# On Mixed Problems for Quasilinear Second-Order Systems

Authors: Rita Cavazzoni
Journal: International Journal of Differential Equations
 (2010)

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## Abstract

The paper is devoted to the study of initial-boundary value problems for quasilinear second-order systems. Existence and uniqueness of the solution in the spaceHs(Ω¯×[0,T]), with s>d/2+3,  is proved in the case where Ω  is a half-space of ℜd. The proof of the main theorem relies on two preliminary results: existence of the solution to mixed problems for linear second-order systems with smooth coefficients, and existence of the solution to initial-boundary value problems for linear second-order operators whose coefficients depend on the variables x and t through a function  v∈Hs(ℜd+1). By means of the results proved for linear operators, the well posedness of the mixed problem for the quasi-linear system is established by studying the convergence of a suitable iteration scheme.


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## Body

## 1. Introduction

The purpose of this paper is the study of initial-boundary value problems for second-order quasi-linear systems. The name initial-boundary value problem or mixed problem refers to the initial data (i.e., at timet=0) and to the boundary data (i.e., on the boundary of the spatial domain). The interest in mixed problems originates from physical issues: in fluid mechanics, for instance, the spatial domain where a phenomenon takes place usually has an entrance, an exit, and walls.I will be concerned with the study of mixed problems for second-order differential operators of the form(1.1)Q=∂t2+S(·)∂t+∑α=1dTα(·)∂t∂α+∑α=1d∑β=1dEα,β(·,∇x·)∂α∂β+∑α=1dFα(·,∇x·)∂α+G(·,∇x·),
where t is a real variable and the variable x denotes a vector of  ℜd, with d > 1. For all α,β=1,…,d, the functions S,Tα,Eα,β,Fα,G, take values in the space of  d×d real matrices with smooth coefficients.The subject of my research will be initial-boundary value problems in a half-spaceΩ=ℜd-1×(0,∞). The second section of the paper deals with the study of the existence and uniqueness of the solution to the following mixed problem in the half-space Ω=ℜd-1×(0,∞)(1.2)Qu=J(u),x∈Ω,t>0,2∑β=1d-1Ed,β(u,∇xu)∂βu+Td(u)∂tu+Fd(u,∇xu)u=0,x∈∂Ω,t>0,u(x,0)=h(x),x∈Ω,∂tu(x,0)=k(x),x∈Ω,
where h and k are functions defined in Ω with values in ℜd, whereas the function J is defined in ℜd; the unknown vector-valued function is denoted by u.In problem (1.2), the initial data h and k will be assumed to lie in Sobolev spaces. The main result of the paper states that if the coefficients of the operator Q are sufficiently regular and the operator satisfies suitable  estimates, then there exists T>0 and a function  u∈Hs(Ω̅×[0,T]),with  s>d/2+3, that provides a unique solution to the mixed problem (1.2).In order to achieve this result, we apply techniques already employed to prove existence and uniqueness in Sobolev spaces of nontrivial solutions to second-order quasi-linear systems, with homogeneous boundary conditions, in the half-spaceΩ=ℜd-1×(0,∞)  (see [1, 2]). The proof of Theorem 2.3 as well as the proof of the main theorems in [1, 2] relies on the assumption that the operator Q and its adjoint satisfy estimates which are analogous to the a priori estimates obtained in [3] in the case of linear Cauchy problems for first-order symmetrizable systems. In fact, in the case where the system under consideration is a first-order linear symmetrizable system with smooth coefficients, according to the results proved in [3], energy estimates imply the well posedness of the Cauchy problem. The existence of a unique solution u∈C([0,T];Hs(ℜd)) for initial data u(0)∈Hs can be obtained