Boundary Layers and Singular Perturbation Lectures 16 and 17 Boundary Layers and Singular Perturbation. x% 0 Ω0æ + Kx% 1 Ω0æ + ` : 0. (9.

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1 Lectues 16 and 17 Bounday Layes and Singula Petubation A Regula Petubation In some physical poblems, the solution is dependent on a paamete K. When the paamete K is vey small, it is natual to expect that the solution not be vey diffeent fom the one with K set to zeo. Conside fo example the motion of a hamonic oscillato of unit mass with the sping constant equal to Ω1 + Ke?t æ. The equation of motion fo this poblem is x6 + Ω1 + Ke?t æx : 0, (9.1) with xω0, Kæ : 1, x%ω0, Kæ : 0, (9.2) whee xωt, Kæ is the displacement of the hamonic oscillato and x% is the time deivative of x. Equation (9.1) can be solved in a closed fom and the exact solution is a Bessel function. But if K is vey small, the sping constant is appoximately equal to unity at all times t ; 0, and to the lowest-ode of appoximation, we expect that we may ignoe its small deviation fom unity. Thus we have xωt, Kæ u xωt, 0æ : cos t. (9.3) To find the small coection due to the small deviation of the sping constant fom unity, one may anticipate that, since the deviation is of the ode of K, this small coection is of the ode of K as well. Indeed, since all tems in (9.1) ae analytic functions of K, we will ty to see if thee is a solution in the fom of a Taylo seies of K: xωt, Kæ u x 0 Ωtæ + Kx 1 Ωtæ + ` + K n x n Ωtæ + `, (9.4) whee x n Ωtæ is independent of K. Substituting (9.4) into (9.1) and (9.2), we get d 2 dt øx 0 Ωtæ + Kx 1 Ωtæ + ` + Ke?t øx 0 Ωtæ + Kx 1 Ωtæ + ` : 0, (9.5) x 0 Ω0æ + Kx 1 Ω0æ + ` : 1, (9.6) and x% 0 Ω0æ + Kx% 1 Ω0æ + ` : 0. (9.7) Setting K in the equations above to zeo, we get x6 0 + x 0 : 0, with x 0 Ω0æ : 1, and x% 0 Ω0æ : 0. The solution x 0 Ωtæ is just xωt, 0æ given by (9.3). Setting to zeo the coefficient of K on the leftside of (9.5), we get x6 1 + x 1 :?e?t cos t. (9.8) Similaly, we get fom (9.6) and (9.7) that x 1 Ω0æ : 0. (9.9) x% 1 Ω0æ : 0. (9.10) The solution of (9.8) satisfying (9.9) and (9.10) is 5 cos t? 3 sin t + e?t 2 sin t x1 Ωtæ : 1? cos t. 5 5 Since both x 1 Ωtæ and x 0 Ωtæ ae OΩ1æ at all times and since K is vey small, we have Kx 1 Ωtæ ò x 0 Ωtæ at all times, except when t is nea the points whee x 0 Ωtæ happens to vanish. One may futhe obtain the n th -ode coection x n Ωtæ fom (9.5) (9.7). We shall find that x n Ωtæ is OΩ1æ at all times. Thus 1

