Local Solvability of Multi Dimensional Initial-Boundary 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 Initial-Boundary 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 n-dimensional box Ω = [0, ω 1 ] [0, ω n ] for the nonlinear hyperbolic equation u m = fx, D m [u] 1 consider the initial-boundary 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 multi-indices, 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 Two-dimensional initial-boundary 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 well-posed Well-posed multi-dimensional 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 well-posedness of two-dimensional 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 well-posedness 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 well-posed 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 initial-boundary 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 initial-boundary 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 Al-Jaber, Multi dimensional boundary value problems for linear hyperbolic equations of higher order Abstracts of the International Workshop on the Qualitative Theory of Differential Equations QUALITDE-2016, pp , Tbilisi, Georgia, December 24 26, 2016; T Al-Jaber workshop 2016pdf; workshop 2016pdf [3] T Kiguradze and R Ben-Rabha, On well-posed 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 QUALITDE-2016, pp ; T Ben-Rabhapdf; workshop 2016pdf [4] T Kiguradze and R Ben-Rabha, On strong well-posedness of initial-boundary 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 well-posedness of initial-boundary value problems for higher-order linear hyperbolic equations with two independent variables Russian Differ Uravn , no 4, ; translation in Differ Equ , no 4,