# Blow-Up Solutions and Global Existence for Quasilinear Parabolic Problems with Robin Boundary Conditions

Authors: Juntang Ding
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We study the blow-up and global solutions for a class of quasilinear parabolic problems
with Robin boundary conditions. By constructing auxiliary functions and using maximum principles,
the sufficient conditions for the existence of blow-up solution, an upper bound for the “blow-up time,”
an upper estimate of the “blow-up rate,” the sufficient conditions for the existence of global solution,
and an upper estimate of the global solution are specified.


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## Body

## 1. Introduction

In this paper, we are going to investigate the blow-up and global solutions of the following quasilinear parabolic problem with Robin boundary conditions:(1)(g(u))t=∇·(a(u,t)b(x)∇u)+h(t)f(u)hhhhhhhhhhhhhhhhhhhhinD×(0,T),∂u∂n+γu=0on⁡∂D×(0,T),u(x,0)=u0(x)>0inD¯,
where D⊂ℝN (N≥2) is a bounded domain with smooth boundary ∂D, ∂/∂n represents the outward normal derivative on ∂D, γ is a positive constant, u0 is the initial value, T is the maximal existence time of u, and D¯ is the closure of D. Set ℝ+≔(0,+∞). We assume, throughout the paper, that g is a C3(ℝ+) function, g′(s)>0 for any s∈ℝ+, a is a positive C2(ℝ+×ℝ+¯) function, b is a positive C1(D¯) function, h is a positive C1(ℝ+¯) function, f is a positive C2(ℝ+) function, and u0(x) is a positive C2(D¯) function. Under the above assumptions, the classical parabolic equation theory [1] assures that there exists a unique classical solution u(x,t) with some T>0 for problem (1) and the solution is positive over D¯×[0,T). Moreover, by regularity theorem [2], u(x,t)∈C3(D×(0,T))∩C2(D¯×[0,T)).Many authors have studied the blow-up and global solutions of nonlinear parabolic problems (see, for instance, [3–14]). Some special cases of the problem (1) have been treated already. Enache [15] investigated the following problem:
(2)ut=∇·(a(u)∇u)+f(u)inD×(0,T),∂u∂n+γu=0on⁡∂D×(0,T),u(x,0)=h(x)≥0inD¯,
where D⊂ℝN (N≥2) is a bounded domain with smooth boundary ∂D. Some conditions on nonlinearities and the initial data were established to guarantee that u(x,t) is global existence or blows up at some finite T. In addition, an upper bound and a lower bound for T were derived. Zhang [16] dealt with the following problem:
(3)ut=∇·(a(u)b(x)∇u)+h(t)f(u)hhhhhhhhhhhhhhhinD×(0,T),∂u∂n+γu=0on⁡∂D×(0,T),u(x,0)=u0(x)>0inD¯,
where D⊂ℝN (N≥2) is a bounded domain with smooth boundary ∂D. By constructing auxiliary functions and using maximum principles, the sufficient conditions characterized by functions a,b,h,f, and u0 were given for the existence of blow-up solution. Ding [17] considered the following problem:
(4)(g(u))t=∇·(a(u)∇u)+f(u)hhhhhhhhhhhhinD×(0,T),∂u∂n+γu=0on⁡∂D×(0,T),u(x,0)=u0(x)>0inD¯,
where D⊂ℝN (N≥2) is a bounded domain with smooth boundary ∂D. By constructing some appropriate auxiliary functions and using a first-order differential inequality technique, the sufficient conditions were specified for the existence of blow-up and global solutions. For the blow-up solution, a lower bound on blow-up time is also obtained. Some authors also discussed blow-up phenomena for parabolic problems with Robin boundary conditions and obtained a lot of interesting results (see [18–24] and the references cited therein).As everyone knows, parabolic equation describes the process of heat conduction. Blow-up and global solutions for parabolic equations reflect the unsteady state and steady state of heat conduction process, respectively. In the problems (2) and (4), the heat conduction coefficient a(u) depends only on the temperature variable u. In the problem (3), the heat conduction coefficient a(u)b(x) depends on the temperature variable u and space variable x. However, in a lot of processes of heat conduction, heat conduction coefficient depends not only on the temperature variable u but also on the space variable x and the time variable t. Therefore, in this paper, we study the problem (1). It seems that the method of [15–17] is not applicable for the problem (1). In this paper, by constructing completely different auxiliary functions with those in [15–17] and technically using maximum principles, we obtain some existence theorems of blow-up solution, an upper bound of “blow-up time,” an upper estimates of “blow-up rate,” the existence theorems of global solution, and an upper estimate of the global solution. Our results extend and supplement those obtained in [15–17].We proceed as follows. In Section2, we study the blow-up solution of (1). Section 3 is devoted to the global solution of (1). A few examples are presented in Section 4 to illustrate the applications of the abstract results.


