TMA4195 Mathematical modelling Autumn 2012

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1 Norwegian University of Science an Technology Department of Mathematical Sciences TMA495 Mathematical moelling Autumn 202 Solutions to exam December, 202 Dimensional matrix A: τ µ u r m - - s kg The, µ, u columns are linearly inepenent, so, µ, u are imensional inepenent an core an possible core variables By Buckingham s Pi-theorem there are 5 rank A = 5 3 = 2 imensional inepenent combinations, eg τ π = a µ b u c = = τ µ u, π 2 = r By Buckingham s Pi-theorem again, any imensionally consistent relation is equivalent to a relation τ = Φ(, µ, u, r) Ψ(π, π 2 ) = 0, an solving for π we fin that π = C(π 2 ), or τ = µu C ( r ) 2 We introuce the scales c = Cc, x = Xx, t = T t The natural scales are C = M an X = L The scale equation is M T c t = DM L 2 c xx + UM L c x rmc By assumption c t, c xx, c x, c, x, t, hence after iviing by M, T D L 2 + U L + r = L 2 ( D + UL + rl 2 ) Case (i) : D UL + rl 2 The time scale is then given by T D L 2, an the scale equation is c t = c xx + ε c x ε 2 c, where ε = UT L = UL D an ε 2 = rt = rl2 D December 9, 203 Page of 5

2 Case (ii) : rl 2 D + UL The time scale is given by T where ε = D rl 2 an ε 2 = U rl r, an the scale equation is c t = ε c xx + ε 2 c x c, 3 (i) coul moel two competing populations of size n an n 2 When n 2 = 0, the n -equation is the logistic equation with boune population, while if n = 0, the n 2 -equation gives exponential growth (ii) coul moel two populations that are mutually beneficial, eg as in symbiosis We fin equilibrium points given by F = ( ) n = t n 2 ( ) n ( n an 2 ) = 0, cn 2 ( n ) that is, (n, n 2 ) {(0, 0), (, 0)} We examine the stability of the equilibrium points To that en consier the Jacobi matrix of F, [ ] 2n an DF(n, n 2 ) = 2 an cn 2 c( n ) By inserting the first equilibrium point we get [ ] 0 DF(0, 0) =, 0 c with at least one positive eigenvalue (λ = ) This implies that (0, 0) is unstable The secon equilibrium point gives [ ] a DF(, 0) =, 0 0 where max{e(λ)} = 0 an no conclusion 4 This is a singular perturbation problem The only bounary point is t = 0, an the bounary layer is there We solve the problem as follows (i) Fin the outer solution (ii) Fin the bounary layer thickness an the rescale equations (iii) Fin the inner solution (iv) Mathcing in the intermeiate region (v) The uniform approximation Implementing the above strategy (i) The outer solution solves the equations with ε = 0, ẋ O = x O + (x O + )(y O ), 0 = x O (x O + )y O December 9, 203 Page 2 of 5

3 The solution is x O (t) = Ce t ( e t ), y O (t) = x O(t) x O (t) +, where we have use the technique of integrating factor to etermine x O (t) (ii) We rescale the equation t = δτ, X(τ) = x(t) an Y (τ) = y(t) Inserting into the equations, X = X + (X + )(Y ), δ τ ε Y = X (X + )Y δ τ If we balance terms in the first equation we get δ = an recover the outer solution If we, on the other han, balance terms in the secon equation we get δ = ε Hence δ = ε is the other time scale (iii) The rescale equations are Ẋ = ε ( X + (X + )(Y )), X(0) = Ẏ = X (X + )Y, Y (0) = 0 We get the inner solution be letting ε = 0, hence X I (t) =, Y I (t) = ( e 2t ) 2 (iv) The matching conition is ( lim xo (t), y O (t) ) ( = lim XI (τ), Y I (τ) ) t 0 + τ We calculate the limits using the solutions an get that C = (v) The uniform approximation is calculate by summing the inner an outer solution an subtracting the intermeiate value Thus x u (t) = 2e t, y u (t) = 2 et 2 e 2 t ε 5 The kinematic spee is c(ρ) = j (ρ) = 2ρ, an the characteristics are given by ẋ = c(z), x(t 0 ) = x 0, ż = 0, z(t 0 ) = ρ(x 0, t 0 ) The solution of this system is z(t) = ρ(x 0, t 0 ), x(t) = x 0 + (t t 0 )c ( ρ(x 0, t 0 ) ) December 9, 203 Page 3 of 5

4 In x 0, inflow bounary correspons to c 0 To fin a bounary conition at x = 0, we convert to Dirichlet conition 3 = j(ρ) = ρ( ρ), 6 which gives ρ = 4 or ρ = 3 4 Since c( 4 ) > 0 an c( 3 4 ) < 0, an we can only impose bounary conitions at inflow, we take ρ = 4 at x = 0, t > 0 The characteristics are x(t) = z(t) = { x 0 + tc ( 8), t0 = 0, x 0 0, t > 0, (t t 0 )c ( 4), x0 = 0, t > t 0 > 0, { 8, t 0 = 0, x 0 0, t > 0, 4, x 0 = 0, t > t 0 > 0 In the ea sector there is a rarefaction wave solution 6 a) Conservation of mass in () t Observe that ρ V = ρ 0 an since is smooth, As v n = ue z n, ρ V = ρv n σ z+ z (z, t) y z 0 z+ z ρ V = ρ t (z, t) z t z ρv n σ = If we ivie () by p z, we fin z+ z z ρ z + {z= z} {z= z+ z} xz, ρ( u)( ) σ ρ( u)( ) σ = ρ ( (u)( z) (u)( z + z) ) t z = ρ( (u)( z) (u)( z + z) ) Let z 0, use efinition of erivative an the funamental theorem of calculus to see that t = z (u) z December 9, 203 Page 4 of 5

5 b) Fix t > 0 an z [0, ), an let z > 0 be such that z + z < If ( z, t) = 0 then ρge z V = C µu e z σ This implies that {x=0} ( u ) ρg( z, t) = Cµ ( z, t), an hence By a), then satisfies u = ρg Cµ 2 = K 2 t ( K 3) z = 0 Metho of characteristics (z(t) = ( x(t), t)) The characteristics are then ẋ = 2Kz 2 = c(z), x(0) = x 0, ż = 0, z(0) = (x 0, 0) x = x 0 + tc ( (x 0, 0) ), z = (x 0, 0) Since z < z 2 implies that (z, 0) < (z 2, 0), which again implies that c ( (z, 0) ) > ( (z2, 0) ), the characteristics will evelop shock The shock spee via the ankine-hugoniot conition is ṡ = 2K ρ2 (z, 0) ρ 2 (z 2, 0) ρ(z, 0) ρ(z 2, 0) < 0 so the shock moves ownwars December 9, 203 Page 5 of 5