Partial Differential Equations

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http://france.elsevier.com/direct/CRASS1/

Partial Differential Equations

Diffusion versus absorption in semilinear parabolic problems

Andrey Shishkov

^a

, Laurent Véron

^b

aInstitute of Applied Mathematics and Mechanics of NAS of Ukraine, R. Luxemburg str. 74, 83114 Donetsk, Ukraine bLaboratoire de mathématiques et physique théorique, CNRS UMR 6083, faculté des sciences, 37200 Tours, France

Received and accepted 20 January 2006 Available online 3 March 2006

Presented by Haïm Brezis

Abstract

We study the limit, whenk→ ∞, of the solutionsu=u_kof (E)∂tu−u+h(t)u^q=0 inR^N×(0,∞),u_k(·,0)=kδ₀, with q >1,h(t) >0. Ifh(t)=e⁻^{ω(t )/t}whereω >0 satisfies to₁

0

√ω(t) t⁻¹dt <∞, the limit functionu_∞is a solution of (E) with a single singularity at(0,0), while ifω(t)≡1,u_∞is the maximal solution of (E). We examine similar questions for equations such as∂_tu−u^m+h(t)u^q=0 withm >1 and∂_tu−u+h(t)e^u=0.To cite this article: A. Shishkov, L. Véron, C. R. Acad. Sci.

Paris, Ser. I 342 (2006).

Résumé

Diffusion versus absorption dans des problèmes paraboliques semi-linéaires.Nous étudions la limite, quandk→ ∞, des solutionsu=u_kde (E)∂tu−u+h(t)u^q=0 dansR^N×(0,∞),u_k(·,0)=kδ₀avecq >1,h(t) >0. Nous montrons que si h(t)=e⁻^{ω(t )/t} oùω >0 vérifie₁

0

√ω(t) t⁻¹dt <∞, la fonction limiteu_∞est une solution of (E) avec une singularité isolée en(0,0), alors que siω(t)≡1,u_∞est la solution maximale de (E). Nous examinons des questions semblables pour des équations des type suivants∂_tu−u^m+h(t)u^q=0 avecm >1 et∂_tu−u+h(t)e^u=0.Pour citer cet article : A. Shishkov, L. Véron, C. R. Acad. Sci. Paris, Ser. I 342 (2006).

Version française abrégée

Soitq >1 eth:R₊→R₊une fonction continue, croissante telle queh(t ) >0 pourt >0. Il est facile de vérifier que toute solution positiveude

∂tu−u+h(t )u^q=0 dansR^N× ]0,+∞ [ (1)

satisfait à

u(x, t )U (t ):=

(q−1)

t

0

h(s)ds

₋1/(q−1)

∀(x, t )∈R^N× ]0,+∞[. (2)

E-mail addresses:[email protected] (A. Shishkov), [email protected] (L. Véron).

doi:10.1016/j.crma.2006.01.021

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Sih∈L¹(0,1, t^{N q/2}dt ), il est classique que pour toutk >0 il existe une unique solution (dite fondamentale)u=u_k de (1) surR^N× ]0,+∞[vérifiantu_k(·,0)=kδ₀. Par le principe du maximumk→u_kest croissant et deux cas peuvent se produire :

(i) ou bienu_∞=limk→∞u_k=U.Explosion initiale complète.

(ii) ou bienu_∞est une solution de (1) singulière en(0,0)vérifiant limt→0u_∞(x, t )=0 pour toutx=0.Explosion initiale ponctuelle.

Théorème 1.(I)Sih(t )=e⁻^σ/tpour unσ >0, alorsu_∞=U.

(II)Sih(t )=e⁻^{ω(t )/t} oùωest monotone croissante sur]0,+∞[et vérifie, pour unα∈ [0,1[,inf{ω(t )/t^α: 0<

t1}>0et₁

0

√ω(t ) t⁻¹dt <∞, alorsu_∞a une explosion initiale ponctuelle.

