y ) − γ. (3.9)By (3.5)-(3.7), f ′ ( y ) = − A ( y ) < . (3.10)Hence f is strictly decreasing. Also f (0+) = µ − γ > , f (+ ∞ ) = − γ < , f has a unique root, which is positive, denote by λ in the rest of this paper. See Figure1 for an illustration of the function f . ✲✻ r µ − γ − γ r λ r Figure 1: The function f is continuous and strictly deceasing, and admits a unique positiveroot λ .On the other hand, owing to the boundary condition v (0 , t ) = 0 and that v t − L v = 0near { x = 0 } , we should have (cid:18) A (cid:18) − v xx v x (cid:19) v xx + B (cid:18) − v xx v x (cid:19) v x − γv x (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x =0 = 0 . (3.11)Dividing by v x , it yields f (cid:16) − v xx v x (cid:17) (cid:12)(cid:12)(cid:12) x =0 = 0. So − v xx v x (cid:12)(cid:12) x =0 = λ by the strictly monotonicity of f . This leads to a boundary condition (cid:16) λv x + v xx (cid:17)(cid:12)(cid:12)(cid:12) x =0 = 0 . (3.12)Combining the above results, we suggest the following variational inequality for u = v x min { u t − T u, u − } = 0 in Q T , (cid:0) λu + u x (cid:1) (0 , t ) = 0 , < t T,u ( x,
0) = 1 , x > . (3.13)This is an obstacle problem for a quasilinear elliptic operator with mixed boundary condi-tions, so we can study it by penalty approximation method. This method gives the followingexistence and uniqueness results. Theorem 3.2
Problem (3.1) (or equivalently (3.8) ) has a unique solution v ∈ C , (cid:0) Q T \ (0 , } (cid:1) T C (cid:0) Q T (cid:1) that satisfies v x > , (3.14) v t > , (3.15) v xx , (3.16) v xxx > in weak sence , (3.17) v xt > , (3.18) λv x + v xx > . (3.19) Moreover, u = v x is the unique solution to Problem (3.13) in W , p, loc (cid:0) Q T (cid:1) T C (cid:0) Q T (cid:1) . Proof:
The proof is given in Appendix B. (cid:3)
From now on, we fix v as in Theorem 3.2. By Verification Theorem 3.1, it completelycharacterizes the value function of the optimal reinsurance and dividend-payout problem(2.5). In the rest part of this paper, we express the optimal reinsurance and dividend-payoutstrategies by this function and study their properties. In the previous section, we have solved the existence and uniqueness issues for the HJBequation (3.1). In the following sections we investigate the optimal strategies for Problem(2.5). We first study the optimal dividend-payout strategy and then the optimal reinsurancestrategy.
To investigate the optimal dividend-payout strategy, we divide the whole domain Q T :=(0 , + ∞ ) × [0 , T ) into a dividend-payout region D = n ( x, t ) ∈ Q T (cid:12)(cid:12)(cid:12) v x ( x, ~t ) = 1 o and a non-dividend-payout region N D = n ( x, t ) ∈ Q T (cid:12)(cid:12)(cid:12) v x ( x, ~t ) > o . Here and hereafter we use the notation ~t := T − t .Because v x > v xx
0, we can express them as D = { x > d ( ~t ) } , N D = { x < d ( ~t ) } . d ( · ) is the dividend-payout boundary , defined by d ( ~t ) = inf { x > | v x ( x, ~t ) = 1 } , ~t > . In the following, we come to show the boundary d ( · ) is uniformly upper bounded by anexplicit given constant. To this end, we will construct a function b u ( x ) such that b u ( x ) > v x ( x, t ), then clearly inf { x > | b u ( x ) = 1 } provides a uniformly upper bound for d ( · ).First, we construct a function to bound v x (0 , t ). For this, we let b v ( x ) := ( C (1 − e − xγ ) , < x x ,C + x − x , x > x where C = µ c + γ > , C = µ c > , x = γ ln C γ > . It is easy to check b v ( x − ) = b v ( x +) and b v x ( x − ) = b v x ( x +), so b v ∈ C (0 , ∞ ). It is easilyseen that b v x is continuous and decreasing, so b v is a concave function. When 0 < x x , b v t − L b v = b v t − A (cid:18) − b v xx b v x (cid:19) b v xx − B (cid:18) − b v xx b v x (cid:19) b v x + γ b v x + c b v = − Z ∞ sup h z (cid:16) h b v xx + h b v x (cid:17) d F ( z ) + γ b v x + c b v > − sup h< ∞ (cid:16) h b v xx + h b v x (cid:17) + γ b v x = b v x b v xx + γ b v x = C e − xγ > , and when x > x , b v t − L b v = − µ + γ + c ( C + x − x ) = γ + c ( x − x ) > . Therefore, b v ∈ W , p ( Q T ) is a super solution to Problem (3.8). Since b v (0 , t ) = v (0 , t ) = 0, weobtain that v x (0 , t ) b v x (0) = C /γ by comparison principle.Now, we are ready to construct a supper solution to Problem (3.13). Define b u ( x ) := ( C ( x − x ) + 1 , < x x , , x > x , where C = c cµ + γ > , x := s C C γ = r γc ( µ + cγ ) ( cµ + γ ) > . b u is convex and b u ∈ W , p ( Q T ). If 0 < x < x , then b u x
0. By (3.5) and using theelementary inequality x − xy > − y , we obtain b u t − T b u > − µ b u xx + γ b u x + c b u = C (cid:0) − µ − γ ( x − x ) + c ( x − x ) (cid:1) + c > C (cid:0) − µ − γ /c (cid:1) + c = 0 . If x > x , then b u t − T b u = c >
0. Because b u (0) = C /γ > v x (0 , t ) = u (0 , t ), by comparisonprinciple, we conclude b u > u . Therefore, x = inf { x > | b u ( x ) = 1 } is a constant upperbound for d ( · ).Summarizing the above results, the dividend-payout boundary is completely character-ized in the following theorem. Theorem 4.1
