Linear stability of contact discontinuities for the nonisentropic - - PDF document

Linear stability of contact discontinuities for the nonisentropic Euler equations in two space dimensions

Alessandro Morando, Paola Trebeschi Department of Mathematics University of Brescia Lausanne, September 19-22, 2006

• 1. The Euler equations

We study the nonisentropic Euler equations of a perfect polytropic ideal gas in the plane R2: ∂tp + u · ∇p + γp∇ · u = 0, ρ(∂tu + (u · ∇)u) + ∇p = 0, ∂tS + u · ∇S = 0. (1) x = (x1, x2) ∈ R2, p = p(t, x) ∈ R the pressure,

u(t, x) ∈ R2 the velocity,

S = S(t, x) ∈ R the entropy, ρ is the density, obeying the constitutive law ρ(p, S) = Ap

1 γe− S γ

with A > 0 given, γ > 1 adiabatic number. Problem: Stability of “contact discontinuities” Result: Under a “supersonic” condition (that precludes violent instability) we prove an ENERGY ESTIMATE for the linearized problem.

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• 2. Discontinuities

2.1. Contact discontinuities Let Γ := {x2 = ϕ(t, x1)} be a smooth hypersurphace and (p, u, S) :=

• (p+, u+, S+)

if x2 > ϕ(t, x1) (p−, u−, S−) if x2 < ϕ(t, x1), a smooth function on either side of Γ. Definition 1. (p, u, S) is a contact discontinuity solution

• f (1) if it is a classical solution of (1) on both sides of

Γ and satisfies the Rankine-Hugoniot jump conditions at Γ: ∂tϕ = u+ · ν = u− · ν , p+ = p− , where ν := (−∂x1ϕ, 1) is a (space) normal vector to Γ. These conditions yield that

• the normal velocity and pressure are continuous

across the interface Γ,

• the only jumps experimented by the solution con-

cern the tangential velocity and the entropy. Thus, a contact discontinuity is a vortex sheet.

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2.2. Planar contact discontinuities A planar contact discontinuity is a piecewise constant solution to (1) (p, u, S) =

• (pr, ur, Sr),

if x2 > σt + nx1, (pl, ul, Sl), if x2 < σt + nx1.

ur,l = (vr,l, ur,l)T are fixed vectors in R2, pr,l > 0, Sr,l, σ, n

are fixed real numbers.

• The previous quantities are related by the Rankine-

Hugoniot jump conditions σ + vrn − ur = 0, σ + vln − ul = 0, pr = pl =: p.

• Without loss of generality, we may assume

σ = ur = ul = 0 , n = 0 , vr + vl = 0 (vr = 0). This corresponds to the following Ur,l = (p, vr,l, 0, Sr,l)T, with vr + vl = 0.

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2.3. Stability of nonisentropic planar contact dis- continuities in dimension d = 2, 3 (Fejer-Miles, 1958-1963 and Coulombel-Morando 2004) Consider a planar contact discontinuity Ur,l = (p, vr,l, 0, Sr,l)T with vr + vl = 0 (vr,l are the tangential velocities) and linearize the Euler equations and the jump conditions around this solution.

• if d = 3, the linearized equations do not satisfy the

Lopatinskii condition ⇒ violent instability;

• if d = 2 and |vr−vl|

2

< 1

2

• c2/3

r

+ c2/3

l

3/2

the linearized equations do not satisfy the Lopatinskii condition ⇒ violent instability;

• if d = 2 and |vr−vl|

2

> 1

2

• c2/3

r

+ c2/3

l

3/2

the linearized equations satisfy the weak Lopatinskii condition ⇒ weak stability, c(ρ) :=

• p′(ρ) is the sound speed and cr, cl are the con-

stant values of c(ρ) in both sides of Γ.

• In any case, the uniform Kreiss-Lopatinskii condition

is never satisfied.

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• 3. Energy estimates in two space dimensions

3.1. Reformulation of the problem in a fixed do- main The interface Γ := {x2 = ϕ(t, x1)} is unknown so that the problem is a free boundary problem.

• In order to work in a fixed domain we introduce the

change of variables (τ, y1, y2) → (t, x1, x2), (t, x1) = (τ, y1), x2 = Φ(τ, y1, y2), where Φ : R3 → R, Φ(τ, y1, 0) = ϕ(τ, y1), ∂y2Φ(τ, y1, y2) ≥ κ > 0.

