Analitic approach to solve a degenerate parabolic PDE for the Heston model
aa r X i v : . [ m a t h . A P ] J un ANALYTIC APPROACH TO SOLVE A DEGENERATE PARABOLICPDE FOR THE HESTON MODEL
A. CANALE, R.M. MININNI, AND A. RHANDI
Abstract.
We present an analytic approach to solve a degenerate parabolic problemassociated to the Heston model, which is widely used in mathematical finance to derivethe price of an European option on an risky asset with stochastic volatility. We give avariational formulation, involving weighted Sobolev spaces, of the second order degenerateelliptic operator of the parabolic PDE. We use this approach to prove, under appropriateassumptions on some involved unknown parameters, the existence and uniqueness of weaksolutions to the parabolic problem on unbounded subdomains of the half-plane. Introduction
Heston in [7] derived the pricing formula of a stock European option when the priceprocess { S t , t ≥ } of the underlying asset satisfies the following stochastic differentialequation (SDE)(1.1) dS t = η S t dt + p Y t S t dW t , t ≥ , where the constant parameter η ∈ R denotes the instantaneous mean return of the under-lying asset, and, contrary to the original Black and Scholes model for European options[3], the non-constant volatility √ Y t is supposed to be stochastic. The variance process Y = { Y t , t ≥ } is assumed to be a diffusion process whose dynamics is described by thefollowing SDE(1.2) dY t = κ ( m − Y t ) dt + σ p Y t dZ t , t ≥ , used in mathematical finance by Cox et al. [4] to model “short-term interest rates” ofzero-coupon bonds. The parameters κ, m and σ are supposed to be positive constants. Theprocess Y is known in literature as CIR process or square-root process . In particular, m is the long-run mean value of Y t as t → ∞ , κ is called the “rate of mean reversion” thatis, κ determines the speed of adjustment of the sample paths of Y toward m , and σ is theconstant volatility of variance (often called the volatility of volatility ). The state space ofthe diffusion Y is the interval [0 , ∞ ). Mathematics Subject Classification.
Key words and phrases.
European option, degenerate parabolic PDE, stochastic volatility process, Hestonmodel, mathematical finance, variational formulation, weighted Sobolev spaces, semigroup of operators.The third author has been supported by the M.I.U.R. research project Prin 2010MXMAJR.
The processes { W t , t ≥ } and { Z t , t ≥ } in (1.1) and (1.2) are standard one-dimensionalBrownian motions. They are supposed to be correlated dW t dZ t = ρ dt, where ρ ∈ [ − ,
1] denotes the instantaneous correlation coefficient.Using the two-dimensional Ito’s formula (cf., for example, [13, Chap. IV.32]), the price U of an European option with a risky underlying asset, fixed maturity date T >
K > (1.3) ∂U∂t + 12 yS ∂ U∂S + 12 yσ ∂ U∂y + ρσyS ∂ U∂S∂y + κ ( m − y ) ∂U∂y + r ( S ∂U∂S − U ) = 0 , in [0 , T ) × [0 , ∞ ) U ( T, S, y ) = h ( S ) in [0 , ∞ ) , with the final pay-off of the option as the terminal condition, namely h ( S ) = ( S − K ) + or h ( S ) = ( K − S ) + corresponding to European call and put options, respectively. The price U := U ( t, S, y )depends on time t , on the stock price variable S and on the variance variable y .The degenerate parabolic problem (1.3) is obtained imposing some assumptions aboutthe financial market, as the no-arbitrage condition i.e., given the evolutions of S t and of Y t ,the European option is priced in such a way that