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A supnorm estimate for one-dimensional porous medium equations with advection

Juliana S. Ziebell1

Instituto de Matem´atica, Estat´ıstica e F´ısica, FURG, Santo Antˆonio da Patrulha, RS

Lucin´eia Fabris2

Coordenadoria Acadˆemica, UFSM, Cachoeira do Sul, RS

Janaina P. Zingano3

Instituto de Matem´atica, UFRGS, Porto Alegre, RS

Lin´eia Sch¨utz4

Instituto de Matem´atica, UFRGS, Porto Alegre, RS

Abstract. We give a short derivation of supnorm estimates for solutions of one-dimensional porous medium equations of the form

ut+ (f(t, u))x= (|u|αux)x,

assuming initial data u(·,0)Lp0(R)L(R) for some 1p0<∞.

Key-words. Porous Medium Equation, Supnorm Estimate, Comparison Theorem

1 Introduction

There are a number of physical applications where the porous medium equation de- scribes processes involving fluid flow, heat transfer or diffusion [5]. The porous medium equation without advection is given by

∂u

∂t = (|u|m−1u)xx+f, m >1, (1) wheref =f(x, t) is a source term.

Here we consider the following initial-value problem

ut+ (f(t, u))x = (|u|αux)x, x∈R, t >0, u(·,0) =u0 ∈Lp0(R)∩L(R), 1≤p0 <∞,

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1julianaziebell@furg.br

2lucineia.fabris@ufsm.br

3jzingano@mat.ufrgs.br

4lineia.schutz@ufrgs.br

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whereα≥0 andf ∈C1([0,∞)×R) are given. The solutions of (2) are known to be defined for all t > 0 and decay as t → ∞ in several spaces. In this work, we derive a supnorm estimate for the solutions of (2) when consideringu(·,0) inLp(R),p=p0+α/2. By solution in some interval [0, T), 0< T ≤ ∞, we mean a measurable functionu:R×[0, T)→R which is bounded in each strip R×[0, T], 0< T < T, and which solves the equation (2) in distributional sense.

2 Preliminary

An important result to obtain a supnorm estimate also for negative solutions is the following Theorem.

Theorem 2.1. (Theorem of comparison:)

Let u(·, t), v(·, t) solutions of the equation (1), with initial value u0, v0 ∈ L(R), respectively, both defined for 0< t < T and limited in the strip R×[0, T]. Also, if

|f(x, t, u)−f(x, t, v)| ≤Kf(M, T)|u−v|, ∀x∈R,∀t, 0≤t≤T, then

u0(x)≤v0(x)a.e.x∈R ⇒ u(x, t)≤v(x, t)∀x∈R, (3) for all t,0< t≤T.

The proof of this Theorem is in [2].

2.1 Some importants inequalities

The following inequalities will be important througout this work.

• For anyp, q and r such that 0< p≤r≤ ∞, 1≤q≤ ∞:

kwkLr(R) ≤K(r, q, p)kwk˜ 1−Lp(R)θ˜ kwxkθL˜q(R) ∀w∈C01(R), (4) where ˜θ= 1+p·(1−1/q)1−p/r , ˜K(r, q, p) = (2θ)θ˜ andθ= 1+p·(1−1/q)1 .

• ∀ β0>0:

kukL(R)

2 +β0 4

kuk1−θ

Lβ0(R)kuxkθL2(R), (5) whereθ= 1

1+β20. .

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2.2 Basic Result

Theorem 2.2. If u(·, t)∈Lloc([0, T), L(R)) solves problem (2) then 1) u(·, t)∈Lp0(R)∩L(R) ∀t, 0< t < T

2) ku(·, t)kLq(R)≤ ku0kLq(R) ∀t, 0< t < T (∀q, p0 ≤q ≤ ∞)

3) ku(·, t)kLq(R)≤ ku(·, t0)kLq(R) ∀t, 0≤t0< t < T, ∀q, p0 ≤q ≤ ∞.

