Exam in TMA4195 Mathematical Modeling Solutions

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1 Norwegian University of Science and Technology Department of Mathematical Sciences Page of 9 Exam in TMA495 Mathematical Modeling Solutions Problem a Here x, y are two populations varying with time t. The equation for ẋ consists of: rx( x K logistic growth term of rate r and capacity K, axy c + x death term of rate a and dependent on both x and y, while the equation for ẏ consists of: my death term, of rate a, + bxy c + x growth term of rate b and dependent on both x and y. Possible models include: x prey, y predators, x humans, y bacteria, x lemmings, y foxes, x fish, y fishermen. b We have [ẋ ] ẏ [ x( x y = F (x, y := ] +x y( + x +x ( Equilibrium points (x e, y e are constant solutions of ( and therefore the solutions of ( x x y = 0, + x ( y + x = 0. + x

2 TMA495 Mathematical Modeling, 6..07, solutions Page of 9 The second equation is satisfied exactly when y = 0 or x =. For y = 0, the first equation is satisfied exactly when x = 0 or x =, while if x = the first equation is satisfied only for y =. Thus the equilibrium points are given by: (0, 0, (, 0, (,. OBS: By carelessly multiplying the equations with ( + x, one might conclude that (, 0 is also an equilibrium point, which is not the case (check!. To calculate the stability of the points we look at the Jacobian of F : DF (x, y := [ ( x y y (+x (+x x +x + x +x We calculate the eigenvalues λ, λ of DF in the different equilibrium points: [ ] 0 DF (0, 0 = where λ 0 =, λ =, [ ] /3 DF (, 0 = where λ 0 /3 =, λ = /3, [ ] / / DF (, = where λ 0 = i 4, λ = 7 4 i 4. The eigenvalues of the last matrix can be find by solving det(df (, λi = 0 ( / λ( λ ( / = 0 λ + λ + = 0. An equilibrium point is stable if max{reλ,reλ } < 0 and unstable if max{reλ,reλ } > 0. Consequently we see that (0, 0 and (, 0 are unstable while (, is stable. ] Problem Performing regular perturbation we write y as y(x = y 0 (x + εy (x + O(ε, where y 0, y are not dependent on ε. As f is smooth we can Taylor expand f about 0 to obtain f(εy = f(0 + f (0εy + O(ε, which in turn can be written as f(εy = f(0 + f (0ε ( y 0 (t + εy (t + O(ε + O(ε = f(0 + f (0εy 0 (t + O(ε.

3 TMA495 Mathematical Modeling, 6..07, solutions Page 3 of 9 Inserting these two representations in the differential equation we obtain while the initial conditions become y 0 + εy + f(0 + f (0εy 0 + O(ε = 0, y 0 (0 + εy (0 + O(ε = 0, y 0 ( + εy ( + O(ε = ε. Equating terms of different order in ε, we obtain the system of equations: O( : y 0 + f(0 = 0, y 0 (0 = 0, y 0 ( = 0, O(ε : y + f (0y 0 = 0, y 0 (0 = 0, y ( =. Solving first for y 0 we obtain y 0 (x = f(0x + Ax + B, where we must take A = f(0 and B = 0 to satisfy the boundary conditions for y 0. We can now insert for y 0 in the expression for y to obtain y + f (0f(0(x x = 0, = y (x = f (0f(0 ( x 4 x3 6 + Cx + D, and the boundary conditions gives C = + 4 f (0f(0 and D = 0. We conclude that y(x = y 0 (x + εy (x + O(ε = f(0(x x + ε ( x + 4 f (0f(0(x 4 x 3 + x + O(ε. Problem 3 We set z(t = ρ(x(t, t and find that the characteristic equations in this case are given by ẋ = j (z = e z, x(0 = x 0 and ż = 0, z(0 = ρ 0 (x 0. Solving we get z(t = ρ 0 (x 0 and x(t = e ρ 0(x 0 t + x 0. a In this first case, we obtain the two family of (x-characteristics x(t = x 0 + et, x 0 < 0, x(t = x 0 + t, x 0 > 0.

