J. Math. Anal. Appl. 423 (2015) 1183–1202

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Journal of Mathematical Analysis and Applications

www.elsevier.com/locate/jmaa

Global solvability for the porous medium equation with boundary

flux governed by nonlinear memory

Jeffrey R. Anderson a,∗, Keng Deng b

a Department of Mathematical Sciences, Indiana University–Purdue University Fort Wayne, Fort Wayne, IN 46805, United Statesb Department of Mathematics, University of Louisiana at Lafayette, Lafayette, LA 70504, United States

a r t i c l e i n f o a b s t r a c t

Article history:Received 29 May 2014Available online 22 October 2014Submitted by P. Sacks

Keywords:Nonlinear diffusion equationGlobal existenceBlow-up in finite timeMemory boundary condition

We introduce the study of global existence and blow-up in finite time for nonlinear diffusion equations with flux at the boundary governed by memory. Via a simple transformation, the memory term arises out of a corresponding model introduced in previous studies of tumor-induced angiogenesis. The present study is also in the spirit of extending work on models of the heat equation with local, nonlocal, and delay nonlinearities present in the boundary flux. Specifically, we establish an identical set of necessary and sufficient conditions for blow-up in finite time as previously established in the case of local flux conditions at the boundary.

Published by Elsevier Inc.

1. Introduction

We initiate a study of global solvability for the nonlinear diffusion model

ut = Δ(um

)on ΩT

∇(um

)· n = 0 on (∂Ω \Σ)T

∇(um

)· n = uqv on ΣT

u = u0 on Ω × {0} (1)

where

vt = up on ΣT

v = 0 on Σ × {0} (2)

* Corresponding author.E-mail address: andersjr@ipfw.edu (J.R. Anderson).

http://dx.doi.org/10.1016/j.jmaa.2014.10.0410022-247X/Published by Elsevier Inc.

1184 J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202

Here, m, p > 0, q ≥ 0, and ΩT = Ω × (0, T ), where Ω is a bounded domain in RN having piecewise smooth boundary ∂Ω with outward pointing unit normal n. Σ is a relatively open subset of ∂Ω. The initial condition u0 is a nonnegative, continuous function on Ω. We establish local existence, continuation, and subsolution comparison for weak solutions of (1)–(2). Our main result is that for m > 1, every solution of (1)–(2) is global if 0 < p + q ≤ 1, whereas if p + q > 1 and u0 > 0, then all maximal solutions blow up in finite time. The analysis herein also applies in the case of the heat equation, m = 1. However, this has already been addressed in a previous paper [3].

Motivation for the study of (1)–(2) comes from both its identification as a subset of previously introduced models of tumor-induced angiogenesis and presenting a memory-type flux condition having had limited treatment in the literature. Formally integrating (2), one obtains the boundary condition

∇(um

)· n = uq

t∫0

up(·, s)ds on ΣT

Our result may thus be viewed in analogy to the localized version of (2), namely ∇(um) ·n = uq+p. As such, we extend results by Wolanski for the localized model [11].

As part of a model for the growth of new capillary networks (“angiogenesis”), initiated by a developing solid tumor, Levine et al. [6] introduced a general transmission condition, similar to (2), of the form

vt = f(x, t, u, v) + G(u)t on ΣT

v = v0 on Σ × {0}.

In the application to angiogenesis, u and v are the concentrations of tumor-released growth factor outside and within the wall of a nearby capillary, respectively. The boundary condition arises in an effort to correctly represent uptake of growth factor within the capillary wall and its transport through the wall. Additionally, Σ represents the capillary wall. (See [6] for the entire system and a complete discussion of the model.)

The general program, beginning with more simplified systems such as (1)–(2), is to establish results on local and global solvability which are applicable to each of the various diffusion models contained within the full angiogenesis system introduced in [6]. Such results are of potential use in allowing future mathematical analyses of solutions for this fairly involved and complex system. Additionally, the local existence and subsolution comparison theory contained in the present work is an extension of known theories on diffusion models to include the case of a relatively new type of delay/memory boundary condition.

In our previous study of (1)–(2) for the case of the heat equation (m = 1) [3], a review of the literature is provided regarding memory conditions as incorporated in diffusion, reaction, and boundary flux in the case of uniformly parabolic models. Concerning models involving nonlinear diffusion and incorporating memory terms into the boundary flux, there appears to be only our previous work [2].

In the context of characterizing conditions under which these models are globally solvable, it is helpful to consider possible comparisons to corresponding localized models. As such, there is the aforementioned work by Wolanski [11], as well as the extension of this work to the reaction–diffusion model ut = Δ(um) + uα

with strictly positive initial data, u0, and nonlinear flux ∇u · n = uβ at the boundary [8]. These authors have established that the model is globally solvable if and only if α ≤ 1, β ≤ min{1, (m +1)/2}, via suitable choices of subsolutions and supersolutions.

Although following from similar arguments applied to localized models, the local existence and comparison theories for (1)–(2) have not been previously addressed. In the next section, we state our results providing a general treatment of local existence for weak solutions along with subsolution comparison. Local solvability is established in Section 3, and subsolution comparison is proven in Section 4. The solution established herein is thus the maximal solution. However, consideration of uniqueness and supersolution comparison

J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202 1185

is not required for the present results and left for a future study. In Section 5, we further develop integral estimates following from the local existence proof, which ultimately prove the main global existence theorem. Finally, in Section 6, appropriately constructed subsolutions allow concluding blow-up in finite time for (1)–(2).

