# On Behavior of Solution of Degenerated Hyperbolic Equation

Authors: Tahir Gadjiev; Rafig Rasulov; Orkhan Aliev
Journal: ISRN Applied Mathematics
 (2012)

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## Abstract

The purpose of this paper is to learn some features of hyperbolic type of nonlinear equations. It is shown that the solution of the equation approaches to the endlessness in the inside of some initial conditions and time of the special marks. The local existence of the equation’s solution has been proved and the problem of unlimited increasing on the solution of nonlinear hyperbolic equations type during the finite time is investigated.


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## Body

## 1. Introduction

In this paper the unbounded increasing solution of the nonlinear hyperbolic-type equation for the finite times is considered. These type equations describe the processes of electron and ionic heat conductivity in plasma, diffusion of neutrons, andα-particles, and so forth. Investigation of unbounded solution or regime of peaking solutions occurs in theory of nonlinear equations where one of the essential ideas is the representation called eigenfunction of nonlinear dissipative surroundings. It is well known that even a simple nonlinearity, subject to critical of exponent, the solution of nonlinear hyperbolic type equation for the finite time may increase unboundedly, that is, there is a number T>0 such that
(1.1)∥u(x,t)∥L∞(Rn)→∞,t→T<∞.
In [1] the existence of unbounded solution for finite time with a simple nonlinearity has been proved. [2] has shown that any nonnegative solution subject to critical exponent is unbounded increasing for the finite time. Similar results were obtained in [3] and corresponding theorems are called Fujita-Hayakawa’s theorems. More detailed reviews can be found in [4–6].The paper is organizing as follows. In Section2 we present some definitions and auxiliary results. In Section 3, we give the main results for type nonlinear equation, where blow up solutions are obtained and the local solution exists. This results  generalize the corresponding result [7].


## 2. Some Definitions and Auxiliary Results

Let us consider the equation(2.1)∂2u∂t2=∑i,j=1n‍∂∂xj(ω(x)|∂u∂xj|p-2∂u∂ui)+f(x,t,u),
in bounded domain Ω⊂Rn,n≥2 with nonsmooth boundary. Let's denote Πa,b={(x,t),x∈Ω,a<t<b},Γa,b={(x,t), x∈∂Ω,a<t<b}. Πa=Πa,∞,Γa=Γa,∞. The functions f(x,t,u),∂f(x,t,u)/∂u are continuous with respect to u uniformly in Π¯0×{u:|u|≤M} at any M<∞,f(x,t,0)≡0,∂f/∂u|u=0≡0. Besides, the function f is measurable by all arguments and does not decrease with respect to u. Let's assume the fulfillment of the following Dirichlet boundary condition:
(2.2)u=0,x∈∂Ω,
with the initial conditions
(2.3)u∣t=0=φ(x),ut′∣t=0=ψ(x),
in some domain Π0,a, where  φ(x) and ψ(x) are smooth functions. Let's assume that ω(x) is measurable, nonnegative function satisfying the conditions ω∈L1,loc(Ω) and for any r>0 and some fixed θ>1(2.4)∫Br‍ω-1/(θ-1)dx<∞,esssupx∈Brω(x)≤C1rn(θ-1)(∫Br‍ω-1/(θ-1)dx)1-θ,
where Br={x∈Ω;|x|<r}. From condition (2.4) it follows that
(2.5)esssupx∈Ωr≤c1r-n∫Br‍ωdx
and that ω∈Aθ, that is,
(2.6)∫Br‍ωdx[∫Br‍ω-1/(θ-1)dx]θ-1≤crnθ.
Condition (2.6) is called Makhenkhoupt condition (see [8]). Besides, as in [8] we will assume that ω∈Dμ,μ<1+p/n, that is,
(2.7)ω(Bs)ω(Bh)≤c1(sh)nμ.Let's introduce the weight Sobolev space(2.8)∥u∥Wp,ω1(Ω)=(∫Ω‍ω(x)(|u|p+|∇u|p)dx)1/p,
for any s≥h>0, where ω(BS)=∫BS‍ω(x)dx with the finite norm Wp,ω1(Ω).As the generalized solution of problem (2.1)–(2.3) in Πa,b we will define the function  u(x,t)∈Wp,ω1,1(Πa,b)  such that
(2.9)∫Πa,b‍utηtdxdt+∫Ω‍[ut(x,b)η(x,b)-ut(x,a)η(x,a)]dx+∫Πa,b‍∑i,j=1n‍ω(x)|∂u∂xi|p-2∂u∂xi∂η∂xjdxdt=∫Πa,b‍f(x,t,u)η(x,t)dxdt,
where η(x,t)∈Wp,ω1,1(Πa,b),η|Γa,b=0,0<a<b.We will study the conditions at some generalized solution fromΠ0(2.10)limt→T-0maxΩ|u(x,t)|=+∞,
at some T=const.Atf(x,t,u)=u|u|p-1,p>1 and at linear main part there are many works, such as [1, 2] on investigation properties of solutions having (2.10). We will show that if φ(x) and ψ(x) are adequately big, then it holds (2.10). For small φ(x) and ψ(x) if limt→+∞u(x,t)=0, then |u|<ce-at,a=const>0, it doesn't depend on u. Let's construct the sufficient condition on u, at which any solution of problem (2.1)–(2.3) at f(x,t,u), φ(x)≥0,ψ(x)≥0,φ(x)≠0,ψ(x)≠0 has “blowup” (without limitation of smallness on φ(x) and ψ(x)).Let's formulate some auxiliary results from [9, 10], and let's determine the harmonic operator Lp(2.11)Lpu=div(|∇u|p-2∇u),p>1.Lemma 2.1 (see [9]).
There exists the positive eigenvalue of spectral problem for the operatorLp to which corresponds the positive in Ω eigenfunction.Lemma 2.2 (see [10]).
Letu,ϑ∈Wp1(Ω),u≤ϑ on ∂Ω and for any nonnegative η∈Wop1(Ω)(2.12)∫Ω‍Lp(u)ηxidx≤∫Ω‍Lp(ϑ)ηxidx,
with η≥0. Then on whole domain Ω,u≤ϑ.


