Periodic solutions for a class of non-autonomous differential delay equations
|
|
- Leo Arntsen
- 4 år siden
- Visninger:
Transkript
1 Nonlinear Differ. Equ. Appl. 16 (2009), c 2009 Birkhäuser Verlag Basel/Switzerland /09/ published online July 4, 2009 DOI /s Nonlinear Differential Equations and Applications NoDEA Periodic solutions for a class of non-autonomous differential delay equations Rong Cheng, Junxiang Xu Abstract. By applying symplectic transformation, Floquet theory and some results in critical point theory, we establish the existence of periodic solutions for a class of non-autonomous differential delay equations, which can be changed to Hamiltonian systems. Mathematics Subject Classification (2000). 34C25, 37J45. Keywords. Hamiltonian system, Floquet theory, symplectic transformation, periodic solution, delay equation. 1. Introduction and main result In 1974, Kaplan and Yorke [1] studied a class of differential delay equations x (t) = f(x(t 1)) (1.1) and x (t) = [f(x(t 1)) + f(x(t 2))], (1.2) where f(x) C(R, R) is odd, xf(x) > 0forx 0,andf satisfies some asymptotically linear conditions both at origin and at infinity. They proved that the Eqs. 1.1 and 1.2 have periodic solutions with period 4 and 6, respectively. They also conjectured that similar results should be true for (1.1) or(1.2) with more than two delays under similar conditions on f. This is called the Kaplan Yorke s conjecture. Since then the two equations and their general cases have been studied by many authors through various methods. In 1978, Nussbaum [3] obtained the existence of a 2n-periodic solution for a class of differential delay equations which include the Eqs. 1.1 and 1.2 as two special cases, where n is a positive integer. This work is supported by the National Natural Science Foundation of China ( ).
2 794 R. Cheng and J. Xu NoDEA Thus, the Kaplan Yorke s conjecture was solved. But the method used in [3] is quite different from that in [1]. Recently many authors [4 9] made use of Kaplan and Yorke s original idea to study some general differential delay equations by changing them into Hamiltonian systems. Li and He [4, 5] studied the existence of multiple periodic solutions of the following equation x (t) = [f(x(t r 1 )) + f(x(t r 2 )) + + f(x(t r n 1 ))] (1.3) where r i (i =1, 2,...,n 1) are positive constants. By variational methods they proved that the system (1.3) has multiple periodic solutions. According to the variational methods used in [4, 5], Cima and his collaborators [6] applied qualitative theory of ordinary differential equations to study the following equation x (t) = [f(x(t r 1 ))g(x(t r 2 )) + f(x(t r 2 ))g(x(t r 1 ))]. They found a suitable condition such that the above equation has periodic solutions. Then Llibre and Tarta [7] generalized the work in [6]. They considered the following system x (t) =af(x(t r 1 ))g(x(t r 2 )) + bf(x(t r 1 ))g(x(t r 3 )) af(x(t r 2 ))g(x(t r 1 )) + cf(x(t r 2 ))g(x(t r 3 )) bf(x(t r 3 ))g(x(t r 1 )) cf(x(t r 1 ))g(x(t r 2 )). They got a sufficient condition for the existence of periodic solutions of the system, where a, b, c take the values 1 or 1. When the delays r i (i = 1, 2,...,n 1) take positive integers, Jekel and Johnston [8] studied the following system x (t) = [f(x(t 1)) + f(x(t 2)) + + f(x(t (n 1)))], (1.4) and by using some known results on the existence of closed orbits for Hamiltonian vector fields they showed that the Eq. 1.4 has a periodic solution. Almost at the same time, Fei [9] took a different approach from those employed in previous papers to study the existence of multiple periodic solutions for the Eq He applied the pseudo index theory established in [10] to obtain the multiple periodic solutions of (1.4) when f(x) is asymptotically linear both at zero and at infinity with respect to x. The previous work is mainly concerned with autonomous differential delay equations. In this paper, we study the existence of periodic solutions for a class of non-autonomous differential delay equations with the following form x (t) = [f(t, x(t τ)) + f(t, x(t 2τ)) + + f(t, x(t (n 1)τ))], (1.5) where τ = π/n and f(t, x) is odd and asymptotically linear both at zero and at infinity with respect to x. We shall apply symplectic transformation and some well known results in critical point theory to obtain the existence of periodic solutions for the Eq. 1.5.
3 Vol. 16 (2009) Periodic solutions for non-autonomous 795 Before introducing our main results, we make the following assumptions. (H 1 ) f(t, x) C(R R, R) is odd with respect to x and periodic in t with period τ. (H 1) f(t, x) C 1 (R R, R) is odd with respect to x and periodic in t with period τ. (H 2 ) For all t [0,τ], f(t, x) f(t, x) lim = β 0 (t), lim = β (t), x 0 x x x i.e., the two limits are uniformly in t. It is obvious that β 0 (t) andβ (t) are periodic functions with period τ. Since there is much difference between n =2N(N Z + )andn =2N +1, we consider the two cases, respectively. In this paper, we always assume that n =2N is even. Let I 2N be the identity matrix in R 2N and α R. We denote the matrix αi 2N by α for convenience. Let A 2N be a 2N 2N skew symmetric matrix defined by A 2N = Let σ(ia 1 2N ) denotes the spectrum of the matrix ia 1 2N. Then from Lemma 4 of [5] and a direct computation, we have σ(ia 1 2N )= ctan 2l 1 } π : l =1, 2,...,2N. 4N Write α 0 = 1 τ τ 0 β 0(t)dt and α = 1 τ τ 0 β (t)dt. Then our main results read as follows. Theorem 1.1. Let (H 1 ) and (H 2 ) hold. Suppose that (H 3 ) α 0 kctan 2l 1 4N π, α kctan 2l 1 4N π, for all k Z+ and 1 l 2N, and (H 4 ) There exist at least an integer l 0 with 1 l 0 2N and some k 0 Z + such that minα 0,α } <k 0 ctan 2l 0 1 4N π<maxα 0,α }. Then the equation 1.5 has at least one nontrivial periodic solution.
4 796 R. Cheng and J. Xu NoDEA Theorem 1.2. Let (H 1) and (H 2 ) hold. Suppose that (H 5 ) α kctan 2l 1 4N π, for all k Z+ and 1 l 2N. Thereexistk i Z + and l i Z + with either 1 l 1 <l 2 < <l n0 N or N +1 l 1 <l 2 < < l n0 2N such that α 0 = k 1 ctan 2l 1 1 4N π = k 2 ctan 2l 2 1 4N π = = k n 0 ctan 2l n 0 1 4N π, where 1 i n 0 and 1 n 0 N. and (H 6 ) There are at least n 0 +1 integers m 1,m 2,...,m n0+1 with 1 m 1 <m 2 < <m n0+1 2N and n 0 +1 integers k 1,k 2,...,k n0+1 such that maxα 0,α } k 1 ctan 2m 1 1 4N π> >k n 0+1 ctan 2m n π>minα 0,α }. 4N Then the equation 1.5 has at least one nontrivial periodic solution. Remark 1.3. In the autonomous case as considered in [4, 5, 9], multiple periodic solutions are obtained. In non-autonomous case, we can only obtain single periodic solution since the functional is not S 1 -invariant. Remark 1.4. As we pointed out above, Theorems 1.1 and 1.2 are concerned with the existence of periodic solutions for non-autonomous differential delay Eq Therefore, our results generalize the results obtained in the references. 2. Proof of the main results We will prove our main results by several steps. We first transform the system (1.5) into an equivalent Hamiltonian system. Then we construct a symplectic transformation which reduces the linear parts of the Hamiltonian system to constant coefficients. Subsequently, some lemmas will be given. Finally, we use two critical point results and Galerkin approximation method to prove Theorems 1.1 and Theorem An equivalent Hamiltonian system We first change the Eq. 1.5 to a form of Hamiltonian system. Suppose that x(t) is a solution for the Eq. 1.5 satisfying x(t) = x(t nτ). Let x 1 (t) =x(t), x 2 (t) =x(t τ),...,x n (t) =x(t (n 1)τ). (2.1) Then X(t) =(x 1 (t),x 2 (t),...,x n (t)) is a solution of the following system d dt X(t) =A 2N F (t, X(t)), (2.2) where F (t, X) =(f(t, x 1 ),f(t, x 2 ),...,f(t, x n )) and A 2N is defined in Sect. 1. Conversely, if a solution X(t) of the system (2.2) has the following symmetric structure x 1 (t) = x n (t τ),x 2 (t) =x 1 (t τ),...,x n (t) =x n 1 (t τ), (2.3)
5 Vol. 16 (2009) Periodic solutions for non-autonomous 797 then x(t) =x 1 (t) is a solution of the Eq. 1.5 satisfying x(t nτ) = x(t). Thus, solutions of the system (2.2) with the symmetric structure (2.3) give solutions of the differential delay equation (1.5). Hence, solving the Eq. 1.5 is equivalent to finding solutions of the Eq. 2.2 with the symmetric structure (2.3). Since A 2N is a nonsingular skew symmetric ( Hamiltonian ) matrix, i.e., it satisfies JA 2N + A 2N J = 0, where J = I2N 0 is the standard symplectic I 2N 0 matrix, the system (2.2) can be written as the following Hamiltonian system y (t) =A 2N y H(t, y), (2.4) where H(t, y) = y 1 0 f(t, x)dx + + y 2N 0 f(t, x)dx, y =(y 1,y 2,...,y 2N ) R 2N,and y H(t, y) denotes the gradient of H(t, y) with respect to y. The matrix A 2N is called the symplectic structure associated to the Hamiltonian system (2.4). Below, we turn to study the Hamiltonian system (2.4) Reducibility of the linear parts The problem now is equivalent to finding periodic solutions of the Hamiltonian system (2.4) with the symmetric structure (2.3). By the assumption (H 1 )and (H 2 ), the system (2.4) is asymptotically linear with respect to y both at origin and at infinity with coefficients of the linear parts depending on t. If we can find a global symplectic transformation in the sense of the symplectic structure A 2N which transforms (2.4) into an equivalent Hamiltonian system with constant coefficients of the linear parts both at zero and at infinity, then we can apply some well known results in [12,13] to prove our theorems. Since the coefficients of the linear parts at origin and infinity may be different, the Floquet theory cannot be applied directly. We want to use Hamiltonian flow mapping to construct such global symplectic transformation. For this purpose, we only need to construct the corresponding Hamiltonian function. By the assumption (H 2 ), we have β0 (t)x + o( x ), as x 0, f(t, x) = β (t)x + o( x ), as x. Then H(t, y) is even with respect to y and satisfies β0 (t)y + o( y ), as y 0, y H(t, y) = β (t)y + o( y ), as y. And the corresponding Hamiltonian system (2.4) satisfies y (t) =A 2N β 0 (t)y + o( y ) as y 0, (2.5) y (t) =A 2N β (t)y + o( y ) as y. (2.6) For the system (2.5), let y = P 1 (t, z), where P 1 (t, z) =e Q1(t) z, Q 1 (t) =A 2N t 0 γ 0 (ξ)dξ and γ 0 (t) =β 0 (t) α 0. Then the transformation y = P 1 (t, z) is symplectic.