2. Local existence, comparison, and global solvability results

We establish the local existence and comparison of solutions for a slightly more general model

ut = Δφ(u) on ΩT

∇φ(u) · n = 0 on (∂Ω \Σ)T∇φ(u) · n = g(u, v) on ΣT

u = u0 ≥ 0 on Ω × {0} (3)

where

vt = f(u) on ΣT

v = v0 ≥ 0 on Σ × {0} (4)

The nonlinear terms φ, g, and f are assumed to be continuous for u, v ≥ 0, differentiable for u, v > 0, with φ′(u) > 0 for u > 0, φ(0) = f(0) = 0, and g(0, v) ≥ 0 for v ≥ 0. We also require monotonicity gv ≥ 0 and f ′ ≥ 0. The initial condition v0 will be assumed to be continuous for purposes of this work, although it is possible to obtain the same results for v0 ∈ L∞(Σ).

The local existence result established here has several possible generalizations. For example, a closely related model is addressed in [2]. Rather than providing the most general existence and comparison theories possible, we are more interested in a method of constructing solutions useful for proving the global solvability of (1)–(2).

The notion of weak solvability used herein is a standard approach obtained upon moving all derivatives to a test function via integration by parts. In order to formulate the definition, it is convenient to introduce the following notation.

IΩ(u, v, ξ, u0) ≡∫Ω

uξ(x, t)dx−∫Ω

u0ξ(x, 0)dx−t∫

0

∫Ω

[uξs + φ(u)Δξ

]dxds

−t∫

0

∫Σ

g(u, v)ξdxds +t∫

0

∫∂Ω

φ(u)∇ξ · ndxds (IΩ)

IΣ(u, v, η, v0) ≡∫Σ

vη(x, t)dx−∫Σ

v0η(x, 0)dx−t∫

0

∫Σ

[vηs + f(u)η(x, s)

]dxds (IΣ)

Here, ξ, η are smooth, nonnegative test functions on ΩT , ΣT .

Definition 2.1. u, v ≥ 0 is a weak solution (subsolution, supersolution) of (3)–(4) on ΩT if u, v are bounded, defined everywhere on ΩT , ΣT , respectively, and IΩ, IΣ = (≤, ≥) 0 for every set of smooth test functions ξ, η ≥ 0.

1186 J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202

Clearly, this definition simply incorporates the integral identities (inequalities) resulting from formal integration by parts on (3)–(4) and classical subsolution/supersolution formulations of the model.

Theorem 2.1 (Local solvability and subsolution comparison). For some T > 0, there exists a solution of (3)–(4) on ΩT , U , V . Furthermore, if u, v is a nonnegative subsolution of (3)–(4) on the same set, with u ≤ U , v ≤ V for t = 0 (in an almost everywhere sense), then u ≤ U , v ≤ V on ΩT , ΣT , respectively.

Either T = ∞ or ‖U(·, t)‖∞,Ω +‖V (·, t)‖∞,Σ is unbounded for t < T . Here, ‖U(·, t)‖∞,Ω is the L∞ norm of U(·, t) on Ω, with a similar definition for ‖V (·, t)‖∞,Σ.

Short of imposing differentiability of the nonlinearities at u = 0, thus limiting the power laws allowed, we are not aware of a general uniqueness and supersolution comparison result for (3)–(4). As we establish global existence by integral estimates and do not resort to supersolution constructions, this does not limit the current work. Specifically, letting φ(u) = um, g(u, v) = uqv, and f(u) = up in (3)–(4), we have the following result for the model (1)–(2).

Theorem 2.2 (Characterization of global solvability). For m > 1, every solution of (1)–(2) is global if p + q ≤ 1, whereas if p + q > 1 and u0 > 0, then all maximal solutions of the model blow up in finite time.

3. Proof of the local existence result

Fix 0 < ε < 1 and define

R ≡ ‖u0‖∞,Ω + ‖v0‖∞,Σ + 1

Let C1 ≡ C1(R) ≥ f(u) for all 0 ≤ u ≤ R. Since f ′ ≥ 0, a suitable choice is C1(R) = f(R). Additionally, define

mg(r) ≡ min{

0, min0≤u≤2r

g(u, 0)}

We construct a solution U , V of (3)–(4) as the limit of un, vn, where the sequences {un}∞n=1, {vn}∞n=0are defined as follows.

v0(x, t) ≡ v0(x) + ε + C1t

Note: In a general case of v0(x) bounded but not necessarily continuous, we instead define v0(x, t) similarly with a continuous function v0(x) having v0 ≥ v0.

With n ≥ 1 and vn−1 defined, un is the solution of

ut = Δφ(u) on ΩT

∇φ(u) · n = 0 on (∂Ω \Σ)T∇φ(u) · n = g(u, vn−1) −mg(ε/n) on ΣT

u = u0 + ε/n on Ω × {0} (3ε,n)

Given un, vn is then defined as the solution of

vt = f(un) on ΣT

v = v0 + ε/n on Σ × {0} (4ε,n)

J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202 1187

Since v0(x, t) ≥ 0, the solution of (3ε,1) exists on ΩT1 for sufficiently small T1 > 0, with u1 ≥ ε, following arguments in [1]. A necessary technicality invoked in this reference is that φ and g(·, v0) be bounded, with φ′(u) ≥ θ > 0 for some constant θ and u ≥ ε/2. This may be satisfied by modifying φ(u) and g(u, v)appropriately for u, v ≥ R + 1. Namely, we may replace φ, g with φR, gR, respectively, such that φ = φR

and g = gR for 0 ≤ u, v ≤ R + 1. Furthermore, we set θ(R) ≤ φ′ for ε/2 ≤ u ≤ R + 1 and φ, g ≤ α(R) for u, v ≤ R + 1.

The standard maximum principle estimate, e.g., [1, Lemma 3.1], yields

ε ≤ u1 ≤ ‖u0‖∞,Ω + ε + Kt1/(N+2)

where K = K(Ω, α, θ, N, T ). Selecting T1 = T1(R) > 0 small enough, we have u1 ≤ R on ΩT1 . Hence, u1 is a solution of (3ε,1).

