# Global and Blow-Up Solutions for a Class of Nonlinear Parabolic Problems under Robin Boundary Condition

Authors: Lingling Zhang; Hui Wang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We discuss the global and blow-up solutions of the following nonlinear parabolic problems with a gradient term under Robin boundary conditions:(b(u))t=∇·(h(t)k(x)a(u)∇u)+f(x,u,|∇u|2,t), in D×(0,T), (∂u/∂n)+γu=0, on ∂D×(0,T), u(x,0)=u0(x)>0, in D¯, where D⊂RN(N≥2) is a bounded domain with smooth boundary ∂D. Under some appropriate assumption on the functions f, h, k, b, and a and initial value u0, we obtain 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.” Our approach depends heavily on the maximum principles.


---

## Body

## 1. Introduction

The study of global and blow-up solutions for nonlinear parabolic equations has received a lot of attention in the past several decades (see [1–4]). In most works, so far, a variety of approaches have been developed in dealing with different nonlinear parabolic problems, such as the existence of global solution, blow-up solution, an upper bound for “blow-up time,” an upper estimate of “blow-up rate,” or global solution. So far, some applications in physics, chemistry, and biology are relevant to blow-up phenomena which can be found in [5–11]. In this paper, we consider the global and blow-up solutions of the following nonlinear parabolic equation with Robin boundary condition:
(1)(b(u))t=∇·(h(t)k(x)a(u)∇u)+f(x,u,q,t),mmmmmmmmmiiiiiiimminD×(0,T),∂u∂n+γu=0,on⁡∂D×(0,T),u(x,0)=u0(x)>0,inD¯,
where q=|∇u|2, D⊂RN  (N≥2) is a bounded domain with smooth boundary ∂D, ∂u/∂n represents the outward normal derivative on ∂D, γ is 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 b(s) is a positive C3(R+) function, b′(s)>0 for any s∈R+, a(s) is a positive C2(R+) function, k(x) is a positive C1(D¯) function, h(t) is a positive C1(R+) function, f(x,s,d,t) is a nonnegative C1(D¯×R+×R+¯×R+) function, and u0(x) is a positive C2(D¯) function. Under these assumptions, the classical parabolic equation theory [12] ensures that there exists a unique classical solution u(x,t) with some T>0 for the problem (1), and the solution is positive over D¯×[0,T). Moreover, by the regularity theorem [13], u(x,t)∈C3(D×(0,T))∩C2(D¯×[0,T)).The problems of the global and blow-up solutions for nonlinear parabolic equations have been investigated extensively by many authors and have got a lot of meaningful results. Some special cases of problem (1) have been treated already. Ding [14] deals with the following problem:
(2)(b(u))t=∇·(a(u)∇u)+f(u),inD×(0,T),∂u∂n+γu=0,on⁡∂D×(0,T),u(x,0)=h(x)>0,inD¯,
where D is a bounded domain of RN  (N≥2) with smooth boundary ∂D. By constructing auxiliary functions and using a first-order differential inequality technique, Ding derives conditions on the data, which guarantee the existence of blow-up or global solution. The following problem is investigated by Enache in [15]:
(3)ut=∇·(a(u)∇u)+f(u),inD×(0,T),∂u∂n+γu=0,on⁡∂D×(0,T),u(x,0)=h(x)>0,inD¯,
where D is a bounded domain of RN  (N≥2) with smooth boundary ∂D. By constructing auxiliary functions and first-order differential inequality technique, Enache establishes some conditions on nonlinearities and the initial date to guarantee that u(x,t) exists for all times t>0 or blows up at some finite time T. Besides, the following problem is investigated by Zhang in [16]:
(4)(b(u))t=Δu+f(u),inD×(0,T),∂u∂n+γu=0,on⁡∂D×(0,T),u(x,0)=h(x)>0,inD¯,
where D is a bounded domain in RN  (N≥2) with smooth boundary. Under appropriate assumptions on the functions f, b, and h, Zhang obtains the conditions under which the solutions may exist globally or blow up in a finite time. Moreover, upper estimates of the “blow-up time,” blow-up rate, and global solutions are obtained also.In this paper, we obtain the existence theorem of global and blow-up solution by constructing completely different auxiliary functions and technically using maximum principles. As a result, the sufficient conditions for the existence of a global solution and an upper estimate of the global solution and 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 assumption on the functionsf, h, k, b, and a and initial value u0. Our results extend and supplement those obtained in [14–16].The content of this paper is organized as follows. In Section2, we study the existence of the global solution of (1). In Section 3, we investigate the blow-up solution of (1). In Section 4, we will give a few examples to explain our results.


## 2. Global Solution

Our main result for the global solution is the following Theorem1.Theorem 1.
Letu be a solution of (1). Suppose that the following conditions (i)–(iv) are satisfied.(i)
For anys∈R+,
(5)(sb′(s))′≥0,sb′(s)-(sb′(s))′≤0,(a(s)b′(s))′≤0,[1a(s)(a(s)b′(s))′+1b′(s)]′+1a(s)(a(s)b′(s))′+1b′(s)≤0.(ii)
For any(x,s,d,t)∈D×R+×R+¯×R+,
(6)(f(x,s,d,t)h(t))t≤0,fd(x,s,d,t)[(1b′(s))′+1b′(s)]≤0,(f(x,s,d,t)b′(s)a(s))s-f(x,s,d,t)b′(s)a(s)+h′(t)(b′(s))2a(s)h(t)≤0.(iii)
Consider the integration(7)∫m0+∞b′(s)esds=+∞,m0=min⁡D¯u0(x).(iv)
Consider(8)α=max⁡D¯{∇·(h(0)k(x)a(u0)∇u0)+f(x,u0,q0,0)eu0}>0,mmmmmmmmmmimmmmmmmmmmmq0=|∇u0|2.
Then the solutionu to problem (1) must be a global solution and
(9)u(x,t)≤H-1(αt+H(u0(x,t))),(x,t)∈D¯×R+¯,
where
(10)H(z)=∫m0z‍b′(s)esds,z≥m0,
and H-1 is the inverse function of H.Proof.
Consider the auxiliary function(11)P(x,t)=b′(u)ut-αeu.
Then, we have
(12)∇P=b′′ut∇u+b′∇ut-αeu∇u,(13)ΔP=b′′′ut|∇u|2+2b′′∇u·∇ut+b′′utΔu+b′Δut-αeu|∇u|2-αeuΔu.
By (1),
(14)(b(u))t=b′ut=∇·(h(t)k(x)a(u)∇u)+f=h(t)k(x)a(u)Δu+h(t)k(x)a′(u)|∇u|2+h(t)a(u)(∇k·∇u)+f.
We have
(15)ut=akhb′Δu+a′khb′|∇u|2+ahb′(∇k·∇u)+fb′,(ut)t=h′(akb′Δu+a′kb′|∇u|2+ab′(∇k·∇u))+h(akb′Δu+a′kb′|∇u|2+ab′(∇k·∇u))t+(fb′)t=(a′′b′-a′b′′(b′)2)hkut|∇u|2+2a′khb′(∇u·∇ut)+(a′b′-ab′′(b′)2)hut(∇k·∇u)+ahb′(∇k·∇ut)+(a′b′-ab′′(b′)2)hkutΔu+akhb′Δut+a′kh′b′|∇u|2+ah′b′(∇k·∇u)+akh′b′Δu+fuut+2fq(∇u·∇ut)+ftb′-fb′′ut(b′)2.
Then
(16)Pt=b′′(ut)2+b′(ut)t-αeuut=b′′(ut)2+(a′-ab′′b′)khutΔu+akhΔut+akh′Δu+(a′′-a′b′′b′)khut|∇u|2+a′kh′|∇u|2+(2a′kh+2fq)(∇u·∇ut)+(a′-ab′′b′)hut(∇k·∇u)+ah(∇k·∇ut)+ah′(∇k·∇u)+(fu-fb′′b′-αeu)ut+ft.
By (13) and (16), it follows that
(17)akhb′ΔP-Pt=(akhb′′′b′+a′khb′′b′-a′′kh)ut|∇u|2+(2akhb′′b′-2a′kh-2fq)(∇u·∇ut)+(2akhb′′b′-a′kh)utΔu-(akhb′αeu+a′h′k)|∇u|2-(akhb′αeu+akh′)Δu-b′′(ut)2+(fb′′b′+αeu-fu)ut+(ahb′′b′-a′h)ut(∇k·∇u)-ah(∇k·∇ut)-ah′(∇k·∇u)-ft.
By (14), we have
(18)Δu=b′akhut-a′a|∇u|2-1k(∇k·∇u)-fakh.
Substitute (18) into (17) to get
(19)akhb′ΔP-Pt=(akhb′′′b′+(a′)2kha-a′khb′′b′-a′′kh)ut|∇u|2+(2akhb′′b′-2a′kh-2fq)(∇u·∇ut)+(b′′-a′b′a)ut2+(a′fa-fb′′b′-fu-b′h′h)ut+(a′khb′αeu-akhb′αeu)|∇u|2-ahb′′b′ut(∇k·∇u)-ah(∇k·∇ut)+ahb′αeu(∇k·∇u)+fb′αeu+fh′h-ft.
By (12), we have
(20)∇ut=1b′∇P-b′′b′ut∇u+αeub′∇u.
Next, we substitute (20) into (19) to obtain
(21)akhb′ΔP-Pt=(2akhb′′(b′)2-2a′khb′-2fqb′)(∇u·∇P)-ahb′(∇k·∇P)+(akhb′′′b′-2akh(b′′)2(b′)2+a′khb′′b′-a′′khmmimm+(a′)2kha+2b′′fqb′)ut|∇u|2+(2akhb′′(b′)2αeu-a′khb′αeu-akhb′αeu-2fqb′αeu)|∇u|2+(b′′-a′b′a)ut2+(a′fa-fb′′b′-fu-b′h′h)ut+fb′αeu+fh′h-ft.
So we have
(22)akhb′ΔP+(2a′khb′+2fqb′-2akhb′′(b′)2)×(∇u·∇P)+ahb′(∇k·∇P)-Pt=(akhb′′′b′-2akh(b′′)2(b′)2+a′khb′′b′-a′′kh+(a′)2khammiim+2b′′fqb′)ut|∇u|2+(2akhb′′(b′)2αeu-a′khb′αeu-akhb′αeu-2fqb′αeu)|∇u|2+(b′′-a′b′a)ut2+(a′fa-fb′′b′-fu-b′h′h)ut+fb′αeu+fh′h-ft.
According to (11), we have
(23)ut=1b′P+αeub′.
Substituting (23) into (22), we have
(24)akhb′ΔP+[2kh(ab′)′+2fqb′](∇u·∇P)+ahb′(∇k·∇P)+[a(b′)2(fb′a)u+h′h]P+[akh(1a(ab′)′)′+2fq(1b′)′]|∇u|2P-Pt=(-αeu)(2fqb′-2b′′fq(b′)2)|∇u|2+(-αeu)kh(a′b′-ab′′(b′)2)|∇u|2+αeu(akhb′′′(b′)2-2akh(b′′)2(b′)3+a′khb′′(b′)2-a′′khb′mmmiiimm+(a′)2khab′)|∇u|2-αeuakh(1b′-b′′(b′)2)|∇u|2+(b′′-a′b′a)ut2+αeu(a′fab′-fb′′(b′)2-fub′+fb′-h′h)+h′hf-ft.
Namely,
(25)akhb′ΔP+[(2kh(ab′)′+2fqb′)∇u+ahb′∇k]·∇P+{a(b′)2(fb′a)u+h′hmiiiiiim+[akh(1a(ab′)′)′+2fq(1b′)′]|∇u|2a(b′)2(fb′a)u+h′h}P-Pt=-αeu{[(1a(ab′)′+1b′)′+1a(ab′)′+1b′]2fq[(1b′)′+1b′]+akhmmiimmmi×[(1a(ab′)′+1b′)′+1a(ab′)′+1b′]}|∇u|2-(b′)2a(ab′)′ut2-αeua(b′)2×[(fb′a)u-fb′a+h′(b′)2ah]-h(fh)t.
The assumptions (5) and (6) guarantee that the right-hand side of (25) is nonnegative; that is,
(26)akhb′ΔP+[(2kh(ab′)′+2fqb′)∇u+ahb′∇k]·∇P+{a(b′)2(fb′a)u+h′hmmmm+[akh(1a(ab′)′)′+2fq(1b′)′]|∇u|2a(b′)2(fb′a)u+h′h}P-Pt≥0,inD×(0,T).
By applying maximum principle (see [17]), it follows from (26) that P can attain its nonnegative maximum only for D¯×{0} or ∂D×(0,T).
ForD¯×{0}, by (8), we have
(27)max⁡D¯P(x,0)=max⁡D¯⁡{b′(u0)(u0)t-αeu0}=max⁡D¯{[∇·(h(0)k(x)a(u0)∇u0)mmmmm+f(x,u0,q0,0)]-αeu0}=max⁡D¯⁡{-α∇·(h(0)k(x)a(u0)∇u0)+f(x,u0,q0,0)eu0]eu0[∇·(h(0)k(x)a(u0)∇u0)+f(x,u0,q0,0)eu0mmmmmimm-α∇·(h(0)k(x)a(u0)∇u0)+f(x,u0,q0,0)eu0]}=0.
For∂D×(0,T), we claim that P cannot take a positive maximum at any point (x,t). In fact, suppose that P can take a positive maximum at one point (x0,t0)∈∂D×(0,T); then
(28)P(x0,t0)>0,∂P∂n|(x0,t0)>0.
Combine (1) and (11) with (23); we have
(29)∂P∂n=b′′ut∂u∂n+b′∂ut∂n-αeu∂u∂n=-γb′′uut+b′(∂u∂n)t+γαueu=-γb′′uut+b′(-γu)t+γαueu=-γ(ub′)′ut+γαueu=-γ(ub′)′(1b′P+α1b′eu)+γαueu=-γ(ub′)′b′P+γαeuub′-(ub′)′b′,mmiimmiimmmion⁡∂D×(0,T).
Next, by using a part condition of (5) (sb′(s))′≥0,  sb′(s)-(sb′(s))′≤0 for any s∈R+, we can obtain
(30)∂P∂n|(x0,t0)≤0,
which contradicts with inequality (28). Thus, we know that the maximum of P in D¯×[0,T) is zero; that is,
(31)P≤0,inD¯×[0,T).
With (11), we know
(32)b′(u)euut≤α.
For each fixed x∈D¯, we integrate (32) from 0 to t:
(33)∫0t‍b′(u)euutdt=∫u0(x)u(x,t)‍b′(s)esds≤αt,
which implies that u must be a global solution of (1). In fact, suppose that u blows up at finite time T; then
(34)lim⁡t→T-u(x,t)=+∞.
Passing to the limit as t→T- in (33) yields
(35)∫u0(x)+∞‍b′(s)esds≤αT,∫m0+∞‍b′(s)esds=∫m0u0(x)‍b′(s)esds+∫u0(x)+∞‍b′(s)esds≤∫m0u0(x)‍b′(s)esds+αT<+∞,
which contradicts with the condition (iii). This shows that u is global solution. Moreover, it follows from (33) that
(36)∫u0(x)u(x,t)‍b′(s)esds=∫m0u(x,t)‍b′(s)esds-∫m0u0(x)‍b′(s)esds=H(u(x,t))-H(u0(x))≤αt.
Since H is an increasing function, we have
(37)u(x,t)≤H-1(αt+H(u0(x))).
The proof is completed.


## 3. Blow-Up Solution

The following theorem is the main result for the blow-up solution of (1).Theorem 2.
Letu be a solution of problem (1). Assume that the following conditions (i)–(iv) are satisfied.(i)
For anys∈R+,
(38)(sb′(s))′≥0,sb′(s)-(sb′(s))′≥0,(a(s)b′(s))′≥0,[1a(s)(a(s)b′(s))′+1b′(s)]′+1a(s)(a(s)b′(s))′+1b′(s)≥0.(ii)
For any(x,s,d,t)∈D×R+×R+¯×R+,
(39)(f(x,s,d,t)h(t))t≥0,fd(x,s,d,t)[(1b′(s))′+1b′(s)]≥0,(f(x,s,d,t)b′(s)a(s))s-f(x,s,d,t)b′(s)a(s)+h′(t)(b′(s))2a(s)h(t)≥0.(iii)
Consider the integration(40)∫M0+∞‍b′(s)esds<+∞,M0=max⁡D¯u0(x).(iv)
Consider(41)β=min⁡D¯{∇·(h(0)k(x)a(u0)∇u0)+f(x,u0,q0,0)eu0}>0,mmmmiimmmmimmmmmminmiimmmq0=|∇u0|2.
Then the solutionu of problem (1) must blow up in finite time T, and
(42)T≤1β∫M0+∞‍b′(s)esds,u(x,t)≤G-1(β(T-t)),(x,t)∈D¯×[0,T),
where
(43)G(z)=∫z+∞‍b′(s)esds,z>0,
and G-1 is the inverse function of G.Proof.
Construct the following auxiliary function:(44)Q(x,t)=b′(u)ut-βeu.
So we have
(45)∇Q=b′′ut∇u+b′∇ut-βeu∇u,ΔQ=b′′′ut|∇u|2+2b′′∇u·∇ut+b′′utΔu+b′Δut-βeu|∇u|2-βeuΔu.
As the previous derivation from (14) to (25), we can obtain
(46)akhb′ΔQ+[(2kh(ab′)′+2fqb′)∇u+ahb′∇k]·∇Q+{a(b′)2(fb′a)u+h′hmmmm+[akh(1a(ab′)′)′+2fq(1b′)′]|∇u|2}Q-Qt=-βeu{2fq[(1b′)′+1b′]mmmiimim+akh[(1a(ab′)′+1b′)′+1a(ab′)′+1b′]}|∇u|2-(b′)2a(ab′)′ut2-βeua(b′)2×[h′(b′)2ah(fb′a)u-fb′a+h′(b′)2ah]-h(fh)t.
It is seen from (38) and (39) that the right-hand side of (46) is nonpositive; that is,
(47)akhb′ΔQ+[(2kh(ab′)′+2fqb′)∇u+ahb′∇k]·∇Q+{a(b′)2(fb′a)u+h′hmmim+[akh(1a(ab′)′)′+2fq(1b′)′]|∇u|2}Q-Qt≤0,inD×(0,T).
By applying maximum principle (see [17]), it follows from (47) that Q can attain its nonpositive minimum only for D¯×{0} or ∂D×(0,T).
ForD¯×{0}, with (41), we have
(48)min⁡D¯Q(x,0)=min⁡D¯{b′(u0)(u0)t-βeu0}=min⁡D¯⁡{∇·(h(0)k(x)a(u0)∇u0)mmmiimm+f(x,u0,q0,0)-βeu0}=min⁡D¯{eu0[∇·(h(0)k(x)a(u0)∇u0)+f(x,u0,q0,0)eu0mimmmiiiiim-β∇·(h(0)k(x)a(u0)∇u0)+f(x,u0,q0,0)eu0]}=0.
For∂D×(0,T), substituting P and α with Q and β in (29), respectively, we have
(49)∂Q∂n=-γ(ub′)′b′Q+γβeuub′-(ub′)′b′,on⁡∂D×(0,T).
Combining (47)–(49) with condition (i), we can apply the maximum principles again to obtain that the minimum of Q in D¯×[0,T) is zero. Thus,
(50)Q≥0,inD¯×[0,T),(51)b′(u)euut≥β.
At the point x*∈D¯, where u0(x*)=M0, we can integrate (51) from 0 to t to get
(52)∫0t‍b′(u)euutdt=∫M0u(x*,t)‍b′(s)esds≥βt,
which implies that u must blow up in finite time. Actually, if u is a global solution of (1), then, for any t>0, (52) shows
(53)∫M0+∞‍b′(s)esds≥∫M0u(x*,t)‍b′(s)esds≥βt.
Letting t→+∞ in (53), we have
(54)∫M0+∞‍b′(s)esds=+∞,
which contradicts with assumption (40). This shows that u must blow up in finite time t=T. Moreover, letting t→T in (52), we have
(55)T≤1β∫M0+∞‍b′(s)esds.
By integrating inequality (51) over [t,s](0<t<s<T), for each fixed x, we obtain
(56)G(u(x,t))≥G(u(x,t))-G(u(x,s))=∫u(x,t)+∞‍b′(s)esds-∫u(x,s)+∞‍b′(s)esds=∫u(x,t)u(x,s)‍b′(s)esds=∫ts‍b′(u)euutdt≥β(s-t).
Hence, by letting s→T, we have
(57)G(u(x,t))≥β(T-t).
Since G is a decreasing function, we obtain
(58)u(x,t)≤G-1(β(T-t)).
The proof is completed.