The dividend-payout boundary d ( ~t ) is continuous and increasing in ~t , andsatisfies d (0+) = 0 < d ( ~t ) d ( ∞ ) x = r γc ( µ + cγ ) ( cµ + γ ) , (4.1) where d (0+) = lim ~t → d ( ~t ) and d ( ∞ ) := lim ~t → + ∞ d ( ~t ) . Furthermore, if Z is a bounded randomvariable, then d ( ~t ) ∈ C ∞ (0 , T ) . See Figure 2 for an illustration of the dividend-payout barrier d ( ~t ) as well as dividend-payoutregion D and non-dividend-payout region N D . ✲✻ ~t xd ( ∞ ) x N D D r r x = d ( ~t ) Figure 2: The dividend-payout barrier x = d ( ~t ) divides the surplus-time space in to adividend-payout region D and a non-dividend-payout region N D . Proof:
The monotone property of d ( · ) is due to (3.18). We now prove the continuity.Suppose on the contrary there exists ~t such that d ( ~t − ) < d ( ~t +). By the continuity of v x ,13e know v x ( d ( ~t − ) , ~t ) = v x ( d ( ~t +) , ~t ) = 1. Since v x is decreasing in x , we have v x ≡ v xx ≡ d ( ~t − ) , d ( ~t +)). By (3.8), we further have v t ( x, ~t ) = − µ + γ − cv ( x, ~t ) for x ∈ ( d ( ~t − ) , d ( ~t +)); and consequently, v tx ( x, ~t ) = − cv x ( x, ~t ) = − c <
0, which contradicts(3.18). In a similar way, we can prove lim ~t → d ( ~t ) = 0.Now we prove d ( · ) >
0. Suppose not, then, by monotonicity, there exists ~t > d ( ~t ) = 0 for all 0 < ~t < ~t . This implies v x ≡ v ≡ x for 0 < ~t < ~t . Denote b v ( x, ~t ) = v ( x, ~t − ~t ), then both b v and v satisfy (3.8). By the uniqueness, we get b v ≡ v .This implies v ≡ x for all t >
0, but this contradicts v t − L v > γ < µ .It is only left to show the smoothness of d ( · ) when Z is bounded. Suppose F (ˆ z ) = 1 forsome ˆ z >
0. Then A ( y ) = 12 µ , B ( y ) = µ , y z . (4.2)For any ~t , because v x and v xx are continuous, we have v x = 1 and v xx = 0 at ( d ( ~t ) , ~t ).This implies v x + ˆ zv xx > B of ( d ( ~t ) , ~t ). By (3.16) and (4.2), wesee u = v x satisfiesmin n u t − µ u xx − ( µ − γ ) u x + cu, u − o = 0 , ( x, ~t ) ∈ B . Because the coefficients are constants in the above equation, using the method in [11], wecan prove d ( ~t ) ∈ C ∞ at a neighborhood of ~t . Since ~t is arbitrarily chosen, we concludethat d ( t ) ∈ C ∞ (0 , T ). (cid:3) In view of the original optimal reinsurance and dividend-payout problem (2.5), by theproof of Theorem 3.1, the optimal dividend-payout policy L ∗ s is the local time of correspond-ing reserve process R ∗ s at the level d ( ~s ), namely ( L ∗ s − L ∗ s − = R ∗ s − − d ( ~s ) , if R ∗ s − > d ( ~s );d L ∗ s = 0 , if R ∗ s − d ( ~s ) . (4.3)The financial meaning is that, when the insurance surplus R ∗ s − is above the threshold d ( ~s ),the insurance company should pay out the reserves of an amount R ∗ s − − d ( ~s ) as dividends toits shareholders at time s ; otherwise, not. We remark that the reserve process R ∗ s is alwayscontinuous and no more than the d ( ~s ), except for the initial time. In this section we study the optimal reinsurance strategy. Recall that we have ascertainedthe value function V ( x, t ) = v ( x, ~t ) by Theorem 3.1.14or the insurer, if its current status is ( x, t ), by (3.4), the corresponding optimal retainedfunction is z b H ( z, x, ~t ) := − v x v xx ( x, ~t ) , if − v xx v x ( x, ~t ) > z ; z, otherwise , (4.4)and the optimal reinsurance function is z b I ( z, x, ~t ) := z − b H ( z, x, ~t ) . These strategies not only depend on the claim z , but also depend on the current insurancesurplus x and time t . In this section, we will discuss the behavior of them.One immediate consequence of (4.4) is that b H ( z, x, ~t ) > z >
0, so b I ( z, x, ~t ) < z. In any case, the insurance company should not cede all risks; it has to bear some risks byitself. This is due to our assumption that the reinsurance is non-cheap, γ >
0. On theother hand, because v x ( x, ~t ) = 1 for ( x, t ) ∈ D , it yields b H ( z, x, ~t ) = z and b I ( z, x, ~t ) = 0.Therefore, the insurance company should not cede risk in the dividend-payout region. Inother words, the insurance company shares its risk with a ceded company only when thesurplus is lower than the dividend-payout boundary d ( ~t ).For each z >
0, depending on whether b I is zero, we divide the surplus-time space into a reinsurance region R z = n ( x, t ) ∈ Q T (cid:12)(cid:12)(cid:12) b I ( z, x, ~t ) > o and a non-reinsurance region N R z = n ( x, t ) ∈ Q T (cid:12)(cid:12)(cid:12) b I ( z, x, ~t ) = 0 o . By (4.4), they can also expressed as R z = (cid:26) ( x, t ) ∈ Q T (cid:12)(cid:12)(cid:12) − v xx v x ( x, ~t ) > z (cid:27) , N R z = (cid:26) ( x, t ) ∈ Q T (cid:12)(cid:12)(cid:12) − v xx v x ( x, ~t ) z (cid:27) . By our above discussion,
D ⊆ N R z and R z ⊆ N D . Lemma 4.2
We have v x , v t ∈ C , (cid:0) N D (cid:1) . (4.5) Furthermore, v xx < if ( x, t ) ∈ N D . And consequently, b I ( z, x, ~t ) = max (cid:8) z + v x /v xx ( x, ~t ) , (cid:9) for z > and ( x, t ) ∈ N D . roof: The proof is given in Appendix C. (cid:3)
This results implies that b I is an increasing function w.r.t z , which of course makesperfect financial meaning that the insurer should get more compensation from the reinsurancecompany when a larger claim arises. Lemma 4.3
We have (cid:18) − v x v xx (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x =0 = 1 λ , (4.6) − v x v xx ( x, ~t ) > λ if ( x, t ) ∈ N D , (4.7) ∂ x (cid:18) − v x v xx (cid:19) ( x, ~t ) > c if ( x, t ) ∈ N D . (4.8) Proof:
The proof is given in Appendix D. (cid:3)
From (4.7) we see that, if ( x, t ) ∈ N D , then b I ( z, x, ~t ) = max (cid:8) z + v x /v xx ( x, ~t ) , (cid:9) max { z − /λ, } . (4.9)So b I ( z, x, ~t ) = 0 when z λ ; the insurance company should bear all arising claims below λ by itself. In other words, any arising claims below λ should not be shared with a cededcompany.By (4.8), − v xx v x is a strictly decreasing function in x , so we have R z = n ( x, t ) ∈ Q T (cid:12)(cid:12)(cid:12) x < K ( z, ~t ) o , N R z = n ( x, t ) ∈ Q T (cid:12)(cid:12)(cid:12) x > K ( z, ~t ) o , where K ( z, ~t ) is the reinsurance boundary, given by K ( z, ~t ) := inf (cid:26) x > (cid:12)(cid:12)(cid:12)(cid:12) − v xx v x ( x, ~t ) z (cid:27) . Because R z ⊆ N D , we see K ( z, ~t ) d ( ~t ). Moreover, when z /λ , we have by (4.9) R z = ∅ , so K ( z, ~t ) = 0.For each z >
0, define the overlapping of the non-reinsurance and non-dividend regionsas non-action region
N A z = N R z \ N D = n ( x, t ) ∈ Q T (cid:12)(cid:12)(cid:12) K (ˆ z, ~t ) x < d ( ~t ) o . In this region the insurer should not cede risk z or pay out dividends. We are now show thisregion is always non-empty. This is equivalent to showing K ( z, ~t ) = d ( ~t ) by recaling K ( z, ~t ) d ( ~t ). Note that d ( ~t ) = inf { x > | v x ( x, ~t ) = 1 } and v ∈ C , (cid:0) Q T \ { (0 , } (cid:1) T C (cid:0) Q T (cid:1) , so( v x + zv xx ) (cid:12)(cid:12) ( d ( ~t ) ,~t ) = 1. But evidently ( v x + zv xx ) (cid:12)(cid:12) ( K ( z,~t ) ,~t ) = 0, so K ( z, ~t ) = d ( ~t ).16ombining the above results, we conclude that, for each risk z >
0, the reinsurance anddividend-payout barriers divide the surplus-time space into three non-overlapping regions:a (possible empty) reinsurance region, a non-action region and a dividend-payout region, inan increasing order of surplus. This is illustrated in Figure 3. ✲✻ ~t xd ( ∞ ) x = K ( z, ~t ) R z N A z D r x = d ( ~t ) Figure 3: The surplus-time space is divided into three non-overlapping regions: R z , N A z and D by the reinsurance barrier x = K ( z, ~t ) and dividend-payout barrier x = d ( ~t ). Moreover, N D = R z S N A z and the t -axis can be regarded as x = K (1 /λ, ~t ) since R /λ = ∅ .Economically speaking, if the surplus level of an insurance company is low ( x < K ( z, ~t )),then the company must cede risk z with a reinsurance company to avoid bankruptcy. Ifthe surplus level is medium ( K ( z, ~t ) x d ( ~t )), then the company can cover the claim z by itself, but has not enough reserves to pay out as dividend, so no actions will be taken.If the surplus level is very high ( x > d ( ~t )), the company should pay out its extra reservesas dividends to its shareholders. It is never optimal for the company to buy reinsurancecontracts and to pay out reserves as dividend, simultaneously.By Lemma 4.3, K ( z, ~t ) is increasing and the reinsurance region R z is getting larger ascede risk z increases. Economically speaking, if a risk is covered by a reinsurance contract,then any larger risks should be covered as well.Define a uniform non-action region N A = \ z> N A z . In this region the insurer should not cede any risk or pay out dividends.If ˆ z = ess sup Z < ∞ , then K ( z, ~t ) K (ˆ z, ~t ). Therefore, [ z> R z = n ( x, t ) ∈ Q T (cid:12)(cid:12)(cid:12) x < K (ˆ z, ~t ) o = R ˆ z . As K (ˆ z, ~t ) < d ( ~t ), we see that N A = n ( x, t ) ∈ Q T (cid:12)(cid:12)(cid:12) K (ˆ z, ~t ) x < d ( ~t ) o = N A ˆ z = ∅ . ✲✻ ~t xd ( ∞ ) x = K ( z, ~t ) R z x = K (ˆ z, ~t ) R ˆ z N A D r x = d ( ~t ) Figure 4: As z is increasing, the region R z is extending to R ˆ z and the curve x = K ( z, ~t ) ismoving right up to x = K (ˆ z, ~t ). The region N A = N A ˆ z is non-empty when Z is bounded.Economically speaking, when the potential claims are bounded ( ˆ z ) and the surpluslevel is relatively high ( x > K (ˆ z, ~t )), then the insurance company can bear them by itselfwithout asking help from a ceded company.Next, by (4.8), we have ∂ x b I ( z, x, ~t ) − c in R z . This means the insurance company should significantly reduce its purchase of reinsurancecontracts as surplus increases. If its surplus level is very high x > z c − cλ , then we claimthere is not need to buy reinsurance for claim z , that is, ( x, t ) / ∈ R z . In fact, if z /λ , thenthere is nothing to prove since R z = ∅ . Otherwise, suppose ( x, t ) ∈ R z . As ( x, t ) ∈ N D , by(4.8), (4.9) and the mean value theorem, we have b I ( z, x, ~t ) b I ( z, , ~t ) − cx z − /λ − cx , contradicting that ( x, t ) ∈ R z . Therefore, we conclude that R z ⊆ (cid:26) ( x, t ) ∈ Q T (cid:12)(cid:12)(cid:12)(cid:12) x < z c − cλ (cid:27) , and consequently, K ( z, ~t ) z c − cλ . This is illustrated in Figure 5. 18 ✻ ~t xd ( ∞ ) x = K ( z, ~t ) R z x = z c − cλ D r x = d ( ~t ) Figure 5: The line x = z c − cλ gives a universal upper bound for the reinsurance barrier K ( z, ~t ) and the reinsurance region R z when z > /λ .Economically speaking, an insurance company only shares sufficiently large claims witha ceded company. A special case: Discrete claims