Dans le cas de l’équation

∂tu−u+h(t )e^u=0 dansR^N× ]0,+∞[, (3)

nous montrons.

Théorème 2.(I)Sih(t )=e⁻^e^σ/t pour unσ >0, alorsu_∞=U.

(II)Sih(t )=e⁻^e^ω(t)/t oùωvérifie les conditions du Théorème1, alorsu_∞a une explosion initiale ponctuelle.

Nos méthodes nous permettent aussi de traiter l’équation des milieux poreux avec absorption.

1. Main results

Letq >1 andh:(0,∞)→(0,∞)be a continuous nondecreasing function. It is easy to prove that any positive solutionuof

∂_tu−u+h(t )u^q=0 inR^N×(0,+∞) (1)

verifies

u(x, t )U (t ):=

(q−1)

t

0

h(s)ds

₋1/(q−1)

∀(x, t )∈R^N×(0,∞). (2)

Ifh∈L¹(0,1, t^{N q/2}dt ), it is classical that, for anyk >0, there exists a unique solution (called fundamental)u=u_k of (1) onR^N×(0,∞)such thatuk(·,0)=kδ0. By the maximum principlek→uk is increasing and the following alternative occurs:

(i) eitheru_∞=limk→∞uk=U.Complete initial blow-up.

(ii) oru_∞is a solution of (1) singular at(0,0)such that limt→0u_∞(x, t )=0 for allx=0. Single-point initial blow-up.

Theorem 1.(I)Ifh(t )=e⁻^σ/t for someσ >0, thenu_∞=U.

(II)Ifh(t )=e⁻^{ω(t )/t} whereωis nondecreasing on(0,+∞)and verifies, for someα∈ [0,1),inf{ω(t )/t^α: 0<

t1}>0and 1 0

√ω(t )dt

t <+∞, (3)

thenu_∞has single-point initial blow-up.

Concerning equation

∂tu−u+h(t )e^u=0 inR^N×(0,+∞), (4)

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any solutionuverifies

u(x, t )U (t ) := −ln t

0

h(s)ds

∀(x, t )∈R^N×(0,+∞) (5)

and the existence of a fundamental solutionu=u_k is ensured ifh(t )=e⁻^{b(t )}where limt→+∞t^N/2b(t )= +∞.

Theorem 2.(I)Ifh(t )=O(e⁻^e^σ/t)for someσ >0, thenu_∞=U.

(II)Ifh(t )=e⁻^e^ω(t)/t whereωsatisfies the conditions of Theorem1, thenu_∞has single-point initial blow-up.

Our methods apply to equations of porous media type

∂tu−u^m+h(t )u^q=0 inR^N×(0,∞), (6)

withm >1, q >1 and h:(0,∞)→(0,∞)is nondecreasing. As above, any positive solution satisfies (2). Ifh∈ L¹((0,1;t⁻^(q⁻^1)/(m⁻¹⁺^2N⁻¹⁾dt ), for anyk >0 there exists a solutionu=uk of (6) such thatuk(·,0)=kδ0. Since k→u_kis increasing, the same alternative as in case of (1) occurs concerningu_∞.

Theorem 3.Assumeq > m >1.(I)Ifh(t )=O(t^(q⁻^m)/(m⁻¹⁾), thenu_∞=U.

(II)Ifh(t )=t^(q⁻^m)/(m⁻¹⁾ω⁻¹(t )whereωis nondecreasing and positive on(0,+∞)and verifies 1

0

ω^θ(t )dt

t <+∞, (7)

whereθ=_{(N (m}₋₁₎₊^m_2(m²⁻¹₊_1))(q₋₁₎, thenu_∞has single-point initial blow-up.

Sketch of the proofs. The complete initial blow-up results are proved by constructing local subsolutions by modify- ing the very singular solutions of some related equations. Since for Eq. (1), the proof is already given in [3] we shall outline the (more complicated) construction for Eq. (4).

Lemma 4.Ifh(t )=σ t⁻²e^{σ t}⁻¹⁻^e⁻^σ/t for someσ >0, complete initial blow-up occurs for Eq.(4).

Proof. Writingh(t )=e⁻^{a(t )}is first observed that fundamental solutionsukof (4) exist for allk >0 if limt→0t^N/2a(t )=

∞. For >1, letv=v_∞,be the very singular solution of

∂_tv−v+ct^αv=0 (8)

inR^N×(0,∞), whereαandcare positive constants. The choice ofα=(N+2)/(−1)/2−1 ensures the existence ofv_∞,. Furthermore, if we write

v_∞,(x, t )= 2c

N+2

1/(−1)

t⁻⁽¹⁺^N/2)f

x/√ t ,

thenf(η)1 forη∈R^N and f+1

2Df·η+N+2

2 f−f=0. (9)

By the maximum principle 0< f< f 1 for > >1. For the particular choice^∗=(N+4)/(N2), we can use the expression of the asymptotic expansion of the very singular solution given in [1],

f∗(η)=C|η|²e^−|^η^|²^/4

1+ ◦(1) as|η| → ∞,

from which followsf(η)f∗(η)δ^∗(|η|²+1)e^−|^η^|²^/4for someδ^∗>0, anyη∈R^Nand^∗. Thus there exists δ >0 depending only onNsuch that

v_∞,(x, t )δc^1/(⁻¹⁾t⁻¹⁻^N/2

|x|²+t e^−|^x^|²^/4t ∀(x, t )∈R^N×(0,∞). (10)

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Because any positive solutionuof (4) satisfies (5), we have to prove that we can fixcandτ >0 such that

ct^α(ρ+1)h(t )e^ρ ∀(t, ρ)∈(0, τ] × [0,V (t ) ]. (11)

Writinghunder the formh(t )= −ω(t )e^{ω(t )}whereω(t )=e^{γ (t )}andγ is a positive decreasingC¹function, infinite att=0, we first notice that it is sufficient to prove this inequality forρ=U (t ), and in that case

ct^α

e^{γ (t )}+1 −γ(t )e^{γ (t )} ∀t∈(0, τ]. (12)

We take nowγ (t )=σ/t, and prove that there existsβ >0, depending only on N such that, for any 0< τ βσ, estimate (11) holds with

c=e⁽¹⁻^)σ/τ⁻²⁻¹^((N⁺²⁾⁻^{N )ln}^τ.

The maximum principle and (11) imply that for any >1 andk >0 the solutionsu=u_kof (4) andv= ˜v_kof

∂_tv−v+ct^α

v+1 =0

with initial datakδ0verifies 0v˜k,uk, on(0, τ]. Thereforev_∞,u_∞+ct^α⁺¹/(α+1)on(0, τ]leads to u_∞(x, τ )δτ⁻¹⁻^N/2

|x|²+τ e

4σ−|x|2

4τ −^(N+2)−N₂₍₁₋₎ lnτ

.

Thus limτ→0u_∞(x, τ )= ∞, locally uniformly inB₂^√_σ, which implies the result. 2

The proof of Theorem 2 follows from the fact that for any σ > σ >0 there exists an interval (0, θ] where σ t⁻²e^σ^t⁻¹⁻^e⁻^σ^/t e⁻^e^σ/t.

The single-point initial blow-up is proved by local energy methods (see [2,5,6] for other types of applications).

Because of their high degree of technicality we shall just give a short sketch of them in the simplest case of Theorem 1.

Fork >0, letu_k=ube the solution of the next result.

