Global and blow-up solutions for nonlinear parabolic problems with a gradient term under Robin boundary conditions

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Abstract

In this paper, we study the global and blow-up solutions of the following nonlinear parabolic problems with a gradient term under Robin boundary conditions:

where $D⊂ R N$ ($N≥2$) is a bounded domain with smooth boundary ∂D. By constructing auxiliary functions and using maximum principles, the sufficient conditions for the existence of a global solution, an upper estimate of the global solution, the sufficient conditions for the existence of a blow-up solution, an upper bound for ‘blow-up time’, and an upper estimate of ‘blow-up rate’ are specified under some appropriate assumptions on the functions f, g, b and initial value $u 0$.

MSC:35K55, 35B05, 35K57.

1 Introduction

In this paper, we study the global and blow-up solutions of the following nonlinear parabolic problems with a gradient term under Robin boundary conditions:

(1.1)

where $q:= | ∇ u | 2$, $D⊂ R N$ ($N≥2$) is a bounded domain with smooth boundary ∂D, $∂/∂n$ represents the outward normal derivative on ∂D, γ is a positive constant, $u 0$ 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 $b(s)$ is a $C 3 ( R + )$ function, $b ′ (s)>0$ for any $s∈ R +$, $g(s)$ is a positive $C 2 ( R + )$ function, $f(x,s,d,t)$ is a nonnegative $C 1 ( D ¯ × R + × R + ¯ × R + )$ function, and $u 0 (x)$ is a positive $C 2 ( D ¯ )$ function. Under the above assumptions, the classical theory  of parabolic equation assures that there exists a unique classical solution $u(x,t)$ with some $T>0$ for problem (1.1) and the solution is positive over $D ¯ ×[0,T)$. Moreover, the regularity theorem  implies $u(x,t)∈ C 3 (D×(0,T))∩ C 2 ( D ¯ ×[0,T))$.

Many papers have studied the global and blow-up solutions of parabolic problems with a gradient term (see, for instance, ). Some authors have discussed the global and blow-up solutions of parabolic problems under Robin boundary conditions and have got a lot of meaningful results (see  and the references cited therein). Some special cases of problem (1.1) have been treated already. Zhang  dealt with the following problem:

where $D⊂ R N$ ($N≥2$) is a bounded domain with smooth boundary ∂D. By constructing auxiliary functions and using maximum principles, the sufficient conditions characterized by functions f, g and $u 0$ were given for the existence of a blow-up solution. Zhang  investigated the following problem:

where $D⊂ R N$ ($N≥2$) is a bounded domain with smooth boundary ∂D. By constructing some auxiliary functions and using maximum principles, the sufficient conditions were obtained there for the existence of global and blow-up solutions. Meanwhile, the upper estimate of a global solution, the upper bound of ‘blow-up time’ and the upper estimate of ‘blow-up rate’ were also given. Ding  considered the following problem:

where $D⊂ R N$ ($N≥2$) 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 obtained for the existence of global and blow-up solutions. For the blow-up solution, an upper and a lower bound on blow-up time were also given.

In this paper, we study problem (1.1). Since the function $f(x,u,q,t)$ contains a gradient term $q= | ∇ u | 2$, it seems that the methods of  are not applicable for problem (1.1). In this paper, by constructing completely different auxiliary functions with those in  and technically using maximum principles, we obtain some existence theorems of a global solution, an upper estimate of the global solution, the existence theorems of a blow-up solution, an upper bound of ‘blow-up time’, and an upper estimates of ‘blow-up rate’. Our results extend and supplement those obtained .

We proceed as follows. In Section 2 we study the global solution of (1.1). Section 3 is devoted to the blow-up solution of (1.1). A few examples are presented in Section 4 to illustrate the applications of the abstract results.

2 Global solution

The main result for the global solution is the following theorem.

