Initial Boundary Value Problems for Scalar and Vector Burgers Equations

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1 Initial Boundary Value Problems for Scalar and Vector Burgers Equations By K. T. Joseph and P. L. Sachdev In this article we stu Burgers equation and vector Burgers equation with initial and boundary conditions. First we consider the Burgers equation in the quarter plane x>, t> with Riemann type of initial and boundary conditions and use the Hopf Cole transformation to linearize the problems and licitly solve them. We stu two its, the small viscosity it and the large time behavior of solutions. Next, we stu the vector Burgers equation and solve the initial value problem for it when the initial data are gradient of a scalar function. We investigate the asymptotic behavior of this solution as time tends to infinity and generalize a result of Hopf to the vector case. Then we construct the exact N-wave solution as an asymptote of solution of an initial value problem extending the previous work of Sachdev et al We also stu the it as viscosity parameter goes to. Finally, we get an licit solution for a boundary value problem in a cylinder.. Introduction The nonlinear parabolic partial differential equation u t + u x = u xx Address for correspondence: Professor P. L. Sachdev, Department of Mathematics, Indian Institute of Science, Bangalore 56, India. STUDIES IN APPLIED MATHEMATICS 6: by the Massachusetts Institute of Technology Published by Blackwell Publishers, 35 Main Street, Malden, MA 48, USA, and 8 Cowley Road, Oxford, OX4 JF, UK.

2 48 K. T. Joseph and P. L. Sachdev was first introduced by J. M. Burgers as the simplest model for fluid flow. This equation describes the interplay between nonlinearity and diffusion, see Sachdev and the references therein for physical interpretation and important solutions. A remarkable feature of this equation is that its solution with initial conditions of the form u x =u x can be licitly written down. In fact Hopf 3 and Cole 4 independently showed that the equation can be linearized through the transformation u = v x 3 v More precisely, assume that the initial data u x are integrable in every finite interval and that x u y = o x then Hopf 3 showed that if v x t satisfies the linear heat equation v t = v xx 4 with initial condition v x = y u z dz 5 then u x t defined by 3 solves and and conversely, if u x t is a solution of and, then v x t defined by 3 is a solution of 4 and 5 up to a time-dependent multiplicative factor that is irrelevent in 3. Solving for v from 4 and 5 and substituting it into 3, he obtained licit formula for the solution of and ; namely, u x t = R xy t R xy y xy u z dz y u z dz and studied the asymptotic behavior of u x t as t. He also constructed licit weak entropy solution of the inviscid Burgers equation. Lighthill 5 discovered the N-wave solution of the Burgers equation ; namely, 6 U x t = t x/t + t c x t 7

3 Boundary Value Problems for Burgers Equations 483 Sachdev et al. 6 showed that this solution can be obtained as time asymptotic of a pure initial value problem and found licitly the constant c in terms of one lobe area of the initial data. In a more recent paper, 7 proposed the vector Burgers equation U t + U U = U 8 They observed that if we seek special solutions of the form U = x φ 9 where φ x t is a scalar function on R n, equation 8 leads to φ t + φ φ = They used this observation to find special solutions of Burgers equation in cylindrical coordinates with axisymmetry. Although there are many results for pure initial value problem, boundary value problem has been studied less. Explicit solutions of the Burgers equation in the quarter plane with integrable initial data and piecewise constant boundary data were constructed by 8 using Hopf Cole transformation. He obtained a formula for its weak it as viscosity parameter goes to. Using maximum principle, this formula for weak it was extended to general boundary data. When the initial and boundary data are two different constants, the problem was considered in some detail by 9 for small. Five distinct cases arose depending on the relative magnitudes of the constants appearing in the initial and boundary conditions. The solutions required the introduction of corner layers, shock layers, boundary layers, and transition layers. In the present article, we write an licit form of the solution and recover various cases in the it. We also consider large time behavior of the solutions. Again, different cases arise depending on the relative magnitudes of the constants that appear in the initial and boundary conditions. It would be desirable to have solution for nonconstant initial boundary problem, but as 9 point out, that leads to an integral equation that seems difficult to solve. Using a generalized Hopf Cole transformation, and, in a series of papers, studied with more general boundary conditions that include the flux condition at the origin. The asymptotic profile at infinity of the solution of with flux condition was obtained by. A table of solutions of the Burgers equation is contained in 3. The aim of the present article is to stu Burgers equation and the vector Burgers equation 8 with initial and boundary conditions and investigate the asymptotic behavior as t tends to and as tends to. In Section, we treat the Burgers equation in the quarter plane with Reimann type initial boundary data, solve it exactly, and stu the asymptotic its as t tends