by means of a duality argument, applying the a priori estimates satisfied by the adjoint operator associated with the system. This existence result can be extended to the case of  a system with Hs-coefficients, provided that s is large enough: in [4], the well posedness of the Cauchy problem for a first-order system whose coefficients depend on the variables x and t through a known function, has been obtained; this result turns out to be a significant tool in non-linear analysis, since it can be employed to state the well posedness of a class of first-order quasi-linear systems (see [3, 5, 6]). As a matter of fact, for general quasi-linear systems, the study of the Cauchy problem turns out to be a difficult task. However, if the system is Friedrichs symmetrizable (see [3]), then Hs well posedness has been proved to hold true, provided that s is large enough: the proof relies on the definition of an iteration scheme and on the well posedness obtained in the linear case.By following the strategy successfully applied to prove existence and uniqueness of the solution to the initial value problem for first-order systems, I will be concerned in this paper with initial-boundary value problems for second-order quasi-linear systems. As a preliminary result, Proposition2.1 states the existence in  Hs(ℜd+1) of the solution to mixed problems for linear second-order operators, whose coefficients depend smoothly on the space variable x and on the time variable t.  Subsequently, by means of a similar procedure, I will prove the existence of the solution to the initial-boundary value problem with homogeneous boundary conditions, in the case where the coefficients of the operator depend on the variables x and t through a function v∈Hs(ℜd+1),with  s>d/2+3. Thus this result enables me to state the main Theorem 2.3 regarding existence and uniqueness of the solution to the quasi-linear problem in the space Hs(Ω̅×[0,T]), in the case where the domain is a half-space of ℜd. An iteration scheme, which consists of a sequence of mixed problems for linear systems, will be defined to approximate the solution. Thus, as a result of the estimates satisfied by the operator Q, one can prove the convergence of the scheme and establish existence of a unique solution to problem (1.2) in the space Hs(Ω̅×[0,T]). The solution turns out to be classical, for this function belongs to Sobolev spaces with sufficiently large exponent.


## 2. Mixed Problem in a Half-Space

Consider the following initial-boundary value problem in the half-spaceΩ=ℜd-1×(0,∞):(2.1)∂t2u+S̅(x,t)∂tu+∑α=1dT̅α(x,t)∂t∂αu+∑α=1d∑β=1dE̅α,β(x,t)∂α∂βu+∑α=1dF̅α(x,t)∂αu+G̅(x,t)u=J̅(x,t),x∈Ω,t>0,2∑β=1d-1E̅d,β(x,t)∂βu+T̅d(x,t)∂tu+F̅d(x,t)u=0,x∈∂Ω,t>0,u(x,0)=h(x),x∈Ω,∂tu(x,0)=k(x),x∈Ω,
where for all α,β=1,…,d, the functions  S̅,T̅α,E̅α,β,F̅α,G̅,J̅ are defined in Ω×ℜ, S̅,T̅α,E̅α,β,F̅α,G̅ belong to the space of   d×d real matrices, whereas J̅ takes values in ℜd. We assume that for every α,β=1,…,d, E̅α,β=E̅β,α.Let us denote byQ̅ the linear second-order operator associated to system (2.1): (2.2)Q̅=∂t2+S̅(x,t)∂t+∑α=1dT̅α(x,t)∂t∂α+∑α=1d∑β=1dE̅α,β(x,t)∂α∂β+∑α=1dF̅α(x,t)∂α+G̅(x,t),
and by Q̅* the corresponding adjoint operator:(2.3)Q̅*=∂t2-∂tS̅T-S̅T∂t+∑α=1d(∂α∂tT̅αT+∂tT̅αT∂α+∂αT̅αT∂t+T̅αT∂α∂t)+∑α=1d∑β=1d(∂β∂αE̅α,βT+2∂αE̅α,βT∂β+E̅α,βT∂α∂β)-∑α=1d(F̅αT∂α+∂α(F̅αT))+G̅T.
Since we are looking for a solution to (2.1) in the functional space Hs(ℜd+1), with s>d/2+2, such a  function will belong to C2(Ω̅), due to Sobolev embedding. Thus, it will be continuous up to the boundary of Ω. As a consequence of this property, one has to require compatibility conditions.Similarly to the proof of Proposition2.1 in [2], we will establish the following result.Proposition 2.1.