Theorem 2.1 Let u be a solution of problem (1.1). Assume that the following conditions (i)-(iv) are satisfied:

1. (i)

for any $s∈ R +$,

$( s b ′ ( s ) ) ′ ≥ 0 , s b ′ ( s ) − ( s b ′ ( s ) ) ′ ≤ 0 , ( g ( s ) b ′ ( s ) ) ′ ≤ 0 , [ 1 g ( s ) ( g ( s ) b ′ ( s ) ) ′ + 1 b ′ ( s ) ] ′ + 1 g ( g ( s ) b ′ ( s ) ) ′ + 1 b ′ ( s ) ≤ 0 ;$
(2.1)
2. (ii)

for any $(x,s,d,t)∈D× R + × R + ¯ × R +$,

$f t ( x , s , d , t ) ≤ 0 , f d ( x , s , d , t ) [ ( 1 b ′ ( s ) ) ′ + 1 b ′ ( s ) ] ≤ 0 , ( f ( x , s , d , t ) b ′ ( s ) g ( s ) ) s − f ( x , s , d , t ) b ′ ( s ) g ( s ) ≤ 0 ;$
(2.2)
3. (iii)
$∫ m 0 + ∞ b ′ ( s ) e s ds=+∞, m 0 := min D ¯ u 0 (x);$
(2.3)
4. (iv)
$α:= max D ¯ ∇ ⋅ ( g ( u 0 ) ∇ u 0 ) + f ( x , u 0 , q 0 , 0 ) e u 0 >0, q 0 := | ∇ u 0 | 2 .$
(2.4)

Then the solution u to problem (1.1) must be a global solution and

$u(x,t)≤ H − 1 ( α t + H ( u 0 ( x , t ) ) ) ,(x,t)∈ D ¯ × R + ¯ ,$
(2.5)

where

$H(z):= ∫ m 0 z b ′ ( s ) e s ds,z≥ m 0 ,$
(2.6)

and $H − 1$ is the inverse function of H.

Proof Consider the auxiliary function

$P(x,t):= b ′ (u) u t −α e u .$
(2.7)

Now we have

$∇P= b ″ u t ∇u+ b ′ ∇ u t −α e u ∇u,$
(2.8)
$ΔP= b ‴ u t | ∇ u | 2 +2 b ″ ∇u⋅∇ u t + b ″ u t Δu+ b ′ Δ u t −α e u | ∇ u | 2 −α e u Δu,$
(2.9)

and

$P t = b ″ ( u t ) 2 + b ′ ( u t ) t − α e u u t = b ″ ( u t ) 2 + b ′ ( g b ′ Δ u + g ′ b ′ | ∇ u | 2 + f b ′ ) t − α e u u t = b ″ ( u t ) 2 + ( g ′ − b ″ g b ′ ) u t Δ u + g Δ u t + ( g ″ − b ″ g ′ b ′ ) u t | ∇ u | 2 + ( 2 g ′ + 2 f q ) ∇ u ⋅ ∇ u t + ( f u − b ″ f b ′ − α e u ) u t + f t .$
(2.10)

It follows from (2.9) and (2.10) that

$g b ′ Δ P − P t = ( b ‴ g b ′ + b ″ g ′ b ′ − g ″ ) u t | ∇ u | 2 + ( 2 b ″ g b ′ − 2 g ′ − 2 f q ) ∇ u ⋅ ∇ u t + ( 2 b ″ g b ′ − g ′ ) u t Δ u − α g b ′ e u | ∇ u | 2 − α g b ′ e u Δ u − b ″ ( u t ) 2 + ( b ″ f b ′ − f u + α e u ) u t − f t .$
(2.11)

By (1.1), we have

$Δu= b ′ g u t − g ′ g | ∇ u | 2 − f g .$
(2.12)

Substitute (2.12) into (2.11), to get

$g b ′ Δ P − P t = ( b ‴ g b ′ − b ″ g ′ b ′ − g ″ + ( g ′ ) 2 g ) u t | ∇ u | 2 + ( 2 b ″ g b ′ − 2 g ′ − 2 f q ) ∇ u ⋅ ∇ u t − ( b ′ ) 2 g ( g b ′ ) ′ ( u t ) 2 + ( f g ′ g − b ″ f b ′ − f u ) u t + ( α g ′ b ′ e u − α g b ′ e u ) | ∇ u | 2 + α f b ′ e u − f t .$
(2.13)

With (2.8), we have

$∇ u t = 1 b ′ ∇P− b ″ b ′ u t ∇u+α e u b ′ ∇u.$
(2.14)

Next, we substitute (2.14) into (2.13) to obtain

$g b ′ Δ P + [ 2 ( g b ′ ) ′ + 2 f q b ′ ] ∇ u ⋅ ∇ P − P t = ( b ‴ g b ′ + b ″ g ′ b ′ − g ″ + ( g ′ ) 2 g − 2 ( b ″ ) 2 g ( b ′ ) 2 + 2 b ″ f q b ′ ) u t | ∇ u | 2 + ( 2 α b ″ g ( b ′ ) 2 e u − α g ′ b ′ e u − α g b ′ e u − 2 α f q b ′ e u ) | ∇ u | 2 − ( b ′ ) 2 g ( g b ′ ) ′ ( u t ) 2 + ( f g ′ g − b ″ f b − f u ) u t + α f b ′ e u − f t .$
(2.15)

In view of (2.7), we have

$u t = 1 b ′ P+α e u b ′ .$
(2.16)

Substituting (2.16) into (2.15), we get

$g b ′ Δ P + [ 2 ( g b ′ ) ′ + 2 f q b ′ ] ∇ u ⋅ ∇ P + { [ g ( 1 g ( g b ′ ) ′ ) ′ + 2 f q ( 1 b ′ ) ′ ] | ∇ u | 2 + g ( b ′ ) 2 ( f b ′ g ) u } P − P t = − α e u { g [ ( 1 g ( g b ′ ) ′ + 1 b ′ ) ′ + 1 g ( g b ′ ) ′ + 1 b ′ ] + 2 f q [ ( 1 b ′ ) ′ + 1 b ′ ] } | ∇ u | 2 − ( b ′ ) 2 g ( g b ′ ) ′ ( u t ) 2 − α g e u ( b ′ ) 2 [ ( f b ′ g ) u − f b ′ g ] − f t .$
(2.17)

The assumptions (2.1) and (2.2) guarantee that the right-hand side of (2.17) is nonnegative, i.e.,

(2.18)

By applying the maximum principle , it follows from (2.18) that P can attain its nonnegative maximum only for $D ¯ ×{0}$ or $∂D×(0,T)$. For $D ¯ ×{0}$, by (2.4), we have

$max D ¯ P ( x , 0 ) = max D ¯ { b ′ ( u 0 ) ( u 0 ) t − α e u 0 } = max D ¯ { ∇ ⋅ ( g ( u 0 ) ∇ u 0 ) + f ( x , u 0 , q 0 , 0 ) − α e u 0 } = max D ¯ { e u 0 [ ∇ ⋅ ( g ( u 0 ) ∇ u 0 ) + f ( x , u 0 , q 0 , 0 ) e u 0 − α ] } = 0 .$

We claim that P cannot take a positive maximum at any point $(x,t)∈∂D×(0,T)$. In fact, suppose that P takes a positive maximum at a point $( x 0 , t 0 )∈∂D×(0,T)$, then

$P( x 0 , t 0 )>0and ∂ P ∂ n | ( x 0 , t 0 ) >0.$
(2.19)

With (1.1) and (2.16), we have

(2.20)

Next, by using the fact that $( s b ′ ( s ) ) ′ ≥0$, $s b ′ (s)− ( s b ′ ( s ) ) ′ ≤0$ for any $s∈ R +$, it follows from (2.20) that

$∂ P ∂ n | ( x 0 , t 0 ) ≤0,$

which contradicts with inequality (2.19). Thus we know that the maximum of P in $D ¯ ×[0,T)$ is zero, i.e.,

and

$b ′ ( u ) e u u t ≤α.$
(2.21)

