# On theBlow-UpSet for Non-Newtonian Equation with a Nonlinear Boundary Condition

Authors: Zhilei Liang
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

We identify the blow-up set of solutions to the problemut=(|ux|p-2ux)x,  x>0, t∈(0,T), -|ux|p-2ux(0,t)=up-1(0,t),  t∈(0,T), and u(x,0)=u0(x)≥0,  x>0, where p>2. We obtain that the blow up set B(u) satisfies [0,p(p-1)/(p-2))⊂B(u)⊂[0,p(p-1)/(p-2)]. The proof is based on the analysis of the asymptotic behavior of self-similar representation and on the comparison methods.


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## Body

## 1. Introduction

Consider a one-dimensional process of diffusion in a medium that occupies the half space{x:x≥0}; that is,  (1.1)ut=(|ux|p-2ux)x,x>0,t∈(0,T),-|ux|p-2ux(0,t)=uq(0,t),t∈(0,T),u(x,0)=u0(x)≥0,x>0,
where p>2,q>0 and u0(x) is an appropriately smooth function with some compatibility conditions. Problem (1.1) describes the non-Newtonian fluid with a power dependence of the tangential stress on the velocity of the displacement under nonlinear condition. It has many applications and has been intensively studied; see [1–3] and the references cited therein. For the local in time existence, we refer to [4]. Also it is known that (1.1) has no classical solution in general due to the possible degeneration at ux=0. So we usually understand the weak solution defined in the following sense.Definition 1.1.
A nonnegative functionu∈C([0,+∞)×[0,t)) with ux∈L∞([0,+∞)×[0,t)) is said to be a weak solution of (1.1), if the integral identity
(1.2)∫0t∫0+∞(uϕt-|ux|p-2uxϕx)dxds+∫0+∞u0(x)ϕ(x,0)dx+∫0tuq(0,s)ϕ(0,s)ds=0
is fulfilled for all ϕ∈C0∞([0,+∞)×[0,t)).An interesting phenomenon is that, due to the boundary effect, the solution of (1.1) may exist for t∈[0,T) and becomes unbounded as t→T for some T<∞. Namely, the solutionoccurs blow-upphenomenon. In this connection, Galaktionov and Levine proved in [5] that the solutions are global in time when 0<q≤2(p-1)/p but occur blow-up for the range 2(p-1)/p<q≤2(p-1), while for q>2(p-1) blow-up happens or not depending on the size of the initial data. The main concern in this work is on the set of points at which solutions becomes unbounded, that is, the blow-up set, which is defined as (1.3)B(u)={x:∃xn→x,tn→T-,suchthatu(xn,tn)→∞asn→∞}.
A problem which has attracted a lot of attention in the literature is the identification of possible blow-up sets. Numerical analysis hints that the blow-up set should be a single point (single point blow-up) when q>p-1, a proper subset of the spatial domain (regional blow-up) when q=p-1, and the whole half line (global blow-up) when 0<q<p-1. In fact, based on the Gilding and Herrero's work in [6], Quirós and Rossi considered the porous medium type equation  (1.4)ut=(um)xx,x>0,t∈(0,T),-(um)x(0,t)=uq(0,t),t∈(0,T),u(x,0)=u0(x)≥0,x>0.
They proved in [1]: the blow-up set B(u)={0} if q>m but B(u)=[0,+∞) if q<m, however, the blow-up set is regional in case of q=m, namely, 0<B(u)<+∞. Afterwards, Cortázar et al. given a detailed description on the regional blow-up set. They proved in [7] that if q=m, then the blow-up set satisfies [0,2m/m-1)⊂B(u)⊂[0,2m/m-1].In the light of previous works, we discuss the blow-up set of solutions for thep-Laplacian equation (1.1). In the current paper, we identify the set of blow-up points in case of q=p-1. So, in the following we consider  (1.5)ut=(|ux|p-2ux)x,x>0,t∈(0,T),-|ux|p-2ux(0,t)=up-1(0,t),t∈(0,T),u(x,0)=u0(x)≥0,x>0,
where p>2. We take u0(x) to be a C1, nonincreasing and compactly supported function with some compatibility conditions. The below theorem is our main result.Theorem 1.2.
Assume thatu0(x)∈C1([0,+∞)) is a nonnegative, nonincreasing and compactly supported function, then all the nontrivial solutions u(x,t) of problem (1.5) occur blow-up; moreover, the blow-up set B(u) satisfies
(1.6)[0,p(p-1)p-2)⊂B(u)⊂[0,p(p-1)p-2].Remark 1.3.
The nonincreasing assumption onu0 makes the proof much simpler (see also [7]).Remark 1.4.
One could expect that if the solutions of (1.1) occur blow-up, then the blow-up set B(u)=[0,+∞) if 0<q<p-1, but B(u)={0} if q>p-1.