6 798 R. Cheng and J. Xu NoDEA Then with the transformation y = P 1 (t, z) the system (2.5) is transformed into the following Hamiltonian system z (t) =α 0 A 2N z + o( z ) as z 0. (2.7) Similarly, let y = P 2 (t, z), where P 2 (t, z) =e Q2(t) z, Q 2 (t) =A 2N t 0 γ (ξ)dξ and γ (t) =β (t) α. In the same way we have that the transformation y = P 2 (t, z) is also symplectic. Then the differential equation (2.6) can be transformed to the following Hamiltonian system z (t) =α A 2N z + o( z ) as z. (2.8) Noting that Q i (t)(i = 1, 2) are τ-periodic, the functions e Qi(t) (i = 1, 2) are bounded. Thus, there exist two positive constants r and R with r < R such that r y e Qi(t) y R y (i =1, 2). (2.9) We now manage to construct a global symplectic transformation Ψ(t, z) such that P1 (t, z), z < r/ y =Ψ(t, z) = R P 2 (t, z), z > R/ r. (2.10) Let Γ 0 (t) = t 0 γ 0(ξ)dξ and Γ (t) = t 0 γ (ξ)dξ. Then Γ 0 (t) and Γ (t) are two τ-periodic functions. Let ρ 1 (w) andρ 2 (w) be two smooth functions satisfying ρ 1 ( w 2 )= 1, as w < r 0, as w > R, ρ 2 ( w 2 )= 0, as w < r 1, as w > R. Let H t (w) = 1 2 Γ 0(t) w 2 ρ 1 ( w 2 )+ 1 2 Γ (t) w 2 ρ 2 ( w 2 ), where t is regarded as a parameter. It is obvious that H t (w) isaτ-periodic function with respect to the parameter t. Consider the following Hamiltonian system dw ds = A 2N w Ht (w). (2.11) Note that the above Hamiltonian system (2.11) is autonomous with respect to s andt is regarded as a parameter. Let Ψ t (s, w) be the general solution of (2.11). By the uniqueness of solutions of ordinary differential equations, Ψ t (s, w) is τ-periodic with respect to the parameter. Now we prove that the flow mapping Ψ t (s, w) is symplectic with respect to the symplectic form A 2N, that is, ( ) Ψ t (s, w) Ψt (s, w) A 2N = A 2N. w w Since A 2N is skew symmetric and non-degenerate, there is a non-degenerate matrix S such that SA 2N S = J,
7 Vol. 16 (2009) Periodic solutions for non-autonomous 799 where J is the standard symplectic matrix. Let w = S 1 z. Then the system (2.11) is transformed to the following standard Hamiltonian system d z ds = J z Ht ( z), (2.12) where Ht ( z) = H t (S 1 z). Let Ψ t (s, z) = SΨ t (s, S 1 z). Then Ψ t (s, z) isthe general solution of the Hamiltonian system (2.12). Thus, we have that Ψ t ( Ψ t z ) = J, i.e., S Ψ ( t w S 1 J S Ψ ) t w S 1 = J S Ψ t w (S 1 J(S 1 ) ) Ψ ( ) t w A Ψt 2N = A 2N. w ( ) Ψt S = J w z J Let y =Ψ(t, z) =Ψ t (s, w) s=1,w=z =Ψ t (1,z). Then we have ( ) Ψ Ψ z A 2N = A 2N. z For w < r, the system (2.11) becomes dw/ds=γ 0 (t)a 2N w.so,w(s)=e Γ0(t)A2N s z. If s < 1, z < r/ R, then w(s) = e Γ0(t)A2N s z R z < r. Thusfor s < 1, z < r/ R, wehavew(s) =e Γ0(t)A2N s z. Therefore, Ψ(t, z) =P 1 (t, z) as z < r/ R. In the same way we can prove that Ψ(t, z) =P 2 (t, z) as z > R/ r. Let z =Φ(t, y) be the inverse mapping of the transformation y =Ψ(t, z). Then the system (2.4) is changed to the following system dz dt = A 2N z Ĥ(t, z)+ Φ t, (2.13) where Ĥ(t, z) =H(t, Ψ(t, z)). By a direct computation, A 1 2N 2 Φ t y is symmetric. Then according to [15], there is a smooth function R such that A 2N z R = Φ t. Then H (t, z) =Ĥ(t, z)+r(t, z) is the Hamiltonian function of the system (2.13) and the Hamiltonian system (2.13) can be rewritten as dz dt = A 2N z H (t, z). (2.14) By (2.7) and (2.8), H (t, z) satisfies the following properties. z H (t, z) =α 0 z + o( z ) as z 0 uniformly in t, (2.15) z H (t, z) =α z + o( z ) as z uniformly in t. (2.16)
8 800 R. Cheng and J. Xu NoDEA Now in order to prove the main results of this paper, we only need to consider the Hamiltonian system (2.14). In the following of this paper, we prove the following two theorems. Theorem 2.1. Under the assumptions of Theorem 1.1, the system 2.14 has at least one nontrivial periodic solution. Theorem 2.2. Under the assumptions of Theorem 1.2, the system 2.14 has at least one nontrivial periodic solution Some lemmas We now have constructed the global symplectic transformation y = Ψ(t, z) which transforms the system (2.4) into the Hamiltonian system (2.14). Note that the system (2.14) satisfies the two properties (2.15) and (2.16). We shall use two well known critical point results given in [12, 13] to prove the two theorems. We work in the Hilbert space E = W 1 2,2 (S 1, R 2N ) with the inner product, and the norm. The space E consists of all z(t) inl 2 (S 1, R 2N )whose Fourier series + z(t) =a 0 + (a k coskt + b k sinkt) satisfies a The inner product on E is defined by z 1,z 2 =(a 1 0,a 2 0)+ 1 2 k( a k 2 + b k 2 ) < +. k[(a 1 k,a 2 k)+(b 1 k,b 2 k)], where z i = a i (ai k coskt + bi ksinkt)(i =1, 2) and (, ) denotes the inner product in R 2N. Recall that A 1 2N is also a nonsingular skew symmetric matrix. For z,y E, we define an operator A on E by extending the bilinear form Az, y = 2π 0 (A 1 2N z (t),y(t))dt. (2.17) It is not difficult to compute that A is a bounded self-adjoint linear operator on E and Az = + (A 1 2N b kcoskt A 1 2N a ksinkt). (2.18)
9 Vol. 16 (2009) Periodic solutions for non-autonomous 801 For each z E, we define a functional ψ on E by ψ(z) = 1 2π 2 Az(t),z(t) H (t, z(t))dt. It is well known that critical points of ψ are solutions of the system (2.14). Hence, to seek periodic solutions of the system (2.14) is equivalent to seeking critical points of ψ. For any α R and z,y E, we define another operator B α on E by 0 2π B α z,y = (αz(t),y(t))dt. (2.19) 0 Then B α is also a bounded self-adjoint linear operator on E. Moreover,B α is compact and for any z E, + 1 B α z = αa 0 + k ( αa kcoskt αb k sinkt). (2.20) For any z(t) E and by (2.19) and (2.20), we have (A + B α)z(t) = αa (( A 1 2N b k 1 ) ( k αa k coskt + A 1 2N a k 1 ) ) k αb k sinkt. (2.21) Lemma 2.3. Suppose that (H 1 ) and (H 2 ) hold. Then the functional ψ satisfies ψ (z) (A + B α0 )z = o( z ) as z 0, (2.22) ψ (z) (A + B α )z = o( z ) as z. (2.23) Proof. From (2.15), for any r >0, there exists a constant C( r) (here and in the following C denotes various constants) such that x H (t, x) α 0 x r x + C( r) x 2, for each x R 2N. Notice that ψ(z) = 1 2π ( 2 (A + B α 0 )z(t),z(t) H (t, z(t)) 1 ) 2 (α 0z(t),z(t)) dt. Thus, we have 0 ψ (z) (A + B α0 )z C z H (t, z) α 0 z C( r z + C( r) z 2 ), which means (2.22). By (2.16), we have x H (t, x) α x r x + C( r), for each x R 2N. Then (2.23) holds similarly.