## 2.  Blow-Up Solution

The main results for the blow-up solution are Theorems1–3. For simplicity, we define the constant
(5)α≔min⁡D¯{1+∇·(a(u0,0)b(x)∇u0)h(0)f(u0)}.
In Theorems 1–3, the three cases 0<α<1,α=1, and α>1 are considered, respectively. In the first case, 0<α<1, we have the following conclusions.Theorem 1.
Letu be a solution of the problem (1). Suppose the following.(i)
Consider(6)0<α<1.(ii)
For(s,t)∈ℝ+×ℝ+,
(7)[1a(s,t)(a(s,t)f(s)g′(s))s]s≥0,(as(s,t)a(s,t))t≥0,(a(s,t)g′(s))s≤0,(h(t)a(s,t))t≥0.(iii)
Fors∈ℝ+,
(8)(sg′(s))′≥0,(f(s)sg′(s))′≥0.(iv)
Consider(9)∫M0+∞‍g′(s)f(s)ds<α∫0+∞‍h(t)dt,M0≔max⁡D¯u0(x).Then, the solution u of the problem (1) must blow up in a finite time T, and
(10)T≤P-1(1α∫M0+∞‍g′(s)f(s)ds),u(x,t)≤H-1(α∫tT‍h(t)dt),
where
(11)P(z)≔∫0z‍h(t)dt,z>0,H(z)≔∫z+∞‍g′(s)f(s)ds,z>0,
and P-1 and H-1 are the inverse functions of P and H, respectively.Proof.
In order to discuss the blow-up solution by using maximum principles, we construct an auxiliary function(12)Q(x,t)≔g′(u)ut-αh(t)f(u),
from which we have
(13)∇Q=g′′ut∇u+g′∇ut-αhf′∇u,(14)ΔQ=g′′′|∇u|2ut+2g′′∇u·∇ut+g′′utΔu+g′Δut-αhf′′|∇u|2-αhf′Δu,(15)Qt=[g′(u)ut-αh(t)f(u)]t=[(g(u))t-h(t)f(u)+(1-α)h(t)f(u)]t=[∇·(a(u,t)b(x)∇u)+(1-α)h(t)f(u)]t=aubutΔu+atbΔu+abΔut+auub|∇u|2ut+autb|∇u|2+2aub∇u·∇ut+auut∇b·∇u+at∇b·∇u+a∇b·∇ut+(1-α)h′f+(1-α)hf′ut.
By (14) and (15), we have
(16)abg′ΔQ-Qt=(abg′′′g′-auub)|∇u|2ut+(2abg′′g′-2aub)∇u·∇ut+(abg′′g′-aub)utΔu-(αabhf′′g′+autb)|∇u|2-(αabhf′g′+atb)Δu-auut∇b·∇u-at∇b·∇u-a∇b·∇ut+(α-1)h′f+(α-1)hf′ut.
It follows from (1) that
(17)Δu=g′abut-aua|∇u|2-1b∇b·∇u-hfab.
Next, we substitute (17) into (16) to obtain
(18)abg′ΔQ-Qt=(abg′′′g′-auub-aubg′′g′+(au)2ba)|∇u|2ut+(2abg′′g′-2aub)∇u·∇ut+(g′′-aug′a)(ut)2-ag′′g′ut∇b·∇u+(auhfa-fg′′hg′-hf′-atg′a)ut+(αaubhf′g′-αabhf′′g′+auatba-autb)∇u|2+αahf′g′∇b·∇u-a∇b·∇ut+αh2ff′g′+athfa+(α-1)h′f.
With (13), it has
(19)∇ut=1g′∇Q-g′′g′ut∇u+αhf′g′∇u.
Substitute (19) into (18) to get
(20)abg′ΔQ+[2b(ag′)u∇u+ag′∇b]·∇Q-Qt=(abg′′′g′-auub+aubg′′g′+(au)2ba-2ab(g′′)2(g′)2)ut×|∇u|2+(2αabhf′g′′(g′)2-αaubhf′g′-αabhf′′g′+auatba-autb)×|∇u|2+(g′′-aug′a)(ut)2+(auhfa-fg′′hg′-hf′-atg′a)ut+αh2ff′g′+athfa+(α-1)h′f.
In view of (12), we have
(21)ut=1g′Q+αhfg′.
Substituting (21) into (20), we get
(22)abg′ΔQ+[2b(ag′)u∇u+ag′∇b]·∇Q+{ab[1a(ag′)u]u|∇u|2+1a(ag′)u×[Q+(2α-1)hf]+hf′g′+ata}Q-Qt=-abh{α[1a(afg′)u]u+1h(aua)t}|∇u|2-α(α-1)h2f2a(ag′)u+(α-1)af(ha)t.
The assumptions (6) and (7) imply that the right-hand side of (22) is nonpositive; that is,
(23)abg′ΔQ+[2b(ag′)u∇u+ag′∇b]·∇Q+{ab[1a(ag′)u]u|∇u|2+1a(ag′)u×[Q+(2α-1)hf]+hf′g′+ata}×Q-Qt≤0,inD×(0,T).
Applying the maximum principle [25], it follows from (23) that Q can attain its nonpositive minimum only for D¯×{0} or ∂D×(0,T). For D¯×{0}, (5) implies