2 K n x n Ωtæ ò K n?1 x n?1 Ωtæ. The inequalities above indicate that keeping moe and moe tems in the seies of (9.4) makes the appoximate solution bette and bette. The seies (9.4) is a petubation seies, and the method given above in obtaining a petubation seies is called that of egula petubation. B Bounday Laye Theoy It may be supising, but not all poblems with a small paamete can be solved by egula petubation. Let us stat the discussion with a simple example. Conside the solution of the diffeential equation K dy + y : 0 (9.11) dx with the initial condition yω0æ : 1. If we set K in the diffeential equation above to zeo, we get yωxæ : 0. But this appoximate solution does not satisfy the initial condition. In ode to undestand why this is so, we solve this diffeential equation in a closed fom. We find that /K yωxæ : e?x. This solution of the diffeential equation satisfies the initial condition fo all values of K not equal to zeo. But if we set K to zeo, the solution above vanishes, and the initial condition is not satisfied. Indeed, consideed as a function of K, the solution of the diffeential equation has an essential singulaity at K : 0. It is theefoe no wonde that it does not have a Maclauin seies expansion. As a compaison, conside the solution of the equation dy + Ky : 0 (9.12) dx with the same initial condition. The exact solution is yωxæ : e?kx. The appoximate solution obtained by setting K in the diffeential equation to zeo is yωxæ : 1 which is a good appoximation of the exact solution when Kx is vey small. What makes the method of egula petubation applicable fo one but not fo the othe? The answe lies in the fact that if we set K to zeo, eq. (9.12) emains a fist-ode diffeential equation, while eq. (9.11) tuns into an algebaic equation. Indeed, while it is always tue that the magnitude of Ky is small compaed to that of y, it is not always tue that the magnitude of Ky is small compaed to that of y. This is because, if yωxæ is a apidly vaying function of x, the magnitude of the deivative of yωxæ is much lage than yωxæ. In such a case, Ky Ωxæ may be of the same ode of magnitude as yωxæ. / Indeed, we may easily veify that this is tue fo the function yωxæ : e?x K, the solution of (9.11). We encounte simila difficulties with bounday laye poblems in fluid mechanics. Let us conside the second-ode diffeential equation Ky Ωxæ + aωxæy Ωxæ + bωxæyωxæ : 0, (9.13) which geneally cannot be solved in a closed fom. We shall choose the inteval in which (9.13) holds to be ø0, 1. Typically, we equie that the solution satisfies the bounday conditions: yω0æ : A, yω1æ : B. (9.14) The paamete K is vey small, while A and B ae OΩ1æ. Note that the small paamete K is the coefficient of y. This poblem is a simplified model of the bounday laye poblem in fluid mechanics. As we know, the highest-ode tem in the Navie-Stokes equation in fluid mechanics is equal to T4 2 v3, whee v3 is the velocity of the fluid and T is the kinematic viscosity. The kinematic viscosity is often a small paamete, and the ode of the Navie-Stokes equation deceases if we ignoe the viscosity. Equation (9.13) can be solved with good accuacy with the use of the WKB method. (See 2