there are no opportunities to make moneyfrom nothing.The PDE in (1.3) has degenerate coefficients in the S variable and possibly also in the y variable. In order to remove the degeneracy with respect to the variable S , we define thestochastic process { X t , t ≥ } as follows X t = ln (cid:18) S t S (cid:19) , t ≥ . Further, consider the following function e u ( t, S, y ) := U ( t, S, y ) − e − r ( T − t ) h ( Se r ( T − t ) ) , which indicates the excess to discounted pay-off. The parameter r ≥ e u decays to zero as S → S → ∞ .Then, by changing the time t → T − t , setting x = ln S (assume S = 1), and using thefollowing transformation(1.4) u ( t, x, y ) := e − ω y e u ( T − t, S, y ) , ω > , we deduce from (1.3) that the function u satisfies the following initial value forward parabolicproblem(1.5) ∂u∂t ( t, x, y ) = − ( L H u )( t, x, y ) + F ( t, y ) , t ∈ (0 , T ] , ( x, y ) ∈ Ω u (0 , x, y ) = 0 , ( x, y ) ∈ Ω , ESTON MODEL 3 where Ω = R × [0 , ∞ ). The operator L H is given by( L H ϕ )( x, y ) = − y ∂ ϕ∂x − σ y ∂ ϕ∂y − ρσy ∂ ϕ∂x∂y − ( ωρσy − y + r ) ∂ϕ∂x − [ ωσ y + κ ( m − y )] ∂ϕ∂y − (cid:20) ωσ y ( ωy + 1) + ωyκ ( m − y ) − r (cid:21) ϕ (1.6)and F ( t, y ) = K ye − rt e − ω y δ ln K − rt . The motivation to consider the transformation (1.4) is explained in [8], taking into accountthat the price U remains bounded for all y (cf. [7]).To our knowledge, the use of a variational approach to prove existence and uniquenessof solutions to these pricing problems is very recent. Achdou et al. [1]-[2] used variationalanalysis using appropriate weighted Sobolev spaces to solve parabolic problems connectedto option pricing when the variance process Y is a function of a mean reverting Ornstein-Uhlenbech (OU) process. Successively, proceedings as in the previous works, Hilber et al. [8]used variational formulation to present numerical solutions by a sparse wavelet finite elementmethod to pricing problems in terms of parabolic PDEs when the volatility is modeled by aOU process or a CIR process. Daskalopoulos and Feehan [5] used variational analysis withthe aid of weighted Sobolev spaces to prove the existence, uniqueness and global regularityof solutions to obstacle problems for the Heston model, which in mathematical financecorrespond to solve pricing problems for perpetual American options on underlying riskyassets.Observe that by applying a space-time transformation, the diffusion Y follows the dy-namics of a squared Bessel process with dimension α = 4 κmσ > α > α -dimensional squared Bessel process starting from a positive initial point stays strictly pos-itive and tends to infinity almost surely as time approaches infinity while, for α = 2 theprocess is strictly positive but gets arbitrarily close to zero and ∞ , and for α > Y , we assume the condition(1.7) κm > σ . Simulation studies to investigate numerically how the effect of varying the correlation ρ (cf.[6]) and the volatility parameter σ (cf. [11]) impacts on the shape of the implied volatilitycurve in the Heston model, clearly show that under the condition (1.7) the volatility √ Y t always remains strictly positive. A. CANALE, R.M. MININNI, AND A. RHANDI
Thus, the above arguments let us to assume y ∈ [ a, ∞ ) with a sufficiently small a >
0, inorder to remove the degeneracy at zero with respect to the variable y and take Ω = R × [ a, ∞ )in (1.5).By using the variational formulation of the parabolic PDE in (1.5) performed in [8], theaim of the present paper is to use form methods to prove the existence and uniqueness ofa weak solution to the problem (1.5) and to study the existence of a positive and analyticsemigroup generated by − L H , with an appropriate domain, in a weighted L -space withsuitable weights φ and ψ .The article is organized as follows. In Section 2 we define the Hilbert and weightedSobolev spaces we shall need throughout this article, describe our assumptions on theHeston operator coefficients and prove the continuity estimate for the sesquilinear formdefined by the operator L H given in (1.6), with Dirichlet boundary conditions. In Section3 we derive Garding’s inequality for the sesquilinear form, and deduce the existence of aunique weak solution to the problem (1.5). We obtain also that the realization of − L H in L with Dirichlet boundary conditions generates an analytic semigroup ( e − t L H ). Thepositivity of ( e − t L H ) can be proved applying the first Beurling-Deny criteria.2. Heston model: the variational formulation
Throughout this article, the coefficients of the operator L H are required to obey theFeller condition (1.7) and Ω = R × [ a, ∞ ) with some positive constant a .We propose to use form methods to solve the parabolic PDE in (1.5). To this purposewe consider the weight functions φ ( x ) = e ν | x | , ψ ( y ) = e µ y , ( x, y ) ∈ Ω , ν, µ > , and define the Hilbert space L φ,ψ (Ω) = { v | ( x, y ) v ( x, y ) φ ( x ) ψ ( y ) ∈ L (Ω) } equipped with the weighted L -norm k v k φ,ψ = (cid:18)Z Ω | v ( x, y ) | φ ( x ) ψ ( y ) dx dy (cid:19) . Furthermore we define the weighted Sobolev space V φ,ψ = n v (cid:12)(cid:12)(cid:12) (cid:16) v, √ y ∂v∂x , √ y ∂v∂y (cid:17) ∈ ( L φ,ψ (Ω)) o . The space V φ,ψ is equipped with the norm k u k V φ,ψ = (cid:16) k u k φ,ψ + (cid:13)(cid:13)(cid:13) √ y ∂u∂x (cid:13)(cid:13)(cid:13) φ,ψ + (cid:13)(cid:13)(cid:13) √ y ∂u∂y (cid:13)(cid:13)(cid:13) φ,ψ (cid:17) . The sesquilinear form associated to L H in L φ,ψ (Ω) is given by(2.1) a φ,ψH ( u, v ) = Z Ω ( L H u )( x, y ) v ( x, y ) φ ( x ) ψ ( y ) dx dy, u, v ∈ C ∞ c (Ω) . We note first the following standard result.
Lemma 2.1.
The following assertions hold:
ESTON MODEL 5 (a) The space of test functions C ∞ c (Ω) is dense in L φ,ψ (Ω) ,(b) the space V φ,ψ equipped with the norm k · k V φ,ψ is a Hilbert space.Proof. Let u ∈ L φ,ψ (Ω). Then uφψ ∈ L (Ω) and so, for any ε > ϕ ∈ C ∞ c (Ω)such that k ϕ − uφψ k L = k φ − ψ − ϕ − u k φ,ψ < ε . Since φ − ψ − ϕ ∈ C c (Ω), we deduce that C c (Ω) is dense in L φ,ψ (Ω). Thus the assertion (a) follows by standard mollifier argument.To prove (b) we have only to show that V φ,ψ equipped with the norm k · k V φ,ψ is complete.Consider a Cauchy sequence ( u n ) in ( V φ,ψ , k · k V φ,ψ ). Since y ≥ a , it follows that V φ,ψ iscontinuously embedded in the classical weighted Sobolev space H φ,ψ (Ω) := n v (cid:12)(cid:12)(cid:12) (cid:16) v, ∂v∂x , ∂v∂y (cid:17) ∈ ( L φ,ψ (Ω)) o . Hence, u n converges to some u ∈ H φ,ψ (Ω). On the other hand, by the convergence of √ y ∂u n ∂x and √ y ∂u n ∂y in L φ,ψ (Ω) (and hence a.e. by taking a subsequence), it follows that u ∈ V φ,ψ and u n converges to u with respect to the norm k · k V φ,ψ . (cid:3) The following lemma shows that a φ,ψH can be extended continuously to a sesquilinear formon V φ,ψ × V φ,ψ , where V φ,ψ denotes the closure of C ∞ c (Ω) in V φ,ψ Lemma 2.2.
There is a positive constant M such that | a φ,ψH ( u, v ) | ≤ M k u k V φ,ψ k v k V φ,ψ , ∀ u, v ∈ V φ,ψ . Proof.
