Estimates of solutions to nonlinear evolution equations

Consider the equation u′(t) = A(t, u(t)), u(0) = u0; u ′ := du dt (1). Under some assumptions on the nonlinear operator A(t, u) it is proved that problem (1) has a unique global solution and this solution satisfies the following estimate ‖u(t)‖ < μ(t)−1 ∀t ∈ R+ = [0,∞). Here μ(t) > 0, μ ∈ C(R+), is a suitable function and the norm ‖u‖ is the norm in a Banach space X with the property ‖u(t)‖′ ≤ ‖u′(t)‖. Mathematics Subject Classification: MSC 2010, 47J05; 47J35; 58D25


Introduction
Let u = A(t, u(t)), u(0) = u 0 ; u : where t ∈ R + = [0, ∞), A(t, u) is a locally continuous map from R + × X into X, where X is a Banach space of functions with the norm · , such that u(t) ≤ u (t) if u(t) is continuously differentiable with respect to t. If u(t) ∈ X is a function then |u(t)| and u(t) make sense. We assume that if |u| ≤ |v| then u ≤ v . For the spaces of continuous functions and L p spaces this assumption holds.
We assume that where k > 0 is a constant which may depend on R, u ≤ R, v ≤ R, and on T , t ∈ [0, T ]. If A(t, u) is a function with values in R and A(t, u) = |A(t, u)|, then (1) is a nonlinear ordinary differential equation and condition (2) guarantees local existence and uniqueness of its solution on an interval [0, T ] where T is a sufficiently small number. If T = ∞ then the solution u(t) is called global.
The map A(t, u) may be of the form where a(t, s, u) is a locally continuous function on R + × R + × X, locally Lipschitz with respect to u.
The following assumptions will be valid throughout this paper: There exists a C 1 (R + ) function µ(t) > 0 such that where w = 1, w ∈ X is an arbitrary element, and Theorem 1. Under the above assumptions the solution to (1) exists globally, is unique, and satisfies the following estimate: Remark 1. Some conditions on A(t, u) of the type (4)-(6) are necessary for the global existence of the solution.
Consider the following example: u = u 2 , u(0) = 1. This problem is equivalent to the equation u = 1 + t 0 u 2 (s)ds. The solution to this problem is where 0 < λ < 1 is arbitrary.

Proofs
The proof of Theorem 1 consists of several parts. We start with the part dealing with the inequality We assume throughout that u(t) is continuously differentiable with respect to t.
where H is a Hilbert space. The inner product in H is denoted as usual (u, v). A simple proof of (8) goes as follows. Start with the inequality and let h → 0. The result is (8). Indeed, the limit of the right side does exist and is equal to u (t) . To calculate the limit of the left side in (9) consider the identity Clearly, the limit of the right side exists and is equal to 2Re(u (t), u(t)). One has lim h→0 ( u(t + h) + u(t) ) = 2 u(t) . Assuming that u(t) > 0 one concludes that If u(t) = 0, then u(t) = lim h→0 h −1 u(t + h) . One has u(t + h) 2 = (u(t + h), u(t + h)) = h 2 u (t) 2 + o(h 2 ). Thus, u(t + h) = |h| u (t) + o(h). Therefore u(t) = lim h→0 h −1 |h| u (t) = sign h u (t) ≤ u (t) . Formula (8) is proved for X = H.
2 If X = R the proof of (8) is left for the reader. One gets | |u(t)| | ≤ |u (t)|. 2.2. Let us study problem (1) assuming that X = R, w = 1 in (4) and u(t) = |u(t)|. Assumption (2) guarantees local existence and uniqueness of the solution to (1). We want to prove that assumptions (4)-(6) guarantee the global existence of the solution u(t) and estimate (7). If (6) holds, then, by continuity, there exists a small δ > 0 such that This and (5) imply Take the absolute value of (1), use (7), (11) and (4) to get Integrating (12) with respect to t one gets This and (6) imply (7) for t ∈ [0, δ]. Define T as follows: Let us prove that T = ∞. Assuming the contrary, i.e., T < ∞, one uses the local existence of the solution to (1) taking as initial value u(T ) and as the interval of the existence of the solution [T, T + h], where h > 0 is a sufficiently small number. Then inequality (7) holds for t ∈ [0, T + h]. This contradicts to the definition (14) of T . So, one gets a contradiction which proves that T = ∞ and estimate (7) holds for all t ∈ R + . Theorem 1 is proved for X = R.
Assume that a(t, s, u) and a t := ∂a ∂t are continuous functions on R + × R + × R, locally Lipschitz with respect to u. Differentiate (15) with respect to t and get u = a(t, t, u(t)) + t 0 a t (t, s, u(s))ds + f (t) := A 1 (t, u(t)). (16) Assume that A 1 (t, u) satisfies conditions (4)-(6) with w = 1, and u(t) = |u(t)|.Then the argument used in scetion 2.2. proves Theorem 1 with A 1 (t, u) replacing A(t, u). Example 1. The aim of this example is to derive sufficient conditions on a(t, s, u) for the assumptions (4)-(6) to hold. Let where c, b > 0 are constants. We assume that a and a t are Lipschitz functions with respect to u. Assume that |a t (t, t, |u|)| ≤ |a t (t, t, |v|)| if |v| ≥ |u|. Let Note that 1 = ac −1 0 e at . If (17) holds, then the following two inequalities and conditions (4)-(5) hold provided that where b is sufficiently large and c is sufficiently small. If in addition (6) holds, i.e., cc 0 < 1, then u(t) exists globally and the estimate |u(t)| < c −1 0 e at ∀t ∈ R + holds.
Here δ > 0 is sufficiently small so that u(t) < 1/µ(t) for 0 ≤ t ≤ δ. Integrate (22) on any interval [0, T ] on which the solution u(t) exists one gets u(t) < 1/µ(t) for t ∈ [0, T ]. As in section 2.3 we prove that T = ∞. Therefore problem (1) has a unique global solution in X and estimate (7) holds. Theorem 1 is proved. 2 The ideas close to the ones used in this paper were developed and used in [1]- [3].