3 homewok poblem 1.) Howeve, we gain moe insights to this poblem by stating afesh. Let us set K : 0. Then (9.13) becomes This equation is eadily solved and we get aωxæy + bωxæy : 0. y s Ωxæ : exp?x bωxæ dx. (9.15) aωxæ The fact is that eq. (9.13) is of the second ode, and has two independent solutions. But the appoximate equation we obtain is of the fist ode, and has only one independent solution. Thus the appoximate method we use yields only one of the independent solutions. This is to say that we lose one of the independent solutions as we set K in the diffeential equation to zeo. It is impossible to satisfy the two bounday conditions of (9.14) with only one solution. How do we find the missing solution appoximately? To get some insight into the answe, let us go to the much simple case in which a and b ae constants. Then the geneal solution of (9.13) is equal to c 1 e 1x + c 2 e 2x, whee c 1 and c 2 ae abitay constants and 1 and 2 ae the oots of the quadatic equation K 2 + a + b : 0. This quadatic equation can be solved easily. But if we set K in this quadatic equation to zeo, we get only one oot u?b/a, which gives the solution e?bx/a. The solution is equal to the one denoted by y s Ωxæ in (9.15) with aωxæ and bωxæ eplaced by the constants a and b, espectively. The othe oot of the quadatic equation is obtained by solving the equation exactly. When K is vey small, this oot is appoximately given by? a/k. This oot tells us that the second solution of the diffeential equation is appoximately e?ax/k. We see that this solution is a apidly vaying function of x. Indeed, it is a apidly deceasing function of x if a is positive, and a apidly inceasing function of x if a is negative. That the missing solution is a apidly vaying function of x also explains why it is not always legitimate to dop the y tem in (9.13). Let us denote the apidly vaying solution by y Ωxæ. Then y Ωxæ is of the ode of K?1 times y Ωxæ, and y Ωxæ is of the ode of K?2 times y Ωxæ. Theefoe, the tem Ky Ωxæ is bigge than y Ωxæ by a lage multiple of K?1. Again, while it is tue that Ky, say, is always much smalle than y, it is not always tue that Ky is much smalle than y. On the othe hand, y s and y s ae both of the same ode of magnitude as y s. Thus it is justified to teat Ky as a small petubation tem. Thus we wite (9.13) as aωxæy s Ωxæ + bωxæy s Ωxæ :?Ky s Ωxæ, (9.16) and seek the solution y s Ωxæ in the petubation seies y s Ωxæ : y Ω s 0æ Ωxæ + Ky Ω s 1æ Ωxæ Substituting this seies into (9.16), we easily find that ys Ω0æ Ωxæ is given by (9.15), and that aωxæ d Ω ys næ Ωxæ + bωxæy Ω s næ Ωxæ :? dx 2 ys Ωn?1æ Ωxæ, n : 1, (9.17) dx d2 Thus ys Ωnæ Ωxæ, n : 1, 2, 666, satisfies a fist-ode ODE with the inhomogeneous tem equal to the Ωn?1æ Ω0æ Ωxæ. Once ys is found, we may then solve (9.17) negative of the second-ode deivative of ys Ω1æ with n : 1 and obtain ys. Similaly, all y Ω s næ Ωxæ, n : 2, 3, 666, can be obtained by successive iteation. We emak that, by teating Ky s " as a small petubation, we have implicitly assumed that the solution is slowly vaying. Since all tems in the petubation seies so obtained ae slowly vaying, the assumption is justified and the method of egula petubation poduces succesive 3

4 appoximations of the slowly vaying solution. On the othe hand, this method cannot poduce the apidly vaying solution. Clealy, if the solution is apidly vaying, it is not justified to teat Ky as a small petubation tem. To find successive appoximations of the apidly vaying solution, we note that, when aωxæ and bωxæ ae constants, the apidly vaying function is expω?ax/kæ. Thus we sumise that, when aωxæ and bωxæ depend on x, the apidly vaying solution is appoximately exp?k?1 X aωxædx. Theefoe, we shall put y Ωxæ : exp?k?1 X aωxædx vωxæ. (9.18) With the apidly vaying behavio of y Ωxæ taken up by the exponential facto, we expect vωxæ to be slowly vaying. Substituting (9.18) into (9.13), we get av + Ωa? bæ v : Kv. (9.19) Teating the tem Kv as a small petubation we may solve (9.19) with egula petubation. We put vωxæ : v 0 Ωxæ + Kv 1 Ωxæ +666, (9.20) and get v X bωxæ 0 Ωxæ : 1 exp dx (9.21) aω xæ aωxæ as well as a d v d 2 n + Ωa? bæv n : v dx 2 n?1. (9.22) dx Note that, just like (9.17) fo ys Ωnæ, (9.22) is a fist-ode diffeential equation fo v n and can be solved in a closed fom. Thus we may obtain all v n by successive iteation. We note that this podecue yields us only the solution which is in the fom of (9.18), whee v is slowly vaying. Hence it does not yield y s. If y Ωxæ is apidly deceasing, we expect that it is appeciable only in a small neighbohood nea the lowe endpoint x : 0, and dops to vey small values as x gets away fom the oigin. The gaph fo the solution (9.23) has a bump nea x : 0. The solution is said to have a bounday laye nea the lowe endpoint. Similaly, if y Ωxæ is apidly inceasing, we expect it to be appeciable only nea the uppe endpoint x : 1. In a specific calculation, we may, instead of pluggin in a and b into the fomulae given above, obtain both y out and y in Ωxæ diectly fom (9.13). Fist of all, the diffeential equation fo y out is obtained by neglecting the tem Ky in (9.13). The esulting equation is of the fist-ode and can be solved in a closed fom. The multiplicative constant in this solution can be fixed with the bounday condition at the endpoint whee y Ωxæ is negligible. The equation fo y in can be obtained by teating the functions a and b as constants. Poblem fo the Reade: Solve appoximately the equation Ky + Ω1 + xæy + y : 0, 0 9 x 9 1, with the bounday conditions yω0æ : yω1æ : 1. Answe We have aωxæ : 1 + x ; 0, thus y Ωxæ is apidly deceasing and thee is a bounday laye nea x : 0 with a width of ode K. Fo x outside of the bounday laye, we have Ω1 + xæy out + y out : 0. Thus we get 4