For each fixed $x∈ D ¯$, integration of (2.21) from 0 to t yields

$∫ 0 t b ′ ( u ) e u u t dt= ∫ u 0 ( x ) u ( x , t ) b ′ ( s ) e s ds≤αt,$
(2.22)

which implies that u must be a global solution. Actually, if that u blows up at finite time T, then

$lim t → T − u(x,t)=+∞.$

Passing to the limit as $t→ T −$ in (2.22) yields

$∫ u 0 ( x ) + ∞ b ′ ( s ) e s ds≤αT$

and

$∫ m 0 + ∞ b ′ ( s ) e s ds= ∫ m 0 u 0 ( x ) b ′ ( s ) e s ds+ ∫ u 0 ( x ) + ∞ b ′ ( s ) e s ds≤ ∫ m 0 u 0 ( x ) b ′ ( s ) e s ds+αT<+∞,$

which contradicts with assumption (2.3). This shows that u is global. Moreover, it follows from (2.22) that

$∫ u 0 ( x ) u ( x , t ) b ′ ( s ) e s ds= ∫ m 0 u ( x , t ) b ′ ( s ) e s ds− ∫ m 0 u 0 ( x ) b ′ ( s ) e s ds=H ( u ( x , t ) ) −H ( u 0 ( x ) ) ≤αt.$

Since H is an increasing function, we have

$u(x,t)≤ H − 1 ( α t + H ( u 0 ( x ) ) ) .$

The proof is complete. □

3 Blow-up solution

The following theorem is the main result for the blow-up solution.

Theorem 3.1 Let u be a solution of problem (1.1). Assume that the following conditions (i)-(iv) are fulfilled:

1. (i)

for any $s∈ R +$,

$( s b ′ ( s ) ) ′ ≥ 0 , s b ′ ( s ) − ( s b ′ ( s ) ) ′ ≥ 0 , ( g ( s ) b ′ ( s ) ) ′ ≥ 0 , [ 1 g ( s ) ( g ( s ) b ′ ( s ) ) ′ + 1 b ′ ( s ) ] ′ + 1 g ( g ( s ) b ′ ( s ) ) ′ + 1 b ′ ( s ) ≥ 0 ;$
(3.1)
2. (ii)

for any $(x,s,d,t)∈D× R + × R + ¯ × R +$,

$f t ( x , s , d , t ) ≥ 0 , f d ( x , s , d , t ) [ ( 1 b ′ ( s ) ) ′ + 1 b ′ ( s ) ] ≥ 0 , ( f ( x , s , d , t ) b ′ ( s ) g ( s ) ) s − f ( x , s , d , t ) b ′ ( s ) g ( s ) ≥ 0 ;$
(3.2)
3. (iii)
$∫ M 0 + ∞ b ′ ( s ) e s ds<+∞, M 0 := max D ¯ u 0 (x);$
(3.3)
4. (iv)
$β:= min D ¯ ∇ ⋅ ( g ( u 0 ) ∇ u 0 ) + f ( x , u 0 , q 0 , 0 ) e u 0 >0, q 0 := | ∇ u 0 | 2 .$
(3.4)

Then the solution u of problem (1.1) must blow up in finite time T, and

$T≤ 1 β ∫ M 0 + ∞ b ′ ( s ) e s ds,$
(3.5)
$u(x,t)≤ G − 1 ( β ( T − t ) ) ,(x,t)∈ D ¯ ×[0,T),$
(3.6)

where

$G(z):= ∫ z + ∞ b ′ ( s ) e s ds,z>0,$
(3.7)

and $G − 1$ is the inverse function of G.