In this part we assume the claims follows a discrete probability distribution given by P ( Z = z j ) = p j > , j = 1 , , ..., N, with 0 < z < z < ... < z N and N X j =1 p j = 1 . Define the j th reinsurance boundary as K j ( ~t ) := inf (cid:26) x > (cid:12)(cid:12)(cid:12)(cid:12) − v xx v x ( x, ~t ) z j (cid:27) , j = 1 , , ..., N. And define the j th reinsurance region as R jz = n ( x, t ) ∈ Q T (cid:12)(cid:12)(cid:12) b I ( z j , x, ~t ) > o . Thanks to Lemma 4.3, the properties of these reinsurance boundaries are given in the fol-lowing result. We leave its proof to the interested readers.19 heorem 4.4
The reinsurance boundaries K j ( · ) , j = 1 , , · · · , N , are all continuously dif-ferentiable in time. Moreover, < K ( ~t ) < z − γ c ;0 < K j ( ~t ) − K j − ( ~t ) < z j − z j − c for each j = 2 , ..., N ;lim ~t → K j ( ~t ) = 0 . Figure 6 illustrates this result. ✲✻ ~t xd ( ∞ ) K ( ~t ) K ( ~t ) K N ( ~t ) · · ·· · ·R z R z N A D r x = d ( ~t ) Figure 6: The reinsurance boundaries K j and reinsurance regions R jz are increasing in j .The uniform non-action region N A is N A z N .Economically speaking, the insurer should only cede the claims z j , z j +1 , · · · z N , if thesurplus level is in R jz . 20 ppendix A Proof of Theorem . Denote b V ( x, t ) = v ( x, T − t ). We first prove b V ( x, t ) > V ( x, t ). For any admissible retainedloss policy H t = { H s ( z ) } s > t and a dividend-payout policy L t = { L s } s > t , assume R s is thesolution to (2.3) with the control variables ( H t , L t ), and τ is the ruin time of R s defined by(2.4). Then by Itˆo’s formula, b V ( x, t ) = E " e − c ( T ∧ τ − t ) b V ( R T ∧ τ , T ∧ τ ) + E " Z T ∧ τt e − c ( s − t ) − b V t − b V xx Z ∞ H s ( z ) d F ( z ) − b V x Z ∞ H s ( z )d F ( z ) + γ b V x + c b V ! ( R s , s )d s + E " Z T ∧ τt e − c ( s − t ) b V x ( R s , s )d L cs − E X t s T ∧ τ e − c ( s − t ) ( b V ( R s , s ) − b V ( R s − , s )) , where L cs is the continuous part of L s . The first two expectations are non-negative since b V > − b V t − L b V > b V x >
1, we have E " Z T ∧ τt e − c ( s − t ) b V x ( R s , s )d L cs > E " Z T ∧ τt e − c ( s − t ) d L cs , and as R s R s − , b V ( R s , s ) − b V ( R s − , s ) R s − R s − = L s − − L s . Thus b V ( x, t ) > E " Z T ∧ τt e − c ( s − t ) d L cs + X t s T ∧ τ e − c ( s − t ) ( L s − L s − ) = E " Z T ∧ τt − e − c ( s − t ) d L s . Since the policies are arbitrary chosen, it implies b V ( x, t ) > V ( x, t ).We now prove the reverse inequality b V ( x, t ) V ( x, t ). Define a boundary d ∗ ( s ) = inf { x > | b V x ( x, s ) = 1 } , s ∈ [ t, T ] . Because b V is concave in x by hypothesis, it yields b V x ( x, s ) > , if x < d ∗ ( s ) . By the equation in (3.1), it follows − b V t − L b V = 0 , if x d ∗ ( s ) , (A.1)21here the C , continuity of b V ensures the above equation holds at ( d ∗ ( s ) , s ). Assume R ∗ s isthe solution to problem (2.3) with the feedback controls H ∗ s ( z, x ) = h ∗ z, − b V xx b V x ! and ( L ∗ s − L ∗ s − = R ∗ s − − d ∗ ( s ) , if R ∗ s − > d ∗ ( s ); L ∗ s − L ∗ s − = 0 , if R ∗ s − d ∗ ( s ) . where h ∗ is defined by (3.4). Under these controls, we see that R ∗ s is continuous and R ∗ s d ∗ ( s ) for s > t . Denote τ ∗ be the corresponding ruin time.Now we show the controls defined above are indeed the optimal controls. By Itˆo’s formula, b V ( x, t ) = E " e − c ( T ∧ τ ∗ − t ) b V ( R ∗ T ∧ τ ∗ , T ∧ τ ∗ ) + E " Z T ∧ τ ∗ t e − c ( s − t ) − b V t − b V xx Z ∞ H ∗ s ( z ) d F ( z ) − b V x Z ∞ H ∗ s ( z )d F ( z ) + γ b V x + c b V ! ( R ∗ s , s )d s + E " Z T ∧ τ ∗ t e − c ( s − t ) b V x ( R ∗ s , s )d L ∗ cs − E X t s T ∧ τ ∗ e − c ( s − t ) ( b V ( R ∗ s , s ) − b V ( R ∗ s − , s )) . (A.2)If τ ∗ < T , then R ∗ T ∧ τ ∗ = R ∗ τ ∗ = 0, so b V ( R ∗ T ∧ τ ∗ , T ∧ τ ∗ ) = b V (0 , τ ∗ ) = 0 by the boundarycondition in (3.1). Otherwise b V ( R ∗ T ∧ τ ∗ , T ∧ τ ∗ ) = b V ( R ∗ T , T ) = 0, again by the boundarycondition in (3.1). Therefore, the first expectation in (A.2) is zero.By our choice of H ∗ , we see that − b V t − b V xx Z ∞ H ∗ s ( z, x ) d F ( z ) − b V x Z ∞ H ∗ s ( z, x )d F ( z ) + γ b V x + c b V = − b V t − sup H ∈H b V xx Z ∞ H ( z ) d F ( z ) + b V x Z ∞ H ( z )d F ( z ) ! + γ b V x + c b V = − b V t − L b V .
Because R ∗ s d ∗ ( s ) for s > t and noticing (A.1), we conclude that the second expectationin (A.2) is also zero.Now we are left with b V ( x, t ) = E " Z T ∧ τ ∗ t e − c ( s − t ) b V x ( R ∗ s , s )d L ∗ cs − E X t s T ∧ τ ∗ e − c ( s − t ) ( b V ( R ∗ s , s ) − b V ( R ∗ s − , s )) . L ∗ cs = 0 unless R ∗ s = d ∗ ( s ), so b V x ( R ∗ s , s )d L ∗ cs = b V x ( d ∗ ( s ) , s )d L ∗ cs = d L ∗ cs . Recall that R ∗ s is continuous for s > t , so E X t s T ∧ τ ∗ e − c ( s − t ) ( b V ( R ∗ s , s ) − b V ( R ∗ s − , s )) = E h b V ( R ∗ t , t ) − b V ( R ∗ t − , t ) i . Putting these together, we have b V ( x, t ) = E " Z T ∧ τ ∗ t e − c ( s − t ) d L ∗ cs − E h b V ( R ∗ t , t ) − b V ( R ∗ t − , t ) i = E " Z T ∧ τ ∗ t − e − c ( s − t ) d L ∗ s − E h ( L ∗ t − L ∗ t − ) + b V ( R ∗ t , t ) − b V ( R ∗ t − , t ) i . If R ∗ t − d ∗ ( t ), then L ∗ t − L ∗ t − = 0 and R ∗ t = R ∗ t − , so( L ∗ t − L ∗ t − ) + b V ( R ∗ t , t ) − b V ( R ∗ t − , t ) = 0If R ∗ t − > d ∗ ( t ), then R ∗ t = d ∗ ( t ). Because b V x ( y, t ) = 1 for y > d ∗ ( t ), we also obtain( L ∗ t − L ∗ t − ) + b V ( R ∗ t , t ) − b V ( R ∗ t − , t ) = ( L ∗ t − L ∗ t − ) + R ∗ t − R ∗ t − = 0 . Now we conclude that b V ( x, t ) = E " Z T ∧ τ ∗ t − e − c ( s − t ) d L ∗ s . The right hand side is by definition no more than V ( x, t ). This completes the proof. Appendix B Proof of Theorem . We first prove the existence of a solution to Problem (3.13), and then construct a solution toProblem (3.1), or equivalently (3.8), from it. As mentioned earlier, we adopt the standardpenalty approximation method; see [9, 14, 30].For sufficiently small ε >
0, let β ε ( · ) be a penalty function satisfying β ε ( · ) ∈ C ( −∞ , + ∞ ) , β ε (0) = − c, β ε ( x ) = 0 for x > ε > ,β ε ( · ) , β ′ ε ( · ) > , β ′′ ε ( · ) , lim ε → β ε ( x ) = , if x > , −∞ , if x < . β ε . ✲✻ xy ε − c r r Figure 7: The penalty function β ε .Because the left boundary condition and initial condition in (3.13) are not consistent at(0 , f ε ( t ) ∈ C ([0 , + ∞ )) to satisfy f ε ( t ) = λ, t = 0 , decreasing , t < ε, , t > ε, and consider the following penalty approximation problem, u εt − T u ε + β ε ( u ε −
1) = 0 in Q L,T := (0 , L ) × (0 , T ] , (cid:0) λu ε + u εx (cid:1) (0 , t ) = f ε ( t ) , < t T,u ε ( L, t ) = 1 , < t T,u ε ( x,
0) = 1 , < x < L. (B.1)where L is any fixed positive constant.The following result is useful for our handling the above problem. Remark B.1
Let F ( y, z ) := A (cid:16) − yz (cid:17) , G ( y, z ) := B (cid:16) − yz (cid:17) . Then they are both uniformly Lipschitz continuous in ( −∞ , ∞ ) × [1 , ∞ ) . In fact, from thedefinitions of A ( z ) and B ( z ) , we see that F and G are continuous in ( −∞ , ∞ ) × [1 , ∞ ) .Moreover, by (3.6) , for any z > , ∂ y F ( y, z ) = A ′ (cid:16) − yz (cid:17) − z = (cid:16) − zy (cid:17) (cid:16) − F (cid:16) − zy (cid:17)(cid:17) z ∈ (cid:2) , µ (cid:3) , if y < , if y > , nd ∂ z F ( y, z ) = A ′ (cid:16) − yz (cid:17) yz = (cid:16) − zy (cid:17) (cid:16) − F (cid:16) − zy (cid:17)(cid:17) z ∈ (cid:2) , µ (cid:3) , if y < , if y > . The proof for G is similar. We first establish a comparison principle for Problem (B.1).