4 484 K. T. Joseph and P. L. Sachdev to or as the viscosity coefficient tends to. In Section 3 we stu the vector Burgers equation, solve exactly the initial value problem for it when the initial data can be written as gradients of a scalar function and carry out the stu of the its as t tends to or tends to. We also stu a boundary value problem for the vector Burgers equation in the same section.. Initial boundary value problem for Burgers equation and its asymptotics In this section, we consider the Burgers equation in x>, t>, with initial condition and the boundary condition u t + u x = u xx u x =u I u t =u B 3 where u I and u B are constants. Using standard Hopf Cole transformation, we reduce this problem to a linear one and solve it licitly. Before the statement of the result we introduce some notations: A x t = A + x t = B x t = x y t x + y t u Iy u Iy x + y + u By t 4 With these notations we prove the following result. A solution of the initial boundary value problem 3 is given by u x t =u B u I u B A + x t A + x t A x t +A + x t +u Bt x + u t B/ u B t x B x t when u I + u B = and u x t =u B + when u I + u B. u I u B A x t A + x t A x t + u Iu B A + x t + u B B x t 5a 5b

5 Boundary Value Problems for Burgers Equations 485 To prove this result, we use Hopf Cole transformation. First, we set w x t = u y t u I 6 x From 3, we get w t + w x + u I w x = w xx w x = w x t =u B u I 7 Now setting v = w 8 we get from 7 v t + u I v x = v xx v x t + u B u I v t = Making the final transformation p = u Ix + u I we get from 9 the following problem for p: v x = 9 t v p t = p xx p x = u Ix p x t +u B p t = This problem has the licit solution see 8 { p x t = πt / u Iy + u B/ πt / u B y z y x y + t x + z t } x + y t u Iy dz

6 486 K. T. Joseph and P. L. Sachdev Now from 6, 8, and, we have In other words u x t =w x + u I = log v x + u I = log p x + u I / +u I = log p x u = p x p To write the formula in a convenient form we note that { } ub z x + z x dz y t { } ub z x + z = y z dz t ub y x + y = u { B ub z t y x + } z dz 3 t Here, the first equality is obtained by just writing the onential of a sum as a product and using the fact that x + z x + z x = z and the second equality is obtained by the use of integration by parts. Similarly, and x { ui y = x t } x + y t { + u I { ui y x t = x u I t x y { ui y } ui y } x + y 4 t } x y 5 t Now, it follows from, 3, 4, and 5 that p x x t = u πt I A x t u IA + x t + u B A + x t +u u B B C x t

7 Boundary Value Problems for Burgers Equations 487 and p x t = πt A x t +A + x t +u B C x t where A, A + and B are given by 4, and C x t = y u B y z x + y u Iy dz t Substituting these formulas for p and p x in, we get u x t = u IA x t u IA + x t +u BA + x t +u B u B C x t A x t +A + x t + u B C x t u I u B A x t A + x t + u B A x t +A + x t + u B C x t = A x t +A + x t + u B C x t u I u B A x t A + x t = A x t +A + x t + u B C x t + u B 6 Now we ress C in terms of A + and B. There are two cases to consider. The first one is when u I + u B =. In this case, C x t = y u B y z ub z = y ub y = y ub y =t y + u B t x x = t t = t x t u Iy x + z t x + y t x + y t x + z t dz dz ub y x + y t ub y x + y + u B t x t + u B t x B x t 7

8 488 K. T. Joseph and P. L. Sachdev When u I + u B, C x t = u B y z u Iy x + z dz y t = u I + u B y ub z x + z y t = u I + u B y u I + u B y = { u I + u B y ub z ub y x + z t x + y t u Iy dz x + y t } dz = u I + u B B x t A + x t 8 Using 7 for the case u I + u B = and 8 for the case u I + u B in 6, the formula 5 for u follows... Stu of the it as Here, we compute the pointwise it of u x t as the viscosity parameter. Let H x be the usual Heaviside function and s = u I + u B /, then for x, t, we have the following formula for the pointwise it function u x t = u x t. Case. u B >u I and u I + u B > u x t =u I H x st +u B H st x Case. u I >, u B > and u I >u B u x t =u B H u B t x + x/t H u I t x H x u B t +u I H x u I t Case 3. u I < and u I + u B u x t =u I Case 4. u I = and u B < oru I > and u B and u I + u B or u I > and u I + u B =, u x t = x/t H u I t x +u I H x u I t

9 Boundary Value Problems for Burgers Equations 489 To stu the u x t, we write 5 for u in terms of the standard erfc and use its asymptotic form. For this, we denote erfc y = u A x t = I t u Ix u = t / I t u Ix y y 9 y x + u It u = t / I t u Ix erfc t y x+u I t t / x + ui t 3 / Similarly, u A + x t = t / I t + u Ix x + ui t erfc 3 / and u B x t = t / B t u Bx x ub t erfc 3 / To stu the asymptotic behavior of u as we stu the behavior of A, A + and B using the asymptotic ansions of the erfc; namely, erfc y = y 4y + o y y 33 3 y 3 and erfc y = π / y 4y + o y y 34 3 y 3 From 3 34, we have the following as. If x + u I t> A x t t x + u I t x 35a and for x + u I t A + x t t x + u I t x 35c

10 49 K. T. Joseph and P. L. Sachdev and for x + u I t B x t t x u B t x and for x u B tu I, u I + u B >. First we take up the subcase u B >, u I <, and u I + u B >. The line x = u I + u B /t divides the quarter plane x>, t> into two regions. In the region x , we get, for x u I + u B /t and x u I + u B t/ and x>u B t, as before, from 5b and 35b 35e, we get u x t u B + u I u B π / O u Ix π / + O u Ix +O / so that u x t =u B + u I u B =u I The other subcases u B >, u I >, u I , u I = are similar.

11 Boundary Value Problems for Burgers Equations 49 Case. u I > u B > u B , and so from 5b and 35a, 35c, and 35f we get u x t u B + Therefore, for xu I t, we have x>u B t so that u x t u B u I u B π / t / x+u I t + π / t / u I u B xu It + t / x+u I t t and we get In the region u B t<x<u I t, we have u x t =u I xu It t xu B t u B xu It t u x t u B + t / x x+u I t and we find that u I u B t / x x+u I t t u x t =u B + t / x x+u I t t + u Iu B t / x t x+u I t t + t / u B x xu B t t u I u B x+u I t x+u I t x+u I t + u Iu B On simplification, we have, for u B t<x<u I t, x+u I t + u B xu B t This completes case. u x t = x t