Proof Construct the following auxiliary function:

$Q(x,t):= b ′ (u) u t −β e u .$
(3.8)

Replacing P and α with Q and β in (2.17), respectively, we get

$g b ′ Δ Q + [ 2 ( g b ′ ) ′ + 2 f q b ′ ] ∇ u ⋅ ∇ Q + { [ g ( 1 g ( g b ′ ) ′ ) ′ + 2 f q ( 1 b ′ ) ′ ] | ∇ u | 2 + g ( b ′ ) 2 ( f b ′ g ) u } Q − Q t = − β e u { g [ ( 1 g ( g b ′ ) ′ + 1 b ′ ) ′ + 1 g ( g b ′ ) ′ + 1 b ′ ] + 2 f q [ ( 1 b ′ ) ′ + 1 b ′ ] } | ∇ u | 2 − ( b ′ ) 2 g ( g b ′ ) ′ ( u t ) 2 − β g e u ( b ′ ) 2 [ ( f b ′ g ) u − f b ′ g ] − f t .$
(3.9)

Assumptions (3.1) and (3.2) imply that the right-hand side in equality (3.9) is nonpositive, i.e.,

(3.10)

With (3.4), we have

$min D ¯ Q ( x , 0 ) = min D ¯ { b ′ ( u 0 ) ( u 0 ) t − β e u 0 } = min D ¯ { ∇ ⋅ ( g ( u 0 ) ∇ u 0 ) + f ( x , u 0 , q 0 , 0 ) − β e u 0 } = min D ¯ { e u 0 [ ∇ ⋅ ( g ( u 0 ) ∇ u 0 ) + f ( x , u 0 , q 0 , 0 ) e u 0 − β ] } = 0 .$
(3.11)

Substituting P and α with Q and β in (2.20), respectively, we have

(3.12)

Combining (3.10)-(3.12) with the fact that $( s b ′ ( s ) ) ′ ≥0$, $s b ′ (s)− ( s b ′ ( s ) ) ′ ≥0$ for any $s∈ R +$, and applying the maximum principles again, it follows that the minimum of Q in $D ¯ ×[0,T)$ is zero. Thus

and

$b ′ ( u ) e u u t ≥β.$
(3.13)

At the point $x ∗ ∈ D ¯$, where $u 0 ( x ∗ )= M 0$, integrate (3.13) over $[0,t]$ to get

$∫ 0 t b ′ ( u ) e u u t dt= ∫ M 0 u ( x ∗ , t ) b ′ ( s ) e s ds≥βt,$
(3.14)

which implies that u must blow up in finite time. Actually, if u is a global solution of (1.1), then for any $t>0$, (3.14) shows

$∫ M 0 + ∞ b ′ ( s ) e s ds≥ ∫ M 0 u ( x ∗ , t ) b ′ ( s ) e s ds≥βt.$
(3.15)

Letting $t→+∞$ in (3.15), we have

$∫ M 0 + ∞ b ′ ( s ) e s ds=+∞,$

which contradicts with assumption (3.3). This shows that u must blow up in finite time $t=T$. Furthermore, letting $t→T$ in (3.14), we get

$T≤ 1 β ∫ M 0 + ∞ b ′ ( s ) e s ds.$

By integrating inequality (3.13) over $[t,s]$ ($0), for each fixed x, we obtain

$G ( u ( x , t ) ) ≥ G ( u ( x , t ) ) − G ( u ( x , s ) ) = ∫ u ( x , t ) + ∞ b ′ ( s ) e s d s − ∫ u ( x , s ) + ∞ b ′ ( s ) e s d s = ∫ u ( x , t ) u ( x , s ) b ′ ( s ) e s d s = ∫ t s b ′ ( u ) e u u t d t ≥ β ( s − t ) .$

Hence, by letting $s→T$, we have

$G ( u ( x , t ) ) ≥β(T−t).$

Since G is a decreasing function, we obtain

$u(x,t)≤ G − 1 ( β ( T − t ) ) .$

The proof is complete. □

4 Applications

When $b(u)≡u$ and $f(x,u,q,t)≡f(u)$, the results stated in Theorem 3.1 are valid. When $g(u)≡1$ and $f(x,u,q,t)≡f(u)$ or $f(x,u,q,t)≡f(u)$, the conclusions of Theorems 2.1 and 3.1 still hold true. In this sense, our results extend and supplement the results of .

In what follows, we present several examples to demonstrate the applications of the abstract results.