## 2. Proof of Theorem1.2

In order to study the solutionu of (1.5) near the blow-up time T, as in [8] we introduce the rescaled function  (2.1)v(x,τ)=(T-t)1/(p-2)u(x,t),
where τ=-ln((T-t)/T)∈[0,+∞). Hence, we obtain the below equations in terms of v by substituting (2.1) into (1.5),  (2.2)∂v∂τ=(|vx|p-2vx)x-1p-2v,x>0,τ>0,-|vx|p-2vx(0,τ)=vp-1(0,τ),τ>0,v(x,0)=T1/(p-2)u0(x)≥0,x>0.Ifv is τ independent, then u is called a self-similar solution. Let w(x)=(T-t)1/(p-2)u(x,t), we have  (2.3)(|wx|p-2wx)x-1p-2w=0,x>0,-|wx|p-2wx(0)=wp-1(0).
A direct integration shows that the non-negative solution of (2.3) is w≡0 or  (2.4)w(x)=(p-2p)(p-1)/(p-2)(12(p-1))1/(p-2)(p(p-1)p-2-x)+p/(p-2),
where a+=max{0,a}.Theorem1.2 is a direct consequence of the following two propositions.Proposition 2.1.
Letv(x,τ) be a solution of (2.2), then there exists a function w(x), such that the limit holds uniformly on [0,+∞),(2.5)v(·,τ)→w(·),asτ→+∞.
Moreover, w(x) has the explicit form (2.4) and solves problem (2.3).Proposition2.1 and the formula (2.1) generate  (2.6)[0,p(p-1)p-2)⊂B(u).Proposition 2.2.
Suppose that the constantsa,b satisfy p(p-1)/(p-2)<a<b<+∞, then there exists a large time point τ0 such that for all τ>τ0(2.7)v(x,τ)≤Ce-τ/(p-2),x∈[a,b],
where constant C depends on a,b and u0.Proposition2.2, along with (2.1), shows that u(x,t) is uniformly bounded for all t∈(0,T) if x>p(p-1)/p-2. This claims  (2.8)B(u)⊂[0,p(p-1)p-2].Hence, Theorem1.2 immediately follows from (2.6) and (2.8). So the remaining task in this paper is to prove the validity of Propositions 2.1 and 2.2. This will be discussed in Section 3.


