Hyperbolic Relaxation of a Fourth Order Evolution Equation

and Applied Analysis 3 Proposition 3. For any constant R ≥ 0 there exists a positive constant K = K(R) such that, for any initial data u 1 (0), u 2 (0) with ‖u i (0)‖ Hη,ε ≤ R, i = 1, 2 one has 󵄩󵄩󵄩󵄩 S ε,η (t)u 1 (0) − S ε,η (t)u 2 (0) 󵄩󵄩󵄩󵄩Hη,ε ≤ e (K 2 /ε 2 )t󵄩󵄩󵄩󵄩u1 (0) − u2 (0) 󵄩󵄩󵄩Hη,ε , (20) where S ε,η (t) is the solution semigroup of the problem (1). Proof. Let u 1 , u 2 , two solutions of (1) with initial data u 1 (0) and u 2 (0). Let, w = u 1 − u 2 then we write ηw tt + w t + ε 2 w xxxx = 1 2 d dx {W 󸀠 (u 1x ) − W 󸀠 (u 2x )} . (21) We multiply the above equation by w t in L(I); then η 2 d dt 󵄩󵄩󵄩󵄩wt 󵄩󵄩󵄩󵄩 2 + ε 2 2 d dt 󵄩󵄩󵄩󵄩wxx 󵄩󵄩󵄩󵄩 2 + 󵄩󵄩󵄩󵄩wt 󵄩󵄩󵄩󵄩 2