10 802 R. Cheng and J. Xu NoDEA In order to obtain solutions of (2.14) with the symmetric structure (2.3), we define a 2N 2N matrix T 2N with the following form T 2N = Then by T 2N, for any z(t) E, define an action on z by δz(t) =T 2N z(t τ). (2.24) Then by a direct computation we have that δ 2N z(t) = z(t 2Nτ),δ 4N z(t) =z(t) and G = δ, δ 2,...,δ 4N } is a compact group action over E. Ifδz(t) =z(t) holds, then through a direct check, we have that z(t) has the symmetric structure (2.3). Write SE = z E : δz(t) =z(t)}. Then from z(t) = z(t 2Nτ) and a direct computation, we have } SE = z E : z(t) = (a k cos(2k 1)t + b k sin(2k 1)t). Let ψ SE be the restriction of ψ on SE. We have the following lemma. Lemma 2.4. Critical points of ψ SE over SE are critical points of ψ on E. Proof. We first check that T 2N Ψ t (s, w) =Ψ t (s, T 2N w), where Ψ t (s, w) is the flow mapping generated by the system (2.11). In fact, by the definition of H, weget w (s) =A 2N Γ 0 (t)wρ 1 ( w 2 )+Γ 0 (t) w 2 ρ 1( w 2 )w +Γ (t)wρ 2 ( w 2 )+Γ (t) w 2 ρ 2( w 2 )w}. Notice that T 2N T2N = I and T 2N commutes with A 2N and J. Set z(s) = Ψ t (s, T 2N w), ỹ(s) =Ψ t (s, w), then d z(s) = A 2N Γ 0 (t)t 2N wρ 1 ( w 2 )+Γ 0 (t) w 2 ρ ds 1( w 2 )T 2N w +Γ (t)t 2N wρ 2 ( w 2 )+Γ (t) w 2 ρ 2( w 2 )T 2N w} = T 2N A 2N Γ 0 (t)wρ 1 ( w 2 )+Γ 0 (t) w 2 ρ 1( w 2 )w +Γ (t)wρ 2 ( w 2 )+Γ (t) w 2 ρ 2( w 2 )w} dỹ(s) = T 2N ds, this, jointly with Ψ t (0,T 2N w)=t 2N w = T 2N Ψ t (0,w) shows that T 2N Ψ t (s, w) = Ψ t (s, T 2N w). The definition of H, together with a direct computation implies that H (t, T 2N z)=h (t, z), z H (t, T 2N z)=t 2N z H (t, z).
11 Vol. 16 (2009) Periodic solutions for non-autonomous 803 Combining these with (2.24) and the fact that any z(t) E is 2π-periodic, we can verify that ψ(δz) =ψ(z), ψ (δz) =δψ (z) =ψ (z). That means ψ is G-invariant and ψ is G-equivariant. Moreover, ψ (z) SE. Therefore, if ψ (z) =0onSE, then ψ (z) =0onE. Hence, the conclusion of Lemma 2.4 holds. Remark 2.5. It follows from Lemma 2.4 that we only need to find critical points of ψ SE on SE. In the following, we use Galerkin approximation method and two critical point results to obtain critical points of ψ SE. We first analyze the spectrum of the operator A + B α. Observing (2.21), we only need to analyze the spectrum of the following matrix ( αi2n ka 1 ) T k (α) = 2N. αi 2N ka 1 2N Let λ 0 be an eigenvalue of A 2N.ByLemma4of[5], Proposition 3.6 of [16] and a direct computation, we have det(λ 0 I 2N A 2N )= 1 2 ((1 λ 0) 2N +(1+λ 0 ) 2N )=0. (2.25) Then (2.25) gives σ(a 2N )= ±itan 2l 1 π : l =1, 2,...,N 4N }, (2.26) i.e., A 2N has N pairs of pure imaginary eigenvalues. Let λ be an eigenvalue of the matrix T k (α). Then det(λi 4N T k (α)) = det((λ + α) 2 I 2N +(ka 1 2N )2 ) = det((λ + α) 2 I 2N (ika 1 2N )2 ) = det((λ + α)i 2N ika 1 2N )det((λ + α)i 2N + ika 1 2N ) = det(λi 2N (ika 1 2N αi 2N ))det(λi 2N ( ika 1 2N αi 2N)). Notice that ika 1 2N αi 2N and ika 1 2N αi 2N have the same spectrum. It is easy to see that if λ takes the eigenvalue of ika 1 2N αi 2N, then det(λi 4N T k (α)) = 0. By (2.26), one has σ(ia 1 2N )= ±ctan 2l 1 4N π : l =1, 2,...,N }, (2.27) i.e. ia 1 2N has N pairs of real eigenvalues. In order to prove our main results we need to use the following index. By M ( ) andm 0 ( ) we denote the negative
12 804 R. Cheng and J. Xu NoDEA and the zero Morse indices of the symmetric matrix, respectively. For a constant diagonal matrix αi 2N, we define our index as i (α) = (M (T k (α)) 2N) and i 0 (α) = M 0 (T k (α)). Since for k large enough, one has M (T k (α)) = 2N and M 0 (T k (α)) = 0. Hence i (α) andi 0 (α) are well defined. The following lemma gives the relation between α R and the indices i 0 (α),i (α). Lemma 2.6. (1) Suppose that (H 3 ) hold. Then i 0 (α 0 )=i 0 (α )=0. (2) Suppose that (H 3 ) and (H 4 ) hold. Then i (α 0 ) i (α ). (3) Suppose that (H 5 ) and (H 6 ) hold. Then i (α ) / [i (α 0 ),i (α 0 )+i 0 (α 0 )]. Proof. Case (1). From (2.27), if α 0 kctan 2l 1 4N π, for all k Z+ and 1 l 2N, then λ 0, where λ is the eigenvalue of T k (α 0 ). That means M 0 (T k (α 0 )) = 0 for k 1. Thus i 0 (α 0 )= M 0 (T k (α 0 )) = 0. Similarly, we have i 0 (α )=0. Case (2). Without loss of generality, we suppose that α 0 <α. Then by the conditions (H 3 ) and (H 4 ), α 0 <k 0 ctan 2l 0 1 4N π<α. Then by a direct check, we have M (T k0 (α 0 )) + 2 M (T k0 (α )). For each k k 0, one can check easily that M (T k (α 0 )) M (T k (α )). Hence, one has (M (T k (α 0 )) 2N) < (M (T k (α )) 2N), since M (T k (α)) = 2N for k large enough. This yields that i (α 0 ) <i (α ). Then case (2) holds. Case (3). By the assumption (H 5 ), i 0 (α 0 )=2n 0. Without loss of generality, let α 0 <α. Then from (H 6 ) and with a similar argument to case (2), we get M (T kj (α )) 2+M (T kj (α 0 )), for j =1, 2,...,n Moreover, M (T k (α )) M (T k (α 0 )) for k k j (j =1, 2,...,n 0 + 1). Thus, i (α ) (i (α 0 )+i 0 (α 0 )) 2. That means i (α ) / [i (α 0 ),i (α 0 )+i 0 (α 0 )]. For k 1, set SE(k) =z E : z(t) =a k cos(2k 1)t + b k sin(2k 1)t} and SE m = SE(1) SE(2) SE(m). Let Ω = P m : m =1, 2,...} be a sequence of orthogonal projections. Ω is called a Galerkin approximation scheme with respect to the operator A + B α,if it satisfies the following properties:
13 Vol. 16 (2009) Periodic solutions for non-autonomous 805 (1) The image of P m, as a subspace of SE, has finite dimension; (2) P m z z as m for any z SE; (3) P m commutes with A + B α. Let P m : SE SE m be the orthogonal projection. One may check easily that P m satisfies the above properties (1 3). Thus, Ω is a Galerkin approximation scheme with respect to the operator A + B α. Let A m and ψ SEm be the restrictions of A and ψ SE on SE m, respectively. We have the following important lemma. Lemma 2.7. Let (H 1 ) and (H 2 ) hold. If α kctan 2l 1 4N π, for all k Z+ and 1 l 2N, then (1) ψ SE satisfies (PS) condition over SE, i.e., every sequence z j } SE with z j SE j, ψ SEj (z j ) 0 and ψ SE (z j ) being bounded, possesses a convergent subsequence. (2) ψ SEj satisfies (PS) condition, i.e., every sequence z i } SE j, ψ SEj (z i ) 0 and ψ SEj (z i ) being bounded, possesses a convergent subsequence. Proof. From Lemma 2.6, α kctan 2l 1 4N π, for all k Z+ and 1 l 2N means that i 0 (α ) = 0 which yields that A + B α has a bounded inverse. Then the proof is standard. Observe that our functional ψ SEm acts on the finite dimensional space SE m. The following two lemmas are well known results in critical point theory. Please consult [2, 12, 13]. Lemma 2.8. Let ψ SEm be a C 1 function satisfying (2.22) and (2.23). IfM 0 (A m + P m B α0 P m )=M 0 (A m +P m B α P m )=0and M (A m +P m B α0 P m ) M (A m + P m B α P m ), then ψ SEm has at least one nontrivial critical point. Lemma 2.9. Let ψ SEm be a C 2 function satisfying (2.22) and (2.23). IfM 0 (A m + P m B α P m )=0and M (A m +P m B α P m ) / [M (A m +P m B α0 P m ),M (A m + P m B α0 P m )+M 0 (A m +P m B α0 P m )], then ψ SEm has at least one nontrivial critical point. To apply Lemmas 2.8 and 2.9 to get critical points of ψ SE, we introduce the following lemma. Lemma If α kctan 2l 1 4N π, for all k Z+ and 1 l 2N, then M (A m + P m B α P m ) M (A m )=M (A k + P k B α P k ) M (A k ) for m, k large enough. Note that α kctan 2l 1 4N π, for all k Z+ and 1 l 2N, means that i 0 (α) = 0. Then the proof is the same as in [13]. For the linear compact operator B α, we define I (B α )=k = M (A m + P m B α P m ) M (A m ), for infinitely many m}, I 0 (B α )=k = M 0 (A m + P m B α P m ), for infinitely many m}.
