Emmrich E., Wittbold P. (editors) Analytical and Numerical Aspects of Partial Differential Equations: Notes of a Lecture Series

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Coupling of a scalar conservation law with a parabolic problem 135

Now we use Green’s formula. From (1.2), we obtain

Ω

∇g(u

) · b(x)dxdt = 0.

It follows the estimate

µ|∇u

dxdt ≤

(Ω)

In view of (3.14), the sequence (u

)

µ>0

is bounded in L

∞

(Q). So we can apply

Theorem 1.5 to deduce the existence of a subsequence, still denoted by (u

)

µ>0

, and a

function π ∈ L

∞

((0, 1) × Q) such that

lim

µ→0

ϕ(t, x, u

)ξdxdt =

ϕ(t, x, π)ξdxdtdα (3.17)

for any continuous function ϕ on Q × [m, M ] and any ξ ∈ L

(Q).

Now we choose in (3.15) the test function v = sgn

− k)ϕ

, where ϕ

∈

∞

([0, T )), ϕ

∈ C

∞

(Ω), ϕ

, ϕ

≥ 0, k ∈ R, and integrate over (0, T ). We use a

generalization of Lemma 1.1 (see [18]) in order to handle the evolution term,

h∂

, sgn

− k)ϕ

idt = −

)ϕ

∂

dxdt −

Ω

)ϕ

(0)dx,

where I

) =

sgn

(τ − k)dτ.

For the diffusion term, we ﬁnd

µ∇u

· ∇(sgn

− k)ϕ

dxdt

µ|∇u

sgn

µ−k

)ϕ

dxdt +

µ∇u

· ∇ϕ

sgn

− k)dxdt.

Note that the ﬁrst term on the right-hand side is nonnegative owing to the deﬁnition of

the function sgn

Using Green’s formula in the convection term, we get (due to (1.2))

b(x)f

)∇u

sgn

− k)ϕ

dxdt

b(x)∇F

)ϕ

dxdt = −

b(x)F

)∇ϕ

dxdt,

where

) =

(τ)sgn

(τ − k)dτ.

136 Julien Jimenez

Therefore, we obtain for µ > 0, η > 0

)ϕ

∂

dxdt +

Ω

)ϕ

(0)dx

b(x)F

)∇ϕ

dxdt −

µ∇u

· ∇ϕ

sgn

− k)dxdt ≥ 0. (3.18)

Now, we take ﬁrst the limit with respect to µ in (3.18). From (3.17), we have

lim

µ→0

)ϕ

∂

dxdt =

(π)ϕ

∂

dxdtdα,

and

lim

µ→0

b(x)F

)∇ϕ

dxdt =

b(x)F

(π)∇ϕ

dxdtdα.

Moreover, the Cauchy–Schwarz inequality and Lemma 3.16 provide

lim

µ→0

µ∇u

· ∇ϕ

sgn

− k)dxdt = 0.

In order to take the limit with respect to η, we remark that

lim

η→0

(π) = |π − k| and lim

η→0

(π) = sgn(π − k)(f(π) − f (k)).

Since C

∞

(0, T ) ⊗ C

∞

(Ω) is dense in C

∞

(Q), we may therefore deduce that π satisﬁes

(3.11). From (3.18), we also deduce that

−

{|π − k|ϕ

∂

+ b(x)F(π, k) · ∇ϕ

}dαdxdt ≤

Ω

− k|ϕ

(0)dx.

Then Lemma 3.13 ensures that π fulﬁlls (3.12).

To show that π satisﬁes (3.13), we use the family of boundary entropy-entropy ﬂux

pairs introduced in Example 3.4. We choose in (3.15) the test function

= ∂

, k)ϕ

, δ > 0,

where ϕ

∈ C

∞

(0, T ), ϕ

∈ C

∞

(

Ω), ϕ

, ϕ

≥ 0.

Note that ∂

, k)ϕ

is an element of L

(0, T ; H

(Ω)). For the convection

term, we have

b(x)f(u

) · ∇(∂

, k)ϕ

)dxdt =

, k)b(x) · ∇(ϕ

)dxdt.

Lemma 1.1 is used to transform the evolution term.

Coupling of a scalar conservation law with a parabolic problem 137

As the function τ 7→ H

(τ, ·) is convex, we get for the diffusion term

∇u

· ∇(∂

, k)ϕ

) ≤ µ

∇u

∂

, k)∇ϕ

dxdt.

Therefore, we obtain

−

, k)ϕ

∂

+ J

, k)b(x) · ∇(ϕ

)dxdt

≤ −µ

∇u

∂

, k)∇ϕ

dxdt.

