A NOTE ON BLASIUS TYPE BOUNDARY VALUE PROBLEMS. Grzegorz Andrzejczak, Magdalena Nockowska-Rosiak, and Bogdan Przeradzki

Opuscula Math. 33, no. 1 213, 5 17 http://dx.doi.org/1.7494/opmath.213.33.1.5 Opuscula Mathematica A NOTE ON BLASIUS TYPE BOUNDARY VALUE PROBLEMS Grzegorz Andrzejczak, Magdalena Nockowska-Rosiak, and Bogdan Przeradzki Communicated by Zdzisław Kamont Abstract. The existence and uniqueness of a solution to a generalized Blasius equation with asymptotic boundary conditions is proved. A new numerical approximation method is proposed. Keywords: Blasius equation, shooting method. Mathematics Subject Classification: 34B15, 34B4, 65L1, 34D5. 1. INTRODUCTION We study the BVP of the form: x + cx p x =, x = = x, lim t x t = β, 1.1 where p 1, c and β are positive constants. The problem is motivated by the classical Blasius equation describing the velocity profile of a fluid in a boundary layer where c = 1 2, p = β = 1. The Blasius equation is a basic equation in fluid mechanics which appears in the study of the flow of an incompressible viscous fluid over a semi infinite plane. Blasius [2] used a similarity transform technique to convert the partial differential equation into his famous ordinary differential equation x + 1 2 x x =, x = = x, lim t x t = 1, 1.2 Ψ where x is the stream function x = 2Uνy1, U is the fluid velocity, ν is the fluid U kinematic viscosity and t is the similarity variable defined as t = y 2 2νy 1, where y 1, y 2 are Cartesian coordinates with y 1 pointing along the free stream direction and y 2 perpendicular to y 1. We refer to [3,4] for an excellent introduction to the problem. c AGH University of Science and Technology Press, Krakow 213 5

6 Grzegorz Andrzejczak, Magdalena Nockowska-Rosiak, and Bogdan Przeradzki A series expansions method was used to solve 1.2 by Blasius. There has been many analytical and numerical methods handling this problem since Blasius s work, [1, 6, 8, 9, 11 15] for instance. The existence and the uniqueness for the Blasius problem was proved by Weyl in 1942 [14] by using an elegant method based on the symmetry of the equation 1.2. If η wη is a solution of 1.2 then η kwkη is as well for any k. It enabled him to work with the initial condition x = x =, x = 1. This problem where the second-order derivative is given at is equivalent to the fixed point problem with an operator possessing some monotonicity property. Weyl solved it as a limit of odd and even successive iterates. For our problem this method cannot work so simply and we shall consider the third initial condition x = a and the dependence of the solution with respect to a >. Moreover, our method gives estimates for errors. It enables us to propose a combined numerical method to solve the problem. For a given ε >, we can find an interval [, T ] such that outside of it the derivative of a solution cannot change more than ε. In the first part of this paper, the existence and uniqueness of 1.1 will be analytically proved by changing the boundary value problem to an initial problem. Using the obtained estimates we will be able to find the value of a = x which guarantees that the solution x a on an initial problem x + cx p x =, x = = x, x = a, 1.3 is the solution of 1.1 we are looking for. In the second part of the article, we shall propose a numerical approximation method and some examples that compare it with many others in the literature. 2. AUXILIARY LEMMAS Let x a stand for the unique solution of x + cx p x = satisfying initial conditions x = = x, x = a. If p N and a <, then x a and x a are negative for small t s thus the solution is concave and negative for all arguments and it cannot solve 1.1. Using the analogous argument about negative values of x a we get that the negative value of a for p > 1, p / N contradicts the assumption of nonnegative x a. For a = we have a trivial solution x a and the seeking solution can be obtained for a >. Lemma 2.1. x a is defined for all t. Proof. Notice that x a is an analytic function. Moreover, x a cannot vanish at any point. If there exists t > such that x at = then from the equation x + cx p x = we get that x k a t = for k 2. So we get that the solution of the ODE is the form

A note on Blasius type boundary value problems 7 x a t = c 1 t+c 2 which contradicts the initial conditions x a = x a =, x a = a. Hence dividing the equation by x a and integrating on [, t] we have which implies x at = a exp c x at = a exp c Integrating once more and applying the Fubini Theorem we get x a t = a t s exp c By 2.1,2.2 and 2.3 we have apriori estimates: x a s p ds, 2.1 x a τ p dτ ds. 2.2 x a τ p dτ ds. 2.3 < x a t < 1 2 at2, < x at < at, < x at < a for any t >. It follows [1, p. 146] that x a is extendable to [,. Lemma 2.2. For any a >, lim t x at = and there exists a finite and positive limit ha := lim x t at. 2.4 Proof. Since x a > and x a >, then lim t + x a t = +. Moreover, since x a >, x a > and x a = cx p ax a, then x a is a decreasing function so lim t x at = g a [, a. Suppose g a >. From x a = cx p ax a we get that On the other hand lim t + x a t =. 2.5 t s t t, t + 1 x at + 1 x at = x a s t and hence lim t + x a t =, which contradicts 2.5. Since x a >, then x a is an increasing function so the limit defining ha exists, possibly infinite. From lim t + x a t = + we get there exists t a > such that cx a t p > 1 for t > t a. From 2.2, we obtain for t > t a, x at a exp c a x a τ p dτ exp t a dτ ds

8 Grzegorz Andrzejczak, Magdalena Nockowska-Rosiak, and Bogdan Przeradzki Thus a a ds + a t a e s ta ds at a + ae ta t a e s ds = at a + 1. for any a >. ha at a + 1 2.6 Lemma 2.3. For any a >, there exists a finite and positive limit µa := lim t hat x a t, 2.7 where the constant ha is defined by 2.4. It means that the graph of x a has a slant asymptote and the following estimates hold: max, hat µa x a t hat. 2.8 Proof. The function t hat x a t is increasing, hence the limit from the assertion exists but it can be infinite. Suppose it equals +. By the arguments from the proof of Lemma 2.2, we have x a t x at for t t a. Integrating this inequality from s to + and using the fact x a+ =, we get x as ha + x as. Next integration from t a to t leads to the following inequality or equivalently Thus x at a x at hat t a + x a t x a t a hat x a t hat a x a t a + ha x at a. < µa hat a x a t a + ha x at a. 2.9 The last part of the assertion is a simple consequence. For an upper bound on ha, µa depending explicitly on a we use x a t at 2 /2 to 2.3. Hence, x a t a t s exp c ap 2 p τ 2p dτ ds = a One can easily show that the function ϕt := t s exp ks α ds, k = a p t s exp c 2p + 12 p s ds. ca p 2 p 2p + 1, α = 2p + 1

A note on Blasius type boundary value problems 9 has a similar behaviour as x a in the sense that its graph has an asymptote x = h t µ, where h = exp ks α ds = Γ1/α αk 1/α, µ = and its graph sits above this line. Hence, x a τ a 1 p Γ 1 2 p c 1 2p τ a 1 2p Γ 2 s exp ks α ds = Γ2/α αk 2/α 2.1 2p + 1 1 2p 2 p c 2. where x a τ c 2 a p+1 τ c3 a 1, 2.11 1 c 2 := Γ 2 p c 1 1 2p 2p, 2 c3 := Γ 2p + 1 2 p c 2. 2.12 Now, we are able to get appropriate estimates for ha. Lemma 2.4. For any a >, c 2 a p+1/ ha c 1 a p+1/, 2.13 where c 1 := c 3 Γ1/p + 1 + c 2 c 1/p+1 c 2 p + 1. p/p+1 Proof. For a lower bound we apply the estimate x a τ 1 2 aτ 2 to the equality This leads to the inequality ha = a ha a exp c x a τ p dτ ds. 2.14 ca p exp 2 p 2p + 1 s ds. 2.1 for k = cap 2 p, α = 2p + 1 gives the lower bound on h.

1 Grzegorz Andrzejczak, Magdalena Nockowska-Rosiak, and Bogdan Przeradzki For an upper bound on h we use the lower estimate of x a 2.11 to the equality 2.14 and we get ha a + a c 3/c 2 a p/ ds+ exp c c 2 a p+1 p τ c3 a 1 dτ ds c 3/c 2 a p/ c 3 c 2 a p+1 + 1 c 2 a p c 3/c 2 a p/ c exp a p+1 t p+1 dt, p + 1c 2 c where we used linear substitutions twice. At last, using 2.1 for k = p+1c 2 a p+1, α = p + 1 we have for any a >, ha c3 c 2 + Γ1/p + 1 c 1/p+1 c 2 p + 1 p/p+1 a p+1. The next lemma presents estimates for µa. Lemma 2.5. For any a >, the constant µa, which is defined by 2.7, satisfies the following estimates: where c 4 a 1/ µa c 5 a 1/, 2 2p/ Γ2/2p + 1 c 4 := 2p + 1 2p 1/ c, 2/ [ c 5 := 1 c 2 c 2 3 2 2 + c2 c 2/p+1 p + 1 1 p/1+p Γ 2 p+1 + c c 3 2 c p+1 1/p+1 Γ 1 p p+1 ]. Proof. From 2.3 and 2.14 we get µa = a s exp c Using the estimate x a τ 1 2 aτ 2 we get µa a x a τ p dτ ds. 2.15 ca p s exp 2 p 2p + 1 s ds.

