Electronic Journal of Qualitative Theory of Differential Equations 2007, No. 12, 1-14; http://www.math.u-szeged.hu/ejqtde/

UNIFORM CONTINUITY OF THE SOLUTION MAP FOR NONLINEAR WAVE EQUATION IN REISSNER-NORDSTRÖM METRIC

Svetlin Georgiev Georgiev University of Sofia, Sofia, Bulgaria email: sgg2000bg@yahoo.com

Abstract

In this paper we study the properties of the solutions to the Cauchy problem

(1)  (u_{tt} − Δu)_{g_{s}} = f(u) + g(|x|), t ∈ [0, 1], x ∈ R^{3},

(2)  u(1, x) = u_{0} ∈ Ḣ^{1}(R^{3}), u_{t}(1, x) = u_{1} ∈ L^{2}(R^{3}),

where g_{s} is the Reissner-Nordström metric (see [2]); f ∈ C^{1}(R^{1}), f(0) = 0, a|u| ≤ f′(u) ≤ b|u|, g ∈ C(R^{+}), g(|x|) ≥ 0, g(|x|) = 0 for |x| ≥ r_{1}, a and b are positive constants, r_{1} > 0 is suitable chosen.

When g(r) ≡ 0 we prove that the Cauchy problem (1), (2) has a nontrivial solution u(t, r) in the form u(t, r) = v(t)ω(r) ∈ C((0, 1]Ḣ^{1}(R^{+})), where r = |x|, and the solution map is not uniformly continuous.

When g(r) ≠ 0 we prove that the Cauchy problem (1), (2) has a nontrivial solution u(t, r) in the form u(t, r) = v(t)ω(r) ∈ C((0, 1]Ḣ^{1}(R^{+})), where r = |x|, and the solution map is not uniformly continuous. Subject classification: Primary 35L10, Secondary 35L50.

1. Introduction

In this paper we study the properties of the solutions to the Cachy problem

(1)  (u_{tt} − Δu)_{g_{s}} = f(u) + g(|x|), t ∈ [0, 1], x ∈ R^{3},

(2)  u(1, x) = u_{0} ∈ Ḣ^{1}(R^{3}), u_{t}(1, x) = u_{1} ∈ L^{2}(R^{3}),

where g_{s} is the Reissner-Nordström metric (see [2])

g_{s} = \frac{r^{2} − Kr + Q^{2}}{r^{2}} dt^{2} − \frac{r^{2}}{r^{2} − Kr + Q^{2}} dr^{2} − r^{2}dϕ^{2} − r^{2} sin^{2} ϕdθ^{2},

K and Q are positive constants, f ∈ C^{1}(R^{1}), f(0) = 0, a|u| ≤ f′(u) ≤ b|u|, g ∈ C(R^{+}), g(|x|) ≥ 0, g(|x|) = 0 for |x| ≥ r_{1}, a and b are positive constants, r_{1} > 0 is suitable chosen.

The Cauchy problem (1), (2) we may rewrite in the form (1)

\frac{r^{2}}{r^{2} − Kr + Q^{2}} u_{tt}− \frac{1}{r^{2}} ∂_{r}((r^{2}−Kr+Q^{2})u_{r})− \frac{1}{r^{2} sin ϕ}∂_{ϕ}(sin ϕu_{ϕ})− \frac{1}{r^{2} sin^{2} ϕ}u_{θθ} = f(u)+g(r),

(2) u(1, r, ϕ, θ) = u_{0} ∈ Ḣ^{1}(R^{+}×[0, 2π]×[0, π]), u_{t}(1, r, ϕ, θ) = u_{1} ∈ L^{2}(R^{+}×[0, 2π]×[0, π]).

When g_{s} is the Minkowski metric; u_{0}, u_{1} ∈ C_{0}^{∞}(R^{3}) in [5](see and [1], section 6.3) is proved that there exists T > 0 and a unique local solution u ∈ C^{2}([0, T) × R^{3}) for the Cauchy problem

(u_{tt} − Δu)_{g_{s}} = f(u), f ∈ C^{2}(R), t ∈ [0, T], x ∈ R^{3},
u|_{t=0} = u_{0}, u_{t}|_{t=0} = u_{1},

for which

sup_{t<T,x∈R^{3}} |u(t, x)| = ∞.

When g_{s} is the Minkowski metric, 1 ≤ p < 5 and initial data are in C_{0}^{∞}(R^{3}), in [5](see and [1], section 6.3) is proved that the initial value problem

(u_{tt} − Δu)_{g_{s}} = u|u|^{p−1}, t ∈ [0, T], x ∈ R^{3},
u|_{t=0} = u_{0}, u_{t}|_{t=0} = u_{1},

admits a global smooth solution.

When g_{s} is the Minkowski metric and initial data are in C_{0}^{∞}(R^{3}), in [4](see and [1], section 6.3) is proved that there exists a number ϵ_{0} > 0 such that for any data (u_{0}, u_{1}) ∈ C_{0}^{∞}(R^{3}) with E(u(0)) < ϵ_{0}, the initial value problem

(u_{tt} − Δu)_{g_{s}} = u^{5}, t ∈ [0, T], x ∈ R^{3},
u|_{t=0} = u_{0}, u_{t}|_{t=0} = u_{1},

admits a global smooth solution.

When g_{s} is the Minkowski metric in [6] is proved that the Cauchy problem

(u_{tt} − Δu)_{g_{s}} = f(u), t ∈ [0, 1], x ∈ R^{3},
u(1, x) = u_{0}, u_{t}(1, x) = u_{1},

has global solution. Here f ∈ C^{2}(R), f(0) = f′(0) = f′′(0) = 0,

|f′′(u) − f′′(v)| ≤ B|u − v|^{q_{1}}

for |u| ≤ 1, |v| ≤ 1, B > 0, \sqrt{2} − 1 < q_{1} ≤ 1, u_{0} ∈ C_{∘}^{5}(R^{3}), u_{1} ∈ C_{∘}^{4}(R^{3}), u_{0}(x) = u_{1}(x) = 0 for |x − x_{0}| > ρ, x_{0} and ρ are suitable chosen.