Now, we take the limit µ → 0. On the right-hand side of the foregoing inequality, we

use Lemma 3.16 while on the left-hand side, we use Theorem 1.5. This yields

−

(π, k)ϕ

∂

+ J

(π, k)b(x) · ∇(ϕ

)dxdtdα ≤ 0.

Then Lemma 3.14 ensures that, for any nonnegative β ∈ L

(Σ),

ess lim

s→0

−

(π(α, σ + sν), k)βb(σ) · νdσdα ≥ 0.

Finally, we observe that the sequence (J

)

δ>0

uniformly converges towards F

as δ

goes to 0

. Thus, π fulﬁlls (3.13).

We conclude that there exists a unique entropy process solution to (3.1). By Theo-

rem 3.12, it follows that there exists a unique entropy solution u ∈ L

∞

(Q) to (3.1).

4 The coupled problem

In this part, we prove an existence and uniqueness result for (1.1). To this end, we

need that the function f is nondecreasing. This implies F (a, b) = |f(a) − f(b)| for

a, b ∈ R. We also have F

(a, b) =

(|f(a) − f (0)| − |f(b) − f(0)| + |f(a) − f(b)|) so

that F

(a, b) ≥ 0.

Moreover, we suppose that b · ν

is nonnegative along the interface Γ

. More

precisely, let

⊂ {σ ∈ Γ

; b(σ) · ν

≥ 0}. (4.1)

Remark 4.1. The aforementioned additional assumptions on f and b ensure that the

interface is included in the set of outwards characteristics for the ﬁrst-order differential

operator posed in the hyperbolic domain. This is a key point that will allow us to prove

uniqueness by considering ﬁrst the behaviour of a solution on the hyperbolic area and

then on the parabolic one.

As meas(Γ

\ Γ

) 6= 0, the function space

V = {v ∈ H

(Ω

); v = 0 a.e. on Γ

\ Γ

}

138 Julien Jimenez

is a Hilbert space when equipped with the standard H

-inner product. The norm in V

is, therefore, still k · k. We denote by h·, ·i

×V

the duality pairing between V and V

We provide a deﬁnition of a weak entropy solution to (1.1) by observing that the

equation set in Q can be viewed as a quasilinear parabolic equation that strongly degen-

erates on a ﬁxed subdomain. We propose a weak formulation relying on a global en-

tropy inequality on the whole space-time cylinder Q, inspired by the one given in [1, 3].

Deﬁnition 4.2. A function u : Q → R is said to be a weak entropy solution to (1.1) if

(i) u ∈ L

∞

(Q), φ(u) ∈ L

(0, T ; V ),

(ii) for all nonnegative ϕ ∈ C

∞

(Q) and all k ∈ R

(|u − k|∂

ϕ + (b(x)F (u, k) − χ

Ω

∇|φ(u) − φ(k)|) · ∇ϕ)dxdt ≥ 0, (4.2)

(iii)

esslim

t→0

Ω

|u(t, x) − u

(x)|dx = 0, (4.3)

(iv) for all nonnegative ϕ ∈ L

(Σ

\ Σ

) and all k ∈ R

ess lim

s→0

−

\Σ

(u(σ + τν

), k)b(σ) · ν

dσ ≥ 0. (4.4)

Remark 4.3. (i) One can choose in (4.2) successively k > kuk

∞

and k < −kuk

∞

, and

compare the two resulting inequalities. We remark that all k-dependent terms vanish.

Therefore, we ﬁnd

(u∂

ϕ + b(x)f(u) − χ

Ω

∇φ(u)) · ∇ϕdxdt = 0 (4.5)

for all ϕ ∈ H

(Q).

(ii) From (4.2), we deduce that for all nonnegative ϕ ∈ C

∞

((0, T ) × Ω

)

(|u − k|∂

ϕ + |f(u) − f (k)|b(x) · ∇ϕ)dxdt ≥ 0. (4.6)

4.1 Uniqueness

The inequalities (4.2) and (4.4) are the starting point to establish the Lipschitz contin-

uous dependence, measured in L

(Ω

), of the weak entropy solution to (1.1) on the

corresponding initial data.

Lemma 4.4. Let u ∈ L

∞

(Q) satisfy (4.2) and (4.4). We then have for all nonnegative

ϕ ∈ L

(Σ

) and all k ∈ R

ess lim

s→0

−

(u(σ + τν

), k)b(σ) · ν

dσ ≥ 0. (4.7)

Coupling of a scalar conservation law with a parabolic problem 139

Proof. Let ϕ ∈ L

(Σ

) with ϕ ≥ 0. Since u satisﬁes (4.6), we can apply Lemma 3.8

to conclude that

ess lim

s→0

−

|f(u(σ + sν

)) − f(k)|b(σ) · ν

ϕdσ exists. (4.8)

This in turn implies that

ess lim

s→0

−

(u(σ + sν

), k)b(σ) · ν

ϕdσ exists.

Moreover, for any s < 0, we have

(u(σ + sν

), k)b(σ) · ν

ϕdσ

\Σ

(u(σ + sν

), k)b(σ) · ν

ϕdσ +

(u(σ + sν

), k)b(σ) · ν

ϕdσ.

Since F

(·, ·) is nonnegative, we have, according to (4.1),

(u(σ + sν

, k)b(σ) · ν

ϕdσ ≥ 0.

We use (4.4) to conclude that (4.7) is fulﬁlled.

We are now able to state uniqueness on the hyperbolic zone.

Theorem 4.5. Let u

and u

be two weak entropy solutions to (1.1) for initial data u

0,1

and u

0,2

, respectively. Then, for almost all t ∈ (0, T ),

Ω

(t, x) − u

(t, x)|dx ≤

Ω

0,1

(x) − u

0,2

(x)|dx.

Proof. Since (4.6) and (4.7) hold, one can apply Theorem 3.7.

We focus now on the parabolic zone. On Q

, we characterize a solution to (1.1)

through a variational equality including the contributions from the hyperbolic part that

enter the parabolic zone.

Proposition 4.6. Let u be a weak entropy solution to (1.1). Then

∂

u ∈ L

(0, T ; V

Furthermore, there holds for all ϕ ∈ L

(0, T ; V )

h∂

u, ϕi

×V

dt +

(∇φ(u) − b(x)f(u)) · ∇ϕdxdt

− ess lim

s→0

−

f(u(σ + sν

))b(σ) · ν

ϕdσ = 0. (4.9)

140 Julien Jimenez

Proof. By virtue of Remark 4.3, u satisﬁes (4.5). Since D(0, T ; H

(Ω)) ⊂ H

(Q),

equation (4.5) is true for ϕ ∈ D(0, T ; H

(Ω)). Let ζ ∈ D(0, T ; V ). We consider an

extension

ζ ∈ D(0, T ; H

(Ω)) of ζ. We also introduce a sequence (ξ

)

ε>0

such that

∈ W

1,∞

(Ω) with 0 ≤ ξ

≤ 1,

(x) =

(

1 if x ∈ Ω

0 if x ∈ Ω

, dist(x, Γ

) ≥ ε,

and εk∇

∞

≤ C for some C > 0 independent of ε. Then, we take ϕ =

ζξ

in (4.5)

and pass to the limit as ε → 0

. We use (4.8) to show that

lim

ε→0

f(u)

ζb(x) · ∇ξ

dxdt = ess lim

s→0

−

f(u(σ + sν

))ζb(σ) · ν

dσdt.

There is no difﬁculty to take the limit with respect to ε in the other terms of (4.5), and

we obtain thereby for all ζ ∈ D(0, T ; V )

u∂

ζdxdt =

(∇φ(u) − f(u)b(x)) · ∇ζdxdt

− ess lim

s→0

−

f(u(σ + sν

))ζb(σ) · ν

dσdt.

Since u is bounded and φ(u) ∈ L

(0, T ; V ), there exists C

> 0 such that



(∇φ(u) − f(u)b(x)) · ∇ζdxdt



≤ C

kζk

(0,T ;V )

The continuity of the trace operator as a mapping of H

(Ω

) into L

(Γ

) ensures, as

long as u is bounded, the existence of a constant C

> 0 such that, for almost all s < 0,



f(u(σ + sν

))b(σ) · ν

ζdσdt



≤ C

kζk

(0,T ;V )

so that



ess lim

s→0

−

f(u(σ + sν

))b(σ) · ν

ζdσdt



≤ C

kζk

(0,T ;V )

Therefore, there exists a constant C > 0 such that for all ζ ∈ D(0, T ; V )



u∂

ζdxdt



≤ Ckζk

(0,T ;V )

Thus, we have ∂

u ∈ L

(0, T ; V

), and for any ζ ∈ D(0, T ; V )

−

u∂

ζdxdt =

h∂

u, ζi

×V

dt.

The assertion follows by density of D(0, T ; V ) in L

(0, T ; V ).

Coupling of a scalar conservation law with a parabolic problem 141

We turn now to the uniqueness result. Let u

and u

be two weak entropy solutions

to (1.1). We suppose that u

and u

have the same initial data on the hyperbolic part.