A note on Blasius type boundary value problems 11 Using 2.1 for k = cap 2 p, α = 2p + 1 we obtain the lower bound. On the other hand, from 2.11, 2.15 we have c 3/c 2 a p/ µa a s ds+ + a s exp c c 2 a p+1 p τ c3 a 1 dτ ds c 3/c 2 a p/ c2 3 2c 2 a 1 1 + 2 c 2 a 1 2 + c 3 c 2 2 c 3/c 2 a p/ c exp a p+1 t p+1 dt, p + 1c 2 c t exp a p+1 t p+1 dt+ p + 1c 2 c where we used linear substitutions twice. At last, using 2.1 for k = p+1c 2 a p+1, α = p + 1 we have for any a > the upper bound. 3. MAIN RESULTS Now, we are able to prove the existence of a solution to 1.1. Theorem 3.1. Let p 1, c >. The BVP x + cx p x =, x = = x, lim t x t = β, has a solution for any β. Proof. The function h : [, R is continuous on, + by the continuous dependence of solutions of ODEs on initial conditions and locally uniform convergence of the integral exp c By the estimates from Lemma 2.4, we have x a τ p dτ ds. lim ha =, lim a + ha = +. a Thus, for any β >, there exists a > such that ha = β. For β =, it is obvious. Finally, the uniqueness of the solution of 1.1 will be proved by using ideas from [5]. For any a > consider the one-to-one function v a : [, ha 2 [,

12 Grzegorz Andrzejczak, Magdalena Nockowska-Rosiak, and Bogdan Przeradzki such that v a x at 2 = x a t for each t. It is well defined since x a and x a are increasing functions and it belongs to C 2, ha 2. Substituting y = x t 2, we shall find an ODE satisfied by v we omit subscript a for simplicity. xt = vy, x t = v y2x tx t hence, x t = 1 2v y, x t = v y 2v y 2 2x tx t = v y y 2v y 3. Put x and its derivatives in our ODE and find From boundary conditions on x we get v y = cvyp v y 2 y. 3.1 v =, v = 1 2a, lim vy = +. 3.2 y ha 2 Now, we are in a position to prove the following theorem. Theorem 3.2. Let p 1, c >. For any β the solution of is unique. x + cx p x =, x = = x, lim t x t = β, Proof. We need to show that the function h is one-to-one. Suppose that ha 1 = ha 2, a 2 > a 1 and take v 1 and v 2 obtained by x a1 and x a2, respectively, that is v i satisfies 3.1 with boundary conditions 3.2 for a = a i, i = 1, 2. Put w = v 1 v 2. Then w =, w = a2 a1 2a 1a 2 >. Notice that w > on the whole interval, ha 1 2 both functions are defined on the same interval. In fact, if it is not true, then there exists s in this interval such that w > on, s and w s =. Hence ws > w = and On the other hand, w w s ξ w s s = lim. ξ + ξ w s = v 1 s v 2 s = cs 1/2 v 1 s p v 2 s p v a i s 2, since w s = implies v a 1 s = v a 2 s. But v 1 s p > v 2 s p from ws > and this gives w s > a contradiction. Thus, we have w > and w > on, ha 1 2. Set V i = 1/v i, i = 1, 2 and W = V 1 V 2. We have, for any y, ha 1 2, W y = V 1y V 2y = v 1 y v 2 y + y2 v 2 = cv 2y p v 1 y p < y2 y v 1

A note on Blasius type boundary value problems 13 from wy >. Hence, W y < W = 1 v 1 1 v 2 = 2a 1 a 2 and lim W y 2a 1 a 2 <. y ha 1 2 On the other hand, V i x a i t 2 = 2x a i t implies and, therefore, lim V iy = lim V i x y ha 1 2 a t i t 2 = 2 lim x a t i t = which contradicts the previous inequality. lim W y =, y ha 1 2 4. NUMERICAL APPROACH All numerical methods cannot work on the infinite interval [, and we do not know the exact value of a = x for the solution. Our earlier results make it possible to find a finite interval [, T ] for any positive value ɛ of the error control tolerance such that x T < ɛ, ha x T < ɛ, xt hat µa < ɛ. Since all these functions decrease, all three inequalities hold for any t > T. First, by using estimates 2.13, we can find an interval [a min, a max ] such that ha min < β < ha max. Next, by 2.1, we need by 2.14, we should have a max exp c a max and by 2.3 and 2.15, we get a max T T exp c T T s exp c x amin τ p dτ < ɛ, x amin τ p dτ ds < ɛ, x amin τ p dτ ds < ɛ.