Upon direct integration of (4ε,1),

ε ≤ v1(x, t)

= v0(x) + ε +t∫

0

f(u1(x, s)

)ds

≤ v0(x) + ε + C1(R)t

Subsequently, ε ≤ u1, v1 ≤ R for t < T2 = T2(R) with 0 < T2 ≤ T1. From this identity, we also have v1(x, t) ≤ v0(x, t).

It will be useful for purposes of establishing the continuation of solutions to observe that, without loss of generality, we can define T2(R) to be decreasing and finite for R ≥ 1. Specifically, for 1 ≤ R1 ≤ R2, the choices θ(R1) ≡ θ(R2), α(R1) ≡ α(R2) are clearly acceptable, giving rise to T1(R1) = T1(R2). As more optimal choices of θ(R1), α(R1) are possible and C1(R1) ≤ C1(R2), we have T2(R1) ≥ T2(R2).

Now, with vn−1 ≥ 0 for n ≥ 1, we may similarly conclude the existence of a local solution for (3ε,n). In order to obtain the lower estimate, un ≥ ε/n, note that

g(ε/n, vn−1) ≥ g(ε/n, 0) ≥ mg(ε/n)

Thus, ε/n is a classical subsolution of (3ε,n), and hence un ≥ ε/n.Due to the monotonicity of g in v and the fact mg(ε/2) ≥ mg(ε), v1 ≤ v0 now implies ε/2 ≤ u2 ≤ u1 ≤ R

on ΩT2 . The monotonicity of f , in turn, implies ε/2 ≤ v2 ≤ v1 ≤ R on ΣT2 . Continuing the process, we obtain

ε/n ≤ un ≤ un−1 ≤ · · · ≤ u1 ≤ R on ΩT2

and

ε/n ≤ vn ≤ vn−1 ≤ · · · ≤ v1 ≤ R on ΣT2

It is now easy to show

IΩ(un, vn−1, ξ, u0 + ε/n) + mg(ε/n)t∫

0

∫Σ

ξdxds = 0

and IΣ(un, vn, η, v0 + ε/n) = 0. Upon letting n → ∞, the limiting functions un → U , vn → V are solutions of (3)–(4) on ΩT2 .

1188 J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202

The continuation result follows from the observation that if U(·, t), V (·, t) remain bounded for 0 ≤t < T ∗, then the process above may be repeated with R ≥ U + V + 1, u0 ≡ U(·, T ∗ − T2(R)/2) and v0 ≡ V (·, T ∗ − T2(R)/2). Since such initial conditions u0, v0 at T ∗ − T2(R)/2 yield

R1 ≡ ‖u0‖∞,Ω + ‖v0‖∞,Σ + 1 ≤ R,

we have T2(R1) ≥ T2(R) and, thus, obtain a solution of (3)–(4) on ΩT for T ≥ T ∗ + T2(R)/2. Therefore, if T ∗ < ∞ is maximal for which (3)–(4) is solvable on ΩT∗ , the result follows.

4. Proof of subsolution comparison and maximality of the constructed solution

If u, v ≥ 0 is a subsolution of (3)–(4) on ΩT , such that u ≤ u0, v ≤ v0 at t = 0 then

∫Ω

(u− un)ξ(x, t)dx ≤t∫

0

∫Ω

(u− un){ξs + φnΔξ}dxds−t∫

0

∫∂Ω

(u− un)φn∇ξ · ndxds

+t∫

0

∫Σ

[g(u, v) − g(un, vn−1) + mg

(ε

n

)]ξdxds

and

∫Σ

(v − vn)η(x, t)dx ≤t∫

0

∫Σ

{(v − vn)ηs +

[f(u) − f(un)

]η}dxds

Here, (u− un)φn ≡ φ(u) − φ(un). Note that

g(u, v) − g(un, vn−1) = g(u, v) − g(un, v) + g(un, v) − g(un, vn−1)

≡ (u− un)gu,n + (v − vn−1)gv,n

We recall the assumption gv ≥ 0, which in the above notation is gv,n ≥ 0.Choose the test function ξn as the solution of

ξs + φnΔξ = 0 x ∈ Ω, s < t

∇ξ · n = 0 x ∈ ∂Ω \Σ, s < t

gu,nξ − φn∇ξ · n = 0 x ∈ Σ, s < t

ξ(x, t) = χ(x) x ∈ Ω

where χ ≥ 0. We also choose η(x, t) ≡ χ(x).This process is technically more correct by introducing appropriate smooth approximations of φn and

gu,n to insure the solution ξ is sufficiently smooth. Then, upon passing to the limit, we ultimately obtain the same results. The details are fairly standard, e.g., see [4], and we omit them here.

With such choices for test functions, ξn ≥ 0,

∫(u− un)χ(x)dx ≤

t∫ ∫ [(v − vn−1)gv,n + mg

(ε

n

)]ξndxds

Ω 0 Σ

J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202 1189

and

∫Σ

(v − vn)χ(x)dx ≤t∫

0

∫Σ

[f(u) − f(un)

]χ(x)dxds

By the definition of subsolution of (3)–(4), we also have IΣ(u, v, η, v0) ≤ 0, that is,

∫Σ

vχ(x)dx ≤t∫

0

∫Σ

f(u)χ(x)dxds +∫Σ

v(·, 0)χ(x)dx

Let R1 ≥ u on ΩT and R1 ≥ R ≡ ‖u0‖∞,Ω + ‖v0‖∞,Σ + 1. Recalling the definitions above, we have f(u) ≤ C1(R1), and C1(R) ≤ C1(R1). Hence, selecting T1 < T so that T1(C1(R1) − C1(R)) ≤ ε, there follows

∫Σ

v(x, t)χ(x)dx ≤∫Σ

v0χdx +t∫

0

∫Σ

f(u)χdx

≤∫Σ

v0χdx + C1(R1)t∫Σ

χdx

≤∫Σ

v0χdx +[(C1(R1) − C1(R)