• We define the new unknowns

(p+

♯ , u+ ♯ , S+ ♯ )(τ, y1, y2) := (p, u, S)(τ, y1, Φ(τ, y1, y2)) ,

(p−

♯ , u− ♯ , S− ♯ )(τ, y1, y2) := (p, u, S)(τ, y1, Φ(t, y1, −y2)) .

The functions p±

♯ , u± ♯ , S± ♯ are smooth on the fixed domain

{y2 > 0}.

• For convenience, we drop the ♯ index and only keep

the + and − exponents.

• Finally, we write (t, x1, x2) instead of (τ, y1, y2).

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• Let us set u = (v, u). The existence of compressible

vortex sheets amounts to prove the existence of smooth solutions of the first order system ∂tp+ + v+∂x1p+ +

• u+ − ∂tΦ+ − v+∂x1Φ+ ∂x2p+

∂x2Φ+

+γp+∂x1v+ + γp+ ∂x2u+

∂x2Φ+ − γp+∂x1Φ+ ∂x2Φ+∂x2v+ = 0,

∂tv+ + v+∂x1v+ +

• u+ − ∂tΦ+ − v+∂x1Φ+ ∂x2v+

∂x2Φ+

+ 1

ρ+∂x1p+ − 1 ρ+ ∂x1Φ+ ∂x2Φ+∂x2p+ = 0,

∂tu+ + v+∂x1u+ + u+ − ∂tΦ+ − v+∂x1Φ+ ∂x2u+

∂x2Φ+

+ 1

ρ+ ∂x2p+ ∂x2Φ+ = 0,

∂tS+ + v+∂x1S+ +

• u+ − ∂tΦ+ − v+∂x1Φ+ ∂x2S+

∂x2Φ+ = 0,

in the fixed domain {x2 > 0}, where Φ±(t, x1, x2) := Φ(t, x1, ±x2) ((p−, v−, u−, S−, Φ−) must solve a similar system ) fulfilling the boundary conditions Φ+

|x2=0 = Φ− |x2=0 = ϕ,

∂tϕ = −v+

|x2=0∂x1ϕ + u+ |x2=0 = −v− |x2=0∂x1ϕ + u− |x2=0,

p+

|x2=0 = p− |x2=0.

The functions Φ± should also satisfy ∂x2Φ+(t, x1, x2) ≥ k, ∂x2Φ−(t, x1, x2) ≤ −k.

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• The equations are not sufficient to determine the un-

knowns U ± := (p±, v±, u±, S±) and Φ±: another equation relating Φ, U and ϕ is needed in order to close the sys- tem. We may prescribe that Φ± solve in the domain {x2 > 0} the eikonal equations ∂tΦ± + v±∂x1Φ± − u± = 0 . With this choice the boundary matrix of the system for U ± has constant rank in the whole domain {x2 ≥ 0} and not only on the boundary {x2 = 0}.

• Matrix form of the system

For all U = (p, v, u, S)T, we get the system (in the interior {x2 > 0}) L(U +, ∇Φ+)U + = 0, L(U −, ∇Φ−)U − = 0, ∂tΦ± + v±∂x1Φ± − u± = 0 where L(U +, ∇Φ+)U + = ∂tU + + A1(U +)∂x1U ++ +

1 ∂x2Φ+

• A2(U +) − ∂tΦ+I4 − ∂x1Φ+A1(U +)
• ∂x2U +

with the boundary conditions (on {x2 = 0}) Φ+

|x2=0 = Φ− |x2=0 = ϕ ,

∂tϕ = −v+

|x2=0 ∂x1ϕ + u+ |x2=0 = −v− |x2=0 ∂x1ϕ + u− |x2=0

p+

|x2=0 = p− |x2=0.

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3.2. The linearized equations

• Consider a planar contact discontinuity

Ur =

  

p vr Sr

   ,

Ul =

  

p vl Sl

   ,

Φr,l = ±x2. We assume that vr + vl = 0, vr > 0.

• Let us consider

Ur,l + ε ˙ U±, ±x2 + ε ˙ ψ± where we have set ˙ U± = ( ˙ p±, ˙

u±, ˙

S±), ˙ ψ± is a small perturbation

• f the exact solution Ur,l, Φr,l.
• Up to second order, ˙

U± = ( ˙ p±, ˙

u±, ˙

S±)T must solve the system L′ ˙ U = 0, in {x2 > 0}, B( ˙ U, ψ) = 0, on {x2 = 0}, where we have set for shortness ˙ U := ( ˙ U+, ˙ U−)T, L′ ˙ U := ∂t ˙ U+

• A1(Ur)

A1(Ul)

• ∂x1 ˙

U+

• A2(Ur)

−A2(Ul)

• ∂x2 ˙

U, and B( ˙ U, ψ) :=

 

(vr − vl)∂x1ψ − ( ˙ u+ − ˙ u−) ∂tψ + vr∂x1ψ − ˙ u+ p+ − p−

  .

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• We have to introduce source terms; we study

L′ ˙ U = f in {x2 > 0}, B( ˙ U, ψ) = g, on {x2 = 0}

• It is useful make another linear change of unknowns

W1 := ˙ v+, W2 := 1

2

• − ˙

p+ γp + ˙ u+ cr

• W3 := 1

2

• ˙

p+ γp + ˙ u+ cr

• , W4 := ˙

S+, W5 := ˙ v−, W6 := 1

2

• − ˙

p− γp + ˙ u− cl

• W7 := 1

2

• ˙

p− γp + ˙ u− cl

• , W8 := ˙

S−

• This change of unknowns W transform the system

L′ ˙ U = f in a symmetric hyperbolic form LW = f, in {x2 > 0} B(W nc, ψ) = g, on {x2 = 0}, with new data f, g, and LW := A0∂tW + A1∂x1W + A2∂x2W B(W nc, ψ) := MW nc + b

• ∂tψ

∂x1ψ

• .

W := (W1, W2, W3, W4, W5, W6, W7, W8)T, W c := (W1, W4, W5, W8)T, W nc := (W2, W3, W6, W7)T = (linear combination of p, u) M :=

 

−cr −cr cl cl −cr −cr −1 1 1 −1

  ,

b :=

 

vr − vl 1 vr

  .

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• We want to find an L2 a priori estimate of the solution

to the linearized problem LW = f, in {x2 > 0} B(W nc, ψ) = g, on {x2 = 0}, in Ω := {(t, x1, x2) ∈ R3 s.t. x2 > 0} = R2 × R+ . The boundary ∂Ω = {x2 = 0} is identified to R2. 3.3. The functional setting

• Define

Hs

γ(R2) := {u ∈ D′(R2) s.t.

exp(−γt)u ∈ Hs(R2)} , equipped with the norm uHs

γ(R2) := exp(−γt)uHs(R2) .

• Define similarly the space Hs

γ(Ω).

• The space L2(R+; Hs

γ(R2) is the space of all functions

v = v(t, x1, x2) in Ω such that the following norm |||v|||2

L2(Hs

γ) :=

+∞

v(·, x2)2

Hs

γ(R2)dx2

is finite.

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3.4. The main result. Theorem 1. Let (Ur,l, Φr,l) be a planar contact discon- tinuity (unperturbed solution). i) If vr − vl >

• c

2 3

r + c

2 3

l

3

2

and vr − vl = √ 2 (cr + cl) then there exists C > 0 such that for all γ ≥ 1 and all (W, ψ) ∈ H2

γ (Ω) × H2 γ (R2)

γ|||W|||2

L2

γ(Ω) + W nc

|x2=02 L2

γ(R2) + ψ2

H1

γ (R2)

≤ C

• 1

γ3|||LW|||2 L2(H1

γ ) + 1

γ2B(W nc, ψ)2 H1

γ (R2)

• .

ii) If vr − vl = √ 2 (cr + cl) then there exists C > 0 such that for all γ ≥ 1 and all (W, ψ) ∈ H3

γ (Ω) × H3 γ (R2)

γ|||W|||2

L2

γ(Ω) + W nc

|x2=02 L2

γ(R2) + ψ2

H1

γ (R2)

≤ C

• 1

γ5|||LW|||2 L2(H2

γ ) + 1

γ4B(W nc, ψ)2 H2

γ (R2)

• .
• If i): loss of one derivative for W; no loss of deriva-

tives for the front function ψ;

• If ii):

loss of two derivatives for W; loss of one derivative for the front function ψ;

• Only the trace of the non-characteristic part of the

solution can be controlled at the boundary since the problem is characteristic. The loss of control regards the tangential velocity and the entropy.

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Further comments

• If we are in the “isentropic case”

cr = cl := c both values (c

2 3

r + c

2 3

l )

3 2 and

√ 2(cr + cl) entailed in i),ii) actually are reduced to 2 √ 2c. In this case, the assumption vr − vl > 2 √ 2c in i) prevents the

• ccurrence of ii);
• If we are in in the “full nonisentropic” situation

cr = cl the value √ 2 (cr + cl) is strictly greater than (c

2 3

r + c

2 3

l )

3 2 so that ii) has to be accounted.

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• 4. Main steps of the proof

4.1. Reformulating the problem Step 1. It is convenient rewriting the system in terms of the new unknown W := exp(−γt)W and ψ := exp(−γt)ψ. Then we consider the following system Lγ W := γA0 W + L W = exp(−γt)f, Bγ( W nc, ψ) := M W nc

|x2=0 + b

• γ

ψ + ∂t ψ ∂x1 ψ

• = exp(−γt)g.

Step 2. It is enough to find an energy estimate for the problem (we drop “tilde”) LγW := 0, Bγ(W nc, ψ) := G (2) where G = G(W nc, ψ, f). Since Aj are symmetric and A0 is positive definite, tak- ing the scalar product of (2)1 with W we get γ|||W|||2

0 ≤ C||W nc |x2=0||2 0.

• Hence it remains to find an estimate for W nc

|x2=0 and the

front ψ.

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4.2. Eliminating the front ψ Step 3. Let W, ψ satisfy

• LγW := 0,

Bγ(W nc, ψ) := G.

• Perform a Fourier transform in (t, x1). Let us denote

the dual variables by (δ, η) and write also τ := δ + iγ.

• The Fourier transform (

W , ψ) of (W, ψ) must solve the following system of algebraic-differential equations (τA0 + iηA1) W + A2 d W dx2 = 0, if x2 > 0 (3) b(τ, η) ψ + M W nc(0) = G, (4) with b(τ, η) defined by b(τ, η) := b

• τ

iη

• =

 

2iηvr τ + iηvr

  .

• It can be shown that b(τ, η) satisfies an elliptic bound,

then we find ψ2

1,γ ≤ C

• W nc

|x2=02 0 + G2

• As a conseguence of the ellipticity of b the boundary

condition (4) can be written as β(τ, η) W nc(0) = h where the front ψ disappears.

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• We focus on the following problem

(τA0 + iηA1) W + A2d

W dx2 = 0,

x2 > 0, β(τ, η) W nc(0) = h. (5) where the new boundary condition involves only the non characteristic part of W. Step 4. Because of the characteristic boundary some equations in (5) do not entail differentiation with respect to x2. We are led to the ODE system

d W nc dx2 = A(τ, η)

W nc, x2 > 0, β(τ, η) W nc = h, x2 = 0, (6) Step 5. We have to prove an energy estimate for the problem (6).

• The technique of the proof is the construction of a

degenerate Kreiss symmetrizer.

• Under the supersonic assumptions

vr − vl > (c

2 3

r + c

2 3

l )

3 2

the problem satisfies the Kreiss Lopatinskii condition in a weak sense; in other words, the Lopatinskii deter- minant associated to (6) vanishes at some boundary frequencies (τ, η) with ℜτ = 0.

• The failure of the uniform Kreiss-Lopatinskii condition

yields a loss of derivatives with respect to the source terms: the loss is strictly related to the order of vanish- ing of the Lopatinskii determinant.

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4.3. Roots of the Lopatinskii determinant Let ∆(τ, η) be the Lopatinskii determinant associated to (6). Let Σ := {(τ, η) ∈ C × R : |τ|2 + v2

r η2 = 1, ℜτ ≥ 0}

be the unit hemi-sphere in the frequency space. Proposition 1. Assume that vr > 1 2(c

2 3

r + c

2 3

l )

3 2,

vr + vl = 0.

• a. If cr = cl =: c (isentropic case) then there exists a

positive number V1 such that for every (τ, η) ∈ Σ, ∆(τ, η) = 0 if and only if τ = 0

• r

τ = ±iV1η. Each of the preceding roots of ∆ is simple; namely if (τ0, η0) is anyone of the points above there exists an open neighborhood V of (τ0, η0) in Σ and a C∞ function h defined on V such that for all (τ, η) ∈ V: ∆(τ, η) = (τ − τ0)h(τ, η), and h(τ0, η0) = 0.

• b. If cr = cl (nonisentropic case) then there exist two

positive numbers X2, X3 satisfying cr − vr < crX2 < crX3 < −cl + vr, (7)

16

such that ∆(τ, η) = 0, for (τ, η) ∈ Σ, if, and only if, τ = iqvrη

• r

τ = icrX2η

• r

τ = icrX3η, (8) where q := cr−cl

cr+cl.

For vr = cr+cl

√ 2 , each of the preceding roots of ∆ is simple

and the same situation occurs as in the isentropic case a. On the contrary, when vr = cr+cl

√ 2

• ne (and only one) of

the two identities below holds true qvr = crX2

• r

qvr = crX3. Hence each of the roots (iqvrη, η) ∈ Σ of ∆ is double. More precisely, this means that to every point (iqvrη0, η0) (with η2

0(1−q2v2 r ) = 1) there correspond an open neigh-

borhood V in Σ and a C∞ function h on V such that ∆(τ, η) = (τ − iqvrη0)2h(τ, η), ∀(τ, η) ∈ V and h(iqvrη0, η0) = 0. The other root of ∆ remains simple, as in the case of vr = cr+cl

√ 2 .

4.4. Construction of the Kreiss symmetrizer The construction of the symmetrizer is microlocal and is performed in a neighborhood of different classes of frequency points (τ, η).

• a. interior points (τ, η) : ℜτ > 0. Here standard Kreiss

symmetrizer exists and L2-estimate without loss of derivatives is obtained.

• b. boundary points (τ, η) : ℜτ = 0. We need to distin-

guish the following classes: b1 Points where A(τ, η) is diagonalizable and the Kreiss-Lopatinskii condition is satisfied ⇒ clas- sical Kreiss techniques apply and give an L2 es- timate with no loss of derivatives. b2 Points where A(τ, η) is diagonalizable and the Lopatinskii condition breaks down (i.e. ∆(τ, η) = 0) ⇒ construction of a degenerate Kreiss’ sym- metrizer: this yields an L2 estimate with loss of

• derivatives. The loss of derivatives depends on

the multiplicity of the roots of the Lopatinskii determinat ∆:

• roots of multiplicity 1 give a loss of 1 deriva-

tive

• roots of multiplicity 2 give a loss of 2 deriva-

tives

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b3 Points where A(τ, η) is not diagonalizable. In those points, the Lopatinskii condition is sat- isfied (i.e. ∆(τ, η) = 0). The construction of the simmetrizer follows as in case a. and an L2-energy estimate without loss of derivatives is obtained. b4 Poles of A(τ, η). At those points, the Lopatin- skii condition is satisfied. We construct a sym- metrizer by working on the original system (5) and the energy estimate is without loss of deriva- tives Work in progress Finding an energy estimate for the variable coefficient problem obtained linearizing the Euler equations around a non constant discontinuity. So far, we have found the energy estimate far from the double roots of the Lopatinskii determinant. Techniques employed:

• paralinearization
• microlocal analysis

Some bibliography No general existence theorem for solutions which allows discontinuities. In the NON CHARACTERISTIC case:

• Complete analysis of existence and stability of a

single shock wave was made by – A. Majda 1983, – G. M´ etivier 2001.

• Existence of rarefaction waves by S. Alinhac 1989.
• Existence of sound waves by G. M´

etivier 1991.

• Uniform existence of shock waves with small strength

by J. Francheteau & G. M´ etivier 2000. In the CHARACTERISTIC case (vortex sheet):

• Existence and stability of vortex sheets for the isentropic

Euler equations by Coulombel-Secchi 2004-06,

• stability of vortex sheets for the nonisentropic Euler

equations by Morando-Trebeschi 2006 (in progress).

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