12 49 K. T. Joseph and P. L. Sachdev Case 3. u I <, u I + u B. First we consider the subcase u I <, u B <, u I u B <. We get from 5b and 35b 35e, u I u B π / O u Ix u x t =u B + π / + O u Iu B u Ix so that, for x>, +O u B xu It u x t =u B + u I u B =u I For the other subcases the analysis is similar and so we omit the details. Case 4. u I = and u B < oru I > and u B and u I + u B non-zero or u I > and u I + u B =. For the first subcase, for x>, we get π / + t / x u x t x t u B u B π / t / x x t + t / x xu B t t so that u x t =u B u B = For the subcase u I >, u B <, u I + u B not zero, following the previous analysis, we have in the region x>u I t, u x t = u B + So we get for x>u I t, u I u B π / t / x+u I t π / + u Iu B t / x+u I t xu It t u x t =u I In the region x<u I t, we have u x t =u B + u I u B xu It t + u B t / x+u B t x+u I t x+u I t x+u I t + u Iu B x+u I t + u B xu B t = x t xu It t The other subcases can be treated similarly. This completes the proof of the result.

13 Boundary Value Problems for Burgers Equations Large time behavior Next, we stu the it of u x t as t where u x t is the solution of 3 given by 5 j j = with a fixed coefficient of viscosity >. We prove the following result.. If u I <, u B < and u I >u B, then u x t =u ui x I coth t + c c = log ub u I 36 u I + u B. If u I <, u I <u B and u B + u I <, then u x t =u ui x I tanh t + c c = log ui u B 37 u I + u B 3. If u I = and u B <, we have 4. For all other cases, u x t = u t B u B x 38 t u x t =u B 39 To prove the result first, we get the asymptotic behavior of A, A +, and B as t tends to infinity. Using the asymptotic ansions 33 and 34 in 3 3 we have, as t, A x t t x + u I t x u I t A x t πt / t x + u I t u I > u Ix x u I < A + x t t x + u I t x u I > u A + x t πt / I t + u Ix B x t t x + u I t x t x u B t x u I < u B < 4a 4b 4c 4d 4e

14 494 K. T. Joseph and P. L. Sachdev and u B x t πt / B t t x u B t x u Bx u B > 4f A x t π/ x t u / I = A x t π/ + x t u / I = 4g 4h B x t π/ + x t u / B = 4i First, we consider the case u I <, u B <. Using 4b, 4d, and 4e in 5b, we get u x t u B + π / u I u B π / u I t So, in this case, we get u I x u I t + π / u I u B u I x u I t π / u I t + u I x + u I x + u B / xu B t u x t =u B + u I u B uix u Ix t u Ix + u Iu B u Ix On simplification we get u x t =u I t u Ix u Iu B u Ix u Ix + u Iu B x u 4 Ix Now, if u I >u B, by rewriting 4 we get 36, the first part of the result. If u I and u I + u B. As before, we have from 5b and 4b, 4d, and 4f the it t u x t =u B + u I u B u Ix u Ix u Ix + u Iu B u Ix + u B u Bx u B u I t 4

15 Boundary Value Problems for Burgers Equations 495 Therefore, u x t =u u Ix u Iu B u Ix I t u Ix + u Iu B u Ix provided u B + u I <, because in this case, u B u I <. On simplification, we get 37. When u I + u B >, we have u B u I >, and so 4 gives t u x t =u B Now, take up the case u I <, u B > and =. From 5a and 4b, 4d, and 4f after dividing numerator and dinominator of the resulting ression by π / πt / u I t we get u x t u B u I u B u Ix + u Ix u Ix + u Ix + u Bt u I t so that π / πt / x t u x t =u B + u B u B tx u Bx When u I = and u B <, using 4f, 4g, and 4h in 5b, we get u B πt / u x t u B πt / x / x / + π/ t xu B t From 43, we get for u I = and u B <, t u x t =u B x x = u B u Bx u B 43 Now, we consider the case u I >. First let u B < and u I + u B. We have from 5b and 4a, 4c, and 4e after cancelling out common terms, t u x t =u B + u I u B u I u B u I x/t u I x/t + u Iu B u I +x/t u I +x/t + u B = u B t xu B t The case u I + u B = is similar; here we use 5a instead of 5b. Now, consider the case u I >, u B >. As before, using the asymptotics