)T1 + C1(R)t

] ∫Σ

χdx

≤∫Σ

(v0 + ε + C1(R)t

)χdx

for any smooth χ ≥ 0 and t ≤ T1. Therefore,

v ≤ v0 + ε + C1(R)t ≤ v0(·, t)

on ΣT1 .Returning to the inequality

∫Ω

(u− un)χ(x)dx ≤t∫

0

∫Σ

[(v − vn−1)gv,n + mg

(ε

n

)]ξndxds

along with the facts gv,n ≥ 0 and mg(ε/n) ≤ 0, it now follows

∫Ω

(u− u1)χ(x)dx ≤ 0

for any smooth χ ≥ 0. Thus, u ≤ u1 on ΩT1 .We now return to the second inequality

∫(v − vn)χ(x)dx ≤

t∫ ∫ [f(u) − f(un)

]χ(x)dxds

Σ 0 Σ

1190 J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202

along with f ′ ≥ 0 to conclude

∫Σ

(v − v1)χ(x)dx ≤ 0

for any smooth χ ≥ 0. Thus, v ≤ v1 on ΣT1 .Continuing the process, we have u ≤ un and v ≤ vn on ΩT1 , ΣT1 , respectively. Passing to the limit

n → ∞ yields u ≤ U , v ≤ V on the same sets. Continuation of the argument to ΩT , ΣT is possible due to u(·, T1) ≤ U(·, T1), v(·, T1) ≤ V (·, T1).

5. Proof of global solvability for p + q ≤ 1

In this section, we show that for any finite T > 0, there is a solution of (1)–(2) on ΩT in the case p +q ≤ 1, i.e., the model is globally solvable. Due to the continuation result above and the form of the differential equation for V , this will follow from an upper bound on the solution U . Our focus is on the power law model (1)–(2), however some further generalizations are certainly possible.

We first establish estimates of ‖U‖r+1,ΩT, ‖∇U (r+m)/2‖2,ΩT

, and ‖U‖r+p+q,ΣTfor sufficiently large r > 0

and any finite T > 0 so long as the solution exists on ΩT . These turn out to be true in all cases m > 0, and the specific result is as follows. Throughout the work, fairly standard notation for integral norms will be used, e.g.,

‖U‖r+1,ΩT≡

[ T∫0

∫Ω

Ur+1(x, s)dxds] 1

r+1

Lemma 5.1. Let U , V denote the maximal solution of (1)–(2) on ΩT . Assume m > 0 and p + q ≤ min{1,(m + 1)/2}. For any r > 0 with r ≥ m − 2(p + q) and all T > 0 such that the solution exists, there is an M > 0 for which

sup0≤t≤T

∥∥U(·, t)∥∥r+1,Ω , ‖U‖r+1,ΩT

,∥∥∇U (r+m)/2∥∥

2,ΩT, ‖U‖r+p+q,ΣT

≤ M

A key step in proving Lemma 5.1 is controlling boundary integral terms by using the standard Sobolev trace estimate

∫Σ

wdx ≤ C

∫Ω

(w + |∇w|

)dx

Resulting constants involved do not permit passing to the limit r → ∞ to obtain a bound on the L∞ norm. Instead, we are able to slightly revise the maximum principle method invoked in Section 3 for proving local solvability and obtain the necessary L∞ bound.

In the power law case, multiplication of (3ε,n) by urn, for r > 0, and integration by parts yields a standard

identity

1r + 1

∫ur+1n dx + 4rm

(r + m)2

t∫ ∫ ∣∣∇u(r+m)/2n

∣∣2dxds = 1r + 1

∫ (u0 + ε

n

)r+1

dx +t∫ ∫

ur+qn vn−1dxds

Ω 0 Ω Ω 0 Σ

J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202 1191

A fact which follows from this identity is existence of the limit

G(t) ≡ limn→∞

t∫0

∫Ω

∣∣∇u(r+m)/2n

∣∣2dxds

For n ≥ 2, integration of (4ε,n−1) provides an identity for vn−1, which is multiplied by ur+qn and then

integrated again to obtain

t∫0

∫Σ

ur+qn vn−1dxds =

t∫0

∫Σ

ur+qn

[ s∫0

upn−1dτ + v0 + ε

n− 1

]dxds

Although v0 ≡ 0 for t = 0, we will derive estimates valid on time steps of finite size. Thus, to continue an estimate from 0 ≤ t ≤ T1 to T1 ≤ t ≤ T2 requires using the updated initial conditions u0 ≡ U(·, T1) and

v0 ≡T1∫0

Up(·, s)ds

on Ω×{T1} and Σ×{T1}, respectively. In order to extend the solution U , V to all of ΩT , it will be essential to proceed on time steps of predetermined size which depend upon only parameters and constants.

To prove Lemma 5.1, we thus begin by establishing an estimate of the boundary integrals, invoking Hölder’s inequality as follows. For 0 < s ≤ t ≤ T and n ≥ 2,

∫Σ

ur+qn

[ s∫0

upn−1dτ

]dx ≤

∫Σ

ur+qn

[ s∫0

up(r+p+q)/pn−1 dτ

] pr+p+q

[ s∫0

1(r+p+q)/(r+q)dτ

] r+qr+p+q

dx

≤ tr+q

r+p+q

[∫Σ

u(r+q)(r+p+q)/(r+q)n dx

] r+qr+p+q

[∫Σ

( t∫0

ur+p+qn−1 dτ

) pr+p+q ·

r+p+qp

dx

] pr+p+q

There follows,

t∫0

∫Σ

ur+qn vn−1dxds ≤ t

r+qr+p+q

t∫0

[∫Σ

ur+p+qn dx

] r+qr+p+q

ds

[ t∫0

∫Σ

ur+p+qn−1 dxds

] pr+p+q

+t∫

0

∫Σ

ur+qn

(v0 + ε

n− 1

)dxds

≤ tr+q

r+p+q · tp

r+p+q

[ t∫0

∫Σ

ur+p+qn dxds

] r+qr+p+q

[ t∫0

∫Σ

ur+p+qn−1 dxds

] pr+p+q

+[ t∫ ∫

ur+p+qn dxds

] r+qr+p+q

[ t∫ ∫ (v0 + ε

n− 1

)(r+p+q)/p

dxds

] pr+p+q

0 Σ 0 Σ

1192 J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202

Invoking the Sobolev trace estimate, and dropping the subscript for the moment as we intend to apply the following analysis to both un−1 and un, we have∫

Σ

ur+p+qdx ≤ C

∫Ω

(ur+p+q +

∣∣∇ur+p+q∣∣)dx

Note that for r ≥ m − 2(p + q)

∇(ur+p+q

)= 2(r + p + q)

r + m

(u(r−m)/2+(p+q))∇(

u(r+m)/2)and

∫Ω

∣∣∇ur+p+q∣∣dx ≤ 2(r + p + q)

√rm

(r + m)√rm

[∫Ω

ur−m+2(p+q)dx

]1/2[∫Ω

∣∣∇u(r+m)/2∣∣2dx]1/2

≤ 12 · 4rm

(r + m)2

∫Ω

∣∣∇u(r+m)/2∣∣2dx + 12 · (r + p + q)2

rm

∫Ω

ur−m+2(p+q)dx

For any l ≤ 1 and r + l > 0, Young’s inequality implies

ur+l ≤ r + l

r + 1ur+1 + 1 − l

r + 1

So, provided p + q ≤ 1, 2(p + q) −m ≤ 1, and r −m + 2(p + q) > 0, the trace estimate yields∫Σ

ur+p+qdx ≤ C

{r + p + q

r + 1

∫Ω

ur+1dx + 1 − p− q

r + 1 |Ω|}

+ C

{2rm

(r + m)2

∫Ω

∣∣∇u(r+m)/2∣∣2dx + (r + p + q)2

2rm

∫Ω

ur−m+2(p+q)dx

}

≤ C

{r + p + q

r + 1

∫Ω

ur+1dx + 1 − p− q

r + 1 |Ω|}

+ 2Crm

(r + m)2

∫Ω

∣∣∇u(r+m)/2∣∣2dx+ C(r + p + q)2

2rm

{r −m + 2(p + q)

r + 1

∫Ω

ur+1dx + 1 + m− 2(p + q)r + 1 |Ω|

}

We observe the condition p + q ≤ min{1, (m + 1)/2} emerges here.In order to exhibit the following estimates more compactly, we define the constants

K1 ≡ C

{r + p + q

r + 1 + (r + p + q)2

2rm · r −m + 2(p + q)r + 1

}

K2 ≡ 2Crm

(r + m)2

K3 ≡ C|Ω|{

1 − p− q

r + 1 + (r + p + q)2

2rm · 1 + m− 2(p + q)r + 1

}

The trace inequality ∫ur+p+qdx ≤ K1

∫ur+1dx + K2

∫ ∣∣∇u(r+m)/2∣∣2dx + K3

Σ Ω Ω

J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202 1193

may now be applied to un and un−1 above to obtain

1r + 1

∫Ω

ur+1n dx + 4rm

(r + m)2

t∫0

∫Ω

∣∣∇u(r+m)/2n

∣∣2dxds

≤ 1r + 1

∫Ω

(u0 + ε

n

)r+1

dx

+ t

{K1

t∫0

∫Ω

ur+1n dxds + K2

t∫0

∫Ω

∣∣∇u(r+m)/2n

∣∣2dxds + K3t

} r+qr+p+q

×{K1

t∫0

∫Ω

ur+1n−1dxds + K2

t∫0

∫Ω

∣∣∇u(r+m)/2n−1

∣∣2dxds + K3t

} pr+p+q

+[ t∫

0

∫Σ

(v0 + ε

n− 1

)(r+p+q)/p

dx

] pr+p+q

×{K1

t∫0

∫Ω

ur+1n dxds + K2

t∫0

∫Ω

∣∣∇u(r+m)/2n

∣∣2dxds + K3t

} r+qr+p+q

for n ≥ 2.Passing to the limit n → ∞, with t ≤ T , and introducing the constant

α ≡ r + p + q

r + qmin

{1

4K1(r + 1) ,rm

K2(r + m)2 , 1}

we have

1r + 1

∫Ω

Ur+1(·, t)dx + 4rm(r + m)2G(t)

≤ 1r + 1

∫Ω

ur+10 dx + t

{K1

t∫0

∫Ω

Ur+1dxds + K2G(t) + K3t

}

+[ t∫

0

∫Σ

v(r+p+q)/p0 dx

] pr+p+q

{K1

t∫0

∫Ω

Ur+1dxds + K2G(t) + K3t

} r+qr+p+q(

α

α

) r+qr+p+q

≤ 1r + 1

∫Ω

ur+10 dx +

(t + r + q

r + p + qα

){K1

t∫0

∫Ω

Ur+1dxds + K2G(t) + K3t

}

+ p

r + p + q

(1α

) r+qp

t

∫Σ

v(r+p+q)/p0 dx

To define the time step size, let

T1 ≤ min{

1,

rm2 , 1, T

}

4K1(r + 1) K2(r + m)

1194 J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202

For all 0 ≤ t ≤ T1, we have

1r + 1

∫Ω

Ur+1(·, t)dx + 4rm(r + m)2G(t) ≤ 1

r + 1

∫Ω

ur+10 dx + 1

2(r + 1)

T1∫0

∫Ω

Ur+1dxds + 2rm(r + m)2G(t)

+ 2K3 + p

r + p + q

(1α

) r+qp

∫Σ

v(r+p+q)/p0 dx

from which follows

12(r + 1)

T1∫0

∫Ω

Ur+1dxds ≤ 1r + 1

∫Ω

ur+10 dx + p

r + p + q

(1α

) r+qp

∫Σ

v(r+p+q)/p0 dx + 2K3

Subsequently, for all 0 ≤ t ≤ T1,

1r + 1

∫Ω

Ur+1(·, t)dx + 2rm(r + m)2G(t) ≤ 2M0

where

M0 ≡ 1r + 1

∫Ω

ur+10 dx + p

r + p + q

(1α

) r+qp

∫Σ

v(r+p+q)/p0 dx + 2K3

Since G(T1) is bounded, it follows that ‖∇u(r+m)/2n ‖2

2,ΩT1is bounded, and the sequence is weakly con-

vergent. We now have∥∥∇U (r+m)/2∥∥2

2,ΩT1≤ G(T1)

and, by the trace inequality,

T1∫0

∫Σ

Ur+p+qdxds ≤ K1

T1∫0

∫Ω

Ur+1dxds + K2G(T1) + K3

Define

M1 ≡ max{

2(r + 1)M0,(r + m)2

rmM0,K1

[2(r + 1)M0

]+ K2

(r + m)2

rmM0 + K3

}

Then, we have established

sup0≤t≤T1

∫Ω

Ur+1(·, t)dx,T1∫0

∫Ω

Ur+1dxds,

T1∫0

∫Ω

∣∣∇U (r+m)/2∣∣2dxds,T1∫0

∫Σ

Ur+p+qdxds ≤ M1

Assuming the solution U , V exists on ΩT , let N0 ≥ 1 be the smallest integer such that N0T1 ≥ T . For 0 ≤ t ≤ T such that T1 ≤ t ≤ 2T1, we repeat the above argument on Ω × (T1, 2T1) with new initial conditions u0 ≡ U(·, T1) and

v0 ≡T1∫Up(·, s)ds

0

J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202 1195

For these initial conditions, define

M1,0 ≡ 1r + 1

∫Ω

ur+10 dx + p

r + p + q

(1α

) r+qp

∫Σ

v(r+p+q)/p0 dx + 2K3

and

M2 ≡ max{

2(r + 1)M1,0,(r + m)2

rmM1,0,K1

[2(r + 1)M1,0

]+ K2

(r + m)2

rmM1,0 + K3

}

We have

supT1≤t≤2T1

∫Ω

Ur+1(·, t)dx,2T1∫T1

∫Ω

Ur+1dxds,

2T1∫T1

∫Ω

∣∣∇U (r+m)/2∣∣2dxds,2T1∫T1

∫Σ

Ur+p+qdxds ≤ M2

It is necessary to be able to estimate M1,0 in terms of M1 and, thus, in terms of M0. To do so, note

∫Σ

[ T1∫0

Up(x, s)ds](r+p+q)/p

dx ≤∫Σ

[T

(r+q)/p1

T1∫0

Ur+p+q(x, s)ds]dx ≤ M1

So,

M1,0 ≤ 1r + 1M1 + p

r + p + q

(1α

) r+qp

M1 + 2K3

The process may be continued in N0 steps to all of ΩT , with all bounds reducible to M0 according to

Mk+1 ≡ max{

2(r + 1)Mk,0,(r + m)2

rmMk,0,K1

[2(r + 1)Mk,0

]+ K2

(r + m)2

rmMk,0 + K3

}

and

Mk,0 ≤ 1r + 1Mk + p

r + p + q

(1α

) r+qp

Mk + 2K3

for k = 1, 2, ..., N0 − 1. It follows that

sup0≤t≤T

∫Ω

Ur+1(·, t)dx,T∫

0

∫Ω

Ur+1dxds,

T∫0

∫Ω

∣∣∇U (r+m)/2∣∣2dxds,T∫

0

∫Σ

Ur+p+qdxds ≤N0∑k=1

Mk

This completes the proof of Lemma 5.1.We now establish the global L∞ estimate on ΩT using estimates in Lr+1(ΩT ), for sufficiently high r

established above. To do so, begin with the standard calculation for (un − k)+ ≡ max{un − k, 0}, with m ≥ 1 and k ≥ ‖u0‖∞ + 1. For fixed t ∈ (0, T ],

12

∫Ω

(un − k)+2(x, t)dx = −t∫

0

∫Ω

mum−1n

∣∣∇(un − k)+∣∣2dxds

+t∫ ∫

(un − k)+uqn

[ s∫upn−1dτ + ε

n− 1

]dxds

0 Σ 0

1196 J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202

≤ −mkm−1t∫

0

∫Ω

∣∣∇(un − k)+∣∣2dxds

+t∫

0

∫Σ

(un − k)+uqn

[ s∫0

upn−1dτ + ε

n− 1

]dxds

The boundedness ‖∇(un−k)+‖22,Ωt

is an instance of r = m in the proof above. Passing to the limit n → ∞, we have

12

∫Ω

(U − k)+2(x, t)dx + mkm−1t∫

0

∫Ω

∣∣∇(U − k)+∣∣2dxds ≤

t∫0

∫Σ

(U − k)+Uq

[ s∫0

Updτ

]dxds

Select λ > 1 so λ < 1 + 1/N and μ > 1 such that 1/λ + 1/μ = 1. Then for r ≥ 1

T∫0

∫Σ

(U − k)+Uq

[ s∫0

Updτ

]dxds =

t∫0

∫Σ

(U − k)+

UrUr+q

[ s∫0

Updτ

]dxds

≤[ t∫

0

∫Σ

((U − k)+

Ur

)λ

dxds

]1/λ[ t∫0

∫Σ

(Ur+q

s∫0

Updτ

)μ

dxds

]1/μ

Similar to our previous calculations

t∫0

∫Σ

(Ur+q

s∫0

Updτ

)μ

dxds ≤t∫

0

∫Σ

Uμr+μq

[( s∫0

Uμpdτ

)1/μ

t1/λ

]μ

dxds

≤ t(λ+μ)/λt∫

0

∫Σ

U r+q+pdxds

where r + p + q ≡ μ(r + p + q), Lemma 5.1 now yields

[ t∫0

∫Σ

(Ur+q

s∫0

Updτ

)μ

dxds

]1/μ

≤ M

for some constant M , which is independent of t ∈ (0, T ].With r ≥ 1 > 1 − 1/λ, we have λr > λ − 1. Since k ≥ 1, U ≥ k implies

U ≤ Uλr/(λ−1) ≤ Uλr/(λ−1) + k

Hence, (U − k)+(λ−1) ≤ Uλr, that is,

((U − k)+

Ur

)λ

≤ (U − k)+

It follows by applying the Sobolev trace estimate that

J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202 1197

12

∫Ω

(U − k)+2dx + mkm−1t∫

0

∫Ω

∣∣∇(U − k)+∣∣2dxds

≤ M1

[ t∫0

∫Σ

(U − k)+dxds]1/λ

≤ CM1

[ t∫0

∫Ω

((U − k)+ +

∣∣∇(U − k)+∣∣)dxds

]1/λ

≤ CM1

[ t∫0

∫Ω

(U − k)+dxds]1/λ

+ CM1

[ t∫0

∫Ω

∣∣∇(U − k)+∣∣dxds

]1/λ

The right hand side of this estimate contains a power which is not normally encountered in this method of establishing an L∞ bound, e.g., see [1,7]. We will thus continue to show the technique still may be successfully applied, due to our choice of λ.

To begin, define Ak(t) ≡ {x ∈ Ω : U(x, t) > k} and

μ(k) ≡t∫

0

∫Ak(s)

dxds

For α > 1,

12

∫Ω

(U − k)+2dx + mkm−1t∫

0

∫Ω

∣∣∇(U − k)+∣∣2dxds ≤ CM1

[( t∫0

∫Ω

(U − k)+αdxds

)1/α

μ(k)1−1/α

]1/λ

+ CM1

[ t∫0

∫Ω

∣∣∇(U − k)+2∣∣dxds]1/2λ

μ(k)1/2λ

Selecting α = (2λ − 1)/(λ − 1) yields 1 − 1/α = λ/(2λ − 1), and

12

∫Ω

(U − k)+2dx + mkm−1t∫

0

∫Ω

∣∣∇(U − k)+∣∣2dxds ≤ CM1‖U‖1/λ

Lα(ΩT )μ(k)1/(2λ−1)

+ 12λmkm−1

t∫0

∫Ω

∣∣∇(U − k)+2∣∣dxds

+(

1 − 12λ

)[CM1

(mkm−1)1/2λμ(k)1/2λ

]2λ/(2λ−1)

Therefore,

12

∫Ω

(U − k)+2dx + 12mkm−1

t∫0

∫Ω

∣∣∇(U − k)+∣∣2dxds ≤ C2μ(k)1/(2λ−1)

The standard argument now proceeds to obtain

(h− k)2μ(h)N/(N+2) ≤ C2μ(k)1/(2λ−1),

1198 J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202

for h > k > k0, e.g., [1, pp. 116–117] for the details. It follows

μ(h) ≤[C2μ(k)1/(2λ−1)

(h− k)2

](N+2)/N

= C3

(h− k)2(N+2)/N μ(k)(N+2)/N(2λ−1)

Since 1 < λ < 1 + 1/N ,

2λ− 1 < 2(1 + 1/N) − 1 = 1 + 2/N = (N + 2)/N

and thus

12λ− 1

(N + 2N

)>

N

N + 2

(N + 2N

)= 1

Subsequently, [5, p. 63] applies to establish μ(k0 + d) = 0 for some d > 0. We thus have U ≤ k0 + d on ΩT .

6. Proof of blow-up in finite time for p + q > 1

Assuming ∂ΩT = ΣT , we rewrite the model (1)–(2) as follows

ut = Δ(um

)on ΩT

∇(um

)· n = uq

t∫0

up(·, s)ds on ΣT

u = u0 on Ω × {0} (5)

Clearly, to establish the blow-up result for (5), it suffices to consider the following problem

ut = Δ(um

)on ΩT

∇(um

)· n = uq

t∫0

up(·, s)ds on ΣT

u = minΩ

u0 on Ω × {0} (6)

Due to the lack of a supersolution comparison result, we first show that solutions of (6) are monotone in time. Let v(x, t) = u(x, t + κ) (κ > 0). Then v satisfies

vt = Δ(vm

)on ΩT

∇(vm

)· n ≥ vq

t∫0

up(·, s)ds on ΣT

v = u(x, κ) on Ω × {0} (7)

Since u(x, κ) ≥ minΩ u0, from the comparison principle for the porous medium equation with a localized boundary condition (cf. [1]), it follows that v(x, t) ≥ u(x, t), that is, u is nondecreasing in t.

J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202 1199

Taking into account the monotonicity of the solution of (6), we now consider the following problem

ut = Δ(um

)on ΩT

∇(um

)· n =

t∫0

up+q(·, s)ds on ΣT

u = minΩ

u0 on Ω × {0} (8)

Since the solution of (8) is a subsolution of (6), as in [9,10], we seek a subsolution of (8) which blows up in finite time. To this end, letting w = um, we rewrite problem (8) as follows

(w

1m

)t= Δw in ΩT

∇w · n =t∫

0

wp+qm (·, s)ds on ΣT

w =(minΩ

u0

)m

on Ω × {0} (9)

By comparison, it is easy to see that w(x, t) ≥ σ = (minΩ u0)m > 0 on ΩT . Let h(x) be the solution of the problem

Δh = 1 on Ω

∇h · n = |Ω||∂Ω| on ∂Ω (10)

Since h(x) + c is also a solution of (10) for any positive constant c, we may assume that h(x) > 0 on Ω. We then consider three cases.