4 TMA495 Mathematical Modeling, 6..07, solutions Page 4 of 9 The left characteristics overtake the right characteristics (collisions in the region t < x < et, so the physical solution ρ is a shock solution, x < s(t, ρ(x, t = 0, x > s(t, where the shock curve s(t starts at s = 0 and satisfies the Rankine-Hugoniot condition: ṡ = j( j(0 0 = e, s(0 = 0 = s(t = (e t. b In the second case, the two families of characteristics are given by x(t = x 0 + t, x 0 < 0, x(t = x 0 + et, x 0 > 0. The left characteristics are slower than the right characteristics so there is a region t < x < et not reached by any characteristic (a dead sector. Hence the physical solution is given by a rarefaction wave ρ(x, t = ϕ(x/t. From the PDE for ρ we find that x t ϕ + e ϕ t ϕ = 0 = φ 0 We then get the solution: e ϕ = x t 0, x < t, ρ(x, t = ln(x/t, t < x < et,, et < x. = ϕ(x/t = ln(x/t. Problem 4 a For simplicity we drop the -notation in this problem and let x, y denote points in R 3. We fix a point x in the rock formation and consider the ball B r := B(x, r, centered at x with radius r > 0, whose volume will be denoted B r. By conservation of mass we have d φρ dy = (j ˆndσ. ( dt B r B r Since we have a bounded smooth domain and smooth integrands, we may change the order of differentiation and integration and use the divergence theorem to get d φρ dy = φρ t dy, dt B r B r (j ˆndσ = ( j dy. B r B r

5 TMA495 Mathematical Modeling, 6..07, solutions Page 5 of 9 Consequently, ( can be rewritten as B r (φρ t + j dy = 0. (3 Let f := φρ t + j, and divide by B r, and add and subtract f(x in (3 to get f(x + (f(y f(x dy = 0. (4 B r B r Since f is continuous f(y f(x for y B r and r, and hence for r, B (f(y f(x dy 0 r dy B r B r B r = φρ t (x, t + j(x, t = f(x 0. Sending r 0 we get equality and the first equation for ρ in the problem. [A rigorous argument which is not required for this exam, it given below: f(y f(x dy B r = ( f(x + s(y x (y x ds dy, B r B r B r 0 max f r ds dy = B r 0 R 3 B r B r max f r 0 as r 0. ] R 3 B r Combining Darcy s law and ideal gas law, we get j = ρ k µ p = ρk RT ρ, µ where we used that R, T are constants. Inserting this into the equation for ρ, we obtain ρ t = krt µφ b Let x = Xx, y = Y y, z = Zz, t = tt, ρ = ρρ, and (ρ ρ =: κ (ρ ρ. (5 x, y, z, t, ρ, ρ x, ρ xx, ρ y, ρ yy, ρ z, ρ zz (scaling assumption. By ideal gas law and max p = p, max ρ = max p RT = p RT. Thus ρ = p RT is a natural scale for ρ. Good scales for x, y, z are X = L, Y = Z = 500 m = εl.

6 TMA495 Mathematical Modeling, 6..07, solutions Page 6 of 9 The time scale is determined introducing the scaled quatities into (5 and balancing, ( ρ ρ t ρ t = κ = ρ t = κρt L L (ρρ x x + ρ ε L (ρρ y y + ρ ε L (ρρ z z ( (ρρ x x + ε [(ρρ y y + (ρρ z z ]. To arrive at the desired equation, we set t = L /κρ. c When x > s(t, ρ F = 0, and we immediately get (ρ F t = 0 = (ρ F (ρ F x x. When x < s(t, ρ F = ct 3 6 x t, and we calculate to conclude that (ρ F t = 3 ct x t, (ρ F x = 3 xt, (ρ F xx = 3 t, (ρ F (ρ F x x = ((ρ F x + ρ F (ρ F xx = 9 x t 3 ct x t = 3 ct x t = (ρ F t. Now we will show that ρ F solves the conservation law in integral form on any 0 < a < s(t < b: The time change of mass inside [a, b] is equal the influx at x = a (i.e. +j(a, t plus the influx at x = b (i.e. j(b, t, and the (scaled flux here is j = ρρ x (see also the hint. Hence we have d b ρ dx = j(b, t + j(a, t = (ρρ x (b, t (ρρ x (a, t. (6 dt a Since ρ F is a (classical solution for x (a, s(t and ρ = 0 for x [s(t, b, we get d ρ F dx = d ( s(t b ρ F dx + ρ F dx dt [a,b] dt a s(t }{{} = ρ F (s(t, ts (t + }{{} = =0 s(t a =0 s(t a (ρ F (ρ F x x dx = ρρ x s(t a = ρρ x b a, (ρ F t dx (Leibniz rule

7 TMA495 Mathematical Modeling, 6..07, solutions Page 7 of 9 and ρ F satisfies (6. Since ρ F = 0 for x > s(t, we first note that s(t s(t ρ F dx = ρ F dx =. Then since also ρ F 0 and lim t 0 s(t = 0, for any continuous function f, s(t ρ F (x, tf(x dx f(0 = ρ F (x, t ( f(x f(0 dx s(t s(t = max f(x f(0 ρ F (x, t dx x s(t s(t = max f(x f(0 0 as t 0. x s(t The function ρ F is a fundamental solution of equation (5 on the exam, if it is a solution with initial condition equal to the delta function. The meaning of the initial condition is that lim t 0 ρ F (x, tf(x = f(0 for all continuous functions f. Both requirements are satisfied by the previous parts of problem c, and hence ρ F is a fundamental solution. d The idea of the method of intermediate asymptotics is to rescale the problem for large (but finite! times in such a way that the rescaled intitial data approximates the δ- function. Then the solution of the rescaled problem will be close to the fundamental solution of this problem. Going back to original variables, we get an approximation valid for large times. Let t = 3 months and t = 5 days. Here we want to rescale the initial value problem such that ρ t + κ(ρ ρ x x = 0, ρ (x, t = ρ 0(x, (i the time scale is t = 3 months (large compared to t, (ii the scaled equation becomes equation (5 on the exam (to be able to use part c, (iii the scaled initial data ρ(x, 0 δ(x, the δ-function. That is, we seek a scaled problem of the form ρ t + (ρρ x x = 0, ρ(x, 0 δ(x, (7 If we can do that, we can conclude (by stability of the PDE that the scaled solution ρ ρ F, where ρ F is the fundamental solution given in part c. Going back to original variables, we find an approximation of the solution ρ at t = t = 3 months.

8 TMA495 Mathematical Modeling, 6..07, solutions Page 8 of 9 Let us now show that (7 holds after scaling. Take ρ = ρρ, t = T t, and x = Xx. The time scale T = t, and we choose X and ρ such that (7 holds. From the equation, we find that ρ t + κ(ρ ρ x x = 0 = ρ t ρ t + κρ X (ρρ x x = 0 = ρ t + κρ t X (ρρ x x = 0, and then we get (7 if κρ t X =. Next we look at the initial condition. Recall that ρ 0 = 0 for x > l := 5 m and of total mass M := ρ dx. We want to choose a scaling such that ρ dx = and ρ(x, 0 = 0 for x > ε for some 0 < ε, or in other words, ρ(x, 0 δ(x. The integral conditions is satisfied if M = ρ dx = ρx ρ dx = ρx. We now have two equations for two unknowns X and ρ, and the solution is ( M 3 ρ =, X = ( κm t 3. κ t Note that ρ(x, t t = ρ ρ (Xx, t = 0 for Xx l or x l X. Since t = 3months s, a quick calculation shows that X = ( 3 κmt ( m = 000 m 5 m = l. Hence l X, and since also t t, it follows that ρ(x, 0 ρ(x, t t δ(x. We conclude that an approximation of ρ at t = t = 3 months, is given by ( x ρ (x, t ρρ F X, t t ( M ( 3 x = ρ F κ t (κm t 3,. As the rock formation is surrounded by impermeable rock, there can be no flux at the boundaries, x = ± L. Thus the boundary conditions are 0 = j (± L, T = ρ (± L, T ρ x (± L, T.

9 TMA495 Mathematical Modeling, 6..07, solutions Page 9 of 9 This is indeed satisfied by our approximation since L X = L (κm t 3 =.5 >.65 (6c = s(, and then since ρ F (x, t = 0 for x > s(t, ρ (± L, t = ρρ F ( L X, = 0.