Example 4.1 Let u be a solution of the following problem:

where $q= | ∇ u | 2$, $D={x=( x 1 , x 2 , x 3 )∣ | x | 2 <1}$ is the unit ball of $R 3$. The above problem can be transformed into the following problem:

Now

$b ( u ) = u e u , g ( u ) = ( 1 + u ) e u , f ( x , u , q , t ) = ( e − u + e q ) ( e − t + | x | 2 ) , u 0 ( x ) = 2 − | x | 2 , γ = 2 .$

In order to determine the constant α, we assume

$s:= | x | 2 ,$

then $0≤s≤1$ and

$α = max D ¯ ∇ ⋅ ( g ( u 0 ) ∇ u 0 ) + f ( x , u 0 , q 0 , 0 ) e u 0 = max D ¯ { 32 | x | 2 − 4 | x | 4 − 18 + ( 1 + | x | 2 ) [ exp ( − 4 + 2 | x | 2 ) + exp ( − 2 + 5 | x | 2 ) ] } = max 0 ≤ s ≤ 1 { 32 s − 4 s 2 − 18 + ( 1 + s ) [ exp ( − 4 + 2 s ) + exp ( − 2 + 5 s ) ] } = 50.4417 .$

It is easy to check that (2.1)-(2.3) hold. By Theorem 2.1, u must be a global solution, and

$u ( x , t ) ≤ H − 1 ( α t + H ( u 0 ( x ) ) ) = − 1 + 50.4417 t + ( 1 + u 0 ( x ) ) 2 = − 1 + 50.4417 t + ( 3 − | x | 2 ) 2 .$

Example 4.2 Let u be a solution of the following problem:

where $q= | ∇ u | 2$, $D={x=( x 1 , x 2 , x 3 )∣ | x | 2 <1}$ is the unit ball of $R 3$. The above problem may be turned into the following problem:

Now we have

$b ( u ) = u + ln u , g ( u ) = 1 + 1 u , f ( x , u , q , t ) = ( e u − e − q ) ( 6 + t | x | 2 ) , u 0 ( x ) = 2 − | x | 2 , γ = 2 .$

By setting

$s:= | x | 2 ,$

we have $0≤s≤1$ and

$β = min D ¯ ∇ ⋅ ( g ( u 0 ) ∇ u 0 ) + f ( x , u 0 , q 0 , 0 ) e u 0 = min D ¯ { − 6 | x | 4 + 26 | x | 2 − 36 ( 2 − | x | 2 ) 2 exp ( 2 − | x | 2 ) + 6 [ 1 − exp ( − 3 | x | 2 − 2 ) ] } = min 0 ≤ s ≤ 1 { − 6 s 2 + 26 s − 36 ( 2 − s ) 2 exp ( 2 − s ) + 6 [ 1 − exp ( − 3 s − 2 ) ] } = 0.0735 .$

Again it is easy to check that (3.1)-(3.3) hold. By Theorem 3.1, u must blow up in finite time T, and

$T ≤ 1 β ∫ M 0 + ∞ b ′ ( s ) e s d s = 1 0.0735 ∫ 2 + ∞ ( 1 + 1 s ) 1 e s d s = 2.5066 , u ( x , t ) ≤ G − 1 ( β ( T − t ) ) = G − 1 ( 0.0735 ( T − t ) ) ,$

where

$G(z)= ∫ z + ∞ b ′ ( s ) e s ds= ∫ z + ∞ ( 1 + 1 s ) 1 e s ds,z≥0,$

and $G − 1$ is the inverse function of G.

Remark 4.1 We can see from Example 4.1 that when the equation has a gradient term with exponential increase, the functions g and b increase exponentially to ensure that the solution of (1.1) blows up. It follows from Example 4.2 that when the equation has a gradient term with exponential decay, the appropriate assumptions on the functions g and b can guarantee the solution of (1.1) to be global.

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Acknowledgements

This work was supported by the National Natural Science Foundation of China (Nos. 61074048 and 61174082) and the Research Project Supported by Shanxi Scholarship Council of China (Nos. 2011-011 and 2012-011).

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Correspondence to Juntang Ding.

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The author declares that he has no competing interests.

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All results belong to Juntang Ding.

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