14 806 R. Cheng and J. Xu NoDEA Let I (ψ, ) =I (B α ), I (ψ, 0) = I (B α0 ); I 0 (ψ, ) =I 0 (B α ), I 0 (ψ, 0) = I 0 (B α0 ). By Lemma 2.10, the index above is singleton and well defined. Now we prove the following lemma. Lemma With respect to the approximation scheme Ω, one has I (ψ SE, 0) = (M (T k (α 0 )) 2N) =i (α 0 ), I (ψ SE, ) = I 0 (ψ SE, 0) = (M (T k (α )) 2N) =i (α ), M 0 (T k (α 0 )) = i 0 (α 0 ). Proof. It follows from the definition of the negative Morse index of T k (α 0 )and (2.21) that the dimension of the negative eigenspace of the operator A m +P m B α0 P m on SE m = m j=1 SE(j) is equal to m (M (T k (α 0 )). For α 0 =0,M (T k (α 0 )= 2N. Therefore, the formula I (ψ SE, 0) = (M (T k (α 0 )) 2N) =i (α 0 ) holds. Similarly, we have the other formulas Proof of Theorem 2.1 We are now ready to prove the main results. We first give the proof of Theorem 2.1. ProofofTheorem2.1 Under the assumptions of Theorem 2.1 and by (2.23) and Lemma 2.6, forɛ = 1 2 (A + B α ) 1 1, there is a large R>0 such that That yields ψ SE (z) (A + B α )(z) <ɛ z for z >R. ψ SE (z) (A + B α ) 1 1 z ψ SE (z) (A + B α )(z) >ɛ z for z >R. This means ψ SE has no critical points outside the ball B R. Since A + B α0 has bounded inverse and (I P m )B α0 0asm 0, there exists a constant C such that (A m + P m B α0 P m ) 1 <C.By(2.22), we can take r small enough with r<rsuch that ψ SE (z) (A + B α0 )(z) < 1 z for z <r. 2C
15 Vol. 16 (2009) Periodic solutions for non-autonomous 807 Thus, for m large enough and for z SE m with z <r, one has ψ SE (z) (A m + P m B α0 P m )(z) ψ SE (z) (A + B α0 )(z) < 1 2C z From the above inequality, we get < 1 2 (A m + P m B α0 P m ) 1 1 z. ψ SE (z) > (A m + P m B α0 P m )(z) 1 2 (A m + P m B α0 P m ) 1 1 z > (A m + P m B α0 P m ) 1 1 z 1 2 (A m + P m B α0 P m ) 1 1 z = 1 2 (A m + P m B α0 P m ) 1 1 z. Therefore, 0 is the unique critical point of ψ SE inside B r. By Lemmas 2.6 and 2.11, M (A m + P m B α0 P m ) M (A m + P m B α P m ). Then Lemmas 2.3 and 2.8 yield ψ SEm has a nontrivial critical point z m with r< z m <R. By Lemma 2.7, ψ SE satisfies the (PS) condition. So, z m } has a subsequence convergent to a point z, which is just a critical point of ψ SE with r< z <R Proof of Theorem 2.2 In this section, we prove Theorem 2.2. ProofofTheorem2.2 Since M 0 (T m (α 0 )) = 0 for m large enough, the null space of (A + B α0 ) SE is included in SE m, where (A + B α0 ) SE is the restriction of A + B α0 on SE. Thus, there exists a constant C such that (A m + P m B α0 P m ) <C for m large enough, where (A m + P m B α0 P m ) denotes the inverse of A m + P m B α0 P m restricted in the range of A m + P m B α0 P m.by(2.22) we can take a small r such that ψ SEm (z) (A + B α0 ) 1 as z <r. 2C As z < 2r, one has ψ SEm (z) (A m + P m B α0 P m ) ψ SEm (z) (A + B α0 ) 1 2C < 1 2 (A m + P m B α0 P m ) 1. Then by a similar argument with [13, Theorem 1.3], we have the fact that if ψ SEm has critical points, then at least one of them is outside B r. By Lemmas 2.6 and
16 808 R. Cheng and J. Xu NoDEA 2.11, we have for m large enough M (A m + P m B α P m ) / [ M (A m + P m B α0 P m ),M (A m + P m B α0 P m ) +M 0 (A m + P m B α0 P m ) ]. Hence, Lemma 2.9, jointly with Lemma 2.3 and the proof of Theorem 2.1 yields that ψ SEm has a nontrivial critical point z m with r< z m <R. By Lemma 2.7, ψ SE satisfies the (PS) condition. So, z m } has a subsequence convergent to a point z, which is just a critical point of ψ SE with r< z <R. References [1] Kaplan, J., Yorke, J.: Ordinary differential equations which yield periodic solutions of differential delay equations. J. Math. Anal. Appl. 48, (1974) [2] Amann, H., Zehnder, E.: Nontrivial solutions for a class of nonresonance problems and applications to nonlinear differential equations. Ann. Scuola Norm. Sup. Pisa Cl. Sci (4) 8, (1980) [3] Nussbaum, R.: Periodic solutions of special differential delay equations: an example in non-linear functional analysis. Proc. R. Soc. Edinb. 81A, (1978) [4] Li, J., He, X., Liu, Z.: Hamiltonian symmetric groups and multiple periodic solutions of differential delay equations. Nonlinear Anal. T.M.A. 35, (1999) [5] Li, J., He, X.: Multiple periodic solutions of differential delay equations created by asymptotically linear Hamiltonian systems. Nonlinear Anal. T.M.A. 31, (1998) [6] Cima, A., Jibin, L., Llibre, J.: Periodic solutions of delay differential equations with two delays via bi-hamiltonian systems. Ann. Diff. Equ. 17, (2001) [7] Llibre, J., Tarta, A.-A.: Periodic solutions of delay equations with three delays via bi-hamiltonian systems. Nonlinear Anal. T.M.A. 64, (2006) [8] Jekel, S., Johnston, C.: A Hamiltonian with periodic orbits having several delays. J. Diff. Equ. 222, (2006) [9] Fei, G.: Multiple periodic solutions of differential delay equations via Hamiltonian systems (I). Nonlinear Anal. T.M.A. 65, (2006) [10] Fannio, L.O.: Multiple periodic solutions of Hamiltonian systems with strong resonance at infinity. Discrete Cont. Dyn. Syst. 3, (1997) [11] Long, Y., Zehnder, E.: Morse theory for forced oscillations of asymptotically linear Hamiltonian systems. In: Albeverio, S. et al. (eds.) Stochatic processes, Physics and Geometry. Proceedings of the Conference in Asconal/Locarno, Switzerland, pp , World Scientific, Singapore, 1990
17 Vol. 16 (2009) Periodic solutions for non-autonomous 809 [12] Chang, K.C.: Solutions of asymptotically linear operator equations via Morse theory. Comm. Pure Appl. Math. 34, (1981) [13] Li, S., Liu, J.: Morse theory and asymptotically linear Hamiltonian systems. J. Diff. Equ. 78, (1989) [14] Mawhin, J., Willem, M.: Critical point theory and Hamiltonian systems, vol. 74. Appl. Math. Sci. Springer, Heidelberg (1989) [15] Meyer, K.R., Hall, G.R.: Introduction to Hamiltonian dynamical systems and the n-body problem. Springer, New York (1992) [16] Liu, Z., Li, J.: Hamiltonian systems and periodic solutions for differential delay equations. Science Publishhouse, Beijing (2000) R. Cheng, J. Xu Department of Mathematics, Southeast University, Nanjing, People s Republic of China mathchr@163.com J. Xu xujun@seu.edu.cn R. Cheng College of Mathematics and Physics, Nanjing University of Information Science and Technology, Nanjing, People s Republic of China Received: 24 December Accepted: 11 June 2009.