Integrating by parts, it follows from (2.1) that a φ,ψH ( u, v ) = 12 Z Ω y ∂u∂x ∂v∂x φ ψ + Z Ω y ∂u∂x v (cid:18) φ ′ φ (cid:19) φ ψ + σ Z Ω y ∂u∂y ∂v∂y φ ψ + σ Z Ω ∂u∂y vφ ψ + µσ Z Ω y ∂u∂y vφ ψ + 2 ρσ Z Ω y ∂u∂y v (cid:18) φ ′ φ (cid:19) φ ψ + ρσ Z Ω y ∂u∂y ∂v∂x φ ψ − Z Ω ( ωρσy − y + r ) ∂u∂x vφ ψ − Z Ω [ ωσ y + κ ( m − y )] ∂u∂y vφ ψ − Z Ω (cid:20) ωσ y ( ωy + 1) + ωyκ ( m − y ) − r (cid:21) uvφ ψ holds for u, v ∈ C ∞ c (Ω). By H¨older’s inequality, and since ya ≥ y ∈ [ a, ∞ ) , a >
0, wehave (cid:12)(cid:12)(cid:12)(cid:12)Z Ω y ∂u∂x ∂v∂x φ ψ (cid:12)(cid:12)(cid:12)(cid:12) ≤ k u k V φ,ψ k v k V φ,ψ , (cid:12)(cid:12)(cid:12)(cid:12)Z Ω y ∂u∂y ∂v∂y φ ψ (cid:12)(cid:12)(cid:12)(cid:12) ≤ k u k V φ,ψ k v k V φ,ψ , (cid:12)(cid:12)(cid:12)(cid:12)Z Ω y ∂u∂y ∂v∂x φ ψ (cid:12)(cid:12)(cid:12)(cid:12) ≤ k u k V φ,ψ k v k V φ,ψ , (cid:12)(cid:12)(cid:12)(cid:12)Z Ω ∂u∂x vφ ψ (cid:12)(cid:12)(cid:12)(cid:12) ≤ √ a k u k V φ,ψ k v k V φ,ψ , and (cid:12)(cid:12)(cid:12)(cid:12)Z Ω ∂u∂y vφ ψ (cid:12)(cid:12)(cid:12)(cid:12) ≤ √ a k u k V φ,ψ k v k V φ,ψ . A. CANALE, R.M. MININNI, AND A. RHANDI
Since ψ ′ ( y ) = µyψ ( y ), it follows that Z Ω yuvφ ψ = − µ (cid:18)Z Ω ∂u∂y vφ ψ + Z Ω u ∂v∂y φ ψ (cid:19) , (2.2) Z Ω y uvφ ψ = − µ (cid:18)Z Ω y ∂u∂y vφ ψ + Z Ω y ∂v∂y uφ ψ + Z Ω uvφ ψ (cid:19) , (2.3) Z Ω y uvφ ψ = − µ (cid:18) Z Ω yuvφ ψ + Z Ω y ∂u∂y vφ ψ + Z Ω y ∂v∂y uφ ψ (cid:19) . (2.4)Thus it suffices to estimate the integrals Z Ω y ∂u∂y vφ ψ , Z Ω y ∂u∂y vφ ψ , Z Ω y ∂u∂x vφ ψ , and Z Ω y ∂u∂x vφ ψ . Applying (2.2) and H¨older’s inequality we have (cid:12)(cid:12)(cid:12)(cid:12)Z Ω y ∂u∂y vφ ψ (cid:12)(cid:12)(cid:12)(cid:12) ≤ k u k V φ,ψ k√ yv k φ,ψ ≤ aµ k u k V φ,ψ k v k V φ,ψ , (cid:12)(cid:12)(cid:12)(cid:12)Z Ω y ∂u∂x vφ ψ (cid:12)(cid:12)(cid:12)(cid:12) ≤ k u k V φ,ψ k√ yv k φ,ψ ≤ aµ k u k V φ,ψ k v k V φ,ψ . On the other hand, applying again H¨older’s inequality we get (cid:12)(cid:12)(cid:12)(cid:12)Z Ω y ∂u∂y vφ ψ (cid:12)(cid:12)(cid:12)(cid:12) ≤ k u k V φ,ψ k y v k φ,ψ and (cid:12)(cid:12)(cid:12)(cid:12)Z Ω y ∂u∂x vφ ψ (cid:12)(cid:12)(cid:12)(cid:12) ≤ k u k V φ,ψ k y v k φ,ψ . It remains to estimate k y v k φ,ψ . It follows from (2.4) that k y v k φ,ψ ≤ µ (cid:12)(cid:12)(cid:12)(cid:12)Z Ω y ∂v∂y vφ ψ (cid:12)(cid:12)(cid:12)(cid:12) ≤ k y v k φ,ψ + 12 µ k√ y ∂v∂y k φ,ψ . Hence, k y v k φ,ψ ≤ µ k√ y ∂v∂y k φ,ψ . This ends the proof of the lemma. (cid:3) Existence and uniqueness of solutions to the variational equation
The following proposition deals with the quasi-accretivity of the sesquilinear form a φ,ψH . Proposition 3.1.