## 3. Proof of Propositions2.1 and 2.2

In this section we prove Propositions2.1 and 2.2. The argument contains several lemmas.Lemma 3.1.
All the nontrivial solutions of (1.5) occur blow-up.The proof is available in [5], (see also [3]).Lemma 3.2.
If the initial functionu0 is nonincreasing, then so does the solution  u(x,t) for all existent times, that is,  ux≤0,a.e.(x,t)∈(0,∞)×(0,T).Proof.
Consider the regularized problem(3.1)∂u∂t=((ux2+ϵ)(p-2)/2ux)x,x>0,t∈(0,T),-(ux2+ϵ)(p-2)/2ux(0,t)=up-1(0,t),t∈(0,T),u(x,0)=u0ϵ(x)≥0,x>0,
where ϵ>0,u0ϵ(x)∈C0∞([0,+∞)) is nonincreasing and converges to u0(x) uniformly as ϵ→0+. By uniqueness, the solution of (1.5) should be the limit function of that of (3.1). Setting z(x,t)=(ux2+ϵ)(p-2)/2ux and then differentiating the equation in t to obtain
(3.2)zt=(ux2+ϵ)(p-2)/2uxt+(p-2)(ux2+ϵ)(p-2)/2-1ux2uxt=[(ux2+ϵ)(p-2)/2+(p-2)(ux2+ϵ)(p-2)/2-1ux2]zxx.
Clearly, (3.2) is a linear parabolic equation with respect to z. On account of the boundary conditions z(0,t)=-u(0,t)p-1≤0 and z(x,0)=(u0ϵx2+ϵ)(p-2)/2u0ϵx≤0, we deduce z(x,t)≤0 via comparison theorem. The proof is completed.Remark 3.3.
Using the maximum principle for problem (3.1), we obtain
(3.3)maxx≥0,0≤t≤T{-(ux2+ϵ)(p-2)/2ux}=max{up-1(0,t),-(u0ϵx2+ϵ)(p-2)/2u0ϵx}.
By sending ϵ→0+, it follows that
(3.4)maxx≥0,0≤t≤T{-|ux|p-2ux}=max{up-1(0,t),-|u0x|p-2u0x}.Lemma 3.4.
There exists a constantC=C(p,u0) such that
(3.5)∫0tup-1(0,s)ds≤C(T-t)-1/(p-2),∀t∈[0,T).Proof.
Integrating (1.5)1, the first equation of (1.5), over [0,+∞)×[0,t) yields
(3.6)∫0∞u(x,t)dx-∫0∞u0dx=∫0tup-1(0,s)ds.
Now multiplying (1.5)1 by x and integrating the resulting expression once more, we have
(3.7)∫0∞xu(x,t)dx-∫0∞xu0dx=∫0t∫0∞x(|ux|p-2ux)xdxds=∫0t∫0∞(x|ux|p-2ux)xdxds-∫0t∫0∞|ux|p-2uxdxds=-∫0t∫0∞|ux|p-2uxdxds.
Thanks to Remark 3.3, we estimate
(3.8)-∫0t∫0∞|ux|p-2uxdxds≤∫0t|ux(0,s)|p-2∫0∞(-ux)dxds+∫0t∫0∞|u0x|p-1=∫0t|ux(0,s)|p-2u(0,s)ds+C(T,u0x)=∫0tup-1(0,s)ds+C(T,u0x),
the last sign comes from the boundary condition. Thus, (3.7) becomes
(3.9)∫0∞xu(x,t)dx-∫0∞xu0dx≤∫0tup-1(0,s)ds+C(T,u0x).
Multiplying (3.6) by a constant K>1 and then subtracting (3.9) produces
(3.10)∫0∞(K-x)u(x,t)dx-∫0∞(K-x)u0dx≥(K-1)∫0tup-1(0,s)ds-C(T,u0x).
If we choose K>1 so large such that suppu0⊂[0,K], then
(3.11)∫0∞(K-x)u(x,t)dx-∫0∞(K-x)u0dx≤∫0K(K-x)u(x,t)dx≤K2u(0,t).
Inequalities (3.10) and (3.11) lead to
(3.12)K2u(0,t)≥(K-1)∫0tup-1(0,s)ds-C(T,u0x).
Putting F(t)=∫0tup-1(0,s)ds and integrating above inequality over (t,T) brings
(3.13)F(t)≤C(T-t)-1/(p-2)+C(T,u0x)≤C1(T-t)-1/(p-2),(tclosetoT).
Because u is bounded when time t is away from the blow-up time T, (3.13) and thus (3.5) is valid for all t∈[0,T).The next lemma proves thatu(x,t) is localized. Namely, the support of u(·,t) is uniformly bounded for all t∈([0,T).Lemma 3.5.
There exists a constantM depending on  p and  u0 such that if  x>M, then
(3.14)u(x,t)≡0,∀t∈[0,T).Proof.
Let us introduce the function(3.15)q(x,t)=D(T-t)-1/(p-2)(L-x)+p/(p-2),
