Adjoints
MEC651 denis.sipp@onera.fr Adjoints 1
Adjoints MEC651 denis.sipp@onera.fr Adjoints 1 Outline - - - PowerPoint PPT Presentation
Adjoints MEC651 denis.sipp@onera.fr Adjoints 1 Outline - Governing equations - Asymptotic development Order 0 : Base-flow Order 1 : Global modes - Bi-orthogonal basis and adjoint global modes Definition of adjoint
MEC651 denis.sipp@onera.fr Adjoints 1
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Incompressible Navier-Stokes equations: ππ’π£ + π£ππ¦π£ + π€ππ§π£ = βππ¦π + π ππ¦π¦π£ + ππ§π§π£ + π ππ’π€ + π£ππ¦π€ + π€ππ§π€ = βππ§π + π ππ¦π¦π€ + ππ§π§π€ + π βππ¦π£ β ππ§π€ = 0 Can be recast into: β¬ππ’π₯+ 1 2 πͺ π₯, π₯ + βπ₯ = π where: π₯ = π£ π π = π β¬ = 1 0 , πͺ π₯1, π₯2 = π£1 β πΌπ£2 + π£2 β πΌπ£1 β = βπΞ() πΌ() βπΌ β () Boundary conditions: Dirichlet, Neumann, Mixed
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a) πͺ π₯1, π₯2 = πͺ π₯2, π₯1 b) 1
2 πͺ π₯0 + πππ₯, π₯0 + πππ₯ = 1 2 πͺ π₯0, π₯0 + π πͺ π₯0, ππ₯
Jacobian
=πͺ
π₯0ππ₯
+
π2 2 πͺ ππ₯, ππ₯
Hessian + β― c) πͺ
π₯0ππ₯ = πͺ π₯0, ππ₯ = ππ£ β πΌπ£0 + π£0 β πΌππ£
d) β¬π₯ = β¬ π£ π = π£ e) ππ’π£ + π£ β πΌπ£ = βπΌπ + ππΌ2π£ β βπΌ2π = πΌ β π£ β πΌπ£ , πππ = ππΌ2π£ β π on solid
freedom of the flow. f) Scalar-product: < π₯1, π₯2 > = β¬ π£1
βπ£2 + π€1 βπ€2 ππ¦ππ§ = β¬ (π₯1 β β¬π₯2)ππ¦ππ§ so
that < π₯, π₯ > is the energy.
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Solution: π₯ π’ = π₯0 + ππ₯1 π’ + β― with Ο΅ βͺ 1 Governing equations: β¬ππ’π₯+ 1 2 πͺ π₯, π₯ + βπ₯ = π Introduce solution into governing eq:: β¬ππ’(π₯0+ππ₯1 + β― )+ 1 2 πͺ π₯0 + ππ₯1 + β― , π₯0 + ππ₯1 + β― + β(π₯0+ππ₯1 + β― ) = π β 1 2 πͺ π₯0, π₯0 + βπ₯0 = π at order π(1) β¬ππ’π₯1+ 1 2 [πͺ π₯1, π₯0 + πͺ π₯0, π₯1 ]
πͺ
π₯0π₯1
+ βπ₯1 = 0 at order π(π) β¬ππ’π₯2+πͺ
π₯0π₯2 + βπ₯2 = β 1
2 πͺ π₯1, π₯1 at order π(π2)
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Definition: π₯ π’ = π₯0 + ππ₯1(π’) + β― Non-linear equilibrium point : 1 2 πͺ π₯0, π₯0 + βπ₯0 = π How to compute a base-flow ? Newton iteration: 1 2 πͺ π₯0 + ππ₯0, π₯0 + ππ₯0 + β(π₯0+ππ₯0) = π Linearization: πͺ π₯0, ππ₯0 + βππ₯0 = π β 1 2 πͺ π₯0, π₯0 β βπ₯0 β ππ₯0 = πͺ
π₯0 + β β1 π β 1
2 πͺ π₯0, π₯0 β βπ₯0 π₯ πΊ π₯ = 1 2 πͺ π₯, π₯ + βπ₯ β π
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π₯0
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ππ = 47 Streamwise velocity field of base-flow.