(24)min⁡D¯Q(x,0)=min⁡D¯{g′(u0)(u0)t-αh(0)f(u0)}=min⁡D¯{(g(u0))t-h(0)f(u0)+(1-α)h(0)f(u0)}=min⁡D¯{∇·(a(u0,0)b(x)∇u0)+(1-α)h(0)f(u0)}=min⁡D¯{h(0)f(u0)[1+∇·(a(u0,0)b(x)∇u0)h(0)f(u0)-α]}=0.
We claim thatQ cannot take a negative minimum at any point (x,t)∈∂D×(0,T). Indeed, if Q take a negative minimum at point (x0,t0)∈∂D×(0,T), then
(25)Q(x0,t0)<0,∂Q∂n|(x0,t0)<0.
It follows from (1) and (21) that
(26)∂Q∂n=g′′ut∂u∂n+g′∂ut∂n-αhf′∂u∂n=-γg′′uut+g′(∂u∂n)t+γαhf′u=-γg′′uut+g′(-γu)t+γαhf′u=-γ(ug′)′ut+γαhf′u=-γ(ug′)′(1g′Q+αhfg′)+γαhf′u=-γ(ug′)′g′Q+γαu2g′h(fug′)on⁡∂D×(0,T).
Next, by using (8) and the fact Q(x0,t0)<0, it follows from (26) that
(27)∂Q∂n|(x0,t0)≥0,
which contradicts inequality (25). Thus, we know that the minimum of Q in D¯×[0,T) is zero. Thus,
(28)Q≥0inD¯×[0,T);
that is,
(29)g′(u)f(u)ut≥αh(t).
At the point x*∈D¯, where u0(x*)=M0, integrate (29) over [0,t] to get
(30)∫0t‍g′(u)f(u)utdt=∫M0u(x*,t)‍g′(s)f(s)ds≥α∫0t‍h(t)dt,
which shows that u must blow up in finite time. In fact, suppose u is a global solution of (1), then, for any t>0, it follows from (30) that
(31)∫M0+∞‍g′(s)f(s)ds>∫M0u(x*,t)‍g′(s)f(s)ds≥α∫0t‍h(t)dt.
Passing to the limit as t→+∞ in (31) yields
(32)∫M0+∞‍g′(s)f(s)ds≥α∫0+∞‍h(t)dt,
which contradicts assumption (9). This shows that u must blow up in a finite time t=T. Furthermore, letting t→T in (30), we have
(33)lim⁡t→T∫M0u(x*,t)‍g′(s)f(s)ds≥lim⁡t→Tα∫0t‍h(t)dt;
that is,
(34)1α∫M0+∞‍g′(s)f(s)ds≥∫0T‍h(t)dt=P(T),
which implies that
(35)T≤P-1(1α∫M0+∞‍g′(s)f(s)ds).
By integrating inequality (29) over [t,s](0<t<s<T), for each fixed x, one gets
(36)H(u(x,t))≥H(u(x,t))-H(u(x,s))=∫u(x,t)u(x,s)‍g′(s)f(s)ds≥α∫ts‍h(t)dt.
Hence, by letting s→T, we obtain
(37)H(u(x,t))≥α∫tT‍h(t)dt.
Since H is a decreasing function, we have
(38)u(x,t)≤H-1(α∫tT‍h(t)dt).
The proof is complete.In the second case,α=1, the following two assumptions (i)a and (ii)a can guarantee that inequality (23) holds.(i)a
Consider(39)α=1.(ii)a
For(s,t)∈ℝ+×ℝ+,
(40)[1a(s,t)(a(s,t)f(s)g′(s))s]s≥0,(as(s,t)a(s,t))t≥0.Hence, by repeating the proof of Theorem 1, we have the following results.Theorem 2.
Letu be a solution of the problem (1). Suppose that (i)a and (ii)a hold and assumptions (iii) and (iv) of Theorem 1 hold. Then, the conclusions of Theorem 1 are valid.In the third case,α>1, the following two assumptions (i)b and (ii)b imply that inequality (23) holds.(i)b Consider
(41)α>1.(ii)b For (s,t)∈ℝ+×ℝ+,
(42)[1a(s,t)(a(s,t)f(s)g′(s))s]s≥0,(as(s,t)a(s,t))t≥0,(a(s,t)g′(s))s≥0,(h(t)a(s,t))t≤0.Theorem 3.
Letu be a solution of the problem (1). Suppose that (i)b and (ii)b hold and assumptions (iii) and (iv) of Theorem 1 hold. Then, the results stated in Theorem 1 still hold.Remark 4.
When(43)∫0+∞‍h(t)dt=+∞,
(9) implies that
(44)∫M0+∞‍g′(s)f(s)ds<+∞,M0=max⁡D¯u0(x).
When
(45)∫0+∞‍h(t)dt<+∞,
(9) implies that
(46)∫M0+∞‍g′(s)f(s)ds<+∞,∫M0+∞‍g′(s)f(s)ds<α∫0+∞‍h(t)dt.


