Transition to turbulence in shear flows Recap on linear matrix algebra Let us imagine that the N eigenvalues are distinct and the eigenvectors are linearly independent (so as to form a basis); at t = 0 we have π πβ²(0) = ΰ· π π π π π=1 π π π β , πβ² = Ο π=1 π β , π π π π = Ο π=1 π π π β , π π = π β π β , π β π β , πβ² π β = π β , π β 43
Transition to turbulence in shear flows Recap on linear matrix algebra Let us imagine that the N eigenvalues are distinct and the eigenvectors are linearly independent (so as to form a basis): at t = 0 we have π π β² 0 = π β²0 = ΰ· π π π π π=1 π π π β , πβ² 0 = Ο π=1 π β , π π π π = Ο π=1 π π π β , π π = π β π β , π β π β , πβ² 0 π β = π β , π β 44
Transition to turbulence in shear flows Recap on linear matrix algebra Let us assume that the eigenvectors are π π = π π , πβ² 0 orthonormalized β , then π πβ² 0 = π± πβ² 0 π π πβ² 0 = Ο π=1 π π π π , πβ² 0 = Ο π=1 π π π π i.e. given that πβ² 0 is any vector, the identity matrix can π π be retrieved from I = Ο π=1 π π π π 45
Transition to turbulence in shear flows Recap on linear matrix algebra The matrix A can thus be represented as π = Ο π=1 l k π π π π π π π π© = π© π± = Ο π=1 π© π π π π i.e. A can be written as the sum of the product of eigenvalues and eigenvectors (of course, under the assumption that eigenvalues are distinct, and eigenvectors are linearly independent) 46
Transition to turbulence in shear flows Recap on linear matrix algebra Let U be the matrix whose columns are the N right eigenvectors and V the matrix with the N left eigenvectors in the colums; assume eigenvalues to be distinct. When e-vectors are orthonormal: π π π = π βπ β ΰ΄₯ πΎ π π½ = π± π β , π π = π βπ β π β π π π = l k π βπ π π© π π = l k π β π β πΎ π π© π½ = L ( the diagonal matrix of eβvalues) β 47
Transition to turbulence in shear flows Recap on linear matrix algebra Let U be the matrix whose columns are the N right eigenvectors and V the matrix with the N left eigenvectors in the colums; assume eigenvalues to be distinct. When e-vectors are orthonormal: π π π = π βπ β ΰ΄₯ πΎ π π½ = π± π β , π π = π βπ β π β π π π = l k π βπ π π© π π = l k π β π β πΎ π π© π½ = L ( the diagonal matrix of eβvalues) β 48
αΆ αΆ Transition to turbulence in shear flows Recap on linear matrix algebra πΎ π π© π½ = L β π© = U L ΰ΄₯ ΰ΄₯ πΎ π ππβ² for the problem let us take ππ’ = π© πβ²(π’) ππβ² π β² = π½ π π’ , ππ’ = π½ αΆ π½ αΆ π , π = π© π½ π , π π’ = π L π’ π(0) π = L π β π = π½ β1 π© π½ π , 49
Transition to turbulence in shear flows Recap on linear matrix algebra π π’ = π½ β1 π β² = π L π’ π 0 = π L π’ π½ β1 π β² 0 ππβ² so that the solution of ππ’ = π© πβ²(π’) π β² = π½ π L π’ π½ β1 π β² 0 = π½ π L π’ ΰ΄₯ πΎ π π β² 0 is π β² = π΄ πβ²(0) propagator of the initial condition 50
Transition to turbulence in shear flows Recap on linear matrix algebra π l k π π π π π π© = ΰ· π=1 π π π ΰ· π = π© 2 = π©π© = ΰ· l k π π π π l h π β π β π=1 β=1 π = π π β π β l k l h π π π π π π Ο π=1 Ο β=1 π = l k l h π π d βπ π β π π Ο π=1 Ο β=1 π π β π β π l h π Ο β=1 51
Transition to turbulence in shear flows Recap on linear matrix algebra π π© π = ΰ· π π π π π l k π π=1 The matrix A π has the same eigenvectors as A , and π . In general, for a linear eigenvalues which are l k combination π of powers of A π π( l k ) π π π π π π(π©) = ΰ· π=1 52
Transition to turbulence in shear flows Recap on linear matrix algebra and in particular π π l k π π π π π π© = ΰ· π π=1 The solution of our linear problem reads also: π l k t π π (π π , π β² 0 ) π β² π’ = π π©π’ π β² 0 = Ο π=1 π with the left eigenvectors weighting the initial condition 53
Transition to turbulence in shear flows Stability conditions π l k t π π π π π β² π’ = Ο π=1 π the eigenvalues l k define the asymptotic growth/decay of the disturbance. Should there be a double eigenvalue, terms of the form t π l k t would appear in the expansion of the solution (resulting in a linear time growth of the disturbance even when Re ( l k ) < 0, for all k ). 54
