4.7 Case studies 175. tank 1 tank 2

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1 47 Case studies 75 s:47c 47 Case studies 47 Two-tank mixing problem s:563 g/lit ct g/lit V lit V lit tank tank 2 Fig 45 color online A cartoon which depicts a two-tank extension of the one-tank problem considered in Chapter 36 Here two identical tanks are connected, except that a solution flows into one tank, but fresh water flows into the other As for the one-tank problem, the incoming concentration ct is not necessarily constant f:twotankmix3 Consider the two-tank mixing problem illustrated in Figure 45 This system is a natural extension of the one-tank problem considered in Chapter 36 Following the same argument as that presented in Chapter 4, if we let x denote the grams of salt in tank and x 2 the grams of salt in tank 2, then the governing equation for the system is the IVP x = Ax + av ct x 2a a, x = ; A = 47 e:343 x 2 a 2a We will be interested in the concentrations in each tank for large time First consider the homogeneous system The eigenvalues and associated eigenvectors for the matrix A are λ = a, v = The homogeneous solution is then x h t = c e at ; λ 2 = 3a, v 2 = + c 2 e 3at Since both eigenvalues are negative, the homogeneous solution x h t satisfies x h t as t + Recalling the discussion leading to Theorem 442, we know that for the full problem the initial data is taken care of via an appropriate choice of the constants c,c 2

2 76 4 Systems of first-order linear differential equations Consequently, the concentration in each tank for large times will be described by the particular solution, and will not depend on the initial concentration in each tank First suppose that ct = c, ie, the incoming concentration is constant Using the method of undetermined coefficients we know that the particular solution will be of the form av c x p t = a Aa + = Solving the linear system yields x p t = 3 V c 2 Since the concentration in each tank is given by we see that after a long time c t = x t V, c 2 t = x 2t V, 472 e:defcab c t 2 3 c, c 2 t 3 c The tank which receives the mixed solution has a higher concentration than that which receives the fresh water Moreover, we have the physically plausible result that the sum of the asymptotic concentrations in each tank is equal to the incoming concentration Now, as in the one-tank problem of Chapter 36 suppose that the incoming concentration is sinusoidal with average c, ie, ct = c cosωt, ω > Using the above result we know that the particular solution will be of the form x p t = 3 V c 2 + cosωta + sinωta Plugging this guess into the system 47 and equating the cosine and sine terms yields the pair of linear systems Aa ωa = av c, ωa + A = In block matrix form this pair of linear systems can be rewritten as the system A ωi2 ωi 2 A a a av c = We will solve the algebraic system using WolframAlpha

3 47 Case studies 77 row reduce 2a,a, b,,a V c, a, 2a,, b,, b,, 2a,a,,,b,a, 2a, Input: row reduce 2 a a b a c V a 2 a b b 2 a a b a 2 a Result: 2 a2 c V 3 a 2 b 2 9 a 4 a 2 b 2 b 4 a2 c V 3 a 2 b 2 9 a 4 a 2 b 2 b 4 a c V 5 a2 b b 3 9 a 4 a 2 b 2 b 4 4 a 3 b c V 9 a 4 a 2 b 2 b 4 Step by step solution In the input Dimensions: we used the substitution b = ω The solution to the linear system is the last column The4 subscripts rows 5 columns for the coefficients were not used in the CAS computation, so we should read the output with c c and V V Upon setting 9a 4 + a 2 b 2 + b 4 = 9a 4 + a 2 ω 2 + ω 4 >, which as WolframAlpha will tell you is the determinant of the coefficient matrix, upon one final simplification the undetermined vectors are a = a2 V c 6a 2 + 2ω 2 3a 2 ω 2, a = av c ω 5a 2 + ω 2 4a 2 Upon collecting terms we see the particular solution is x p t = 2 3 V c a2 V c 6a 2 + 2ω 2 3a 2 ω 2 cosωt av c ω 5a 2 + ω 2 4a 2 sinωt In terms of the individual components we have x,p t = 2 3 V c av c x 2,p t = 3 V c av c [ a6a 2 + 2ω 2 cosωt + ω5a 2 + ω 2 sinωt ] [ a3a 2 ω 2 cosωt + 4ωa 2 sinωt ] Generated by Wolfram Alpha wwwwolframalphacom on March 27, 24 from Champaign, IL Wolfram Alpha LLC A Wolfram Research Company