Let s be an integer number so thats>d/2+2. Consider the initial-boundary value problem (2.1) and assume the following conditions are fulfilled:(i)
for allα,β=1,…,d, the matrix-valued functions S̅,T̅α,E̅α,β,F̅α,G̅belong to C∞(Ω̅×[0,T]) and are bounded as well as their derivatives, in addition E̅α,β=E̅β,α and E̅d,d is the null matrix;(ii)
the functionJ̅ belongs to the space Hs(ℜd+1); moreover, h∈Hs+1(Ω̅); k∈Hs(Ω̅);(iii)
there exists a positive real constantλ such that for every vector function φ∈L2(ℜd+1)∩C2(Ω̅×[0,T[), with supp∥φ∥ compact subset of Ω̅×[0,T[(2.4)∥∂tφ(0)∥H-s2+∥φ(0)∥H-s2+∥φ∥H-s2≤λ∥Q̅*φ∥H-s2;(iv)
let us setr=[s-d/2], if s-d/2∉IN;r=[s-d/2]-1, if s-d/2∈IN; and assume the following compatibility conditions to be satisfied:  the function 2∑β=1d-1E̅d,β∂βh+T̅dk+F̅dh, and the derivatives of B[u]=2∑β=1d-1E̅d,β∂βu+T̅d∂tu+F̅du with respect to the time variable t of order less than or equal to  r-1, vanish when evaluated at t=0 and x∈∂Ω.
Then there exists a functionu∈Hs(ℜd+1), which provides a solution to the mixed problem (2.1).Proof.
Let us denote byE the space
(2.5)E={φ:ℜd×ℜ→ℜd:φ∈C2(Ω̅×[0,T[),supp∥φ∥⊆Ω̅×[0,T[andcompact}.
Let φ∈E and define the linear functional L on the space Q̅*(E)⊆H-s(ℜd+1) as follows:
(2.6)L(Q̅*φ)=〈k+S̅(0)h+∑α=1dT̅α(0)∂αh,φ(0)〉Hs,H-s-〈h,∂tφ(0)〉Hs,H-s+〈J̅,φ〉Hs,H-s.
Thanks to (iii), the operator Q̅* is one to one in E. Hence the functional L turns out to be well defined. Furthermore, it turns out to be continuous, due to the assumptions. Thus L can be extended to the space H-s(ℜd+1).Let us denote by M this functional. Thanks to Riesz theorem, there exists a unique function u∈Hs(ℜd+1),such that for all v∈H-s(ℜd+1), M(v)=〈u,v〉Hs,H-s. If φ∈E, then
(2.7)M(Q̅*φ)=〈u,Q̅*φ〉Hs,H-s=〈k+S̅(0)h+∑α=1dT̅α(0)∂αh,φ(0)〉Hs,H-s-〈h,∂tφ(0)〉Hs,H-s+〈J̅,φ〉Hs,H-s.
Let φ∈C0∞(Ω×(0,T)). Thus 〈u,Q̅*φ〉Hs,H-s=〈J̅,φ〉Hs,H-s; as a consequence,  for all φ∈C0∞(Ω×(0,T)), 〈Q̅u,φ〉L2,L2=〈J̅,φ〉L2,L2, and the function u turns out to be solution of system (2.1) in Ω×(0,T).
Consider a functionχ∈C0∞(A×]0,T[), where A  is an open subset of  ℜd, and define the function χT∈E in the following way:
(2.8)χT(x,t)={χ(x,t),ifx∈Ω̅,t∈]0,T[,0,ifx∈A∖Ω̅,t∈]0,T[.
Thus 〈u,Q̅*χT〉Hs,H-s=〈J̅,χT〉Hs,H-s.
Integrating by parts, we deduce(2.9)∫0T∫ℜd-1〈2∑β=1d-1E̅d,β∂βu+T̅d∂tu+F̅du,χ〉(y,0,t)dydt=0.
By means of a standard argument, one proves that the function u satisfies the boundary conditions of problem (2.1).
Let us consider now a functionψ∈C0∞(Ω×]-∞,T[) and define the function ψT∈E, as
(2.10)ψT(x,t)={ψ(x,t),ifx∈Ω,0≤t<T,0,ifx∈Ω,t<0.
We have 〈u,Q̅*ψT〉Hs,H-s=M(Q̅*ψT).
Integrating by parts, we obtain the following identity:(2.11)-〈u(0),∂tψ(0)〉L2,L2+〈∂tu(0)+∑α=1dT̅α(0)∂αu(0)+S̅(0)u(0),ψ(0)〉L2,L2+〈Q̅u,ψ〉L2,L2=〈k+S̅(0)h+∑α=1dT̅α(0)∂αh,ψ(0)〉L2,L2-〈h,∂tψ(0)〉L2,L2+〈J̅,ψ〉L2,L2.