where D satisfies where Dp-2=(1/2(p-1))(p-2/p)p-1 and L is an arbitrary positive constant. It is easy to check that q(x,t) satisfies (1.5)1. By differentiating (3.5), we have u(0,t)≤C(T-t)-1/p-2. This, together with the monotonicity of u, deduces
(3.16)u(x,t)≤C(T-t)-1/(p-2),∀x≥0,t∈(0,T).
Because the initial support is bounded, we choose a suitably large point x0<+∞ to guarantee suppu0⊂[0,x0]. Setting ũ(x,t)=u(x+x0,t), it has
(3.17)ũ(x,0)=u(x+x0,0)≡0≤q(x,0),∀x>0.
By virtue of (3.16), there exists some large constant L0 such that for all L>L0, it holds
(3.18)ũ(0,t)=u(x0,t)≤C(T-t)-1/(p-2)≤DLp/(p-1)(T-t)-1/(p-2)=q(0,t),∀t∈(0,T).
So comparison theorem concludes
(3.19)ũ(x,t)≤q(x,t),∀x≥0,t∈(0,T).
Hence, for every constant L satisfying L>L0, (3.19) ensures suppũ(·,t)⊂[0,L]. Thus suppu(·,t)⊂[0,x0+L]. Denoted by M=x0+L, it follows that for all x>M,(3.20)u(x,t)≡0,∀t∈(0,T).
This completes the proof.Corollary 3.6.
Lemma3.5 and (2.1) claim that v(x,τ) is localized as well. Moreover, (2.1) and (3.16) guarantee the boundedness of v(x,τ) for all x≥0 and τ≥0. By a similar argument as that in Lemma 3.2, one can prove that |vx|p-2vx also is bounded.In what follows, we analyze the large time behavior of the solutionv(x,τ) of problem (2.2).Multiplying (2.2) by v and then integrating by parts, we deduce that  (3.21)∫0∞∫τ1τ2vτ2dxds+J[v](τ2)≤J[v](τ1),a.e.,0≤τ1<τ2<∞,
where J[v](τ) is the Liapunov functional and  (3.22)J[v](τ)=1p∫0∞|vx|pdx-1p∫0∞vp(0,τ)+12(p-2)∫0∞v2(x,τ)dx.
Based on the estimates  (3.23)v,vx∈L∞([0,+∞)×[0,+∞)),vτ∈L2([0,+∞)×[0,+∞)),
the ω-limit set(3.24)ω(v0)={w(x):∃τn→∞,s.t.v(·,τn)→w(x)inL2}
consists of solutions of the problem (2.3). Indeed, using the estimates (3.23) in passing to the limit in (2.2), we have that, given a monotone sequence τn→∞, v(·,τn+s)→w(x;s)inL2, where w(x;s) depending on s solves (2.3) in a weak sense, and w(x;0)∈ω(v0). It follows from (3.23) that uniformly in s∈[0,1](3.25)∥v(·,τn+s)-v(·,τn)∥≤∫τnτn+s∥vt∥L22dτ→0(τn→∞).
This means that the limit function w does not rely on s and is a weak solution for the (2.3). Finally, the independence of the choice of the sequence τn→∞ follows from the nonincreasing of J[v](τ) in time. Furthermore, the convergence is uniform in x due to the boundness of vx. In conclusion we arrive to the following lemma.Lemma 3.7.
The limit holds asτ→∞(3.26)v(x,τ)→w(x),uniformlyforx≥0,
where w(x) satisfies the problem (2.3).To finish Proposition2.1, it remains to confirm that w(x) is not null and thus to be of the form (2.4). The proof will be given at the end of the paper. Now let us turn to Proposition 2.2.Consider the initial-boundary-value problem(3.27)hτ=(|hx|p-2hx)x-1p-2h,x∈(a-δ,b+δ),τ>0,h(a-δ,τ)=h(b+δ,τ)=ϵ,τ>0,h(x,0)=ϵ,x∈(a-δ,b+δ),
where a,b,ϵ,δ are positive constants and ϵ,δ are suitably small. We have the following.Lemma 3.8.
LetDδp/(p-2)=ϵ suitably small with D that satisfies Dp-2=(1/2(p-1))((p-1)/p)p-1. Then there exists a constant C>0 such that for all τ≥0,
(3.28)h(x,τ)≤Ce-τ/(p-2),∀x∈[a,b].Proof.
It is easy to check that the function of the form(3.29)j(x)={D(a-x)p/(p-2),x∈[a-δ,a-δ2),0,x∈[a-δ2,b+δ2],D(x-b)p/(p-2),x∈(b+δ2,b+δ].
solves
(3.30)0=(|jx|p-2jx)x-1p-2j,x∈(a-δ,b+δ),j(a-δ)=j(b+δ)=Dδp/(p-2).
Moreover, it has j(x)≤Dδp/(p-2) since Dδp/(p-2) is a super solution to problem (3.30). Choosing Dδp/(p-2)=ϵ and subtracting (3.30) from (3.27), we obtain
(3.31)(h-j)τ=(|hx|p-2hx-|jx|p-2jx)x-1p-2(h-j),x∈(a-δ,b+δ),τ>0,h(a-δ,τ)-j(a-δ)=h(b+δ,τ)-j(b+δ)=0,τ>0,h(x,0)-j(x)≥0,x∈(a-δ,b+δ).
Integrating (3.31) over (a-δ,b+δ)×(0,τ) gives rise to
(3.32)∫a-δb+δ(h-j)(x,τ)dx=∫a-δb+δ(h-j)(x,0)dx+∫0τ(|hx|p-2hx-|jx|p-2jx)(b+δ,s)ds-∫0τ(|hx|p-2hx-|jx|p-2jx)(a-δ,s)ds-1p-2∫0τ∫a-δb+δ(h-j)(x,s)dxds.
Noting that (h-j)(x,τ) is symmetric with respect to (a+b)/2 and nonincreasing on [a+b/2,b+δ], thus we have for all τ>0,(3.33)sign{(|hx|p-2hx-|jx|p-2jx)(b+δ,τ)}=sign{(hx-jx)(b+δ,τ)}≤0,
where sign is the sign function. Then,
(3.34)∫0τ(|hx|p-2hx-|jx|p-2jx)(b+δ,s)ds≤0.
Similarly,
(3.35)∫0τ(|hx|p-2hx-|jx|p-2jx)(a-δ,s)ds≥0.
Therefore, (3.32) becomes
(3.36)∫a-δb+δ(h-j)(x,τ)dx≤∫a-δb+δ(h-j)(x,0)dx-1p-2∫0τ∫a-δb+δ(h-j)(x,s)dxds.
Applying the Gronwall inequality yields
(3.37)∫a-δb+δ(h-j)(x,τ)dx≤Ce-(1/(p-2))τ.
On the other hand, since h is symmetric and nondecreasing on [(a+b)/2,b+δ] with x-variable, for all x∈[a,b] it has
(3.38)h(x,τ)≤1δ∫a-δ/2b+δ/2h(y,τ)dy=1δ∫a-δ/2b+δ/2(h-j)(y,τ)dy≤1δ∫a-δb+δ(h-j)(x,τ)dx≤C1e-(1/(p-2))τ,
where we have used the fact suppj(·)⊂[a-δ,a-δ/2]∪[b+δ/2,b+δ]. That is,
(3.39)h(x,τ)≤C1e-(1/(p-2))τ.
This completes Lemma 3.8.We are now in a position to prove Proposition2.2. From Lemma 3.7 we observe that v tends to the limit function w(x) which is none for all points x>p(p-1)/(p-2). Taking a>p(p-1)/(p-2) and a small constant δ>0 to satisfy a-δ>p(p-1)/(p-2). Then, for a given ϵ>0, the inequality v(x,τ)≤ϵ for all x≥a-δ holds provided τ>τ0, where τ0 is a large point which depends on ϵ. Using comparison theorem, one has
(3.40)v(x,τ+τ0)≤h(x,τ),forτ>τ0,x>a-δ.
The proof ends up with (3.40) and Lemma 3.8.Finally, We prove that the functionw(x) appeared in Lemma 3.7 takes the form (2.4). For this purpose, it is enough to illuminate that w(x)≡0 is impossible. We argue by contradiction. Assuming that w(x)≡0 for a moment. Then u(x,t) blows up only at the boundary point x=0. By this we can find a sequence tn→∞ as n→∞ such that u(0,tn)→∞asn→∞. Moreover, By the maximin principle (see Remark 3.3), one has
(3.41)maxx≥0{-|ux|p-2ux(x,tn)}=max{-|u0x|p-2u0x,up-1(0,tn)}=up-1(0,tn),
as long as n is chosen large enough. Recalling that u is nonincreasing with respect to x-variable, from (3.41) we conclude
(3.42)-ux(x,tn)≤u(0,tn),∀x≥0.
Integrating (3.42) gives rise to
(3.43)(1-x)u(0,tn)≤u(x,tn).
This shows that u occurs blow-up at least in the interval [0,1), which contradicts with the assumption that blow-up happens only at x=0.Now the proof of Theorem1.2 is completed.



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