Modeling with first order equations (Sect. 2.3). The mathematical modeling of natural processes.

Transcription

1 Modeling with first order equations (Sect. 2.3). The mathematical modeling of natural processes. Main example: Salt in a water tank. The experimental device. The main equations. Analysis of the mathematical model. Predictions for particular situations. The mathematical modeling of natural processes. Remarks: Physics describes natural processes with mathematical constructions, called physical theories. More often than not these physical theories contain differential equations. Natural processes are described through solutions of differential equations. Usually a physical theory, constructed to describe all known natural processes, predicts yet unknown natural processes. If the prediction is verified by an experiment or observation, one says that we have unveiled a secret from nature.

2 Salt in a water tank. Problem: Study the mass conservation law. Particular situation: Salt concentration in water. Main ideas of the test: Assuming the mass of salt and water is conserved, we construct a mathematical model for the salt concentration in water. We study the predictions of this mathematical description. If the description agrees with the observation of the natural process, then we conclude that the conservation of mass law holds for salt in water. Modeling with first order equations (Sect. 2.3). The mathematical modeling of natural processes. Main example: Salt in a water tank. The experimental device. The main equations. Analysis of the mathematical model. Predictions for particular situations.

3 The experimental device. water salt pipe r i q (t) i V (t) tank Q (t) r o q (t) o instantaneously mixed The experimental device. Definitions: r i (t), r o (t): Rates in and out of water entering and leaving the tank at the time t. q i (t), q o (t): Salt concentration of the water entering and leaving the tank at the time t. V (t): Water volume in the tank at the time t. Q(t): Salt mass in the tank at the time t. Units: [ ri (t) = [ r o (t) = Volume Time, [ qi (t) = [ q o (t) = Mass Volume. [ V (t) = Volume, [ Q(t) = Mass.

4 Modeling with first order equations (Sect. 2.3). The mathematical modeling of natural processes. Main example: Salt in a water tank. The experimental device. The main equations. Analysis of the mathematical model. Predictions for particular situations. The main equations. Remark: The mass conservation provides the main equations of the mathematical description for salt in water. Main equations: d dt V (t) = r i(t) r o (t), Volume conservation, (1) d dt Q(t) = r i(t) q i (t) r o (t) q o (t), Mass conservation, (2) q o (t) = Q(t), Instantaneously mixed, (3) V (t) r i, r o : Constants. (4)

5 The main equations. Remarks: [ dv dt = Volume Time = [ r i r o, [ dq dt = Mass Time = [ r i q i r o q o, [r i q i r o q o = Volume Time Mass Volume = Mass Time. Modeling with first order equations (Sect. 2.3). The mathematical modeling of natural processes. Main example: Salt in a water tank. The experimental device. The main equations. Analysis of the mathematical model. Predictions for particular situations.

6 Analysis of the mathematical model. Eqs. (4) and (1) imply V (t) = (r i r o ) t + V, (5) where V () = V is the initial volume of water in the tank. Eqs. (3) and (2) imply Eqs. (5) and (6) imply d dt Q(t) = r Q(t) i q i (t) r o V (t). (6) d dt Q(t) = r i q i (t) r o (r i r o ) t + V Q(t). (7) Analysis of the mathematical model. Recall: d dt Q(t) = r i q i (t) r o (r i r o ) t + V Q(t). Notation: a(t) = r o (r i r o ) t + V, and b(t) = r i q i (t). The main equation of the description is given by Q (t) = a(t) Q(t) + b(t). Linear ODE for Q. Solution: Integrating factor method. Q(t) = 1 [ t Q + µ(s) b(s) ds µ(t) with Q() = Q, where µ(t) = e A(t) and A(t) = t a(s) ds.

7 Modeling with first order equations (Sect. 2.3). The mathematical modeling of natural processes. Main example: Salt in a water tank. The experimental device. The main equations. Analysis of the mathematical model. Predictions for particular situations. Predictions for particular situations. Assume that r i = r o = r and q i are constants. If r, q i, Q and V are given, find Q(t). Solution: Always holds Q (t) = a(t) Q(t) + b(t). In this case: a(t) = r o (r i r o ) t + V a(t) = r V = a, We need to solve the IVP: b(t) = r i q i (t) b(t) = rq i = b. Q (t) = a Q(t) + b, Q() = Q.

8 Predictions for particular situations. Assume that r i = r o = r and q i are constants. If r, q i, Q and V are given, find Q(t). Solution: Recall the IVP: Q (t) = a Q(t) + b, Q() = Q. Integrating factor method: A(t) = a t, µ(t) = e a t, Q(t) = 1 µ(t) t [ t Q + µ(s) b ds. µ(s) b ds = b a ( e a t 1 ) Q(t) = e a t [ Q + b a ( e a t 1 ). So: Q(t) = ( Q b a ) e a t + b a. But b V = rq i a r We conclude: Q(t) = ( Q q i V ) e rt/v + q i V. = q i V. Predictions for particular situations. Assume that r i = r o = r and q i are constants. If r, q i, Q and V are given, find Q(t). Solution: Recall: Q(t) = ( Q q i V ) e rt/v + q i V. Particular cases: Q V > q i ; Q V = q i, so Q(t) = Q ; Q V < q i. Q Q Q Q = q i V Q Q t

9 Predictions for particular situations. Assume that r i = r o = r and q i are constants. If r = 2 liters/min, q i =, V = 2 liters, Q /V = 1 grams/liter, find t 1 such that q(t 1 ) = Q(t 1 )/V (t 1 ) is 1% the initial value. Solution: This problem is a particular case q i = of the previous. Since Q(t) = ( Q q i V ) e rt/v + q i V, we get Q(t) = Q e rt/v. Since V (t) = (r i r o ) t + V and r i = r o, we obtain V (t) = V. So q(t) = Q(t)/V (t) is given by q(t) = Q V e rt/v. Therefore, 1 1 Q V = q(t 1 ) = Q V e rt 1/V e rt 1/V = 1 1. Predictions for particular situations. Assume that r i = r o = r and q i are constants. If r = 2 liters/min, q i =, V = 2 liters, Q /V = 1 grams/liter, find t 1 such that q(t 1 ) = Q(t 1 )/V (t 1 ) is 1% the initial value. Solution: Recall: e rt 1/V = 1 1. Then, r ( 1 ) t 1 = ln = ln(1) V 1 r t 1 = ln(1). V We conclude that t 1 = V r In this case: t 1 = 1 ln(1). ln(1).

10 Predictions for particular situations. Assume that r i = r o = r are constants. If r = 5x1 6 gal/year, q i (t) = 2 + sin(2t) grams/gal, V = 1 6 gal, Q =, find Q(t). Solution: Recall: Q (t) = a(t) Q(t) + b(t). In this case: a(t) = r o (r i r o ) t + V a(t) = r V = a, b(t) = r i q i (t) b(t) = r [ 2 + sin(2t). We need to solve the IVP: Q (t) = a Q(t) + b(t), Q() =. Q(t) = 1 µ(t) t µ(s) b(s) ds, µ(t) = e a t, We conclude: Q(t) = re rt/v t e rs/v [ 2 + sin(2s) ds.