∂_tu−u+h(t )|u|^q⁻¹u=0 inR^N×(0,∞),

u(x,0)=u0,k(x)=M_k^1/2k⁻^N/2ηk(x) ∀x∈R^N, (13)

whereη_k∈C(R^N)is nonnegative, has compact support inB_k₋1, converges weakly toδ₀ask→ ∞, and{M_k}satisfies limk→∞k⁻^N/2M_k= ∞. Furthermore it can be assumed thatη_kL²c₀k^N/2. The single-point initial blow-up will be a consequence of

Lemma 5.For anyδ >0there existsC=C(δ)such that:

sup

t∈[0,1]

|x|δ

u²_k(x, t )dx+ 1

0

|x|δ

|∇uk|²+u²_k dxdtC ∀k∈N. (14)

Proof. Forr∈(0,1),τ0 we setΩ(τ )= {x∈R^N: |x|> τ},Q^r(τ )=Ω(τ )×(0, r],Q_r(τ )=Ω(τ )×(r,1)and Q_r =R^N×(r,1), and denoteI₁(r)=

Qr|∇u|²dxdt,I₂(r)=

Qru²dxdt,I₃(r)=

Qrh(t )|u|^q⁺¹dxdt. If we multiply the equation byu(x, t )e^(r⁻^{t )/(2}⁻^r), integrate onQ_r and use Hölder’s inequality, we get,

R^N

u²(x,1)dx+I₁(r)+I₂(r)+I₃(r)c

R^N

u²(x, r)dx

cτ

N (q−1)

q+1 h(r)^q+⁻²¹

−I₃(r)

q+12 +c

Ω(τ )

u²(x, r)dx. (15)

Letτ→µ(τ )be a smooth decreasing function, we define

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E₁^µ(r, τ )=

Q^r(τ )

|∇u|²+µ²u²(x, t ) e⁻^µ²^tdxdt,

E₂(r, τ )=

Q^r(τ )

u²dxdt and f_µ(r, τ )=sup

e⁻^µ²^t

Ω(τ )

u²(x, t )dx: 0tr

andf (r)=f0(r,0). Then we introduce a parameter in the equation as in [4] by multiplying it byu(x, t )exp(−µ²(τ )t ) and integrating in the domainQ^r(τ )withτ > k⁻¹Q^r(τ )andτ > k⁻¹. After some simple computations we deduce:

f_µ(r, τ )+2E₁^µ(r, τ ) 2 µ

r

0

∂Ω(τ )

|∇u|²+µ²u²(x, t ) e⁻^µ²^tdSdt ∀τ > k⁻¹.

Assuming 1−2µ/µ²>1/2, we deduce from last inequality:

f_µ(r, τ )+E₁^µ(r, τ )− 2 µ(τ )

dE₁^µ(r, τ )

dτ ∀τ > k⁻¹, and by integration

f_µ(r, τ₂) e

τ2 τ1

µ(τ )dτ

2 −1 +E₁^µ(r, τ₂)e

τ2 τ1

µ(τ )dτ

2 E^µ₁(r, τ₁) ∀τ₂> τ₁> k⁻¹. The choiceµ(τ )=r⁻¹(τ−k⁻¹)/8 (τ > k⁻¹) yields to

Ω(τ )

u²(x, r)dx+

Q^r(τ )

|∇xu|²+(τ−k⁻¹)² 64r² u²

dxdt

c₁e⁻^(τ⁻^k−

1)2

64r ×

Q^r(τ₀^k)

|∇xu|²+u² 2r

dxdt ∀ττ˜₀^k:=k⁻¹+8√

r > τ₀^k:=k⁻¹+4√

2r. (16)

We will need standard global energy estimate of solution of problem (13) too:

R^N

u(x, r)²dx+

Q^r

|∇xu|²+ |u|²+h(t )|u|^q⁺¹ dxdtcu_0,k²_L

2(R^N)cM¯ _k ∀r >0. (17) Estimating the right-hand side terms in (15) and (16) by (17), we derive:

(i) 3

i=1

Ii(r)c1τ^{N (q−1)}^q⁺¹ h(r)^q⁻²⁺¹

−I₃(r)

