TMA 4195 Mathematical Modeling, 30 November 2006 Solution with additional comments

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1 TMA 495 Mathematical Modeling, 30 November 006 Solution with additional comments Problem A cookbook states a cooking time of 3.5 hours at the oven temperature T oven For a 0 kg (American) turkey, 7 hours is suggested. (a) How can we derive the heat equation for a 3 kg = kr T; () where the density, the heat capacity c, and the heat transmission k, are constant? We assume that the turkey s heating follows the equation. The heat transfer through the turkey s surface follows Newton s law of heating, where is a constant. krt j surface = (T oven T surface ) ; () (b) The cooking time, t s, has to be dependent on, in addiction to the parameters above, the turkey s diameter D, the temperature di erence between the oven and the turkey before ( T 0 ) and after ( T ) the cooking (We assume that all turkeys are geometrically similar!). Use dimension analysis to nd an epression for t s (Hint: combine the parameters in Eqns. and before you use them in the dimension analysis). (c) After short time, the turkey s surface reach the same temperature of the oven. This implies that equation simpli es to T oven = T surface, and the parameter with disappears. Use this information to simplify dimension analysis and show that t s is proportional to the turkey s weight to the power /3. How does this match with what has been indicated in the cookbook? (a) The heat density is given by ct, and the heat u is j = we obtain the integral conservation law Z Z d ct dv + j nd = 0; krt. Without source terms, This implies - by moving the derivative into the integral, using the divergence theorem, and letting R vary - the di erential formulation c dt + r ( krt ) = 0: (b) We follow the hint in the Problem and assume that t s = t s D; ; k ; T 0; T ; where the heat di usion coe cient is = k= (c).

2 Digression: One could think that T surface and rt j surface should also be considered. However, these quantities are dependent on time and dependent on the temperature variation. This variation is dependent on =k and other parameters. From the equations, we derive that [] = s/m and [=k] = m. Thus, we are able to state the dimension matri as t s D =k T 0 T m s K The matri has rank 3 and, hence we have 3 dimensionless combinations which are easily derived: Thus = ( ; 3 ), or The alternative epression is equally correct. = t s D ; = D k ; 3 = T 0 T ; t s = D D k ; T 0 : T t s = D k ; T 0 T The most elegant solution (found by one of the students) is however to scale the equations and using D for length and t s for time. Then and drop out immediately. (c) If we get rid of, we nd t s = D T0 Since the weight can be written as W = GD 3, where G is a constant dimensionless geometry factor, t s = (W=G)=3 T0 _ W =3 : T With cooking time for the 3 kg turkey is 3.5 hours, the cooking time for a 0 kg turkey should be (assuming that T 0 and T remain the same) t s (0kg) = t s (3kg) T : 0 =3 0 =3 = 3:5 hour = 7:8 hour. 3 3 Eight hours for a 0 kg turkey would therefore be a better rule than seven hours. Problem Let us consider the problem " d y + dy = t; 0 t ; 0 < " ; y (0) = y () = :

3 What do we call such problems? Find, to leading order in ", the outer, inner, and uniform solutions to the problem (Hint: the boundary layer is near t = 0). Since this is an equation with a small parameter in front of the highest derivative, it is called a singular perturbation problem. The leading order outer solution, y 0 (t), is found by setting " = 0, dy 0 = t: The general solution is A + t, but it is impossible to ful l both the boundary conditions, since y (0) = implies that y 0 (t) = + t, and y () = implies that y 0 (t) = t. Given the hint, it is reasonable to try a new time scale, t = around 0. This leads to By choosing = ", the equation becomes " Y + Y = : Y + Y = " ; with a general solution to the leading order, Y 0 () = A + Be (One could think that = " = was a possibility, but this would give an equation Y 0 = 0 which does not help us). If we ful l the boundary condition Y 0 (0) =, we obtain A + B =, or Y 0 () = + A e : The solution would satisfy both boundary conditions if we set A = 0, but we then apply Y 0 outside its admissible region (this misuse of singular perturbation is sometimes seen in science). The matching condition is, in its simplest form, given by and this leads to, by using y 0 () =, lim y 0 (t) = lim Y 0 () ; t!0! 0 = A; or A =. Thus, the uniform solution to the leading order is y (u) t 0 = y 0 (t) + Y 0 y 0 (0) = t + e t=" : " The error around t = is totally negligible. Digression (not part of the eam): The eact solution to the equation is y e (t) = A + Be t " + t t"; and, thus, with a negligible error O e =", y e = t + e t=" + " e t " t : Note that to the leading order in " means the O () term and not O ("). 3

