Local Solvability of Multi Dimensional InitialBoundary Value Problems for Higher Order Nonlinear Hyperbolic Equations


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1 International Workshop QUALITDE 2017, December 24 26, 2017, Tbilisi, Georgia 97 Local Solvability of Multi Dimensional InitialBoundary Value Problems for Higher Order Nonlinear Hyperbolic Equations Tariel Kiguradze and Audison Beaubrun Florida Institute of Technology, Melbourne, USA Let m 1,, m n be positive integers In the ndimensional box Ω = [0, ω 1 ] [0, ω n ] for the nonlinear hyperbolic equation u m = fx, D m [u] 1 consider the initialboundary conditions h ik ux 1,, x i 1,, x i+1,, x n = ϕ ik x i k = 1,, m i, i = 1,, n 1, u 0,,0,k 1 x 1,, x n 1, 0 = ϕ nk x n k = 1,, m n 2 Here x = x 1,, x n, x i = x 1,, x i 1, x i+1,, x n, α = α 1,, α n, m i = m m i and m i = 0,, m i,, 0 are multiindices, u α x = α 1+ +α n ux x α 1 1, xαn n D m [u] = u α α m, Dm [u] = u α α<m, Ω i = [0, ω 1 ] [0, ω i 1 ] [0, ω i+1 ] [0, ω n ], f CΩ R m 1 m n, h ik : C m i 1 [0, ω i ] R k = 1,, m i ; i = 1,, n 1 are bounded linear functionals, and ϕ ik C m i Ω i k = 1,, m i ; i = 1,, n Furthermore, it is assumed that the functions ϕ ik satisfy the following consistency conditions: h ik ϕ jl x ij h jl ϕ ik x ij k = 1,, m i ; l = 1,, m j ; i, j = 1,, n, where x ij = x x i x j Set: fx, Z Z = z α α<m ; f α x, Z = z α α = α 1,, α n Υ m α i = m i for some i = 1,, n The variables z α α Υ m are called principal phase variables of the function fx, Z By a solution of problem 1, 2 we understand a classical solution, ie, a function u C m Ω satisfying equation 1 and boundary conditions 2 Twodimensional initialboundary value problems were studied in [4, 5] Definition Let n = n 1,, n r, Ω = [0, ω 1 ] [0,, ω r ], y = y 1,, y r, and let the function g : CΩ R n 1 n r be continuously differentiable with respect to the phase variables A solution v 0 C n Ω of the problem v n = gy, D n [v], 3 h ij vy1,, y i 1,, y i+1,, y r = ψ ij ŷ i j = 1,, n i ; i = 1,, r 4
2 98 International Workshop QUALITDE 2017, December 24 26, 2017, Tbilisi, Georgia is called strongly isolated if the linearized problem v n = α<n p α yv α, h ij vy1,, y i 1,, y i+1,, y r = 0 j = 1,, n i ; i = 1,, r, where p α y = g α y, D k [v 0 y], is wellposed Wellposed multidimensional boundary value problems for higher order linear hyperbolic equations were studied in [2] The concept of a strongly isolated solution is closely related to the concept of strongly wellposedness Strong wellposedness of twodimensional boundary value problems for higher order nonlinear hyperbolic equations were introduced in [3] Theorem 1 Let the function f be continuously differentiable with respect to the phase variables, and let v 0 be a strongly isolated solution of the problem where v mn = p x n, D mn [v], 5 h ik ux1,, x i 1,, x i+1,, x n 1 = ϕ mn ik x ni k = 1,, m i ; i = 1,, n 1, 6 p x n, D mn [v] = f x 1,, x n 1, 0, D m 1n [u 0 ]x 1,, x n 1, 0, D mn [v], x ni = x 1,, x i 1, x i+1,, x n 1 and 1 n = 0,, 0, 1 Then there exists δ 0, ω n ] such that in the set Ω δ = [0, ω 1 ] [0, ω n 1 ] [0, δ] problem 1, 2 has a unique solution u satisfying the condition u mn x 1,, x n 1, 0 = v 0 x 1,, x n 1 7 Consider the perturbed equation u m = f x, D m [u] + x n q x, D m [u] 8 Theorem 2 Let the conditions of