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Hindawi Publishing Corporation Boundary Value Problems Volume 2007, Article ID 42954, 51 pages doi:10.1155/2007/42954 Research Article Blow up of the Solutions of Nonlinear Wave Equation Svetlin Georgiev Georgiev Received 14 March 2007; Accepted 26 May 2007 Recommended by Peter Bates We construc t for ever y fixed n ≥ 2 the metric g s = h 1 (r)dt 2 − h 2 (r)dr 2 − k 1 (ω)dω 2 1 − ···− k n−1 (ω)dω 2 n −1 ,whereh 1 (r), h 2 (r), k i (ω), 1 ≤ i ≤ n −1, are continuous functions, r =|x|, for which we consider the Cauchy problem (u tt −Δu) gs = f (u)+g(|x|), where x ∈ R n , n ≥ 2; u(1,x) = u ◦ (x) ∈ L 2 (R n ), u t (1,x) = u 1 (x) ∈ ˙ H −1 (R n ), where f ∈ Ꮿ 1 (R 1 ), f (0) = 0, a|u|≤f  (u) ≤ b|u|, g ∈ Ꮿ(R + ), g(r) ≥ 0, r =|x|, a and b are positive con- stants. When g(r) ≡ 0, we prove that the above Cauchy problem has a nontrivial solution u(t,r)intheformu(t,r) = v(t)ω(r) for which lim t→0 u L 2 ([0,∞)) =∞.Wheng(r) = 0, we prove that the above Cauchy problem has a nontrivial solution u(t,r)intheform u(t,r) = v(t)ω(r) for which lim t→0 u L 2 ([0,∞)) =∞. Copyright © 2007 Svetlin Georgiev Georgiev. This is an open access article distributed under the Creative Commons Attribution License, which p ermits unrestricted use, dis- tribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction In this paper, we study the properties of the solutions of the Cauchy problem  u tt −Δu  g s = f (u)+g  | x|  , x ∈ R n , n ≥2, (1) u(1,x) = u ◦ (x) ∈ L 2  R n  , u t (1,x) = u 1 (x) ∈ ˙ H −1  R n  ,(2) where g s is the met ric g s = h 1 (r)dt 2 −h 2 (r)dr 2 −k 1 (ω)dω 2 1 −···−k n−1 (ω)dω 2 n −1 , (1.1) 2 Boundary Value Problems the functions h 1 (r), h 2 (r) satisfy the conditions h 1 (r),h 2 (r) ∈Ꮿ 1  [0,∞)  , h 1 (r) > 0, h 2 (r) ≥0 ∀r ∈[0,∞),  ∞ 0  h 2 (s) h 1 (s)  ∞ s  h 2 (τ) h 1 (τ) dτ < ∞,  ∞ 0  h 2 (s) h 1 (s)  ∞ s  h 1 (τ)h 2 (τ)dτds < ∞,  ∞ 0   ∞ r  h 2 (s) h 1 (s)   ∞ s   h 2 (τ) h 1 (τ) C 1 + C 2  h 1 (τ)h 2 (τ)  2 dτ  1/2 ds  2 dr < ∞, C 1 ,C 2 are arbitrary nonnegative constants,  ∞ 0   ∞ r  h 2 (s) h 1 (s)   ∞ s   h 2 (τ) h 1 (τ) C 1 + C 2  h 1 (τ)h 2 (τ)  dτ  ds  2 dr < ∞, C 1 ,C 2 are arbitrary nonnegative constants, max r∈[0,∞)  h 1 (r)h 2 (r) < ∞,  ∞ 0   ∞ r  h 2 (s) h 1 (s)  ∞ s  h 2 (τ) h 1 (τ) dτ ds  2 dr < ∞,  ∞ 0   ∞ r  h 2 (s) h 1 (s)   ∞ s h 1 (τ)h 2 (τ)dτ  1/2 ds  2 dr < ∞,  ∞ 0   ∞ r  h 2 (s) h 1 (s)   ∞ s h 2 (τ) h 1 (τ) dτ  1/2 ds  2 dr < ∞,  ∞ 0   ∞ r  h 2 (s) h 1 (s) ds  2 dr < ∞, (i1) k i (ω) ∈ Ꮿ 1 ([0,2π] ×···×[0,2π]), i = 1, ,n −1, f ∈ Ꮿ 1 (᏾ 1 ), f (0) = 0, a|u|≤f  (u) ≤ b|u|, a and b are p ositive constants, g ∈ Ꮿ(R 1 ), g(|x|) ≥ 0for|x|∈[0,∞). (In Section 2 we will give example for