4 78 4 Systems of first-order linear differential equations π 6 tank 3π/4 tank 2 A * ω 4 φ ω π/2 2 tank 2 π/4 tank ω ω Fig 46 color online A plot of the amplitudes A j ω left panel and φ j ω right panel for the variation about the mean concentration in each tank when a = 45 The curves for tank are given by a dashed blue curve, and those for tank 2 by a solid red curve Note that the variation about the mean in the first tank - that closest to the incoming solute - is always greater than that in the second tank However, there is more of a phase-lag for the concentration in the tank which is furthest from the incoming solute f:twotankamplitudephas In order to better interpret the solution, we will now use the identity c cosωt + c 2 sinωt = c 2 + c2 2 cosωt φ, tanφ = c 2 c After some simplification this identity yields av c av c with the amplitudes being and the phases satisfying [ a6a 2 + 2ω 2 cosωt + ω5a 2 + ω 2 sinωt ] = V c A ωcosωt φ ω [ a3a 2 ω 2 cosωt + 4ωa 2 sinωt ] = V c A 2 ωcosωt φ 2 ω, A ω a a 2 6a 2 + 2ω ω 2 5a 2 + ω 2 2 A 2 ω a a 2 3a 2 ω a 4 ω 2,

5 47 Case studies 79 tanφ ω ω5a2 + ω 2 a6a 2 + 2ω 2, tanφ 4ωa2 2 ω a3a 2 ω 2 It can be checked that A = 2 3, A 2 = 3, and moreover for fixed a each amplitude function is a strictly decreasing function of the frequency ω Using 472 it is that case that after a long time the concentration in each tank is c t 2 3 c c A cosωt φ, c 2t 3 c c A 2 cosωt φ 2 The mean concentration in each tank is the same as for the previous case of constant incoming concentration Just as in the one-tank example of Chapter 36 there is a variation about the mean; moreover, the variation decreases as the frequency increases 2c_ 5c_/3 4c_/3 tank 2 tank ct c_ 2c_/3 c_/ t Fig 47 color online A plot of the solution to 47 for large t when a = 45,ω = 3, and c = The incoming concentration is given by the thin green curve, the concentration in tank is given by the dashed blue curve, and the concentration in tank 2 is denoted by the solid red curve Note that the incoming concentration varies about its mean c, the concentration in tank varies about its mean 2c /3, and the concentration in tank 2 varies about its mean c /3 f:twotanksolution A sample plot of the amplitudes and phases is given in Figure 46 for the case that a = 45 Note that in tank there will more of a variation about the mean concentration in that tank, but that in tank 2 the larger phase value implies there will be more lag between the variation in the incoming solute and that in the tank A solution curve for the concentration in each tank for t is given in Figure 47 when a = 45,ω = 3, and c = Here we clearly see that the concentration in each tank varies about its mean, and that the amount of variation is markedly different for each tank Further note the phase lag: the maximum concentration in tank 2 occurs at a later time than that in tank, which in turn occurs at a later time than that for the incoming solute For large frequency

6 8 4 Systems of first-order linear differential equations the lag in tank is approximately T /4, whereas the lag for tank 2 is approximately T /2 Here T refers to the period of oscillation, ie, T = 2π/ω 472 Lead in the human body s:lead Lead poisoning is a medical condition which is caused by increased levels of the heavy metal lead in the body Lead interferes with a variety of body processes and is toxic to many organs and tissues including the heart, bones, intestines, kidneys, and reproductive and nervous systems It interferes with the development of the nervous system; therefore, it is particularly toxic to children, as it causes potentially permanent learning and behavior disorders Routes of exposure to lead include contaminated air, water, soil, food, and consumer products One of the largest threats to children is lead paint that exists in many homes, especially older ones No safe threshold for lead exposure has been discovered; in other words, as far as is known, any amount of lead whatsoever will cause harm to the body In Borelli and Coleman [5, Chapter 6] a mathematical model is derived for the ingestion and absorption of lead in the body for a healthy male volunteer in Los Angeles: x = 3 36 x ,875 x , x 3 + I L t x 2 = 9 x 35 x 2 x 3 = 7 8 x 7 2, x e:lead The variable x corresponds to the amount of lead in the blood, x 2 is the amount of lead in tissue, and x 3 is the amount of lead in the bones The amount is measured in micrograms, and the rate of change is in micrograms/day The forcing term I L represents the rate of ingestion of lead into the blood We wish to determine how the amount of lead in each component of the body changes as a function of time Written as a first-order system the ODEs become I L t x = Ax +, A = 3/36 272/2, 875 7/2, /9 /35 7/8 7/2, First consider the homogeneous system Using WolframAlpha the eigenvalues and associated eigenvectors of A are