5 y(x) Bounday Layes and Singula Petubation y out Ωxæ : c. 1 + x The constant c is detemined fom the bounday condition at x : 1 and we have 2 y out Ωxæ : 1 + x. In paticula, y out Ω0æ : 2. To obtain y in Ωxæ nea x : 0, we make the appoximation Thus Theefoe we have The bounday condition equies that Thus We also find that aωxæ u aω0æ : 1. Ky in + y in u 0. /K y in Ωxæ : 2 + c e?x. yω0æ : 1 c :?1. /K y in Ωxæ : 2? e?x. y unifom : 2? e?x/k. 1 + x The solution is plotted in the figue below. We see that it is given by 2/Ω1 + xæ almost eveywhee. But the function 2/Ω1 + xæ does not satisfy the bounday condition at x : 0. Thus the apidly vaying solution y kicks in nea the point x : 0 and the solution y goes though a apid tansition inside the bounday laye e = 0.01 exact solution unifom solution 1.6 e = x Figue 9.1. C Tuning Points In the peceding section, we assume that aωxæ is of the same sign thoughout the inteval in which the diffeential equation (9.13) holds. The solutions y Ωxæ and y s Ωxæ, given by (9.24) and (9.15) espectively, ae good appoximations of the two independent solutions of the diffeential equation. But if x 0 is a simple zeo of aωxæ, the function aωxæ changes sign at x 0, and the appoximations 5

6 we have made in the peceding section fail in the neighbohood of x 0. Boowing the teminology in quantum mechanics, we shall call x 0 a tuning point of eq. (9.13). To obtain appoximate solutions nea a tuning point, we appoximate aωxæ and bωxæ in (9.13) by algebaic functions of x, and solve the esulting equation exactly. This is the same appoach as the one used to handle the tuning point of the wave equation discussed in Chapte 7. The only diffeence is that the appoximate solutions fo (9.13) nea a tuning point ae paabolic cylinde functions athe than Aiy functions, as we shall show. It will be convenient get id of the y tem in (9.13). Let us wite (9.13) as ΩD + a K ædy + b K y : 0. This tells us that, to eliminate the y tem, we put y : exp? X x aωx æ dx YΩxæ. (9.42) 2K This is because, as we move the exponential facto to the left of D, we make the eplacement D î D? a. 2K Thus eq. (9.13) becomes D + a D? a Y + 2K 2K b K Y : 0, o D 2 2bΩxæ? a Y + Ωxæ a 2 Ωxæ? Y : 0. (9.43) 2K 4K 2 Since 1/Ω4K 2 æ is vey lage, we see that an altenative way to solve eq. (9.43) is to use the WKB method, appoximating NΩxæ in (7.16) by a 2 + 2KΩa? 2bæ P a P p + a? 2b. 4K 2 2K 2 P a P The WKB appoximation is valid povided that aωxæ does not vanish. This appoximation yields fo Y two WKB solutions, one of them a apidly inceasing function of x, while the othe a apidly deceasing function of x. To obtain the solutions of (9.13), we multiply these two WKB solutions by the exponential facto of (9.42), getting the slowly vaying solution y s and the apidly vaying solution y. (See homewok poblem 1). The WKB appoximation fails at a tuning point. Let x 0 be a tuning point. We shall assume that x0 is a simple zeo of aωxæ. Then we have, nea x : x 0, aωxæ u FΩx? x 0 æ and bωxæ u G. By (7.18), the WKB appoximation holds only if P x? x 0 P;; K. Fotunately, when x is close to x 0, o moe pecisely P x? x 0 P99 1, eq. (9.43) is appoximately d 2 Y 2G? F F 2 Ωx? x + 0 æ? 2 Y u 0, (9.44) dx 2 2K 4K 2 which we will be able to solve in a closed fom. Befoe we show how to do this, we fist note that, when x is nea x 0, we have x aωx æ exp? X dx p exp? FΩx? x 0æ 2. x 0 2K 4K Thus, fo x close to x 0, the solution y is elated to Y by y u e?fωx?x 0æ 2 /Ω4Kæ Y. (9.45) We hee take note that, if F is positive, the facto multiplying Y in (9.45) dops exponentially as x? x 0 inceases; on the othe hand, if F is negative, this facto gows exponentially as x? x 0 inceases. 6