Assume that (1.7) is satisfied. Then, under appropriate conditions on ρ, ν, µ and ω , there are constants c > and c ∈ R such that (3.1) ℜ a φ,ψH ( v, v ) ≥ c k v k V φ,ψ + c k v k φ,ψ , ∀ v ∈ V φ,ψ . ESTON MODEL 7
Proof.
The real part of the quadratic form a φ,ψH ( v, v ) is given by ℜ a φ,ψH ( v, v ) = 12 Z Ω y (cid:12)(cid:12)(cid:12) ∂v∂x (cid:12)(cid:12)(cid:12) φ ψ + σ Z Ω y (cid:12)(cid:12)(cid:12) ∂v∂y (cid:12)(cid:12)(cid:12) φ ψ + ℜ (cid:18)Z Ω y ∂v∂x v (cid:18) φ ′ φ (cid:19) φ ψ (cid:19) + ℜ (cid:18)Z Ω (cid:16) y − ωρσy − r (cid:17) ∂v∂x vφ ψ (cid:19) + ℜ (cid:18)Z Ω (cid:16) σ − κm ) (cid:17) ∂v∂y vφ ψ (cid:19) + κ ℜ (cid:18)Z Ω y ∂v∂y v φ ψ (cid:19) − ωσ ℜ (cid:18)Z Ω y ∂v∂y vφ ψ (cid:19) + σ µ ℜ (cid:18)Z Ω y ∂v∂y vφ ψ (cid:19) + ρ σ ℜ (cid:18)Z Ω y ∂v∂x ∂v∂y φ ψ (cid:19) + 2 ρσ ℜ (cid:18)Z Ω y ∂v∂y v (cid:18) φ ′ φ (cid:19) φ ψ (cid:19) − ω σ Z Ω y | v | φ ψ − ( ωκm + ωσ Z Ω y | v | φ ψ + ωκ Z Ω y | v | φ ψ + r Z Ω | v | φ ψ = I + I + I + I + I + I + I + I + I + I + I + I + I + I . (3.2)By the definition of the L φ,ψ - norm(3.3) I + I = 12 (cid:13)(cid:13)(cid:13) √ y ∂v∂x (cid:13)(cid:13)(cid:13) φ,ψ + σ (cid:13)(cid:13)(cid:13) √ y ∂v∂y (cid:13)(cid:13)(cid:13) φ,ψ . To estimate the next integrals we use H¨older’s and Young’s inequalities as well as integrationby parts taking in mind that ℜ (cid:0) ∂v∂x v (cid:1) = ∂ | v | ∂x , ℜ (cid:16) ∂v∂y v (cid:17) = ∂ | v | ∂y , φ ′ = ( sign x ) νφ and ψ ′ = µyψ . • Estimate of I :(3.4) | I | ≤ ǫ (cid:13)(cid:13)(cid:13) √ y ∂v∂x (cid:13)(cid:13)(cid:13) φ,ψ + ν ǫ k√ yv k φ,ψ , ǫ > . • Estimate of I :(3.5) | I | ≤ ν k√ yv k φ,ψ + ωρσν Z Ω y | v | φ ψ + rν k v k φ,ψ . • Estimate of I :(3.6) I = (cid:0) κm − σ (cid:1) µ k√ yv k φ,ψ . • Estimate of I and I :(3.7) I + I ≥ − (cid:16) κ ρσν (cid:17) k v k φ,ψ − ( κµ + 2 ρσνµ ) Z Ω y | v | φ ψ . A. CANALE, R.M. MININNI, AND A. RHANDI • Estimate of I and I :(3.8) I + I = σ ( µ − ω ) ℜ (cid:18)Z Ω y ∂v∂y vφ ψ (cid:19) . • Estimate of I :(3.9) | I | ≤ ǫ (cid:13)(cid:13)(cid:13) √ y ∂v∂x (cid:13)(cid:13)(cid:13) φ,ψ + ρ σ ǫ (cid:13)(cid:13)(cid:13) √ y ∂v∂y (cid:13)(cid:13)(cid:13) φ,ψ , ǫ > . On the other hand, it follows from (2.4) that(3.10) k√ yv k φ,ψ = −ℜ (cid:18)Z Ω y ∂v∂y vφ ψ (cid:19) − µ k y v k φ,ψ . It follows from (3.2)-(3.10) that ℜ a φ,ψH ( v, v ) ≥ α (cid:13)(cid:13)(cid:13) √ y ∂v∂x (cid:13)(cid:13)(cid:13) φ,ψ + α (cid:13)(cid:13)(cid:13) √ y ∂v∂y (cid:13)(cid:13)(cid:13) φ,ψ + α k v k φ,ψ + α Z Ω y | v | φ ψ + α ℜ (cid:18)Z Ω y ∂v∂y vφ ψ (cid:19) + α k y v k φ,ψ , where α = (cid:16) − ǫ − ǫ (cid:17) , α = σ (cid:0) − ρ ǫ (cid:1) =: σ τ , α = ( − rν − κ − ρσν + r ), α = ( ωκ − κµ − ωρσν − ρσνµ ), α = ω (cid:16) κm − σ (cid:17) + σ µ + β − (cid:16) κm − σ (cid:17) µ , α = ωµ (cid:16) κm + σ (cid:17) − ω σ + µ (cid:16) β − (cid:16) κm − σ (cid:17) µ (cid:17) = µα + ωµσ − σ µ − ω σ and β = (cid:16) ν ǫ + ν (cid:17) . In order to ensure that the coefficients α , α are nonnegative we use the assumption | ρ | < ǫ and ǫ such that ρ < ǫ < − ǫ . Furthermore we take ω > µ , and(3.11) ν ≤ κ ( ω − µ ) ρσ ( ω + 2 µ ) , when 0 < ρ < α ≥ | ρ | < ESTON MODEL 9
To prove the lemma, we need first to show that (cid:12)(cid:12)(cid:12) R Ω y ∂v∂y vφ ψ (cid:12)(cid:12)(cid:12) can be estimated by (cid:13)(cid:13)(cid:13) √ y ∂v∂y (cid:13)(cid:13)(cid:13) φ,ψ . Indeed, by means of H¨older’s and Young’s inequalities, (cid:12)(cid:12)(cid:12) Z Ω y ∂v∂y vφ ψ (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) Z Ω √ y ∂v∂y y vφ ψ (cid:12)(cid:12)(cid:12) ≤ ǫ (cid:13)(cid:13)(cid:13) √ y ∂v∂y (cid:13)(cid:13)(cid:13) φ,ψ + 12 ǫ k y v k φ,ψ (3.12)with any ǫ > κm > σ and ω > µ , we deduce α > ω > (cid:16) κm − σ (cid:17) µ − βκm − σ . So, by (3.12), we obtain ℜ a φ,ψH ( v, v ) ≥ α (cid:13)(cid:13)(cid:13) √ y ∂v∂x (cid:13)(cid:13)(cid:13) φ,ψ + (cid:16) α − α ǫ (cid:17)(cid:13)(cid:13)(cid:13) √ y ∂v∂y (cid:13)(cid:13)(cid:13) φ,ψ + α k v k φ,ψ + (cid:16) α − α ǫ (cid:17) k y v k φ,ψ . (3.13)Choosing(3.14) ǫ < α α , we deduce that α − α ǫ > α − α ǫ ≥ . This is equivalent to show that ω satisfies the inequality σ ω − h(cid:16) κm − σ (cid:17)(cid:16) µ − ǫ (cid:17) + σ µ i ω + σ µ − β (cid:16) µ − ǫ (cid:17) ++ (cid:16) κm − σ (cid:17)(cid:16) µ − ǫ (cid:17) µ − σ µ (cid:16) µ − ǫ (cid:17) ≤ . (3.16)So we need to have∆ ω := (cid:16) κm − σ (cid:17) (cid:16) µ − ǫ (cid:17) + µσ (cid:16) µ − ǫ (cid:17) + 2 βσ (cid:16) µ − ǫ (cid:17) ≥ . Let us observe that (3.15) can be rewritten in the following way (cid:18) µ − ǫ (cid:19) α + ωµσ − σ µ − ω σ ≥ , from which we can deduce that ǫ > µ , since ω σ − ωµσ + σ µ = σ (cid:0) ( ω − µ ) + µ (cid:1) >