When g_{s} is the Reissner - Nordström metric, n = 3, p > 1, q ≥ 1, γ ∈ (0, 1) are fixed constants, f ∈ C^{1}(R^{1}), f(0) = 0, a|u| ≤ f′(u) ≤ b|u|, g ∈ C(R^{+}), g(|x|) ≥ 0, g(|x|) = 0 for |x| ≥ r_{1}, a and b are positive constants, r_{1} > 0 is suitable chosen, in [7] is proved that the initial value problem (1), (2), has nontrivial solution u ∈ C((0, 1]Ḃ_{p,q}^{γ}(R^{+})) in the form

u(t, r) = \{ \begin{matrix}v(t)ω(r) for r ≤ r_{1}, t ∈ [0, 1], \\ 0 for r ≥ r_{1}, t ∈ [0, 1], \end{matrix}

where r = |x|, for which lim_{t⟶0} ||u||_{Ḃ_{p,q}^{γ}(R^{+})} = ∞.

In this paper we will prove that the Cauchy problem (1), (2) has nontrivial solution u = u(t, r) ∈ C((0, 1]Ḣ^{1}(R^{+})) and the solution map is not uniformly continuous. When we say that the solution map (u_{∘}, u_{1}, g) ⟶ u(t, r) is uniformly continuous we understand: for every positive constant ϵ there exist positive constants δ and R such that for any two solutions u, v of the Cauchy problem (1), (2), with right hands g = g_{1}, g = g_{2} of (1), so that

(2′)  E(1, u − v) ≤ δ, ||g_{1}||_{L^{2}(R^{+})} ≤ R, ||g_{2}||_{L^{2}(R^{+})} ≤ R, ||g_{1} − g_{2}||_{L^{2}(R^{+})} ≤ δ,

the following inequality holds

(2′′)  E(t, u − v) ≤ ϵ for ∀t ∈ [0, 1],

where

E(t, u) := ||∂_{t}u(t, ·)||_{L^{2}(R^{+})}^{2} + ||\frac{∂}{∂r} u(t, ·)||_{L^{2}(R^{+})}^{2}.

Our main results are

Theorem 1.1. Let K, Q are positive constants for which

\{ \begin{matrix}K^{2} > 4Q^{2}, \frac{1}{1−K+Q^{2}} ≥ 1, \\ 1 − K + Q^{2} > 0 is enough small such that ^{\frac{K−\sqrt{K^{2}−4Q^{2}}}{2}} − 2\sqrt{1 − K + Q^{2}} > 0. \end{matrix}

Let also g ≡ 0, f ∈ C^{1}(R^{1}), f(0) = 0, a|u| ≤ f′(u) ≤ b|u|, a and b are positive constants. Then the homogeneous Cauchy problem (1), (2) has nontrivial solution u(t, r) = v(t)ω(r) ∈ C((0, 1]Ḣ^{1}(R^{+})). Also there exists t_{∘} ∈ [0, 1) for which exists constant ϵ > 0 such that for every positive constant δ exist solutions u, v of (1), (2), so that

E(1, u − v) ≤ δ,

and

E(t_{∘}, u − v) ≥ ϵ.

Theorem 1.2. Let K, Q are positive constants for which

\{ \begin{matrix}K^{2} > 4Q^{2}, \frac{1}{1−K+Q^{2}} ≥ 1, \\ 1 − K + Q^{2} > 0 is enough small such that ^{\frac{K−\sqrt{K^{2}−4Q^{2}}}{2}} − 2\sqrt{1 − K + Q^{2}} > 0. \end{matrix}

Let also g ≠ 0, g ∈ C(R^{+}), g(r) ≥ 0 for r ≥ 0, g(r) = 0 for r ≥ r_{1}, f ∈ C^{1}(R^{1}), f(0) = 0, a|u| ≤ f′(u) ≤ b|u|, a and b are positive constants. Then the nonhomogeneous Cauchy problem (1), (2) has nontrivial solution u(t, r) = v(t)ω(r) ∈ C((0, 1]Ḣ^{1}(R^{+})). Also there exists t_{∘} ∈ [0, 1) for which exists constant ϵ > 0 such that for every pair positive constants δ and R exist solutions u, v of (1), (2), with right hands g = g_{1}, g = g_{2} of (1), so that

E(1, u − v) ≤ δ, ||g_{1}||_{L^{2}(R^{+})} ≤ R, ||g_{2}||_{L^{2}(R^{+})} ≤ R, ||g_{1} − g_{2}||_{L^{2}(R^{+})} ≤ δ,

and

E(t_{∘}, u − v) ≥ ϵ.

The paper is organized as follows. In section 2 we prove theorem 1.1. In section 3 we prove theorem 1.2.

2. Proof of theorem 1.1.

For fixed q ≥ 1 and γ ∈ (0, 1) we put

C = (\frac{qγ2^{qγ}}{2^{qγ} − 1})^{\frac{1}{q}}

For fixed p > 1, q ≥ 1, γ ∈ (0, 1) and g ∈ C(R^{+}), g(r) ≥ 0 for r ≥ 0 we suppose that the constants A > 0, Q > 0, a > 0, b > 0, B > 0, K > 0, 1 < β < α satisfy the conditions i1)\frac{1}{1−K+Q^{2}} (\frac{1}{1−K+Q^{2}} \frac{2a}{A^{2}} + \frac{b}{AB} ) ≤ 1, A > 1, \frac{A^{2}}{a} > 1;

i2) \{ \begin{matrix}\frac{1}{α^{2}(1−αK+α^{2}Q^{2})} \frac{a}{2A^{4}} − \frac{br_{1}^{2}}{AB} ≥ 0, \\ \frac{a}{2A^{6}α^{2}(1−αK+α^{2}Q^{2})} − \frac{2br_{1}^{2}}{A^{2}B^{2}} ≥ 0, \\ \frac{a}{2A^{4}(1−αK+α^{2}Q^{2})}(\frac{1}{β} − \frac{1}{α}) − \frac{4ar_{1}}{A^{3}B(1−K+Q^{2})^{2}} ≥ 0, \\ (\frac{1}{β} − \frac{1}{α})^{2} \frac{1}{(1−αK+α^{2}Q^{2})^{2}} \frac{a}{4A^{6}} − r_{1}^{2} \frac{2b}{(1−K+Q^{2})A^{2}B^{2}} − r_{1}^{2} \frac{1}{1−K+Q^{2}} max_{r∈[0,r_{1}]} g(r) ≥ \frac{1}{A^{2}} , \\ (\frac{1}{β} − \frac{1}{α})^{2} \frac{1}{(1−αK+α^{2}Q^{2})^{2}} \frac{a}{2A^{4}} − \frac{2ar_{1}^{4}}{AB(1−K+Q^{2})} > 0, \end{matrix}

i3)C(\frac{1}{(1−K+Q^{2})^{2}} \frac{2a}{A^{2}} + \frac{2b}{AB(1−K+Q^{2})} + \frac{1}{A^{2}} )\frac{2^{2−γ}}{(q(1−γ))^{\frac{1}{q}}} + C \frac{2^{1−γ}}{A^{2}(1−K+Q^{2})^{2}(q(1−γ))^{\frac{1}{q}}} < 1,

i4) \{ \begin{matrix}(\frac{1}{α^{2}} \frac{1}{1−αK+α^{2}Q^{2}} \frac{a}{4A^{6}} − \frac{1}{β^{2}} \frac{a}{4A^{4}} ) > 0, \\ (\frac{1}{α^{2}} \frac{1}{1−αK+α^{2}Q^{2}} \frac{a}{2A^{6}} − \frac{1}{β^{2}} \frac{a}{2A^{4}} ) > 0, \end{matrix}

i5) \{ \begin{matrix}K^{2} > 4Q^{2}, A ≥ \frac{1}{1−K+Q^{2}} > 1, \\ 1 > \frac{2Q^{2}}{K} > ^{\frac{K−\sqrt{K^{2}−4Q^{2}}}{2}} , \\ 1 − K + Q^{2} > 0 is enough small such that \\ 1 > ^{\frac{K−\sqrt{K^{2}−4Q^{2}}}{2}} − 2\sqrt{1 − K + Q^{2}} > 0, \\ \frac{2}{K−\sqrt{K^{2}−4Q^{2}}−2(1−K+Q^{2})} < β < α ≤ 3, \\ \frac{a}{4A^{6}α^{2}(1−αK+α^{2}Q^{2})} − \frac{2br_{1}^{2}}{A^{2}B^{2}} − r_{1}^{2} max_{r∈[0,r_{1}]} g(r) ≥ 0, \end{matrix}

i6)\frac{1}{1−K+Q^{2}} (\frac{2}{AB} (\frac{1}{1−K+Q^{2}} \frac{2a}{A^{2}} + \frac{b}{AB} ) + r_{1}^{2} max_{s∈[0,r_{1}]} g(s)) ≤ \frac{2}{AB}; i7) max_{s∈[0,r_{1}]} g(s) ≤ (\frac{1}{β} − \frac{1}{α})^{2} \frac{1}{A^{4}} , where

r_{1} = \frac{K − \sqrt{K^{2} − 4Q^{2}}}{2} − \sqrt{1 − K + Q^{2}}.

Example. Let 0 < ϵ << \frac{1}{3} is enough small, A = \frac{1}{ϵ^{4}} , B = \frac{1}{ϵ}, p = \frac{3}{2}, q = \frac{3}{2}, γ = \frac{1}{3}, α = 3, \frac{1}{β} = ^{\frac{K−\sqrt{K^{2}−4Q^{2}}}{2}} − 2\sqrt{1 − K + Q^{2}},

g(r) = \{ \begin{matrix}ϵ^{11}(r − r_{1})^{2} for r ≤ r_{1}, \\ 0 for r ≥ r_{1}, \end{matrix}

K = \frac{4}{3} + \frac{1}{6}ϵ^{20} − \frac{3}{2}ϵ^{2}, Q^{2} = \frac{1}{3} + \frac{1}{6}ϵ^{20} − \frac{1}{2}ϵ^{2}, a = ϵ^{4}, b = ϵ^{3}. Then

\begin{matrix}1 − αK + α^{2}Q^{2} = 1 − 3K + 9Q^{2} = ϵ^{20}, \\ 1 − K + Q^{2} = ϵ^{2} • . \end{matrix}

When g(r) ≡ 0 we put

(1′)  u_{0} := v(1)ω(r) =

EJQTDE, 2007 No. 12, p. 5

= \{ \begin{matrix}∫_{r}^{r_{1}} \frac{1}{τ^{2}−Kτ+Q^{2}} ∫_{τ}^{r_{1}}(\frac{s^{4}}{s^{2}−Ks+Q^{2}} v′′(1)ω(s) − s^{2}f(v(1)ω(s)))dsdτ for r ≤ r_{1}, \\ 0 for r ≥ r_{1}, \end{matrix}

and u_{1} ≡ 0. Here v(t) is fixed function which satisfies the conditions

(H1) v(t) ∈ C^{3}[0, ∞), v(t) > 0 for ∀t ∈ [0, 1];

(H2) v′′(t) > 0 for ∀t ∈ [0, 1], v′(1) = v′′′(1) = 0, v(1) ≠ 0;

(H3) \{ \begin{matrix}min_{t∈[0,1]} \frac{v′′(t)}{v(t)} ≥ \frac{a}{2A^{4}} , max_{t∈[0,1]} \frac{v′′(t)}{v(t)} ≤ \frac{2a}{A^{2}} ; \\ v′′(t) − \frac{a}{2A^{4}} v(t) ≥ 0 for t ∈ [0, 1]. \end{matrix}

Bellow we will prove that the equation (1′) has unique nontrivial solution ω(r) for which ω(r) ∈ C^{2}[0, r_{1}], ω(r) ∈ Ḣ^{1}[0, r_{1}], |ω(r)| ≤ \frac{2}{AB} for r ∈ [0, r_{1}], ω(r) ≥ \frac{1}{A^{2}} for r ∈ [\frac{1}{α}, \frac{1}{β} ], ω(r_{1}) = ω′(r_{1}) = ω′′(r_{1}) = 0.

Example.

There exists function v(t) for which (H1)-(H3) are hold. Really, let us consider the function

(3)  v(t) = \frac{(t − 1)^{2} + \frac{4A^{4}}{a} − 1}{\frac{A^{3}}{a}},

where the constants A and a satisfy the conditions A > 1, \frac{A^{2}}{a} > 1. Then 1) v(t) ∈ C^{3}[0, ∞) and v(t) > 0 for all t ∈ [0, 1], i.e. (H1) is hold. 2)

v′(t) = _{\frac{2(t−1)}{\frac{A^{3}}{a}}}, v′(1) = 0,
v′′(t) = _{\frac{2}{\frac{A^{3}}{a}}} ≥ 0 ∀ t ∈ [0, 1],
v′′′(t) = 0, v′′′(1) = 0,

consequently (H2) is hold. On the other hand we have

\frac{v′′(t)}{v(t)} = \frac{2}{(t − 1)^{2} + \frac{4A^{4}}{a} − 1}.

From here

min_{t∈[0,1]} \frac{v′′(t)}{v(t)} ≥ \frac{a}{2A^{4}} ,
max_{t∈[0,1]} \frac{v′′(t)}{v(t)} ≤ \frac{2a}{A^{2}} ,
v′′(t) − \frac{a}{2A^{4}} v(t) = _{\frac{1}{\frac{2A^{7}}{a^{2}}}} (2 − t)t,

i.e. (H3) is hold.

2.1. Local existence of nontrivial solutions of homogeneous Cauchy problem (1), (2)

Let v(t) is fixed function which satisfies the hypothesis (H1) − (H3).

In this section we will prove that the homogeneous Cauchy problem (1), (2) has nontrivial solution in the form

u(t, r) = \{ \begin{matrix}v(t)ω(r) for r ≤ r_{1}, t ∈ [0, 1], \\ 0 for r ≥ r_{1}, t ∈ [0, 1]. \end{matrix}

Let us consider the integral equation (⋆)

u(t, r) = \{ \begin{matrix}∫_{r}^{r_{1}} \frac{1}{τ^{2}−Kτ+Q^{2}} ∫_{τ}^{r_{1}}(\frac{s^{4}}{s^{2}−Ks+Q^{2}} \frac{v′′(t)}{v(t)}u(t, s) − s^{2}f(u(t, s)))dsdτ, 0 ≤ r ≤ r_{1}, t ∈ [0, 1], \\ 0 for r ≥ r_{1}, t ∈ [0, 1], \end{matrix}

where u(t, r) = v(t)ω(r).

Theorem 2.1. Let v(t) is fixed function which satisfies the hypothesis (H1)-(H3). Let also p > 1, q ∈ [1, ∞) and γ ∈ (0, 1) are fixed and the positive constants A, a, b, B, Q,K, α > β > 1 satisfy the conditions i1)-i6) and f ∈ C^{1}(R^{1}), f(0) = 0, a|u| ≤ f′(u) ≤ b|u|. Then the equation (⋆) has unique nontrivial solution u(t, r) = v(t)ω(r) for which w ∈ C^{2}[0, r_{1}], u(t, r) = u_{r}(t, r) = u_{rr}(t, r) = 0 for r ≥ r_{1}, u(t, r) ∈ C((0, 1]Ḣ^{1}[0, r_{1}]), for r ∈ [\frac{1}{α}, \frac{1}{β} ] and t ∈ [0, 1] u(t, r) ≥ \frac{1}{A^{2}} , for r ∈ [0, r_{1}] and t ∈ [0, 1] |u(t, r)| ≤ \frac{2}{AB}.

Proof. In [7, p. 303-309,theorem 3.1] is proved that the equation (⋆) has solution u(t, r) in the form u(t, r) = v(t)ω(r) for which

u(t, r) ∈ C([0, 1] × [0, r_{1}]);
u(t, r) = u_{r}(t, r) = u_{rr}(t, r) = 0 for r ≥ r_{1} and t ∈ [0, 1],
u(t, r) ∈ C((0, 1]Ḃ_{p,q}^{γ}[0, r_{1}]);
for r ∈ [\frac{1}{α}, \frac{1}{β} ] and t ∈ [0, 1] u(t, r) ≥ \frac{1}{A^{2}} ;
u(t, r) ≥ 0 for t ∈ [0, 1] and r ∈ [\frac{1}{α}, r_{1}];
for r ∈ [0, r_{1}] and t ∈ [0, 1] |u(t, r)| ≤ \frac{2}{AB} .

In [7] is used the following definition of the Ḃ_{p,q}^{γ}(M)-norm (γ ∈ (0, 1), p > 1, q ≥ 1) (see [3, p.94, def. 2], [1])

||u||_{Ḃ_{p,q}^{γ}(M)} = (∫_{0}^{2} h^{−1−qγ}||Δ_{h}u||_{L^{p}(M)}^{q}dh)^{\frac{1}{q}} ,

where

Δ_{h}u = u(x + h) − u(x).

Let t ∈ [0, 1] is fixed. Then

\begin{matrix}||\frac{∂}{∂r} u||_{L^{2}([0,∞))}^{2} = \\ = ∫_{0}^{r_{1}}(\frac{1}{r^{2}−Kr+Q^{2}} ∫_{r}^{r_{1}}(\frac{s^{4}}{s^{2}−Ks+Q^{2}} \frac{v′′(t)}{v(t)}u(t, s) − s^{2}f(u(t, s)))dsdτ)^{2}dr ≤ \end{matrix}

here we use that from i5) we have that r_{1} < 1, \frac{s^{4}}{s^{2}−Ks+Q^{2}} ≤ \frac{1}{1−K+Q^{2}} for s ∈ [0, r_{1}], \frac{1}{r^{2}−Kr+Q^{2}} ≤ \frac{1}{1−K+Q^{2}} for r ∈ [0, r_{1}](see [7, Remark, p.300])

≤ ∫_{0}^{r_{1}}(\frac{1}{1 − K + Q^{2}} ∫_{r}^{r_{1}}(\frac{1}{1 − K + Q^{2}} max_{t∈[0,1]} \frac{v′′(t)}{v(t)}|u(t, s)| + s^{2}|f(u(t, s))|)dsdτ)^{2}dr ≤

here we use that f(0) = 0, |f(u)| ≤ \frac{b}{2}|u|^{2},

≤ ∫_{0}^{r_{1}}(\frac{1}{1 − K + Q^{2}} ∫_{r}^{r_{1}}(\frac{1}{1 − K + Q^{2}} max_{t∈[0,1]} \frac{v′′(t)}{v(t)}|u(t, s)| + s^{2} \frac{b}{2}|u(t, s)|^{2})dsdτ)^{2}dr ≤

here we use that |u(t, r)| ≤ \frac{2}{AB} for r ∈ [0, r_{1}], t ∈ [0, 1],

≤ ∫_{0}^{r_{1}}(\frac{1}{1 − K + Q^{2}} ∫_{r}^{r_{1}}(\frac{1}{1 − K + Q^{2}} max_{t∈[0,1]} \frac{v′′(t)}{v(t)} \frac{2}{AB} + r_{1}^{2} \frac{b}{2} \frac{4}{A^{2}B^{2}} )dsdτ)^{2}dr ≤

here we use that max_{t∈[0,1]} \frac{v′′(t)}{v(t)} ≤ \frac{2a}{A^{2}} ,

\begin{matrix}≤ ∫_{0}^{r_{1}}(\frac{1}{1−K+Q^{2}} ∫_{r}^{r_{1}}(\frac{1}{1−K+Q^{2}} \frac{2a}{A^{2}} \frac{2}{AB} + r_{1}^{2} \frac{b}{2} \frac{4}{A^{2}B^{2}} )dsdτ)^{2}dr ≤ \\ ≤ r_{1}^{3}(\frac{1}{1−K+Q^{2}} (\frac{1}{1−K+Q^{2}} \frac{4a}{A^{3}B} + \frac{2r_{1}^{2}b}{A^{2}B^{2}} ))^{2} < ∞. \end{matrix}