Let us consider a vector function ψ, which, in a neighbourhood of the origin t=0, is expressed by the product  ψ(x,t)=tμ(x), where μ∈C0∞(Ω). Thus, we prove, by means of a standard argument, that u(x,0)=h(x), in Ω.  As a result, we deduce similarly that ∂tu(x,0)=k(x), in Ω.Thanks to the result established in Proposition2.1, we can prove the existence of the solution to a mixed problem for a second-order linear operator with coefficients that depend on the space variable and on the time-variable through a given function. Let  v∈Hs(ℜd+1) with s>d/2+3 and denote by Qv the operator
(2.12)Qv=∂t2+S(v)∂t+∑α=1dTα(v)∂t∂α+∑α=1d∑β=1dEα,β(v,∇xv)∂α∂β+∑α=1dFα(v,∇xv)∂α+G(v,∇xv).
Let us consider the mixed problem
(2.13)Qvu=J(v),x∈Ω,t>0,2∑β=1d-1Ed,β(v,∇xv)∂βu+Td(v)∂tu+Fd(v,∇xv)u=0,x∈∂Ω,t>0,u(x,0)=h(x),x∈Ω,∂tu(x,0)=k(x),x∈Ω,
where for all α,β=1,…,d, the functions Eα,β,Fα,G, are defined in ℜd×ℜd2, with values in the space of d×d real matrices with smooth coefficients, the matrix valued functions S,Tα are defined in  ℜd, the function J is also defined in ℜd, with values in the space ℜd.Proposition 2.2.
Let s be an integer number,s>d/2+3 and v∈Hs(ℜd+1). Assume the following conditions are satisfied:(i)
for allα,β=1,…,d, the matrix valued functions S,Tα,Eα,β,Fα,G, have C∞-entries, which are bounded as well as their derivatives; moreover, Eα,β=Eβ,α; Ed,d as well as S(0) and Tα(0) are null matrices;(ii)
J∈C∞(ℜd), and J(0) is the null matrix;(iii)
h∈Hs+1(Ω̅), k∈Hs(Ω̅);(iv)
there exists a positive real constantλ such that for every φ∈L2(ℜd+1)∩C2(Ω̅×[0,T[), with supp∥φ∥ compact subset of Ω̅×[0,T[(2.14)∥∂tφ(0)∥H-s2+∥φ(0)∥H-s2+∥φ∥H-s2≤λ∥Q̅v*φ∥H-s2;(v)
the initial values h and k satisfy compatibility conditions as in (iv) of the previous proposition.Then the initial-boundary value problem (2.13) admits a solution u∈Hs(ℜd+1).Proof.
Thanks to Moser estimates [3], the functions S(v) and Tα(v) belong to the space Hs(ℜd+1), for v∈Hs(ℜd+1), S,Tα∈C∞, S(0) and Tα(0) are null matrices. As a consequence, (S(v(·,0))h+∑α=1dTα(v(·,0))∂αh)∈Hs(Ω̅). Therefore, by defining a linear functional L on the space Qv*(E), as in the proof of Proposition 2.1, one can prove the existence of a solution u∈Hs(ℜd+1) to the mixed problem (2.13).The existence result of Proposition2.2 allows us to prove the existence of the solution to problem (1.2).Theorem 2.3.