14 Grzegorz Andrzejczak, Magdalena Nockowska-Rosiak, and Bogdan Przeradzki We do not know the function x amin but we can use estimate 2.11 to get a max a max T T a max exp c T exp T s exp c c p+1 p c 2 amin τ c 3a 1 min dτ < ɛ, 4.1 p+1 p c 2 amin τ c 3a 1 min dτ p+1 c 2 amin τ c 3a 1 We start with a family of initial value problems min ds < ɛ, 4.2 pdτ ds < ɛ. 4.3 x a t, t [, T ], x a = = x a, x a = a, a [a min, a max ]. 4.4 If we approximate this solution in [, T ] with an error less than ɛ, then the best approximation of x in [T, is xt = βt + x a T βt. The lower and upper bounds for the second derivative describe the shooting window - for each a the only solution in this direction exists at t = T, and the computed value of x T is more and more close to the expected limit β. As long as β is contained between the computed values x T of the best two shots, we apply the classical bisection method: If y and z are solutions such that y < z, and there is y T < β < z T, then the next problem to solve is 4.4 with a = y + z /2. For any value ɛ > we determine T by solving the inequality 4.1. After that we check if inequalities 4.2, 4.3 hold for the obtained T. Otherwise, we increase T to the moment where 4.2, 4.3 hold. 4.1. EXAMPLES While solving the initial value problems we apply an adaptive Runge-Kutta-Fehlberg method RK45 [7], in which a tolerance parameter ɛ controls the local error of the method. The values of ɛ range from 1 8 to 1 14. For representing real values we use the standard 16 17-digits double data type. Numerical results for the classical Blasius equation p = 1, c=1/2, and β =1. Here a min =.2 694 86 459, a max =.3 42 953 216. For T = 14 we get the all three inequalities 4.1, 4.2 and 4.3 for ɛ = 1.e 14. Below N stands for the number of steps in RK45 average Tab. 1.

A note on Blasius type boundary value problems 15 Table 1. ɛ N a x a 1 x T xt 1.e 8 112.332 57 33 68 21 6.15e 9 1.2e 8 12.279 212 18 321 1.e 9 197.332 57 335 646 357 5.69e 1 1.1e 9 12.279 212 327 474 1.e 1 338.332 57 336 149 237 6.6e 11 1.3e 1 12.279 212 34 74 1.e 11 611.332 57 336 21 248 4.95e 12 5.7e 13 12.279 212 342 472 1.e 13 1831.332 57 336 215 154 4.19e 14 6.5e 14 12.279 212 342 479 1.e 14 3346.332 57 336 215 186 1.6e 14 5.5e 16 12.279 212 342 48 The last value of a differs in the two last digits from the one cited in [3]: a =.332 57 336 215 196 3. The last two columns of Table 1 have been computed for ɛ = 1 14. As there is x T = 7.68e 13, for t > T the straight line approximation of the solution is the most effective. Numerical results for the equation with p = 7, c = 1/2 and β = 1. Here, a min =.373 397 838 8, a max =.38 548 242 7. As above for the tolerance ɛ = 1.e 14, the interval [, 4] is sufficiently large and we get the following results by the RK45 method Tab. 2. Table 2. ɛ N a x a 1 x T xt 1.e 8 189.379 398 164 451 122 2.47e 8 3.5e 8 2.673 55 448 977 1.e 9 316.379 398 187 63 634 2.4e 9 2.9e 9 2.673 55 57 853 1.e 1 549.379 398 189 5 442 1.3e 1 1.4e 1 2.673 55 581 319 1.e 11 961.379 398 189 86 642 2.2e 11 3.1e 11 2.673 55 581 757 1.e 12 1688.379 398 189 16 95 1.69e 12 2.3e 12 2.673 55 581 866 1.e 13 2827.379 398 189 18 438 1.62e 13 1.9e 13 2.673 55 581 874 1.e 14 4634.379 398 189 18 571 2.91e 14.e 14 2.673 55 581 875 Remarks as above. Here x T = 9.3e 18. The value of a =.379398189186 here, all digits are true. Numerical experiment for the equation with p =.1, c = 1/2, β = 1. Taking ɛ = 1 14 we get T = 5. The proof of the existence and uniqueness result for p < 1 fails, since we cannot claim that the initial value problem 1.3 has a unique solution and that it depends continuously on a. Hence, function h can be multivalued. If one will prove the uniqueness, then, due to [1, p. 172], h will be continuous and all results of this paper will be true also for p < 1. The stability of numerical experiments cited below suggests it is the fact. Numerical results presented below in Table 3 for this case suggest that the method works also wihout the uniqueness of solution to initial value problems.