## 3.  Global Solution

We define the constant(47)β≔max⁡D¯{1+∇·(a(u0,0)b(x)∇u0)h(0)f(u0)}.
The following Theorems 5–7 are the main results for the global solution. In Theorems 5–7, we study the three cases 0<β<1,β=1, and β>1, respectively. In the first case, 0<β<1, we have the following results.Theorem 5.
Letu be a solution of the problem (1). Suppose the following.(i)
Consider(48)0<β<1.(ii)
For(s,t)∈ℝ+×ℝ+,
(49)[1a(s,t)(a(s,t)f(s)g′(s))s]s≤0,(as(s,t)a(s,t))t≤0,(a(s,t)g′(s))s≥0,(h(t)a(s,t))t≤0.(iii)
Fors∈ℝ+,
(50)(sg′(s))′≥0,(f(s)sg′(s))′≤0.(iv)
Consider(51)∫M0+∞‍g′(s)f(s)ds≥β∫0+∞‍h(t)dt,M0:=max⁡D¯u0(x).Then, the solution u to the problem (1) must be a global solution and
(52)u(x,t)≤F-1(β∫0t‍h(t)dt+F(u0(x))),hhhhhhhhhhhhhhhhhh(x,t)∈D¯×ℝ+¯,
where
(53)F(z)≔∫m0z‍g′(s)f(s)ds,z≥m0,m0:=min⁡D¯u0(x),
and F-1 is the inverse function of F.Proof.
In order to study the global solution by using maximum principles, we construct an auxiliary function(54)G(x,t)≔g′(u)ut-βh(t)f(u).
Substituting Q and α with G and β in (22), respectively, gives
(55)abg′ΔG+[2b(ag′)u∇u+ag′∇b]·∇G+{ab[1a(ag′)u]u|∇u|2+1a(ag′)u×[G+(2β-1)hf]+hf′g′+ata}G-Gt=-abh{β[1a(afg′)u]u+1h(aua)t}|∇u|2-β(β-1)h2f2a(ag′)u+(β-1)af(ha)t.
Assumptions (48) and (49) guarantee that the right side in equality (55) is nonnegative; that is,
(56)abg′ΔG+[2b(ag′)u∇u+ag′∇b]·∇G+{ab[1a(ag′)u]u|∇u|2+1a(ag′)u×[G+(2β-1)hf]+hf′g′+ata}G-Gt≥0aaaaaaaaaaaainD×(0,T).
It follows from (47) that
(57)max⁡D¯G(x,0)=max⁡D¯{g′(u0)(u0)t-βh(0)f(u0)}=max⁡D¯{(g(u0))t-h(0)f(u0)+(1-β)h(0)f(u0)}=max⁡D¯{∇·(a(u0,0)b(x)∇u0)+(1-β)h(0)f(u0)}=max⁡D¯{h(0)f(u0)[1+∇·(a(u0,0)b(x)∇u0)h(0)f(u0)-β]}=0.
Replacing Q and α with G and β in (26), respectively, we have
(58)∂G∂n=-γ(ug′)′g′G+γβu2g′h(fug′)′on⁡∂D×(0,T).