Case 1 (1 < p + q < m − 1). We choose ϕ(ζ) to satisfy the following

ϕ(ζ) ≡ σ for 0 ≤ ζ ≤ 1 and ϕ′(ζ) =ζ∫

1

ϕ−α(η)dη for ζ > 1 (11)

where α = 1 − p+q+1m . Since p + q < m − 1, 0 < α < 1. Clearly,

ϕ′′(ζ) ≡ 0 for 0 ≤ ζ < 1 and ϕ′′(ζ) = ϕ−α(ζ) for ζ > 1 (12)

Multiplying the second equation in (12) by ϕ′(ζ) and integrating over (1, ζ), we obtain

ϕ′(ζ) =[

21 − α

(ϕ1−α − σ1−α

)] 12

for ζ > 1

Then it follows that ϕ(ζ) exists for 0 ≤ ζ < ∞ and ϕ(ζ) → ∞ as ζ → ∞.Let g(t) be the solution of the following problem

g′(t) = mδϕ1− 1m

(g(t)

)t > 0

g(0) = 1 (13)

2

1200 J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202

where δ is chosen so that 0 < δ ≤ |∂Ω||Ω| and δmaxΩ h(x) < 1

2 . Since g′(t) ≥ mδσ1− 1m and ∫∞

ϕ−1+ 1m− 1−α

2 dϕ < ∞, g must blow up in finite time. We now construct a subsolution w(x, t) of (9)as follows

w(x, t) = ϕ(g(t) + δh(x)

).

By (12)–(13) and the fact that ϕ(ζ) is nondecreasing for ζ > 0, we have that

(w

1m

)t= 1

mϕ

1m−1(g + δh)ϕ′(g + δh)g′(t)

≤ δϕ′(g + δh)

≤ δϕ′(g + δh)Δh + δ2ϕ′′(g + δh)|∇h|2 = Δw

a.e. on ΩT . On the other hand, by (11) and (13), we find that

∇w · n = δϕ′(g + δh)∇h · n

≤g(t)+δh(x)∫

1

ϕ−α(η)dη

≤t∫

0

ϕ−α(g(s) + δh(x)

)g′(s)ds

≤t∫

0

wp+qm (x, s)ds

on ΣT . Moreover, w(x, 0) = ϕ(g(0) + δh(x)) = σ ≤ w(x, 0). Thus, w(x, t) is indeed a subsolution of (9). Since g(t) blows up in finite time and limζ→∞ ϕ(ζ) = ∞, w(x, t) blows up in finite time, and so does w(x, t).

Case 2 (m − 1 ≤ p + q ≤ m). We choose ϕ(ζ) to satisfy the following

ϕ(ζ) ≡ σ for 0 ≤ ζ ≤ 1 and ϕ′(ζ) =ζ∫

1

ϕβ(η)dη for ζ > 1 (14)

where β = p+q+1m −1. Since m −1 ≤ p +q ≤ m, 0 ≤ β ≤ 1

m . We then let g(t) be the solution of problem (13). If 0 < β ≤ 1

m , proceeding as for the case 1 < p +q < m −1, we can show that ϕ(ζ) exists globally, ϕ(ζ) → ∞as ζ → ∞, and g(t) blows up in finite time. Thus, w(x, t) = ϕ(g(t) + δh(x)) is a desired subsolution of (9). If β = 0, i.e., m = p + q + 1, ϕ(ζ) = ζ2

2 − ζ + 12 + σ for ζ > 1, and hence g′(t) = mδ[ g

2(t)2 − g(t) + 1

2 + σ]1− 1m

for t ≥ 1mδσ1− 1

m. It then follows that g(t) can only exist locally, and so w(x, t) = ϕ(g(t) + δh(x)) blows up

in finite time.

Case 3 (p + q > m). We choose ϕ(ζ) to satisfy the following

ϕ(ζ) ≡ σ for 0 ≤ ζ ≤ 1 and ϕ′(ζ) =ζ∫ϕ

p+qm (η)dη for ζ > 1 (15)

1

J.R. Anderson, K. Deng / J. Math. Anal. Appl. 423 (2015) 1183–1202 1201

Thus we have that

ϕ′′(ζ) ≡ 0 for 0 ≤ ζ < 1 and ϕ′′(ζ) = ϕp+qm (ζ) for ζ > 1 (16)

Multiplying the second equation in (16) by ϕ′(ζ) and integrating over (1, ζ), we find

ϕ′(ζ) =[

2mp + q + m

(ϕ

p+qm +1 − σ

p+qm +1)] 1

2

for ζ > 1

Since p+qm > 1, ϕ(ζ) must blow up in finite time. We now construct a subsolution w(x, t) of (9) as follows

w(x, t) = ϕ(εt + δh(x)

)where ε is chosen so that 0 < ε ≤ min{1, mδσ1− 1

m }.By (16) and the monotonicity of ϕ(ζ), we have that

(w

1m

)t= ε

mϕ

1m−1(εt + δh)ϕ′(εt + δh)

≤ δϕ′(εt + δh)Δh + δ2ϕ′′(εt + δh)|∇h|2 = Δw

a.e. on ΩT . On the other hand, in view of (15), we find that

∇w · n = δϕ′(εt + δh)∇h · n

≤εt+δh(x)∫

1

ϕp+qm (η)dη

≤t∫

0

ϕp+qm

(εs + δh(x)

)ds

=t∫

0

wp+qm (x, s)ds

on ΣT . Moreover, w(x, 0) = ϕ(δh(x)) = σ ≤ w(x, 0). Thus, w(x, t) of (9) blows up in finite time.

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