## 3. Main Results

Letu0(x)>0 be an eigenfunction of spectral problem for the operator Lp corresponding to λ=λ1>0,∫Ω‍ω(x)u0(x)dx=1. Denote g(t)=∫Ω‍ω(x)u0(x)u(x,t)dx. Let's assume the fulfillment of the conditions:
(3.1)I=∫Ω‍ω(x)(|∂u∂xi|p-2∂u∂xi-|∂u0∂xi|p-2∂u0∂xi)∂(u0ω)∂xidx≥0(3.2)(g′(0))2+λ1g2(0)-2c1σ+1gσ+1(0)>0.Theorem 3.1.
Letf(x,t,u)≥α0|u|σ-1u at  (x,t)∈Π0,u≥0,σ>1,α0=const>0. There exists k=const such that if, u(x,0)≥0,ut′(x,0)≥0,∫Ω‍u(x,0)u0dx≥k,∫Ω‍ut′(x,0)u0dx≥k or condition (3.2) is fulfilled. Then
(3.3)limt→TmaxΩ(ω(x)·u0(x)u(x,t))=∞,
where T=const>0.Proof.
Let's assume the opposite. Thenu(x,t) is a solution of (2.1) in Π0 and condition (2.2) is fulfilled on Γ0. By the virtue of Lemma 2.2 in Π0,u(x,t)>0.  Substitute in (2.9) η=ɛ-1u0(x)ω(x),b=a+ɛ,a>0,ɛ>0, where u0(x)>0 in Ω is an eigenfunction of spectral problem for the operator Lp, corresponding to eigenvalue λ1>0. Such eigenvalue exists by virtue of Lemma 2.1. As a result, we will obtain
(3.4)ɛ-1[∫Ωω(x)ut(x,a+ɛ)u0(x)dx-∫Ωω(x)ut(x,a)u0(x)dx]+ɛ-1∫Πa,a+ɛω(x)∑i,j=1n‍|∂u∂xi|p-2∂u∂xi∂η∂xjdxdt=ɛ-1∫Πa,a+ɛu0ωf(x,t,u)dxdt.
Let’s formulate some transformations. After some simple manipulation we get(3.5)ɛ-1∫Πa,a+ɛ‍ω(x)|∂u0∂xi|p-2∂u0∂xi∂η∂xjdxdt.
Using condition (3.1), tending ɛ to zero at all t>0, we obtain
(3.6)∂∂t∫Ω‍u0(x)ω(x)ut(x,t)dx≥-λ1∫Ω‍u02ω2(x)dx+∫Ω‍u0ω|u|σdx.
Here denoting
(3.7)g(t)=∫Ω‍u0(x)ω(x)u(x,t)dx,
we have
(3.8)g′′(t)≥-λ1∫Ω‍u02ω2(x)dx+a0∫Ω‍u0ω|u|σdx.
Let's estimate first integral on the right-hand side of (3.8). Using the Cauchy inequality with   ɛ>0 we get
(3.9)∫Ω‍ωuu0dx≤ɛ2∫Ω‍u2dx+12ɛ∫Ω‍ω2u02dx.
Hence
(3.10)∫Ω‍ωuu0dx-ɛ2∫Ω‍u2dx<12ɛ∫Ω‍ω2u02dx,
then
(3.11)λ1∫Ω‍ω2u02dx>-λ1∫Ω‍ωuu0+λ1ɛ2∫Ω‍u2dx>-λ1∫Ω‍ωuu0dx.
So, from (3.8) we have
(3.12)g′′(t)≥-λ1∫Ω‍ωu02udx+a0∫Ω‍u0ωuσdx.
By virtue of Holder inequality we obtain
(3.13)(∫Ω‍uu0ωdx)σ≤[(∫Ω‍uσu0ωdx)1/σ(∫Ω‍ωu0dx)(σ-1)/σ]σ≤c1∫Ω‍uσu0ωdx.
Thus, we have
(3.14)g′′(t)≥-λ1g(t)+c1gσ(t),c=const>0.
Let's multiply (3.14) by g', then
(3.15)12((g′)2)′≥c1gσg′-λ1gg′=c1gσg′-12λ1(g2)′.
Hence
(3.16)((g′)2)′≥c1gσg′-λ1(g2)′.
Let's integrate by t from 0 to t, then from (3.16) we have
(3.17)∫0t‍((g')2)′dt≥c1∫0t‍gσg′dt-λ1∫0t‍(g2)′dt,(g′(t))2-(g′(0))2≥2c1σ+1[gσ+1(t)-gσ+1(0)]-λ1[g2(t)-g2(0)],
or
(3.18)(g′(t))2≥2c1σ+1gσ+1(t)-λ1g2(t)-2c1σ+1gσ+1(0)+λ1g2(0)-g2(0).
Denote
(3.19)A=(g′(0))2+λ1g2(0)-2c1σ+1gσ+1(0).
If the condition (3.2) is fulfilled, that is,