4 3 Problem Let (t ) be the cod population in an ocean region as a function of time t. The region may hold a maimum sustainable amount of sh equal to K, and, as long as shing is prohibited and K, will grow with rate r, d = = r. We assume that the amount of caught sh per time unit is B, where B is the number of participating boats, and is a constant. (a) State a model for the amount of sh such that, using a suitable scale, we obtain the form d (t) = (t) (t) (t) : (b) Discuss the equilibrium points for di erent values of. Sketch the possible trajectories path for the amount of sh. (c) Find an epression for the number of boats giving an optimal management of the sh resources. (a) The equation suggests we should assume a logistic model in absence of shing activities. Thus, we nd immediately that d = r B K We scale the model by setting = K; t = r t Thus, or Kr d = rk ( ) KB; _ = ( ) ; = B r : (b) The equilibrium solutions are = 0; = : Using straightforward linear stability and d ( ( ) ) d = ; we obtain that is stable for > and unstable for <, while is stable for < and unstable for > (the physical acceptable points are clearly, 0). 4

5 If =, the equation becomes _ = ; and both equilibrium solutions coincide at = 0. This equation can be solved generally: i.e. = t + C; = t + C ; and also = 0, as obviously is a solution. If we start in (0) > 0 and close to 0, we have (t)! 0 when t!. If, on the contrary and un-physically, (0) = =C < 0, we have (t)! when t! C. The equilibrium point is stable for =, since we have that 0, and the answer is as follows: s = ma f ; 0g ; 0 is stable, while is unstable. u = 0; 0 < ; (c) A constant outtake per time unit can be epressed as Q = (B) = (B) K( ) = (r) K( ) = Kr( ); (3) and this has a maimum for = =, that is B = r=(), and = K=. 4 Problem In this problem we study the tra c along a road (a one-way street towards + and without in- or out- ow for the moment). We further assume that all variables are scaled so that the car density is between 0 and, and the car speed v is. (a) Show how one derives an epression for the speed U of a shock in the car density, and that we in this case obtain U =, where and are the densities on each side of the shock. Assume that the car density along the road for t < 0 is equal to =. Between t = 0 and t = the cars get a red light due to a pedestrian crossing placed in = 0. For t >, the light is green. (b) Find the solution (; t) for t 0. (Hint: make a sketch of the situation in an -t-diagram. Show that the solution for has to be determined in 5 di erent regions, where the values in 4 of them is obvious. In order to nd the regions, one has to nd their borders). A second road (with similar properties as the rst one) is now merging from the side with the rst one. (c) Which conservation law has to be ful lled at the merging point? Assume that the u on the rst road is constant, j = =8, and that the density is less than =. Describe (without further calculations) the evolution of the car density on the rst road when the density on the second road increases from 0 to. The cars on the rst road are eible and let entering cars merge whenever it is possible. Consider in particular what happens when the u on the second road reaches =8. 5

6 C t D A Epansion wave B G ρ = ρ =0 ρ =/ ρ =/ O Figur : Sketch of the situation around the crossing. This is the so-called standard model: 0 ; v = ; j = ( ). The conservation law can be written as Z d b d + j (b) a j (a) = 0; (4) or in di erential form, t + ( ( )) = t + ( ) = 0: (a) We nd out the shock speed by considering a shock with speed U in the interval [a; b]. The density to the left of the shock is, while the one on the right. By using the equation 4, we obtain ( ) U = ( ) ( ) ; or,. U = : (b) For = = the characteristics are vertical, and we have a situation as sketched in gure For the shock OA, = and = =, in other words, the shock speed is U OA = =. In an analogous way we nd that U OB = =, U GB = and U GA =. The point A is at =, t =, and B is at =, t =. Within the rarefaction wave, the solution is given by or = =, while = e (; t) = = c () (t ) = ( ) (t ) ; e (; t) = : t It remains to nd the shocks AC and BD. In a point (; t) on the shock AC we have that t. Therefore, U (; t) = 6 = t (t ) :

7 j j j out Figur : The second road merges with the rst road. The equation for the shock becomes _ = () = : (t ) ; The equation is separable and the solution is AC (t) = p t ; t. In a equivalent way we nd BD = p t. Thus, the solution is completely determined. (c) The situation is sketched in gure. Since the crossing is not a parking lot, we must have We know that j = =8, but the equation j + j = j out : j = 8 = ( ) has to possible solutions, ; = r 8 : Here, since we also know that < =, the density is equal to As long as j < =8 the u j out < =4, the maimum value. When j passes =8, cars will pile up on the second road in front of the crossing. Since the out ow from the second way can at most be =8, the cars density just before the crossroad becomes f = + r 8 : In the back of the line we have a shock moving backward with speed U = f : up to = f. When increases further from f, the u of the cars on the second road will be so small (</8) that all may enter without problems! q 8. 7