From here

||\frac{∂}{∂r} u||_{L^{2}([0,∞))} < ∞

for every fixed t ∈ (0, 1]. Therefore u(t, r) ∈ C((0, 1]Ḣ^{1}([0, ∞))). •

Let ũ is the solution from the theorem 2. 1, i.e ũ is the solution to the equation (⋆). From proposition 2.1([7]) ũ satisfies the equation (1). Then ũ is solution to the Cauchy problem (1), (2) with initial data

u_{0} = \{ \begin{matrix}∫_{r}^{r_{1}} \frac{1}{τ^{2}−Kτ+Q^{2}} ∫_{τ}^{r_{1}}(\frac{s^{4}}{s^{2}−Ks+Q^{2}} v′′(1)ω(s) − s^{2}f(v(1)ω(s)))dsdτ for r ≤ r_{1}, \\ 0 for r ≥ r_{1}, \end{matrix}

u_{1} = \{ \begin{matrix}∫_{r}^{r_{1}} \frac{1}{τ^{2}−Kτ+Q^{2}} ∫_{τ}^{r_{1}}(\frac{s^{4}}{s^{2}−Ks+Q^{2}} v′′′(1)ω(s) − s^{2}f′(u)v′(1)ω(s))dsdτ = 0 for r ≤ r_{1}, \\ 0 for r ≥ r_{1}, \end{matrix}

u_{0} ∈ Ḣ^{1}(R^{+}), u_{1} ∈ L^{2}(R^{+}), ũ ∈ C((0, 1]Ḣ^{1}[0, r_{1}]).

2.2. Uniformly continuity of the solution map for the homogeneous Cauchy problem (1), (2)

Let v(t) is same function as in Theorem 2.1.

Theorem 2.2. Let p > 1, q ≥ 1 and γ ∈ (0, 1) are fixed and the positive constants a, b, A, B, Q, K, 1 < β < α satisfy the conditions i1)-i6). Let f ∈ C^{1}(R^{1}), f(0) = 0, a|u| ≤ f′(u) ≤ b|u|. Then there exists t_{∘} ∈ [0, 1) for which there exists constant ϵ > 0 such that for every positive constant δ exist solutions u_{1} and u_{2} so that

E(1, u_{1} − u_{2}) ≤ δ

and

E(t_{∘}, u_{1} − u_{2}) ≥ ϵ.

Proof. Let us suppose that the solution map (u_{∘}, u_{1}, g) ⟶ u(t, r) is uniformly continuous.

Let

(4) 0 < ϵ < (\frac{1}{β} − \frac{1}{α})^{3}(\frac{α^{2}}{1 − αK + α^{2}Q^{2}} (\frac{a}{2A^{6}α^{2}(1 − αL + α^{2}Q^{2})} − \frac{2b}{β^{2}A^{2}B^{2}} ))^{2}.

Let also

u_{1} = ũ, u_{2} = 0.

Then there exists positive constant δ such that

E(1, u_{1} − u_{2}) ≤ δ,

and

E(t, u_{1} − u_{2}) ≤ ϵ for ∀t ∈ [0, 1].

From here, for t ∈ [0, 1)

ϵ ≥ ||\frac{∂}{∂r} ũ||_{L^{2}([0,∞))}^{2} =

= ∫_{0}^{r_{1}}(\frac{1}{r^{2}−Kr+Q^{2}} ∫_{r}^{r_{1}}(\frac{s^{4}}{s^{2}−Ks+Q^{2}} \frac{v′′(t)}{v(t)} ũ(t, s) − s^{2}f(ũ(t, s)))dsdτ)^{2}dr ≥

≥ ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{r^{2}−Kr+Q^{2}} ∫_{r}^{r_{1}}(\frac{s^{4}}{s^{2}−Ks+Q^{2}} \frac{v′′(t)}{v(t)} ũ(t, s) − s^{2}f(ũ(t, s)))dsdτ)^{2}dr ≥

here we use that for s ∈ [\frac{1}{α}, r_{1}] and for t ∈ [0, 1] we have that \frac{s^{4}}{s^{2}−Ks+Q^{2}} \frac{v′′(t)}{v(t)} ũ(t, s) − s^{2}f(ũ(t, s)) ≥ 0(see [7, p. 305-306]) and \frac{1}{r^{2}−Kr+Q^{2}} ≥ 0 for r ∈ [0, r_{1}],

\begin{matrix}≥ ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{r^{2}−Kr+Q^{2}} ∫ \frac{1}{β} ; \frac{1}{α} (\frac{s^{4}}{s^{2}−Ks+Q^{2}} \frac{v′′(t)}{v(t)} ũ(t, s) − s^{2}f(ũ(t, s)))dsdτ)^{2}dr ≥ \\ ≥ ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{r^{2}−Kr+Q^{2}} ∫ \frac{1}{β} ; \frac{1}{α} (\frac{s^{4}}{s^{2}−Ks+Q^{2}} min_{t∈[0,1]} \frac{v′′(t)}{v(t)} ũ(t, s) − s^{2}f(ũ(t, s)))dsdτ)^{2}dr ≥ \end{matrix}

from (H3) we have that min_{t∈[0,1]} \frac{v′′(t)}{v(t)} ≥ \frac{a}{2A^{4}}

≥ ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{r^{2} − Kr + Q^{2}} ∫ \frac{1}{β} ; \frac{1}{α} (\frac{s^{4}}{s^{2} − Ks + Q^{2}} \frac{a}{2A^{4}} ũ(t, s) − s^{2}f(ũ(t, s)))dsdτ)^{2}dr ≥

here we use that f(0) = 0, f(ũ) ≤ \frac{b}{2}ũ^{2}

≥ ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{r^{2} − Kr + Q^{2}} ∫ \frac{1}{β} ; \frac{1}{α} (\frac{s^{4}}{s^{2} − Ks + Q^{2}} \frac{a}{2A^{4}} ũ(t, s) − s^{2} \frac{b}{2}ũ^{2}(t, s))dsdτ)^{2}dr ≥

here we use that ũ(t, s) ≥ \frac{1}{A^{2}} for t ∈ [0, 1] and s ∈ [\frac{1}{α}, \frac{1}{β} ], ũ^{2}(t, s) ≤ \frac{4}{A^{2}B^{2}} for t ∈ [0, 1] and s ∈ [0, r_{1}]