such metric g s .) We search a solution u = u(t, r) to the Cauchy problem (1), (2). Therefore, if the Cauchy problem (1), (2) has such solution, it will satisfy the Cauchy problem 1 h 1 (r) u tt − 1  h 1 (r)h 2 (r) ∂ r  h 1 (r)  h 1 (r)h 2 (r) u r  = f (u)+g(r), (1.2) u(1,r) = u ◦ ∈ L 2  [0,∞)  , u t (1,r) =u 1 ∈ ˙ H −1  [0,∞)  . (1.3) In this paper, we will prove that the Cauchy problem (1), (2) has nontrivial solution u = u(t,r) for which lim t→0 u L 2 ([0,∞)) =∞. (1.4) Our main results are the following. Svetlin Georgiev Georgiev 3 Theorem 1.1. Suppose n ≥ 2 is fixed, h 1 (r), h 2 (r) satisfy the conditions (i1), g ≡ 0, f ∈ Ꮿ 1 (R 1 ), f (0) =0, a|u|≤f  (u) ≤ b|u|, a and b are positive constants. Then the homoge- neous problem of Cauchy (1), (2) has nontrivial solution u = u(t,r) ∈ Ꮿ((0,1]L 2 ([0,∞))) for which lim t→0 u L 2 ([0,∞)) =∞. (1.5) Theorem 1.2. Suppose n ≥ 2 is fixed, h 1 (r), h 2 (r) satisfy the conditions (i1). Suppose also that a and b are fixed positive constants, a ≤ b, f ∈Ꮿ 1 (R 1 ), f (0) = 0, a|u|≤f  (u) ≤ b|u|, b/2 ≥ f (1) ≥ a/2, g = 0, g ∈ Ꮿ([0,∞)), g(r) ≥ 0 for every r ≥ 0, g(r) ≤ b/2 − f (1) for ev- ery r ∈ [0,∞).ThenthenonhomogeneousproblemofCauchy(1), (2) has nontrivial solut ion u = u(t,r) ∈ Ꮿ((0,1]L 2 ([0,∞))) for which lim t→0 u L 2 ([0,∞)) =∞. (1.6) When g s is the Minkowski metric and u 0 ,u 1 ∈ Ꮿ ∞ 0 (R 3 )in[1] (see also [2, Section 6.3]), it is proved that there exists T>0 and a unique local solution u ∈ Ꮿ 2 ([0,T) ×R 3 )forthe Cauchy problem  u tt −Δu  g s = f (u), f ∈ Ꮿ 2 (R), t ∈ [0, T], x ∈ R 3 , u   t=0 = u 0 , u t   t=0 = u 1 , (1.7) for which sup t<T,x∈R 3   u(t,x)   =∞ . (1.8) When g s is the Minkowski metric, 1 ≤ p<5 and initial data are in Ꮿ ∞ 0 (R 3 )in[1] (see also [2, Section 6.3]), it 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 (1.9) admits a global smooth solution. When g s is the Minkowski metric and initial data are in Ꮿ ∞ 0 (R 3 )in[3](seealso[2, Section 6.3]) it is proved that there exists a number  0 > 0 such that for any data (u 0 ,u 1 ) ∈ Ꮿ ∞ 0 (R 3 )withE(u(0)) <  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 (1.10) admits a global smooth solution. When g s is the Reissner-Nordstr ¨ om metric in [4], it is proved that the Cauchy problem  u tt −Δu  g s + m 2 u = f (u), t ∈[0,1], x ∈ R 3 , u(1,x) = u 0 ∈ ˙ B γ p,p  R 3  , u t (1,x) = u 1 ∈ ˙ B γ−1 p,p  R 3  , (1.11) 4 Boundary Value Problems where m = 0isconstantand f ∈Ꮿ 2 (R 1 ), a|u|≤|f (l) (u)|≤b|u|, l =0,1, a and b are pos- itive constants, has unique nontrivial solution u(t, r) ∈ Ꮿ((0,1] ˙ B γ p,p (R + )), r =|x|, p>1, for which lim t→0 u ˙ B γ p,p (R + ) =∞. (1.12) When g s is the Minkowski metric in [5], it 