16 496 K. T. Joseph and P. L. Sachdev 4a, 4c, and 4f in 5b and dividing both numerator and dinominator by π / t x / in the resulting ression, we get t u x t = u B + u I u B u I u B u I tx + u Iu B u I tx u I t+x + u B π / xu Bt u I t+x = u B The case u I > and u B = is similar. This completes the proof of the result. Consider the stationary problem for <x< q xx = q x q =u B q = u I It can be easily checked that this problem has solution if u I < and u B <u I or u I < and u I + u B < oru I = and u B <. Solving the problem licitly, we get the functions on the right-hand side of Here, we have shown that they are time asymptotes of boundary value problems Higher dimensional extensions Consider vector equivalent of Burgers equation studied by 7; namely, U t + U U = U 44 7 observed that if we seek a solution U of 44, which is gradient of some scalar function φ, then equation 44 becomes x φ t + φ φ = This leads to φ t + φ U = x φ 45 φ = f t 46 where f t is an arbitrary function of t. Because we are interested in the space derivative x φ, and this is independent of f t, weletf t =. If we are given initial data for U that are gradients of some scalar function φ of the form, U x = x φ x 47

17 Boundary Value Problems for Burgers Equations 497 it is enough to find a solution φ of φ t + φ φ = 48 with initial condition φ x =φ x 49 We may then use 45 to get the solution U of 44 and 47. To solve 48 and 49 we use the Hopf Cole transformation θ x t = φ 5 Using 5 in 48 and 49, we get the linear problem Solving 5 and 5, we have θ t = θ 5 θ x = φ x 5 θ x t = πt n R n x y φ y 53 From 45, 5, and 5, we see that the solution to 44 and 47 is given by From 53 and 54, it follows that U = θ 54 θ U x t = R n xy t R n xy xy φ y φ y 55 First, we stu the asymptotic behavior of this solution as t.

18 498 K. T. Joseph and P. L. Sachdev 3.. Asymptotic behavior as t, a general result Here, we assume initial data of the form 47 U x = x φ x where φ satisfies the estimate φ x = n φ i x i +o x 56 where φ i x i = n are differentiable functions from R to R. Assume further that both the its exist. With the notations x i φi x i =φ i x i φi x i =φ + i 57 ξ = x k t i = φ+ i φ i 58 and g i ξ i = k ξi i ki + ξ i z i dz i z i dz i 59 for i = n, we prove the following result. Let U x t be the solution of 44 and 47 given by the formula 53 and 54 with φ satisfying the conditions 56 and 57, then we have the following it as t uniformly in ξ in bounded subsets of R n : t g U t ξ t = ξ t g ξ g ξ g ξ g n ξ n 6 g n ξ n To prove this result, we follow 3. Consider θ given by 53, where φ satisfies the conditions 56 and 57. After a change of variable, the ression for θ becomes θ x t = π n = π n z R n φ { z R n φ x t z dz t ξ z } dz

19 Boundary Value Problems for Burgers Equations 499 Now, using 56 we get θ x t { z π n R n n φ i t ξi z i } dz i n = z i i π φi t ξi z i dz i 6 as t uniformly on bounded sets of ξ in R n. Now take the i-th term in this product. Following the argument of Hopf 3, we get { z i φi t ξi z i } dz i φ+ i ξi z i dz i + φ i z i dz i 6 Thus we have from 6 and 6, n π n θ ξ t t = ξi t φ+ i i + φ i Similarly, we get π n t θ t x l ξ t t = ξi φ+ i i l ξ i z i dz i z i φ+ l ξ i ξ i dz i + φ i φ l z i z i dz i dz i 63 ξ l 64 Because t/ U = t x θ /θ and θ is bounded away from, we have, from 63 and 64, t/ U θ t ξ t = t x t t θ θ x θ θ x n θ g = ξ g ξ g n ξ n g n ξ n This completes the proof. Here, we observe that for Burgers equation; that is, when n =, the parameter k can be computed in terms of the mass of the initial data, k = φ + φ = u y x x u y = u y

20 5 K. T. Joseph and P. L. Sachdev and then formula 6 with n = is exactly Hopf s 3 result. Also, we note that the it function obtained in the result written in the x t variable; namely, { U x t = log v x t } x { log v x t } { log v n x n t } x x n where for i = n, v i x i t = k xi / t i ki + is an exact solution of 44. x i / t z i z i dz i dz i 3.. N-wave solution of vector Burgers equation We start with the solution θ x t = + t n x c t of the heat equation 5 where c is a constant. Its space gradient is x θ x t = x n t t x t Consider the function U = θ/θ. By earlier discussion, this is an exact solution of the vector Burgers equation 44. On simplification, we get U x t = x t / t / + t n c x t 65 For n =, this exact solution of the Burgers equation was discovered by 5. Sachdev et al. 6 showed that it can be obtained as time asymptotic of a pure initial value problem. This solution is called the N-wave solution. Here, we generalize it for the vector equation 44. We choose special initial data for 44 whose mass is zero and that is antisymmetric with respect to the origin. To solve the problem licitly, we need these data to be written as gradients. To construct such initial data, we consider the ball in R n of radius l with center ; namely, B l = x x l. We take the initial condition as U x =xχ x l x 66

21 Boundary Value Problems for Burgers Equations 5 where χ x l x is the characteristic function of the set x x l. Note that this initial data can be written as the gradient x χ x l x + l χ x l x So the earlier analysis holds, and we get the following formula for the solution of 44 and 66: U x t = Q Q 67 where Q is given by Q x t = πt n/ + πt y l n/ y x y + t l y >l x y 68 We shall prove the following result. Let U x t be the solution of 44 and 66 as given by 67 and 68, then we have U x t =U x t t uniformly in the variable ξ = x/ t belonging to a bounded subset of R n, where U is given by 65 with l c = z dz l B l π n y l 69 Here, B l denotes the volume, if space dimension n 3, area if n =, and length if n =, of B l. To prove this result, first we note that Q x t can be written as Q x t =I + l I 7 where I = I = πt n πt n y l y >l y + x y x y t

22 5 K. T. Joseph and P. L. Sachdev It can be easily checked by making use of error function and its asymptotics that, as t, x I z dz I πt n z l x B l πt n 7 Substituting these asymptotics in 7, we get Q x t l + x πt n z dz l B l z l 7 Similarly, x x x Q x t t t πt z z l Using 7 and 73 in 67, we get U x t x t t l x πt n z z l + x πt n On rearranging the terms, we get U x t dz l B l 73 l dz B l z l dz B l z l t + t n c x/t x t where c is given by 69. This completes the proof Stu of the it First, we remark that the following analysis gives an licit formula for φ, the solution of φ t φ = φ φ x =φ x 74

23 Boundary Value Problems for Burgers Equations 53 namely, φ x t =log πt R n x y φ y 75 Following the analysis of Hopf 3 and Lax 4, see also Joseph 8, we get an licit formula for the solution of the initial value problem for the Hamilton Jacobi equation φ t + φ = 76 φ x =φ x 77 namely, φ x t = φ x t =min φ y + t x y 78 y Note that this is the licit formula derived for the viscosity solution of 77 derived by other methods, see 5. Furthermore, it was shown by 5 that φ x t is Lipschitz continuous when φ x is and for almost every x t there is a unique minimizer for 78, which we call y x t. Now, to stu the it U x t we note that 54 can be written as U x t = R n x y t dµ x t y 79 where, for each x t and, dµ x t y is a probability measure given licitly by xy dµ x t y = R n xy φ y φ y 8 Following the argument of Hopf 3 and Lax 4, it can easily be seen that this measure tends to the δ-measure concentrated at y x t, the minimizer of 78, which is unique for almost every x t. So, for almost all x t, we get from 79 and 8 U x t = x y x t t 8 where y x t is as before a minimizer of 78.