In view of Theorem 4.5, we already know that u

= u

a.e. on Q

. On the parabolic

zone, the Holmgren-type duality method, described in Section 2.2, cannot be applied

as we lack information about the traces of u

and u

along Σ

. For this reason, we use

the method of doubling only the time variable. Moreover, to deal with the convection

term, we suppose that f ◦ φ

−1

is Hölder continuous on R with an exponent greater or

equal than 1/2, i.e., we assume there exist θ ∈ [1/2, 1], K > 0 such that for all x, y ∈ R

|(f ◦ φ

−1

)(x) − (f ◦ φ

−1

)(y)| ≤ K|x − y|

. (4.10)

Theorem 4.7. Under the assumption (4.10), (1.1) admits at most one weak entropy

solution. Moreover, if u

and u

are two weak entropy solutions corresponding to

initial data u

0,1

and u

0,2

with u

0,1

= u

0,2

a.e. on Ω

then, for almost all t ∈ (0, T ),

Ω

(t, x) − u

(t, x)|dx ≤

Ω

0,1

(x) − u

0,2

(x)|dx.

Proof. For any j ∈ N and t,

t ∈ (0, T ) , we set

(t,

t) = γ



t +





t −



where γ ∈ C

∞

(0, T ), γ ≥ 0, and (ρ

)

is a sequence of molliﬁers. We notice that

≥ 0 and, for j large enough, α

∈ C

∞

((0, T ) × (0, T )).

In (4.9), written for the weak entropy solution u

and in variables (t, x), we choose

the test function

ϕ(t,

t, x) = sgn

(φ(u

(t, x)) − φ(u

(

t, x)))α

(t,

and integrate with respect to

t. In (4.9), written for the weak entropy solution u

and in

variables (

t, x), we take the same test function and integrate with respect to t. Taking

then the difference yields

h∂

− ∂

˜u

, sgn

(φ(u

) − φ( ˜u

))i

×V

dtd

]0,T [×Q

∇(φ(u

) − φ( ˜u

)) · ∇sgn

(φ(u

) − φ( ˜u

))α

dxdtd

−

]0,T [×Q

(f(u

) − f( ˜u

))b(x) · ∇sgn

(φ(u

) − φ( ˜u

))α

dxdtd

ess lim

s→0

−

f(u

(σ + sν

))b(σ) · ν

sgn

(φ(u

) − φ( ˜u

))α

dσd

−

ess lim

s→0

−

f(u

(˜σ + sν

))b(σ) · ν

sgn

(φ(u

) − φ( ˜u

))α

d ˜σdt.

(4.11)

142 Julien Jimenez

To simplify the notation, we add a tilde to any function in

t. We want to pass to the

limit in (4.11), ﬁrst as η goes to 0

and then as j tends to ∞.

In the ﬁrst integral on the left-hand side, we use an integration-by-parts formula

based on a generalization of Lemma 1.1 (see [18]) to obtain

h∂

− ∂

˜u

, sgn

(φ(u

) − φ( ˜u

))i

×V

dtd

= −



˜u

sgn

(φ(r) − φ( ˜u

))dr



∂



dxdtd

−



˜u

sgn

(φ(u

) − φ(r))dr



∂



dxdtd

In the third integral of (4.11), we write f(u

) = f ◦ φ

−1

(φ(u

)), i = 1, 2. Then, due to

(4.10) and Young’s inequality, we derive

−

(f(u

) − f( ˜u

))b(x) · ∇sgn

(φ(u

) − φ( ˜u

))α

dxdtd

≤

kbk

∞

]0,T [×Q

|φ(u

) − φ( ˜u

2θ

sgn

(φ(u

) − φ( ˜u

))α

dxdtd

sgn

(φ(u

) − φ( ˜u

))|∇(φ(u

) − φ( ˜u

))|

dxdtd

By deﬁnition of sgn

, the ﬁrst term on the right-hand side of the previous inequality is

bounded by

|φ(u

) − φ( ˜u

2θ−1

{−η<φ(u

)−φ( ˜u

)<η}

dxdtd

where C > 0. As θ ≥

, we obtain

lim

η→0

|φ(u

) − φ( ˜u

2θ−1

{−η<φ(u

)−φ( ˜u

)<η}

dxdtd

t = 0.