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π₯ π’ = π₯0 + ππ₯1(π’) + β― Linear governing equation: β¬ππ’π₯1 + πͺ
π₯0π₯1 + βπ₯1 = 0
Solution π₯1 under the form: π₯1 = πππ’π₯ + c.c This leads to :
MEC651 denis.sipp@onera.fr
Eigenvalue: π = π + ππ Eigenvector: π₯ = π₯ r + iw π Real solution: π₯1 = πππ’π₯ + c.c = 2πππ’(cos ππ’ π₯ π β sin ππ’ π₯ π) πβ¬π₯ + πͺ
π₯0 + β π₯
= 0
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Eigenvalue problem solved with shift-invert strategy:
π΅β1: π΅β1π¦ = πβ1π¦. For this, evaluate π΅β1 ππ¦0
π΅ β π‘π½ β1: π΅ β π‘π½ β1π¦ = π β π‘ β1π¦. For this, evaluate π΅ β π‘π½ β1 ππ¦0
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Spectrum ππ = 47 Real part of cross-stream velocity field Marginal eigenmode
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We consider the linear Ginzburg-Landau equation ππ’π₯1 + βπ₯1 = 0 where β = πππ¦ β π π¦ β πΏππ¦π¦, π π¦ = ππ0 + π0 β π2 π¦2 2 . Here π, πΏ, π0, π0 and π2 are positive real constants. The state π₯(π¦, π’) is a complex variable on ββ < π¦ < +β such that |π₯| β 0 as π¦ β β. In the following, π₯π, π₯π = π₯π π¦ βπ₯π π¦ ππ¦
+β ββ
. 1/ What do the different terms in the Ginzburg Landau equation represent?
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2/ Show that π₯ (π¦) = ππ
π 2πΏπ¦βπ2π¦2 2 with π =
π2 2πΏ
1 4 and π =
π π
1 4π 1 8 π2 πΏ2π2
verifies ππ₯ + βπ₯ = 0. What is the eigenvalue π associated to this eigenvector? The constant π has been selected so that π₯ , π₯ = 1. 3/ Show that the flow is unstable if the constant π0 is chosen such that: π0 > ππ, where ππ =
π2 4πΏ + πΏπ2 2 .
Nota: ππ = iπ0 + π0 β π2
4πΏ β 2π + 1 πΏπ2 2 , π₯
π = πππΌπ ππ¦ π
π 2πΏπ¦βπ2π¦2 2
are all the eigenvalues/eigenvectors of β, πΌπ being Hermite polynomials.
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Global mod
π΅π₯ π = πππ₯ π The eigenvectors π₯ π form a basis: π₯ = π½πw i
π
Definition of adjoint global modes: with <> as a given scalar-product (say < π₯1, π₯2 > = π₯1
βπ₯2), there exists for each π½π a unique π₯
π such that π½π =< π₯ π, π₯ > for all π₯. The adjoint global modes are the structures π₯ π. In the following: π₯ π, π₯ π = 1. Properties:
π and w j are bi-orthogonal bases: they verify π₯ π = < π₯ π, π₯ π > w i
π
and so < π₯ π, w j > = πππ (in matrix notations π βπ = π½)
π, w i > β€< π₯ π, π₯ π >
1 2< π₯
π, π₯ π >
1 2
Hence: < π₯ π, π₯ π >
1 2β₯ 1 and cos angle π₯
π, π₯ π =
1 <π₯ π,π₯ π>
1 2
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π₯ 1 π₯ 2 π₯ 1 π₯ 2 Def of π₯ 1: π₯ 1 β π₯ 1 = 1 π₯ 1 β π₯ 2 = 0 Def of π₯ 2: π₯ 2 β π₯ 2 = 1 π₯ 2 β π₯ 1 = 0 π βπ = π½ Method 1 : π = W ββ1 Method 2 : π = π π β πβπ βπ = π½ β π = π βπ
β1 β π
= π π βπ
β1
Method 3 : adjoint global modes π₯ = (π₯ 1β π₯)π₯ 1 + (π₯ 2β π₯)π₯ 2
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Global mod
ππβ¬π₯ π + πͺ
π₯0 + β π₯
π = 0 The eigenvectors π₯ π form a basis: π₯ = π½πw i
π
Definition of adjoint global modes: with <> as a given scalar-product, there exists for each π½π a unique π₯ π such that π½π =< π₯ π, β¬π₯ > for all π₯. The adjoint global modes are the structures π₯ π. In the following: < π₯ π, β¬π₯ π > = 1. Properties:
π and w j are bi-orthogonal bases: they verify π₯ π = < π₯ π, β¬π₯ π > w i
π
and so < π₯ π, β¬w j > = πππ
π, β¬w i > β€< π₯ π, β¬π₯ π >
1 2< π₯
π, β¬π₯ π >
1
< π₯ π, β¬π₯ π >
1 2β₯ 1 and cos angle π₯
π, π₯ π =
1 <π₯ π,β¬π₯ π>
1 2
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Definition of optimal initial condition Initial-value problem: β¬ππ’π₯1 + πͺ
π₯0 + β π₯1 = 0,
π₯1 π’ = 0 = π₯π½ Solution: π₯1 π’ = < π₯ π, β¬π₯π½ > ππππ’π₯ π
π
If (π₯ 1, π1) is the global mode which displays largest growth rate, at large times: π₯1 π’ β< π₯ 1, β¬π₯π½ > ππ1π’π₯ 1 We look for unit-norm π₯π½ (< π₯π½, β¬wI >= 1) which maximizes the amplitude of the response at large times. π₯π½ is the optimal initial condition.
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If direct global mode as initial condition: π₯π½ = π₯ 1 In this case, at large time: π₯1 π’ β ππ1π’π₯ 1 If adjoint global mode as initial condition: π₯π½ = π₯ 1 < π₯ 1, β¬π₯ 1 >
1 2
Then, at large time: π₯1 π’ β< π₯ 1, β¬π₯ 1 >
1 2 ππ1π’π₯
1 This is optimal since: < π₯ 1, β¬π₯π½ > β€< π₯ 1, β¬π₯ 1 >
1 2< π₯π½, β¬π₯π½ > 1 2 1
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Estimation of gain: From Causchy-Lifschitz: < π₯ 1, β¬π₯ 1 >
1 2β₯ 1
Amplitude gain: < π₯ 1, β¬π₯ 1 >
1 2=
1 cos angle π₯ 1, π₯ 1
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π₯1
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π₯ 1 π₯ 2 π₯ 2 π₯ 1 π₯2 π₯2 β β1.2π₯ 1 + 1.8π₯ 2 π₯2 = 1 π₯1 β 1.9π₯ 1 β 1.3π₯ 2 π₯1 = 1 π₯ 2 = 0π₯ 1 + 1π₯ 2 π₯ 2 = 1 π₯ 1 = 1π₯ 1 + 0π₯ 2 π₯ 1 = 1
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Problem: β¬ππ’π₯+ 1 2 πͺ π₯, π₯ + βπ₯ = πβ¬π
1
π₯ = π₯0 + ππ₯1 At first order: β¬ππ’π₯1 + πͺ
π₯0 + β π₯1 = π 1
In frequency domain: π₯1 = ππππ’π₯ and π
1 = ππππ’π
Governing equation: ππβ¬π₯ + πͺ
π₯0 + β π₯
= β¬f Where to force (π ) and at which frequency (π) to obtain strongest response (π₯ )?
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Introducing global mode basis: π₯ = < π₯ π, β¬π₯ >
π
π₯ π and β¬π = < π₯ π, β¬π > β¬
π
π₯ π: ππ < π₯ π, β¬π₯ > β¬π₯ π β ππ < π₯ π, β¬π₯ > β¬π₯ π = < π₯ π, β¬π > β¬π₯ π
π π
Scalar-product with π₯ π and using bi-orthogonality:< π₯ π, β¬π₯ π > = πππ < π₯ π, β¬π₯ > ππ β ππ =< π₯ π, β¬f > < π₯ π, β¬π₯ >= < π₯ π, β¬f > ππ β ππ Solution: π₯1 π’ = ππππ’ < π₯ π, β¬f > ππ β ππ β¬π₯ π
π
To maximize response: a/ force at frequencies iπ closest to ππ b/ force with f =
π₯ i <π₯ π,β¬π₯ π>
1 2
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Definition of adjoint operator: Let β¨π₯1, π₯2β© be a scalar product and π a linear operator. The adjoint operator of π verifies β¨π₯1, ππ₯2β© = β¨π π₯1, π₯2β© whatever π₯1 and π₯2.