0. Thus,(3.17) ∆ ω ≥ ⇐⇒ (cid:16) κm − σ (cid:17) ≥ ǫ µ ǫ µ − (cid:20) − ǫ µ ǫ µ − − βµσ (cid:21) σ =: g (2 ǫ µ ) σ , where g ( t ) = (2 + c ) t − (1 + c ) t ( t − with c = βµσ . On the other hand, by (1.7), there exists δ > κm > (1 + 2 √ δ ) σ .Thus, it follows that(3.18) (cid:18) κm − σ (cid:19) > δσ . Hence, (3.17) holds if g (2 ǫ µ ) ≤ δ . An easy computation shows that if(3.19) 2 ǫ µ > t := 1 + 1 √ δ then g (2 ǫ µ ) < δ and therefore ∆ ω >
0. On the other hand, it follows from (3.14) and(3.19) that α < µα t and therefore, using (3.11),(3.20) µ < ω < (cid:16) κm − σ (cid:17) µ + t µα − βκm − σ = µ + γσ µ − βκm − σ , where γ = τt −
1. This implies in particular that γ > µ > βγσ . Thus, using conditions (3.11) and (3.20), we deduce that (3.16) holds if ω ∈ ( M, N ), where M = max (cid:16) κm − σ (cid:17)(cid:16) µ − ǫ (cid:17) + σ µ − √ ∆ ω σ , µ and N = min (cid:16) κm − σ (cid:17)(cid:16) µ − ǫ (cid:17) + σ µ + √ ∆ ω σ , µ + γσ µ − βκm − σ . Let us observe that (cid:16) κm − σ (cid:17)(cid:16) µ − ǫ (cid:17) + σ µ > √ ∆ ω if and only if(3.22) β < µ (cid:16) κm − σ (cid:17) + µσ ǫ µ − . ESTON MODEL 11
Moreover, it is easy to see that µ ≤ N .To get (cid:16) κm − σ (cid:17)(cid:16) µ − ǫ (cid:17) + σ µ − √ ∆ ω σ < µ + γσ µ − βκm − σ or, equivalently,(3.23) (cid:16) κm − σ (cid:17)(cid:16) µ − ǫ (cid:17) − σ κm − σ (cid:0) γσ µ − β (cid:1) < √ ∆ ω , we firstly require that(3.24) (cid:16) κm − σ (cid:17) ≥ ǫ µ ǫ µ − (cid:18) γ − βµσ (cid:19) σ =: f (2 ǫ µ ) σ to have that the left side in (3.23) is nonnegative.It follows from (3.19) and (3.21) that 0 < f (2 ǫ µ ). Thus, from (3.18) we obtain (3.24) if f (2 ǫ µ ) ≤ δ . From the definition of t and since τ < δ > γ . Using again τ <
1, we obtain 2 τ − < < √ δ = δ √ δ + √ δ and so, γ − c < γ = 2 τt − < δt √ δ = δ √ δ . Hence, √ δ < δγ − ( c/ − . This implies that t > δδ − γ +( c/ . This together with (3.19) imply that f (2 ǫ µ ) ≤ δ. Thus,(3.24) holds.Using now the definition of ∆ ω , one can see that proving (3.23) is equivalent to show(3.25) (cid:16) κm − σ (cid:17) > ǫ µ ǫ µ (1 + 2 γ ) − (2 + 2 γ ) (cid:18) γ − βµσ (cid:19) σ = ˜ f (2 ǫ µ ) σ . Since t < t < inf γ ∈ (0 , t − (cid:18) γ (cid:19) = 44 − t , one deduces that ˜ f (2 ǫ µ ) < t < ǫ µ < γ . Therefore, if ω ∈ ( M, N ) , ν satisfies (3.11) when ρ >
0, and β < min (cid:26) µγσ , µ (cid:18) κm − σ (cid:19) + µσ ǫ µ − (cid:27) from (3.21) and (3.22), with 0 < γ < δ . Then (3.13) can be written as ℜ a φ,ψH ( v, v ) ≥ c k v k V φ,ψ + α k v k φ,ψ , ∀ v ∈ C ∞ c (Ω) , provided that2 ρ − t < ǫ < − ǫ and ǫ ∈ (cid:18) t µ , min (cid:26) α α , µ (cid:18) γ (cid:19)(cid:27)(cid:19) , where c := min { α , α − α ǫ , a ( α − α ǫ ) } >
