The method used here is one of the most efficient functional analysis methods in solving partial differential equations with integral conditions, the so-called a priori estimate method or the energy-integral method. This method is essentially based on the construction of multiplicators for each specific given problem, which provides the a priori estimate from which it is possible to establish the solvability of the posed problem. More precisely, the proof is based on an energy inequality and the density of the range of the operator generated by the abstract formulation of the stated problem. But here we use the energy inequality method for the equivalent problem (PR) given in Lemma 1; so to investigate the posed problem, we introduce the needed function spaces.
We introduce function spaces needed in our investigation. We denote by Lρ2(Ω) the weighted Lebesgue space that consists of all measurable functions u equipped with the finite norm
(14)∥u∥Lρ2(Ω)=(∫Ωρ(x)|u(x,·)|2dx)1/2,
where ρ(x)=(x2+x). If ρ(x)=1, Lρ2(Ω) are identified with the standard spaces L2(Ω).
In this paper, we prove the existence and the uniqueness for solution of the problem (1)–(4) as a solution of the operator equation
(15)Lu=ℱ,
where L=(ℒ,ℓ), with domain of definition B consisting of functions u∈Lρ2(Ω) such that ∂u/∂t, ∂u/∂x, ∂2u/∂x2, ∂2u/∂t∂x∈Lρ2(Ω), and u satisfy condition (4); the operator L is considered from B to F, where B is the Banach space consisting of all functions u(x,t) having a finite norm
(16)∥u∥B2=∫Ω(x2+x)(∂u∂t)2dx dt +sup0≤τ≤T∫01(x2+x)(∂u(x,τ)∂x)2dx,
and F is the Hilbert space consisting of all elements ℱ=(f,φ) for which the norm
(17)∥ℱ∥F2=∫Ωf2dx dt+∫01(x2+x)(∂φ∂x)2dx
is finite.
Proof.
Multiplying (1) by the following Mu:
(19)Mu={(x2-x)∂u∂x+(x2+x)∂u∂t,0≤x≤α,-x∂u∂x+(x2+x)∂u∂t,α≤x≤β,(x2-x)∂u∂x+(x2+x)∂u∂t,β≤x≤1,
and integrating over Ωτ, where Ωτ=(0,1)×(0,τ),
(20)∫Ωτℒu·Mudx dt ={∫Ωατ(x2-x)∂u∂x∂u∂tdx dt +∫Ωατ(1-x)(∂u∂x)2dx dt +∫Ωατ(x2+x)(∂u∂t)2dx dt -∫Ωατ(x-x2)∂u∂x∂2u∂x2dx dt -∫Ωατ(x+1)∂∂x(x∂u∂x)∂u∂tdx dt =∫Ωατf·((x2-x)∂u∂x +(x2+x)∂u∂t(x2-x)∂u∂x)dx dt, Ωατ,∫Ωα,βτ(x2+x)(∂u∂t)2dx dt -∫Ωα,βτx∂u∂x∂u∂tdx dt+∫Ωα,βτ(∂u∂x)2dx dt +∫Ωα,βτx∂u∂x∂2u∂x2dx dt -∫Ωα,βτ(x+1)∂∂x(x∂u∂x)∂u∂tdx dt =∫Ωα,βτf·(-x∂u∂x +(x2+x)∂u∂t)dx dt, Ωα,βτ,∫Ωβτ(x2+x)(∂u∂t)2dx dt +∫Ωβτ(x2-x)∂u∂x∂u∂tdx dt +∫Ωβτ(1-x)(∂u∂x)2dx dt +∫Ωβτ(x-x2)∂u∂x∂2u∂x2dx dt -∫Ωβτ(x+1)∂∂x(x∂u∂x)∂u∂tdx dt =∫Ωβτf·((x2-x)∂u∂x +(x2+x)∂u∂t)dx dt, Ωβτ.