7 We shall scale the vaiable Ωx? x 0 æ by Then (9.44) becomes whee d 2 Y + dx 2 x? x 0 : K F G? signωfæ F 2 signωfæ : F/ F. X. (9.46)? X2 Y : 0, 4 We note that, with the independent vaiable scaled this way, the esulting equation (9.47) is independent of the small paamete K. As a esult, the solution of (9.47) vaies by an amount OΩ1æ when X vaies by an amount OΩ1æ. Thus we say that the scale of X is unity. By (9.46), the scale of Ωx? x 0 æ is K. ( We have made the implicit assumption that F is of the ode of unity.) Theefoe, thee is a bounday laye with width K nea the tuning point x 0. Thee is a way to see diectly fom (9.13) that the width of the bounday laye nea x : x 0 is of the ode of K. Let us choose x 0 : 0 fo convenience and wite eq. (9.13) nea the tuning point as Ky + Fxy + Gy u 0. To detemine the scale of x, we make the change of vaiable Then we have We see that if we choose x : JX. K d 2 y + Gy : 0. J 2 dx y + FX 2 dx d J : K, (9.47) the diffeential equation above is independent of the small paamete K. Thus both solutions expessed as functions of X ae independent of K. This means that the scale of X fo these solutions is unity, o the scale of x fo these solutions is K as concluded ealie. Equation (9.47) is a paabolic cylinde equation, which is usually witten in the geneic fom d 2` + ΩT + 1/2? z 2 /4æ` : 0. (9.48) dz 2 One of the solutions of eq. (9.48) is the paabolic cylinde function denoted by D T Ωzæ. Since equation (9.48) is unchanged with the eplacement of z î?z, D T Ω?zæ also satisfies equation (9.48). In addition, (9.48) is unchanged if we make the eplacements z î iz as well as T î?t? 1. Thus D?T?1 Ωizæ and D?T?1 Ω?izæ ae also solutions of eq. (9.48). We shall choose the two independent solutions of y nea the tuning point to be and whee y 1 : e?fx2/ω4kæ F D T x, (9.49) K y 2 : e?fx2 /Ω4Kæ D T? G? signωfæ + 1 T : F 2. F x, (9.50) K The asymptotic foms of the paabolic cylinde function ae given by D T ΩXæ u X T e?x2 /4, X î K, 2Z u X?T?1 e X2 /4, X î?k. <Ω?Tæ We shall eplace the paabolic cylinde functions in (9.49) and (9.50) by these asymptotic foms when X is lage. 7

8 The qualitative behavios of y 1 and y 2 depend on the sign of F. This is because the exponential facto in (9.49) and (9.50) ae eithe gowing exponentially o vanishing exponentially as X becomes lage, depending on the sign of F. Thus thee ae two sepaate cases to discuss. Case of F ; 0 Fo F ; 0, we have G T : F? 1. Also, the exponential function vanishes as X inceases. Thus we have and y 1 î ΩXæ?1+G/F e?x2 /2, X î K, î 2Z Ω X æ?g/f, X î?k (9.51) <Ω1? G/Fæ 2Z y 2 Ωxæ î <Ω1? G/Fæ X?G/F, X î K The behavios of these two functions ae schematically illustated below: y 1 î P X P?1+G/F e?x2 /2, X î?k. (9.52) y 2 a < 0 a > 0 a < 0 a > 0 Figue 9.2. In the egion x ; 0, y 1 is the apidly vaying function, and y 2 is the slowly vaying solution, while in the egion x 9 0, y 1 is the slowly vaying solution and y 2 is the apidly vaying solution. The ole of y 1 intechanges with that of y 2 as one cosses the tuning point. Case of F 9 0 We have G T :? F. Also, the exponential function blows up as P X P becomes lage. Thus and y 1 î X?G/F, X î K, 2Z î <ΩG / Fæ X?1+G/F e X2 /2, X î?k, (9.53) 2Z y 2 î <ΩG / Fæ X?1+G/F e X2 /2, X î K, We illustate in the figue below the behavios of y 1 and y 2 fo F 9 0.?G/F î X, X î?k. (9.54) 8

9 y 1 y 2 Figue 9.3. In the egion x ; 0, y 2 is the apidly vaying solution and y 1 the slowly vaying solution. And in the egion x 9 0, y 1 is the apidly vaying solution and y 2 is the slowly vaying solution. In the next thee sections, we shall show how to match these appoximate solutions valid inside the neighbohood of the tuning point with the appoximate solutions valid outside of the neighbohood of the tuning point. D Tuning Point at an Endpoint Poblem fo the Reade: Solve appoximately with the bounday conditions Ky + 2xy + Ω1 + 3x 3 æy : 0, 0 9 x 9 1, K ò 1, yω0æ : 0, yω1æ : 1. Answe Since aωxæ : 2x ì 0, the apidly vaying solution is a deceasing function of x. Thee is a bounday laye at x : 0. Since x : 0 is also a tuning point, the width of the bounday laye nea x : 0 is K. The solution y is negligible outside of the bounday laye. Thus, when x ô K, the solution is equal to y out which satisfies This equation yields 2xy out + Ω1 + 3x 3 æy out : 0. y out Ωxæ : 1 e?x 3 /2, x whee we have made use of the bounday condition at x : 1. Inside the bounday laye nea x : 0, we have 2 / y in Ωxæ : e?x /2 Ω2Kæ c 1 D?1 2 x + c 2 D?1/ 2? 2 x K K. These solutions ae good appoximations if x Thee ae two abitay constants, c 1 and c 2. We detemine one of these constants by matching y in Ωxæ with y out Ωxæ in the ovelapping egion whee both appoximations ae valid. The paabolic cylinde function solution is valid when x 99 1, while by (7.18), the WKB solutions ae valid when P d K x P99 1, dx o x ;; K. Thus the ovelapping egion is K 99 x When x ô K we have y /4 x?1/2 in Ωxæ u c 2 Ω2Kæ 1. 9

10 Also, when x ò 1, we have /2 y out Ωxæ u x?1. / We note that both y in Ωxæ and y out Ωxæ ae equal to a constant times x?1 2 in the egion 1 ô x ô K. This is an indication that this egion is indeed the ovelapping egion in which both appoximations hold. Joining y in Ωxæ with y out in this egion, we get Fom the bounday condition at x : 0, we get Thus we have /4 c 2 : Ω2Kæ?1. c1 :?c 2. y in Ωxæ : e?x2/ω2kæ Ω2Kæ?1?D 2?1/2 K x + D?1/2? 2 K x /4. As a final obsevation, we note that youtω0æ is infinite. But as we continue youtωxæ into the egion of the bounday laye, it tuns into e?x2 /Ω2Kæ Ω2Kæ?1/4 D?1/2? 2 K x / At x : 0, the expession above is equal to Ω2Kæ?1 4 D?1/2 Ω0æ, which is a lage numbe but is not infinity.. 10

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