## 4. Applications

Whenh(t)≡1, k(x)≡1, f(x,u,q,t)=f(u) or b(u)=u, h(t)≡1, k(x)≡1, f(x,u,q,t)=f(u), or h(t)≡1, k(x)≡1, a(u)≡1, f(x,u,q,t)=f(u), the conclusions of Theorems 1 and 2 still hold true. In this sense, our results extend and supplement the results in [14–16]. In what follows, we present several examples to demonstrate the applications of the abstract results.Example 1.
Letu be a solution of the following problem:
(59)(ueu)t=∇·(11+te|x|2(1+u)eu∇u)+11+t(e-u+eq)(e-t+|x|2),inD×(0,T),∂u∂n+2u=0,on⁡∂D×(0,T),u(x,0)=2-|x|2,inD¯,
where q=|∇u|2, D={x=(x1,x2,x3)∣|x|2<1} is the unit ball of R3. Now we have
(60)b(u)=ueu,h(t)=11+t,k(x)=e|x|2,a(u)=(1+u)eu,γ=2,f(x,u,q,t)=11+t(e-u+eq)(e-t+|x|2),u0(x)=2-|x|2.
In order to determine the constant α, we assume
(61)s=|x|2,
and then 0≤s≤1 and
(62)α=max⁡D¯∇·(h(0)k(x)a(u0)∇u0)+f(x,u0,q0,0)eu0=max⁡D¯{e|x|2(10|x|2-18)+(1+|x|2)mmmim×(e-4+2|x|2+e-2+5|x|2)}=max⁡0≤s≤1{[e-4+2s+e-2+5s]es(10s-18)+(1+s)mmimm×[e-4+2s+e-2+5s]}=18.6955.
It is easy to check that (5)–(7) hold. By Theorem 1, u must be a global solution, and
(63)u(x,t)≤H-1(αt+H(u0(x)))=-1+18.6955t+(1+u0(x))22=-1+18.6955t+(3-|x|2)22.Example 2.
Letu be a solution of the following problem:
(64)(u+ln⁡u)t=∇·((1+t)e-|x|2(1+1u)∇u)+(1+t)(eu-e-q)(6+t|x|2),mmmmmmmmiiiiminD×(0,T),∂u∂n+2u=0,on⁡∂D×(0,T),u(x,0)=2-|x|2,inD¯,
where q=|∇u|2, D={x=(x1,x2,x3)∣|x|2<1} is the unit ball of R3. Now we have
(65)b(u)=u+ln⁡u,h(t)=1+t,k(x)=e-|x|2,a(u)=1+1u,γ=2,f(x,u,q,t)=(1+t)(eu-e-q)(6+t|x|2),u0(x)=2-|x|2.
In order to determine the constant β, we assume
(66)s=|x|2,
and then 0≤s≤1 and
(67)β=min⁡D¯∇·(h(0)k(x)a(u0)∇u0)+f(x,u0,q0,0)eu0=min⁡D¯{4|x|6-26|x|4+50|x|2-36(2-|x|2)2e2mmmmm+6(1-e-3|x|2-2)4|x|6-26|x|4+50|x|2-36(2-|x|2)2e2}=min⁡0≤s≤1{4s3-26s2+50s-36(2-s)2e2mmmmm+6(1-e-3s-2)4s3-26s2+50s-36(2-s)2e2}=3.96997.
It is easy to check that (38)–(40) hold. By Theorem 2, u must blow up in finite time T, and
(68)T≤1β∫M0+∞‍b′(s)esds=13.96997∫2+∞‍(1+1s)1esds=0.04641,u(x,t)≤G-1(β(T-t))=G-1(3.96997(T-t)),
where
(69)G(z)=∫z+∞‍b′(s)esds=∫z+∞‍(1+1s)1esds,z≥0,
and G-1 is the inverse function of G.



---

*Source: 241650-2014-08-14.xml*