Theorem 1 hold, and let the function qx, Z be continuously differentiable with respect to the principal phase variables z α α Υ m Then there exists δ 0, ω n ] such that in the set Ω δ = [0, ω 1 ] [0, ω n 1 ] [0, δ] problem 8, 2 has a at least one solution u satisfying condition 7 Moreover, if the function q is locally Lipschitz continuous with respect to the rest of the phase variables, then such solution is unique The following is a particular case of conditions 2: h 1k ux1,, x i 1,, x i+1,, x n = ϕ 1k x i k = 1,, m i, u 0,,k i 1,,0 x 1,, 0,, x n = ϕ iki x i k i = 1,, m i ; i = 2,, n 9 Corollary Let the function f be continuously differentiable with respect to the phase variables, and let v 0 be a strongly isolated solution of the problem where v m 1 = px 1, v, v,, v m 1 1, h 1k v = ϕ m 1 1k 0 k = 1,, m 1, px 1, v, v,, v m1 1 = f x 1, 0,, 0, D m [ 1 D m 1 [u 0 ] ] x 1, 0,, 0, v, v,, v m 1 1 Then there exist δ i 0, ω i ] i = 2,, n such that in the set Ω δ2 δ n = [0, ω 1 ] [0, δ 2 ] [0, δ n ] problem 1, 7 has a unique solution u satisfying the condition u m 1 x 1, 0,, 0 = v 0 x 1
3 International Workshop QUALITDE 2017, December 24 26, 2017, Tbilisi, Georgia 99 Remark In Theorem 1 the requirement of strong isolation of the solution v 0 cannot be replaced by wellposedness of problem 5, 6 In order to illustrate this, consider the problem u 1,1 = u 0,1 3 y 2 u 0,1, 10 uω 1, y u0, y = For this case problem 5, 6 is the following one: y 0 t sin 1 dt, ux, 0 = 0 11 t v = v 3, vω 1 v0 = 0 12 By Corollary 42 and Theorem 44 from [1], problem 12 has a unique solution v 0 y 0 and is wellposed On the other hand, it is clear, that v 0 y 0 is not strongly isolated Our goal is to show that problem 10, 11 has no solution in the rectangle Ω δ = [0, ω 1 ] [0, δ] no matter how small δ > 0 is Assume the contrary that problem 10, 11 has a solution u in Ω δ for some δ > 0 Then for an arbitrarily fixed y 0, δ], the function v = u 0,1, y is a solution of the problem v = v 3 y 2 v, 13 vω 1 v0 = y sin 1 y 14 containing the parameter y [0, ω 2 ] Moreover, if problem 10, 11 has a solution, then v is a solution 13, 14 depending continuously on the parameter y For every fixed y 0, δ] equation 13 has three constant solutions: v 0 x = 0, v 1 x = y and v 2 x = y Due to the existence and uniqueness theorem, a nonconstant solution v of equation 13 intersects v 0, v 1 or v 2, and thus v x 0 for x [0, ω 1 ] Let k > 1 2πδ and x 1 π + 2πk, 1 2πk Then vω 1 > v0 and v x > 0 for x [0, ω 1 ] Therefore, either or vx > y for x [0, ω 1 ], vx y, 0 for x [0, ω 1 ] If y = 1 π 2 +2πk, then vω 1 v0 = y, and consequently, vx y, 0 for x [0, ω 1 ] From the aforesaid, in view of continuity of u 0,1 in Ω δ, it follows that u 0,1 1 x, y > y for y π + 2πk, 1 2πk Similarly, one can show that u 0,1 x, y < y for y 1 2πk + 1, 1 π + 2πk However, the latter two inequalities imply that u 0,1 x, y is discontinuous along the lines y = 1 πk k = 1, 2, Thus we have proved that problem 10, 11 has no solution in Ω δ for any δ > 0