Lemma B.2
Suppose u , u ∈ C , (cid:0) Q L,T (cid:1) T C (cid:0) Q L,T (cid:1) satisfy ∂ t u − T u + β ε ( u − ∂ t u − T u + β ε ( u −
1) in Q L,T , (cid:0) λu + ∂ x u (cid:1) (0 , t ) > (cid:0) λu + ∂ x u (cid:1) (0 , t ) , < t T,u ( L, t ) u ( L, t ) , < t T,u ( x, u ( x, , < x < L. If u , u > , then u ( x, t ) u ( x, t ) , ( x, t ) ∈ Q L,T . (B.2) Proof:
Let w ( x, t ) = e x/λ u ( x, t ) , w ( x, t ) = e x/λ u ( x, t ) . Then ∂ t w − e x/λ T ( e − x/λ w ) + e x/λ β ε ( e − x/λ w − ∂ t w − e x/λ T ( e − x/λ w ) + e x/λ β ε ( e − x/λ w − ,∂ x w (0 , t ) > ∂ x w (0 , t ) , < t T,w ( L, t ) w ( L, t ) , < t T,w ( x, w ( x, , < x < L. By Remark B.1, the assumption that u , u > F ( · , · ) and G ( · , · )in T ( e − x/λ w i ) be Lipschitz continuous on w i , w ix , i = 1 ,
2. By the comparison principle fornon linear equations (see [21]), we obtain w w in Q L,T . (cid:3) Lemma B.3
There exists a solution u ε ∈ C , (cid:0) Q L,T (cid:1) to Problem (B.1). Moreover, forsufficiently small ε > , u ε satisfies u ε Ke Λ t x + 1 /λ , in Q L,T , (B.3) where K = L + 1 /λ + 1 , Λ = µ γ + γλ > . roof: Using the Leray-Schauder fixed point theorem (see [13]) and the embedding theorem(see [10]), we get the existence of a C α, α (cid:0) Q L,T (cid:1) (0 < α <
1) solution u ε to Problem (B.1).By the Schauder estimation (see [26]), we also have u ε ∈ C α, α (cid:0) Q L,T (cid:1) .It is left to establish (B.3). Let φ ≡
1, then φ t − A (cid:16) − u εx u ε (cid:17) φ xx − B (cid:16) − u εx u ε (cid:17) φ x + γφ x + cφ + β ε ( φ −
1) = 0 , (cid:0) λφ + φ x (cid:1) (0 , t ) = λ > f ε ( t ) , < t T,φ ( L, t ) = 1 , < t T,φ ( x,
0) = 1 , < x < L, By a similar argument to prove Lemma B.2 and using the comparison principle for linearequations, we have u ε > φ = 1.Let Φ = Ke Λ t / ( x + 1 /λ ). Then, in Q L,T , Φ > ε for sufficiently small ε >
0, so wehave β ε (Φ −
1) = 0. Notice thatΦ t = ΛΦ , Φ x = − Φ x + 1 /λ < , Φ xx = 2Φ( x + 1 /λ ) > , so, by (3.5),Φ t − A (cid:18) − ΦΦ x (cid:19) Φ xx − B (cid:18) − ΦΦ x (cid:19) Φ x + γ Φ x + c Φ + β ε (Φ − > Φ t − µ Φ xx + γ Φ x + c Φ = (cid:18) Λ − µ ( x + 1 /λ ) − γx + 1 /λ + c (cid:19) Φ > . Together with boundary conditions (cid:0) λ Φ + Φ x (cid:1) (0 , t ) = 0 f ε ( t ) , < t T, Φ( L, t ) > , < t T, Φ( x, > , < x L, applying Lemma B.2, we obtain u ε Φ. (cid:3) Before passing to the limit, we give some properties of u ε . Lemma B.4
We have u εt > , (B.4) u εx , (B.5)26 roof: We first prove (B.4). For any 0 < ∆ < T , let e u ε ( x, t ) = u ε ( x, t + ∆), then both e u ε and u ε satisfy the equation in (B.1) in the domain (0 , L ) × (0 , T − ∆]. Moreover, (cid:16) λ e u ε + e u εx (cid:17) (0 , t ) = f ε ( t + ∆ t ) f ε ( t ) = (cid:16) λu ε + u εx (cid:17) (0 , t ) in (0 , L ) × (0 , T − ∆] , e u ε ( L, t ) = u ε ( L, t ) = 1 , < t T − ∆ , e u ε ( x,
0) = u ε ( x, ∆) > u ε ( x, , < x < L. Applying Lemma B.2 we have e u ε > u ε in (0 , L ) × (0 , T − ∆], which implies (B.4).To prove (B.5), we differentiate the equation in (B.1) w.r.t. x and obtain ∂ t u εx − ∂ x (cid:20) A (cid:18) − u εx u ε (cid:19) ∂ x u εx (cid:21) − B (cid:18) − u εx u ε (cid:19) ∂ x u εx − B ′ (cid:18) − u εx u ε (cid:19) ∂ x (cid:18) − u εx u ε (cid:19) u εx + cu εx + β ′ ε ( u ε − u εx = 0 . (B.6)Note that ∂ x (cid:18) − u εx u ε (cid:19) = − u εxx u ε + (cid:18) − u εx u ε (cid:19) , so (B.6) can be written as ∂ t u εx − ∂ x (cid:20)(cid:26) A (cid:18) − u εx u ε (cid:19)(cid:27) ∂ x u εx (cid:21) − (cid:26) B (cid:18) − u εx u ε (cid:19) + B ′ (cid:18) − u εx u ε (cid:19) − u εx u ε (cid:27) ∂ x u εx + ( − B ′ (cid:18) − u εx u ε (cid:19) (cid:18) − u εx u ε (cid:19) + c + β ′ ε ( u ε − ) u εx = 0 . (B.7)It is a linear equation about u εx in divergence form if we regard the terms in {· · · } as knowncoefficients. By (3.5)-(3.7), | A ( y ) | µ , | B ( y ) | µ , | B ′ ( y ) y | | y − (1 − F ( y − )) | µ , (cid:12)(cid:12) B ′ ( y ) y (cid:12)(cid:12) | − F ( y − ) | , | β ′ ε ( u ε − | c, so all the coefficients are bounded. Moreover, since u εx (0 , t ) = ( f ε ( t ) − λu ε (0 , t )) , together with f ε λ λu ε , it implies that u εx (0 , t ) . From u ε > u ε ( L, t ) = 1 we have u εx ( L, t ) . u εx ( x,
