aa r X i v : . [ m a t h . A P ] J a n ON SOME NONLINEAR EVOLUTION EQUATION OF SECONDORDER
KAMAL N. SOLTANOV
Abstract.
Here we study the abstract nonlinear differential equation of sec-ond order that in special case is the equation of the type of equation of trafficflow. We prove the solvability theorem for the posed problem under the ap-propriate conditions and also investigate the behaviour of the solution. Introduction
In this article we study the following nonlinear evolution equation(1.1) x tt + A ◦ F ( x ) = g (cid:16) x, A − x t (cid:17) , t ∈ (0 , T ) , < T < ∞ under the initial conditions(1.2) x (0) = x , x t (0) = x here A is a linear operator in a real Hilbert space H , F : X −→ X ∗ and g : D ( g ) ⊆ H × H −→ H are a nonlinear operators, X is a real Banach space. For example,operator A denotes − ∆ with Dirichlet boundary conditions (such as homogeneousor periodic) and f, g are functions such as above, that in the one space dimensioncase, we can formulate in the form(1.3) u tt − ( f ( u ) u x ) x = g ( u ) , ( t, x ) ∈ R + × (0 , l ) , l > , (1.4) u (0 , x ) = u ( x ) , u t (0 , x ) = u ( x ) , u ( t,
0) = u ( t, l ) , where u ( x ), u ( x ) are known functions, f ( · ) , g ( · ) : R −→ R are a continouosfunctions and l > Mathematics Subject Classification.
Primary 46T20, 47J35; Secondary 35L65, 58F10.
Key words and phrases.
Abstract nonlinear hyperbolic equation, solvability, behaviour. Solvability of problem (1.1) - (1.2)
Let A is a symmetric linear operator densely defined in a real Hilbert space H andpositive, A has a self-adjoint extension, moreover there is linear operator B definedin H such that A ≡ B ∗ ◦ B , here f : R −→ R is continuous as function, X is a realreflexive Banach space and X ⊂ H , g : D ( g ) ⊆ H × H −→ H , where g : R −→ R is a continuous as function and x : [0 , T ) −→ X is an unknown function. Let F ( r )as a function is defined as F ( r ) = r R f ( s ) ds . Let the inequation k x k H ≤ k Bx k H is valid for any x ∈ D ( B ). We denote by V , W and by Y the spaces defined as V ≡ { y ∈ H | By ∈ H } , W = { x ∈ H | Ax ∈ H } and as Y ≡ { x ∈ X | Ax ∈ X } ,respectively, for which inclusions W ⊂ V ⊂ H are compact and Y ⊂ W .Let H is the real separable Hilbert space, X is the reflexive Banach space and X ⊂ H ⊂ X ∗ ; V is the previously defined space. It is clear that W ⊂ V ⊂ H ⊂ V ∗ ⊂ W ∗ are a framed spaces by H , these inclusions are compact and X ⊂ V ∗ then one can define the framed spaces Y ⊂ V ⊂ H ⊂ V ∗ ⊂ Y ∗ then X ⊂ V ∗ ⊂ Y ∗ are compact, with use the property of the operator A . Assume that operator A such as A : V B −→ V ∗ B and A : X ∗ −→ Y ∗ . Consequently, we get A ◦ F : X −→ Y ∗ and A ◦ F ◦ A : Y −→ Y ∗ . Moreover we assume that [ X ∗ , Y ] ⊆ V .Since operator A is invertible, here one can set the function y ( t ) = A − x ( t ) forany t ∈ (0 , T ), in the other words one can assume the denotation x ( t ) = Ay ( t ).We will interpret the solution of the problem (1.1) - (1.2) by the following manner. Definition 1.
A function x : (0 , T ) −→ X , x ∈ C (0 , T ; X ) ∩ C (0 , T ; V ∗ ) ∩ C (0 , T ; Y ∗ ) , x = Ay , is called a weak solution of problem (1.1) - (1.2) if x a. e. t ∈ (0 , T ) satisfies the following equation (2.1) d dt h x, z i + h A ◦ F ( x ) , z i = h g ( x, By t ) , z i for any z ∈ Y and the initial conditions (1.2) (here and farther the expression h· , ·i denotes the dual form for the pair: the Banach space and his dual). Consider the following conditions( i ) Let A : W ⊂ H −→ H is the selfadjoint and positive operator, moreover A : V −→ V ∗ , A : X ∗ −→ Y ∗ , there exists an linear operator B : V −→ H that satisfiesthe equation Ax ≡ ( B ∗ ◦ B ) x for any x ∈ D ( A ) and k x k H ≤ k Bx k H = k x k V .( ii ) Let F : X −→ X ∗ is the continuously differentiable and monotone operatorwith the potential Φ that is the functional defined on X (his Frechet derivative isthe operator F ). Moreover for any x ∈ X the following inequalities hold k F ( x ) k X ∗ ≤ a k x k p − X + a k x k H ; h F ( x ) , x i ≥ b k x k pX + b k x k H , where a , b > a , b ≥ p > iii ) Assume g : H × V −→ H is a continuous operator that satisfies the condition |h g ( x, y ) − g ( x , y ) , z i| ≤ g |h x − x , z i| + g |h y − y , z i| , for any ( x, y ) , ( x , y ) ∈ H × H , z ∈ H and consequently for any ( x, y ) ∈ H × H the inequation k g ( x, y ) k H ≤ g k x k H + g k y k H + g , g ≥ k g (0 , k H holds, where g is a number. N SOME NONLINEAR EVOLUTION EQUATION OF SECOND ORDER 3
In the beginning for the investigation of the posed problem we set the followingexpression in order to obtain of the a priori estimations h x tt , y t i + h A ◦ F ( x ) , y t i = h g ( x, By t ) , y t i here element y is defined as the solution of the equation Ay ( t ) = x ( t ), i.e. y ( t ) = A − x ( t ) for any t ∈ (0 , T ) as was already mentioned above.Hence follow h By tt , By t i + h F ( x ) , x t i = h g ( x, By t ) , y t i , or(2.2) 12 ddt k By t k H + ddt Φ ( x ) = h g ( x, By t ) , y t i , where Φ ( x ) is the functional that defined as Φ ( x ) = R h F ( sx ) , x i ds (see, [6]).Then using condition ( iii ) on g ( x, By t ) in (2.2) one can obtain12 ddt k By t k H + ddt Φ ( x ) ≤ k g ( x, By t ) k H + k y t k H ≤ (cid:16) g k x k H + g k By t k H + g (cid:17) + k y t k H ≤ e C (cid:18) k x k H + 12 k By t k H + g (cid:19) here one can use the estimation k x k H ≤ e c (Φ ( x ) + 1) (if b > k x k H ≤ e c Φ ( x ))as 2 < p by virtue of the condition (ii) . Consequently we get to the Cauchy problemfor the inequation(2.3) ddt (cid:18) k By t k H ( t ) + Φ ( x ( t )) (cid:19) ≤ C (cid:18) k By t k H ( t ) + Φ ( x ( t )) (cid:19) + C with the initial conditions(2.4) x ( t ) | t =0 = x ; y t ( t ) | t =0 = A − x t | t =0 = A − x where C j ≥ x ( t ). From here follows12 k By t k H ( t ) + Φ ( x ( t )) ≤ e tC h k By k + 2Φ ( x ) i + C C (cid:0) e tC − (cid:1) . This give to we the following estimations for every T ∈ (0 , ∞ )(2.5) k By t k H ( t ) ≤ C ( x , x ) e C T , Φ ( x ( t )) ≤ C ( x , x ) e C T , for a. e. t ∈ (0 , T ), i.e. y = A − x is contained in the bounded subset of the space y ∈ C (0 , T ; V ) ∩ C (0 , T ; Y ), consequently we obtain that if the weak solution x ( t )exists then it belong to a bounded subset of the space C (0 , T ; X ) ∩ C (0 , T ; V ∗ ).Hence one can wait, that the following inclusion y ∈ C (0 , T ; X ∗ ∩ H ) ∩ C (0 , T ; X ∗ ∩ V ) ∩ C (0 , T ; Y )holds by virtue of (2.1) in the assumption that x = Ay is a solution of the posedproblem in the sense of Definition 1.In order to prove of the solvability theorem we will use the Galerkin approach.Let the system (cid:8) y k (cid:9) ∞ k =1 ⊂ Y be total in Y such that it is complete in the spaces Y, V , and also in the spaces
X, H . We will seek out of the approximative solutions y m ( t ), consequently and x m ( t ), in the form x m ( t ) ≡ Ay m ( t ) = m X k =1 c i ( t ) Ay k or x m ( t ) ∈ span (cid:8) y , ..., y m (cid:9) KAMAL N. SOLTANOV as the solutions of the problem locally with respect to t , where c i ( t ) are as theunknown functions that will be defined as solutions of the following Cauchy problemfor system of ODE d dt (cid:10) x m , y j (cid:11) + (cid:10) F ( x m ) , Ay j (cid:11) = (cid:10) g ( x m , By mt ) , y j (cid:11) , j = 1 , , ..., mx m (0) = x m , x tm (0) = x m , where x m and x m are contained in span (cid:8) y , ..., y m (cid:9) , m = 1 , , ... , moreover x m −→ x in [ X, Y ] ⊆ V ; x m −→ x in X, m ր ∞ . Thus we obtain the following problem(2.6) d dt (cid:10) x m , y j (cid:11) + (cid:10) F ( x m ) , Ay j (cid:11) = (cid:10) g ( x m , By mt ) , y j (cid:11) , j = 1 , , ..., m (cid:10) x m ( t ) , y j (cid:11) | t =0 = (cid:10) x m , y j (cid:11) , ddt (cid:10) x m ( t ) , y j (cid:11) | t =0 = (cid:10) x m , y j (cid:11) that solvable by virtue of estimates (2.5) on (0 , T ) for any m = 1 , , ... , j = 1 , , ... and T > . Hence we set(2.7) d dt h x m , z i + h F ( x m ) , Az i = h g ( x m , By mt ) , z i for any z ∈ Y and m = 1 , , ... .Consequently with use of the known procedure ([7], [8], [10]) we obtain, y mt ∈ C (0 , T ; V ), y m ∈ C (0 , T ; Y ) and x m ∈ C (0 , T ; X ), x mt ∈ C (0 , T ; V ∗ ), more-over they are contained in the bounded subset of these spaces for any m = 1 , , ... .Hence from (2.5) we get x mtt ∈ C (0 , T ; Y ∗ ) or x m ∈ C (0 , T ; Y ∗ ) , ( V ∗ ⊂ Y ∗ ).Thus we obtain, that the sequence { x m } ∞ m =1 of the approximated solutions of theproblem is contained in a bounded subset of the space C (0 , T ; X ) ∩ C (0 , T ; V ∗ ) ∩ C (0 , T ; Y ∗ )or { x m } ∞ m =1 such that for a. e. t ∈ (0 , T ) takes place the following inclusions { y m ( t ) } ∞ m =1 ⊂ Y ⊂ X ⊂ H , { y mt ( t ) } ∞ m =1 ⊂ V , { y mtt ( t ) } ∞ m =1 ⊂ X ∗ . So we have { y m ( t ) } ∞ m =1 ⊂ C (0 , T ; Y ) ∩ C (0 , T ; V ) ∩ C (0 , T ; X ∗ )therefore { y m ( t ) } ∞ m =1 possess a precompact subsequence in C (cid:16) , T ; [ X ∗ , Y ] (cid:17) and in C (0 , T ; V ), as [ X ∗ , Y ] ⊆ V by virtue of conditions on X and A (byvirtue of well known results, see, e. g. [2], [9] etc.). From here follows y m ( t ) −→ y ( t ) in C (0 , T ; V ) for m ր ∞ (Here and hereafter in order to abate the num-ber of index we don’t changing of indexes of subsequences). Then the sequence { F ( Ay m ( t )) } ∞ m =1 ⊂ X ∗ and bounded for a. e. t ∈ (0 , T ); the sequence { g ( x m ( t ) , x mt ( t )) } ∞ m =1 ≡ { g ( Ay m ( t ) , By mt ( t )) } ∞ m =1 ⊂ H and bounded for a. e. t ∈ (0 , T ) also, by virtue of the condition (iii). Indeed, forany m the estimation k g ( Ay m , By mt ) k H ( t ) ≤ k Ay m ( t ) k H + k By mt ( t ) k H + k g (0 , k H holds and therefore { g ( Ay m ( t ) , By mt ( t )) } ∞ m =1 is contained in a bounded subset of H for a. e. t ∈ (0 , T ). Consequently { F ( Ay m ) } ∞ m =1 and { g ( Ay m ( t ) , By mt ( t )) } ∞ m =1 have an weakly converging subsequences to η ( t ) and θ ( t ) in X ∗ and H , respectively, N SOME NONLINEAR EVOLUTION EQUATION OF SECOND ORDER 5 for a. e. t ∈ (0 , T ). Hence one can pass to the limit in (2.7) with respect to m ր ∞ .Then we obtain the following equation(2.8) d dt h x, z i + h Aη ( t ) , z i = h θ ( t ) , z i . So for us is remained to show the following: if the sequence { x m ( t ) } ∞ m =1 ≡{ Ay m ( t ) } ∞ m =1 is weakly converging to x ( t ) = Ay ( t ) then η ( t ) = F ( x ( t )) and θ ( t ) = g ( x ( t ) , By t ( t )). In order to show these equations are fulfilled we will usethe monotonicity of F and the condition (iii). We start to show θ ( t ) = g (( t ) , By t ( t )) as x ∈ X ⊂ H and y t ∈ V , By t ∈ H therefore g ( x, By t ) is defined for a. e. t ∈ (0 , T ). Consequently one can considerof the expression h g ( Ay m ( t ) , By mt ( t )) − g ( Ay ( t ) , By t ( t )) , b y i for any b y ∈ C (0 , T ; Y ) ∩ C (0 , T ; V ). So we set this expression and investigatethis for any b y ∈ C (0 , T ; Y ) ∩ C (0 , T ; V ) then we have |h g ( Ay m ( t ) , By mt ( t )) − g ( Ay ( t ) , By t ( t )) , b y i| ≤ (2.9) g |h Ay m ( t ) − Ay ( t ) , b y ( t ) i| + g |h By mt ( t ) − By t ( t ) , b y ( t ) i| that takes place by virtue of the condition (iii). Using here the weak convergencesof Ay m ( t ) ⇀ Ay ( t ) and By mt ( t ) −→ By t ( t ) and by passing to the limit in theinequatin (2.9) with respect to m : m ր ∞ we get |h θ ( t ) − g ( Ay ( t ) , By t ( t )) , b y i| ≤ b y ∈ C (0 , T ; H ). Consequently the equation θ ( t ) = g (( t ) , By t ( t )) holds,then the following equation is valid d dt h x, z i + h Aη ( t ) , z i = h g ( x ( t ) , By t ( t )) , z i for any z ∈ Y , as (cid:8) y k (cid:9) ∞ k =1 is complete in Y that display fulfilling of equation(2.10) Aη ( t ) = g ( x ( t ) , By t ( t )) − d xdt in the sense of Y ∗ .In order to show the equation η ( t ) = F ( x ( t )) one can use the monotonicity of F . So the following inequation holds h A ◦ F ( Az ) − A ◦ F ( Ay ) , z − y i = h F ( Az ) − F ( Ay ) , Az − Ay i = h F ( e x ) − F ( x ) , e x − x i ≥ y, z ∈ Y , Ay = x and Az = e x by condition (i) . Then one can write0 ≤ h F ( x m ) − F ( e x ) , x m − e x i = h F ( Ay m ) − F ( Az ) , Ay m − Az i =take account here the equation (2.5) h F ( Ay m ) , Ay m i − (cid:28) d dt x m − g ( x m , By mt ) , z (cid:29) − h F ( Az ) , Ay m − Az i =(2.11) h F ( x m ) , x m i − (cid:28) d dt x m − g ( x m , By mt ) , z (cid:29) − h F ( e x ) , x m − e x i . KAMAL N. SOLTANOV
Here one can use the well-known inequationlim sup h F ( x m ) , x m i ≤ h η, x i = h η, Ay i = h Aη, y i . Then passing to the limit in (2.11) with respect to m : m ր ∞ we obtain0 ≤ h Aη, y i − (cid:28) d dt x − g ( x, By t ) , z (cid:29) − h F ( e x ) , x − e x i = h Aη, y i − h
Aη, z i − h F ( e x ) , Ay − Az i = h Aη − A ◦ F ( e x ) , y − z i by virtue of (2.10)Consequently we obtain, that the equation Aη ( t ) = A ◦ F ( x ) holds since z isarbitrary element of Y .Now for us is remained to show the obtained function x ( t ) = Ay ( t ) satisfy theinitial conditions. Consider the following equation h y mt , Ay m i ( t ) = t Z (cid:28) d ds Ay m , y m (cid:29) ds + t Z (cid:28) dds By m , dds By m (cid:29) ds + h y m , Ay m i = t Z (cid:28) d ds y m , Ay m (cid:29) ds + t Z (cid:13)(cid:13)(cid:13)(cid:13) dds By m (cid:13)(cid:13)(cid:13)(cid:13) H ds + h y m , Ay m i for m = 1 , , ... , here x m ( t ) = Ay m ( t ). Hence we get: the left side is bounded asfar as all addings items in the right side are bounded by virtue of the obtainedestimations. Therefore one can pass to limit with respect to m as here y mt arecontinous with respect to t for any m then y mt strongly converges to y t and Ay m weakly converges to Ay in H . It must be noted the equationlim m −→∞ t Z (cid:13)(cid:13)(cid:13)(cid:13) dds By m (cid:13)(cid:13)(cid:13)(cid:13) H dxds = t Z (cid:13)(cid:13)(cid:13)(cid:13) dds By (cid:13)(cid:13)(cid:13)(cid:13) H ds holds by virtue of the above reasonings that { y m ( t ) } ∞ m =1 is a precompact subset in C (0 , T ; V ). Consequently the left side converges to the expression of such type,i.e. to h y t , Ay i ( t ).The obtained results shows that the following convergences are just: x m ( t ) = Ay m ( t ) ⇀ Ay ( t ) = x ( t ) in X , x mt ( t ) = Ay mt ⇀ Ay t = x t ( t ) in V ∗ . From herefollows, that the initial conditions are fulfilled in the sense of X and V ∗ , respectively.Thus the following existence resut is proved. Theorem 1.
Let spaces
H, V, W, X, Y that are defined above satisfy all above men-tioned conditions and condition (i)-(iii) are fulfilled then problem (1.1) - (1.2)is solvable in the space C (0 , T ; X ) ∩ C (0 , T ; V ) ∩ C (0 , T ; Y ∗ ) for any x ∈ V ∩ [ X ∗ , Y ] and x ∈ H in the sense of Definition 1. Remark 1.
This theorem shows that there exist a flow S ( t ) defined in V × X andthe solution of the problem (1.1) - (1.2) one can represent as x ( t ) = S ( t ) ◦ ( x , x ) . N SOME NONLINEAR EVOLUTION EQUATION OF SECOND ORDER 7 Behaviour of solutions of problem (1.1) - (1.2)
Here we consider problem under the following complementary conditions:( iv ) Let g ( x, By t ) = 0 and k x k pH ( t ) ≤ c Φ ( x ( t )) for some c > E ( t ) = k Bw k H ( t ) and consider this function on the solutionof problem (1.1) - (1.2), then for E ( t ) = k By k H ( t ) we have(3.1) E ´( t ) = 2 h By t , By i ≤ k By t k ( t ) + k By k ( t ) , where y = A − x . Here we will use equation (2.5). For this we lead the followingequation 12 k By s k H ( s ) + Φ ( x ( s )) (cid:12)(cid:12) t = 0as g ( x, By t ) = 0. Hence12 k By s k H ( t ) + Φ ( x ( t )) = 12 k By k H ( t ) + Φ ( x )and k By t k H ( t ) = −
2Φ ( x ( t )) + k By k H + 2Φ ( x ) . Granting this in (3.1) we get E ´( t ) ≤ E ( t ) − E r ( t ) + k By k H + 2Φ ( x )by virtue of the condition Φ ( x ) ≥ c k x k pX and of the continuity of embedding X ⊂ H , r = p/ z ( t ) = E ( t ) we have the Cauchy problem for differential inequality(3.2) z ´( t ) ≤ z ( t ) − cz r ( t ) + C ( x , x ) , z (0) = k By k H , that we will investigate. Inequation (3.2) one can rewrite in the form( z ( t ) + kC ( x , x ))´ ≤ z ( t ) + kC ( x , x ) − δ [ z ( t ) + kC ( x , x )] r , where k > δ = δ ( c, C, k, r ) > z ( t ) + kC ( x , x ) ≤ h e (1 − r ) t ( z + kC ( x , x )) − r + δ (cid:16) − e (1 − r ) t (cid:17)i − r or E ( t ) ≤ (cid:20) e (1 − r ) t (cid:16) k By k H + kC ( x , x ) (cid:17) − r + δ (cid:16) − e (1 − r ) t (cid:17)(cid:21) − r − kC ( x , x )(3.3) k By k H ( t ) ≤ e t (cid:16) k By k H + kC ( x , x ) (cid:17)(cid:20) δ (cid:16) k By k H + kC ( x , x ) (cid:17) r − (cid:0) e ( r − t − (cid:1)(cid:21) r − − kC ( x , x ) . here the right side is greater than zero, because δ ≤ k − k r C r and 2 r = p > Lemma 1.
Under conditions (i), (ii), (iv) the function y ( t ) , defined by the solutionof problem (2.1)-(2.2), for any t > is contained in ball B X ∩ Vl (0) ⊂ X ∩ V depending from the initial values ( x , x ) ∈ ( X ∩ V ) × H , here l = l ( x , x , p ) > . We would like to note that this equation shows the stability of the energy of the consideredsystem in this case.
KAMAL N. SOLTANOV
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Institute of Math. and Mech. Nat. Acad. of Sci. , Baku, AZERBAIJAN; Dep. ofMath., Fac. of Sci., Hacettepe University, Ankara, TURKEY
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