4 100 International Workshop QUALITDE 2017, December 24 26, 2017, Tbilisi, Georgia In conclusion, as examples, consider the following initialboundary value problems Example 1 u 2,2,1 = u 2 u 2,0,1 + u 1,1,0 4 u 0,2,1 u 0,0,1 6 + q x 1, x 2, x 3, u, u 1,0,0, u 0,1,0, u 1,1,0, 15 u0, x 2, x 3 = 0, uω 1, x 2, x 3 = 0; ux 1, 0, x 3 = 0, ux 1, ω 2, x 3 = 0; ux 1, x 2, 0 = ψx 1, x 2 16 Let the function q be continuous Then, by Corollary 4 from [2] and Theorem 2, there exists δ 0, ω 3 ] such that in the set Ω δ = [0, ω 1 ] [0, ω 2 ] [0, δ] problem 15, 16 has a at least one solution Moreover, if the function q is locally Lipschitz continuous with respect to the phase variables, then problem 15, 16 is uniquely solvable Example 2 u 2,2,1 = u 2,0,1 + u 2,0,1 5 + u 0,2,1 u 0,0,1 + q x 1, x 2, x 3, u, u 1,0,0, u 0,1,0, u 1,1,0, 17 u i,0,0 0, x 2, x 3 = u i,0,0 ω 1, x 2, x 3 ; u 0,i,0 x 1, 0, x 3 = u 0,i,0 x 1, ω 2, x 3 i = 0, 1; ux 1, x 2, 0 = ψx 1, x 2 18 Let the function q be continuous Then, by Corollary 5 from [2] and Theorem 2, there exists δ 0, ω 3 ] such that in the set Ω δ = [0, ω 1 ] [0, ω 2 ] [0, δ] problem 17, 18 has a at least one solution Moreover, if the function q is locally Lipschitz continuous with respect to the phase variables, then problem 17, 18 is uniquely solvable Example 3 Let p α be smooth functions, q be a continuous function, m = m 1,, m n, 0 and 1 n+1 = 0,, 0 For the equation u 2m+1 n+1 = α<m pα x, uu α+1 n+1 α + q x, D 2m  1 [u] 19 consider the initialboundary value problems with the Dirichlet and periodic boundary conditions and u k1 i x 1,, 0,, x n+1 = 0, u k1 i x 1,, ω i,, x n+1 = 0 u k1 i x 1,, 0,, x n+1 = u k1 i x 1,, ω i,, x n+1 k = 0,, m i 1; i = 1,, n; ux 1,, x n, 0 = ϕx, 20 k = 0,, 2m i 1; i = 1,, n; ux 1,, x n, 0 = ϕx 21 Let 1 m + α p α 0 1 m + α p α < 0 for α < m Then, by Theorem 2 from [2] and Theorem 2, there exists δ 0, ω n+1 ] such that in the set Ω δ = [0, ω 1 ] [0, ω n ] [0, δ] problem 19, 20 problem 19, 21 has a at least one solution Moreover, if the function q is locally Lipschitz continuous with respect to the phase variables, then problem 19, 20 problem 19, 21 is uniquely solvable References [1] I T Kiguradze, Boundary value problems for systems of ordinary differential equations Russian Translated in J Soviet Math , no 2, Itogi Nauki i Tekhniki, Current problems in mathematics Newest results, Vol 30 Russian, 3 103, 204, Akad Nauk SSSR, Vsesoyuz Inst Nauchn i Tekhn Inform, Moscow, 1987
5 International Workshop QUALITDE 2017, December 24 26, 2017, Tbilisi, Georgia 101 [2] T Kiguradze and N AlJaber, Multi dimensional boundary value problems for linear hyperbolic equations of higher order Abstracts of the International Workshop on the Qualitative Theory of Differential Equations QUALITDE2016, pp , Tbilisi, Georgia, December 24 26, 2016; T AlJaber workshop 2016pdf; workshop 2016pdf [3] T Kiguradze and R BenRabha, On wellposed boundary value problems for higher order nonlinear hyperbolic equations with two independent variables Abstracts of the International Workshop on the Qualitative Theory of Differential Equations QUALITDE2016, pp ; T BenRabhapdf; workshop 2016pdf [4] T Kiguradze and R BenRabha, On strong wellposedness of initialboundary value problems for higher order nonlinear hyperbolic equations with two independent variables Georgian Math J , no 3, [5] T I Kiguradze and T Kusano, On the wellposedness of initialboundary value problems for higherorder linear hyperbolic equations with two independent variables Russian Differ Uravn , no 4, ; translation in Differ Equ , no 4,
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