Combining (56)–(58) with (50) and applying the maximum principles again, it follows that the maximum of G in D¯×[0,T) is zero. Thus,
(59)G≤0inD¯×[0,T),(60)g′(u)f(u)ut≤βh(t).
For each fixed x∈D¯, integration of (60) from 0 to t yields
(61)∫0t‍g′(u)f(u)utdt=∫u0(x)u(x,t)‍g′(s)f(s)ds≤β∫0t‍h(t)dt,
which implies that u must be a global solution. Actually, if u blows up at a finite time T, then
(62)lim⁡t→T-u(x,t)=+∞.
Letting t→T- in (61), we have
(63)∫u0(x)+∞‍g′(s)f(s)ds≤β∫0T‍h(t)dt<β∫0+∞‍h(t)dt,∫M0+∞‍g′(s)f(s)ds=∫M0u0(x)‍g′(s)f(s)ds+∫u0(x)+∞‍g′(s)f(s)ds≤∫u0(x)+∞‍g′(s)f(s)ds<β∫0+∞‍h(t)dt,
which contradicts with assumption (51). This shows that u is global. Moreover, it follows from (61) that
(64)∫u0(x)u(x,t)‍g′(s)f(s)ds=∫m0u(x,t)‍g′(s)f(s)ds-∫m0u0(x)‍g′(s)f(s)ds=F(u(x,t))-F(u0(x))≤β∫0t‍h(t)dt.
Since F is an increasing function, we have
(65)u(x,t)≤F-1(β∫0t‍h(t)dt+F(u0(x))).
The proof is complete.In the second caseβ=1 and the third case β>1, we have the following results.Theorem 6.
Letu be a solution of the problem (1). Suppose that assumptions (i)c and (ii)c hold.(i)c Consider
(66)β=1.(ii)c For (s,t)∈ℝ+×ℝ+,
(67)[1a(s,t)(a(s,t)f(s)g′(s))s]s≤0,(as(s,t)a(s,t))t≤0.And assumptions (iii) and (iv) of Theorem 5 hold. Then, the results of Theorem 5 are valid.Theorem 7.
Letu be a solution of the problem (1). Suppose that assumptions (i)d and (ii)d hold.(i)d
Consider(68)β>1.(ii)d
For(s,t)∈ℝ+×ℝ+,
(69)[1a(s,t)(a(s,t)f(s)g′(s))s]s≤0,(as(s,t)a(s,t))t≤0,(a(s,t)g′(s))s≤0,(h(t)a(s,t))t≥0.And assumptions (iii) and (iv) of Theorem 5 hold. Then, the conclusions stated in Theorem 5 still hold.Remark 8.
When(70)∫0+∞‍h(t)dt=+∞,
(51) implies that
(71)∫M0+∞‍g′(s)f(s)ds=+∞,M0=max⁡D¯u0(x).
When
(72)∫0+∞‍h(t)dt<+∞,
(51) implies that
(73)∫M0+∞‍g′(s)f(s)ds=+∞
or
(74)∫M0+∞‍g′(s)f(s)ds<+∞,∫M0+∞‍g′(s)f(s)ds≥β∫0+∞‍h(t)dt.