UNIVERSITETET I OSLO
UNIVERSITETET I OSLO Det matematisk-naturvitenskapelige fakultet Eksamen i MAT2400 Analyse 1. Eksamensdag: Onsdag 15. juni 2011. Tid for eksamen: 09.00 13.00 Oppgavesettet er på 6 sider. Vedlegg: Tillatte
DetaljerUnit Relational Algebra 1 1. Relational Algebra 1. Unit 3.3
Relational Algebra 1 Unit 3.3 Unit 3.3 - Relational Algebra 1 1 Relational Algebra Relational Algebra is : the formal description of how a relational database operates the mathematics which underpin SQL
DetaljerTrigonometric Substitution
Trigonometric Substitution Alvin Lin Calculus II: August 06 - December 06 Trigonometric Substitution sin 4 (x) cos (x) dx When you have a product of sin and cos of different powers, you have three different
DetaljerUNIVERSITY OF OSLO DEPARTMENT OF ECONOMICS
UNIVERSITY OF OSLO DEPARTMENT OF ECONOMICS Postponed exam: ECON420 Mathematics 2: Calculus and linear algebra Date of exam: Tuesday, June 8, 203 Time for exam: 09:00 a.m. 2:00 noon The problem set covers
DetaljerGraphs similar to strongly regular graphs
Joint work with Martin Ma aj 5th June 2014 Degree/diameter problem Denition The degree/diameter problem is the problem of nding the largest possible graph with given diameter d and given maximum degree
DetaljerSlope-Intercept Formula
LESSON 7 Slope Intercept Formula LESSON 7 Slope-Intercept Formula Here are two new words that describe lines slope and intercept. The slope is given by m (a mountain has slope and starts with m), and intercept
DetaljerDynamic Programming Longest Common Subsequence. Class 27
Dynamic Programming Longest Common Subsequence Class 27 Protein a protein is a complex molecule composed of long single-strand chains of amino acid molecules there are 20 amino acids that make up proteins
DetaljerSVM and Complementary Slackness
SVM and Complementary Slackness David Rosenberg New York University February 21, 2017 David Rosenberg (New York University) DS-GA 1003 February 21, 2017 1 / 20 SVM Review: Primal and Dual Formulations
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
UNIVERSITETET I OSLO ØKONOMISK INSTITUTT Exam: ECON320/420 Mathematics 2: Calculus and Linear Algebra Eksamen i: ECON320/420 Matematikk 2: Matematisk analyse og lineær algebra Date of exam: Friday, May
DetaljerPhysical origin of the Gouy phase shift by Simin Feng, Herbert G. Winful Opt. Lett. 26, (2001)
by Simin Feng, Herbert G. Winful Opt. Lett. 26, 485-487 (2001) http://smos.sogang.ac.r April 18, 2014 Introduction What is the Gouy phase shift? For Gaussian beam or TEM 00 mode, ( w 0 r 2 E(r, z) = E
DetaljerUniversitetet i Bergen Det matematisk-naturvitenskapelige fakultet Eksamen i emnet Mat131 - Differensiallikningar I Onsdag 25. mai 2016, kl.
1 MAT131 Bokmål Universitetet i Bergen Det matematisk-naturvitenskapelige fakultet Eksamen i emnet Mat131 - Differensiallikningar I Onsdag 25. mai 2016, kl. 09-14 Oppgavesettet er 4 oppgaver fordelt på
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
UNIVERSITETET I OSLO ØKONOMISK INSTITUTT Utsatt ksamen i: ECON3120/4120 Matematikk 2: Matematisk analyse og lineær algebra Postponed exam: ECON3120/4120 Mathematics 2: Calculus and linear algebra Eksamensdag:
DetaljerMoving Objects. We need to move our objects in 3D space.
Transformations Moving Objects We need to move our objects in 3D space. Moving Objects We need to move our objects in 3D space. An object/model (box, car, building, character,... ) is defined in one position
DetaljerStationary Phase Monte Carlo Methods
Stationary Phase Monte Carlo Methods Daniel Doro Ferrante G. S. Guralnik, J. D. Doll and D. Sabo HET Physics Dept, Brown University, USA. danieldf@het.brown.edu www.het.brown.edu Introduction: Motivations
DetaljerGeneralization of age-structured models in theory and practice
Generalization of age-structured models in theory and practice Stein Ivar Steinshamn, stein.steinshamn@snf.no 25.10.11 www.snf.no Outline How age-structured models can be generalized. What this generalization
DetaljerNeural Network. Sensors Sorter
CSC 302 1.5 Neural Networks Simple Neural Nets for Pattern Recognition 1 Apple-Banana Sorter Neural Network Sensors Sorter Apples Bananas 2 Prototype Vectors Measurement vector p = [shape, texture, weight]
DetaljerOppgave 1. ( xφ) φ x t, hvis t er substituerbar for x i φ.
Oppgave 1 Beviskalklen i læreboka inneholder sluttningsregelen QR: {ψ φ}, ψ ( xφ). En betingelse for å anvende regelen er at det ikke finnes frie forekomste av x i ψ. Videre så inneholder beviskalklen
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
UNIVERSITETET I OSLO ØKONOMISK INSTITUTT Eksamen i: ECON30/40 Matematikk : Matematisk analyse og lineær algebra Exam: ECON30/40 Mathematics : Calculus and Linear Algebra Eksamensdag: Tirsdag 0. desember
DetaljerMathematics 114Q Integration Practice Problems SOLUTIONS. = 1 8 (x2 +5x) 8 + C. [u = x 2 +5x] = 1 11 (3 x)11 + C. [u =3 x] = 2 (7x + 9)3/2
Mathematics 4Q Name: SOLUTIONS. (x + 5)(x +5x) 7 8 (x +5x) 8 + C [u x +5x]. (3 x) (3 x) + C [u 3 x] 3. 7x +9 (7x + 9)3/ [u 7x + 9] 4. x 3 ( + x 4 ) /3 3 8 ( + x4 ) /3 + C [u + x 4 ] 5. e 5x+ 5 e5x+ + C
DetaljerDatabases 1. Extended Relational Algebra
Databases 1 Extended Relational Algebra Relational Algebra What is an Algebra? Mathematical system consisting of: Operands --- variables or values from which new values can be constructed. Operators ---
DetaljerSplitting the differential Riccati equation
Splitting the differential Riccati equation Tony Stillfjord Numerical Analysis, Lund University Joint work with Eskil Hansen Innsbruck Okt 15, 2014 Outline Splitting methods for evolution equations The
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
UNIVERSITETET I OSLO ØKONOMISK INSTITUTT Eksamen i: ECON320/420 Matematikk 2: Matematisk analyse og lineær algebra Exam: ECON320/420 Mathematics 2: Calculus and Linear Algebra Eksamensdag: Tirsdag 7. juni
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
UNIVERSITETET I OSLO ØKONOMISK INSTITUTT Eksamen i: ECON3120/4120 Matematikk 2: Matematisk analyse og lineær algebra Exam: ECON3120/4120 Mathematics 2: Calculus and Linear Algebra Eksamensdag: Tirsdag
DetaljerGradient. Masahiro Yamamoto. last update on February 29, 2012 (1) (2) (3) (4) (5)
Gradient Masahiro Yamamoto last update on February 9, 0 definition of grad The gradient of the scalar function φr) is defined by gradφ = φr) = i φ x + j φ y + k φ ) φ= φ=0 ) ) 3) 4) 5) uphill contour downhill
DetaljerCall function of two parameters
Call function of two parameters APPLYUSER USER x fµ 1 x 2 eµ x 1 x 2 distinct e 1 0 0 v 1 1 1 e 2 1 1 v 2 2 2 2 e x 1 v 1 x 2 v 2 v APPLY f e 1 e 2 0 v 2 0 µ Evaluating function application The math demands
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
UNIVERSITETET I OSLO ØKONOMISK INSTITUTT Eksamen i: ECON20/420 Matematikk 2: Matematisk analyse og lineær algebra Exam: ECON20/420 Mathematics 2: Calculus and Linear Algebra Eksamensdag: Fredag 2. mai
DetaljerHan Ola of Han Per: A Norwegian-American Comic Strip/En Norsk-amerikansk tegneserie (Skrifter. Serie B, LXIX)
Han Ola of Han Per: A Norwegian-American Comic Strip/En Norsk-amerikansk tegneserie (Skrifter. Serie B, LXIX) Peter J. Rosendahl Click here if your download doesn"t start automatically Han Ola of Han Per:
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
UNIVERSITETET I OSLO ØKONOMISK INSTITUTT Eksamen i: ECON320/420 Matematikk 2: Matematisk analyse og lineær algebra Exam: ECON320/420 Mathematics 2: Calculus and Linear Algebra Eksamensdag: Onsdag 6. desember
DetaljerSolutions #12 ( M. y 3 + cos(x) ) dx + ( sin(y) + z 2) dy + xdz = 3π 4. The surface M is parametrized by σ : [0, 1] [0, 2π] R 3 with.
Solutions #1 1. a Show that the path γ : [, π] R 3 defined by γt : cost ı sint j sint k lies on the surface z xy. b valuate y 3 cosx dx siny z dy xdz where is the closed curve parametrized by γ. Solution.
DetaljerTFY4170 Fysikk 2 Justin Wells
TFY4170 Fysikk 2 Justin Wells Forelesning 5: Wave Physics Interference, Diffraction, Young s double slit, many slits. Mansfield & O Sullivan: 12.6, 12.7, 19.4,19.5 Waves! Wave phenomena! Wave equation
DetaljerRingvorlesung Biophysik 2016
Ringvorlesung Biophysik 2016 Born-Oppenheimer Approximation & Beyond Irene Burghardt (burghardt@chemie.uni-frankfurt.de) http://www.theochem.uni-frankfurt.de/teaching/ 1 Starting point: the molecular Hamiltonian
DetaljerSecond Order ODE's (2P) Young Won Lim 7/1/14
Second Order ODE's (2P) Copyright (c) 2011-2014 Young W. Lim. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or
DetaljerUNIVERSITETET I OSLO
UNIVERSITETET I OSLO Det matematisk-naturvitenskapelige fakultet Eksamen i INF 3230 Formell modellering og analyse av kommuniserende systemer Eksamensdag: 4. juni 2010 Tid for eksamen: 9.00 12.00 Oppgavesettet
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
UNIVERSITETET I OSLO ØKONOMISK INSTITUTT Eksamen i: ECON320/420 Matematikk 2: Matematisk analyse og lineær algebra Exam: ECON320/420 Mathematics 2: Calculus and Linear Algebra Eksamensdag: Mandag 8. desember
DetaljerVerifiable Secret-Sharing Schemes
Aarhus University Verifiable Secret-Sharing Schemes Irene Giacomelli joint work with Ivan Damgård, Bernardo David and Jesper B. Nielsen Aalborg, 30th June 2014 Verifiable Secret-Sharing Schemes Aalborg,
DetaljerTrust region methods: global/local convergence, approximate January methods 24, / 15
Trust region methods: global/local convergence, approximate methods January 24, 2014 Trust region methods: global/local convergence, approximate January methods 24, 2014 1 / 15 Trust-region idea Model
DetaljerExistence of resistance forms in some (non self-similar) fractal spaces
Existence of resistance forms in some (non self-similar) fractal spaces Patricia Alonso Ruiz D. Kelleher, A. Teplyaev University of Ulm Cornell, 12 June 2014 Motivation X Fractal Motivation X Fractal Laplacian
DetaljerKneser hypergraphs. May 21th, CERMICS, Optimisation et Systèmes
Kneser hypergraphs Frédéric Meunier May 21th, 2015 CERMICS, Optimisation et Systèmes Kneser hypergraphs m, l, r three integers s.t. m rl. Kneser hypergraph KG r (m, l): V (KG r (m, l)) = ( [m]) l { E(KG
DetaljerSTILLAS - STANDARD FORSLAG FRA SEF TIL NY STILLAS - STANDARD
FORSLAG FRA SEF TIL NY STILLAS - STANDARD 1 Bakgrunnen for dette initiativet fra SEF, er ønsket om å gjøre arbeid i høyden tryggere / sikrere. Både for stillasmontører og brukere av stillaser. 2 Reviderte
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
UNIVERSITETET I OSLO ØKONOMISK INSTITUTT Utsatt eksamen i: ECON320/420 Matematikk 2: Matematisk analyse og lineær algebra Postponed exam: ECON320/420 Mathematics 2: Calculus and Linear Algebra Eksamensdag:
Detaljer0:7 0:2 0:1 0:3 0:5 0:2 0:1 0:4 0:5 P = 0:56 0:28 0:16 0:38 0:39 0:23
UTKAST ENGLISH VERSION EKSAMEN I: MOT100A STOKASTISKE PROSESSER VARIGHET: 4 TIMER DATO: 16. februar 2006 TILLATTE HJELPEMIDLER: Kalkulator; Tabeller og formler i statistikk (Tapir forlag): Rottman: Matematisk
DetaljerHØGSKOLEN I NARVIK - SIVILINGENIØRUTDANNINGEN
HØGSKOLEN I NARVIK - SIVILINGENIØRUTDANNINGEN EKSAMEN I FAGET STE 6243 MODERNE MATERIALER KLASSE: 5ID DATO: 7 Oktober 2005 TID: 900-200, 3 timer ANTALL SIDER: 7 (inklusiv Appendix: tabell og formler) TILLATTE
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
UNIVERSITETET I OSLO ØKONOMISK INSTITUTT Eksamen i: ECON3120/4120 Mathematics 2: Calculus an linear algebra Exam: ECON3120/4120 Mathematics 2: Calculus an linear algebra Eksamensag: Tirsag 3. juni 2008
DetaljerMaple Basics. K. Cooper
Basics K. Cooper 2012 History History 1982 Macsyma/MIT 1988 Mathematica/Wolfram 1988 /Waterloo Others later History Why? Prevent silly mistakes Time Complexity Plots Generate LATEX This is the 21st century;
DetaljerTMA4329 Intro til vitensk. beregn. V2017
Norges teknisk naturvitenskapelige universitet Institutt for Matematiske Fag TMA439 Intro til vitensk. beregn. V17 ving 4 [S]T. Sauer, Numerical Analysis, Second International Edition, Pearson, 14 Teorioppgaver
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
UNIVERSIEE I OSLO ØKONOMISK INSIU Eksamen i: ECON320/420 Mathematics 2: Calculus and Linear Algebra Exam: ECON320/420 Mathematics 2: Calculus and Linear Algebra Eksamensdag:. desember 207 Sensur kunngjøres:
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
UNIVERSITETET I OSLO ØKONOMISK INSTITUTT Utsatt eksamen i: ECON420 Matematikk 2: Matematisk analyse og lineær algebra Postponed exam: ECON420 Mathematics 2: Calculus and Linear Algebra Eksamensdag: Mandag
DetaljerEksamensoppgave i TMA4320 Introduksjon til vitenskapelige beregninger
Institutt for matematiske fag Eksamensoppgave i TMA432 Introduksjon til vitenskapelige beregninger Faglig kontakt under eksamen: Anton Evgrafov Tlf: 453 163 Eksamensdato: 8. august 217 Eksamenstid (fra
DetaljerSpeed Racer Theme. Theme Music: Cartoon: Charles Schultz / Jef Mallett Peanuts / Frazz. September 9, 2011 Physics 131 Prof. E. F.
September 9, 2011 Physics 131 Prof. E. F. Redish Theme Music: Speed Racer Theme Cartoon: Charles Schultz / Jef Mallett Peanuts / Frazz 1 Reading questions Are the lines on the spatial graphs representing
DetaljerLie 2-Groups, Lie 2-Algebras, and Loop Groups
Lie 2-Groups, Lie 2-Algebras, and Loop Groups Alissa S. Crans Joint work with: John Baez Urs Schreiber & Danny Stevenson in memory of Saunders Mac Lane April 8, 2006 Internalization Often a useful first
DetaljerEndelig ikke-røyker for Kvinner! (Norwegian Edition)
Endelig ikke-røyker for Kvinner! (Norwegian Edition) Allen Carr Click here if your download doesn"t start automatically Endelig ikke-røyker for Kvinner! (Norwegian Edition) Allen Carr Endelig ikke-røyker
DetaljerDen som gjør godt, er av Gud (Multilingual Edition)
Den som gjør godt, er av Gud (Multilingual Edition) Arne Jordly Click here if your download doesn"t start automatically Den som gjør godt, er av Gud (Multilingual Edition) Arne Jordly Den som gjør godt,
DetaljerECON3120/4120 Mathematics 2, spring 2004 Problem solutions for the seminar on 5 May Old exam problems
Department of Economics May 004 Arne Strøm ECON0/40 Mathematics, spring 004 Problem solutions for the seminar on 5 May 004 (For practical reasons (read laziness, most of the solutions this time are in
DetaljerEksamen i TMA4190 Mangfoldigheter Onsdag 4 juni, Tid :
Norges teknisk-naturvitenskapelige universitet Institutt for matematiske fag SOLUTIONS Eksamen i TMA4190 Mangfoldigheter Onsdag 4 juni, 2013. Tid : 09.00 13.00 Oppgave 1 a) La U R n være enhetsdisken x
Detaljermelting ECMI Modelling week 2008 Modelling and simulation of ice/snow melting Sabrina Wandl - University of Linz Tuomo Mäki-Marttunen - Tampere UT
and and ECMI week 2008 Outline and Problem Description find model for processes consideration of effects caused by presence of salt point and numerical solution and and heat equations liquid phase: T L
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
UNIVERSITETET I OSLO ØKONOMISK INSTITUTT Utsatt eksamen i: ECON420 Matematikk 2: Matematisk analyse og lineær algebra Postponed exam: ECON420 Mathematics 2: Calculus and Linear Algebra Eksamensdag: Mandag
Detaljer32.2. Linear Multistep Methods. Introduction. Prerequisites. Learning Outcomes
Linear Multistep Methods 32.2 Introduction In the previous Section we saw two methods (Euler and trapezium) for approximating the solutions of certain initial value problems. In this Section we will see
DetaljerUNIVERSITETET I OSLO
UNIVERSITETET I OSLO Det matematisk-naturvitenskapelige fakultet Eksamen i INF 3230 Formell modellering og analyse av kommuniserende systemer Eksamensdag: 4. april 2008 Tid for eksamen: 9.00 12.00 Oppgavesettet
DetaljerEmneevaluering GEOV272 V17
Emneevaluering GEOV272 V17 Studentenes evaluering av kurset Svarprosent: 36 % (5 av 14 studenter) Hvilket semester er du på? Hva er ditt kjønn? Er du...? Er du...? - Annet PhD Candidate Samsvaret mellom
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
UNIVERSITETET I OSLO ØKONOMISK INSTITUTT Eksamen i: ECON360/460 Samfunnsøkonomisk lønnsomhet og økonomisk politikk Exam: ECON360/460 - Resource allocation and economic policy Eksamensdag: Fredag 2. november
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
1 UNIVERSITETET I OSLO ØKONOMISK INSTITUTT BOKMÅL Utsatt eksamen i: ECON2915 Vekst og næringsstruktur Eksamensdag: 07.12.2012 Tid for eksamen: kl. 09:00-12:00 Oppgavesettet er på 5 sider Tillatte hjelpemidler:
DetaljerKROPPEN LEDER STRØM. Sett en finger på hvert av kontaktpunktene på modellen. Da får du et lydsignal.
KROPPEN LEDER STRØM Sett en finger på hvert av kontaktpunktene på modellen. Da får du et lydsignal. Hva forteller dette signalet? Gå flere sammen. Ta hverandre i hendene, og la de to ytterste personene
Detaljer5 E Lesson: Solving Monohybrid Punnett Squares with Coding
5 E Lesson: Solving Monohybrid Punnett Squares with Coding Genetics Fill in the Brown colour Blank Options Hair texture A field of biology that studies heredity, or the passing of traits from parents to
DetaljerNumerical Simulation of Shock Waves and Nonlinear PDE
Numerical Simulation of Shock Waves and Nonlinear PDE Kenneth H. Karlsen (CMA) Partial differential equations A partial differential equation (PDE for short) is an equation involving functions and their
DetaljerResolvable Mendelsohn Triple Systems with Equal Sized Holes F. E. Bennett Department of Mathematics Mount Saint Vincent University Halifax, Nova Scoti
Resolvable Mendelsohn Triple Systems with Equal Sized Holes F. E. Bennett Department of Mathematics Mount Saint Vincent University Halifax, Nova Scotia, Canada B3M 2J6 R. Wei Department of Mathematics
DetaljerExam in Quantum Mechanics (phys201), 2010, Allowed: Calculator, standard formula book and up to 5 pages of own handwritten notes.
Exam in Quantum Mechanics (phys01), 010, There are 3 problems, 1 3. Each problem has several sub problems. The number of points for each subproblem is marked. Allowed: Calculator, standard formula book
DetaljerSmart High-Side Power Switch BTS730
PG-DSO20 RoHS compliant (green product) AEC qualified 1 Ω Ω µ Data Sheet 1 V1.0, 2007-12-17 Data Sheet 2 V1.0, 2007-12-17 Ω µ µ Data Sheet 3 V1.0, 2007-12-17 µ µ Data Sheet 4 V1.0, 2007-12-17 Data Sheet
DetaljerOppgave 1. Norges teknisk-naturvitenskapelige universitet NTNU Institutt for fysikk EKSAMEN I: MNFFY 245 INNFØRING I KVANTEMEKANIKK
Norges teknisk-naturvitenskapelige universitet NTNU Institutt for fysikk EKSAMEN I: MNFFY 45 INNFØRING I KVANTEMEKANIKK DATO: Fredag 4 desember TID: 9 5 Antall vekttall: 4 Antall sider: 5 Tillatte hjelpemidler:
DetaljerMA2501 Numerical methods
MA250 Numerical methods Solutions to problem set Problem a) The function f (x) = x 3 3x + satisfies the following relations f (0) = > 0, f () = < 0 and there must consequently be at least one zero for
DetaljerSolutions of nonlinear differential equations
Nonlinear Differ. Equ. Appl. 17 (21), 249 27 c 29 Birkhäuser Verlag Basel/Switzerland 121-9722/1/2249-22 published online December 11, 29 DOI 1.17/s3-9-52-7 Nonlinear Differential Equations and Applications
Detaljer1 Aksiomatisk definisjon av vanlige tallsystemer
Notat XX for MAT1140 1 Aksiomatisk definisjon av vanlige tallsystemer 1.1 Aksiomer Vi betrakter en mengde R, utstyrt med to avbild- Algebraiske aksiomer. ninger: addisjon { R R R, (x, y) x + y. { R R R,
DetaljerHvor mye praktisk kunnskap har du tilegnet deg på dette emnet? (1 = ingen, 5 = mye)
INF247 Er du? Er du? - Annet Ph.D. Student Hvor mye teoretisk kunnskap har du tilegnet deg på dette emnet? (1 = ingen, 5 = mye) Hvor mye praktisk kunnskap har du tilegnet deg på dette emnet? (1 = ingen,
DetaljerUNIVERSITETET I OSLO ØKONOMISK INSTITUTT
UNIVERSITETET I OSLO ØKONOMISK INSTITUTT Eksamen i: ECON320/420 Mathematics 2: Calculus and linear algebra Exam: ECON320/420 Mathematics 2: Calculus and linear algebra Eksamensdag: Tirsdag 30. mai 207
DetaljerGir vi de resterende 2 oppgavene til én prosess vil alle sitte å vente på de to potensielt tidskrevende prosessene.