0. We note that the above first inequalitysatisfied by ǫ is a consequence of γ >
0. On the other hand, by assuming | ρ | < q − √ δ ,there exists a ǫ satisfying the above condition, since q − √ δ = q − t . (cid:3) Remark 3.2.
It follows from Lemma 2.2 and Proposition 3.1 that the form norm definedby k u k a H := q ℜ a φ,ψH ( u, u ) + (1 − c ) k u k φ,ψ , is equivalent to the norm k·k V φ,ψ . So, by Lemma 2.1, the sesquilinear form a φ,ψH with domain V φ,ψ is closed.We define the operator associated to a φ,ψH by D ( A ) = { u ∈ V φ,ψ s.t. ∃ v ∈ L φ,ψ (Ω) : a φ,ψH ( u, ϕ ) = Z Ω vϕφ ψ , ∀ ϕ ∈ C ∞ c (Ω) } Au = v. The estimate (3.1) is known as Garding’s inequality. Applying [10, Section 4.4, Theorem4.1] we obtain the existence of a unique weak solution to the problem (1.5).
Theorem 3.3.
Assume the same conditions as in Proposition 3.1. Then, there is a uniqueweak solution u ∈ L ([0 , T ] , V φ,ψ ) ∩ C ([0 , T ] , L φ,ψ (Ω)) to the parabolic problem (1.5) . Applying the Lumer-Phillips theorem we obtain the following generation result.
Theorem 3.4.
Assume the same conditions as in Proposition 3.1. Then, the operator − A defined above generates a positivity preserving and quasi-contractive analytic semigroup on L φ,ψ (Ω) .Proof. It follows form Lemma 2.1, Lemma 2.2, Proposition 3.1 and Remark 3.2 that the form a φ,ψH with domain V φ,ψ is densely defined, closed, continuous and quasi-accretive sesquilinearform on L φ,ψ (Ω). Thus, − A generates a quasi-contractive analytic semigroup ( e − tA ) t ≥ on L φ,ψ (Ω) (cf. [12, Theorem 1.52]).For the positivity, we note first that the semigroup ( e − tA ) t ≥ is real and one can see thatfor every u ∈ D ( a φ,ψH ) ∩ L φ,ψ (Ω , R ) , u + ∈ D ( a φ,ψH ) and a φ,ψH ( u + , u − ) = 0, since u − = ( − u ) + and ∇ u + = χ { u> } ∇ u (cf. [12, Proposition 4.4]). Thus, by the first Beurling-Deny criteria,( e − tA ) t ≥ is a positivity preserving semigroup on L φ,ψ (Ω) (cf. [12, Theorem 2.6]). (cid:3) ESTON MODEL 13
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E-mail address : [email protected] Dipartimento di Matematica Universit`a degli Studi di Bari A. Moro, Via E. Orabona 4,70125 Bari, Italy.
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