24 54 K. T. Joseph and P. L. Sachdev 3.4. A boundary value problem The method described before can be used to solve some boundary value problems as well. Let be a bounded connected open set with a smooth boundary. Consider the cylindrical domain D =. Consider the problem U t + U U = U U x = φ x x n x t U x t = α 8 Here, φ is a smooth function from to R, α is real constant, and n x t is the unit outward normal of the boundary points of D. We note that n x t U x t is the normal component of U x t at the boundary point x t and when α =, 8 says that U x t is tangential to the boundary point x t. Here again, we seek a solution of the form U x t = φ x t and, as before, we get U x t = θ θ where θ satisfy the linear problem 83 The licit solution of 84 is θ t = θ θ x = φ x n θ + αθ = 84 θ x t = c n λ n t φ n x 85 where c n = φ x φ n x dx and λ n and φ n are eigenvalues and normalized eigenfunctions of the eigenvalue problem in : φ = λφ n φ + αφ =

25 Boundary Value Problems for Burgers Equations 55 Using 85 in 83, we get U x t = Letting t in 86 we get U x t = φ t c n λ n t φ n x c 86 n λ n t φ n x References. J. M. Burgers, A mathematical model illustrating the theory of turbulance, in Advances in Applied Mechanics, Vol. R. von Mises and T. von Karman, Eds., pp. 7 99, Academic Press, New York, P. L. Sachdev, Nonlinear Diffusive Waves, Cambridge University Press, New York, E. Hopf, The partial differential equation u t + uu x = u xx, Comm. Pure Appl. Math. 3: J. D. Cole, On a quasilinear parabolic equation occurring in aeronamics, Quart. Appl. Math. 9: M. J. Lighthill, Viscosity effects in sound waves of finite amplitude, in Surveys in mechanics, G. K. Batchelor and R. M. Davis, Eds., pp. 5 35, Cambridge University Press, New York, P. L. Sachdev, K. T. Joseph, and K. R. C. Nair, Exact solutions for the nonplanar Burgers equations, Proc. Roy. Soc. Lon. 7: S. Nerney, E. D. Schmahl, and Z. E. Musielak, Analytic solutions of the vector Burgers equation, Quart. Appl. Math., 56: K. T. Joseph, Burgers equation in the quarter plane, a formula for the weak it, Comm. Pure Appl. Math. 4: J. Kevorkian and J. D. Cole, Perturbation Methods in Applied Mathematics, Springer Verlag, New York, 98.. F. Calogero and S. De Lillo, The Burgers equation on the semiline and finite intervals, Nonlinearity : F. Calogero and S. De Lillo, The Burgers equation on the semiline with general boundary condition at the origin, J. Math. Phys. 3: K. T. Joseph and P. L. Sachdev, Exact analysis of Burgers equation on the semiline with flux condition at the origin, Int. J. Nonlinear Mech 8: E. R. Benton and G. W. Platzman, A table of solutions of the one-dimensional Burgers equation, Quart. Appl. Math. 3: P. D. Lax, Hyperbolic systems of conservation laws II, Comm. Pure Appl. Math. : P. L. Lions, Generalized Solutions of Hamilton Jacobi equations, Pitman Lecture Notes, Pitman, 98. φ TIFR Mumbai Indian Institute of Science Received April 3,

A CLASSICAL SOLUTION OF THE PROBLEM OF SEEPAGE IN TWO LAYERED SOIL WITH AN INCLINED BOUNDARY

International Journal of Mathematical Sciences Vol., No. 3-4, July-December, pp. 69-74 Serials Publications A CLASSICAL SOLUTION OF THE PROBLEM OF SEEPAGE IN TWO LAYERED SOIL WITH AN INCLINED BOUNDARY