≥ ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{r^{2} − Kr + Q^{2}} ∫ \frac{1}{β} ; \frac{1}{α} (\frac{s^{4}}{s^{2} − Ks + Q^{2}} \frac{a}{2A^{4}} \frac{1}{A^{2}} − \frac{1}{β^{2}} \frac{b}{2} \frac{4}{A^{2}B^{2}} )dsdτ)^{2}dr ≥

here we use that \frac{s^{2}}{s^{2}−Ks+Q^{2}} is increase function for s ∈ [0, r_{1}]. Therefore, for s ∈ [\frac{1}{α}, \frac{1}{β} ] we have \frac{s^{4}}{s^{2}−Ks+Q^{2}} ≥ \frac{1}{α^{2}(1−αK+α^{2}Q^{2})}

≥ ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{r^{2} − Kr + Q^{2}} ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{α^{2}(1 − αK + α^{2}Q^{2})} \frac{a}{2A^{6}} − \frac{2b}{β^{2}A^{2}B^{2}} )dsdτ)^{2}dr ≥

here we use that for r ∈ [0, r_{1}] the function \frac{1}{r^{2}−Kr+Q^{2}} is increase function. Therefore for r ∈ [\frac{1}{α}, \frac{1}{β} ] we have \frac{1}{r^{2}−Kr+Q^{2}} ≥ \frac{α^{2}}{1−αK+α^{2}Q^{2}}

≥ (\frac{1}{β} − \frac{1}{α})^{3}(\frac{α^{2}}{1 − αK + α^{2}Q^{2}} (\frac{a}{2A^{6}α^{2}(1 − αL + α^{2}Q^{2})} − \frac{2b}{β^{2}A^{2}B^{2}} ))^{2}

which is contradiction with (4). •

3. Proof of Theorem 1.2.

3.1. Local existence of nontrivial solutions for nonhomogenious Cauchy problem (1), (2)

Let v(t) is fixed function which satisfies the conditions (H1), (H2) and (H4), where

(H4) min_{t∈[0,1]} \frac{v′′(t)}{v(t)} ≥ \frac{a}{4A^{4}} , max_{t∈[0,1]} \frac{v′′(t)}{v(t)} ≤ \frac{2a}{A^{2}} .

For instance, the function

v(t) = \frac{(t − 1)^{2} + \frac{8A^{4}}{a} − 1}{\frac{A^{3}}{a}},

satisfies the hypothesis (H1), (H2) and (H4).

Let us consider the equation (⋆′)

u(t, r) = \{ \begin{matrix}∫_{r}^{r_{1}} \frac{1}{τ^{2}−Kτ+Q^{2}} ∫_{τ}^{r_{1}}(\frac{s^{4}}{s^{2}−Ks+Q^{2}} \frac{v′′(t)}{v(t)}u(t, s) − s^{2}f(u(t, s)) − s^{2}g(s))dsdτ, 0 ≤ r ≤ r_{1}, \\ 0 for r ≥ r_{1}, \end{matrix}

t ∈ [0, 1], where u(t, r) = v(t)ω(r).

Theorem 3.1. Let v(t) is fixed function which satisfies the hypothesis (H1), (H2), (H4). Let also p > 1, q ∈ [1, ∞) and γ ∈ (0, 1) are fixed and the positive constants A, a, b, B, Q,K, α > β > 1 satisfy the conditions i1)-i7) and f ∈ C^{1}(R^{1}), f(0) = 0, a|u| ≤ f′(u) ≤ b|u|, g ∈ C(R^{+}), g(r) ≥ 0 for ∀r ∈ R^{+}, g(r) = 0 for r ≥ r_{1}. Then the equation (⋆′) has unique nontrivial solution u(t, r) = v(t)ω(r) for which w ∈ C^{2}[0, r_{1}], u(t, r) = u_{r}(t, r) = u_{rr}(t, r) = 0 for r ≥ r_{1}, u(t, r) ∈ C((0, 1]Ḣ^{1}[0, r_{1}]), for r ∈ [\frac{1}{α}, \frac{1}{β} ] and

t ∈ [0, 1] u(t, r) ≥ \frac{1}{A^{2}} , for r ∈ [0, r_{1}] and t ∈ [0, 1] |u(t, r)| ≤ \frac{2}{AB}.

Proof. In [7, p. 313-316,theorem 4.1] is proved that the equation (⋆′) has solution u(t, r) in the form u(t, r) = v(t)ω(r) for which

u(t, r) ∈ C([0, 1] × [0, r_{1}]);
u(t, r) = u_{r}(t, r) = u_{rr}(t, r) = 0 for r ≥ r_{1} and t ∈ [0, 1],
u(t, r) ∈ C((0, 1]Ḃ_{p.q}^{γ}[0, r_{1}]);
for r ∈ [\frac{1}{α}, \frac{1}{β} ] and t ∈ [0, 1] u(t, r) ≥ \frac{1}{A^{2}} ;
u(t, r) ≥ 0 for t ∈ [0, 1] and r ∈ [\frac{1}{α}, r_{1}];
for r ∈ [0, r_{1}] and t ∈ [0, 1] |u(t, r)| ≤ \frac{2}{AB} .

Let t ∈ [0, 1] is fixed. Then

\begin{matrix}||\frac{∂}{∂r} u||_{L^{2}([0,∞))}^{2} = \\ = ∫_{0}^{r_{1}}(\frac{1}{r^{2}−Kr+Q^{2}} ∫_{r}^{r_{1}}(\frac{s^{4}}{s^{2}−Ks+Q^{2}} \frac{v′′(t)}{v(t)}u(t, s) − s^{2}(f(u(t, s)) + g(s)))dsdτ)^{2}dr ≤ \end{matrix}

here we use that from i5) we have that r_{1} < 1, \frac{s^{4}}{s^{2}−Ks+Q^{2}} ≤ \frac{1}{1−K+Q^{2}} for s ∈ [0, r_{1}], \frac{1}{r^{2}−Kr+Q^{2}} ≤ \frac{1}{1−K+Q^{2}} for r ∈ [0, r_{1}](see [7, Remark, p.300])

≤ ∫_{0}^{r_{1}}(\frac{1}{1 − K + Q^{2}} ∫_{r}^{r_{1}}(\frac{1}{1 − K + Q^{2}} max_{t∈[0,1]} \frac{v′′(t)}{v(t)}|u(t, s)|+s^{2}(|f(u(t, s))|+g(s)))dsdτ)^{2}dr ≤

here we use that f(0) = 0, |f(u)| ≤ \frac{b}{2}|u|^{2},

≤ ∫_{0}^{r_{1}}(\frac{1}{1 − K + Q^{2}} ∫_{r}^{r_{1}}(\frac{1}{1 − K + Q^{2}} max_{t∈[0,1]} \frac{v′′(t)}{v(t)}|u(t, s)|+s^{2}(\frac{b}{2}|u(t, s)|^{2}+g(s)))dsdτ)^{2}dr ≤

here we use that |u(t, r)| ≤ \frac{2}{AB} for r ∈ [0, r_{1}], t ∈ [0, 1], from i7) we have max_{r∈[0,r_{1}]} g(r) ≤

(\frac{1}{β} − \frac{1}{α})^{2} \frac{1}{A^{4}}

≤ ∫_{0}^{r_{1}}(\frac{1}{1 − K + Q^{2}} ∫_{r}^{r_{1}}(\frac{1}{1 − K + Q^{2}} max_{t∈[0,1]} \frac{v′′(t)}{v(t)} \frac{2}{AB} +r_{1}^{2} \frac{b}{2} \frac{4}{A^{2}B^{2}} +r_{1}^{2}(\frac{1}{β} −\frac{1}{α})^{2} \frac{1}{A^{4}} )dsdτ)^{2}dr ≤

here we use that max_{t∈[0,1]} \frac{v′′(t)}{v(t)} ≤ \frac{2a}{A^{2}} ,

\begin{matrix}≤ ∫_{0}^{r_{1}}(\frac{1}{1−K+Q^{2}} ∫_{r}^{r_{1}}(\frac{1}{1−K+Q^{2}} \frac{2a}{A^{2}} \frac{2}{AB} + r_{1}^{2} \frac{b}{2} \frac{4}{A^{2}B^{2}} + r_{1}^{2}(\frac{1}{β} − \frac{1}{α})^{2} \frac{1}{A^{4}} )dsdτ)^{2}dr ≤ \\ ≤ r_{1}^{3}(\frac{1}{1−K+Q^{2}} (\frac{1}{1−K+Q^{2}} \frac{4a}{A^{3}B} + \frac{2r_{1}^{2}b}{A^{2}B^{2}} + r_{1}^{2}(\frac{1}{β} − \frac{1}{α})^{2} \frac{1}{A^{4}} ))^{2} < ∞. \end{matrix}

From here

||\frac{∂}{∂r} u||_{L^{2}([0,∞))} < ∞

for every fixed t ∈ (0, 1]. Therefore u(t, r) ∈ C((0, 1]Ḣ^{1}([0, ∞))). •

Let ū is the solution from the theorem 3. 1. , i.e ¯ ū is the solution to the equation (⋆′). From proposition 2.3[7] we have that ū satisfies the equation (1). Then ū is solution to the Cauchy problem (1), (2) with initial data

¯ū_{0} = \{ \begin{matrix}∫_{r}^{r_{1}} \frac{1}{τ^{2}−Kτ+Q^{2}} ∫_{τ}^{r_{1}}(\frac{s^{4}}{s^{2}−Ks+Q^{2}} v′′(1)ω(s) − s^{2}f(v(1)ω(s)) − s^{2}g(s))dsdτ for r ≤ r_{1}, \\ 0 for r ≥ r_{1}, \end{matrix}

¯ū_{1} = \{ \begin{matrix}∫_{r}^{r_{1}} \frac{1}{τ^{2}−Kτ+Q^{2}} ∫_{τ}^{r_{1}}(\frac{s^{4}}{s^{2}−Ks+Q^{2}} v′′′(1)ω(s) − s^{2}f′(u)v′(1)ω(s))dsdτ ≡ 0 \\ for r ≤ r_{1}, \\ 0 for r ≥ r_{1}, \end{matrix}

¯ū_{0} ∈ Ḣ^{1}(R^{+}), ¯ ū_{1} ∈ L^{2}(R^{+}), ¯ ū ∈ C((0, 1]Ḣ^{1}[0, r_{1}]).

3.2. Uniformly continuity of the solution map for the nonhomogeneous Cauchy problem (1), (2)

Let v(t) is same function as in Theorem 3.1.

Theorem 3.2. Let p > 1, q ≥ 1 and γ ∈ (0, 1) are fixed and the positive constants a, b, A, B, Q, K, 1 < β < α satisfy the conditions i1)-i7). Let f ∈ C^{1}(R^{1}), f(0) = 0, a|u| ≤ f′(u) ≤ b|u|, g ∈ C(R^{+}), g(r) ≥ 0 for r ≥ 0, g(r) = 0 for r ≥ r_{1}. Then there exists t_{∘} ∈ [0, 1) for which there exists constant ϵ > 0 such that for every positive constants δ and R exist solutions u_{1}, u_{2} of (1), (2) with right hands g = g_{1}, g = g_{2} of (1), so that

E(1, u_{1} − u_{2}) ≤ δ, ||g_{1}||_{L^{2}(R^{+})} ≤ R, ||g_{2}||_{L^{2}(R^{+})} ≤ R, ||g_{1} − g_{2}||_{L^{2}(R^{+})} ≤ δ,

and

E(t_{∘}, u_{1} − u_{2}) ≥ ϵ.

Proof. Let us suppose that the solution map (u_{∘}, u_{1}, g) ⟶ u(t, r) is uniformly continuous. Let (5)

0 < ϵ < (\frac{1}{β} −\frac{1}{α})^{3}(\frac{α^{2}}{1 − αK + α^{2}Q^{2}} (\frac{a}{4A^{6}α^{2}(1 − αK + α^{2}Q^{2})}−\frac{2b}{β^{2}A^{2}B^{2}} −r_{1}^{2}(\frac{1}{β} −\frac{1}{α})^{2} \frac{1}{A^{4}} ))^{2}.

Let also

u_{1} = ū, u_{2} = 0, g_{2} ≡ 0, g_{1} ≡ g,

where g is the function from theorem 3.1. Then there exist positive constants δ and R such that

||g_{1}||_{L^{2}(R^{+})} ≤ R, ||g_{2}||_{L^{2}(R^{+})} ≤ R, ||g_{1} − g_{2}||_{L^{2}(R^{+})} ≤ δ,
E(1, u_{1} − u_{2}) ≤ δ,

and

E(t, u_{1} − u_{2}) ≤ ϵ for ∀t ∈ [0, 1].

From here, for t ∈ [0, 1)

ϵ ≥ ||\frac{∂}{∂r} ū||_{L^{2}([0,∞))}^{2} =

= ∫_{0}^{r_{1}}(\frac{1}{r^{2}−Kr+Q^{2}} ∫_{r}^{r_{1}}(\frac{s^{4}}{s^{2}−Ks+Q^{2}} \frac{v′′(t)}{v(t)} ū(t, s) − s^{2}(f(ū(t, s)) + g(s)))dsdτ)^{2}dr ≥

≥ ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{r^{2}−Kr+Q^{2}} ∫_{r}^{r_{1}}(\frac{s^{4}}{s^{2}−Ks+Q^{2}} \frac{v′′(t)}{v(t)} ū(t, s) − s^{2}(f(ū(t, s)) + g(s)))dsdτ)^{2}dr ≥

here we use that for r ∈ [\frac{1}{α}, r_{1}] we have

\begin{matrix}\frac{s^{4}}{s^{2}−Ks+Q^{2}} \frac{v′′(t)}{v(t)} ū(t, s) − s^{2}f(ū(t, s)) − s^{2}g(s) ≥ \\ ≥ \frac{a}{4A^{6}α^{2}(1−αK+α^{2}Q^{2})} − \frac{2br_{1}^{2}}{A^{2}B^{2}} − r_{1}^{2} max_{r∈[0,r_{1}]} g(r) ≥ 0 \end{matrix}

(see i5)) and for r ∈ [0, r_{1}] we have \frac{1}{r^{2}−Kr+Q^{2}} > 0

\begin{matrix}≥ ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{r^{2}−Kr+Q^{2}} ∫ \frac{1}{β} ; \frac{1}{α} (\frac{s^{4}}{s^{2}−Ks+Q^{2}} \frac{v′′(t)}{v(t)} ū(t, s) − s^{2}(f(ū(t, s)) + g(s)))dsdτ)^{2}dr ≥ \\ ≥ ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{r^{2}−Kr+Q^{2}} ∫ \frac{1}{β} ; \frac{1}{α} (\frac{s^{4}}{s^{2}−Ks+Q^{2}} min_{t∈[0,1]} \frac{v′′(t)}{v(t)} ū(t, s) − s^{2}(f(ū(t, s)) + g(s)))dsdτ)^{2}dr ≥ \end{matrix}

from (H4) we have that min_{t∈[0,1]} \frac{v′′(t)}{v(t)} ≥ \frac{a}{4A^{4}}

≥ ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{r^{2} − Kr + Q^{2}} ∫ \frac{1}{β} ; \frac{1}{α} (\frac{s^{4}}{s^{2} − Ks + Q^{2}} \frac{a}{4A^{4}} ū(t, s) − s^{2}(f(ū(t, s)) + g(s)))dsdτ)^{2}dr ≥

here we use that f(0) = 0, f(ū) ≤ \frac{b}{2}ū^{2}

≥ ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{r^{2} − Kr + Q^{2}} ∫ \frac{1}{β} ; \frac{1}{α} (\frac{s^{4}}{s^{2} − Ks + Q^{2}} \frac{a}{4A^{4}} ū(t, s) − s^{2} \frac{b}{2}ū^{2}(t, s) − s^{2}g(s))dsdτ)^{2}dr ≥

here we use that ū(t, s) ≥ \frac{1}{A^{2}} for t ∈ [0, 1] and s ∈ [\frac{1}{α}, \frac{1}{β} ], ū^{2}(t, s) ≤ \frac{4}{A^{2}B^{2}} for t ∈ [0, 1] and

s ∈ [0, r_{1}], max_{r∈[0,r_{1}]} g(r) ≤ (\frac{1}{β} − \frac{1}{α})^{2} \frac{1}{A^{4}} ,

≥ ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{r^{2} − Kr + Q^{2}} ∫ \frac{1}{β} ; \frac{1}{α} (\frac{s^{4}}{s^{2} − Ks + Q^{2}} \frac{a}{4A^{4}} \frac{1}{A^{2}} − \frac{1}{β^{2}} \frac{b}{2} \frac{4}{A^{2}B^{2}} −r_{1}^{2}(\frac{1}{β} − \frac{1}{α})^{2} \frac{1}{A^{4}} )dsdτ)^{2}dr ≥

here we use that \frac{s^{2}}{s^{2}−Ks+Q^{2}} is increase function for s ∈ [0, r_{1}]. Therefore, for s ∈ [\frac{1}{α}, \frac{1}{β} ] we

have \frac{s^{4}}{s^{2}−Ks+Q^{2}} ≥ \frac{1}{α^{2}(1−αK+α^{2}Q^{2})}

≥ ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{r^{2} − Kr + Q^{2}} ∫ \frac{1}{β} ; \frac{1}{α} (\frac{1}{α^{2}(1 − αK + α^{2}Q^{2})} \frac{a}{4A^{6}} −\frac{2b}{β^{2}A^{2}B^{2}} −r_{1}^{2}(\frac{1}{β} −\frac{1}{α})^{2} \frac{1}{A^{4}} )dsdτ)^{2}dr ≥

here we use that for r ∈ [0, r_{1}] the function \frac{1}{r^{2}−Kr+Q^{2}} is increase function. Therefore for

r ∈ [\frac{1}{α}, \frac{1}{β} ] we have \frac{1}{r^{2}−Kr+Q^{2}} ≥ \frac{α^{2}}{1−αK+α^{2}Q^{2}}

≥ (\frac{1}{β} − \frac{1}{α})^{3}(\frac{α^{2}}{1 − αK + α^{2}Q^{2}} (\frac{a}{4A^{6}α^{2}(1 − αK + α^{2}Q^{2})} − \frac{2b}{β^{2}A^{2}B^{2}} − r_{1}^{2}(\frac{1}{β} − \frac{1}{α})^{2} \frac{1}{A^{4}} ))^{2}

which is contradiction with (5). •

References

[1] Shatah, J., M. Struwe. Geometric wave equation. Courant lecture notes in mathematics 2(1998).

[2] Taylor, M. Partial differential equations III. Springer-Verlag, 1996.

[3] Runst, T., W. Sickel. Sobolev spaces of fractional order, Nemytskij operators and nonlinear partial differential equations, 1996, New York.

[4] Rauch, J. The u^{5}− Klein - Gordon equation, Brezis. and Lions. eds., Pilman Research Notes in Math. 53, pp. 335-364.

[5] Jörgens, K. Das Anfangswertproblem im Grossen für eine Klasse nicht-linear Wellengleichungen, Math. Z., 77, 1961, pp. 295-308.

[6] John, F. Blow up of solutions of nonlinear wave equations in three space dimensions, Manuscripta Math. 28, 1979, 235-268.

[7] Georgiev, S. Blow - up of the solutions of nonlinear wave equation in Reissner - Nordström metric, Dynamics of PDE, Vol. 3, No. 4, 295-329, 2006.

(Received April 23, 2007)