Blowing up Solution of Initial-Boundary Value Problem for a Kind of Nonlinear Evolution Equations

This dissertation is to discuss the initial-boundary value problem under the third nonlinear boundary condition for a kind of nonlinear evolution equations. To apply the maximum value theory and convex functional method£it is proved that the blowing up of solution in the definite time under some assumed conditions. The conclusion popularizes the results of references [Zhongwei Zha(2003)-Zhongwei Zha(2010)].


Introduction
This paper discusses the initial-boundary value problem under the third nonlinear boundary condition for a kind of nonlinear evolution equations. (I) where D is a smooth boundary area in ) is a gradient operator and ∂u ∂v is the exterior normal derivative for ∂D.
The solution losing regularity and occuring blow condition in finite time study for nonlinear partial differential equations (Nonlinear Evolution Equations) related to time variable t has important practical significance [Chaohao Gu(1993), Daqian Li(1989)], which are studied in [Pao C.V.(1980)-Mizoguchi N.(1997)].However, due to the wide equation range and various characteristics of nonlinear equations, the existed results are obtained for finite solutions of specific physical problems, such as papers [Zhongwei Zha(2003)-Zhongwei Zha(2010)] are special cases of (I).
For convenience, denote ∂t denotes inner product.For initial-boundary value problem (I), the followings are supposed: , f and f t are non-negative and the inner product f q • q t ≥ 0. Denote R) when p ≥ 0 and g ≤ 0. And when p < 0 and g > 0, g t and the inner product g q • q t are non-negative, while g p < 1.

Non-negative property of solutions for problem (I)
In order to discuss the blowing up solutions for finite problem (I), we first prove two properties of solutions, that is Proof.If it isn't, then suppose the solution u(x, t) of (I) can be taken negative number, there exists a point P 0 (x 0 , t 0 ) ∈ D × [0, T ) such that u(P 0 ) is the negative minimum.From the initial condition (1.3) and condition (i), t 0 0. Noting the expression of H(x, t) in condition (ii), it is obviously that H(x, t) takes the negative minimum in P 0 too.
If P 0 ∈ D × (0, T ) and since u(P 0 ) and H(P 0 ) are the negative minimums, then Express the definite solution problem (I) by the following form: Take the derivation for variable t, denoted by ∂u ∂t = V(x, t), then we obtain the initial boundary problem what V(x, t) satisfies: (II) Similar with Lemma 1, we can prove that for any (x, t) ∈ D × [0, T ), the solution of mixed problem (II) V(x, t) ≥ 0, therefore, The proof is proven.

Blow-up of solution in problem (I)
When the hypothesis conditions (i)-(iv) satisfy, the mixed problem (I) doesn't exist global smooth solution, i.e. the solution must appear blow-up within definite time, which is the following theorem: Theorem If the conditions (i)-(iv) hold and u(x, t) is the smooth solution of (I), then there exists time T 0 (0 < T 0 < +∞), such that lim Noting that the definite problem (I) can be expressed the following: So we only need to prove the solutions of mixed problem (III) appear blow-up within definite time under the assumed condition. Denote where β is a constant in condition (ii), then G(u) > 0, F(t) < 0 and Take equation (3.1) into equation ( 6), we have Apply integration by parts to the first integral of the right side in equation ( 8) and noting the boundary condition (3.2) in (III), we have where ds is the area element of ∂D.Derivative the two sides of (9) about t, we have: On the other side, if derivativing the two sides of (6) about t directly, we have: We derivative the equation (3.1) about t , take the results into (11) and apply integration by parts, then Double( 12) and then use it to subtract(10): Noting the last integral in the right side of ( 13) can be expressed : In fact, since h(u) is monotone decreasing positive continuous function about u, and from the expression of h(u) and H(u) > 0, H (u) = h(u) > 0, while H (u) = h (u) < 0, therefore, in the above equation H 2 − HH > 0.