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 (1.13) has global solution. Here f ∈ Ꮿ 2 (R), f (0) = f  (0) = f  (0) = 0,   f  (u) − f  (v)   ≤ B|u −v| q 1 (1.14) for |u|≤1, |v|≤1, B>0, √ 2 −1 <q 1 ≤ 1, u 0 ∈ Ꮿ 5 ◦ (R 3 ), u 1 ∈ Ꮿ 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-Nordstr ¨ om metric, n = 3, p>1, q ≥ 1, γ ∈ (0,1) are fixed constants, f ∈ Ꮿ 1 (R 1 ), f (0) = 0, a|u|≤f  (u) ≤ b|u|, g ∈ Ꮿ(R + ), g(|x|) ≥ 0, g(|x|) = 0 for |x|≥r 1 , a and b are positive constants, r 1 > 0 is suitable chosen, in [6], it is proved that the initial value problem (1), (2) has nontriv ial solution u ∈ Ꮿ((0,1] ˙ B γ p,q (R + )) in the form u(t,r) = ⎧ ⎨ ⎩ v(t)ω(r), for r ≤ r 1 , t ∈ [0, 1], 0, for r ≥ r 1 , t ∈ [0, 1], (1.15) where r =|x|, for which lim t→0 u ˙ B γ p,q (R + ) =∞. The paper is organized as follows. In Section 2, we will prove some preliminary results. In Section 3,wewillproveTheorem 1.1.InSection 4,wewillproveTheorem 1.2.Inthe appendix we will prove some results which are used for the proof of Theorems 1.1 and 1.2. 2. Preliminary results Proposition 2.1. Let h 1 (r), h 2 (r) satisfy the conditions (i1), f ∈ Ꮿ(−∞,∞), g ≡ 0.If for every fixed t ∈ [0,1] the function u(t,r) = v(t)ω(r),wherev(t) ∈ Ꮿ 4 ([0,1]), v(t) = 0 for every t ∈ [0,1], ω(r) ∈ Ꮿ 2 ([0,∞)), ω(∞) =ω  (∞) = 0,satisfies(1 ), then the function u(t,r) = v(t)ω(r) satisfies the integral equation u(t,r) =  ∞ r  h 2 (s) h 1 (s)  ∞ s   h 2 (τ) h 1 (τ) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ) f (u)  dτ ds (1 ∗ ) for every fixed t ∈ [0,1]. Proof. Suppose that t ∈ [0,1]isfixedandthefunctionu(t,r)=v(t)ω(r), v(t) ∈ Ꮿ 4 ([0,1]), v(t) = 0foreveryt ∈[0,1], ω(r) ∈Ꮿ 2 ([0,∞)), ω(∞) = ω  (∞) = 0, satisfies (1). Then for Svetlin Georgiev Georgiev 5 every fixed t ∈ [0,1] and for r ∈[0,∞)wehave u tt (t,r) = v  (t) v(t) u(t,r), 1 h 1 (r) v  (t) v(t) u(t,r) − 1  h 1 (r)h 2 (r) ∂ r  h 1 (r)  h 1 (r)h 2 (r) u r (t,r)  = f (u), 1  h 1 (r)h 2 (r) ∂ r  h 1 (r)  h 1 (r)h 2 (r) u r (t,r)  = 1 h 1 (r) v  (t) v(t) u(t,r) − f (u), ∂ r  h 1 (r)  h 1 (r)h 2 (r) u r (t,r)  =  h 2 (r) h 1 (r) v  (t) v(t) u(t,r) −  h 1 (r)h 2 (r) f (u). (2.1) Now we integrate the last equality from r to ∞ here we suppose that u r (t,r) = v(t)ω  (r), u r (t,∞) = v(t)ω  (∞) = 0, then we get − h 1 (r)  h 1 (r)h 2 (r) u r (t,r) =  ∞ r   h 2 (τ) h 1 (τ) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ) f (u)  dτ, −  h 1 (r) h 2 (r) u r (t,r) =  ∞ r   h 2 (τ) h 1 (τ) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ) f (u)  dτ, −u r (t,r) =  h 2 (r) h 1 (r)  ∞ r   h 2 (τ) h 1 (τ) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ) f (u)  dτ. (2.2) Now we integrate the last equality from r to ∞;weusethatu(t, ∞) = v(t)ω(∞) = 0, then we get u(t,r) =  ∞ r  h 2 (s) h 1 (s)  ∞ s   h 2 (τ) h 1 (τ) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ) f (u)  dτ ds, (2.3) that is, for every fixed t ∈ [0,1] if the function u(t,r) = v(t)ω(r) satisfies (1), then the function u(t, r) = v(t)ω(r) satisfies the integral equation (1 ∗ ). Here v(t) ∈ Ꮿ 4 ([0,1]), v(t) = 0foreveryt ∈ [0,1], ω(r) ∈Ꮿ 2 ([0,∞)), ω(∞) = ω  (∞) = 0.  Proposition 2.2. Let h 1 (r), h 2 (r) satisfy the conditions (i1), f ∈ Ꮿ(−∞,∞), g ≡ 0.Iffor every fixed t ∈ [0,1] the function u(t,r) = v(t)ω(r),wherev(t) ∈ Ꮿ 4 ([0,1]), v(t) = 0 for every t ∈ [0,1], ω(r) ∈ Ꮿ 2 ([0,∞)), ω(∞) = ω  (∞) = 0, satisfies the integral equation (1 ∗ ) then the function u(t,r) = v(t)ω(r) satisfies (1)foreveryfixedt ∈ [0,1]. Proof. Let t ∈ [0,1] be fixed and let the function u(t,r)=v(t)ω(r), where v(t) ∈Ꮿ 4 ([0,1]), v(t) = 0foreveryt ∈ [0,1], ω(r) ∈ Ꮿ 2 ([0,∞)), ω(∞) = ω  (∞) = 0, satisfy the integral equation (1 ∗ ). From here and from f ∈ Ꮿ(−∞,∞), for every fixed t ∈ [0,1] we have 6 Boundary Value Problems u(t,r) ∈ Ꮿ 2 ([0,∞)) and u r (t,r) =−  h 2 (r) h 1 (r)  ∞ r   h 2 (r) h 1 (r) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ) f (u)  dτ,  h 1 (r) h 2 (r) u r (t,r) =−  ∞ r   h 2 (r) h 1 (r) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ) f (u)  dτ, h 1 (r)  h 1 (r)h 2 (r) u r (t,r) =−  ∞ r   h 2 (r) h 1 (r) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ) f (u)  dτ, ∂ r  h 1 (r)  h 1 (r)h 2 (r) u r (t,r)  =  h 2 (r) h 1 (r) v  (t) v(t) u(t,r) −  h 1 (r)h 2 (r) f (u),  h 2 (r) h 1 (r) v  (t) v(t) u(t,r) −∂ r  h 1 (r)  h 1 (r)h 2 (r) u r (t,r)  =  h 1 (r)h 2 (r) f (u), 1 h 1 (r) v  (t) v(t) u(t,r) − 1  h 1 (r)h 2 (r) ∂ r  h 1 (r)  h 1 (r)h 2 (r) u r (t,r)  = f (u). (2.4) Since for every fixed t ∈ [0,1] we have v  (t) v(t) u(t,r) = u tt (t,r), (2.5) we get 1 h 1 (r) u tt (t,r) − 1  h 1 (r)h 2 (r) ∂ r  h 1 (r)  h 1 (r)h 2 (r) u r (t,r)  = f (u), (2.6) that is, for every fixed t ∈ [0,1] if the function u(t,r) =v(t)ω(r), where v(t) ∈ Ꮿ 4 ([0,1]), v(t) = 0foreveryt ∈ [0,1], ω(r) ∈ Ꮿ 2 ([0,∞)), ω(∞) = ω  (∞) = 0, satisfies (1 ∗ ), then it satisfies (1)foreveryfixedt ∈ [0,1].  Proposition 2.3. Let h 1 (r), h 2 (r) satisfy the conditions (i1), f ∈Ꮿ(−∞,∞), g ∈Ꮿ([0,∞)), g(r) ≥ 0 for every r ≥ 0.Ifforeveryfixedt ∈ [0,1] the function u(t,r) = v(t)ω(r),where v(t) ∈ Ꮿ 4 ([0,1]), v(t) = 0 for every t ∈ [0,1], ω(r) ∈ Ꮿ 2 ([0,∞)), ω(∞) = ω  (∞) = 0,sat- isfies (1), then the function u(t,r) = v(t)ω(r) satisfies the integral equation u(t,r) =  ∞ r  h 2 (s) h 1 (s)  ∞ s   h 2 (τ) h 1 (τ) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ)  f (u)+g(r)   dτ ds (1 ∗∗ ) for every fixed t ∈ [0,1]. Proof. Let t ∈ [0,1] be fixed and let the function u(t, r) = v(t)ω(r), v(t) ∈ Ꮿ 4 ([0,1]), v(t) = 0foreveryt ∈[0,1], ω(r) ∈ Ꮿ 2 ([0,∞)), ω(∞) = ω  (∞) = 0, satisfy (1). Then for Svetlin Georgiev Georgiev 7 every fixed t ∈ [0,1] and for r ∈[0,∞)wehave u tt (t,r) = v  (t) v(t) u(t,r), 1 h 1 (r) v  (t) v(t) u(t,r) − 1  h 1 (r)h 2 (r) ∂ r  h 1 (r)  h 1 (r)h 2 (r) u r (t,r)  =  f (u)+g(r)  , 1  h 1 (r)h 2 (r) ∂ r  h 1 (r)  h 1 (r)h 2 (r) u r (t,r)  = 1 h 1 (r) v  (t) v(t) u(t,r) −  f (u)+g(r)  , ∂ r  h 1 (r)  h 1 (r)h 2 (r) u r (t,r)  =  h 2 (r) h 1 (r) v  (t) v(t) u(t,r) −  h 1 (r)h 2 (r)  f (u)+g(r)  . (2.7) Now we integrate the last e quality from r to ∞;herewesupposethatu r (t,r) =v(t)ω  (r), u r (t,∞) = v(t)ω  (∞) = 0, then we get − h 1 (r)  h 1 (r)h 2 (r) u r (t,r) =  ∞ r   h 2 (τ) h 1 (τ) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ)  f (u)+g(r)   dτ, −  h 1 (r) h 2 (r) u r (t,r) =  ∞ r   h 2 (τ) h 1 (τ) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ)  f (u)+g(r)   dτ, −u r (t,r) =  h 2 (r) h 1 (r)  ∞ r   h 2 (τ) h 1 (τ) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ)  f (u)+g(r)   dτ. (2.8) Now we integrate the last equality from r to ∞; we suppose that u(t,∞) = v(t)ω(∞) = 0, then we get u(t,r) =  ∞ r  h 2 (s) h 1 (s)  ∞ s   h 2 (τ) h 1 (τ) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ)  f (u)+g(r)   dτ ds, (2.9) that is, for every fixed t ∈ [0,1] if the function u(t,r) = v(t)ω(r) satisfies (1), then the function u(t,r) = v(t)ω(r) satisfies the integral equation (1 ∗∗ ). Here v(t) ∈ Ꮿ 4 ([0,1]), v(t) = 0foreveryt ∈ [0,1], ω(r) ∈Ꮿ 2 ([0,∞)), ω(∞) = ω  (∞) = 0.  Proposition 2.4. Let h 1 (r), h 2 (r) satisfy the conditions (i1), f ∈Ꮿ(−∞,∞), g ∈Ꮿ([0,∞)), g(r) ≥ 0 for every r ≥ 0.Ifforeveryfixedt ∈ [0,1] the function u(t,r) = v(t)ω(r),where 8 Boundary Value Problems v(t) ∈ Ꮿ 4 ([0,1]), v(t) = 0 for every t ∈ [0,1], ω(r) ∈ Ꮿ 2 ([0,∞)), ω(∞) = ω  (∞) = 0,sat- isfies the integral equation (1 ∗∗ ), then the function u(t,r) = v(t)ω(r) satisfies (1)forevery fixed t ∈ [0,1]. Proof. Let t ∈ [0,1] be fixed and let the function u(t,r)=v(t)ω(r), where v(t) ∈Ꮿ 4 ([0,1]), v(t) = 0foreveryt ∈ [0,1], ω(r) ∈ Ꮿ 2 ([0,∞)), ω(∞) = ω  (∞) = 0, satisfy the integral equation (1 ∗∗ ). From here and from f ∈ Ꮿ(−∞, ∞), g ∈ Ꮿ([0, ∞)), for every fixed t ∈ [0,1] we have u(t,r) ∈Ꮿ 2 ([0,∞)) and u r (t,r) =−  h 2 (r) h 1 (r)  ∞ r   h 2 (r) h 1 (r) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ)  f (u)+g(r)   dτ,  h 1 (r) h 2 (r) u r (t,r) =−  ∞ r   h 2 (r) h 1 (r) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ)  f (u)+g(r)   dτ, h 1 (r)  h 1 (r)h 2 (r) u r (t,r) =−  ∞ r   h 2 (r) h 1 (r) v  (t) v(t) u(t,τ) −  h 1 (τ)h 2 (τ)  f (u)+g(r)   dτ, ∂ r  h 1 (r)  h 1 (r)h 2 (r) u r (t,r)  =  h 2 (r) h 1 (r) v  (t) v(t) u(t,r) −  h 1 (r)h 2 (r)  f (u)+g(r)  ,  h 2 (r) h 1 (r) v  (t) v(t) u(t,r) −∂ r  h 1 (r)  h 1 (r)h 2 (r) u r (t,r)  =  h 1 (r)h 2 (r)  f (u)+g(r)  , 1 h 1 (r) v  (t) v(t) u(t,r) − 1  h 1 (r)h 2 (r) ∂ r  h 1 (r)  h 1 (r)h 2 (r) u r (t,r)  = f (u)+g(r). (2.10) Since for every fixed t ∈ [0,1] we have v  (t) v(t) u(t,r) = u tt (t,r), (2.11) we get 1 h 1 (r) u tt (t,r) − 1  h 1 (r)h 2 (r) ∂ r  h 1 (r)  h 1 (r)h 2 (r) u r (t,r)  = f (u)+g(r), (2.12) that is, for every fixed t ∈ [0,1] if the function u(t,r) =v(t)ω(r), where v(t) ∈ Ꮿ 4 ([0,1]), v(t) = 0foreveryt ∈ [0, 1], ω(r) ∈Ꮿ 2 ([0,∞)), ω(∞) = ω  (∞) = 0, satisfies (1 ∗∗ ), then it satisfies (1)foreveryfixedt ∈ [0,1].  Svetlin Georgiev Georgiev 9 For fixed n ≥ 2, h 1 (r), h 2 (r) which satisfy the conditions (i1) and fixed positive con- stants a and b, we suppose that the positive constants c, d, A, B, A 1 , A 2 satisfy the condi- tions c ≤ d, A ≥ B, A 1 ≤ A 2 , A 1 − b 2B > 0,  h 2 (r) h 1 (r) A 1 − b B  h 1 (r)h 2 (r) ≥0foreveryr ∈[0,∞),  ∞ r  h 2 (s) h 1 (s)  ∞ s   h 2 (τ) h 1 (τ) A 1 − b 2B  h 1 (τ)h 2 (τ)  dτ ds ≥ 1forr ∈[c,d],  ∞ 1  h 2 (s) h 1 (s)  ∞ s  A 1  h 2 (τ) h 1 (τ) − b B  h 1 (τ)h 2 (τ)  dτ ds ≥ A 1 10 10 (H1) max r∈[0,∞)  ∞ r  h 2 (s) h 1 (s)  ∞ s   h 2 (τ) h 1 (τ) A 2 +  h 1 (τ)h 2 (τ) b 2B  dτ ds ≤ 1, max r∈[0,∞)  h 2 (r) h 1 (r)  ∞ r   h 2 (τ) h 1 (τ) A 2 +  h 1 (τ)h 2 (τ) b 2B  dτ ≤ 1, (H2)  ∞ 0   ∞ r  h 2 (s) h 1 (s)   ∞ s   h 2 (τ) h 1 (τ) A 2 +  h 1 (τ)h 2 (τ) b 2B  2 dτ  1/2 ds  2 dr < 1 7 ,  ∞ 0   ∞ r  h 2 (s) h 1 (s)  ∞ s   h 2 (τ) h 1 (τ) A 2 + b B  h 1 (τ)h 2 (τ)  dτ ds  2 dr < ∞, (H3)  ∞ 0   ∞ r  h 2 (s) h 1 (s)   ∞ s   h 2 (τ) h 1 (τ) A 2 +  h 1 (τ)h 2 (τ) b B  2 dτ  1/2 ds  2 dr < 1. (H4) Example 2.5. Let 0 <   1/3 be enough small, n ≥ 2isfixed.Wechoosec>0, d>0, c ≤ d<∞ such that for every r ∈ [c,d]wehave π 4 ≤ arctg(d +1−r) 3 ,arctgd 3 < π 3 . (2.13) Let also b = 8 3 , a = 4 3 , A = 60, B =40, A 1 =  3 , A 2 = 2 3 .Let h 1 (r) =  B b  − 1+  1+2 A 1 b B  2 , h 2 (r) = 144(d +1−r) 4  (d +1−r) 6 +1  2 . (2.14) 10 Boundary Value Problems We note that the functions h 1 (r)andh 2 (r) satisfy all conditions of (i1)and A 1  h 1 (r) − b 2B  h 1 (r) =1,  ∞ r  h 2 (s) h 1 (s)  ∞ s   h 2 (τ) h 1 (τ) A 1 − b 2B  h 1 (τ)h 2 (τ)  dτ ds ≥  d+1 r  h 2 (s) h 1 (s)  d+1 s   h 2 (τ) h 1 (τ) A 1 − b 2B  h 1 (τ)h 2 (τ)  dτ ds ≥ 1forr ∈[c,d] . (2.15) We note that  h 1 (r) ∼ 1,6. For fixed n ≥ 2, h 1 (r), h 2 (r), which satisfy the conditions (i1), the constants a, b, c, d, A, B, A 1 , A 2 are fixed which satisfy the conditions (H1), ,(H4), then we suppose that the function v(t) is fixed function and satisfies the conditions v(t) ∈ Ꮿ 4  [0,1]  , v  (t) v(t) > 0, v(t) > 0, ∀t ∈ [0,1], (H5) A 1 ≤ v  (t) v(t) ≤ A 2 , v  (1) = 0, v  (1) = 0, (H6) lim t→0  v  (t) v(t) − a 2  = +0. (H7) Example 2.6. Let a, b, c, d, A 1 , A 2 , B, A be the constants from the above example. Then a/2 = A 2 and v(t) = C  e √ A 2 (t−1) + e − √ A 2 (t−1)  , (2.16) where C is arbitrary positive constant, satisfiing the hypotheses (H5), (H6), (H7). Here and below we suppose that v(t) is fixed function which satisfies the conditions (H5), ,(H7). When g(r) ≡ 0weput u ◦ :=v(1)ω(r)=  ∞ r  h 2 (s) h 1 (s)  ∞ s   h 2 (τ) h 1 (τ) v  (1)ω(τ)−  h 1 (τ)h 2 (τ) f  v(1)ω(τ)   dτ ds, u 1 ≡ 0. (1  ) In Section 3,wewillprovethat(1  ) has unique nontrivial solution ω(r) ∈L 2 ([0,∞)). [...]... u2 < u1 − u2 L2 ([0,∞)) (3.27) Consequently, the operator L : M → M is contractive operator We note that the set M is closed subset of the space Ꮿ((0,1]L2 ([0, ∞))) (for the proof see Lemma A.1 in the appendix of this paper) Therefore, (1 ) has unique nontrivial solution in the set M Let u be the solution from Theorem 3.1, that is, u is a solution to the integral equation (1 ) From Proposition 2.2,... ([0,∞)) < u1 − u2 L2 ([0,∞)) (4.32) Consequently, the operator R : M → M is contractive operator We note that the set M is closed subset of the space Ꮿ((0,1]L2 ([0, ∞))) (for the proof see Lemma A.1 in the appendix of this paper) Therefore (1 ) has unique nontrivial solution in the set M Let u be the solution from Theorem 4.1, that is, u is a solution to the integral equation (1 ) From Proposition 2.4,... Consequently, u is solution to the Cauchy problem (1), (2) with initial data u◦ = v(1)ω(r) = ∞ r h2 (s) h1 (s) ∞ s h2 (τ) v (1)ω(τ) − h1 (τ)h2 (τ) f v(1)ω(τ) h1 (τ) dτ ds, u1 ≡ 0 (3.28) ˙ We have u ∈ Ꮿ((0,1]L2 ([0, ∞))), u◦ ∈ L2 ([0, ∞)), u1 ∈ H −1 ([0, ∞)) 3.2 Blow up of the solutions of homogeneous Cauchy problem (1), (2) Let v(t) be the same function as in Theorem 3.1 Theorem 3.2 Let n ≥ 2 be fixed,... (H7) we have lim F = +0 t →0 (3.61) Therefore lim u t →0 L2 ([0,∞)) = ∞ (3.62) Svetlin Georgiev Georgiev 23 4 Proof of Theorem 1.2 4.1 Local existence of nontrivial solutions of nonhomogeneous Cauchy problem (1), (2) In this section we will prove that the nonhomogeneous Cauchy problem (1), (2) has nontrivial solution in the form u(t,r) = v(t)ω(r) Let us consider the integral equation u(t,r) = ∞ h2... nontrivial solution ω(r) ∈ L2 ([0,1)) 3 Proof of Theorem 1.1 3.1 Local existence of nontrivial solutions of homogeneous Cauchy problem (1), (2) In this section, we will prove that the homogeneous Cauchy problem (1), (2) has nontrivial solution in the form u(t,r) = v(t)ω(r) For fixed function v(t), which satisfies the conditions (H5), (H6), and (H7) we consider the integral equation u(t,r) = ∞ h2 (s) h1... ,u is solution to the nonhomogeneous Cauchy problem (1), (2) with initial data u◦ = v(1)ω(r) = ∞ r h2 (s) h1 (s) ∞ h2 (τ) v (1)ω(τ) − h1 (τ)h2 (τ) f v(1)ω(τ) + g(τ) h1 (τ) s dτ ds, u1 ≡ 0 (4.33) ˙ We have u ∈ Ꮿ((0,1]L2 ([0, ∞))), u◦ ∈ L2 ([0, ∞)), u1 ∈ H −1 ([0, ∞)) 4.2 Blow up of the solutions of nonhomogeneous Cauchy problem (1), (2) Let v(t) be the same function as in Theorem 4.1 Theorem 4.2 Let n... fixed, let h1 (r), h2 (r) fixed, which satisfy the conditions (i1), be the positive constants a, b be fixed, let the positive constants c, d, A, B, A1 , A2 be fixed which satisfy the conditions (H1), , (H4) and f ∈ Ꮿ1 ((−∞, ∞)), f (0) = 0, a|u| ≤ f (u) ≤ b|u| Then for the solution u of the Cauchy problem (1), (2) one has lim u t →0 L2 ([0,∞)) = ∞ (3.29) Proof For every fixed t ∈ (0,1] and for every r ∈... (r) be fixed, which satisfy the conditions (i1), let the positive constants a, b, a ≤ b be fixed, and let the positive constants c, d, A, B, A1 , A2 be fixed which satisfy the conditions (H1), , (H4) and f ∈ Ꮿ1 ((−∞, ∞)), f (0) = 0, a|u| ≤ f (u) ≤ b|u|, b/2 ≥ f (1) ≥ a/2, g(r) ∈ Ꮿ([0, ∞)), g(r) ≥ 0 for every r ≥ 0, g(r) ≤ b/2 − f (1) for every r ≥ 0 Then for the solution u of the Cauchy problem (1), (2)... prove that the operator L : M → M is contractive operator Let u1 and u2 be two elements of the set M Then, for every fixed t ∈ [0,1] we have L u1 − L u2 ∞ = ≤ r ∞ r ∞ h2 (s) h1 (s) h2 (s) h1 (s) s ∞ s h2 (τ) v (t) u1 − u2 − h1 (τ)h2 (τ) f u1 − f u2 h1 (τ) v(t) h2 (τ) v (t) u1 − u2 + h1 (τ)h2 (τ) f u1 − f u2 h1 (τ) v(t) dτ ds dτ ds, (3.20) 16 Boundary Value Problems then from the middle-point theorem we... h1 (τ) v(t) s (1 ) Theorem 3.1 Let n ≥ 2 be fixed, let h1 (r), h2 (r) fixed, which satisfy the conditions (i1), let the positive constants a, b be fixed, a ≤ b, let the positive constants c, d A, B, A1 , A2 be fixed which satisfy the conditions (H1), , (H4) and f ∈ Ꮿ1 ((−∞, ∞)), f (0) = 0, a|u| ≤ f (u) ≤ b|u| Let also v(t) be fixed function which satisfies the conditions (H5), , (H7) Then (1 ) has unique . Corporation Boundary Value Problems Volume 2007, Article ID 42954, 51 pages doi:10.1155/2007/42954 Research Article Blow up of the Solutions of Nonlinear Wave Equation Svetlin Georgiev Georgiev Received. u 1 ∈ ˙ H −1 ([0,∞)). 3.2. Blow up of the solutions of homogeneous Cauchy problem (1), (2). Let v(t)bethe same function as in Theorem 3.1. Theorem 3.2. Let n ≥ 2 be fixed, let h 1 (r), h 2 (r) fixed, which satisfy the. (3.27) Consequently, the operator L : M → M is contractive operator. We note that the set M is closed subset of the space Ꮿ((0,1]L 2 ([0,∞))) (for the proof see Lemma A.1 in the appen- dix of this paper). Therefore,