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Space: π₯ β βπ Scalar-product: < π₯1, π₯2 > = π₯1
βπ π₯2
with π a Hermitian matrix π β = π . Linear operator: π matrix. Adjoint operator: < π₯1, ππ₯2 > = π₯1
βπ ππ₯2 = π₯1 βπ ππβ1ππ₯2 = πβ1πβππ₯1 βππ₯2 =< π
π₯1, π₯2 > with π = πβ1πβπ If π = π½, then π = πβ
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4/ Determine the operator β adjoint to β, considering the scalar product β ,β .
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Space: Functions π¦ β 0,1 β β such that π£ 0 = ππ¦π£ 1 = 0. Scalar-product: < π£1, π£2 > = π£1
βπ£2ππ¦ 1
Linear operator π: ππ£ = πππ¦π£ β π½π£ β πππ¦π¦π£ Adjoint operator: < π£1, ππ£2 > = π£1
β πππ¦π£2 β π½π£2 β πππ¦π¦π£2 ππ¦ 1
= π£1
βπππ¦π£2 β π½π£1 βπ£2 β ππ£1 βππ¦π¦π£2 ππ¦ 1
= π£1
βππ£2 β ππ£1 βππ¦π£2 0 1 +
βππ¦ π£1
βπ π£2 β π½π£1 βπ£2 + πππ¦π£1 βππ¦π£2 ππ¦ 1
= π£1
βππ£2 β ππ£1 βππ¦π£2 + π(ππ¦π£1 β)π£2 0 1 +
βππ¦ ππ£1 β π½π£1 β πππ¦π¦π£1 βπ£2ππ¦
1
= < π π£1, π£2 > Hence: π π£ = βππ¦ ππ£ β π½π£ β πππ¦π¦π£ = βπππ¦π£ βπ£ππ¦ π β π½π£ β πππ¦π¦π£
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Boundary integral term: π£1
βππ£2 β ππ£1 βππ¦π£2 + π(ππ¦π£1 β)π£2 0 1 = 0
At π¦ = 0: π£2= 0 and ππ¦π£2 β 0, so that π£1 = 0 At π¦ = 1: ππ¦π£2 = 0 and π£2 β 0, so that π£1
βπ + π(ππ¦π£1 β) = 0, or π£1π + πππ¦π£1 = 0
π£1 should be in the following space: Functions π¦ β 0,1 β β such that π£ 0 = π£1(1)π + πππ¦π£ 1 = 0.
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Theorem: Let w i, ππ be eigenvalues/eigenvectors of π΅w i = ππw
π, ππ
β
solution of the adjoint eigenproblem π΅βπ₯ i = ππ
βπ₯
global modes and may be scaled such that π₯ π
βw
j = πππ. The vectors π₯ i are bi-
j.
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Proof: ππw i = π΅w i ππ
βπ₯
π = π΅βπ₯ π πππ₯ π
βw
i = π₯ π
βπ΅w
i = π΅βπ₯ π
βw
i = ππ
βπ₯
π
βw
i = πππ₯ π
βw
i ππ β ππ π₯ π
βw
i = 0 If ππ β ππ, then π₯ π
βw
i = 0 If π₯ π
βw
i β 0, then ππ = ππ. Conclusion: π₯ π can be chosen such that π₯ π
βw
i = π
ππ
MEC651 denis.sipp@onera.fr
Theorem: Let w i, ππ be eigenvalues/eigenvectors of ππβ¬w i + πͺ
π₯0 + β w
i = 0. Then there exists π₯ π, ππ
β solution of the adjoint eigenproblem ππ ββ¬π₯
π + πͺ π₯0 + β π₯ π =
< π₯ π, β¬w j >= πππ. The vectors π₯ i are bi-orthogonal with respect to the vectors w j.