0) = 0 , by the maximum principle for weak solution (see [21]), we deduce that u εx (cid:3) Lemma B.5
We have λu ε + u εx > . (B.8) Proof: If u ε (0 , t ) = 1, since u ε > u εx
0, we have u ε ( x, t ) ≡ u εx ( x, t ) ≡ x >
0, so ( λu ε + u εx )( x, t ) ≡ λ >
0, then (B.8) holds.Otherwise u ε (0 , t ) > cu ε + β ε ( u ε − , t ) > c + β ε (0) = 0 . (B.9)Consider the equation in (B.1), i.e. u εt − A (cid:18) − u εx u ε (cid:19) u εxx − B (cid:18) − u εx u ε (cid:19) u εx + γu εx + cu ε + β ε ( u ε −
1) = 0 . Dividing both sides by u εx and using the identity − u εxx u εx = (cid:20) ∂ x (cid:18) − u ε u εx (cid:19) + 1 (cid:21) (cid:18) − u εx u ε (cid:19) , it follows u εt u εx + A (cid:18) − u εx u ε (cid:19) (cid:20) ∂ x (cid:18) − u ε u εx (cid:19) + 1 (cid:21) (cid:18) − u εx u ε (cid:19) − B (cid:18) − u εx u ε (cid:19) + γ + cu ε + β ε ( u ε − u εx = 0 . Denote ν ε = − u ε /u εx . Then A (cid:18) ν ε (cid:19) ν εx + 1 ν ε − B (cid:18) ν ε (cid:19) + γ − cu ε + β ε ( u ε − u ε ν ε = − u εt u εx > ν εx > A (cid:0) ν ε (cid:1) (cid:20) f (cid:18) ν ε (cid:19) ν ε + cu ε + β ε ( u ε − u ε ( ν ε ) (cid:21) , where f ( · ) is defined in (3.9). Notice that cu ε + β ε ( u ε − > c + β ε (0) = 0 , and f ( z ) > z > /λ , so ν εx > ν ε > /λ. (B.10)28oreover, by f ( λ ) = 0, (B.9) and ν ε (0 , t ) = 1 /λ, (B.11)it yields ν εx (0 , t ) > . (B.12)Combining with (B.11), (B.12) and (B.10) we get ν ε > /λ , which implies (B.8). (cid:3) By Lemma B.3, Lemma B.4 and Lemma B.5, we see that | u εx | λu ε Kλ e Λ T in Q L,T . This provides an upper bound for | u εx | , independent of ε , so u ε are uniformly Lipschitzcontinuous in x . Moreover, by Lemma B.4 and Lemma B.5 and the monotonicity of A , A (cid:18) − u εx u ε (cid:19) > A ( λ ) > . (B.13)This confirms the uniform parabolic condition in (B.1) about the equation of u ε (if we regard A ( − u εx /u ε ) and B ( − u εx /u ε ) as known coefficients in the operator T ). Lemma B.6
We have u εxx > . (B.14) Proof:
By the equation in (B.1), Lemma B.3, Lemma B.4, f ( λ ) = 0, and (B.13), we have A (cid:18) − u εx u ε (cid:19) u εxx = u εt − B (cid:18) − u εx u ε (cid:19) u εx + γu εx + cu ε + β ε ( u ε − > (cid:20) − B (cid:18) − u εx u ε (cid:19) + γ (cid:21) u εx > [ − B ( λ ) + γ ] u εx = − λA ( λ ) u εx > . This gives (B.14). (cid:3)
Now we give some uniform norm estimates for u ε . Applying C α, α estimate (see [5] fornon-divergence form) to (B.1), we have | u ε | α, Q L,T C ( | u ε | , Q L,T + | β ε ( · ) | L p ( Q L,T ) + 1) C. Rewrite (B.7) as ∂ t u εx − ∂ x (cid:20) A (cid:18) − u εx u ε (cid:19) ∂ x u εx (cid:21) − (cid:26) B (cid:18) − u εx u ε (cid:19) + B ′ (cid:18) − u εx u ε (cid:19) − u εx u ε (cid:27) ∂ x u εx + ( − B ′ (cid:18) − u εx u ε (cid:19) (cid:18) − u εx u ε (cid:19) + c ) u εx = − ∂ x " β ε ( u ε − . C α, α interior estimate (see [21] for divergence form), we obtain | u εx | α, Q r/ C ( | u εx | , Q r/ + | β ε ( · ) | L p ( Q r/ ) + 1) C. where Q r := Q L,T \ { ( x, t ) | x + t r } . According to Remark B.1, A ( − u εx /u ε ) and B ( − u εx /u ε ) are uniform C α, α in Q r/ . So we canapply W , p interior estimate (see [21]) to (B.1) in Q L,T to obtain | u ε | W , p ( Q r ) C ( | u ε | L p ( Q r/ ) + | β ε ( u ε − | L p ( Q r/ ) + 1) C. We emphasize that C s in the above estimates are independent of ε , so there exits u ∈ W , p, loc ( Q L,T ) T C ( Q L,T ) and a subsequence of u ε (still denoted by u ε ) such that u ε −→ u weakly in W , p ( Q r ) and uniformly in C ( Q L,T ) . Now, set v ( x, t ) = Z x u ( y, t )d y, we come to prove v is a solution to Problem (3.8) in Q L,T . The initial and boundaryconditions are clearly satisfied. Owing to v (0 , t ) = 0 and v t (0 , t ) = 0, together with theboundary condition in (3.13) that ( λu + u x )(0 , t ) = 0, we have ( v t − L v )(0 , t ) = 0. Therefore,( v t − L v )( x, t ) = ( v t − L v )(0 , t ) + Z x ∂ x ( v t − L v )( y, t )d t = Z x ( u t − T u )( y, t )d t > . (B.15)On the other hand, if v x ( x, t ) = u ( x, t ) >