16 Grzegorz Andrzejczak, Magdalena Nockowska-Rosiak, and Bogdan Przeradzki Table 3. ɛ N a x a 1 x T xt 1.e 8 142.443 643 25 985 427 2.16e 7 4.5e 7 48.54 26 863 24 1.e 9 256.443 643 43 844 98 1.78e 8 3.7e 8 48.54 28 581 2 1.e 1 466.443 643 42 192 529 1.49e 9 3.1e 9 48.54 282 21 34 1.e 11 839.443 643 421 42 885 2.8e 1 5.8e 1 48.54 282 33 96 1.e 12 155.443 643 421 66 499 2.25e 11 4.7e 11 48.54 282 356 64 1.e 13 2669.443 643 421 681 56 1.49e 12 3.6e 12 48.54 282 358 73 1.e 14 4922.443 643 421 683 245 2.45e 13 2.2e 16 48.54 282 358 9 Remarks as above. Here x T = 1.2e 15. The value of a =.443 643 421 683 here, all digits are true. 5. CONCLUSIONS The authors know that our computation of the value of the second derivative of the solution are not more exact than others. However, the proposed method gives a possibility of controlling errors and it is very simple. We hope a similar approach can be applied for more general equations as x + fx gx = with qualitative assumptions on functions f and g. REFERENCES [1] F.M. Allan, M.I. Syam, On the analytic solutions of tht nonhomogneous Blasius problem, J. Comput. Appl. Math. 182 25, 362 371. [2] H. Blasius, Grenzschichten in Flüssigkeiten mit kleiner Reibung, Math. Phys. 56 198, 1 37. [3] J.P. Boyd, The Blasius function in the complex plane, Exp. Math. 8 1999, 381 394. [4] J.P. Boyd, The Blasius function: computations before computers, the value of tricks, undergraduate projects and open research problems, SIAM Review 5 28, 791 84. [5] B. Brighi, J.-D. Hoernel, On the concave and convex solutions of a mixed convection boundary layer approximation in a porous medium, Appl. Math. Lett. 19 26 69 74. [6] T. Fang, W. Liang, C.-F.F. Lee, A new solution branch for the Blasius equation A shrinking sheet problem, Comput. Math. Appl. 56 28 12, 388 395. [7] J.H. Mathews, Numerical Methods for Computer Science, Engineering and Mathematics, Prentice Hall Inc., New Jersey, 1992. [8] K. Parand, M. Dehghan, A. Pirkhedri, Sinc-collocation method for solving the Blasius equation, Phys. Lett. 373 29, 46 465. [9] K. Parand, A. Taghavi, Rational scaled generalized Laguerre function collocation method for solving the Blasius equation, J. Comp. and Appl. Math. 233 29, 98 989.

A note on Blasius type boundary value problems 17 [1] L.C. Piccinini, G. Stampacchia, G. Vidossich, Ordinary Differential Equations in R n, Applied Mathematical Sciences 39, Springer-Verlag, New York-Berlin-Heidelberg, 1984. [11] A.I. Ranasinghe, Solution of Blasius equation by decomposition, Appl. Math. Sci. 3 29, 65 611. [12] L. Wang, A new algorithm for solving classical Blasius equation, Appl. Math. Comput. 157 24 1, 1 9. [13] A.-M. Wazwaz, The variational iteration method for solving two forms of Blasius equation on a half-infinite domain, Appl. Math. Comput. 188 27 1, 485 491. [14] H. Weyl, On the differential equations of the simplest boundary-layer problems, Ann. Math. 43 1942 2, 381 47. [15] L.-T. Yu, C.C. Kuang, The solution of the Blasius equation by the differential transformation method, Math. Comput. Modelling 28 1998 1, 11 111. Grzegorz Andrzejczak grzegorz.andrzejczak@p.lodz.pl Lodz University of Technology Institute of Mathematics Wolczanska 215, 9-924 Lodz, Poland Magdalena Nockowska-Rosiak magdalena.nockowska@p.lodz.pl Lodz University of Technology Institute of Mathematics Wolczanska 215, 9-924 Lodz, Poland Bogdan Przeradzki bogdan.przeradzki@p.lodz.pl Lodz University of Technology Institute of Mathematics Wolczanska 215, 9-924 Lodz, Poland Received: June 7, 212. Accepted: June 23, 212.