Numerical solution of a critical Sobolev exponent problem with weight on 𝕊3

In this paper, we prove the existence of a positive solution for elliptic nonlinear partial differential equation with weight involving a critical exponent of Sobolev imbedding on . Moreover, we discuss numerically the influence of the weight on the radius of the domain for which the given PDE has a positive solution.


Introduction
Let D be a geodesic ball with radius θ 1 , centred at the north pole, on S 3 . We study the following elliptic nonlinear partial differential equation with weight involving the critical exponent of Sobolev imbedding on S 3 ⎧ ⎨ ⎩ − div S 3 ((α + β |x| k )∇ S 3 u) = u 5 + λ u, in D, where the exponent 5 + 1 = 6 is critical in the sense of Sobolev embedding, and the constants α, k, β and the parameter λ are assumed to be positive.
In this work, we treat the general case of α and β in a bounded domain of S 3 . More precisely, we study the influence of the function α + β |x| k on the existence of solutions of a weighted elliptic PDE. Our method combined two directions: first, we study the existence of a positive solution. This procedure is not obvious and presents many difficulties due to the presence of critical Sobolev exponent which generate a lack of compactness. We overcome this problem using minimizing technique and variational approach. We obtain the existence of a positive solution only in the case k > 1 and for λ in a well-determined interval. Therefore, second, in order to obtain a complete result of our problem, we use Newton iteration method with classical fourth-order Runge-Kutta procedure and we carry out a numerical solution in the cases that we have no theoretical results.
Using the stereographic transformation, D is mapped onto a ball B(0, R) ⊂ R 3 and we write (1) as By a result of Padilla [16] (extending the classical result of Gidas, Ni and Nirenberg [13] to domain on manifolds of constant curvature), a solution u of (2) is symmetric, i.e. it only depends on the azimuthal angle, then we write (2) as the boundary problem with ρ(r) = 2 1+r 2 . For large dimensions N ≥ 5, there is a little difference between studying the problem for a domain in S N and a domain in R N . However, the results differ considerably for N = 4 and N = 3, see [4] for S 3 and [10] for R N . So, in this work we will study the case N = 3 where we prove the existence of a positive solution for λ > 0 and k > 1. We have no theoretical results for λ > 0 and 0 < k ≤ 1 or λ < 0 and k > 1. Nevertheless, we obtain some numerical results for the existence of a positive solution. This approach is motivated by the results of [4] and [7] for α = 1 and β = 0, where the authors found numerical solutions for λ negative enough when the geodesic radius θ 1 > π 2 . The rest of the paper is organized as follows: in Section 2, we present some theoretical results. In Section 3, we present numerical results for existence that complete the results announced in Section 2. Furthermore, in Section 4 we give some interpretations on the influence of the parameters k, α, β and λ, in a ball with radius R where the problem has a positive solution. Finally, in Section 5 we summarize our results and describe future work.

Theoretical results
Let λ div 1 be the first eigenvalue of − div S 3 ((α + β |x| k ) ∇ S 3 u) on D with zero Dirichlet boundary condition.
We define The main result is: Let S be the best Sobolev constant for the injection of . We consider the associate minimizing problem to (2) S λ,α,β = inf The proof of the first part of Theorem 2.1 is based on the two following lemmas.

Proof of Theorem 2.1:
(1) Let (u j ) be a minimizing sequence of S λ,α,β . More precisely, From Lemma 6, we know that S λ,α,β < αS for λ > λ * (k) and k > 1. Thus we have a minimizing sequence of S λ,α,β and S λ,α,β < αS. Consequently, from Lemma 2.1, we deduce that S λ,α,β is achieved by a function u such that u j → u strongly in H 1 0 ( ) and We may assume that u ≥ 0 (otherwise we replace u by |u|). Since u is a minimizer of (5) then there exists a Lagrange multiplier μ ∈ R such that In fact μ = S λ,α,β and S λ,α,β > 0 since λ < λ div 1 . Then u > 0 on by the strong maximum principle. Finally, there exists a positive constant k > 0 (more precisely k = (S λ,α,β ) 1 4 ) such that k u is a solution of problem 2.
(2) There is no solution of (2) when λ > λ div 1 . Indeed, let equivalently, under the stereographic projection, to Suppose that u is a solution of (2), then we have Thus λ < λ div 1 , since the functions ρ, u and ϕ 1 are positive which concludes the proof of Theorem 2.1.
Using Newton iteration method to approximate u 0 : and differentiating the differential Equation (13) with respect to u 0 , yields where By substituting in Equations (13) and (15), we get the following system of first-order differential equations: subjected to the initial conditions which can be written in a matrix form as Using the classical fourth order Runge-Kutta method for the system (16a)-(16b), and for a given step-size which gives the solution u i = u i (u 0 ) in terms of u 0 for i = 0, 1, 2, 3, . . . , m, and the last nonlinear equation is and we have ⎛ The last equation is nonlinear, which will be solved for u 0 by Newton iterative method:  Next, we solve the boundary value problem (12) for different values of α, β, k, λ, and the corresponding R with the solutions are presented in Figures 1, 2, 3. Figure 1 presents a numerical solution of (12) in the case λ > 0 and 0 < k ≤ 1, where α = 1, β = 1, λ = 2, and R = 0.981918, k = 0.1 (Figure 1 a), R = 1.08295,  (Figure 1 d). We notice that the numerical solution is decreasing from u(0) to u(R) = 0.

Variation of the problem parameters
Next, for a given u(0) = u 0 > 0, we solve numerically the boundary value problem (12) for different values of the parameters in both cases when λ positive and negative and choose the minimum radius R so that the obtained numerical solution stay positive.

Example 4.2:
We solve the boundary value problem (12) for β = 1,k = 0.9, λ = 2 with different values of α and the obtained minimum radius R satisfied the boundary condition u(R) = 0 are presented in Table 2.  We remark that by increasing β the minimum radius R satisfied the boundary condition u(R) = 0 increases significantly compared when increasing the parameter α.

Example 4.3:
We solve the boundary value problem (12) for α = 1,k = 1.3, λ = −6 with different values of β and the obtained minimum radius R satisfied the boundary condition u(R) = 0 are presented in Table 3.

Example 4.4:
We solve the boundary value problem (12) for β = 0.1,k = 1.3, λ = −6 with different values of α and the obtained minimum radius R satisfied the boundary condition u(R) = 0 are presented in Table 4.
Similarly as in the first case, the minimum radius R satisfied the boundary condition u(R) = 0 increases considerably by changing the parameters α and β.

Conclusion
In this work, we proved the existence of a positive solution for the boundary value problem (3) for all λ = 0 and for different positive values of k. Theoretically, we proved the result for only λ > 0 and k > 1, and we completed the other cases numerically. More precisely, we obtained the existence of solutions using numerical methods when λ is positive and k is between 0 and 1, also when λ is negative and k is positive. We concluded that the radius of a ball in S 3 for which the problem (5) has a numerical solution is depending on the parameters α, β and k. Future work will include proving the existence of a positive solution in S 4 especially for k = 2 and λ strictly positive close to zero or λ strictly negative.