1, then, by (B.5), u ( y, t ) > y ∈ [0 , x ],which implies ( u t −T u )( y, t ) = 0 for y ∈ [0 , x ], thus the inequality in (B.15) becomes equality.Hence v satisfies the variational inequality in Problem (3.8) in Q L,T .Moreover, the estimates (3.14)-(3.19) follow from (B.3), (B.5), (B.14), (B.4) and (B.8).We come to ascertain the order of smoothness of v . Now, we already proved v x = u ∈ W , p ( Q r ) T C (cid:0) Q L,T (cid:1) . The Sobolev embedding theorem implies that v xx = u x ∈ C (cid:0) Q r (cid:1) .Moreover, using the method in [11], we can prove v xt = u t is continuous passing through thefree boundary.Next, we prove the uniqueness. Suppose v , v are two solutions to (3.8). Set N = { ∂ x v > ∂ x v } , then ∂ t v − L v = 0 , ∂ t v − L v > , ( x, t ) ∈ N ,v = v = 0 , ( x, t ) ∈ ∂ N ∩ { x = 0 } ,∂ x v = ∂ x v , ( x, t ) ∈ ∂ N \ ( { x = 0 } ∪ { t = 0 } ∪ { t = T } ) ,v = v = x, ( x, t ) ∈ ∂ N ∩ { t = 0 } . v > v in N ,which implies { ∂ x v > ∂ x v } ⊂ { v > v } , i.e. C := { v < v } ⊂ { ∂ x v ∂ x v } . If C is nonempty, using the fact that v = v on the left boundary of C and ∂ x v ∂ x v in C , we get v v in C , which is impossible.Let x be defined in (4.1) and choose L > x . Using a similar argument in Section4.1 leads to v x ( x, t ) = 1 for x ∈ [ x , L ], so we can extend our solution to the unboundeddomain Q T by setting v ( x, t ) = v ( L, t ) + ( x − L ) for x > L . Then after extension, v ∈ C , ( Q T \ { (0 , } ) T C ( Q T ) is a unique solution to (3.8) in Q T . Moreover, the properties(3.14)-(3.19) remain true in Q T .Furthermore, Remark B.1 and the equation in { v x > } implies v xt = u t ∈ C (cid:0) { v x > } \ { (0 , } (cid:1) , so v xt ∈ C (cid:0) Q T \ { (0 , } (cid:1) . Hence we have v, v x ∈ C (cid:0) Q T (cid:1) , v xx , v t , v xt ∈ C (cid:0) Q T \ { (0 , } (cid:1) . This completes the proof.
Appendix C Proof of Lemma . The equation (3.8) gives v t − A (cid:18) − v xx v x (cid:19) v xx − B (cid:18) − v xx v x (cid:19) v x + γv x + cv = 0 in N D . (C.1)Differentiating (C.1) w.r.t. x and t , respectively, using (3.6) we obtain v tx − (cid:20) A (cid:18) − v xx v x (cid:19) v xxx + B (cid:18) − v xx v x (cid:19) v xx (cid:21) + γv xx + cv x = 0 in N D , (C.2)and v tt − (cid:20) A (cid:18) − v xx v x (cid:19) v xxt + B (cid:18) − v xx v x (cid:19) v xt (cid:21) + γv xt + cv t = 0 in N D . (C.3)Since 0 < A ( λ ) A (cid:18) − v xx v x (cid:19) µ , B (cid:18) − v xx v x (cid:19) µ and A (cid:18) − v xx v x (cid:19) , B (cid:18) − v xx v x (cid:19) ∈ C α,α/ ( Q T )31owing to that v x , v xx ∈ C α,α/ ( Q T ) and Remark B.1), we can apply the Schauder estimateto (C.2) and (C.3), respectively, to obtain v x , v t ∈ C α, α/ ( N D ) . Suppose v xx ( x , t ) = 0 for some ( x , t ) ∈ N D . Because v xx
0, ( x , t ) is a maximizerpoint for v xx . Hence the first order condition gives v xxx ( x , t ) = 0. By (C.2), v tx ( x , t ) + cv x ( x , t ) = 0 , but because of (3.14) and (3.18), this is impossible. Therefore v xx < N D . Appendix D Proof of Lemma . Equation (4.6) is derived from the boundary condition in (3.13). And (4.7) is an immediateconsequence of (4.6) and (4.8). So it is only left to prove (4.8). In
N D , the equation in(3.13) holds, i.e., v xt − A (cid:18) − v xx v x (cid:19) v xxx − B (cid:18) − v xx v x (cid:19) v xx + γv xx + cv x = 0 . Because v xxx = (cid:20) ∂ x (cid:18) − v x v xx (cid:19) + 1 (cid:21) v xx v x , it follows v xt − A (cid:18) − v xx v x (cid:19) (cid:20) ∂ x (cid:18) − v x v xx (cid:19) + 1 (cid:21) v xx v x − B (cid:18) − v xx v x (cid:19) v xx + γv xx + cv x = 0 . Dividing v x yields − A (cid:18) − v xx v x (cid:19) (cid:20) ∂ x (cid:18) − v x v xx (cid:19) + 1 (cid:21) v xx v x − B (cid:18) − v xx v x (cid:19) v xx v x + γ v xx v x + c = − v xt v x , by (3.14) and (3.18). Denote ν = − v x /v xx . Then the above inequality reads A (cid:0) ν (cid:1) ( ν x + 1)2 ν − B (cid:0) ν (cid:1) ν + γν − c > , so ν x > A (1 /ν ) (cid:20)(cid:18) − ν A (cid:18) ν (cid:19) + B (cid:18) ν (cid:19) − γ (cid:19) ν + cν (cid:21) = 1 A (1 /ν ) (cid:20) f (cid:18) ν (cid:19) ν + cν (cid:21) . (D.1)By (3.19), ν > /λ >
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