Construction of the solution

Here the primes denote differentiation with respect to h. The initial condition for s yields the boundary conditions f ¹ ` ¼ 1 and f ` ¼ 22 At h ¼ 0, the solution has to satisfy the flux continuity condition lim h↑0 ¹ f l h l D f f 9 ¼ lim h↓0 ¹ f r h r D f f 9 23 and the extended pressure condition 15 with s r and s l replaced by f r ¼ lim h↓0 f h and f l ¼ lim h↓0 f h .

3.1 Construction of the solution

To construct the similarity solution, we first solve equations 20, 21 and 22, and then match the corresponding solu- tions at h ¼ 0 so that the interface conditions 15 and 23 are satisfied. Thus, we start with the subproblems 1 2 hf 9 þ h r D f f 9 9 ¼ 0 , 0 , h , ` f ¼ f r , f ` ¼ 24 and 1 2 hf 9 þ h l D f f 9 9 ¼ 0 , ¹ ` , h , 0 , f ¹ ` ¼ 1 , f ¼ f l 25 where 0 f l , f r 1 have to be determined from the inter- face conditions. It is well-known e.g. 1,2,19 that problem 24 has a unique solution f þ ¼ f þ h for every f r [ [0,1]. If f r ¼ then f þ h ¼ 0 for all h 0; if f r . 0 then there is a positive a r ` such that f þ h . for 0 , h , a r ¼ for h a r 26 and f 9 þ h , 0 for h between zero and a r . The behaviour of the diffusion coefficient near f ¼ 0 determines whether a r ¼ ` or a r , ` . The precise condition is given in Section 3.2. Similarly, problem 25 has a unique solution f - ¼ f - h for every f l [ [0,1]. If f l ¼ 1 then f - h ¼ 1 for all h 0; if f l , 1 then there is a negative a l ¹` such that f ¹ h ¼ 1 for h a l , 1 for a l , h , 0 27 and f 9 - h , 0 for h between a l and zero. Here, the beha- viour of the diffusion coefficient near f ¼ 1 determines whether a l ¼ ¹` or a l . ¹` . For the precise conditions were refer to Section 3.2. To apply the interface conditions we need to know the fluxes at h ¼ 0. Let F r ¼ ¹ f r h r D f r f 9 þ 28 and F l ¼ ¹ f l h l D f l f 9 ¹ 29 For every value of f r [ [0,1], which determines the solution f þ of problem 24, a unique F r results. We denote this dependence by writing F r ¼ F r f r . This function is continuous and strictly increasing in f r [ [0,1] with F r 0 ¼ 0. An analytic proof of these statements is given in ref. 19 , a computational result is shown in Fig. 1, where the flux function F r is given for the Brooks–Corey and Van Genuchten model. In a similar fashion, F l can be considered as a function of f l . This function, which is continuous and strictly decreasing in f l [ [0,1] with F l 1 ¼ 0, is also shown in Fig. 1. Having discussed these preliminary results, we can now outline the matching procedure. 3.1.1 Existence of a unique pair f l ,f r matching the interface conditions We consider in detail the case h l . h r . In Fig. 1 the graphs of the capillary pressure and fluxes are shown, both as func- tions of the saturation at each side of the origin. Note that here the lower capillary pressure curves correspond to the left-hand side of the origin, the upper curves to the right- hand side. The fluxes were obtained by numerically solving transformed versions of problems 24 and 25 for different f l and f r . The details of the transformation and computation are given in Section 3.1.2. We treat the cases with and with- out entry pressure separately. 3.1.1.1 Zero entry pressure. If the entry pressure is zero, then the saturations at the origin have to satisfy condition 13, reflecting continuity of the capillary pressure. Since the capillary pressure functions are strictly decreasing, it follows from continuity of the capillary pressure that the right saturation depends monotonically on the left saturation [cf. Fig. 1 left]: for instance, when we increase the left saturation f l , then the right saturation f r increases as well. Furthermore, for increasing f l , the left flux F l decreases while the corresponding right flux F r increases. Now, using continuity and monotonicity of the graphs in Fig. 1 left, we find, for f l increasing from zero, a unique pair f l , f r such that both pressure and flux are equal. The continuity of the fluxes combined with F l 0 . F r 0 and F l 1 , F r 1 yields the existence of such a pair. The moton- icity of the fluxes implies the uniqueness. 3.1.1.2 Positive entry pressure. If the entry pressure is positive, as in Fig. 1 right, then the situation is different in the sense that now the saturations at the discontinuity are related to each other through the extended pressure condition 15. Increasing the left saturation f l , we see that the right saturation increases only if f l s , but is constant f r ¼ 1 if f l . s . Moreover, for increasing f l , the left flux F l decreases while the correspondiing right flux F r increases only if f l s , but is constant F r ¼ F r 1 if f l . s . So, if F l s . F r 1, then f l must be greater than s in order to have continuity of the flux. In that case, f r ¼ 1 and F r ¼ F r 1. Since F l f l is a strictly decreasing function of f l with F l 1 ¼ 0, it