Figure over viser 5 arbeidsoppgaver som hver tar 0 miutter å utføre av e arbeider. (E oppgave ka ku utføres av é arbeider.) Hver pil i figure betyr at oppgave som blir pekt på ikke ka starte før oppgave
DetaljerOn the Existence of Strong Solutions to a Fluid Structure Interaction Problem with Navier Boundary Conditions
J. Math. Fluid Mech. 2019 21:36 c 2019 Springer Nature Switzerland AG https://doi.org/10.1007/s00021-019-0440-7 Journal of Mathematical Fluid Mechanics On the Existence of Strong Solutions to a Fluid Structure
DetaljerQi-Wu-Zhang model. 2D Chern insulator. León Martin. 19. November 2015
Qi-Wu-Zhang model 2D Chern insulator León Martin 19. November 2015 Motivation Repeat: Rice-Mele-model Bulk behavior Edge states Layering 2D Chern insulators Robustness of edge states Motivation topological
DetaljerFIRST LEGO League. Härnösand 2012
FIRST LEGO League Härnösand 2012 Presentasjon av laget IES Dragons Vi kommer fra Härnosänd Snittalderen på våre deltakere er 11 år Laget består av 4 jenter og 4 gutter. Vi representerer IES i Sundsvall
DetaljerLevel Set methods. Sandra Allaart-Bruin. Level Set methods p.1/24
Level Set methods Sandra Allaart-Bruin sbruin@win.tue.nl Level Set methods p.1/24 Overview Introduction Level Set methods p.2/24 Overview Introduction Boundary Value Formulation Level Set methods p.2/24
DetaljerExercise 1: Phase Splitter DC Operation
Exercise 1: DC Operation When you have completed this exercise, you will be able to measure dc operating voltages and currents by using a typical transistor phase splitter circuit. You will verify your
DetaljerSCE1106 Control Theory
Master study Systems and Control Engineering Department of Technology Telemark University College DDiR, October 26, 2006 SCE1106 Control Theory Exercise 6 Task 1 a) The poles of the open loop system is
DetaljerDagens tema: Eksempel Klisjéer (mønstre) Tommelfingerregler
UNIVERSITETET I OSLO INF1300 Introduksjon til databaser Dagens tema: Eksempel Klisjéer (mønstre) Tommelfingerregler Institutt for informatikk Dumitru Roman 1 Eksempel (1) 1. The system shall give an overview
DetaljerElektronisk innlevering/electronic solution for submission:
VIKINGTIDSMUSEET Plan- og designkonkurranse/design competition Elektronisk innlevering/electronic solution for submission: Det benyttes en egen elektronisk løsning for innlevering (Byggeweb Anbud). Dette
DetaljerLøsning til deleksamen 2 i SEKY3322 Kybernetikk 3
Høgskolen i Buskerud. Finn Haugen (finn@techteach.no). Løsning til deleksamen 2 i SEKY3322 Kybernetikk 3 Tid: 7. april 28. Varighet 4 timer. Vekt i sluttkarakteren: 3%. Hjelpemidler: Ingen trykte eller
DetaljerFlows and Critical Points
Nonlinear differ. equ. appl. 15 (2008), 495 509 c 2008 Birkhäuser Verlag Basel/Switzerland 1021-9722/040495-15 published online 26 November 2008 DOI 10.1007/s00030-008-7031-2 Nonlinear Differential Equations
DetaljerSolvability and Regularity for an Elliptic System Prescribing the Curl, Divergence, and Partial Trace of a Vector Field on Sobolev-Class Domains
J. Math. Fluid Mech. 19 (2017), 375 422 c 2016 Springer International Publishing 1422-6928/17/030375-48 DOI 10.1007/s00021-016-0289-y Journal of Mathematical Fluid Mechanics Solvability and Regularity
DetaljerDean Zollman, Kansas State University Mojgan Matloob-Haghanikar, Winona State University Sytil Murphy, Shepherd University
Dean Zollman, Kansas State University Mojgan Matloob-Haghanikar, Winona State University Sytil Murphy, Shepherd University Investigating Impact of types of delivery of undergraduate science content courses
DetaljerMa Linær Algebra og Geometri Øving 5
Ma20 - Linær Algebra og Geometri Øving 5 Øistein søvik 7. oktober 20 Excercise Set.5.5 7, 29,.6 5,, 6, 2.7, A = 0 5 B = 0 5 4 7 9 0-5 25-4 C = 0 5 D = 0 0 28 4 7 9 0-5 25 F = 6 2-2 0-5 25 7. Find an elementary
DetaljerGEF2200 Atmosfærefysikk 2017
GEF2200 Atmosfærefysikk 2017 Løsningsforslag til sett 3 Oppgaver hentet fra boka Wallace and Hobbs (2006) er merket WH06 WH06 3.18r Unsaturated air is lifted (adiabatically): The rst pair of quantities
DetaljerOn time splitting for NLS in the semiclassical regime
On time splitting for NLS in the semiclassical regime Rémi Carles CNRS & Univ. Montpellier Rémi Carles (Montpellier) Splitting for semiclassical NLS 1 / 21 Splitting for NLS i t u + 1 2 u = f ( u 2) u,
DetaljerAn oscillatory bifurcation from infinity, and
Nonlinear differ. equ. appl. 15 (28), 335 345 121 9722/8/3335 11, DOI 1.17/s3-8-724-1 c 28 Birkhäuser Verlag Basel/Switzerland An oscillatory bifurcation from infinity, and from zero Philip Korman Abstract.
DetaljerOppgave 1a Definer følgende begreper: Nøkkel, supernøkkel og funksjonell avhengighet.
TDT445 Øving 4 Oppgave a Definer følgende begreper: Nøkkel, supernøkkel og funksjonell avhengighet. Nøkkel: Supernøkkel: Funksjonell avhengighet: Data i en database som kan unikt identifisere (et sett
DetaljerEKSAMENSOPPGAVE I FAG TKP 4105
EKSAMENSOPPGAVE I FAG TKP 4105 Faglig kontakt under eksamen: Sigurd Skogestad Tlf: 913 71669 (May-Britt Hägg Tlf: 930 80834) Eksamensdato: 08.12.11 Eksamenstid: 09:00 13:00 7,5 studiepoeng Tillatte hjelpemidler:
DetaljerInformation search for the research protocol in IIC/IID
Information search for the research protocol in IIC/IID 1 Medical Library, 2013 Library services for students working with the research protocol and thesis (hovedoppgaven) Open library courses: http://www.ntnu.no/ub/fagside/medisin/medbiblkurs
DetaljerUNIVERSITY OF OSLO DEPARTMENT OF ECONOMICS
UNIVERSITY OF OSLO DEPARTMENT OF ECONOMICS English Exam: ECON2915 Economic Growth Date of exam: 25.11.2014 Grades will be given: 16.12.2014 Time for exam: 09.00 12.00 The problem set covers 3 pages Resources
DetaljerSolution for INF3480 exam spring 2012
Solution for INF3480 exam spring 0 June 6, 0 Exercise Only in Norwegian a Hvis du har en robot hvor ikke den dynamiske modellen er kjent eller spesielt vanskelig å utlede eksakt, kan en metode liknende
DetaljerUNIVERSITY OF OSLO. Faculty of Mathematics and Natural Sciences
Page 1 UNIVERSITY OF OSLO Faculty of Mathematics and Natural Sciences Exam in BIO4210/9210 Classification and Phylogeny Day of exam: 13. December 2011 Exam hours: 9.00-12.00 (3 hours) This examination
DetaljerEmnedesign for læring: Et systemperspektiv
1 Emnedesign for læring: Et systemperspektiv v. professor, dr. philos. Vidar Gynnild Om du ønsker, kan du sette inn navn, tittel på foredraget, o.l. her. 2 In its briefest form, the paradigm that has governed
DetaljerEKSAMENSOPPGAVE I BI2034 Samfunnsøkologi EXAMINATION IN: BI Community ecology
Norges teknisk-naturvitenskapelige universitet Institutt for Biologi EKSAMENSOPPGAVE I BI2034 Samfunnsøkologi EXAMINATION IN: BI2034 - Community ecology - Faglig kontakt under eksamen/contact person/subject
DetaljerEksamen ENG1002/1003 Engelsk fellesfag Elevar og privatistar/elever og privatister. Nynorsk/Bokmål
Eksamen 22.11.2012 ENG1002/1003 Engelsk fellesfag Elevar og privatistar/elever og privatister Nynorsk/Bokmål Nynorsk Eksamensinformasjon Eksamenstid Hjelpemiddel Eksamen varer i 5 timar. Alle hjelpemiddel
DetaljerUNIVERSITY OF OSLO DEPARTMENT OF ECONOMICS
UNIVERSITY OF OSLO DEPARTMENT OF ECONOMICS English Postponed exam: ECON2915 Economic growth Date of exam: 11.12.2014 Time for exam: 09:00 a.m. 12:00 noon The problem set covers 4 pages Resources allowed:
Detaljer