Integration by parts, we obtain(21)-∫Ωατ(x-x2)∂u∂x∂2u∂x2dx dt =∫0τ(α2-α22)(∂u∂x(α,t))2dt -∫Ωατ(12-x)(∂u∂x)2dx dt,-∫Ωατ(x+1)∂∂x(x∂u∂x)∂u∂tdx dt =-∫0τ(α2+α)∂u∂x(α,t)∂u∂t(α,t)dt +∫Ωατx∂u∂x∂u∂tdx dt+12∫0α(x2+x)(∂u(·,τ)∂x)2dx -12∫0α(x2+x)(∂φ∂x)2dx,∫Ωα,βτx∂u∂x∂2u∂x2dx dt =β2∫0τ(∂u∂x(β,t))2dt-α2∫0τ(∂u∂x(α,t))2dt -12∫Ωα,βτ(∂u∂x)2dx dt,-∫Ωα,βτ(x+1)∂∂x(x∂u∂x)∂u∂tdx dt =-∫0τ(β2+β)∂u∂x(β,t)∂u∂t(β,t)dt +∫0τ(α2+α)∂u∂x(α,t)∂u∂t(α,t)dt +∫Ωα,βx∂u∂x∂u∂tdx dt +12∫αβ(x2+x)(∂u(·,τ)∂x)2dx -12∫αβ(x2+x)(∂φ∂x)2dx,∫Ωβτ(x-x2)∂u∂x∂2u∂x2dx dt =∫0τ(β22-β2)(∂u∂x(β,t))2dt -∫Ωβτ(12-x)(∂u∂x)2dx dt,-∫Ωβτ(x+1)∂∂x(x∂u∂x)∂u∂tdx dt =∫0τ(β2+β)∂u∂x(β,t)∂u∂t(β,t)dt +∫Ωβτx∂u∂x∂u∂tdx dt +12∫β1(x2+x)(∂u(·,τ)∂x)2dx -12∫β1(x2+x)(∂φ∂x)2dx,
and by using the Cauchy's ε-inequality, we get
(22)∫Ωατx2∂u∂x∂u∂tdx dt ≤ε2∫Ωατx2(∂u∂x)2dx dt+12ε∫Ωατx2(∂u∂t)2dx dt ≤14∫Ωατ(∂u∂x)2dx dt +14(α2+α)∫Ωατ(x2+x)(∂u∂x)2dx dt +(α2+α)2∫Ωατ(x2+x)(∂u∂t)2dx dt,∫Ωατf·((x2-x)∂u∂x+(x2+x)∂u∂t)dx dt ≤(α2+α)2∫Ωατf2dx dt+14∫Ωατ(∂u∂x)2dx dt +14(α2+α)∫Ωατ(x2+x)(∂u∂x)2dx dt +12(α2+α)∫Ωατ(x2+x)(∂u∂t)2dx dt,∫Ωα,βτf·(-x∂u∂x+(x2+x)∂u∂t)dx dt ≤((β-α)((β2-α2)+(β-α))2+12) ×∫Ωα,βτf2dx dt+14∫Ωα,βτ(∂u∂x)2dx dt +14((β2-α2)+(β-α)) ×∫Ωα,βτ(x2+x)(∂u∂x)2dx dt +12∫Ωα,βτ(x2+x)(∂u∂t)2,-∫Ωβτx2∂u∂x∂u∂tdx dt ≤ε2∫Ωβτx(∂u∂x)2dx dt+12ε∫Ωβτx(∂u∂t)2dx dt ≤ε(1-β)2∫Ωβτ(∂u∂x)2dx dt +12ε∫Ωβτ(x2+x)(∂u∂t)2dx dt ≤14∫Ωβτ(∂u∂x)2dx dt +(1-β)∫Ωβτ(x2+x)(∂u∂t)2dx dt,∫Ωβτf·((x2-x)∂u∂x+(x2+x)∂u∂t)dx dt ≤((1-β)2+(1-β))2∫Ωβτf2+14∫Ωβτ(∂u∂x)2 +14((1-β)2+(1-β))∫Ωβτ(x2+x)(∂u∂x)2 +12((1-β)2+(1-β))∫Ωβτ(x2+x)(∂u∂t)2.
Substituting (21)–(22) into ∫Ωτℒu·Mudx dt, we obtain
(23)(1-α2+α2-12(α2+α))∫Ωατ(x2+x)(∂u∂t)2dx dt +12∫0α(x2+x)(∂u(·,τ)∂x)2dx∫0τ(α2-α22)(∂u∂x(α,t))2dt -∫0τ(α2+α)∂u∂x(α,t)∂u∂t(α,t)dt ≤(α2+α)2∫Ωατf2dx dt +12∫0α(x2+x)(∂φ∂x)2dx +12(α2+α)∫Ωατ(x2+x)(∂u∂x)2dx dt, Ωατ,(24)12∫Ωα,βτ(x2+x)(∂u∂t)2 +12∫αβ(x2+x)(∂u(·,τ)∂x)2dx -∫0τ(β2+β)∂u∂x(β,t)∂u∂t(β,t)dt +∫0τ(α2-α22)(∂u∂x(α,t))2dtβ2∫0τ(∂u∂x(β,t))2dt-α2∫0τ(∂u∂x(α,t))2dt ≤((β-α)2(β+α+1)2+12)∫Ωα,βτf2dx dt +12∫αβ(x2+x)(∂φ∂x)2dx +14((β2-α2)+(β-α)) ×∫Ωα,βτ(x2+x)(∂u∂x)2dx dt, Ωα,βτ,(25)(β-12((1-β)2+(1-β)))∫Ωβτ(x2+x)(∂u∂t)2dx dt +12∫β1(x2+x)(∂u(·,τ)∂x)2dx∫0τ(β22-β2)(∂u∂x(β,t))2dt +∫0τ(β2+β)∂u∂x(β,t)∂u∂t(β,t)dt ≤((1-β)2+(1-β))2∫Ωβτf2dx dt +12∫β1(x2+x)(∂φ∂x)2dx +14((1-β)2+(1-β)) ×∫Ωβτ(x2+x)(∂u∂x)2dx dt, Ωβτ.
Now, by using the conditions (PR) in (23), (24), and (25), respectively, we obtain
(26)(1-α2+α2-12(α2+α))∫Ωατ(x2+x)(∂u∂t)2dx dt +12∫0α(x2+x)(∂u(·,τ)∂x)2dx ≤(α2+α)2∫Ωατf2dx dt+12∫0α(x2+x)(∂φ∂x)2dx +12(α2+α)∫Ωατ(x2+x)(∂u∂x)2dx dt, Ωατ,(27)12∫Ωα,βτ(x2+x)(∂u∂t)2 +12∫αβ(x2+x)(∂u(·,τ)∂x)2dx ≤((β-α)2(β+α+1)2+12)∫Ωα,βτf2dx dt +12∫αβ(x2+x)(∂φ∂x)2dx +14((β2-α2)+(β-α)) ×∫Ωα,βτ(x2+x)(∂u∂x)2dx dt, Ωα,βτ,(28)(β-12((1-β)2+(1-β)))∫Ωβτ(x2+x)(∂u∂t)2dx dt +12∫β1(x2+x)(∂u(·,τ)∂x)2dx ≤((1-β)2+(1-β))2∫Ωβτf2dx dt +12∫β1(x2+x)(∂φ∂x)2dx +14((1-β)2+(1-β)) ×∫Ωβτ(x2+x)(∂u∂x)2dxdt. Ωβτ.
Or in Ωτ, by combining (26), (27), and (28), we get
(29)min(1-α2+α2-12(α2+α),12, β-12((1-β)2+(1-β))) ×∫Ωτ(x2+x)(∂u∂t)2dx dt +12∫01(x2+x)(∂u(·,τ)∂x)2dx ≤max((α2+α)2,((β-α)2(β+α+1)2+12), ((1-β)2+(1-β))2((β-α)2(β+α+1)2+12)) ×∫Ωτf2dx dt+12∫01(x2+x)(∂φ∂x)2dx +max(12(α2+α),14((β2-α2)+(β-α)), 14((1-β)2+(1-β))) ×∫Ωτ(x2+x)(∂u∂x)2dx dt.
Using Lemma 1 in [20], one has
(30)(∫Ω(x2+x)(∂u∂t)2dx dt+∫01(x2+x)(∂u(·,τ)∂x)2dx) ≤k(∫Ωf2dx dt+∫01(x2+x)(∂φ∂x)2dx),
where
(31)k=max((α2+α)2,((β-α)2(β+α+1)2+12), ((1-β)2+(1-β))2)((β-α)2(β+α+1)2+12)) ×(12,β-(1/2((1-β)2+(1-β))))min(121-α2+α2-12(α2+α),12, β-12((1-β)2+(1-β))))-1 ×e(max(1/2(α2+α),1/4((β2-α2)+(β-α)),1/4((1-β)2+(1-β)))T).
The right-hand side of (30) is independent of τ, hence replacing the left-hand side by its upper bound with respect to τ from 0 to T, we obtain the desired inequality, where c=(k)1/2.