follows that there is a unique f l such that continuity of the flux is satisfied. Note that the capillary pressure is discontinuous in this case. Similarity solutions for capillary redistribution 455 If F l s F r 1, then it is necessary that f l s in order to have continuity of flux. Consequently, the capillary pres- sure is continuous, and again as in the case of zero entry pressure, there is a unique pair f l , f r , such that the interface conditions are satisfied. Remark: The case h l h r can be treated in a similar manner. In that case too, the pair f l ,f r is uniquely deter- mined. However, now the capillary pressure is always continuous, since f l must be less than one in order to have a positive flux F l , and therefore f r , s . Note that in this case f l . f r . 3.1.2 Computation of f l ,f r , the flux functions and the solution To obtain a solution, we first determine the pair f l ,f r that satisfies the interface conditions, and then use these values to solve problems 24 and 25. It is not necessary to solve these problems separately: their solutions follow directly from the method that will be used to compute the fluxes F l and F r . We first explain how to obtain the pair f l ,f r . For the time being, let us assume that the functions F l f l and F r f r are known: further on we discuss how they can be determined numerically. We distinguish between the cases h l . h r capillary pressure possibly discontinuous and h l h r capillary pressure continuous. If h l . h r we first have to check whether the capillary pressure is continuous. For that purpose we need the values of F l s and F r 1. If F l s . F r 1, then the capillary pressure is discontinuous, and hence f r ¼ 1. In that case we have to find the root f l [ s ,1 of F l f l ¼ F r 1 30 If F l s F r 1, then the capillary pressure is continuous, and hence f l s . In that case, determine f r as a function of f l using continuity of capillary pressure. Then find the root f l s that solves the equation F l f l ¼ F r f r f l 31 If h l h r , then the capillary pressure is continuous and we proceed as above: determine f r as a function of f l using continuity of capillary pressure, and find the root f l [ 0,1 that solves equation 31. To find the root of equations 30 or 31 we use the bisection method. Crucial in the construction are the flux functions F l f l and F r f r . Of course it is not necessary to determine the entire graphs of F l and F r . The functions F l and F r only have to be evaluated at the iteration points resulting from the algorithm that is used to find the root of equations 30 or 31. We discuss below how F r f r , with 0 f r 1, can be solved numerically. The function F l f l is found in an entirely analogous way. Therefore, those details are omitted. To determine the right flux F r f r we need to solve prob- lem 24 and compute the flux at h ¼ 0. The complication here is the boundary condition f` ¼ 0 which is not always easy to verify. Fortunately, there is a more direct way to obtain the flux-saturation relation at h ¼ 0. The idea is to transform equation 21 into a differential equation for the flux with the saturation as independent variable. 2 Since f þ is strictly decreasing on 0,a r we can invert f þ ¼ f þ h for 0 h a r 32 to obtain h ¼ j þ f for 0 f f r 33 where j þ is the inverse of f þ with j þ 0 ¼ a r and j þ f r ¼ 0. Next, consider the scaled flux up to the porosity as a function of saturation, i.e. y f : ¼ ¹ h r D f f 9 þ j þ f for 0 f f r 34 Note that yf . 0 for 0 , f , f r , because f þ is monotoni- cally decreasing. Using equation 21, one easily verifies that dy df f ¼ 1 2 j þ f for 0 , f , f r 35 and y d 2 y df 2 ¼ ¹ 1 2 h r D f for 0 , f , f r 36 Since the flux vanishes whenever the saturation vanishes we also have y0 ¼ 0. Thus, for given f r [ [0,1], we want to solve the boundary value problem y d 2 y df 2 ¼ ¹ 1 2 h r D f for 0 , f , f r , dy df f r ¼ , y ¼ , 37 such that y . 0 on 0, f r . Having established the solution y ¼ yf, we know the flux at h ¼ 0 through the relation F r f r ¼ f r y f r 38 Problem 37 is solved by a shooting technique. 3 In its specific application to problem 37, we replace the bound- ary condition y0 ¼ 0 by yf r ¼ y r , where y r is the shooting parameter. The objective is to find y r such that the boundary condition y0 ¼ 0 is satisfied. We use the bisection method to obtain convergence to the required value of y r . The method to compute yf can be conveniently employed to determine the solution f þ of problem 24. In the shooting procedure, the second-order differential equa- tion in y is rewritten into a system of first-order differential equations in the dependent variables y and dydf. From equation 35 we have h ¼ j þ f ¼ 2 dy df f for 0 , f , f r 39 Hence, using equation 39, the algorithm directly gives for every value of f the corresponding value of h.

3.2 Structure of the solution