7 47 Case studies 8 eigenvalues 3 36, ,7 2, 9, 35,, 7 8,, 7 2 Input: Eigenvalues Results: Λ Λ Λ More digits Corresponding eigenvectors: Exact forms More digits v 482, 79636, v , , v 3 237, 43667, In scientific Alternate notation form: the eigenvalues are root of x x x near x 44687, λ 447 2, λ 2 2 2, λ root of x x x near x 99998, Since all of the eigenvalues are negative, upon arguing as in the previous example of the two-tank mixing problem it will be the case that the homogeneous solution satisfies x h t as t + However, the smallness of the eigenvalues in absolute value implies 3 in this case that large means times of 2 O , which is2 on the 37order 8 of years recall that t is in terms of days! Thus, while it is mathematically the case that for large times the amount of lead in each part of the body will not depend upon the initial amount of lead in the body, from a practical perspective this is simply not true Now consider the particular solution under the assumption of a constant ingestion rate of I L t = 493 The particular solution will be of the form 493 x p t = a Aa + =, so that upon solving using WolframAlpha,, 848, 75/27, 84 x p t = 2, 56, 875/ , 25, /38 2, 623 Generated by Wolfram Alpha wwwwolframalphacom on April 4, 24 from Champaign, IL Wolfram Alpha LLC A Wolfram Research Company

8 82 4 Systems of first-order linear differential equations blood bone 5 tissue t years Fig 48 color online A plot of the solution to the system 473 with the initial condition x = x 2 = x 3 = The ingestion rate is given by I L = 493 for t < 365, and I L = for t 365; in other words, there is a steady intake of lead in the first year, and afterwards the environment is lead-free The amount of lead in the bone is given by the solid red curve, the amount of lead in the blood by the dashed blue curve, and the amount of lead in the tissue is given by the solid green curve Note that the lead in the blood and tissue clears out after a year or so of living in a lead-free environment, but that there is still a significant amount of lead in the bone f:leadbook From zero initial data, and after approximately one year, the amount of lead in the blood and tissue, respectively, is given by x t,57778, x 2 t 63 These values are relatively close to the equilibrium values On the other hand, after 8 years x 3 t 9,8627: the amount of lead in the blood at this time is roughly only 6% of the equilibrium value In other words, the amount of lead in the bones continually increases until death, and never reaches an equilibrium value Now let us briefly consider the problem of what happens to the amount of lead in the body after there is no more lead in the environment to be ingested In particular, suppose that 493, t < 365 I L t =,, 365 t and further suppose that at time t = the body contains no lead In other words, we are supposing that the body is continually exposed to lead for one year, and is afterward placed in a lead-free environment The solution is plotted in Figure 48 Here we see that after one year of living in a lead-free environment there will no longer be any significant amounts of lead in the blood or tissue On the other hand, there will still be a significant amount of lead in the bone Indeed, while it is not shown in this figure, there will be 93 micrograms of lead in the bone after 8 years of living in a lead-free environment Unfortunately, exposure to lead for a relatively short time means that according to this mathematical model you will carry that lead with you for the rest of your life Why does it take so long for the lead to clear out of the bones? The disparity in the size of the eigenvalues means that for large time the solution will approximately be

9 47 Case studies 83 In this case, since xt c 3 e λ 3t v 3, t t > 58 e λ 2t,e λ 3t < 5 large time is on the order of about two years Since an eigenvector is unique only up to scalar multiplication, we can rescale it so that each entry corresponds to a percentage this was done in Chapter 232 during our study of the Northern spotted owl We have 237, 43667, Input interpretation: 237, 43667, Result: 863, , ie, the rescaled Total: eigenvector is v Vector length: Thus, of the remaining lead in the body, approximately % is in the blood, 4% is in the tissue, and 9984% is in the bones As a final remark, the size of the eigenvalue λ 3 means that Normalized it will vector: take at least t = O 5 days which is O 3 years! before this final term becomes 237, negligible 43667, In other words, once the lead has been ingested, it will never really leave your body Exercises Spherical coordinates radial, polar, azimuthal : r , Θ , Φ 2273 Exercise 47 Consider the system of two interconnected tanks given below: