Dimensional Analysis - Concepts

Transcription

1 Dimensional Analysis - Concepts Physical quantities: R j = v(r j )[R j ] = value unit, j = 1,..., m. Units: Dimension matrix of R 1,, R m : A = Change of units change of values: [R j ] = F a 1j 1 F a nj n, F 1,..., F n fundamental units. a a 1m..... a n1 a nm Lemma 1: F i = x i ˆF i, x i > 0 ˆv(R j ) = x a 1j 1... x a nj n v(r j ) Dimensionless combination: π = R λ1 1 Rλm m if λ 0, [π] = 1 Dimensionally independent R 1,..., R s if no dimensionless comb ns exist Physical relations Φ(R 1,..., R m ) = 0 are dimensionally consistent, i.e. Φ(v(R 1 ),..., v(r m )) = 0 Φ(ˆv(R 1 ),..., ˆv(R m )) = 0 for all changes of units ˆF i. (consistent under change of units) ERJ (NTNU) TMA4195 Mathematical Modeling August 22, / 2

2 It remains to prove Lemma 2 and the Pi-theorem. ERJ (NTNU) TMA4195 Mathematical Modeling August 22, / 2 Dimensional Analysis - Buckingham s pi-theorem (A1) F 1,..., F n are fundamental units (A2) R 1,..., R m are physical quantities (A3) Φ(R 1,..., R m ) = 0 is dimensionally consistent. Lemma 2: Let r = rank A, then R 1,..., R m have m r independent dimensionless combinations. OBS: The rank = number of linearly independent collumns in the matrix. Buckingham s pi-theorem: If (A1) (A3) hold, then there are m r independent dimensionless combinations, and for any set of m r independent dimensionless combinations π 1,..., π m r, there is a relation Ψ such that Φ(R 1,..., R m ) = 0 Ψ(π 1,..., π m r ) = 0, where r = rank A and A is the n m dimension matrix of R 1,..., R m.

3 Scaling and non-dimensionalizing Produce dimensionless O(1) variabels and dim.less eq ns with terms 1 Scaling a variable u : u = Uu where scaled variable: u 1, [u] = 1 scaling constant: U max u, [U] = [u ] Scaling/nondimensionalizing an equation: 1 scaling all variables in the equation 2 dividing the resulting equation by biggest coefficient. Finding scales: - look for combinations of the parameters - balance 2 dominating ( biggest ) terms in the equation (using that all scaled variables should be O(1)) - solve a reduced problem to find estimates - typical time scale for u (t ): T = max u max du dt ERJ (NTNU) TMA4195 Mathematical Modeling August 29, / 2

4 Remarks on scaling Remark 1: 2 dominating terms balanced their coefficients are equal and biggest in equation. dividing scaled equation by this coefficient: All variables and coefficients become dimensionless, 1, and 2 coefficients = 1. Remark 2: Different situations different scales max x, max t, max u, etc., and the dominating terms in the equation depend on the situation. Remark 3: Advantages of scaling: - minimize the number of parameters/coefficients (experiments!!), - normalize all variables and coefficients, - reduce round-off errors in subsequent numerical calculations, - make small terms visible easy to do approximations/perturbation. ERJ (NTNU) TMA4195 Mathematical Modeling August 29, / 2

5 Regular Perturbation Given scaled(!) equation: 1 ẍ = (1 + εx) 2, 0 < ε Perturbation Assumption: x = x 0 + εx 1 + ε 2 x Insert into equation, expand as power series in ε: ẍ 0 + εẍ 1 + ε 2 1 ẍ 2 + = ( ) ε(x 0 + εx ) ( ) = 1 + 2ε x 0 + εx ε 2( ) 2 x 0 + εx = 1 + ε2x 0 + ε 2 (2x 1 3x 2 0 ) Equate terms of same order in ε equations for x 0, x 1,... : O(1) : ẍ 0 = 1 O(ε) : ẍ 1 = 2x 0 O(ε 2 ) : ẍ 2 = 2x 1 3x Solve these equations recursively for x 0, x 1, x 2,.... ERJ (NTNU) TMA4195 Mathematical Modeling September 3, / 1

6 Singular Perturbation Signs: Multiple time/space scales Initial/boundary layers Small paramater multiplying principal term Naive approximation changes problem completely Facts: No single scale is good for complete resolution of problem Different regions, different scales, different (re)scaled equations Scales found by balancing terms in equation In each region regular perturbation works Matching conditions between perturbation sol ns of different regions ERJ (NTNU) TMA4195 Mathematical Modeling September 12, / 2

7 Singular Perturbation first approximation εy + 2y + y = 0, 0 < x < 1; y(0) = 0, y(1) = 1; 0 < ε Guess where boundary layer is: x = a. Here a = Outer solution y O. Set ε = 0 and solve equation and boundary condition outside boundary layer: 2y O + y O = 0; y O (1) = 1 = y O (x) = e 1 2 x Find length of boundary layer δ (the other consistent space scale) by balancing terms... δ = ε. 4. Rescale equation: (x, y) = (δξ, Y ) Y (ξ) + 2Y (ξ) + εy (ξ) = 0 5. Inner solution y I. Set ε = 0 and solve rescaled equation and boundary condition inside boundary layer. y I + 2y I = 0, y I (0) = 0 = y I (ξ) = C(1 e 2ξ ). 6. Matching. y O y I in intermediate region... C = e 1 2 ε 0 7. Uniform solution: y U (x) = y O (x) + y I ( x δ ) lim y O(Θη) ε 0 (approx n!) ERJ (NTNU) TMA4195 Mathematical Modeling September 12, / 2

8 Equilibrium points 1. Equilibrium point = constant solution u e (e.g. of ODEs or PDEs) 2. An equilibrium point u e is stable if all solutions starting near u e, remain near u e for all t Linear stability analysis 1 Write solution u = u e + ũ, ũ small perturbation 2 Linearize equation(s) about u e: insert u = u e + ũ into equation(s) drop small(=non-linear in ũ) terms Result: linear equation(s) for ũ, with equilibrium point ũ e = 0. 3 Check stability of ũ e = 0 (linearized equation(s)!!) 4 Conclusion: ũ e = 0 stable/unstable indicate that u e stable/unstable. 4. Over time all physical systems tend to be at their stable equilibrium solutions! (... always small disturbances...) ERJ (NTNU) TMA4195 Mathematical Modeling September 19, / 3

9 Aggregation of Amoeba Background: Lack of food amoeba produce attractant and aggregate. Question: Can onset of aggregation be caused by simple, uninteligent mechanism? Model near onset of aggregation: Physical quantities: a(x, t), c(x, t) = amoeba, attractant densities; parameters Modelling (conservation+diffusion+attraction+production+decay): a t = ) (1) (ka x lac x, c t = Dc xx + q 1 a q 2 c. x Equilibrium points (=constant solutions): Constants (a 0, c 0 ) such that q 1 a 0 = q 2 c 0. Linearize equation around (a 0, c 0 ): a = a 0 + ã, c = c 0 + c; ã, c small; drop small terms ã t = ) (2) (kã x la 0 c x, c t = D c xx + q 1 ã q 2 c. x ERJ (NTNU) TMA4195 Mathematical Modeling September 19, / 3

10 Aggregation of Amoeba (2) ã t = ) (kã x la 0 c x, c t = D c xx + q 1 ã q 2 c. x Particular solutions of (2): Fourier modes/eigenfunctions solve (2) iff 1 α 2 + bα + c = 0 for (ã, c) = e αt cos(βx)(c 1, C 2 ) b = kβ 2 + Dβ 2 + q 2 and c = kq 2β 2 + kdβ 4 q 1la 0β 2, 2 (C 1, C 2) satisfy two linear equations (last time). Every β R two real α + linear subspace of solutions (C 1, C 2 ) Boundedness of solutions of (2): (ã, c) bounded α 0 c 0 kdβ 2 + kq 2 q 1 la 0 Unboundedness: kq 2 < q 1 la 0 (ã, c) blow up at for β 1 Take C 1, C 2 arbitrarily small (0, 0) unstable equilibrium pt. of (2) Indicate that (a 0, c 0) ( (ã, c) = (0, 0)) unstable equi. pt. of (1) ERJ (NTNU) TMA4195 Mathematical Modeling September 19, / 3

11 Convervation in Continuum Mechanics The transport theorem: R(t) = {x(t) : ẋ = v(x, t), x(t 0 ) = x 0 R(t 0 )} d dt R(t) f (x, t) dx t=t0 = d dt R(t f (x, t) dx 0) t=t0 + R(t f (x, t 0) 0) ( v n) dσ Conservation of mass and momentum: (1) d dt R ρ dx + R ρ ( v n) dσ = R q dx. d (2) ρ v dx + ρ v ( v n) dσ dt R R Transp. d = ρ v dx Newton s = fb dx thm. dt 2nd law R(t) R(t) body forces + fs dσ R(t) surface forces Newtonian fluid: f S = T n, T ij = (p µ v)δ ij + µ( v i x j Differential form - the Navier-Stokes equations: (1 ) (2 ) ρ t + (ρ v) = q v j x i ) (ρv i ) t + (ρv i v) = f B,i p x i + µ ( 2 v i x i ( v) ), i = 1, 2, 3 ERJ (NTNU) TMA4195 Mathematical Modeling October 17, / 1

12 Flow in rivers the shallow water equations Assumptions: z a h 1 v, h depend only on x, t; a small; ρ = const. 2 Dominant forces in x-direction: 1 Gravity f g e x = ρg sin a 2 Hydrostatic pressure f p e x ρg(h z)( n e x) 3 Bottom friction f f e x = ρc f v 2 v Control volume: R = {(x, y, z) : x [x 0, x 0 + x], y [0, B], z [0, h(x, t 0 )]} Conservation of mass and momentum in x-direction: d dt R ρ dx + ρ ( v n) dσ = 0, R d dt R ρv dx + R ρv ( v n) dσ = f R g e x dx + R ( f p + f f ) e x dσ. ERJ (NTNU) TMA4195 Mathematical Modeling October 22, / 2 x h z B y

13 Flow in rivers the shallow water equations Compute all integrals, divide by common factor ρb: d x0+ x dt x 0 h dx + [(vh)(x 0 + x) (vh)(x 0 )] = 0, d x0+ x dt x 0 vh dx + [(v 2 h)(x 0 + x) (v 2 h)(x 0 )] = x 0+ x x 0 gh sin a dx g 2 [h2 (x 0 + x) h 2 (x 0 )] x 0+ x x 0 C f v 2 dx. Divide by x, let x 0: h t + (vh) = 0, x ( ) vh + t x ( v 2 h + g 2 h2) = gh sin a C f v 2. This is the shallow water equations or St. Venant system. ERJ (NTNU) TMA4195 Mathematical Modeling October 22, / 2