2

q+1 +c₂M_k

r e⁻^(τ⁻^k⁻¹⁾²^/64r ∀ττ˜₀^k(r), (ii) f0(r, τ )+E₁⁰(r, τ )+τ−k⁻¹

64r² E2(r, τ )c₂M_k

r e⁻^(τ⁻^k⁻¹⁾²^/64r ∀τ τ˜₀^k(r). (18) Next we chooseM_k=e^e^k, fix₀∈(0,e⁻¹)and define a pair(r_k, τ_k)by the following relations:r_k=sup{r: I₁(r)+ I₂(r)+I₃(r) >2M_k⁰}; c₂r_k⁻¹exp(−_64r^τ^k²_k)M_k =M_k⁰ ⇔ τ_k =8

r_k(1−₀)e^k+ln(c2/r_k). Taking τ =τ_k +k⁻¹ in (18)(i) and solving the corresponding O.D.E. yields the estimate:

3 i=1

I_i(r)c₃

τ_k+k⁻¹ H (r) ⁻^2/(q⁻¹⁾ ∀rr_k, H (r)= r

0

h(s)ds. (19)

If we writeh(t )=e⁻^{ω(t )/t}, the assumption inf{ω(t )/t^α: 0< t1}>0 implies thatH (r)c₀e⁻^ω(r)/rr²/ω(r)and, replacingτ_k by its expression, (19) turns into

3 i=1

I_i(r)c₄

r_k(1−₀)e^k+ln(c2/r_k)+k⁻¹ N

ω(r)e⁻^ω(r)/r r²

2/(q−1)

∀rr_k.

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Thusr_kb_k, whereb_k is solution of equation:

c4

rk(1−0)e^k+ln(c2/bk) +k⁻¹ N

ω(b_k)e⁻^ω(b^k^)/b^k b_k²

2/(q−1)

=2M_k⁰=2e⁰^e^k.

From this inequality, using additionally assumption onω(t ), we obtain:c₅e^kω(b_k)/b_kc₆e^k,c₆>0;b_ke⁻^c⁷^k, c₇>0. These inequalities yield:

τ_kc₈

ω

c₉e⁻^k . (20)

Using the definition ofr_k, inequality (18)(ii), the fact that 3M_k⁰cM¯ _k₋1∀kk0(c)¯ (c¯is from (17), 0< 0<e⁻¹), we deduce the main result of first round of computations:

3 i=1

Ii(rk)+f0

rk, τk+k⁻¹ + 2

i=1

Ei

rk, τk+k⁻¹ 3M_k⁰cM¯ k−1. (21)

Next we organize the second round of estimates withµ(τ )=(τ−τk−k⁻¹)/8,rk−1andτk−1be defined similarly as rkandτk, up to the change of indices, using obtained estimate (21) instead of (17). As result we derive:

3 i=1

I_i(r_k₋₁)+f₀

r_k₋₁, τ_k+τ_k₋₁+k⁻¹ + 2

i=1

E_i

r_k₋₁, τ_k+τ_k₋₁+k⁻¹ cM¯ _k₋₂. (22) Fixing arbitraryn > k0(c)¯ and repeating the above described round of computationsk−ntimes, we obtain:

3 i=1

Ii(rn)+f0

rn,

k−n

j=0

τk−j+k⁻¹

+ 2 i=1

Ei

rn,

k−n

j=0

τk−j+k⁻¹

cM¯ n−1, (23)

and, since by inductionτk−j satisfies (20) withkreplaced byk−j, we obtain

k−n

j=0

τk−jc8 k−n

j=0

ω

c9e⁻^(k⁻^{j )} c10 c₉e⁻ⁿ c₉e⁻^k

√ω(s)ds

s . (24)

We denoteτ^∗(n)=limk→∞c8k−n j=0

ω(c9e⁻^(k⁻^{j )}). We derive from (23) by lettingk→ ∞,

sup

0<trn

|x|τ^∗(n)

u²(x, t )dx+

r_n

0

|x|τ^∗(n)

|Du|²+u² dxdtcM¯ _n₋1. (25)

Due to assumption (4)τ^∗(n)→0 asn→ ∞, therefore inequality (25) implies the result.

References

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[3] M. Marcus, L. Véron, Initial trace of positve solutions to semilinear parabolic inequalities, Adv. Nonlinear Stud. 2 (2002) 395–436.

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