Transition to turbulence in shear flows Stability conditions An autonomous system, with evolution matrix A equipped with N distict eigenvalues is: β’ Asymptotically stable is all eigenvalues of A have negative real part β’ Marginally stable if one (or more) eigenvalues have real part equal to zero (and the others have negative real part) β’ Unstable if at least one eigenvalue has real part larger than zero 55
Transition to turbulence in shear flows Stability conditions Near the marginal stability conditions, typically a single (1!) eigenvalue crosses the stability boundary, i.e. for π’ β β we have π β² π’ ~ π l 1 t π 1 (π 1 , π 0 ) = π l 1 t π 1 π 1 all other modes ( k = 2, 3, 4, β¦ N ) being damped. 56
Transition to turbulence in shear flows Stability conditions l k = π π + π π π growth rate angular frequency The eigenvalue problem is π© π π = l π π π and in hydrodynamic stability analysis the π π βs are called normal modes 57
Transition to turbulence in shear flows Stability conditions πΉ π’ = (π β² , πβ²) E unstable region r E r G r c r 58
Transition to turbulence in shear flows Stability conditions Stable E Conditionally stable (III) Globally r E r G r c r stable (II) Monotonically stable (I) 59
Transition to turbulence in shear flows A simple example Consider the very simple linear system ( A in the chosen example is called a Jordan block): ππβ² ππ’ = π© π β² π’ π© = A has a double eigenvalue ( l 1 = l 2 = 0) to which is 1 associated the double eigenvector π 1 = π 2 = 0 60
Transition to turbulence in shear flows Example π¦ 10 If π β² (0) = is the initial condition (at t = 0) π¦ 20 then the solution is π¦ 10 + π’ π¦ 20 π β² = π¦ 20 i.e. the disturbance vector grows linearly in time ( algebraic growth ), despite the fact that the eigenvalues have vanishing real part. 61
Transition to turbulence in shear flows Example Consider now the perturbed system with A = l 1 = βΟ΅ , l 2 = β2Ο΅ eigenvalues: 1 0 1 1 e-vectors: π 1 = 0 , π 2 = , π 1 = 1/Ο΅ , π 2 = β1/Ο΅ βΟ΅ and the solution is: π β² = π 1 π βππ’ π 1 + π 2 π β2ππ’ π 2 62
Transition to turbulence in shear flows Example The question is: how do the two solutions (linear growth and exponential decrease) match as Ο΅ decreases to become Ο΅ = 0 ? 63
Transition to turbulence in shear flows Example The question is: how do the two solutions (linear growth and exponential decrease) match as Ο΅ decreases to become Ο΅ = 0 ? To answer we must focus on the energy of the disturbance, E(t) = ( π β² , π β² ). For large times (when Ο΅ t >> 1) the exponential behavior of the previous slide holds and eventually at large times the solution goes like π β² ~ π βππ’ π 1 . What about short times? 64
Transition to turbulence in shear flows Example For short times ( π t << 1) a Taylor series of the energy gives: πΉ π’ π 1 + π 2 2 + π 2 π 2 2 = 2 + 4 1 + π 2 π 2 2 + 6π 1 π 2 ππ’ + O π 2 π’ 2 β 2π 1 and a linear growth in time is possible if the factor of ( π t ) is negative. This growth is related to the fact that the two eigenvectors π 1 and π 2 are not orthogonal to one another (in fact, they are almost parallel !) 65
Transition to turbulence in shear flows Example growth, then decay monotonic decay π t 66
Transition to turbulence in shear flows Example The optimal initial condition growth, then decay It is easy to see that the initial condition which yields the largest gain, ratio of final to initial energy, for t large enough, is the first monotonic decay left eigenvector. π t 67
Transition to turbulence in shear flows Another simple example 68
Transition to turbulence in shear flows Another simple example 69
Transition to turbulence in shear flows Another simple example 70
Transition to turbulence in shear flows Another simple example 71
Transition to turbulence in shear flows Shear flow problems 72
Transition to turbulence in shear flows Shear flow problems For the simple problems above, the flow is parallel or quasi- parallel and it is a good approximation to consider the velocity profile as π 0 = π π§ π 73
Transition to turbulence in shear flows Hydrodynamic stability of // shear flows πΌ . π = 0 ππ 1 ππ πΌ 2 π ππ’ + (π . πΌ)π = β πΌ π + π π, 0 assigned π π, π’ = 0 on solid boundaries π = π 0 + π π 1 + π 2 π 2 + β¦ π βͺ 1 π = π 0 + π π 1 + π 2 π 2 + β¦ 74
Transition to turbulence in shear flows Hydrodynamic stability πΌ . π 0 = 0 1 ππ πΌ 2 π 0 (π 0 . πΌ)π 0 = β πΌ π 0 + πΌ . π 1 = 0 ππ 1 1 ππ πΌ 2 π 1 ππ’ + (π 1 . πΌ)π 0 + (π 0 . πΌ)π 1 = β πΌ π 1 + 75
Transition to turbulence in shear flows Hydrodynamic stability 76
Transition to turbulence in shear flows Hydrodynamic stability into the equations and collecting terms of O ( π ): 77
Transition to turbulence in shear flows Hydrodynamic stability The solution of the linear O ( π ) problem relies crucially on the streamlines being parallel, i.e. U is independent of x . This makes the solution separable , e.g. β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ . More particularly, it can be shown that the solutions are the sum of exponential functions of the invariant directions, x and t (Fourier modes). A normal mode has the form: ; this is a x-travelling wave , with a the streamwise wavenumber and w the circular frequency. 78
Transition to turbulence in shear flows Hydrodynamic stability 79
Transition to turbulence in shear flows Hydrodynamic stability Boundary conditions 80
Transition to turbulence in shear flows Hydrodynamic stability 81
Transition to turbulence in shear flows Hydrodynamic stability . 82
Transition to turbulence in shear flows Stability conditions 83
Transition to turbulence in shear flows The inviscid problem Rayleigh inflection point theorem 84
Transition to turbulence in shear flows The inviscid problem Rayleigh inflection point theorem The imaginary part of the equation above ( a real) is ad this relation is satisfied for π π β 0 only when the integral vanishes, which occurs only if π β²β² β‘ 0 or π β²β² changes sign at least once in 85
Transition to turbulence in shear flows The inviscid problem Rayleigh inflection point theorem: a necessary, but not sufficient, condition for instability is that the velocity profile have an inflection point 86
Transition to turbulence in shear flows The inviscid problem FjΓΈrtoftβs theorem Let there be an inflection point at and let If π π β 0 then The real part of the expression derived previously is 87
Transition to turbulence in shear flows The inviscid problem FjΓΈrtoftβs theorem Adding up leads to: a necessary, but not sufficient, condition for instability is < 0, somewhere in the flow that 88
Transition to turbulence in shear flows The inviscid problem FjΓΈrtoftβs theorem 89
Transition to turbulence in shear flows The inviscid problem Howardβs semi-circle theorem the complex phase velocity lies inside, or on, the semi- π½ πππ + π½ πππ π½ πππ β π½ πππ circle centred on of radius π π 90
Transition to turbulence in shear flows The viscous problem 91
Transition to turbulence in shear flows The viscous problem 92
Transition to turbulence in shear flows The viscous problem: PPF Re = 10000 , a = 1 93
Transition to turbulence in shear flows The viscous problem: Blasius Re = 500, a = 0.2 94
Transition to turbulence in shear flows The viscous problem: Blasius 95
Transition to turbulence in shear flows The viscous problem Typical wind tunnel experiments say that transition occurs around Re = 2000 in PPF and around Re = 400 in the Blasius boundary layer β¦ Could this be a 3D effect? Or something else? 96
Transition to turbulence in shear flows The viscous problem: 3D disturbances Consider 3D disturbances and replace by ( a x + b z β w t ) π£, ΰ· ΰ· π€, ΰ· π₯, ΰ· π = π£ π§ , π€ π§ , π₯ π§ , π(π§) to end up with the OS and Squire equations 97
Transition to turbulence in shear flows The viscous problem: 3D disturbances οΎ ο¨ = π πΎ ΰ·€ ΰ·₯ π£ β π π½ ΰ·₯ π₯ mode shape of the normal vorticity 98
Transition to turbulence in shear flows The 3D viscous problem In discrete form the temporal problem is a generalized eigenvalue problem of the form: π π π© π π πͺ π π ο¨ = π ο¨ π« π© π π πͺ π π© πβ² = π πͺ πβ² 99
Transition to turbulence in shear flows The 3D viscous problem There are two families of solutions of the Orr-Sommerfeld and Squire problems OS modes: Squire modes: 100
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