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Proof: ππβ¬w i + πͺ
π₯0 + β w
i = 0 ππ
ββ¬π₯
π + πͺ π₯0 + β π₯ π = 0 < π₯ π, πͺ
π₯0 + β w
i > = βππ < π₯ π, β¬w i> < π₯ π, πͺ
π₯0 + β w
i > =< πͺ π₯0 + β π₯ π, w i > =< βππ
ββ¬π₯
π, w i > = βππ < π₯ π, β¬w i > ππ β ππ < π₯ π, β¬w i >= 0 If ππ β ππ, then < π₯ π, β¬w i > = 0 If < π₯ π, β¬w i >β 0, then ππ = ππ. Conclusion: π₯ π can be chosen such that < π₯ π, β¬w i > = π
ππ
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5/ Show that: π₯ (π¦) = ππ
β π
2πΏπ¦βπ2π¦2 2 with π =
ππβ1
4 is solution of πβπ₯
+ β π₯ = 0. Note that the normalization constant π has been chosen so that: π₯ , π₯ = 1. Can you qualitatively represent π₯ (π¦) and π₯ π¦ ? 6/ Noting that: π₯ , π₯ = π
1 2 2 π2 πΏ
3 2π2 1 2,
what does π₯ , π₯ represent? What is the effect of the advection velocity π and viscosity πΏ
Nota: π₯ π(π¦) = πππΌπ(ππ¦)π
β π
2πΏπ¦βπ2π¦2 2 are all the adjoint eigenvectors.
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Theorem: Let πͺ
π₯0π₯ = π£ β πΌπ£0 + π£0 β πΌπ£
be an operator acting on w = (π£, π€, π) such that π£ = π€ = 0 on boundaries. If < π₯1, π₯2 > = β¬ π£1
βπ£2 + π€1 βπ€2 + π1 βπ2 ππ¦ππ§, the adjoint operator of πͺ π₯0 is
πͺ π₯0 = ππ¦u0 ππ§π£0 ππ¦π€0 ππ§π€0
β
+ βπ£0
βππ¦ β π€0 βππ§
βπ£0
βππ¦ β π€0 βππ§
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< π₯1, πͺ
π₯0π₯2 >=< πͺ
π₯0π₯1, π₯2 > π£1
β π£0ππ¦π£2 + π€0ππ§π£2 + π£2ππ¦π£0 + π€2ππ§π£0
+π€1
β π£0ππ¦π€2 + π€0ππ§π€2 + π£2ππ¦π€0 + π€2ππ§π€0
ππ¦ππ§ = π£1
βπ£0ππ¦π£2 + π£1 βπ€0ππ§π£2 + π€1 βπ£0ππ¦π€2 + π€1 βπ€0ππ§π€2
+ π£1
βππ¦π£0π£2 + π£1 βππ§π£0π€2 + π€1 βππ¦π€0π£2 + π€1 βππ§π€0π€2
ππ¦ππ§ = π£1
βπ£0ππ¦π£2 + π£1 βπ€0ππ§π£2 + π€1 βπ£0ππ¦π€2 + π€1 βπ€0ππ§π€2 ππ¦ππ§ β
+ π£1ππ¦π£0
β + π€1ππ¦π€0 β β π£2 + π£1ππ§π£0 β + π€1ππ§π€0 β βπ€2 ππ¦ππ§
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β = π£1
βπ£0ππ¦π£2 + π£1 βπ€0ππ§π£2 + π€1 βπ£0ππ¦π€2 + π€1 βπ€0ππ§π€2 ππ‘
β ππ¦ π£1
βπ£0 π£2 + ππ§ π£1 βπ€0 π£2 + ππ¦ π€1 βπ£0 π€2 + ππ§ π€1 βπ€0 π€2 ππ¦ππ§
= β ππ¦ π£1π£0
β + ππ§ π£1π€0 β β π£2 + ππ¦ π€1π£0 β + ππ§ π€1π€0 β β π€2 ππ¦ππ§
πͺ π₯0π₯1 = π£1ππ¦π£0
β + π€1 ππ¦π€0 β
π£1ππ§π£0
β + π€1ππ§π€0 β
+ βππ¦ π£1π£0
β β ππ§ π£1π€0 β
βππ¦ π€1π£0
β β ππ§ π€1π€0 β βπ£0
βππ¦π£1βπ€0 βππ§π£1
βπ£0
βππ¦π€1βπ€0 βππ§π€1
(using ππ¦π£0 + ππ§π€0 = 0 )
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Conclusion: πͺ π₯0π₯1 = ππ¦u0 ππ§π£0 ππ¦π€0 ππ§π€0
β π£1
π€1 π1 + βπ£0
βππ¦ β π€0 βππ§
βπ£0
βππ¦ β π€0 βππ§
π£1 π€1 π1 πͺ
π₯0π₯2 =
ππ¦u0 ππ§π£0 ππ¦π€0 ππ§π€0 π£2 π€2 π2 + π£0ππ¦ + π€0ππ§ π£0ππ¦ + π€0ππ§ π£2 π€2 π2 πͺ
π₯0 β πͺ
π₯0 because of:
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Theorem: Let β = βπΞ() πΌ() βπΌ β () be an operator acting on w = (π£, π€, π) such that π£ = π€ = 0 on boundaries. If π₯1, π₯2 = β¬ π£1
βπ£2 + π€1 βπ€2 + π1 βπ2 ππ¦ππ§, the
= β.
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< π₯1, βπ₯2 >=< β π₯1, π₯2 > π£1
β βπππ¦π¦π£2 β πππ§π§π£2 + ππ¦π2 + π€1 β βπππ¦π¦π€2 β πππ§π§π€2 + ππ§π2
+ π1
β βππ¦π£2 β ππ§π€2 ππ¦ππ§
= βππ£1
βππ¦π¦π£2 β ππ£1 βππ§π§π£2 + π£1 βππ¦π2 β ππ€1 βππ¦π¦π€2 β ππ€1 βππ§π§π€2
+ π€1
βππ§π2 β π1 βππ¦π£2 β π1 βππ§π€2 ππ¦ππ§
β = βπΞ() πΌ() βπΌ β () = βππ£1
βππ¦ππ¦π£2 β ππ£1 βππ§ππ§π£2 + π£1 βππ¦π2 β ππ€1 βππ¦ππ¦π€2 β ππ€1 βππ§ππ§π€2 + π€1 βππ§π2
β π1
βππ¦π£2 β π1 βππ§π€2 ππ‘
β βπππ¦π£1
βππ¦π£2 β πππ§π£1 βππ§π£2 + ππ¦π£1 βπ2 β πππ¦π€1 βππ¦π€2 β πππ§π€1 βππ§π€2
+ ππ§π€1
βπ2 β ππ¦π1 βπ£2 β ππ§π1 βπ€2 ππ¦ππ§
π£ = π€ = 0 on boundaries
Adjoints 45 MEC651 denis.sipp@onera.fr
= β ππ¦π£1
βπ2 + ππ§π€1 βπ2 β ππ¦π1 βπ£2 β ππ§π1 βπ€2 ππ¦ππ§
+ β βπππ¦π£1
βππ¦π£2 β πππ§π£1 βππ§π£2 β πππ¦π€1 βππ¦π€2 β πππ§π€1 βππ§π€2 ππ¦ππ§ (β)
β = β βπππ¦π£1
βππ¦π£2 β πππ§π£1 βππ§π£2 β πππ¦π€1 βππ¦π€2 β πππ§π€1 βππ§π€2 ππ‘
+ βπππ¦π¦π£1
βπ£2 β πππ§π§π£1 βπ£2 β πππ¦π¦π€1 βπ€2 β πππ§π§π€1 βπ€2 ππ¦ππ§
β π₯1 = βπππ¦π¦ β πππ§π§ ππ¦ βπππ¦π¦ β πππ§π§ ππ§ βππ¦ βππ§ π£1 π€1 π1
Adjoints 46 MEC651 denis.sipp@onera.fr
β π₯1 = βπππ¦π¦ β πππ§π§ ππ¦ βπππ¦π¦ β πππ§π§ ππ§ βππ¦ βππ§ π£1 π€1 π1 βπ₯2 = βπππ¦π¦ β πππ§π§ ππ¦ βπππ¦π¦ β πππ§π§ ππ§ βππ¦ βππ§ π£2 π€2 π2 β = β
Adjoints 47 MEC651 denis.sipp@onera.fr
MEC651 denis.sipp@onera.fr Adjoints 48
Adjoints 49 MEC651 denis.sipp@onera.fr
Real part of cross-stream velocity field. Marginal adjoint global mode