(3.20)(∫Ω‍ω(x)u0(x)ψ(x)dx)2+λ1(∫Ω‍ω(x)u0(x)φ(x)dx)2-2c1σ+1(∫Ω‍ω(x)u0(x)φ(x)dx)στ+1>0
and if g(0)>c2=(λ1/2c1/(σ+1))2/(σ+1), then limt→T-0g(t)=+∞. Hence we have
(3.21)limt→T-0maxΩ(ω(x)u0(x)u(x,t))=∞.Remark 3.2.
This theorem generalizes the corresponding result [7].
That is why, (2.1) is not a solution in Π0, satisfying the boundary condition (2.2), if u(x,0)≥0,ut′(x,0)≥0 is not smaller.
Now we will show that at small initial functions the solution of problem (2.1)–(2.3) exists in small in Π0.Theorem 3.3.
Let one assume that|f(x,t,u)|≤(c3+c4tm)|u|σ,σ>1. There exists δ>0 such that if |ψ(x)|<δ,|φ(x)|<δ, then the solution of problem (2.1)–(2.3) exists in  small in Π0 and |u(x,t)|≤c5e-αt,α=const=>0 does not depend on u.Proof.
LetΩ¯⊂BR, where BR={x:|x|≤R}. Let ϑ>0 in BR be an eigenfunction, corresponding to positive eigenvalue λ1 of the boundary-value problem
(3.22)Lpu+λ1u=0,x∈Ω,u=0,x∈∂Ω.
Let's consider the function V(x,t)=ɛe-λ1t/2ϑ(x). We have
(3.23)Vtt-LpV-f(x,t,V)=ɛ(14λ12+c2)e-λ1t/2ϑ(x)-(c3+c4tm)ɛσe-λ1t/2ϑσ≥0,(x,t)∈Π0,V>0,(x,t)∈Γ0,
if ɛ>0 is sufficiently small. Inequality (3.23) is understood in weak sense (see [11]). From (3.23) and Lemma 2.2, it follows that |u|≤V≤c5e-λ1t/2. Let us determine the class of functions |φ(x)|≤δ,|ψ(x)|≤δ=ɛ·minΩ-ϑ(x) consisting from g(x,t) continuous in Π¯-∞,+∞ equal to zero at t≤T and such that |g(x,t)|≤Ke-ht is a subset at Banach space of continuous functions in Π¯-∞,+∞ with the norm |g|=supΠ-∞,+∞|geht|.
Let(3.24)θ(t)∈C∞(R1),θ(t)≡1,ift≤T,θ(t)=0,ift>T+1.
Let's determine the operator H on K putting Hg=θ(t)z,g∈K, where z is a solution of linearizing problem. By virtue of obtained estimation above, H maps K to K. This follows from the obtained estimation and theorem on the solution of the hyperbolic problems in Π-a,a at the small [11].
From Lerey-Shaudeer theorem, it follows that the operatorH has a fixed point z. This shows the existence of solution in the small. The theorem is proved.Note that the sufficient condition at which any nonnegative solution of problem (2.1)–(2.3) has “blow-up” is
(3.25)limt→T-0maxΩ¯(ω(x)u0(x)u(x,t))=+∞,
where T=const>0.Theorem 3.4.
Letf(x,t,u)≥cσeλ1σtuσ at (x,t)∈Π0,u≥0,σ=const>1, and λ1 be positive eigenvalue of problem (3.16) in Ω which corresponds to the positive in Ω eigenfunction. If u(x,0)≥0,ut′(x,0)≥0,u(x,0)≠0, where ut′(x,0)≠0, u(x,t) is a solution of problem (2.1)–(2.3) then it holds (3.25).Proof.
Analogously, as it is constructed in inequality (3.14), we will obtain
(3.26)g′′(t)≥-λ1g+Cgeλ1σgσ(t).
Letg(t)=ψ(t)e-λ1t. From (3.25) it follows that ψ′′≥c8ψτ. Hence ψ(t)→+∞  at  t→T-0. Thus it tends g(t) to k+∞ and t→T-0. Consequently, maxΩ¯(ω(x)u(x,t)) also tends to infinity.Corollary 3.5.
Letf(x,t,u)≥c8eλ1σuσ at (x,t)∈Π0,u≥0,σ>1. Then there is no positive in Π0 solution of the problem (2.1)–(2.3).
Let one consider the following equation:(3.27)∂2u∂t2=∑i,j=1n‍∂∂xi(ω(x)uσ∂u∂xj)+f(x,t,u),
in bounded domain Ω∈Rn,n≥2, and let u0(x)>0,∫Ω‍u0(x)dx=1, then one will have eigenfunction of the problem Δu+λu=0 in Ω,u=0 on ∂Ω, and λ1>0 the corresponding eigenvalue.Theorem 3.6.
Letf(x,t,u)≥α0|u|σ+1 at α0=const≥0 at (x,t)∈Π0,u≥0,σ>1. There exists  k=const>0  such that if u(x,0)≥0,ut′(x,0)≥0,∫Ω‍u(x,0)u0dx≥k,∫Ω‍ut′(x,0)u0dx≥k or condition (3.2) is fulfilled, then
(3.28)limt→TmaxΩ(ω(x)u0(x)u(x,t))=∞,
where T=const>0.Proof.
Let's assume a contrary. Thenu(x,t) is a solution of (3.27) in Π0 and condition (2.2) is fulfilled on Γ0. Substituting in corresponding integral identity Γ0  η=ɛ-1u0(x)ω(x),b=a+ɛ,ɛ>0 to get
(3.29)ɛ-1[∫Ω‍ω(x)ut(x,a+ɛ)u0(x)dx-∫Ω‍ω(x)ut(x,a)u0(x)dx]+ɛ-1∫Πa,a+ɛ‍∑i,j=1n‍∂∂xi(ω(x)uσ∂u∂xj)u0(x)ω(x)dx=ɛ-1∫Πa,a+ɛ‍u0ωf(x,t,u)dxdt,
where u0(x)>0. Tending ɛ to zero, we'll obtain
(3.30)∂2∂t2g(t)≥∫Ω‍u0(x)ω(x)∂∂xi(ωuσ∂u∂xj)dx+∫Ω‍u0ωuσ+1dx,
where g(t)=∫Ω‍ω(x)u0(x)u(x,t)dx. Let’s consider first the integral on the right-hand side in (3.30). Let φ(x)>0,ψ(x)>0 in Ω and suppφ(x)⊂Ω,supψ(x)⊂Ω. Denote the surface of degeneration of (3.27), that is, the boundary of the solution supu(t,x)≡θ(t). Since u(t,x)≡0 on ∂θ(t) and Ω/θ(t) from the Green's formula, we get
(3.31)∫Ω‍ωu0∂∂xi(ω4σ∂u∂xj)dx≡1σ+1∫θ(t)‍ωu0Δuσ+1dx≡1σ+1[∫θ(t)‍Δu0(0)ω(x)σ+1udx+∫∂θ(t)‍u0ω∂uσ+1∂ntds-∫∂θ(t)‍∂(u0ω)∂ntuσ+1dx],
where ∂/∂nt is a derivative on direction of external norms to ∂θ(t). Since uσ+1=0 is on ∂θ(t) and by virtue of continuity of flow ∂uσ+1/∂ut≡∇uσ+1·nt=0 at x∈∂θ(t). Therefore, the last two integrals in (3.31) are equal to zero. Then we will obtain
(3.32)∫Ω‍ωu0(x)∂∂xi(ωuσ∂u∂xj)dx=1σ+1∫Ω‍uσ+1Δωu0(x)dx.
Using Δu0(x)ω(x)=-λ1u0(x)ω(x) we will get
(3.33)∂2g(t)∂t2≥(1-λ1σ+1)∫Ω‍uσ+1u0(x)ω(x)dx,t>0.
Using the results of paper [12] we will obtain the required result.



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