Proof of (1). For simplicity, we will consider the case of positive solutions, which are known to be smooth. Let ζ ∈C2(R) be such that ζ(x) = 1∀ |x| ≤1, ζ(x) = 0∀ |x| ≥2, 0≤ζ(x)≤1 ∀x∈R. GivenR >0, letζRbe the cut-off function given byζR(x) =ζx

R

. Letp0 ≤q <∞. Multipliyng the PDE at the initial value problem (2) by quq−1ζR(x) we have

∂tuqζR(x) +f(t, u)xquq−1ζR(x) =quq−1(uαux)xζR(x).

Integrating onR×[0, t], Z

|x|<2R

u(x, t)ζR(x)dx+q(q−1) Z t

0

Z

|x|<2R

uq+α−2u2xζR(x)dxdτ = Z

|x|<2R

u0(x)qζR(x)dx

+ q

q+α Z t

0

Z

|x|<2R

uq+αζR′′(x)dxdτ− Z t

0

Z

|x|<2R

f(t, u)xquq−1ζR(x)dxdτ.

Next, integrating by parts and then, lettingR → ∞, we get the result.

Proof of (2) and (3). Again, we consider the simpler case of positive solutions. Defining F(t, U) =

Z U

0

f(t, v)vq−1dv, then equation (4) can be written as Z

|x|<2R

u(x, t)ζR(x)dx+q(q−1) Z t

0

Z

|x|<2R

uq+α−2u2xζR(x)dxdτ = Z

|x|<2R

u0(x)qζR(x)dx

+ q

q+α Z t

0

Z

|x|<2R

uq+αζR′′(x)dxdτ+q Z t

0

Z

|x|<2R

F(t, u)ζR (x)dxdτ.

Observe that Z t

0

Z

R<|x|<2R

F(t, u)ζR (x)dxdτ ≤ Z t

0

Z

R<|x|<2R

|F(t, u)||ζR (x)|dxdτ

≤ M R

Z t

0

Z

R<|x|<2R

|u(x, τ)|qdxdτ→0, when R→ ∞, where M is a constant. Then

Z

R

u(x, t)dx ≤ Z

R

u(x, t)qdx+q(q−1) Z t

0

Z

R

u(x, τ)q+α−2u2xdxdτ

≤ Z

R

u0(x)qdx.

(4)

Therefore, we get

ku(·, t)kLq(R)≤ ku0kLq(R) ∀q, p0 ≤q <∞, ∀t,0< t < T, and

ku(·, t)kL(R)≤ ku0kL(R) ∀t, 0< t < T.

as claimed. In particular, solutions of the initial-value problem (2) are globally defined (i.e.,T =∞).

3 Main Theorems

Theorem 3.1. If u(·, t)∈Lloc([0,∞), L(R)) solves problem (2) with u0>0, then ku(·, t)kL(R)≤K(α, p0)ku(·, t0)k

2p0 α+2p0

Lp0(R)(t−t0)

1

α+2p0, ∀t, 0≤t0 < t, (6) where K(α, p0) is a constant that only depends on α and p0.

Proof. Let ψ∈C1(R) be monotonically increasing such that ψ(u) = 1∀ u≥1,ψ(0) = 0 and ψ(u) =−1, ∀u≤ −1. Takingδ >0, let us define ψδ(u) =ψ(uδ) andφδ(u) =Lδ(u)q, q ≥2, where Lδ(u) =Ru

0 ψδ(v)dv, Lδ ∈ C2(R). Let γ > 0. Multiplying the equation in (2) above by (t−t0)γφδ(u) and integrating inR×[t0, t], we get

Z t t0

Z

R

(τ −t0)γφδ(u(x, τ))u(x, τ)τdxdτ+ Z t

t0

Z

R

(t−t0)γφδ(u(x, τ))(f(τ, u))xdxdτ

= Z t

t0

Z

R

(τ −t0)γφδ(u(x, τ))(|u|αux)xdxdτ

By Fubini’s theorem, integrating by parts, using an appropriate cut-off function and taking δ→0, this gives

(t−t0)γku(x, t)kqLq(R)+q(q−1) Z t

t0

(τ −t0)γ Z

R

|u(x, τ)|α+q−2(ux)2dxdτ

≤ γ Z t

t0

(τ −t0)γ−1ku(x, τ)kqLq(R)dτ Introducing

v[q](x, t) :=

u(x, t) seσ=α+q= 2,

|u(x, t)|σ/2, σ=α+q >2, we then have

(t−t0)γkv[q](·, t)k2q/σ

L2q/σ(R)+ 4q(q−1) (α+q)2

Z t t0

(τ−t0)γkvx|q|(·, τ)k2L2(R)

≤ γ Z t

t0

(τ−t0)γ−1kv[q](·, τ)k2q/σL2q/σ(R)

(5)

Using H¨older, Nirenberg-Sobolev-Gagliardo II (5), with β0 = 2q/σ and q = 2p0, and Nirenberg-Sobolev-Gagliardo I (4) inequalities, we obtain the supnorm estimate (6).

Letw(·, t) be the solution of (2) with initial condition w0 = u+0 +ǫζ for some ǫ >0, where u+0 denotes the positive part of u0 and ζ ∈ C0(R)∩Lp0(R)∩L(R). That is, w0 ≥u0. Then, by the Theorem of Comparison (2.1), u(·, t) ≤w(·, t), for all 0 ≤t < T and

kw(·, t)kL(R)≤K(α, p0)kw(·, t0)k

2p0 α+2p0

Lp0(R)(t−t0)

1

α+2p0, ∀t, 0≤t0 < t, (7) Now let z(·, t) be the solution of (2) with initial condition z0 = −u0 −ǫζ for some ǫ >0, whereu0 denotes the negative part ofu0. That is,z0 ≤u0. Then, by the Theorem of Comparison (2.1),u(·, t)≥w(·, t), for allt, 0≤t < T and

kz(·, t)kL(R)≤K(α, p0)kz(·, t0)k

2p0 α+2p0

Lp0(R)(t−t0)

1

α+2p0, ∀t, 0≤t0 < t, (8) By (7) and (8), we have

ku(·, t)kL(R) ≤K(α, p0) max{ku+0k,ku0k}

2p0 α+2p0

Lp0(R)(t−t0)

1

α+2p0, ∀t, 0≤t0 < t.

This proves the following theorem:

Theorem 3.2. If u(·, t)∈Lloc(R, L(R)) solves problem (2), then ku(·, t)kL(R) ≤K(α, p0) max{ku+0k,ku0k}

2p0 α+2p0

Lp0(R)(t−t0)

1

α+2p0, ∀t, 0≤t0 < t, where K(α, p0) is a constant that only depends on α and p0 and u+0 and u0 denote the positive and negative part of u0, respectively.

4 Conclusions

We derived a supnorm estimate for the solution of the porous medium equation (2) with no restriction on the sign of u0.

Acknowledgments

We thank Prof. Paulo Zingano for comments and fruitful discussions.

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References

[1] E. DiBenedetto.Partial Differential Equations. Birkhauser, Boston, 2000.

[2] L. Fabris. Sobre a existˆencia global e limita¸c˜ao uniforme de solu¸c˜oes da equa¸c˜ao dos meios porosos com termos advectivos arbitr´arios. Tese de Doutorado, PPG- MAT/UFRGS, 2013.

[3] L. Sch¨utz, J. S. Ziebell, J. P. Zingano and P. P. Zingano. A new proof of a fundamen- tal supnorm estimate for one-dimensional advection-diffusion equations.Advances in Differential Equations and Control Processes, Vol. 11, Number 1, pages 41-51. Pushpa Publishing House, India, 2013.

[4] J. M. Urbano.The Method of Intrinsic Scaling. Springer, Berlin, 2000.

[5] J. L. V´azquez. The Porous Medium Equation - Mathematical Theory. Oxford Math- ematical Monographs. Oxford Science Publication, New York, 2007.

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