## 4.  Applications

Wheng(u)≡u,a(u,t)≡a(u),b(x)≡1 and h(t)≡1 or g(u)≡u and a(u,t)≡a(u) or a(u,t)≡a(u),b(x)≡1 and h(t)≡1, the conclusions of Theorems 1–3 and 5–7 still hold true. In this sense, our results extend and supplement the results of [15–17].In what follows, we present several examples to demonstrate the applications of the abstract results.Example 9.
Letu be a solution of the following problem:
(75)(up)t=∇·(un∇u)+uqinD×(0,T),∂u∂n+γu=0on⁡∂D×(0,T),u(x,0)=u0(x)>0inD¯,
where D⊂ℝN(N≥2) is a bounded domain with smooth boundary ∂D, p>0,-∞<n<+∞,-∞<q<+∞. Here,
(76)g(u)=up,a(u,t)=un,b(x)=1,h(t)=1,f(u)=uq.
Assume that(77)α=min⁡D¯{1+∇·(u0n∇u0)u0q}>0
and one of the following three assumptions holds.(i)
In the case0<α<1,p-q<0≤n+1≤p or p-q≤n+1<0.(ii)
In the caseα=1,p-q<0≤n+1 or p-q≤n+1<0.(iii)
In the caseα>1,p-q<0<p≤n+1.It follows from Theorems1–3 that u blows up in a finite time T, and
(78)T≤P-1(1α∫M0+∞‍g′(s)f(s)ds)=pα(q-p)M0q-p,u(x,t)≤H-1(α∫0T‍h(t)dt)=(pα(q-p)(T-t))1/(q-p).
Assume that
(79)β=max⁡D¯{1+∇·(u0n∇u0)u0q}>0
and one of the following three assumptions holds.(i)
In the case0<β<1,0≤p-q≤p≤n+1 or p<p-q≤n+1.(ii)
In the caseβ=1,0≤p-q≤n+1.(iii)
In the caseβ>1,0≤p-q≤n+1≤p.By Theorems5–7, u must be a global solution and
(80)u(x,t)≤F-1(β∫0t‍h(t)dt+F(u0(x)))={[β(p-q)pt+(u0(x))p-q]1/(p-q),p-q>0,u0(x)eαt/p,p-q=0.Example 10.
Letu be a solution of the following problem:
(81)(ueu)t=∇·(eu+t(2-∑i=13‍xi2)∇u)+u(1+u)2eu+thhhhhhhhhhhhhhhhhhhhhhhhhhhhhinD×(0,T),∂u∂n+2u=0on⁡∂D×(0,T),u(x,0)=2-∑i=13‍xi2inD¯,
where D={x=(x1,x2,x3)∣∑i=13‍xi2<1} is the unit ball of ℝ3. Now, we have
(82)g(u)=ueu,a(u,t)=eu+t,b(x)=2-∑i=13‍xi2,h(t)=et,f(u)=u(1+u)2eu,u0(x)=2-∑i=13‍xi2,γ=2.In order to determine the constantα, we assume
(83)w≔∑i=13‍xi2.
Then, 0≤w≤1 and
(84)α=min⁡D¯{1+∇·(a(u0,0)b(x)∇u0)h(0)f(u0)}=min⁡D¯{1+-12+18∑i=13xi2-4(∑i=13xi2)2(2-∑i=13xi2)(3-∑i=13xi2)2}=min⁡0≤w≤1{1+-12+18w-4w2(2-w)(3-w)2}=13.
It is easy to check that (6)-(9) hold. By Theorem 1, u must blow up in a finite time T, and
(85)T≤P-1(1α∫M0+∞‍g′(s)f(s)ds)=ln⁡(13ln⁡32+1),u(x,t)≤H-1(α∫tT‍h(t)dt)=1e(1/3)(eT-et)-1.Example 11.
Letu be a solution of the following problem:
(86)(ueu)t=∇·(eu-t(2-∑i=13‍xi2)∇u)+(1+u)eu-thhhhhhhhhhhhhhhhhhhhhhhhhhhinD×(0,T),∂u∂n+2u=0on⁡∂D×(0,T),u(x,0)=2-∑i=13‍xi2inD¯,
where D={x=(x1,x2,x3)∣∑i=13‍xi2<1} is the unit ball of ℝ3. Now,
(87)g(u)=ueu,a(u,t)=eu-t,b(x)=2-∑i=13‍xi2,h(t)=e-t,f(u)=(1+u)eu,u0(x)=2-∑i=13‍xi2,γ=2.By setting(88)w≔∑i=13‍xi2,
we have 0≤w≤1 and
(89)β=max⁡D¯{1+∇·(a(u0,0)b(x)∇u0)h(0)f(u0)}=max⁡D¯{1+-12+18∑i=13xi2-4(∑i=13xi2)23-∑i=13xi2}=max⁡0≤w≤1{1+-12+18w-4w2(3-w)}=2.
It is easy to check that (68)-(69) and (50)-(51) hold. By Theorem 7, u must be a global solution, and
(90)u(x,t)≤F-1(β∫0t‍h(t)dt+F(u0(x)))=2(2-e-t)-∑i=13‍xi2.



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*Source: 324857-2014-03-30.xml*