Lets be an integer number, with s>d/2+3 and assume that the second-order quasi-linear operator Q (1.1) satisfies the following conditions:(i)
for allα,β=1,…,d, the matrix-valued functions S,Tα,Eα,β,Fα,G, have C∞-entries, which are bounded as well as their derivatives, moreover, Eα,β=Eβ,α, Ed,d as well as Eα,β(0,0),Fα(0,0),G(0,0), S(0) and Tα(0) are null matrices;(ii)
J∈C∞(ℜd), and J(0) is the null matrix, in addition, the function J has bounded derivatives;(iii)
h∈Hs+1(Ω̅);k∈Hs(Ω̅);(iv)
the initial valuesh and k satisfy compatibility conditions as in (iv) of Proposition 2.1;(v)
ifv∈Hs(ℜd+1), there exists a positive real constant λ such that for every φ∈L2(ℜd+1)∩C2(Ω̅×[0,T[), with supp∥φ∥ compact subset of Ω̅×[0,T[(2.15)∥∂tφ(0)∥H-s2+∥φ(0)∥H-s2+∥φ∥H-s2≤λ∥Q̅v*φ∥H-s2;(vi)
there exists a  positive constantΓ such that for any T>0 and v∈Hs(ℜd+1) it holds true that  for every φ∈Hs(ℜd+1)(2.16)∥φ∥Hs(Ω̅×[0,T])2≤ΓT∥Qvφ∥Hs-2(Ω̅×[0,T])2+∥φ(0)∥Hs-1(Ω̅)2+∥∂tφ(0)∥Hs-1(Ω̅)2.
Then there exist T̃>0 and a function u∈Hs(Ω̅×[0,T̃]) which provides a classical solution, to the initial-boundary value problem (1.2), with initial data u(x,0)=h(x) and ∂tu(x,0)=k(x). Moreover, the solution turns out to be unique.Proof.
According to the notation introduced above, we denote byQn-1 the second-order operator Qun-1(2.17)Qn-1=∂t2+S(un-1)∂t+∑α=1dTα(un-1)∂t∂α+∑α=1d∑β=1dEα,β(un-1,∇xun-1)∂α∂β+∑α=1dFα(un-1,∇xun-1)∂α+G(un-1,∇xun-1).
Let us address the study of the well posedness of the initial-boundary value problem (1.2), by means of an iteration scheme, which consists of the following sequence of mixed problems:
(2.18)Qn-1un=J(un-1),x∈Ω,t>0,2∑β=1d-1Ed,β(un-1,∇xun-1)∂βun+Td(un-1)∂tun+Fd(un-1,∇xun-1)un=0,x∈∂Ω,t>0,un(x,0)=h(x),x∈Ω,∂tun(x,0)=k(x),x∈Ω.
Let δ be a positive real number, such that ||h||Hs(Ω̅)≤δ/4 and ||k||Hs(Ω̅)≤δ/4.  Let u0 be a function in the space Hs(ℜd+1) whose norm satisfies ||u0||Hs≤δ. We initialize the scheme by means of the function u0. Due to the result of Proposition 2.2, since un-1∈Hs(ℜd+1) for all n∈N, the mixed problem (2.18) admits a solution un∈Hs(ℜd+1).This solution turns out to be unique, for the operator Qn-1 satisfies condition (vi).
We will prove that the sequence of functions(un)n∈IN is convergent in the space  Hs(Ω̅×[0,T]), to a function u, which provides the solution of the mixed problem (1.2). The result can be achieved similarly to the existence result of Theorem 3.1 in [2]. For the sake of completeness, let us outline the proof.
At first, let us prove by induction that for alln∈ℕ, ∥un∥Hs≤δ. In accordance with the choice of the function u0, we have  ||u0||Hs≤δ. Assume that  for all n<m, ||un||Hs≤δ, and prove that, as long as T is small enough, the estimate is true in the case where n≥m. Since s>d/2+3  and the functions un∈Hs, there exists a positive constant μ, which depends on δ, so that  for every α,β=1,…,d, for all  n<m(2.19)∥un∥L∞(ℜd+1)≤μ,∥∇xun∥L∞(ℜd+1)≤μ,∥∂α∂βun∥L∞(ℜd+1)≤μ,∥∂tun∥L∞(ℜd+1)≤μ,∥∂t∂αun∥L∞(ℜd+1)≤μ.
Thanks to Moser estimates (see [3]) and to the conditions satisfied by the matrices S,Tα,Eα,β,Fα,G, for all α,β=1,…,d, we obtain
(2.20)∥Qm-1um∥Hs-2(ℜ(d+1))≤ζ(s,μ)∥um∥Hs(ℜ(d+1))+ρ(s,μ)∥∂tum∥Hs-1(ℜd+1),
where ζ(s,μ) and ρ(s,μ) are  suitable positive constants  that depend on s,μ and on the matrices S,Tα,Eα,β,Fα,G. Thus,
(2.21)∥Qm-1um∥Hs-2(Ω̅×[0,T])≤ζ(s,μ)∥um∥Hs(Ω̅×[0,T])+ρ(s,μ)∥∂tum∥Hs-1(Ω̅×[0,T]).
Because of (vi), we deduce
(2.22)∥um∥Hs(Ω̅×[0,T])2≤ΓT(ζ(s,μ)∥um∥Hs(Ω̅×[0,T])+ρ(s,μ)∥∂tum∥Hs-1(Ω̅×[0,T]))2+∥h∥Hs(Ω̅)+∥k∥Hs(Ω̅)≤2ΓTmax(ζ2(s,μ),ρ2(s,μ))∥um∥Hs(Ω̅×[0,T])2+δ24.
If T<1/(4Γmax(ζ2(s,μ),ρ2(s,μ))), then ∥um∥Hs(Ω̅×[0,T])2≤δ2.
As a consequence of the previous estimate, by induction, we get for alln∈ℕ,  ||un||Hs(Ω̅×[0,T])≤δ.
The subsequent step of the proof consists in proving that the sequence(un)n∈ℕ is convergent in the space Hs(Ω̅×[0,T]).
LetWn be the difference un-un-1, for every n∈ℕ; hence,
(2.23)Qn-1Wn=Qn-1un-Qn-1un-1=J(un-1)-Qn-1un-1±Qn-2un-1=(Qn-2-Qn-1)un-1+J(un-1)-J(un-2).
Thanks to the mean value theorem,
(2.24)∥(S(un-2)-S(un-1))∂tun-1∥Hs-1≤const(δ)∥Wn-1∥Hs.
Similar estimates can be computed for the other terms. Moreover,
(2.25)∥J(un-2)-J(un-1)∥Hs≤∥DJ(ζ)·Wn-1∥Hs≤const∥Wn-1∥Hs
Thus, for all n∈ℕ,
(2.26)∥Qn-1Wn∥Hs-2≤const(δ)∥Wn-1∥Hs.
Let us notice that if  Qn-1Wn was the null function, then the function un-1, would provide a solution to the mixed problem (1.2) and the proof of the theorem would be completed. In fact, we would obtain Qn-1un-1=J(un-1) and because of the initial conditions Wn(x,0)=0, ∂tWn(x,0)=0, and the estimate (2.16), we would have ||Wn||Hs(Ω̅×[0,T])2≤ΓT||Qn-1Wn||Hs-2(Ω̅×[0,T])2. Thus un≡un-1.
If we assume thatQn-1Wn does not vanish identically, thanks to (vi), we obtain
(2.27)∥Wn∥Hs(Ω̅×[0,T])2≤ΓT(const(δ)2∥Wn-1∥Hs(Ω̅×[0,T])2).
The constant in (2.27) is independent of n; hence, provided that T is sufficiently small, the sequence of functions (un)n∈ℕ turns out to be a Cauchy sequence in the space Hs(Ω̅×[0,T]). Let u be the limit of (un)n∈ℕ. Therefore, the limit function u belongs to the space C2(Ω̅×[0,T]).
The study of the convergence of the iteration scheme (2.18) is the last step of the proof.
The sequence(un)n∈ℕ is convergent to the function u in Hs(Ω̅×[0,T]), with s>d/2+3, thus as n→∞, the sequence (Qn-1un(x,t))n turns out to be convergent to Qu(x,t), for every  (x,t)∈Ω×(0,T). Therefore, thanks to Sobolev embedding, the function u provides a classical solution to system (1.2),Qu=J, in Ω×(0,T).
Furthermore, by continuity of the trace operator, the solutionu turns out to satisfy the boundary condition: 2∑β=1d-1Ed,β(u,∇xu)∂βu+Td(u)∂tu+Fd(u,∇xu)u=0, where x∈∂Ω, t∈[0,T].
As far as the initial datum is concerned, we getu(x,0)=h and ∂tu(x,0)=k(x), as x∈Ω.
By means of the same estimates obtained for the sequence of functions(Wn)n∈ℕ, we can prove the uniqueness, in the space Hs(Ω̅×[0,T]), of the solution u to the problem (1.2), with initial conditions u(x,0)=h(x) and ∂tu(x,0)=k(x), x∈Ω.



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