Let us now study the right-hand side of (4.11). Since u

= u

a.e. on Q

, we have

(σ + sν

) = u

(σ + sν

) for almost all negative s. Thus we may refer to (4.8) to

show (as in the proof of Lemma 3.10) the existence of θ ∈ L

∞

(Σ

) such that for any

β ∈ L

(Σ

)

ess lim

s→0

−

f(u

(σ + sν

))b(σ) · ν

β(σ)dtdσ =

θ(σ)β(σ)dtdσ

and

ess lim

s→0

−

f(u

(˜σ + sν

))b(σ) · ν

β( ˜σ)dtdσ =

θ( ˜σ)β( ˜σ)dtd ˜σ.

Coupling of a scalar conservation law with a parabolic problem 143

Therefore, the right-hand side of (4.11) is equal to

(θ(t, σ) − θ(

t, σ))sgn

(φ(u

)(σ) − φ(u

)(

t, σ))α

(

t, t)dtdσd

Finally, when η goes to 0

in (4.11), by Lebesgue’s theorem on dominated conver-

gence, we obtain

−

(0,T )×Q

− ˜u

|(∂

+ ∂

)dxdtd

t ≤

|θ(t, σ) − θ(

t, σ)|α

dtdσd

With the deﬁnition of α

, we see that ∂

+∂

= γ

(

)ρ

(t −

t). Now we are able

to take the limit with respect to j and ﬁnd

−

− u

|γ

(t)dxdt ≤ 0.

The conclusion follows by choosing a suitable test function γ (see the proof of Theo-

rem 3.7).

4.2 Existence

We approximate a weak entropy solution to (1.1) through a sequence of solutions to

viscous problems linked to (1.1) by adding a diffusion term in accordance with the pro-

posed physical modelling of two layers in the subsoil with different geological charac-

teristics. Thus, for any positive real number µ, we are ﬁrst interested in the uniqueness

and existence of a measurable and essentially bounded function u

satisfying in the

sense of distributions











∂

+ ∇ · (b(x)f(u

)) = ∇ · (λ

(x)∇φ(u

)) in Q,

(0, ·) = u

on Ω,

= 0 on Σ,

(4.12)

with

(x) = χ

Ω

(x) + µχ

Ω

(x).

We state ﬁrst

Proposition 4.8. There exists a unique weak solution u

∈ W (0, T ) to (4.12), i.e.,

m ≤ u

(t, x) ≤ M for almost all (t, x) ∈ Q, (4.13)

for almost all t ∈ (0, T ) and all v ∈ H

(Ω)

h∂

, vi +

Ω

(λ

(x)∇φ(u

) − b(x)f(u

)) · ∇vdx = 0, (4.14)

and

(0) = u

in L

(Ω).

144 Julien Jimenez

Proof. We remark that, for any ﬁxed µ the partial differential operator is not degener-

ated. So one may apply the Schauder–Tikhonov ﬁxed point theorem and the Holmgren-

type duality method presented in Section 2.

We look now for a priori estimates fulﬁlled by the sequence (u

)

µ>0

in order to

study its limit as µ tends to 0

Proposition 4.9. There exists a constant C > 0, independent of µ, such that

kλ

1/2

∇

φ(u

(Q)

≤ C, (4.15)

k∂

(0,T ;H

−1

(Ω))

≤ C, (4.16)

where

φ(u

) =

(τ)dτ.

Proof. We choose v = u

in (4.14) and integrate over (0, T ). We observe that

h∂

, u

idt =

(T )k

(Ω)

−

(Ω)

and

Ω

f(u

)b(x) · ∇u

dxdt = 0.

The diffusion term can be written as

(λ

(x)

1/2

)∇u

)

dxdt = kλ

1/2

∇

φ(u

(Q)

and (4.15) follows. The second assertion then follows from the ﬁrst one together with

the equation.

Our aim is now to describe the behaviour of the sequence (u

)

µ>0

when µ goes

to 0

. On the hyperbolic domain we take advantage of (4.13) and Theorem 1.5. On

the parabolic area, estimates (4.13), (4.15) and (4.16) are not sufﬁcient to study the

behaviour of (u

)

µ>0

. For this reason, we use an additional assumption on φ:

−1

is Hölder continuous with an exponent τ ∈ (0, 1). (4.17)

Proposition 4.10. Under (4.17), there exists a function u ∈ L

∞

(Q) with φ(u) in

(0, T ; V ), and a subsequence, still denoted by (u

)

µ>0

, such that

∗

* u in L

∞

(Q), u

→ u in L

) for any 1 ≤ q < ∞ and a.e. on Q

Moreover, we also have

∇φ(u

) * ∇φ(u) in L

)

, µ∇φ(u

) → 0 in L

)

In order to prove this proposition, we need the following lemma: